Simple Harmonic Motion
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- 1.1 Intro
- 1.2 Easy Examples
- 1.3 Mass On Table Attached To Hanging Mass
- 1.4 Constant Forces And Shifting The Point Of Equilibrium
- 1.5 Rod Hanging Attached To Spring
- 1.6 Mathematical Pendulum (torque)
- 1.7 Explnation About A Physical Pendulum
- 1.8 Example- Physical Pendulum
- 1.9 Energy Analysis
- 1.10 Example- Mathematical Pendulum (Energy)

Damped Harmonic Motion
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Driven Harmonic Motion
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Double Masses
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Exercises
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Exercises for Advanced
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{"Free":0,"Sample":1,"Paid":2}

[{"Name":"Simple Harmonic Motion","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"1.1 Intro","Duration":"12m 40s","ChapterTopicVideoID":9140,"CourseChapterTopicPlaylistID":5365,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.650","Text":"Hello. This lesson is an introduction to harmonic motion."},{"Start":"00:04.650 ","End":"00:06.945","Text":"Now when we\u0027re dealing with harmonic motion,"},{"Start":"00:06.945 ","End":"00:08.444","Text":"we have a spring,"},{"Start":"00:08.444 ","End":"00:10.710","Text":"which is attached to something that doesn\u0027t move,"},{"Start":"00:10.710 ","End":"00:12.788","Text":"so in this case, a wall,"},{"Start":"00:12.788 ","End":"00:15.585","Text":"and then on the other end of the spring,"},{"Start":"00:15.585 ","End":"00:18.180","Text":"we have some kind of mass attached to it."},{"Start":"00:18.180 ","End":"00:21.959","Text":"Now, we say that the spring has a constant,"},{"Start":"00:21.959 ","End":"00:25.620","Text":"which is called k. When we\u0027re writing out"},{"Start":"00:25.620 ","End":"00:29.415","Text":"an equation for the sum of the forces when dealing with this,"},{"Start":"00:29.415 ","End":"00:32.865","Text":"so we\u0027ll say that the sum of all of the forces is equal to"},{"Start":"00:32.865 ","End":"00:37.313","Text":"negative our spring constant, negative k,"},{"Start":"00:37.313 ","End":"00:42.060","Text":"multiplied by x minus x_0,"},{"Start":"00:42.060 ","End":"00:48.045","Text":"where x_0 is the point at which our spring is limp."},{"Start":"00:48.045 ","End":"00:49.320","Text":"It\u0027s not being stretched,"},{"Start":"00:49.320 ","End":"00:50.699","Text":"and it\u0027s not being compressed,"},{"Start":"00:50.699 ","End":"00:54.390","Text":"so it has no forces acting on it. That\u0027s our x_0."},{"Start":"00:54.390 ","End":"00:57.360","Text":"Our x is our displacement,"},{"Start":"00:57.360 ","End":"01:01.495","Text":"so our distance pulled from a certain point of reference."},{"Start":"01:01.495 ","End":"01:07.340","Text":"This is, of course, going to be equal to our mass multiplied by acceleration."},{"Start":"01:07.340 ","End":"01:08.855","Text":"Now because we\u0027re dealing with,"},{"Start":"01:08.855 ","End":"01:10.340","Text":"over here, we have our variable,"},{"Start":"01:10.340 ","End":"01:13.389","Text":"x, which represents our displacement,"},{"Start":"01:13.389 ","End":"01:17.105","Text":"so we\u0027re going to use the second derivative of our x,"},{"Start":"01:17.105 ","End":"01:18.793","Text":"x double dot,"},{"Start":"01:18.793 ","End":"01:20.546","Text":"and this is our acceleration."},{"Start":"01:20.546 ","End":"01:25.700","Text":"You\u0027ll see why we write it like this."},{"Start":"01:26.150 ","End":"01:30.640","Text":"This equation is the equation for harmonic motion."},{"Start":"01:30.640 ","End":"01:33.370","Text":"It\u0027s very important. Please write this down."},{"Start":"01:33.370 ","End":"01:42.270","Text":"Another name for the type of this equation is that it\u0027s an equation of motion."},{"Start":"01:42.270 ","End":"01:49.375","Text":"Now,"},{"Start":"01:49.375 ","End":"01:54.874","Text":"an equation of motion is any equation that has the variables of acceleration,"},{"Start":"01:54.874 ","End":"01:57.265","Text":"so here we have x double dot, that\u0027s acceleration,"},{"Start":"01:57.265 ","End":"02:00.480","Text":"or velocity, or displacement."},{"Start":"02:00.480 ","End":"02:02.160","Text":"x is displacement."},{"Start":"02:02.160 ","End":"02:04.700","Text":"It can also have 2 of them,"},{"Start":"02:04.700 ","End":"02:06.950","Text":"or if you find something with 3 of them, it doesn\u0027t matter."},{"Start":"02:06.950 ","End":"02:08.930","Text":"Those are equations of motion,"},{"Start":"02:08.930 ","End":"02:12.555","Text":"and this equation describes harmonic motion."},{"Start":"02:12.555 ","End":"02:14.045","Text":"So if we have a harmonic system,"},{"Start":"02:14.045 ","End":"02:18.080","Text":"this equation will describe its starting position,"},{"Start":"02:18.080 ","End":"02:22.380","Text":"its velocity, and so on and so forth."},{"Start":"02:22.690 ","End":"02:28.415","Text":"Another useful trait of this equation is that if you get a question in which you\u0027re"},{"Start":"02:28.415 ","End":"02:33.814","Text":"asked to prove that some system is working with harmonic motion,"},{"Start":"02:33.814 ","End":"02:38.330","Text":"if you can get the equation of your system into this type of format,"},{"Start":"02:38.330 ","End":"02:42.540","Text":"then you can know that the system is in harmonic motion."},{"Start":"02:44.380 ","End":"02:51.825","Text":"Our equation over here has to be in this exact format. What does that mean?"},{"Start":"02:51.825 ","End":"02:53.370","Text":"That means that, Number 1,"},{"Start":"02:53.370 ","End":"02:58.980","Text":"this negative must always be here, always."},{"Start":"02:58.980 ","End":"03:05.520","Text":"Number 2, that means that our k and our m are positive constants."},{"Start":"03:05.520 ","End":"03:09.630","Text":"This is const,"},{"Start":"03:09.630 ","End":"03:13.804","Text":"and it\u0027s a positive, and also this."},{"Start":"03:13.804 ","End":"03:15.990","Text":"This is also a constant,"},{"Start":"03:15.990 ","End":"03:18.278","Text":"and it\u0027s positive,"},{"Start":"03:18.278 ","End":"03:22.755","Text":"and we must always have our x_0 here."},{"Start":"03:22.755 ","End":"03:27.275","Text":"Now our x_0 can be positive or negative,"},{"Start":"03:27.275 ","End":"03:30.000","Text":"but it\u0027s also a constant."},{"Start":"03:30.370 ","End":"03:33.463","Text":"But this can be positive or negative."},{"Start":"03:33.463 ","End":"03:35.305","Text":"It\u0027s the only difference,"},{"Start":"03:35.305 ","End":"03:39.240","Text":"and it must always look exactly like this."},{"Start":"03:40.760 ","End":"03:44.210","Text":"You\u0027ll notice that we have here our x,"},{"Start":"03:44.210 ","End":"03:46.400","Text":"and here we have an x double dot."},{"Start":"03:46.400 ","End":"03:49.535","Text":"So that means that this is a differential equation,"},{"Start":"03:49.535 ","End":"03:52.415","Text":"and the answer to this differential equation,"},{"Start":"03:52.415 ","End":"03:54.995","Text":"which, if you want you can solve on your own,"},{"Start":"03:54.995 ","End":"03:57.670","Text":"is going to be our x as a function of t,"},{"Start":"03:57.670 ","End":"03:59.975","Text":"so our displacement is a function of time,"},{"Start":"03:59.975 ","End":"04:07.145","Text":"is going to be equal to A cosine of Omega, angular velocity,"},{"Start":"04:07.145 ","End":"04:14.805","Text":"multiplied by t plus Phi and then plus x_0."},{"Start":"04:14.805 ","End":"04:19.110","Text":"Now a lot of professors will omit this x_0."},{"Start":"04:19.110 ","End":"04:22.040","Text":"I think that it makes it more clear,"},{"Start":"04:22.040 ","End":"04:23.795","Text":"and it\u0027s a more accurate answer,"},{"Start":"04:23.795 ","End":"04:26.490","Text":"so you can write it if you wish."},{"Start":"04:26.870 ","End":"04:34.125","Text":"Now, this equation is called the general solution to this differential equation."},{"Start":"04:34.125 ","End":"04:36.571","Text":"So this is the answer."},{"Start":"04:36.571 ","End":"04:39.120","Text":"Now let\u0027s take a look at what this means."},{"Start":"04:39.120 ","End":"04:43.260","Text":"So we can see that our variable is t,"},{"Start":"04:43.260 ","End":"04:47.080","Text":"so we have a t over here and a t over here."},{"Start":"04:47.080 ","End":"04:49.535","Text":"This is the variable of the equation."},{"Start":"04:49.535 ","End":"04:52.955","Text":"Now we have all sorts of constants in here."},{"Start":"04:52.955 ","End":"04:56.135","Text":"We have our A, we have our Omega,"},{"Start":"04:56.135 ","End":"04:57.650","Text":"we have our Phi,"},{"Start":"04:57.650 ","End":"05:00.860","Text":"and we have our x_0."},{"Start":"05:00.860 ","End":"05:04.235","Text":"Now what we\u0027re going to do is we\u0027re going to take a look"},{"Start":"05:04.235 ","End":"05:08.120","Text":"at what each of these values means."},{"Start":"05:08.720 ","End":"05:11.925","Text":"Let\u0027s deal first with our x_0."},{"Start":"05:11.925 ","End":"05:14.520","Text":"Now, our x_0, over here,"},{"Start":"05:14.520 ","End":"05:18.750","Text":"we said that this is where our spring is."},{"Start":"05:18.750 ","End":"05:21.410","Text":"It doesn\u0027t have any forces acting on it."},{"Start":"05:21.410 ","End":"05:25.145","Text":"It means it\u0027s limp. But here, when we\u0027re seeing it,"},{"Start":"05:25.145 ","End":"05:28.250","Text":"we\u0027re speaking about the point of equilibrium,"},{"Start":"05:28.250 ","End":"05:30.620","Text":"where the forces are balanced."},{"Start":"05:30.620 ","End":"05:32.300","Text":"So it\u0027s not necessarily limp,"},{"Start":"05:32.300 ","End":"05:33.800","Text":"but the forces are balanced,"},{"Start":"05:33.800 ","End":"05:37.500","Text":"so point of equilibrium."},{"Start":"05:43.480 ","End":"05:47.870","Text":"Now, the definition for finding your point of equilibrium,"},{"Start":"05:47.870 ","End":"05:53.030","Text":"it\u0027s where you get that your sum of all of the forces is equal to 0."},{"Start":"05:53.030 ","End":"05:55.160","Text":"That\u0027s the sum of all of the forces,"},{"Start":"05:55.160 ","End":"05:59.590","Text":"not just the force of the wire or of the spring."},{"Start":"05:59.590 ","End":"06:03.360","Text":"So the sum of all of the forces is equal to 0."},{"Start":"06:05.060 ","End":"06:10.835","Text":"This, our x_0, is very easy to find because we\u0027ll know what this is in the question."},{"Start":"06:10.835 ","End":"06:14.365","Text":"We just take this x_0 and plunk it over here."},{"Start":"06:14.365 ","End":"06:16.635","Text":"The next thing I want to know is,"},{"Start":"06:16.635 ","End":"06:19.180","Text":"what is my A?"},{"Start":"06:20.560 ","End":"06:25.640","Text":"My A describes my amplitude, A for amplitude."},{"Start":"06:25.640 ","End":"06:27.590","Text":"What in fact is my amplitude?"},{"Start":"06:27.590 ","End":"06:33.890","Text":"My amplitude is the farthest distance away I can get from my point of equilibrium."},{"Start":"06:33.890 ","End":"06:37.160","Text":"Let\u0027s say my point of equilibrium is here, x_0."},{"Start":"06:37.160 ","End":"06:39.185","Text":"That means that, from this point,"},{"Start":"06:39.185 ","End":"06:42.765","Text":"I can go this distance, A,"},{"Start":"06:42.765 ","End":"06:46.700","Text":"up until here, and similarly,"},{"Start":"06:46.700 ","End":"06:50.584","Text":"I can go the exact same distance up until here,"},{"Start":"06:50.584 ","End":"06:57.570","Text":"but minus A. I can go A to the right and minus A,"},{"Start":"06:57.570 ","End":"06:59.520","Text":"which means it\u0027s going to the left,"},{"Start":"06:59.520 ","End":"07:00.920","Text":"and that is my amplitude,"},{"Start":"07:00.920 ","End":"07:04.620","Text":"the distance I go away from my x_0."},{"Start":"07:04.930 ","End":"07:07.400","Text":"Now, in harmonic motion,"},{"Start":"07:07.400 ","End":"07:10.775","Text":"this is very important because the whole point is that"},{"Start":"07:10.775 ","End":"07:16.890","Text":"my mass is going to be moving in harmonic motion between these 2 points."},{"Start":"07:17.400 ","End":"07:21.895","Text":"I\u0027m moving periodically between A and minus A,"},{"Start":"07:21.895 ","End":"07:24.370","Text":"so let\u0027s just show you what that means."},{"Start":"07:24.370 ","End":"07:26.800","Text":"If I\u0027m starting, for instance, at A,"},{"Start":"07:26.800 ","End":"07:32.491","Text":"and I go from A to minus A and then back,"},{"Start":"07:32.491 ","End":"07:35.120","Text":"so that is 1 period."},{"Start":"07:36.980 ","End":"07:41.600","Text":"This blue line starting from here,"},{"Start":"07:41.600 ","End":"07:47.600","Text":"and going, and also ending at the same point is called 1 period."},{"Start":"07:48.060 ","End":"07:50.620","Text":"If that is 1 period,"},{"Start":"07:50.620 ","End":"07:55.900","Text":"then we can say that my frequency is"},{"Start":"07:55.900 ","End":"08:01.690","Text":"going to be equal to 1/T,"},{"Start":"08:01.690 ","End":"08:07.845","Text":"which is my time for 1 period to be completed."},{"Start":"08:07.845 ","End":"08:14.250","Text":"My frequency is the amount of periods I have in 1 second."},{"Start":"08:14.330 ","End":"08:17.375","Text":"Now notice this is a capital T,"},{"Start":"08:17.375 ","End":"08:21.150","Text":"so it\u0027s different to this t over here."},{"Start":"08:21.260 ","End":"08:26.430","Text":"Our frequency is the number of periods per second,"},{"Start":"08:26.430 ","End":"08:29.085","Text":"and it\u0027s equal to 1 divided by T,"},{"Start":"08:29.085 ","End":"08:33.070","Text":"which is our T this over here."},{"Start":"08:35.200 ","End":"08:41.180","Text":"This capital T is the time I wrote with a capital T to differentiate from"},{"Start":"08:41.180 ","End":"08:47.415","Text":"this time over here to complete a period, so 1 period."},{"Start":"08:47.415 ","End":"08:51.430","Text":"Now, the next thing that we have,"},{"Start":"08:52.010 ","End":"08:56.385","Text":"our Omega, so I write this up here."},{"Start":"08:56.385 ","End":"08:59.800","Text":"Now our Omega is equal to 2Pif,"},{"Start":"09:02.920 ","End":"09:06.975","Text":"where f is our frequency over here."},{"Start":"09:06.975 ","End":"09:13.830","Text":"This is frequency in radians."},{"Start":"09:14.260 ","End":"09:20.315","Text":"I\u0027ve highlighted this f over here so that you can connect it to what is over here."},{"Start":"09:20.315 ","End":"09:25.735","Text":"Now, the next thing that we have is our Phi over here."},{"Start":"09:25.735 ","End":"09:28.545","Text":"It has something to do with our initial conditions,"},{"Start":"09:28.545 ","End":"09:31.550","Text":"so where our system starts its motion,"},{"Start":"09:31.550 ","End":"09:33.850","Text":"and so on and so forth."},{"Start":"09:33.850 ","End":"09:39.200","Text":"Now over here, we have our equation for our Omega over here."},{"Start":"09:39.200 ","End":"09:42.785","Text":"Now let\u0027s see how we find what our Omega is."},{"Start":"09:42.785 ","End":"09:44.389","Text":"We have our equation,"},{"Start":"09:44.389 ","End":"09:53.250","Text":"which is that Omega is equal to the square root of k divided by m,"},{"Start":"09:53.250 ","End":"10:02.270","Text":"where our k over here is the coefficient of our term x minus x_0, so this over here,"},{"Start":"10:02.270 ","End":"10:04.865","Text":"this green, this is our k,"},{"Start":"10:04.865 ","End":"10:10.100","Text":"and our m is the coefficient of this term where we have the x double dot,"},{"Start":"10:10.100 ","End":"10:12.145","Text":"so it\u0027s our greens."},{"Start":"10:12.145 ","End":"10:18.365","Text":"This green highlighter over here is to remind you that it\u0027s these coefficients."},{"Start":"10:18.365 ","End":"10:22.855","Text":"Then our next thing that we can look at is our A and our Phi."},{"Start":"10:22.855 ","End":"10:28.325","Text":"Now our A and our Phi are both dependent on our initial conditions."},{"Start":"10:28.325 ","End":"10:34.610","Text":"The way that we do this is that we find our position at t=0,"},{"Start":"10:34.610 ","End":"10:38.940","Text":"and we find our velocity at t=0,"},{"Start":"10:38.940 ","End":"10:42.080","Text":"so our initial position and our initial velocity."},{"Start":"10:42.080 ","End":"10:49.055","Text":"Then through this, we\u0027ll be able to find out what our A and what our Phi is equal to."},{"Start":"10:49.055 ","End":"10:53.720","Text":"Now notice that our Omega is dependent on constants,"},{"Start":"10:53.720 ","End":"11:01.010","Text":"on our k and on our m. That\u0027s the physics or the building blocks of our system."},{"Start":"11:01.010 ","End":"11:05.130","Text":"Our A and our Phi is dependent on what we do with our system,"},{"Start":"11:05.130 ","End":"11:11.000","Text":"if we begin with the spring stretched a lot or stretched slightly,"},{"Start":"11:11.000 ","End":"11:13.600","Text":"and this is given to change."},{"Start":"11:13.600 ","End":"11:17.240","Text":"With that, depending on how I begin my system,"},{"Start":"11:17.240 ","End":"11:19.453","Text":"if it\u0027s stretched a lot or a little,"},{"Start":"11:19.453 ","End":"11:21.455","Text":"my A and my Phi will change."},{"Start":"11:21.455 ","End":"11:25.980","Text":"However, my Omega will always stay the same."},{"Start":"11:26.420 ","End":"11:31.340","Text":"My Omega, even if I stretch my mass up until here,"},{"Start":"11:31.340 ","End":"11:33.350","Text":"or I stretch my mass up until here,"},{"Start":"11:33.350 ","End":"11:34.940","Text":"or I compress my mass,"},{"Start":"11:34.940 ","End":"11:39.130","Text":"my Omega is constant."},{"Start":"11:39.130 ","End":"11:45.740","Text":"To sum up how to work out questions dealing with harmonic motion,"},{"Start":"11:45.740 ","End":"11:49.480","Text":"so the first thing that we do, force equations."},{"Start":"11:49.480 ","End":"11:51.765","Text":"Then once we\u0027ve written that,"},{"Start":"11:51.765 ","End":"11:55.685","Text":"we sort out all of our variables into"},{"Start":"11:55.685 ","End":"12:00.260","Text":"this format for the equation of motion when dealing with harmonic motion."},{"Start":"12:00.260 ","End":"12:05.615","Text":"Then after we\u0027ve done step Number 2,"},{"Start":"12:05.615 ","End":"12:11.840","Text":"then we write out our general solution."},{"Start":"12:11.840 ","End":"12:16.140","Text":"We solve for x(t),"},{"Start":"12:16.140 ","End":"12:18.545","Text":"and then once we\u0027ve done that,"},{"Start":"12:18.545 ","End":"12:21.290","Text":"then we have to find out what our A, Omega,"},{"Start":"12:21.290 ","End":"12:23.960","Text":"Phi is and our x_0 is, obviously,"},{"Start":"12:23.960 ","End":"12:26.011","Text":"this x_0 over here,"},{"Start":"12:26.011 ","End":"12:29.250","Text":"so then we find the constants."},{"Start":"12:29.330 ","End":"12:31.605","Text":"So Omega A and Phi,"},{"Start":"12:31.605 ","End":"12:33.675","Text":"which we saw our Omega is this,"},{"Start":"12:33.675 ","End":"12:37.025","Text":"and our A and Phi come from our initial conditions."},{"Start":"12:37.025 ","End":"12:40.320","Text":"That\u0027s the end of the introduction."}],"ID":9410},{"Watched":false,"Name":"1.2 Easy Examples","Duration":"17m 43s","ChapterTopicVideoID":9141,"CourseChapterTopicPlaylistID":5365,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.055","Text":"Hello. In this lesson,"},{"Start":"00:02.055 ","End":"00:04.605","Text":"we\u0027re going to be dealing with a quick and easy example,"},{"Start":"00:04.605 ","End":"00:08.355","Text":"which will help you understand what I spoke about in the introduction."},{"Start":"00:08.355 ","End":"00:11.490","Text":"Let\u0027s say we have the exact same system."},{"Start":"00:11.490 ","End":"00:15.405","Text":"Here I have my spring and attached to it,"},{"Start":"00:15.405 ","End":"00:17.535","Text":"I have my mass m,"},{"Start":"00:17.535 ","End":"00:19.815","Text":"and it\u0027s located over here."},{"Start":"00:19.815 ","End":"00:22.590","Text":"It\u0027s at x_0."},{"Start":"00:22.590 ","End":"00:28.680","Text":"Let\u0027s say that over here we have the origin of my axis,"},{"Start":"00:28.680 ","End":"00:31.095","Text":"which is where my x=0."},{"Start":"00:31.095 ","End":"00:38.594","Text":"Now say I stretch my spring and move my mass so that now its position is over here,"},{"Start":"00:38.594 ","End":"00:45.515","Text":"and this distance is a distance of d. Here\u0027s my system."},{"Start":"00:45.515 ","End":"00:47.630","Text":"Now, the first thing that I\u0027m going to do is"},{"Start":"00:47.630 ","End":"00:50.030","Text":"I\u0027m going to write out like in the previous video."},{"Start":"00:50.030 ","End":"00:52.940","Text":"This is already step number 2,"},{"Start":"00:52.940 ","End":"00:57.020","Text":"my sum of all of the forces is equal to negative k x"},{"Start":"00:57.020 ","End":"01:04.215","Text":"minus x_0 which is equal to mx double-dot."},{"Start":"01:04.215 ","End":"01:06.950","Text":"Notice that now when my mass is over here,"},{"Start":"01:06.950 ","End":"01:11.540","Text":"I released it and it now starts bouncing back and forth in harmonic motion."},{"Start":"01:11.540 ","End":"01:13.620","Text":"This was step number 2."},{"Start":"01:13.620 ","End":"01:15.740","Text":"My step number 1 was drawing this out."},{"Start":"01:15.740 ","End":"01:22.265","Text":"Now my step number 3 is putting this all into this equation, my general solution."},{"Start":"01:22.265 ","End":"01:28.190","Text":"My x as a function of t is going to be equal to A cosine of"},{"Start":"01:28.190 ","End":"01:35.370","Text":"Omega t plus Phi plus my x_0."},{"Start":"01:35.370 ","End":"01:39.470","Text":"Now I have to find out what my variables are."},{"Start":"01:39.470 ","End":"01:41.915","Text":"First I\u0027ll have my Omega,"},{"Start":"01:41.915 ","End":"01:45.995","Text":"remember it\u0027s constant and it equals to the square root of k"},{"Start":"01:45.995 ","End":"01:52.200","Text":"divided by m. Now I have to find out what my A and my Phi is."},{"Start":"01:52.200 ","End":"01:55.275","Text":"Now let\u0027s see what I\u0027m doing."},{"Start":"01:55.275 ","End":"01:56.880","Text":"I\u0027m going to set out."},{"Start":"01:56.880 ","End":"01:59.450","Text":"This is my last step from the previous video."},{"Start":"01:59.450 ","End":"02:03.410","Text":"I\u0027m going to say that my x at time t=0."},{"Start":"02:03.410 ","End":"02:09.870","Text":"Now notice quickly before I carry on with this equation that my x at"},{"Start":"02:09.870 ","End":"02:16.990","Text":"t=0 is not the same as my x_0, this over here."},{"Start":"02:16.990 ","End":"02:19.170","Text":"These 2 are not the same."},{"Start":"02:19.170 ","End":"02:23.045","Text":"My x_0 is at my point of equilibrium,"},{"Start":"02:23.045 ","End":"02:27.770","Text":"which means that at this point my sum of all my forces is equal to 0."},{"Start":"02:27.770 ","End":"02:29.675","Text":"That\u0027s what this means."},{"Start":"02:29.675 ","End":"02:35.030","Text":"This means that my x at t=0."},{"Start":"02:35.030 ","End":"02:38.510","Text":"These 2 don\u0027t mean the same thing."},{"Start":"02:38.510 ","End":"02:41.775","Text":"Now it\u0027s important to note in a lot of other exercises,"},{"Start":"02:41.775 ","End":"02:45.465","Text":"my x at 0 is the same as my x_0."},{"Start":"02:45.465 ","End":"02:47.810","Text":"However, when we\u0027re dealing with harmonic motion,"},{"Start":"02:47.810 ","End":"02:50.020","Text":"these are not the same."},{"Start":"02:50.020 ","End":"02:52.110","Text":"Let\u0027s work this out."},{"Start":"02:52.110 ","End":"02:59.450","Text":"My x at t=0 means what is my position right when I\u0027m beginning at t=0."},{"Start":"02:59.450 ","End":"03:02.380","Text":"We can see that this was my starting position."},{"Start":"03:02.380 ","End":"03:05.765","Text":"Before I released, before harmonic motion began,"},{"Start":"03:05.765 ","End":"03:07.910","Text":"my mass was located over here."},{"Start":"03:07.910 ","End":"03:10.745","Text":"From the origin, it\u0027s located a distance of"},{"Start":"03:10.745 ","End":"03:15.920","Text":"x_0 plus d. That\u0027s this total distance over here."},{"Start":"03:15.920 ","End":"03:19.449","Text":"I write over here x_0 plus"},{"Start":"03:19.449 ","End":"03:25.835","Text":"d. This has to equal this equation when I substitute here that my t=0,"},{"Start":"03:25.835 ","End":"03:31.980","Text":"so that\u0027s going to be equal to A cosine of Omega t,"},{"Start":"03:31.980 ","End":"03:33.480","Text":"where my t=0,"},{"Start":"03:33.480 ","End":"03:41.655","Text":"so this is equal to 0 plus my Phi plus my x_0."},{"Start":"03:41.655 ","End":"03:47.285","Text":"Now what I\u0027m looking to do is I did my x at my time 0."},{"Start":"03:47.285 ","End":"03:50.450","Text":"Now I want my v at t=0."},{"Start":"03:50.450 ","End":"03:55.965","Text":"My velocity at t=0 if I\u0027m releasing this,"},{"Start":"03:55.965 ","End":"04:01.355","Text":"that means that my velocity at t=0 is 0 because it starts from rest."},{"Start":"04:01.355 ","End":"04:03.410","Text":"That\u0027s the equivalent of this."},{"Start":"04:03.410 ","End":"04:06.260","Text":"However, I have my x as a function of t,"},{"Start":"04:06.260 ","End":"04:10.740","Text":"and I need this to equal my v as a function of t. Let\u0027s see."},{"Start":"04:10.740 ","End":"04:15.470","Text":"My v as a function of t is simply going to be"},{"Start":"04:15.470 ","End":"04:17.600","Text":"the derivative of this because we know that"},{"Start":"04:17.600 ","End":"04:21.230","Text":"my velocity is my first derivative of my position,"},{"Start":"04:21.230 ","End":"04:28.234","Text":"so my x dot t. Let\u0027s take the derivative of this."},{"Start":"04:28.234 ","End":"04:31.085","Text":"We\u0027re going to have that this is equal to"},{"Start":"04:31.085 ","End":"04:39.360","Text":"negative A Omega sine of Omega t"},{"Start":"04:39.360 ","End":"04:42.150","Text":"plus Phi."},{"Start":"04:42.150 ","End":"04:46.460","Text":"This is something that\u0027s very important to remember."},{"Start":"04:46.460 ","End":"04:48.635","Text":"I\u0027ve put it in a square,"},{"Start":"04:48.635 ","End":"04:51.050","Text":"excuse that my writing is slanted."},{"Start":"04:51.050 ","End":"04:55.415","Text":"Now we have our v as a function of t. That means that our v at time 0,"},{"Start":"04:55.415 ","End":"04:58.935","Text":"so it equals 0 and this has to equal this."},{"Start":"04:58.935 ","End":"05:06.575","Text":"Then we have negative A Omega sine of Omega t,"},{"Start":"05:06.575 ","End":"05:08.330","Text":"and again our t=0,"},{"Start":"05:08.330 ","End":"05:12.400","Text":"so this is 0 plus Phi."},{"Start":"05:12.400 ","End":"05:14.760","Text":"Now I have my 2 equations."},{"Start":"05:14.760 ","End":"05:20.000","Text":"I have my x at t=0 and I have my v at t=0 over here."},{"Start":"05:20.000 ","End":"05:22.280","Text":"Now what I have to do is I have to try and find"},{"Start":"05:22.280 ","End":"05:24.935","Text":"out what my A and what my Phi is equal to."},{"Start":"05:24.935 ","End":"05:28.795","Text":"Let\u0027s rewrite this over here."},{"Start":"05:28.795 ","End":"05:39.360","Text":"Here I\u0027ll have that negative A Omega sine of Phi is equal to 0,"},{"Start":"05:39.360 ","End":"05:41.550","Text":"so my minus can cross off,"},{"Start":"05:41.550 ","End":"05:44.300","Text":"my amplitude I know is never going to be equal to"},{"Start":"05:44.300 ","End":"05:48.155","Text":"0 because that means that my body isn\u0027t in emotion."},{"Start":"05:48.155 ","End":"05:50.195","Text":"That means that it\u0027s not moving."},{"Start":"05:50.195 ","End":"05:52.540","Text":"My Omega, as we know is a constant,"},{"Start":"05:52.540 ","End":"05:55.465","Text":"it\u0027s the square root of k divided by m,"},{"Start":"05:55.465 ","End":"05:59.390","Text":"which my k is never going to be equal to 0 and my mass isn\u0027t going to be equal to 0,"},{"Start":"05:59.390 ","End":"06:01.200","Text":"so that isn\u0027t,"},{"Start":"06:01.200 ","End":"06:03.470","Text":"so that means that the only thing that can make"},{"Start":"06:03.470 ","End":"06:06.790","Text":"this equation equal to 0 is my sine of Phi."},{"Start":"06:06.790 ","End":"06:11.030","Text":"That means that my sign of Phi is equal to 0,"},{"Start":"06:11.030 ","End":"06:16.250","Text":"which means when is sine of something equal to 0,"},{"Start":"06:16.250 ","End":"06:17.690","Text":"when the something here,"},{"Start":"06:17.690 ","End":"06:21.650","Text":"the Phi is equal to 0 or Pi."},{"Start":"06:21.650 ","End":"06:25.160","Text":"Now, obviously, there\u0027s different multiplications of these values,"},{"Start":"06:25.160 ","End":"06:27.415","Text":"but these are the most basic."},{"Start":"06:27.415 ","End":"06:34.810","Text":"My Phi in order for this equation to be correct has to be equal to either 0 or Pi."},{"Start":"06:34.810 ","End":"06:38.210","Text":"I\u0027m going to choose the 0 because that\u0027s the easiest."},{"Start":"06:38.210 ","End":"06:41.885","Text":"If I say that my Phi is equal to 0,"},{"Start":"06:41.885 ","End":"06:46.095","Text":"then I can substitute that into my x at t=0,"},{"Start":"06:46.095 ","End":"06:50.580","Text":"and we can see that my x_0 over here and my x_0 over here cross off."},{"Start":"06:50.580 ","End":"06:56.120","Text":"That means that my x at t=0 is going to be equal to d,"},{"Start":"06:56.120 ","End":"07:00.755","Text":"which is going to be equal to my A cosine."},{"Start":"07:00.755 ","End":"07:02.550","Text":"Now this is 0 and this is 0,"},{"Start":"07:02.550 ","End":"07:06.935","Text":"so 0 plus 0 is 0, cosine of 0."},{"Start":"07:06.935 ","End":"07:11.035","Text":"Now as we know, cosine of 0 is equal to 1."},{"Start":"07:11.035 ","End":"07:20.160","Text":"That means that my A is equal to d. Now we\u0027ve found what our Omega is,"},{"Start":"07:20.160 ","End":"07:21.720","Text":"what our Phi is equal to,"},{"Start":"07:21.720 ","End":"07:25.520","Text":"and what our A is equal to."},{"Start":"07:25.520 ","End":"07:28.295","Text":"Now, specifically in this type of question,"},{"Start":"07:28.295 ","End":"07:31.760","Text":"if we have some mass which has stretched from"},{"Start":"07:31.760 ","End":"07:35.750","Text":"its point of equilibrium at certain distance and released from rest,"},{"Start":"07:35.750 ","End":"07:39.560","Text":"we can always say that our Phi is equal to 0 and that our amplitude"},{"Start":"07:39.560 ","End":"07:44.070","Text":"is equal to the distance it was stretched from its point of equilibrium."},{"Start":"07:44.070 ","End":"07:47.720","Text":"Specifically, only in these types of questions where it\u0027s stretched at"},{"Start":"07:47.720 ","End":"07:51.365","Text":"certain distance from its point of equilibrium and released from rest,"},{"Start":"07:51.365 ","End":"07:56.285","Text":"I can straight away say that my Phi is equal to 0 and my A is equal to the distance."},{"Start":"07:56.285 ","End":"08:00.815","Text":"But don\u0027t always take this as a surefire rule"},{"Start":"08:00.815 ","End":"08:05.935","Text":"because you\u0027ll see that there are other questions where these will turn out different."},{"Start":"08:05.935 ","End":"08:09.350","Text":"I hope this explained it well and now we\u0027re going to go"},{"Start":"08:09.350 ","End":"08:13.315","Text":"on to a slightly more complicated example."},{"Start":"08:13.315 ","End":"08:18.635","Text":"This is our second example and it\u0027s a little bit more complicated."},{"Start":"08:18.635 ","End":"08:22.130","Text":"Let\u0027s say we have the same starting system."},{"Start":"08:22.130 ","End":"08:27.620","Text":"We have our wall with our spring with a mass m attached to it,"},{"Start":"08:27.620 ","End":"08:30.665","Text":"and the spring is of constant k. Now,"},{"Start":"08:30.665 ","End":"08:34.645","Text":"what\u0027s different in this question is that we have another mass,"},{"Start":"08:34.645 ","End":"08:36.965","Text":"and we\u0027ll say it\u0027s also of mass m,"},{"Start":"08:36.965 ","End":"08:41.560","Text":"which is moving in this direction with a velocity of v_0."},{"Start":"08:41.560 ","End":"08:47.030","Text":"Now, once the mass hits this mass over here,"},{"Start":"08:47.030 ","End":"08:49.915","Text":"so we have plastic collision."},{"Start":"08:49.915 ","End":"08:52.415","Text":"PC, plastic collision."},{"Start":"08:52.415 ","End":"08:55.100","Text":"Then our harmonic motion begins and"},{"Start":"08:55.100 ","End":"09:00.640","Text":"the 2 masses are stuck together and they move together in harmonic motion."},{"Start":"09:00.640 ","End":"09:04.325","Text":"What we\u0027re being asked to find over here is"},{"Start":"09:04.325 ","End":"09:09.095","Text":"an equation expressing our position as a function of time."},{"Start":"09:09.095 ","End":"09:13.440","Text":"This is our question. Let\u0027s see how we go about this."},{"Start":"09:15.750 ","End":"09:18.290","Text":"First, what we\u0027re going to do,"},{"Start":"09:18.290 ","End":"09:22.330","Text":"is we\u0027re going to write that the sum of all of our forces is equal"},{"Start":"09:22.330 ","End":"09:27.440","Text":"to the sum of forces for a spring in harmonic motion,"},{"Start":"09:27.440 ","End":"09:33.265","Text":"so that means that we have negative k multiplied by x minus x_0,"},{"Start":"09:33.265 ","End":"09:39.985","Text":"and that this is going to be equal to our mass multiplied by our x double dot."},{"Start":"09:39.985 ","End":"09:41.365","Text":"Now, our mass is over here,"},{"Start":"09:41.365 ","End":"09:46.150","Text":"because once the system over here starts moving in harmonic motion,"},{"Start":"09:46.150 ","End":"09:49.044","Text":"so we have 2 masses attached."},{"Start":"09:49.044 ","End":"09:54.740","Text":"They\u0027re moving together, so it\u0027s going to be 2mx double dot."},{"Start":"09:56.340 ","End":"09:59.215","Text":"From here, now, obviously,"},{"Start":"09:59.215 ","End":"10:02.815","Text":"over here, we\u0027re going to have our x_0,"},{"Start":"10:02.815 ","End":"10:05.800","Text":"so our point of equilibrium,"},{"Start":"10:05.800 ","End":"10:10.570","Text":"and we can say somewhere over here that this is where our x=0."},{"Start":"10:10.570 ","End":"10:12.655","Text":"This is our origin."},{"Start":"10:12.655 ","End":"10:18.069","Text":"Now I\u0027m going to say that my positive direction is in this direction."},{"Start":"10:18.069 ","End":"10:21.010","Text":"Now notice that in our previous example,"},{"Start":"10:21.010 ","End":"10:26.115","Text":"we didn\u0027t even have to say which direction our positive direction was in."},{"Start":"10:26.115 ","End":"10:28.365","Text":"In general, we also don\u0027t."},{"Start":"10:28.365 ","End":"10:32.620","Text":"I could have also said that my positive direction is in this direction."},{"Start":"10:33.420 ","End":"10:35.575","Text":"Why does it not matter,"},{"Start":"10:35.575 ","End":"10:37.495","Text":"especially, at this stage?"},{"Start":"10:37.495 ","End":"10:41.260","Text":"When I\u0027m substituting in what will go into my force,"},{"Start":"10:41.260 ","End":"10:43.795","Text":"my force is constantly changing."},{"Start":"10:43.795 ","End":"10:45.970","Text":"It\u0027s constantly changing direction."},{"Start":"10:45.970 ","End":"10:47.545","Text":"One moment it\u0027s moving in this direction,"},{"Start":"10:47.545 ","End":"10:49.915","Text":"the next is pushing in this direction."},{"Start":"10:49.915 ","End":"10:53.425","Text":"That\u0027s because my x is constantly changing."},{"Start":"10:53.425 ","End":"10:56.695","Text":"Because my force is always changing direction,"},{"Start":"10:56.695 ","End":"11:00.250","Text":"it doesn\u0027t matter if it\u0027s positive or negative,"},{"Start":"11:00.250 ","End":"11:03.770","Text":"and it will still work out with this equation."},{"Start":"11:05.220 ","End":"11:11.740","Text":"Now I\u0027m going to substitute this into my general solution for my x as a function of time,"},{"Start":"11:11.740 ","End":"11:16.405","Text":"which is equal to my A cosine of"},{"Start":"11:16.405 ","End":"11:22.825","Text":"Omega t plus Phi plus my x_0,"},{"Start":"11:22.825 ","End":"11:25.090","Text":"which is over here."},{"Start":"11:25.090 ","End":"11:27.358","Text":"As we know, of course,"},{"Start":"11:27.358 ","End":"11:32.080","Text":"that my Omega is going to be the square root of my coefficient over here,"},{"Start":"11:32.080 ","End":"11:34.120","Text":"which is simply k,"},{"Start":"11:34.120 ","End":"11:37.285","Text":"divided by my coefficient of my x double dot,"},{"Start":"11:37.285 ","End":"11:39.865","Text":"which over here is 2m."},{"Start":"11:39.865 ","End":"11:41.650","Text":"Pay attention to that."},{"Start":"11:41.650 ","End":"11:46.315","Text":"Remember when I said it\u0027s the coefficients, so pay attention."},{"Start":"11:46.315 ","End":"11:52.130","Text":"Now, what I have to find is what is my A and what is my Phi."},{"Start":"11:52.290 ","End":"11:58.100","Text":"I\u0027m going to start with my x at t=0."},{"Start":"11:59.100 ","End":"12:02.245","Text":"When I start my harmonic motion,"},{"Start":"12:02.245 ","End":"12:06.595","Text":"it\u0027s when this mass has collided with this mass over here,"},{"Start":"12:06.595 ","End":"12:12.050","Text":"which means that at exactly this point over here, at my point x_0."},{"Start":"12:12.360 ","End":"12:15.895","Text":"Before any movement has started in this direction,"},{"Start":"12:15.895 ","End":"12:19.060","Text":"at the point of collision it\u0027s at the point x_0."},{"Start":"12:19.060 ","End":"12:21.240","Text":"Now, what I have to do,"},{"Start":"12:21.240 ","End":"12:25.400","Text":"is I have to find my velocity at t=0."},{"Start":"12:26.250 ","End":"12:30.740","Text":"This is slightly more complicated."},{"Start":"12:31.080 ","End":"12:36.505","Text":"What I\u0027m going to use here in order to work out what my initial velocity is,"},{"Start":"12:36.505 ","End":"12:40.880","Text":"is I\u0027m going to use conservation of momentum."},{"Start":"12:41.850 ","End":"12:47.365","Text":"Let\u0027s see. Before the collision,"},{"Start":"12:47.365 ","End":"12:50.050","Text":"we have our mass."},{"Start":"12:50.050 ","End":"12:52.495","Text":"This is stationary. this mass over here."},{"Start":"12:52.495 ","End":"12:54.760","Text":"It\u0027s mass times velocity,"},{"Start":"12:54.760 ","End":"12:57.010","Text":"Its velocity is 0, so it\u0027s going to be equal to 0."},{"Start":"12:57.010 ","End":"13:00.380","Text":"Then we have mass multiplied by our v_0,"},{"Start":"13:00.380 ","End":"13:03.850","Text":"because this is what\u0027s moving."},{"Start":"13:03.850 ","End":"13:06.190","Text":"That\u0027s our initial momentum,"},{"Start":"13:06.190 ","End":"13:09.715","Text":"and then it has to be equal to our momentum after."},{"Start":"13:09.715 ","End":"13:12.996","Text":"Then we\u0027re going to have that it equals to 2m,"},{"Start":"13:12.996 ","End":"13:17.135","Text":"because we have this mass and this mass,"},{"Start":"13:17.135 ","End":"13:18.760","Text":"multiplied by their velocity,"},{"Start":"13:18.760 ","End":"13:20.905","Text":"which we don\u0027t know, so let\u0027s say it\u0027s u."},{"Start":"13:20.905 ","End":"13:24.670","Text":"Now, all I have to do is isolate out u to find out what it is,"},{"Start":"13:24.670 ","End":"13:29.000","Text":"so it\u0027s going to be equal to 1/2 of v_0."},{"Start":"13:31.680 ","End":"13:38.290","Text":"If this is our u, that means that this is the velocity at the point of collision."},{"Start":"13:38.290 ","End":"13:44.600","Text":"That means that this is equal to our v at t=0."},{"Start":"13:46.770 ","End":"13:57.370","Text":"Now, I can say that my v at t=0 is equal to my 1/2v_0 from over here."},{"Start":"13:57.370 ","End":"14:02.469","Text":"Then I know that this is equal to the derivative of this equation,"},{"Start":"14:02.469 ","End":"14:04.669","Text":"which we did a few minutes ago,"},{"Start":"14:04.669 ","End":"14:06.085","Text":"so this is equal to"},{"Start":"14:06.085 ","End":"14:15.670","Text":"my negative Omega A sine of Omega t,"},{"Start":"14:15.670 ","End":"14:20.000","Text":"where my t is equal to 0 plus Phi."},{"Start":"14:20.000 ","End":"14:22.650","Text":"Of course, not forgetting over here,"},{"Start":"14:22.650 ","End":"14:27.450","Text":"my x_0 is equal to my x_0,"},{"Start":"14:27.450 ","End":"14:29.835","Text":"which is equal to this."},{"Start":"14:29.835 ","End":"14:32.658","Text":"I have my A cosine,"},{"Start":"14:32.658 ","End":"14:34.360","Text":"and again, my t is equal to 0,"},{"Start":"14:34.360 ","End":"14:36.366","Text":"so I\u0027ll just write cosine of Phi,"},{"Start":"14:36.366 ","End":"14:41.000","Text":"because this will cross out, plus my x_0."},{"Start":"14:41.000 ","End":"14:47.920","Text":"Now, what I have, is I have my first equation and I have my second equation."},{"Start":"14:47.920 ","End":"14:50.780","Text":"Right now, what do we have to do,"},{"Start":"14:50.780 ","End":"14:55.615","Text":"is we have to find out what our A is equal to and what our Phi is equal to."},{"Start":"14:55.615 ","End":"15:02.155","Text":"If we take a look over here in our equation number 1,"},{"Start":"15:02.155 ","End":"15:07.585","Text":"so we can see that we can cross off our x over here and our x over here."},{"Start":"15:07.585 ","End":"15:16.075","Text":"That means that 0 is equal to A cosine of Phi."},{"Start":"15:16.075 ","End":"15:19.540","Text":"Now, my A is not equal to 0 ever,"},{"Start":"15:19.540 ","End":"15:22.915","Text":"so that means that my cosine of Phi is equal to 0."},{"Start":"15:22.915 ","End":"15:25.930","Text":"Now, if my cosine of Phi is equal to 0,"},{"Start":"15:25.930 ","End":"15:32.770","Text":"so that means that my Phi can equal Pi over 2 or 3Pi over 2,"},{"Start":"15:32.770 ","End":"15:35.290","Text":"but we\u0027re going to choose Pi over 2."},{"Start":"15:35.290 ","End":"15:38.605","Text":"It can obviously be other multiples."},{"Start":"15:38.605 ","End":"15:41.770","Text":"Then when I substitute that into over here,"},{"Start":"15:41.770 ","End":"15:43.180","Text":"so into my v_0,"},{"Start":"15:43.180 ","End":"15:48.820","Text":"so I\u0027ll have that 1/2v_0 is going to be equal"},{"Start":"15:48.820 ","End":"15:54.970","Text":"to negative Omega A sine of Phi."},{"Start":"15:54.970 ","End":"15:59.374","Text":"Now, sine of Pi over 2, 1,"},{"Start":"15:59.374 ","End":"16:04.500","Text":"so this is going to be equal to negative Omega A."},{"Start":"16:04.500 ","End":"16:08.125","Text":"Because our Phi is Pi over 2."},{"Start":"16:08.125 ","End":"16:10.120","Text":"Now, all I have to do,"},{"Start":"16:10.120 ","End":"16:12.400","Text":"is I have to isolate out my A."},{"Start":"16:12.400 ","End":"16:17.005","Text":"That means that I\u0027ll get that my A is going to be equal to"},{"Start":"16:17.005 ","End":"16:24.775","Text":"negative 1 over 2 Omega multiplied by v_0."},{"Start":"16:24.775 ","End":"16:27.280","Text":"Now we have our Omega over here,"},{"Start":"16:27.280 ","End":"16:28.780","Text":"which we know what this is,"},{"Start":"16:28.780 ","End":"16:34.945","Text":"so we\u0027re going to have negative 1 divided by 2 Omega,"},{"Start":"16:34.945 ","End":"16:39.475","Text":"which is the square root of k divided by 2m,"},{"Start":"16:39.475 ","End":"16:43.645","Text":"and all of this multiplied by v_0."},{"Start":"16:43.645 ","End":"16:45.550","Text":"Now what do we have to do,"},{"Start":"16:45.550 ","End":"16:48.070","Text":"because we were told to find what our x is,"},{"Start":"16:48.070 ","End":"16:49.240","Text":"a function of t is,"},{"Start":"16:49.240 ","End":"16:54.385","Text":"so we\u0027re just going to substitute this into our equation."},{"Start":"16:54.385 ","End":"16:56.130","Text":"That means that our x,"},{"Start":"16:56.130 ","End":"16:57.530","Text":"as a function of t,"},{"Start":"16:57.530 ","End":"16:59.545","Text":"is going to be equal to our A,"},{"Start":"16:59.545 ","End":"17:09.310","Text":"which is negative v_0 divided by 2 over square root of k divided by 2m."},{"Start":"17:09.310 ","End":"17:17.360","Text":"This is going to be multiplied by cosine of our Omega t. Our Omega, again,"},{"Start":"17:17.360 ","End":"17:22.562","Text":"is our square root of k divided by 2m,"},{"Start":"17:22.562 ","End":"17:25.945","Text":"multiplied by t plus our Phi,"},{"Start":"17:25.945 ","End":"17:27.730","Text":"which we said is our Pi over 2,"},{"Start":"17:27.730 ","End":"17:30.460","Text":"so plus Pi over 2."},{"Start":"17:30.460 ","End":"17:33.880","Text":"Then plus our x_0,"},{"Start":"17:33.880 ","End":"17:35.545","Text":"which is a given."},{"Start":"17:35.545 ","End":"17:37.120","Text":"Now, obviously, if you wanted,"},{"Start":"17:37.120 ","End":"17:40.210","Text":"you could rearrange this to make it look slightly neater,"},{"Start":"17:40.210 ","End":"17:44.060","Text":"but this is the answer to our question."}],"ID":9411},{"Watched":false,"Name":"1.3 Mass On Table Attached To Hanging Mass","Duration":"19m 52s","ChapterTopicVideoID":9142,"CourseChapterTopicPlaylistID":5365,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.350 ","End":"00:03.225","Text":"Hello. In this question,"},{"Start":"00:03.225 ","End":"00:09.300","Text":"we\u0027re being given this system where we have a spring with its coefficient being k,"},{"Start":"00:09.300 ","End":"00:12.255","Text":"we have our m_1 over here,"},{"Start":"00:12.255 ","End":"00:13.800","Text":"and because there are wheels here,"},{"Start":"00:13.800 ","End":"00:16.410","Text":"there\u0027s no friction over here."},{"Start":"00:16.410 ","End":"00:22.235","Text":"Then there\u0027s a string attached to m_1 which goes over a pulley system."},{"Start":"00:22.235 ","End":"00:25.305","Text":"Again, there\u0027s no friction over here and is"},{"Start":"00:25.305 ","End":"00:29.685","Text":"attached on its other end to another mass, m_2."},{"Start":"00:29.685 ","End":"00:32.190","Text":"Now, in our first question,"},{"Start":"00:32.190 ","End":"00:40.080","Text":"so at a, we\u0027re being asked to find what our point of equilibrium is."},{"Start":"00:40.080 ","End":"00:44.498","Text":"What is our value for x_0?"},{"Start":"00:44.498 ","End":"00:48.165","Text":"There are 2 ways to do this."},{"Start":"00:48.165 ","End":"00:55.250","Text":"Our first way is to say that sum of all of the forces is equal to 0,"},{"Start":"00:55.250 ","End":"00:58.460","Text":"because that\u0027s at a point of equilibrium or"},{"Start":"00:58.460 ","End":"01:01.940","Text":"to build out our equation and to find it from there."},{"Start":"01:01.940 ","End":"01:06.450","Text":"Let\u0027s do this first method first."},{"Start":"01:08.120 ","End":"01:11.715","Text":"Let\u0027s draw our force diagram."},{"Start":"01:11.715 ","End":"01:14.630","Text":"What we can do is we can say that this is"},{"Start":"01:14.630 ","End":"01:20.180","Text":"the positive x-direction and then we can take a look at this,"},{"Start":"01:20.180 ","End":"01:24.770","Text":"so in our rope we\u0027re going to have tension going in"},{"Start":"01:24.770 ","End":"01:31.190","Text":"this direction and then we\u0027re also going to have in 1 of the directions,"},{"Start":"01:31.190 ","End":"01:33.470","Text":"the force from our spring."},{"Start":"01:33.470 ","End":"01:34.790","Text":"The force of my spring,"},{"Start":"01:34.790 ","End":"01:36.365","Text":"I\u0027m going to call F_k."},{"Start":"01:36.365 ","End":"01:40.565","Text":"Now of course this isn\u0027t frictional force because we said that there\u0027s no friction."},{"Start":"01:40.565 ","End":"01:43.350","Text":"It\u0027s the force from the spring."},{"Start":"01:43.550 ","End":"01:49.700","Text":"Now if I write down my force equation for my mass number 1,"},{"Start":"01:49.700 ","End":"01:52.460","Text":"all I need is going in this direction,"},{"Start":"01:52.460 ","End":"01:56.630","Text":"because it\u0027s moving along the x-axis and what\u0027s happening for"},{"Start":"01:56.630 ","End":"02:02.095","Text":"m_1 on the y-axis doesn\u0027t factor into my harmonic motion."},{"Start":"02:02.095 ","End":"02:04.875","Text":"It\u0027s not on the same axes."},{"Start":"02:04.875 ","End":"02:08.405","Text":"I\u0027m going to have t and then, as we know,"},{"Start":"02:08.405 ","End":"02:12.500","Text":"when we write out our equation for our harmonic motion,"},{"Start":"02:12.500 ","End":"02:14.990","Text":"when we write out the sum of all of the forces,"},{"Start":"02:14.990 ","End":"02:20.635","Text":"so it always equals to negative k multiplied by x minus x_0."},{"Start":"02:20.635 ","End":"02:28.405","Text":"There\u0027s always a negative over here and then there\u0027s k multiplied by x minus x_0."},{"Start":"02:28.405 ","End":"02:32.325","Text":"This is the equation for F_k."},{"Start":"02:32.325 ","End":"02:34.655","Text":"Then as we said,"},{"Start":"02:34.655 ","End":"02:39.535","Text":"that the sum of all of the forces is going to be equal to 0."},{"Start":"02:39.535 ","End":"02:46.385","Text":"Now we\u0027re going to take a look at our m_2 and write out the force equation for that."},{"Start":"02:46.385 ","End":"02:51.994","Text":"We\u0027re going to say that our positive y-direction is in this direction going downwards."},{"Start":"02:51.994 ","End":"03:00.710","Text":"In that case, we have our force t in the rope going upwards and going downwards,"},{"Start":"03:00.710 ","End":"03:03.620","Text":"we have our m_2_g."},{"Start":"03:03.620 ","End":"03:07.730","Text":"I notice over here our motion for mass number"},{"Start":"03:07.730 ","End":"03:11.269","Text":"2 is only in the y-direction and not in the x-direction."},{"Start":"03:11.269 ","End":"03:14.250","Text":"That means we don\u0027t have to factor that in."},{"Start":"03:15.350 ","End":"03:17.955","Text":"Then writing this ours,"},{"Start":"03:17.955 ","End":"03:20.620","Text":"we\u0027ll have that our m_2_g,"},{"Start":"03:20.620 ","End":"03:26.735","Text":"because that\u0027s going in the positive direction minus our tension is equal to 0,"},{"Start":"03:26.735 ","End":"03:30.940","Text":"because the sum of all of our forces at the point of equilibrium is equal to 0."},{"Start":"03:30.940 ","End":"03:34.110","Text":"Now I can join these 2 equations."},{"Start":"03:34.110 ","End":"03:36.405","Text":"I can cross out the t\u0027s on both sides,"},{"Start":"03:36.405 ","End":"03:43.125","Text":"and we\u0027ll see that I get my m_2_g minus k,"},{"Start":"03:43.125 ","End":"03:49.900","Text":"x minus x_0 has to equal to 0."},{"Start":"03:50.030 ","End":"03:56.700","Text":"Now I can see that my x minus x_0 is equal to"},{"Start":"03:56.700 ","End":"04:03.550","Text":"m_2_g divided by k. Now,"},{"Start":"04:03.550 ","End":"04:05.795","Text":"I made a little bit of a mistake over here,"},{"Start":"04:05.795 ","End":"04:10.865","Text":"so what we\u0027re going to do is we\u0027re going to say that over here is going to be our origin."},{"Start":"04:10.865 ","End":"04:15.430","Text":"This is where the spring isn\u0027t being stretched or compressed."},{"Start":"04:15.430 ","End":"04:23.420","Text":"We can say that this is called x_0 tag and if we say that this is at the origin,"},{"Start":"04:23.420 ","End":"04:25.085","Text":"so it\u0027s at 0."},{"Start":"04:25.085 ","End":"04:26.990","Text":"That means that over here,"},{"Start":"04:26.990 ","End":"04:30.215","Text":"this is also a tag and this is also a tag."},{"Start":"04:30.215 ","End":"04:33.080","Text":"Notice that mistake and that means that this is also a tag and"},{"Start":"04:33.080 ","End":"04:36.545","Text":"what I\u0027m trying to find is just my x."},{"Start":"04:36.545 ","End":"04:38.555","Text":"That\u0027s this."},{"Start":"04:38.555 ","End":"04:43.370","Text":"If I say that my spring is limp,"},{"Start":"04:43.370 ","End":"04:48.225","Text":"isn\u0027t being stretched or compressed at this 0 at the origin."},{"Start":"04:48.225 ","End":"04:51.585","Text":"That means that this becomes 0,"},{"Start":"04:51.585 ","End":"04:53.939","Text":"so that means that my x,"},{"Start":"04:53.939 ","End":"04:55.835","Text":"this, which is what I\u0027m looking for,"},{"Start":"04:55.835 ","End":"04:57.410","Text":"my point of equilibrium,"},{"Start":"04:57.410 ","End":"05:06.500","Text":"so that\u0027s going to be at m_2_g divided by k. That\u0027s going to be somewhere."},{"Start":"05:06.500 ","End":"05:08.315","Text":"If I draw this on the graph,"},{"Start":"05:08.315 ","End":"05:12.865","Text":"it\u0027s going to be somewhere over here."},{"Start":"05:12.865 ","End":"05:15.735","Text":"This will be my point x."},{"Start":"05:15.735 ","End":"05:19.920","Text":"In order to get to this point x our point of equilibrium,"},{"Start":"05:19.920 ","End":"05:23.870","Text":"what has to happen is that our spring has to stretch a little bit"},{"Start":"05:23.870 ","End":"05:28.680","Text":"in order to balance out this mg over here."},{"Start":"05:28.680 ","End":"05:32.545","Text":"That\u0027s why our point is a little bit further along."},{"Start":"05:32.545 ","End":"05:35.985","Text":"Now we\u0027re moving on to b."},{"Start":"05:35.985 ","End":"05:41.515","Text":"In b, I\u0027m being asked to find what is Omega equal to."},{"Start":"05:41.515 ","End":"05:44.840","Text":"Here it\u0027s a little bit more complicated because we have"},{"Start":"05:44.840 ","End":"05:49.040","Text":"my m_1 and my m_2 and they\u0027re moving on different axes."},{"Start":"05:49.040 ","End":"05:53.080","Text":"What I\u0027m going to do in order to find my Omega?"},{"Start":"05:53.080 ","End":"05:58.880","Text":"In order to find my Omega and going to have to find the equation for motion."},{"Start":"05:58.880 ","End":"06:01.460","Text":"Now how am I going to find the equation for motion?"},{"Start":"06:01.460 ","End":"06:06.685","Text":"I\u0027m going to start by finding my equation for the force."},{"Start":"06:06.685 ","End":"06:09.405","Text":"Let\u0027s write out our equations."},{"Start":"06:09.405 ","End":"06:13.940","Text":"We have the sum of all of the forces on our first body."},{"Start":"06:13.940 ","End":"06:18.905","Text":"Those are in the x-direction on the first body and that\u0027s going to be equal to,"},{"Start":"06:18.905 ","End":"06:23.060","Text":"so we\u0027re going to have our T and then again as per usual, our F_k."},{"Start":"06:23.060 ","End":"06:26.490","Text":"That\u0027s going to be negative k,"},{"Start":"06:26.490 ","End":"06:30.630","Text":"and then because we chose our x_0 over here is equal to 0,"},{"Start":"06:30.630 ","End":"06:33.270","Text":"so I\u0027m just going to write out my x."},{"Start":"06:33.270 ","End":"06:38.030","Text":"Then this is going to be equal to my m_1,"},{"Start":"06:38.030 ","End":"06:39.560","Text":"the mass of body number 1,"},{"Start":"06:39.560 ","End":"06:41.735","Text":"multiplied by its acceleration,"},{"Start":"06:41.735 ","End":"06:46.405","Text":"so multiplied by x_1 double dot."},{"Start":"06:46.405 ","End":"06:50.975","Text":"Now, the sum of all of the forces on my second body,"},{"Start":"06:50.975 ","End":"07:01.020","Text":"so they\u0027re in the y-direction and they are m_2_g minus t,"},{"Start":"07:01.020 ","End":"07:05.270","Text":"because the positive y-direction is going downwards and this is going to be equal"},{"Start":"07:05.270 ","End":"07:10.145","Text":"to m_2 and then because it\u0027s acceleration is on the y-axis,"},{"Start":"07:10.145 ","End":"07:14.730","Text":"so multiplied by y double dot."},{"Start":"07:14.730 ","End":"07:20.584","Text":"Because we can say that the distance between every single point on the rope is constant,"},{"Start":"07:20.584 ","End":"07:23.055","Text":"our rope isn\u0027t changing length,"},{"Start":"07:23.055 ","End":"07:25.950","Text":"that means that the 2 bodies move together,"},{"Start":"07:25.950 ","End":"07:29.660","Text":"m_1 and m_2 move at the same rate,"},{"Start":"07:29.660 ","End":"07:31.940","Text":"which means that they accelerate at the same rate."},{"Start":"07:31.940 ","End":"07:40.970","Text":"That means that we can write this x_1 double dot is equal to y_2 double dot."},{"Start":"07:40.970 ","End":"07:46.880","Text":"Now notice if my green arrows over here representing the positive x and y directions,"},{"Start":"07:46.880 ","End":"07:49.610","Text":"were pointing in different directions,"},{"Start":"07:49.610 ","End":"07:52.100","Text":"then we might have negative signs over here,"},{"Start":"07:52.100 ","End":"07:55.190","Text":"so the relationship between our x_1 double dot and"},{"Start":"07:55.190 ","End":"07:59.835","Text":"our y_2 double dot might involve negative signs."},{"Start":"07:59.835 ","End":"08:03.905","Text":"Now what we can do is we can join up our 2 equations."},{"Start":"08:03.905 ","End":"08:06.530","Text":"As we can see, our T\u0027s cancel out and because"},{"Start":"08:06.530 ","End":"08:09.545","Text":"my x_1 double dot and my y_2 double dot are equal,"},{"Start":"08:09.545 ","End":"08:17.690","Text":"so I can write that I have my m_2g minus my k_x and this"},{"Start":"08:17.690 ","End":"08:25.950","Text":"is going to be equal to my m_1 plus my m_2 and then multiplied by,"},{"Start":"08:25.950 ","End":"08:29.330","Text":"I\u0027ll just write x_1 double dot."},{"Start":"08:29.330 ","End":"08:31.985","Text":"They\u0027re being multiplied by the same thing."},{"Start":"08:31.985 ","End":"08:36.680","Text":"Now what I have to do is I have to turn this equation over here into"},{"Start":"08:36.680 ","End":"08:41.810","Text":"the format for the force equation in harmonic motion,"},{"Start":"08:41.810 ","End":"08:43.700","Text":"which is this over here in red."},{"Start":"08:43.700 ","End":"08:49.860","Text":"I\u0027ve put it in red because it\u0027s good if you remember this and write this down."},{"Start":"08:50.390 ","End":"08:54.545","Text":"What we have to do is we have to put this into this format."},{"Start":"08:54.545 ","End":"08:55.760","Text":"Now it\u0027s very similar,"},{"Start":"08:55.760 ","End":"08:59.210","Text":"however here we have this edition of our m_2_g."},{"Start":"08:59.210 ","End":"09:04.505","Text":"I have to somehow reformat this,"},{"Start":"09:04.505 ","End":"09:06.755","Text":"so that it will look like this."},{"Start":"09:06.755 ","End":"09:11.275","Text":"Let\u0027s see how we do this and it\u0027s a little bit of a trick."},{"Start":"09:11.275 ","End":"09:14.510","Text":"Because I want my coefficient over here,"},{"Start":"09:14.510 ","End":"09:18.250","Text":"my k to be on the outside of my brackets,"},{"Start":"09:18.250 ","End":"09:22.240","Text":"what I can do is take out the common factor over here."},{"Start":"09:22.240 ","End":"09:25.910","Text":"Negative k outside because I want my negative k"},{"Start":"09:25.910 ","End":"09:30.995","Text":"out and that means that I\u0027m going to be multiplying this by x,"},{"Start":"09:30.995 ","End":"09:40.924","Text":"and then minus m_2g divided by k. Then you can see if you open up this bracket,"},{"Start":"09:40.924 ","End":"09:42.410","Text":"you\u0027ll get this exact thing,"},{"Start":"09:42.410 ","End":"09:46.430","Text":"and then this is going to be equal to my m_1 plus"},{"Start":"09:46.430 ","End":"09:53.510","Text":"my m_2 multiplied by my x_1 double dot."},{"Start":"09:53.510 ","End":"10:00.510","Text":"Then this is obviously just my m over here from this equation."},{"Start":"10:01.310 ","End":"10:06.820","Text":"Now we can see that we\u0027re in this exact same format over here."},{"Start":"10:06.820 ","End":"10:10.140","Text":"You can see that here my x_0,"},{"Start":"10:10.140 ","End":"10:12.765","Text":"which over here is just my x."},{"Start":"10:12.765 ","End":"10:15.060","Text":"Let\u0027s write this, my x_0,"},{"Start":"10:15.060 ","End":"10:18.510","Text":"which is equal to my x over here, what we found."},{"Start":"10:18.510 ","End":"10:22.025","Text":"We\u0027ll see we have here my m_2_g divided by k,"},{"Start":"10:22.025 ","End":"10:25.915","Text":"which is exactly what I have over here from my point of equilibrium."},{"Start":"10:25.915 ","End":"10:27.710","Text":"Remember in this equation,"},{"Start":"10:27.710 ","End":"10:30.980","Text":"this is my point of equilibrium and that\u0027s exactly what we have over here,"},{"Start":"10:30.980 ","End":"10:33.770","Text":"which we found from a. Alternatively,"},{"Start":"10:33.770 ","End":"10:36.520","Text":"we could have just simply started answering b,"},{"Start":"10:36.520 ","End":"10:40.720","Text":"and through that, got our answer for a."},{"Start":"10:41.990 ","End":"10:45.585","Text":"But we were asked to find what our Omega is."},{"Start":"10:45.585 ","End":"10:55.330","Text":"As we know, our Omega is equal to the square root of the coefficient of our x minus x_0."},{"Start":"10:55.730 ","End":"10:59.510","Text":"We can see that it\u0027s k just without"},{"Start":"10:59.510 ","End":"11:04.355","Text":"the minus divided by our coefficient of our x double dot."},{"Start":"11:04.355 ","End":"11:10.150","Text":"Here our coefficient of our x double dot is m_1 plus m_2."},{"Start":"11:13.140 ","End":"11:17.020","Text":"This is the answer."},{"Start":"11:17.020 ","End":"11:21.040","Text":"In my third question; question c,"},{"Start":"11:21.040 ","End":"11:25.750","Text":"I\u0027m being asked to find the maximum amplitude."},{"Start":"11:25.750 ","End":"11:30.760","Text":"I\u0027m trying to find that such that my tension in the rope;"},{"Start":"11:30.760 ","End":"11:34.135","Text":"my T, does not equal 0."},{"Start":"11:34.135 ","End":"11:38.240","Text":"It can equal anything else, but not 0."},{"Start":"11:38.730 ","End":"11:43.480","Text":"Let\u0027s go back 1 second to our diagram."},{"Start":"11:43.480 ","End":"11:46.285","Text":"We\u0027ll see over here up until now,"},{"Start":"11:46.285 ","End":"11:50.680","Text":"I\u0027ve written out my equations for the general motion of the system."},{"Start":"11:50.680 ","End":"11:53.590","Text":"I\u0027ve said how it\u0027s built and everything,"},{"Start":"11:53.590 ","End":"11:56.965","Text":"however, I don\u0027t know yet how it\u0027s going to be moving."},{"Start":"11:56.965 ","End":"11:58.720","Text":"What is its amplitude?"},{"Start":"11:58.720 ","End":"12:00.490","Text":"Because I can stretch out,"},{"Start":"12:00.490 ","End":"12:04.720","Text":"I can pull down my mass too downwards which will stretch"},{"Start":"12:04.720 ","End":"12:10.885","Text":"my spring and my mass will move past it\u0027s point of equilibrium."},{"Start":"12:10.885 ","End":"12:16.270","Text":"Then it will stretch to some amplitude and the whole system will move according to that."},{"Start":"12:16.270 ","End":"12:21.190","Text":"I\u0027m trying to find what is my maximum amplitude such"},{"Start":"12:21.190 ","End":"12:27.350","Text":"that my force is over here and my T aren\u0027t equal to 0."},{"Start":"12:27.890 ","End":"12:32.655","Text":"What will happen when my tensions are equal to 0?"},{"Start":"12:32.655 ","End":"12:34.560","Text":"That means that there\u0027s no tension in"},{"Start":"12:34.560 ","End":"12:38.385","Text":"the rope which means that my rope isn\u0027t going to be torts,"},{"Start":"12:38.385 ","End":"12:42.010","Text":"it\u0027s going to be loose and limp"},{"Start":"12:42.010 ","End":"12:46.390","Text":"which means that my harmonic motion will cease because all"},{"Start":"12:46.390 ","End":"12:50.320","Text":"of these equations that we\u0027ve built in order to find out what our Omega is and what"},{"Start":"12:50.320 ","End":"12:54.490","Text":"our x is and everything or at least Omega,"},{"Start":"12:54.490 ","End":"12:57.880","Text":"it\u0027s based on the fact that our tension exists."},{"Start":"12:57.880 ","End":"12:59.755","Text":"There\u0027s tension in the rope."},{"Start":"12:59.755 ","End":"13:05.170","Text":"In actual fact, this question is checking what the limits are."},{"Start":"13:05.170 ","End":"13:09.820","Text":"We\u0027re being asked what the limits are that I can pull my masses and"},{"Start":"13:09.820 ","End":"13:15.680","Text":"play around with them such that my system will still move in harmonic motion."},{"Start":"13:15.870 ","End":"13:19.300","Text":"If my tension can equal 0,"},{"Start":"13:19.300 ","End":"13:27.680","Text":"so in actual fact I need my tension to be bigger than 0."},{"Start":"13:28.580 ","End":"13:31.230","Text":"How I\u0027m I going to find the amplitude?"},{"Start":"13:31.230 ","End":"13:36.390","Text":"Let\u0027s go back to what we have over here with our equations for our force."},{"Start":"13:36.390 ","End":"13:38.475","Text":"I have two equations."},{"Start":"13:38.475 ","End":"13:44.744","Text":"I have my forces on my first body and forces on my second body."},{"Start":"13:44.744 ","End":"13:48.970","Text":"If I look at my first equation what I can do is I can"},{"Start":"13:48.970 ","End":"13:53.440","Text":"solve this by working out what my acceleration is;"},{"Start":"13:53.440 ","End":"13:57.955","Text":"so substituting this into over here what my acceleration is,"},{"Start":"13:57.955 ","End":"14:04.720","Text":"and then from that trying to isolate out my T or what I can do is use"},{"Start":"14:04.720 ","End":"14:07.675","Text":"the second equation and isolate out"},{"Start":"14:07.675 ","End":"14:12.250","Text":"my T. What I\u0027m going to do is I\u0027m going to use my second equation,"},{"Start":"14:12.250 ","End":"14:13.825","Text":"so let\u0027s copy it out."},{"Start":"14:13.825 ","End":"14:22.540","Text":"I have my m_2g minus T is equal to my m2_y_2-double dot."},{"Start":"14:22.540 ","End":"14:27.070","Text":"As we know from what we wrote over here,"},{"Start":"14:27.070 ","End":"14:33.200","Text":"so this is also equal to my x_1 doubl- dot multiplied by m_2."},{"Start":"14:34.440 ","End":"14:39.415","Text":"I\u0027m going to isolate out my T and I\u0027m going to use my x-double dot."},{"Start":"14:39.415 ","End":"14:44.185","Text":"From now on, I\u0027m going to say that my x_1-double dot is just equal to x-double dot."},{"Start":"14:44.185 ","End":"14:53.980","Text":"I\u0027ll have that my T is equal to m_2g minus m_2x- double-dot."},{"Start":"14:53.980 ","End":"14:57.260","Text":"Just isolate out my T from here."},{"Start":"14:58.140 ","End":"15:03.970","Text":"What I\u0027m going to do now is to remind you that the equation that is written"},{"Start":"15:03.970 ","End":"15:09.950","Text":"in red over here is my general solution for my harmonic motion."},{"Start":"15:10.200 ","End":"15:13.015","Text":"We already have my x_0,"},{"Start":"15:13.015 ","End":"15:14.650","Text":"we already found what this is."},{"Start":"15:14.650 ","End":"15:17.928","Text":"That was just this over here;"},{"Start":"15:17.928 ","End":"15:21.025","Text":"our x, so that we found already."},{"Start":"15:21.025 ","End":"15:23.845","Text":"What do we want to do is we want to somehow"},{"Start":"15:23.845 ","End":"15:27.354","Text":"write this out because I have my amplitude over here,"},{"Start":"15:27.354 ","End":"15:32.440","Text":"so this is the variable I\u0027m looking for in order to substitute it into here."},{"Start":"15:32.440 ","End":"15:40.420","Text":"I need to take the second derivative of this equation in order to substitute into here."},{"Start":"15:40.420 ","End":"15:43.075","Text":"I\u0027m just going to write it out."},{"Start":"15:43.075 ","End":"15:45.466","Text":"Forget the mathematics for a second,"},{"Start":"15:45.466 ","End":"15:48.260","Text":"you can do this alone if you want."},{"Start":"15:48.260 ","End":"15:53.859","Text":"It\u0027s going to be equal to negative Omega squared A multiplied"},{"Start":"15:53.859 ","End":"16:00.490","Text":"by cosine of Omega T plus Phi."},{"Start":"16:00.490 ","End":"16:03.820","Text":"All I\u0027ve done here is I\u0027ve taken the derivative of"},{"Start":"16:03.820 ","End":"16:07.570","Text":"this equation twice in order to get my x-double dot."},{"Start":"16:07.570 ","End":"16:12.685","Text":"All I have to do is I have to substitute this into my x-double dot over here."},{"Start":"16:12.685 ","End":"16:16.690","Text":"I\u0027m going to have that my T is equal to"},{"Start":"16:16.690 ","End":"16:22.930","Text":"m_2g minus m_2 and then we have a minus and a minus"},{"Start":"16:22.930 ","End":"16:25.630","Text":"so this is plus Omega^2"},{"Start":"16:25.630 ","End":"16:33.850","Text":"A cosine of Omega T plus Phi."},{"Start":"16:33.850 ","End":"16:36.250","Text":"Then from what I said over here,"},{"Start":"16:36.250 ","End":"16:40.405","Text":"this all has to be bigger than 0."},{"Start":"16:40.405 ","End":"16:43.370","Text":"This was my condition."},{"Start":"16:43.950 ","End":"16:47.980","Text":"My key phrase over here is that my T;"},{"Start":"16:47.980 ","End":"16:50.677","Text":"my tension, has to always;"},{"Start":"16:50.677 ","End":"16:52.320","Text":"the keyword being always,"},{"Start":"16:52.320 ","End":"16:54.490","Text":"be bigger than 0."},{"Start":"16:54.490 ","End":"17:02.290","Text":"That means that at every single time no matter what my value for time is over here,"},{"Start":"17:02.290 ","End":"17:08.605","Text":"this expression is always going to be a positive number which is bigger than 0."},{"Start":"17:08.605 ","End":"17:11.305","Text":"How I\u0027m I going to do that?"},{"Start":"17:11.305 ","End":"17:17.710","Text":"An easy way is to work out what my minimum tension is, so T_min."},{"Start":"17:17.710 ","End":"17:22.600","Text":"If my T_min my minimum tension is bigger than 0,"},{"Start":"17:22.600 ","End":"17:25.900","Text":"then any tension that is bigger than my minimum tension,"},{"Start":"17:25.900 ","End":"17:31.315","Text":"so any other tension will also therefore be bigger than 0."},{"Start":"17:31.315 ","End":"17:35.950","Text":"That way my tension will always be bigger than 0 no matter"},{"Start":"17:35.950 ","End":"17:41.605","Text":"what value for a t for time I plug in over here."},{"Start":"17:41.605 ","End":"17:45.820","Text":"One way in order to find when my tension will be at"},{"Start":"17:45.820 ","End":"17:50.170","Text":"a minimum is to take the derivative of this expression."},{"Start":"17:50.170 ","End":"17:52.874","Text":"That\u0027s one way taking the derivative,"},{"Start":"17:52.874 ","End":"17:56.290","Text":"but an easier way is to check what"},{"Start":"17:56.290 ","End":"18:00.505","Text":"my minimum value for cosine of Omega T plus Phi can be."},{"Start":"18:00.505 ","End":"18:09.355","Text":"As we know the minimum value that my sine or cosine can be is to be equal to negative 1."},{"Start":"18:09.355 ","End":"18:15.050","Text":"That\u0027s the minimum value of this expression over here that it can possibly be."},{"Start":"18:15.630 ","End":"18:18.205","Text":"Then my T_min;"},{"Start":"18:18.205 ","End":"18:19.830","Text":"my minimum tension,"},{"Start":"18:19.830 ","End":"18:26.080","Text":"is going to be equal to my m_2g and then plus"},{"Start":"18:26.080 ","End":"18:30.550","Text":"my m_2 Omega^2 multiplied by"},{"Start":"18:30.550 ","End":"18:36.310","Text":"A and then multiply it by this minimum value over here which is negative 1."},{"Start":"18:36.310 ","End":"18:40.225","Text":"Then all of this must be bigger than 0."},{"Start":"18:40.225 ","End":"18:45.220","Text":"All we have to do is we have to isolate out our A."},{"Start":"18:45.220 ","End":"18:47.455","Text":"We can write this out."},{"Start":"18:47.455 ","End":"18:49.408","Text":"Here we have a minus;"},{"Start":"18:49.408 ","End":"18:51.310","Text":"this expression is a negative,"},{"Start":"18:51.310 ","End":"18:53.395","Text":"so we can move it to the other side."},{"Start":"18:53.395 ","End":"19:03.310","Text":"We\u0027re going to have that our m_2g is going to be bigger than m_2 Omega^2"},{"Start":"19:03.310 ","End":"19:10.360","Text":"A. I can divide both sides by my m_2 and now I just isolate out"},{"Start":"19:10.360 ","End":"19:19.340","Text":"my A. I get that my A has to be smaller than my g divided by my Omega^2."},{"Start":"19:20.190 ","End":"19:24.453","Text":"That means that we can say that my A_max;"},{"Start":"19:24.453 ","End":"19:29.120","Text":"so what we were trying to find over here,"},{"Start":"19:29.120 ","End":"19:33.550","Text":"is when it\u0027s smaller than g divided by Omega^2 when"},{"Start":"19:33.550 ","End":"19:38.370","Text":"the limit from the left-hand side of our amplitude"},{"Start":"19:38.370 ","End":"19:42.705","Text":"when the limit is equal to g divided by Omega^2"},{"Start":"19:42.705 ","End":"19:48.295","Text":"then we have our maximum amplitude and our left limit we\u0027re taking."},{"Start":"19:48.295 ","End":"19:52.190","Text":"That\u0027s the end of this question."}],"ID":9412},{"Watched":false,"Name":"1.4 Constant Forces And Shifting The Point Of Equilibrium","Duration":"11m 41s","ChapterTopicVideoID":9143,"CourseChapterTopicPlaylistID":5365,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.055","Text":"Hello. In this lesson,"},{"Start":"00:02.055 ","End":"00:07.710","Text":"I want to speak to you about a certain technique that will appear in a lot of questions,"},{"Start":"00:07.710 ","End":"00:11.085","Text":"and also a lot of professors like working with it."},{"Start":"00:11.085 ","End":"00:16.875","Text":"We\u0027re going to have a similar type of situation that we\u0027ve seen up until now."},{"Start":"00:16.875 ","End":"00:19.650","Text":"We have a wall, at the base of the wall,"},{"Start":"00:19.650 ","End":"00:22.035","Text":"this is where our origin is at point 0."},{"Start":"00:22.035 ","End":"00:23.790","Text":"We have some spring,"},{"Start":"00:23.790 ","End":"00:27.420","Text":"with a spring constant of k and the mass of mass m,"},{"Start":"00:27.420 ","End":"00:32.220","Text":"which is located a distance of L_0 away from the origin."},{"Start":"00:32.220 ","End":"00:33.870","Text":"There is wheels over here,"},{"Start":"00:33.870 ","End":"00:37.305","Text":"so we know that there\u0027s no friction in the system."},{"Start":"00:37.305 ","End":"00:43.245","Text":"This L_0 represents also where our spring is limp."},{"Start":"00:43.245 ","End":"00:45.135","Text":"No forces are acting,"},{"Start":"00:45.135 ","End":"00:47.640","Text":"our spring isn\u0027t being stretched or compressed,"},{"Start":"00:47.640 ","End":"00:49.770","Text":"it\u0027s limp, it\u0027s resting."},{"Start":"00:49.770 ","End":"00:52.095","Text":"Now we have, very important,"},{"Start":"00:52.095 ","End":"00:58.455","Text":"some constant force pushing our mass in this direction."},{"Start":"00:58.455 ","End":"01:00.240","Text":"This is very important."},{"Start":"01:00.240 ","End":"01:02.940","Text":"This is our positive x-direction."},{"Start":"01:02.940 ","End":"01:07.380","Text":"Now, when we\u0027re writing out our equation for the sum of the forces,"},{"Start":"01:07.380 ","End":"01:11.505","Text":"so right now because all of the forces are acting in the x-direction,"},{"Start":"01:11.505 ","End":"01:13.260","Text":"so we have that."},{"Start":"01:13.260 ","End":"01:18.150","Text":"That equals, or all of the forces that are relevant right now to our example."},{"Start":"01:18.150 ","End":"01:21.525","Text":"That equals to our negative k,"},{"Start":"01:21.525 ","End":"01:26.080","Text":"and then we have our x minus instead of x_0 here it\u0027s L_0."},{"Start":"01:27.350 ","End":"01:31.560","Text":"Then we add in our constant force,"},{"Start":"01:31.560 ","End":"01:34.380","Text":"remembering that our force has to be constant."},{"Start":"01:34.380 ","End":"01:37.080","Text":"This is incorrect if the force is not constant."},{"Start":"01:37.080 ","End":"01:41.445","Text":"This is equal to our mass times acceleration."},{"Start":"01:41.445 ","End":"01:46.530","Text":"We can see that we have our same equation for harmonic motion that we\u0027re used to,"},{"Start":"01:46.530 ","End":"01:51.730","Text":"but we have this added F over here representing this force."},{"Start":"01:52.460 ","End":"01:54.975","Text":"Now what I\u0027m going to do,"},{"Start":"01:54.975 ","End":"01:58.920","Text":"I\u0027m going to rearrange this so that this whole expression,"},{"Start":"01:58.920 ","End":"02:00.525","Text":"the side of the equation,"},{"Start":"02:00.525 ","End":"02:04.710","Text":"is going to be equal to what we usually want it to be equal to,"},{"Start":"02:04.710 ","End":"02:09.915","Text":"which is negative k open brackets and then some expression inside,"},{"Start":"02:09.915 ","End":"02:13.065","Text":"close brackets, equals to ma."},{"Start":"02:13.065 ","End":"02:18.720","Text":"Let\u0027s see what we do. We\u0027re going to keep my negative k out,"},{"Start":"02:18.720 ","End":"02:20.445","Text":"and then our x,"},{"Start":"02:20.445 ","End":"02:26.580","Text":"and then we have negative my L_0."},{"Start":"02:26.580 ","End":"02:31.245","Text":"But then what we want to do is we want to put this F inside over here."},{"Start":"02:31.245 ","End":"02:38.460","Text":"Then over here we\u0027re going to have negative F divided"},{"Start":"02:38.460 ","End":"02:45.455","Text":"by my k. Then you\u0027ll see that when I multiply this out,"},{"Start":"02:45.455 ","End":"02:49.580","Text":"I\u0027m going to get negative k(x minus l) over here,"},{"Start":"02:49.580 ","End":"02:51.350","Text":"and then plus F,"},{"Start":"02:51.350 ","End":"02:59.445","Text":"and this is going to be equal to ma of course."},{"Start":"02:59.445 ","End":"03:02.715","Text":"Now because I want it in the form for my harmonic motion,"},{"Start":"03:02.715 ","End":"03:09.250","Text":"which means I need to have negative k in brackets and some x minus x_0."},{"Start":"03:09.800 ","End":"03:14.175","Text":"What I\u0027m going to do is I want to make this term"},{"Start":"03:14.175 ","End":"03:19.815","Text":"equal to my x_0 term in my general equation."},{"Start":"03:19.815 ","End":"03:23.760","Text":"What I\u0027m going to do is I\u0027m going to add in brackets over here,"},{"Start":"03:23.760 ","End":"03:27.075","Text":"so my x_0 term is in these smaller brackets."},{"Start":"03:27.075 ","End":"03:28.500","Text":"Then because I added the brackets,"},{"Start":"03:28.500 ","End":"03:29.640","Text":"the negative and the negative,"},{"Start":"03:29.640 ","End":"03:33.315","Text":"so here I have to put a positive to equal that out."},{"Start":"03:33.315 ","End":"03:34.680","Text":"If you multiply this out,"},{"Start":"03:34.680 ","End":"03:38.320","Text":"you\u0027ll see that we get the exact same answer over here."},{"Start":"03:38.900 ","End":"03:42.225","Text":"Now we\u0027re in the format that we want."},{"Start":"03:42.225 ","End":"03:44.625","Text":"What we found over here,"},{"Start":"03:44.625 ","End":"03:51.315","Text":"x_0 is our point of equilibrium."},{"Start":"03:51.315 ","End":"03:55.875","Text":"That means that when we\u0027re dealing with a system like this,"},{"Start":"03:55.875 ","End":"04:01.080","Text":"where we have not just our mass moving in a spring,"},{"Start":"04:01.080 ","End":"04:03.780","Text":"but rather we have a constant force acting,"},{"Start":"04:03.780 ","End":"04:07.170","Text":"that means that our point of equilibrium is going to change."},{"Start":"04:07.170 ","End":"04:09.480","Text":"It\u0027s no longer going to be over here,"},{"Start":"04:09.480 ","End":"04:14.400","Text":"but the force is driving our point of equilibrium to be at some other point,"},{"Start":"04:14.400 ","End":"04:17.970","Text":"and remembering that at my point of equilibrium,"},{"Start":"04:17.970 ","End":"04:21.820","Text":"the sum of all of the forces is equal to 0."},{"Start":"04:22.960 ","End":"04:26.600","Text":"At my x_0, it\u0027s my point of equilibrium,"},{"Start":"04:26.600 ","End":"04:30.020","Text":"which means that the sum of all my forces there is equal to 0."},{"Start":"04:30.020 ","End":"04:34.280","Text":"My x_0 is going to be something around over here."},{"Start":"04:34.280 ","End":"04:37.310","Text":"My forces at this point are equal to 0,"},{"Start":"04:37.310 ","End":"04:40.000","Text":"which means that around this point,"},{"Start":"04:40.000 ","End":"04:43.260","Text":"at every point that\u0027s close to my x_0."},{"Start":"04:43.260 ","End":"04:45.750","Text":"I know that around there,"},{"Start":"04:45.750 ","End":"04:49.875","Text":"my harmonic motion is going to occur through this point,"},{"Start":"04:49.875 ","End":"04:52.360","Text":"up and down, up and down like this."},{"Start":"04:52.580 ","End":"04:57.375","Text":"This is where my amplitudes will have an a to here and then a to there."},{"Start":"04:57.375 ","End":"05:02.265","Text":"The size of my a is obviously something to do with my initial conditions."},{"Start":"05:02.265 ","End":"05:05.760","Text":"But my harmonic motion is occurring around this point,"},{"Start":"05:05.760 ","End":"05:11.295","Text":"not at this point where the spring was loose,"},{"Start":"05:11.295 ","End":"05:12.390","Text":"but rather at this point,"},{"Start":"05:12.390 ","End":"05:14.775","Text":"at my point of equilibrium."},{"Start":"05:14.775 ","End":"05:17.580","Text":"Now we\u0027ve found our point of equilibrium,"},{"Start":"05:17.580 ","End":"05:20.880","Text":"we also could have found it by saying that the sum of all of our forces is equal to"},{"Start":"05:20.880 ","End":"05:24.675","Text":"0 and then we also would have found our x_0,"},{"Start":"05:24.675 ","End":"05:28.805","Text":"our point of equilibrium to be this same value,"},{"Start":"05:28.805 ","End":"05:31.980","Text":"like what we did in our previous question."},{"Start":"05:32.540 ","End":"05:36.990","Text":"Now I\u0027m going to speak about some technique which was pretty"},{"Start":"05:36.990 ","End":"05:41.445","Text":"common and it might also help you to make the equations a little bit easier to solve."},{"Start":"05:41.445 ","End":"05:45.420","Text":"That means that instead of having the origin over here."},{"Start":"05:45.420 ","End":"05:47.115","Text":"To make it a little bit easier,"},{"Start":"05:47.115 ","End":"05:48.750","Text":"I\u0027ll move my origin,"},{"Start":"05:48.750 ","End":"05:53.625","Text":"I\u0027ll shift it to where my point of equilibrium is."},{"Start":"05:53.625 ","End":"05:57.660","Text":"This will be my 0 point instead of over here."},{"Start":"05:57.660 ","End":"06:00.885","Text":"It will be easier to solve a lot of questions like that."},{"Start":"06:00.885 ","End":"06:09.030","Text":"What I will do is I will make some variable x tag,"},{"Start":"06:09.030 ","End":"06:17.910","Text":"let\u0027s say, and say that that equals to what\u0027s over here inside my negative k bracket."},{"Start":"06:17.910 ","End":"06:24.570","Text":"That\u0027s going to be equal to my x minus my L_0 plus"},{"Start":"06:24.570 ","End":"06:32.190","Text":"F divided by k. This is my new position that\u0027s shifted,"},{"Start":"06:32.190 ","End":"06:35.805","Text":"so my origin is now at my point of equilibrium."},{"Start":"06:35.805 ","End":"06:39.225","Text":"Then if I want to have my acceleration,"},{"Start":"06:39.225 ","End":"06:41.430","Text":"that will be my a tag,"},{"Start":"06:41.430 ","End":"06:46.725","Text":"so that\u0027s going to be equal to my x tag double-dots,"},{"Start":"06:46.725 ","End":"06:51.060","Text":"and my x tag double-dot because all of this is a constant."},{"Start":"06:51.060 ","End":"06:53.280","Text":"Remember everything here is a constant,"},{"Start":"06:53.280 ","End":"06:55.170","Text":"my L_0 is a constant, it\u0027s given,"},{"Start":"06:55.170 ","End":"06:58.950","Text":"it\u0027s a certain position plus my F,"},{"Start":"06:58.950 ","End":"07:01.710","Text":"which is a constant force divided by my k,"},{"Start":"07:01.710 ","End":"07:03.720","Text":"which is also a constant."},{"Start":"07:03.720 ","End":"07:08.430","Text":"When I take the derivative or the second derivative,"},{"Start":"07:08.430 ","End":"07:11.370","Text":"so all my constant terms fall off."},{"Start":"07:11.370 ","End":"07:14.505","Text":"I\u0027m going to get that my x tag double-dot,"},{"Start":"07:14.505 ","End":"07:18.345","Text":"is simply going to be because these terms will fall off,"},{"Start":"07:18.345 ","End":"07:20.445","Text":"it will just be equal to x double-dot,"},{"Start":"07:20.445 ","End":"07:21.705","Text":"which was as we know,"},{"Start":"07:21.705 ","End":"07:25.470","Text":"equal to my regular acceleration."},{"Start":"07:25.470 ","End":"07:30.945","Text":"When I shift my origin over here,"},{"Start":"07:30.945 ","End":"07:35.145","Text":"so only my x values will change."},{"Start":"07:35.145 ","End":"07:37.560","Text":"I need a new variable, it will be x tag,"},{"Start":"07:37.560 ","End":"07:43.780","Text":"but my a over here is going to remain the same as you can see over here."},{"Start":"07:44.120 ","End":"07:49.260","Text":"What that means is that I can rewrite my equation."},{"Start":"07:49.260 ","End":"07:52.815","Text":"This complex thing as simply"},{"Start":"07:52.815 ","End":"07:59.715","Text":"negative k. Then this whole mass over here is simply my x tag,"},{"Start":"07:59.715 ","End":"08:01.770","Text":"so multiply by x tag."},{"Start":"08:01.770 ","End":"08:06.150","Text":"This is going to be equal to my mass times acceleration."},{"Start":"08:06.150 ","End":"08:09.285","Text":"My ma or my ma tag, it doesn\u0027t really matter."},{"Start":"08:09.285 ","End":"08:12.280","Text":"My a is equal to a tag."},{"Start":"08:12.830 ","End":"08:15.540","Text":"If we\u0027re taking a look at this,"},{"Start":"08:15.540 ","End":"08:22.050","Text":"what I could have done is just said that originally that my point at x_0=0."},{"Start":"08:22.050 ","End":"08:23.770","Text":"That its position is 0,"},{"Start":"08:23.770 ","End":"08:25.935","Text":"and in that case,"},{"Start":"08:25.935 ","End":"08:28.290","Text":"what I could have written, so I\u0027m going to write it over here,"},{"Start":"08:28.290 ","End":"08:33.965","Text":"I could have instead of writing negative kx tag is equal to ma tag,"},{"Start":"08:33.965 ","End":"08:43.880","Text":"I can just write that negative kx is equal to ma, simply."},{"Start":"08:43.880 ","End":"08:46.220","Text":"Instead of going through all of this,"},{"Start":"08:46.220 ","End":"08:48.290","Text":"because it\u0027s just a shift in position."},{"Start":"08:48.290 ","End":"08:53.015","Text":"It doesn\u0027t matter, as long as my force acting over here is a constant force,"},{"Start":"08:53.015 ","End":"08:56.180","Text":"so I can shift my origin wherever I want,"},{"Start":"08:56.180 ","End":"09:00.390","Text":"it doesn\u0027t really matter and then I\u0027m going to end up with the same equation."},{"Start":"09:00.390 ","End":"09:07.010","Text":"A lot of the time you\u0027ll find that your professors and in questions,"},{"Start":"09:07.010 ","End":"09:10.730","Text":"you\u0027ll get that the equation for your harmonic motion"},{"Start":"09:10.730 ","End":"09:14.600","Text":"when a constant force is acting on the mass is equal to this."},{"Start":"09:14.600 ","End":"09:16.160","Text":"Because as you can see,"},{"Start":"09:16.160 ","End":"09:17.495","Text":"when you write it like that,"},{"Start":"09:17.495 ","End":"09:20.615","Text":"you don\u0027t need to include your F in."},{"Start":"09:20.615 ","End":"09:26.585","Text":"If we\u0027re going to go according to steps on how we would do this,"},{"Start":"09:26.585 ","End":"09:30.720","Text":"so what we would do is step number 1,"},{"Start":"09:30.720 ","End":"09:33.090","Text":"we would write out"},{"Start":"09:33.090 ","End":"09:40.860","Text":"our force equations and we\u0027ll say that at the point of equilibrium they equal to 0."},{"Start":"09:40.860 ","End":"09:46.250","Text":"That\u0027s our first thing. Our equation and then say that it\u0027s equal to 0."},{"Start":"09:46.250 ","End":"09:53.630","Text":"Then we\u0027ll say from that that we can work out what our x_0 is equal to."},{"Start":"09:53.660 ","End":"09:57.795","Text":"Then once we found what our x_0 is equal to,"},{"Start":"09:57.795 ","End":"10:02.760","Text":"our number 3 will be to write out this."},{"Start":"10:02.760 ","End":"10:05.760","Text":"We have a change in variable."},{"Start":"10:05.760 ","End":"10:09.150","Text":"We have that our x tag is equal to,"},{"Start":"10:09.150 ","End":"10:13.325","Text":"and then we have our brackets over here including our x_0,"},{"Start":"10:13.325 ","End":"10:15.745","Text":"so our x minus our x_0."},{"Start":"10:15.745 ","End":"10:21.380","Text":"Then once we\u0027ve gotten to this point,"},{"Start":"10:21.380 ","End":"10:25.760","Text":"instead of saying that our x_0 was our point of equilibrium,"},{"Start":"10:25.760 ","End":"10:30.750","Text":"so now we can just shift our origin to our point of equilibrium."},{"Start":"10:30.750 ","End":"10:34.730","Text":"Say that our point of equilibrium is at the origin, so it equals to 0."},{"Start":"10:34.730 ","End":"10:38.494","Text":"Then we can just write out our fourth step,"},{"Start":"10:38.494 ","End":"10:43.640","Text":"which means that we don\u0027t have to include any of this with our F and this whole mess."},{"Start":"10:43.640 ","End":"10:47.720","Text":"We can just say that our equation is equal to negative kx,"},{"Start":"10:47.720 ","End":"10:51.360","Text":"which is equal to ma."},{"Start":"10:52.070 ","End":"10:59.435","Text":"The conclusion that we get from this whole thing is that the constant forces,"},{"Start":"10:59.435 ","End":"11:05.070","Text":"so this, and very important that it\u0027s a constant."},{"Start":"11:05.070 ","End":"11:08.540","Text":"Simply shift the system. What does that mean?"},{"Start":"11:08.540 ","End":"11:13.660","Text":"That means that the origin is shifted to the point of equilibrium,"},{"Start":"11:13.660 ","End":"11:15.035","Text":"and so its value,"},{"Start":"11:15.035 ","End":"11:19.700","Text":"the value of the point of equilibrium may be ignored because that is where it equals 0."},{"Start":"11:19.700 ","End":"11:25.400","Text":"Now remember, if the force acting isn\u0027t constant,"},{"Start":"11:25.400 ","End":"11:28.475","Text":"so let\u0027s say we have another spring added to the system,"},{"Start":"11:28.475 ","End":"11:30.110","Text":"then, of course,"},{"Start":"11:30.110 ","End":"11:31.970","Text":"none of this will be correct."},{"Start":"11:31.970 ","End":"11:38.605","Text":"Then we have to add in our spring like we did a few questions ago."},{"Start":"11:38.605 ","End":"11:41.470","Text":"That\u0027s the end of this lesson."}],"ID":9413},{"Watched":false,"Name":"1.5 Rod Hanging Attached To Spring","Duration":"31m 59s","ChapterTopicVideoID":10288,"CourseChapterTopicPlaylistID":5365,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.220","Text":"Hello. In this question,"},{"Start":"00:02.220 ","End":"00:07.170","Text":"we\u0027re being told that a rod of length L and of mass M hangs from"},{"Start":"00:07.170 ","End":"00:13.140","Text":"the ceiling and is free to rotate about its point of attachment like a pendulum."},{"Start":"00:13.140 ","End":"00:21.554","Text":"The other end of the rod over here is attached via a spring of constant K to the wall."},{"Start":"00:21.554 ","End":"00:24.105","Text":"Now we have 2 questions to answer."},{"Start":"00:24.105 ","End":"00:25.965","Text":"Let\u0027s start with question number 1."},{"Start":"00:25.965 ","End":"00:29.219","Text":"We\u0027re being asked to show that the motion of the rod at"},{"Start":"00:29.219 ","End":"00:33.825","Text":"small angles is harmonic and to find our Omega."},{"Start":"00:33.825 ","End":"00:36.070","Text":"Let\u0027s see how we do this."},{"Start":"00:36.070 ","End":"00:41.730","Text":"The first thing to notice about this question is that we\u0027re dealing with a rigid body."},{"Start":"00:41.730 ","End":"00:45.124","Text":"Up until now, in our previous lessons,"},{"Start":"00:45.124 ","End":"00:48.515","Text":"in this chapter, we\u0027ve been dealing with point masses."},{"Start":"00:48.515 ","End":"00:50.555","Text":"When dealing with a point mass,"},{"Start":"00:50.555 ","End":"00:53.030","Text":"I\u0027m going to say PM, point mass."},{"Start":"00:53.030 ","End":"00:57.379","Text":"The way that we began to solve it is to say that the sum of all of"},{"Start":"00:57.379 ","End":"01:04.785","Text":"our forces was equal to mx double dot and to write out our equation."},{"Start":"01:04.785 ","End":"01:09.140","Text":"Now over here, because we\u0027re dealing with a rigid body."},{"Start":"01:09.140 ","End":"01:11.795","Text":"This is also a very common question that they can ask,"},{"Start":"01:11.795 ","End":"01:15.505","Text":"to include rigid bodies in harmonic questions."},{"Start":"01:15.505 ","End":"01:18.530","Text":"There\u0027s a slightly different way to solve it."},{"Start":"01:18.530 ","End":"01:21.030","Text":"When dealing with a rigid body,"},{"Start":"01:21.030 ","End":"01:26.855","Text":"R.B what we want to be doing is to write out our equation."},{"Start":"01:26.855 ","End":"01:31.790","Text":"Also, that the sum of all of the forces is equal to mx."},{"Start":"01:31.790 ","End":"01:35.929","Text":"But then, we also remember that in order to solve rigid body questions,"},{"Start":"01:35.929 ","End":"01:39.844","Text":"we also use the equation that the sum of all of the torques,"},{"Start":"01:39.844 ","End":"01:44.815","Text":"or sum of all of the moments is equal to I Alpha."},{"Start":"01:44.815 ","End":"01:50.120","Text":"Then what we also did is we wrote the connection between I acceleration,"},{"Start":"01:50.120 ","End":"01:51.935","Text":"our linear acceleration,"},{"Start":"01:51.935 ","End":"01:53.840","Text":"and our angular acceleration."},{"Start":"01:53.840 ","End":"02:00.785","Text":"That was that, our x double dot or a is equal to Omega Alpha."},{"Start":"02:00.785 ","End":"02:03.065","Text":"Now, for this question specifically,"},{"Start":"02:03.065 ","End":"02:05.165","Text":"I don\u0027t need this third equation."},{"Start":"02:05.165 ","End":"02:08.260","Text":"I\u0027m going to rub this out right now."},{"Start":"02:08.260 ","End":"02:13.600","Text":"I\u0027m only going to be using these 2 equations in order to solve this question."},{"Start":"02:13.600 ","End":"02:18.049","Text":"I\u0027m going to start with writing out the sum of all of the torques."},{"Start":"02:18.049 ","End":"02:20.375","Text":"Now, how am I going to do this?"},{"Start":"02:20.375 ","End":"02:21.770","Text":"I\u0027m not going to write out"},{"Start":"02:21.770 ","End":"02:27.935","Text":"my torque equations when my system is in this type of position."},{"Start":"02:27.935 ","End":"02:31.460","Text":"Because right now I\u0027m at my point of equilibrium."},{"Start":"02:31.460 ","End":"02:41.734","Text":"What I\u0027m going to do is I\u0027m going to imagine that I\u0027m taking my rod over to here."},{"Start":"02:41.734 ","End":"02:45.120","Text":"I\u0027ve pulled it at some kind of angle Theta."},{"Start":"02:45.890 ","End":"02:48.945","Text":"The spring has been stretched as well."},{"Start":"02:48.945 ","End":"02:51.680","Text":"Now what I\u0027m going to do is I\u0027m going to work out the sum of"},{"Start":"02:51.680 ","End":"02:55.955","Text":"all my torques as a function of Theta."},{"Start":"02:55.955 ","End":"02:59.270","Text":"That\u0027s going to be the most general way to solve this."},{"Start":"02:59.270 ","End":"03:03.725","Text":"Now, a tip in order to avoid getting confused with your signs,"},{"Start":"03:03.725 ","End":"03:05.809","Text":"with your plus and minus."},{"Start":"03:05.809 ","End":"03:13.530","Text":"If I say that my Theta angle is positive when I\u0027ve pulled the rod in this direction,"},{"Start":"03:13.530 ","End":"03:16.255","Text":"so I\u0027ll define this direction,"},{"Start":"03:16.255 ","End":"03:18.124","Text":"my anticlockwise direction,"},{"Start":"03:18.124 ","End":"03:20.600","Text":"to be the positive direction."},{"Start":"03:20.600 ","End":"03:23.930","Text":"Because if I said that this anticlockwise direction was"},{"Start":"03:23.930 ","End":"03:27.815","Text":"the positive direction but I moved my rod in this direction,"},{"Start":"03:27.815 ","End":"03:29.510","Text":"because my Theta would be here."},{"Start":"03:29.510 ","End":"03:31.430","Text":"I would have a negative Theta,"},{"Start":"03:31.430 ","End":"03:34.684","Text":"and then later on when solving this question,"},{"Start":"03:34.684 ","End":"03:38.855","Text":"it\u0027s very easy to make calculation errors and lose points."},{"Start":"03:38.855 ","End":"03:41.390","Text":"Whenever you start, you say,"},{"Start":"03:41.390 ","End":"03:44.070","Text":"if this is the positive angle,"},{"Start":"03:44.070 ","End":"03:46.640","Text":"so then following in that direction,"},{"Start":"03:46.640 ","End":"03:50.185","Text":"that\u0027s going to be my positive direction of rotation."},{"Start":"03:50.185 ","End":"03:53.765","Text":"Now let\u0027s draw the forces acting on our diagram."},{"Start":"03:53.765 ","End":"03:55.760","Text":"Because we\u0027re dealing with a rigid body,"},{"Start":"03:55.760 ","End":"03:59.064","Text":"so we\u0027re going to say that from its center of mass,"},{"Start":"03:59.064 ","End":"04:02.745","Text":"that\u0027s where our mg is going to be coming out."},{"Start":"04:02.745 ","End":"04:08.730","Text":"We have our mg where our mass is given in the question."},{"Start":"04:08.730 ","End":"04:15.379","Text":"Then of course we have the force going in this direction due to the spring."},{"Start":"04:15.379 ","End":"04:18.679","Text":"Now, the force of the spring is defined as"},{"Start":"04:18.679 ","End":"04:23.860","Text":"K multiplied by the change in length of the spring."},{"Start":"04:23.860 ","End":"04:26.610","Text":"If this was its point of equilibrium,"},{"Start":"04:26.610 ","End":"04:27.890","Text":"so it\u0027s this distance,"},{"Start":"04:27.890 ","End":"04:29.420","Text":"this will be our delta X."},{"Start":"04:29.420 ","End":"04:32.779","Text":"Soon we\u0027re going to speak about how we work that out."},{"Start":"04:32.779 ","End":"04:35.420","Text":"Now, of course, at the axis of rotation,"},{"Start":"04:35.420 ","End":"04:38.105","Text":"there\u0027s also a force acting over here."},{"Start":"04:38.105 ","End":"04:40.850","Text":"However, when I\u0027m going to write the sum"},{"Start":"04:40.850 ","End":"04:43.790","Text":"of all of my torques or the sum of all of my moments,"},{"Start":"04:43.790 ","End":"04:46.520","Text":"it won\u0027t interest me because its moment will be"},{"Start":"04:46.520 ","End":"04:51.380","Text":"0 because it\u0027s being applied from the axis of rotation."},{"Start":"04:51.380 ","End":"04:55.534","Text":"Its value for I is going to be equal to 0."},{"Start":"04:55.534 ","End":"05:00.080","Text":"Its distance from the axis of rotation is going to be 0,"},{"Start":"05:00.080 ","End":"05:04.504","Text":"which means that its torque is going to be equal to 0,"},{"Start":"05:04.504 ","End":"05:08.014","Text":"which means that we don\u0027t have to include it in these calculations."},{"Start":"05:08.014 ","End":"05:10.879","Text":"In general, in these types of questions,"},{"Start":"05:10.879 ","End":"05:13.414","Text":"we\u0027re going to only want to work with this equation."},{"Start":"05:13.414 ","End":"05:15.469","Text":"We can also work with this equation."},{"Start":"05:15.469 ","End":"05:19.250","Text":"But usually, it\u0027s best to try and work solely with this."},{"Start":"05:19.250 ","End":"05:23.930","Text":"Because in every axes of rotation we have a force acting."},{"Start":"05:23.930 ","End":"05:25.730","Text":"Because in this question specifically,"},{"Start":"05:25.730 ","End":"05:27.605","Text":"we aren\u0027t given what that force is,"},{"Start":"05:27.605 ","End":"05:31.009","Text":"so it\u0027s going to be a lot more complicated to work out"},{"Start":"05:31.009 ","End":"05:34.609","Text":"the answer by using or utilizing this equation over here,"},{"Start":"05:34.609 ","End":"05:36.080","Text":"the sum of all of the forces."},{"Start":"05:36.080 ","End":"05:38.390","Text":"That\u0027s why we\u0027re going to want to use the sum of all of"},{"Start":"05:38.390 ","End":"05:42.080","Text":"the torques because the torque of this force over here will be 0."},{"Start":"05:42.080 ","End":"05:45.980","Text":"Then we don\u0027t even have to include it or mess around with it."},{"Start":"05:45.980 ","End":"05:47.525","Text":"Hopefully, for this question,"},{"Start":"05:47.525 ","End":"05:49.660","Text":"that will be enough for us to solve."},{"Start":"05:49.660 ","End":"05:53.975","Text":"Let\u0027s start writing out our equation for the sum of all of the torques."},{"Start":"05:53.975 ","End":"05:56.135","Text":"We\u0027re answering question number 1."},{"Start":"05:56.135 ","End":"06:00.087","Text":"We have that the sum of all torques,"},{"Start":"06:00.087 ","End":"06:03.365","Text":"and now we\u0027re going to be working out both the size and direction."},{"Start":"06:03.365 ","End":"06:05.675","Text":"I have my arrow over here."},{"Start":"06:05.675 ","End":"06:08.605","Text":"This is going to be equal to,"},{"Start":"06:08.605 ","End":"06:12.575","Text":"so let\u0027s start with working out the torque for my mass,"},{"Start":"06:12.575 ","End":"06:15.245","Text":"my mg, this force over here."},{"Start":"06:15.245 ","End":"06:19.335","Text":"Firstly, I\u0027m going to write the force."},{"Start":"06:19.335 ","End":"06:23.254","Text":"That\u0027s the size of the force, mass times gravity,"},{"Start":"06:23.254 ","End":"06:28.294","Text":"multiplied by its distance away from the axis of rotation."},{"Start":"06:28.294 ","End":"06:32.390","Text":"Now, we know that my rod is of length L,"},{"Start":"06:32.390 ","End":"06:37.580","Text":"which means that the center of mass where my mg is acting is going to be L over 2,"},{"Start":"06:37.580 ","End":"06:40.359","Text":"so multiplied by L over 2,"},{"Start":"06:40.359 ","End":"06:42.604","Text":"its distance from the axis of rotation,"},{"Start":"06:42.604 ","End":"06:47.675","Text":"and then multiplied by sine of the angle."},{"Start":"06:47.675 ","End":"06:49.534","Text":"Now the angle is Theta,"},{"Start":"06:49.534 ","End":"06:52.345","Text":"and let\u0027s take a look."},{"Start":"06:52.345 ","End":"06:55.950","Text":"As we remember when dealing with vectors,"},{"Start":"06:55.950 ","End":"06:58.610","Text":"so as long as we keep their size and direction,"},{"Start":"06:58.610 ","End":"07:00.665","Text":"we can move them to anywhere we want."},{"Start":"07:00.665 ","End":"07:07.585","Text":"If we take a look and we copy our mg arrow over here onto the axis of rotation,"},{"Start":"07:07.585 ","End":"07:12.056","Text":"so then we\u0027ll see we have this going like this,"},{"Start":"07:12.056 ","End":"07:14.825","Text":"and then here we have the continuation of the rod."},{"Start":"07:14.825 ","End":"07:20.430","Text":"That\u0027s the exact same situation as what is going on over here."},{"Start":"07:20.430 ","End":"07:26.050","Text":"Then we can see that really the angle between the 2 is in fact Theta."},{"Start":"07:26.050 ","End":"07:30.070","Text":"We have mg, the force multiplied by"},{"Start":"07:30.070 ","End":"07:33.880","Text":"its distance from the axis of rotation because we\u0027re dealing with the center of mass."},{"Start":"07:33.880 ","End":"07:37.280","Text":"Then multiply by sine of the angle between the 2."},{"Start":"07:37.280 ","End":"07:40.194","Text":"So far, we have I size,"},{"Start":"07:40.194 ","End":"07:43.390","Text":"but now we have to find the direction because this is a vector."},{"Start":"07:43.390 ","End":"07:46.930","Text":"If we go back to this and we see that this is our mg,"},{"Start":"07:46.930 ","End":"07:48.730","Text":"let me just write it over here."},{"Start":"07:48.730 ","End":"07:50.010","Text":"This is our mg,"},{"Start":"07:50.010 ","End":"07:51.239","Text":"this red arrow over here,"},{"Start":"07:51.239 ","End":"07:56.770","Text":"and now we can see if we take a look in order to find the direction,"},{"Start":"07:56.770 ","End":"07:59.110","Text":"it\u0027s like when we do the cross-product."},{"Start":"07:59.110 ","End":"08:05.789","Text":"When we do that, we go from our radius to our mg,"},{"Start":"08:05.789 ","End":"08:08.175","Text":"so from the radius until the force."},{"Start":"08:08.175 ","End":"08:11.540","Text":"Then we can see if we draw the angle of the direction that that would be"},{"Start":"08:11.540 ","End":"08:14.989","Text":"in that\u0027s going in the leftwards direction,"},{"Start":"08:14.989 ","End":"08:17.554","Text":"which if we carry that on as a circle,"},{"Start":"08:17.554 ","End":"08:20.420","Text":"then that means that we\u0027re going in the clockwise direction."},{"Start":"08:20.420 ","End":"08:25.040","Text":"We already said that the anticlockwise direction is the positive direction."},{"Start":"08:25.040 ","End":"08:30.319","Text":"We know that we have to put a minus sign over here."},{"Start":"08:30.319 ","End":"08:34.459","Text":"Another way of looking at this in order to know and understand why we have"},{"Start":"08:34.459 ","End":"08:39.379","Text":"a minus over here is that without looking at what\u0027s going on over here,"},{"Start":"08:39.379 ","End":"08:44.925","Text":"we can just see that our mg is a restoration for us."},{"Start":"08:44.925 ","End":"08:48.980","Text":"If I\u0027ve moved my rod and angle of Theta outwards like this,"},{"Start":"08:48.980 ","End":"08:52.610","Text":"so my mg is always going to try and push"},{"Start":"08:52.610 ","End":"08:58.475","Text":"my rod in order to be back at its position where it\u0027s at equilibrium."},{"Start":"08:58.475 ","End":"09:03.779","Text":"It\u0027s trying to restore it to its point of equilibrium."},{"Start":"09:03.779 ","End":"09:05.959","Text":"Every force which is the restoring force"},{"Start":"09:05.959 ","End":"09:08.585","Text":"which tries to bring you back to your point of equilibrium,"},{"Start":"09:08.585 ","End":"09:12.560","Text":"there\u0027s going to be a negative sign in front of the expression."},{"Start":"09:12.560 ","End":"09:14.959","Text":"Now onto my next force,"},{"Start":"09:14.959 ","End":"09:17.450","Text":"which is my force from the spring."},{"Start":"09:17.450 ","End":"09:20.030","Text":"I\u0027m going to leave a blank over here where I\u0027m going"},{"Start":"09:20.030 ","End":"09:22.774","Text":"to put either a positive or a negative sign,"},{"Start":"09:22.774 ","End":"09:26.725","Text":"and I\u0027m going to go on with writing the sum of all of the torques."},{"Start":"09:26.725 ","End":"09:29.345","Text":"First I write my force."},{"Start":"09:29.345 ","End":"09:34.430","Text":"Which is my k multiplied by my Delta x."},{"Start":"09:34.430 ","End":"09:36.505","Text":"Now, what is my Delta x?"},{"Start":"09:36.505 ","End":"09:41.565","Text":"My Delta x is this distance over here."},{"Start":"09:41.565 ","End":"09:44.025","Text":"That is my Delta x."},{"Start":"09:44.025 ","End":"09:46.309","Text":"From my point of equilibrium,"},{"Start":"09:46.309 ","End":"09:48.740","Text":"the amount that my spring has been"},{"Start":"09:48.740 ","End":"09:54.379","Text":"stretched extra from the point of equilibrium, that\u0027s my Delta x."},{"Start":"09:54.379 ","End":"09:55.985","Text":"Soon we\u0027re going to work this out."},{"Start":"09:55.985 ","End":"10:04.835","Text":"I\u0027m just going to leave that as Delta x. I\u0027m not yet looking at the signs."},{"Start":"10:04.835 ","End":"10:12.040","Text":"You can imagine that I have the lines over here representing my absolute value."},{"Start":"10:12.040 ","End":"10:16.095","Text":"Now that we have my force,"},{"Start":"10:16.095 ","End":"10:19.625","Text":"now I have to multiply it by"},{"Start":"10:19.625 ","End":"10:25.740","Text":"the length from where it is happening from the axis of rotation."},{"Start":"10:25.740 ","End":"10:32.930","Text":"Because my rod is attached at its opposite end from the axis of rotation to my spring,"},{"Start":"10:32.930 ","End":"10:38.750","Text":"so I know that this total distance is L. The distance that"},{"Start":"10:38.750 ","End":"10:44.345","Text":"this force is from my axis of rotation is a distance of L. Now,"},{"Start":"10:44.345 ","End":"10:48.235","Text":"I have to multiply it by sine of the angle."},{"Start":"10:48.235 ","End":"10:52.590","Text":"Now let\u0027s take a look at what the angle is."},{"Start":"10:52.590 ","End":"10:59.920","Text":"Let\u0027s draw it. I have here my point of equilibrium."},{"Start":"10:59.920 ","End":"11:03.355","Text":"Here I have my rod where it\u0027s located."},{"Start":"11:03.355 ","End":"11:05.604","Text":"This angle over here is 90 degrees."},{"Start":"11:05.604 ","End":"11:08.365","Text":"This angle here is Theta."},{"Start":"11:08.365 ","End":"11:14.620","Text":"My force, my K x is acting in this direction,"},{"Start":"11:14.620 ","End":"11:19.255","Text":"so this is my K Delta x."},{"Start":"11:19.255 ","End":"11:21.295","Text":"What does that mean?"},{"Start":"11:21.295 ","End":"11:25.780","Text":"That means that in order to find the angle, what do I have to find?"},{"Start":"11:25.780 ","End":"11:28.479","Text":"I have to write out the angle"},{"Start":"11:28.479 ","End":"11:35.600","Text":"between where my force is happening to the direction that it\u0027s pointing in."},{"Start":"11:35.850 ","End":"11:40.480","Text":"That means this rod going from here to there."},{"Start":"11:40.480 ","End":"11:44.665","Text":"That means I have to find this angle over here."},{"Start":"11:44.665 ","End":"11:46.765","Text":"Let\u0027s call it Alpha."},{"Start":"11:46.765 ","End":"11:49.074","Text":"In order to find out what my Alpha is,"},{"Start":"11:49.074 ","End":"11:53.725","Text":"I can say that the total angles in the triangle are 180,"},{"Start":"11:53.725 ","End":"12:00.265","Text":"and then I have to minus 90 and minus my Theta,"},{"Start":"12:00.265 ","End":"12:05.110","Text":"that\u0027s going to be equal to my Alpha angle."},{"Start":"12:05.110 ","End":"12:07.450","Text":"This is what I\u0027m trying to find,"},{"Start":"12:07.450 ","End":"12:12.445","Text":"so that\u0027s simply going to be equal to 90 minus Theta,"},{"Start":"12:12.445 ","End":"12:14.800","Text":"so that\u0027s my Alpha."},{"Start":"12:14.800 ","End":"12:20.410","Text":"I have to work out sine of 90 minus Theta."},{"Start":"12:20.410 ","End":"12:26.650","Text":"Now as you know, sine of 90 minus Theta or sine of 90 plus Theta,"},{"Start":"12:26.650 ","End":"12:31.539","Text":"because sines and our cosine are periodical."},{"Start":"12:31.539 ","End":"12:36.760","Text":"I can just change this into instead of sine of 90 minus Theta,"},{"Start":"12:36.760 ","End":"12:40.975","Text":"it will become cosine of Theta."},{"Start":"12:40.975 ","End":"12:45.745","Text":"That\u0027s why there\u0027s a cosine over here and I wrote over here the rule,"},{"Start":"12:45.745 ","End":"12:51.865","Text":"and now what we have to do is we have to decide what the sign is going to be over here."},{"Start":"12:51.865 ","End":"12:55.000","Text":"As we know, we work out the sign by the cross-product,"},{"Start":"12:55.000 ","End":"12:56.469","Text":"and from the cross products,"},{"Start":"12:56.469 ","End":"13:01.615","Text":"we have to go from the radius until the direction of the force."},{"Start":"13:01.615 ","End":"13:05.095","Text":"I\u0027m going to rub all of this out over here,"},{"Start":"13:05.095 ","End":"13:07.165","Text":"and I\u0027m going to draw that out."},{"Start":"13:07.165 ","End":"13:12.715","Text":"My radius, my rod is going in this direction, that\u0027s my radius."},{"Start":"13:12.715 ","End":"13:16.344","Text":"Then the direction of my force, my K Delta x,"},{"Start":"13:16.344 ","End":"13:19.975","Text":"is going in this straight line over here,"},{"Start":"13:19.975 ","End":"13:22.225","Text":"so this is my force."},{"Start":"13:22.225 ","End":"13:27.235","Text":"Now in order to get from my r vector to my force vector,"},{"Start":"13:27.235 ","End":"13:29.740","Text":"I\u0027m going in this direction."},{"Start":"13:29.740 ","End":"13:33.654","Text":"Which is in the clockwise direction,"},{"Start":"13:33.654 ","End":"13:36.040","Text":"which as we can see over here,"},{"Start":"13:36.040 ","End":"13:38.800","Text":"is in our negative direction because we said that"},{"Start":"13:38.800 ","End":"13:42.340","Text":"our positive direction is in the anticlockwise direction."},{"Start":"13:42.340 ","End":"13:45.205","Text":"Here we\u0027re going to write a negative,"},{"Start":"13:45.205 ","End":"13:51.279","Text":"then this is of course equal to our I."},{"Start":"13:51.279 ","End":"13:52.840","Text":"Because our axis of rotation,"},{"Start":"13:52.840 ","End":"13:53.979","Text":"we\u0027re saying it\u0027s at the origin,"},{"Start":"13:53.979 ","End":"13:59.380","Text":"so we can symbolize it with an O over here multiplied by our Alpha."},{"Start":"13:59.380 ","End":"14:02.214","Text":"Our moment of inertia over here,"},{"Start":"14:02.214 ","End":"14:07.644","Text":"we\u0027re dealing with a rod whose axis of rotation is at its end."},{"Start":"14:07.644 ","End":"14:11.245","Text":"As we know, the moment of inertia for that is going to be"},{"Start":"14:11.245 ","End":"14:16.735","Text":"1/3 the mass multiplied by the length of the rod squared,"},{"Start":"14:16.735 ","End":"14:18.850","Text":"and then my Alpha."},{"Start":"14:18.850 ","End":"14:20.560","Text":"This is our equation."},{"Start":"14:20.560 ","End":"14:25.465","Text":"Now notice this isn\u0027t yet a harmonic equation,"},{"Start":"14:25.465 ","End":"14:29.304","Text":"and also we can see that I have too many unknowns."},{"Start":"14:29.304 ","End":"14:31.390","Text":"I have my Theta which is unknown,"},{"Start":"14:31.390 ","End":"14:33.655","Text":"I have my Delta x which is unknown,"},{"Start":"14:33.655 ","End":"14:37.430","Text":"and also my Delta x is dependent on my Theta."},{"Start":"14:38.130 ","End":"14:43.150","Text":"I\u0027m going to want to change 1 of these unknowns."},{"Start":"14:43.150 ","End":"14:46.989","Text":"Now we\u0027re being told to show that the motion"},{"Start":"14:46.989 ","End":"14:50.575","Text":"of the rod at small angles is harmonic and find Omega."},{"Start":"14:50.575 ","End":"14:52.629","Text":"What is a small angle?"},{"Start":"14:52.629 ","End":"14:55.735","Text":"A small angle is when my angle,"},{"Start":"14:55.735 ","End":"14:58.555","Text":"my Theta is much smaller."},{"Start":"14:58.555 ","End":"15:02.830","Text":"This is the symbol for a much smaller, then Pi."},{"Start":"15:02.830 ","End":"15:05.005","Text":"We\u0027re working in radians."},{"Start":"15:05.005 ","End":"15:06.940","Text":"Now what we\u0027re going to do,"},{"Start":"15:06.940 ","End":"15:08.440","Text":"I\u0027ll scroll down a little bit,"},{"Start":"15:08.440 ","End":"15:15.310","Text":"is to try and change my Delta x in order to be something in terms of Theta,"},{"Start":"15:15.310 ","End":"15:18.685","Text":"so that I get rid of my unknowns."},{"Start":"15:18.685 ","End":"15:21.299","Text":"As we know from our SOHCAHTOA,"},{"Start":"15:21.299 ","End":"15:27.085","Text":"sine is our opposite divided by hypotenuse."},{"Start":"15:27.085 ","End":"15:30.640","Text":"We notice that if this is our angle Theta over here,"},{"Start":"15:30.640 ","End":"15:33.219","Text":"then I\u0027m trying to find my Delta x,"},{"Start":"15:33.219 ","End":"15:37.495","Text":"which is my opposite side in the triangle."},{"Start":"15:37.495 ","End":"15:39.580","Text":"Then if we rearrange this,"},{"Start":"15:39.580 ","End":"15:45.009","Text":"we\u0027ll get that our Delta x is approximately equal"},{"Start":"15:45.009 ","End":"15:51.940","Text":"to so L multiplied by sine of the angle Theta."},{"Start":"15:51.940 ","End":"15:57.460","Text":"Then, another way that we can write this is if we remember,"},{"Start":"15:57.460 ","End":"16:05.590","Text":"tan of Theta is equal to sine of Theta divided by cosine of Theta."},{"Start":"16:05.590 ","End":"16:07.330","Text":"Now at small angles,"},{"Start":"16:07.330 ","End":"16:11.440","Text":"cosine of Theta is equal to 1,"},{"Start":"16:11.440 ","End":"16:13.660","Text":"and we\u0027re dealing with small angles,"},{"Start":"16:13.660 ","End":"16:21.715","Text":"so we can say that our sine of Theta is around about equal to tan of Theta."},{"Start":"16:21.715 ","End":"16:26.230","Text":"That means that we can also say that this is around about"},{"Start":"16:26.230 ","End":"16:31.660","Text":"equal to L multiplied by tan of Theta."},{"Start":"16:31.660 ","End":"16:35.110","Text":"Then another way that we can also say this,"},{"Start":"16:35.110 ","End":"16:37.614","Text":"at small angles or so,"},{"Start":"16:37.614 ","End":"16:45.385","Text":"sine of theta is around about equal to Theta at small angles."},{"Start":"16:45.385 ","End":"16:52.105","Text":"Then we can also say that that is equal to L multiplied by Theta."},{"Start":"16:52.105 ","End":"16:57.700","Text":"These are all different ways that we can express our Delta x in terms of Theta,"},{"Start":"16:57.700 ","End":"17:00.114","Text":"and they all mean around about the same things,"},{"Start":"17:00.114 ","End":"17:05.140","Text":"specifically when we\u0027re dealing with small angles."},{"Start":"17:05.140 ","End":"17:08.560","Text":"Let\u0027s highlight that over here."},{"Start":"17:08.560 ","End":"17:12.279","Text":"Now what we\u0027re going to do is we\u0027re going to substitute all of"},{"Start":"17:12.279 ","End":"17:16.975","Text":"this back into this equation to get a simplified expression."},{"Start":"17:16.975 ","End":"17:20.754","Text":"I\u0027m just going to write the key points of my substitution over here."},{"Start":"17:20.754 ","End":"17:23.290","Text":"When I\u0027m dealing with small angles,"},{"Start":"17:23.290 ","End":"17:26.904","Text":"which means that Theta is significantly smaller than Pi,"},{"Start":"17:26.904 ","End":"17:30.985","Text":"then that means that my sine Theta,"},{"Start":"17:30.985 ","End":"17:36.625","Text":"I can just say that it equals to Theta inequations,"},{"Start":"17:36.625 ","End":"17:43.390","Text":"and that also means that my cosine of Theta in small angles,"},{"Start":"17:43.390 ","End":"17:46.735","Text":"that\u0027s when my Theta is significantly smaller than Pi,"},{"Start":"17:46.735 ","End":"17:50.440","Text":"is around about equal to 1."},{"Start":"17:50.440 ","End":"17:54.400","Text":"Now I\u0027m going to substitute everything back in."},{"Start":"17:54.400 ","End":"17:56.650","Text":"Let\u0027s scroll down a little bit more."},{"Start":"17:56.650 ","End":"18:05.215","Text":"I have my negative Mg multiplied by my L divided by 2,"},{"Start":"18:05.215 ","End":"18:08.530","Text":"and then sine Theta at small angles is Theta."},{"Start":"18:08.530 ","End":"18:13.945","Text":"I have my Theta over here and then minus my K multiplied by"},{"Start":"18:13.945 ","End":"18:21.940","Text":"my Delta x. I\u0027m going to say that my Delta x is around about equal to my L Theta."},{"Start":"18:21.940 ","End":"18:26.380","Text":"I\u0027m going to have K multiplied by L Theta multiply by L,"},{"Start":"18:26.380 ","End":"18:33.520","Text":"so that\u0027s going to be KL^2 Theta and then multiplied by cosine Theta,"},{"Start":"18:33.520 ","End":"18:36.580","Text":"which at small angles is equal to 1,"},{"Start":"18:36.580 ","End":"18:40.315","Text":"or round about equal to 1 multiplied by 1,"},{"Start":"18:40.315 ","End":"18:48.460","Text":"and that is going to be equal to my 1/3 multiplied by M L^2,"},{"Start":"18:48.460 ","End":"18:49.914","Text":"and then my Alpha."},{"Start":"18:49.914 ","End":"18:52.900","Text":"Now, my Alpha is my angular acceleration,"},{"Start":"18:52.900 ","End":"18:55.629","Text":"which when dealing with this variable,"},{"Start":"18:55.629 ","End":"19:00.145","Text":"my Theta is equal to Theta double dot."},{"Start":"19:00.145 ","End":"19:03.715","Text":"Just like my linear position is x,"},{"Start":"19:03.715 ","End":"19:07.675","Text":"so my linear acceleration in x double-dot."},{"Start":"19:07.675 ","End":"19:12.025","Text":"When dealing with angular acceleration,"},{"Start":"19:12.025 ","End":"19:14.665","Text":"if my position is in Theta,"},{"Start":"19:14.665 ","End":"19:19.550","Text":"then my angular acceleration is going to be Theta double dot."},{"Start":"19:20.070 ","End":"19:24.084","Text":"Now something quick to add."},{"Start":"19:24.084 ","End":"19:26.095","Text":"When we\u0027re dealing with small angles."},{"Start":"19:26.095 ","End":"19:31.330","Text":"I said that my cosine of Theta is around about equal to 1."},{"Start":"19:31.330 ","End":"19:32.965","Text":"Now, why is this a roundabout?"},{"Start":"19:32.965 ","End":"19:39.984","Text":"Because we open up our cosine Theta via the Taylor expansion."},{"Start":"19:39.984 ","End":"19:44.660","Text":"What we actually get is 1 plus, sorry,"},{"Start":"19:44.660 ","End":"19:50.875","Text":"1 minus Theta^2 divided by 2,"},{"Start":"19:50.875 ","End":"19:54.174","Text":"and then with lots of other terms afterwards,"},{"Start":"19:54.174 ","End":"19:57.190","Text":"depending on how much we expand this out."},{"Start":"19:57.190 ","End":"20:00.684","Text":"Now, in physics, it\u0027s very important and this is"},{"Start":"20:00.684 ","End":"20:03.970","Text":"really important in mechanics and also in electricity,"},{"Start":"20:03.970 ","End":"20:06.744","Text":"your next course, electricity and magnetism,"},{"Start":"20:06.744 ","End":"20:12.849","Text":"is that if we\u0027re dealing with an equation where our 1 will cancel out,"},{"Start":"20:12.849 ","End":"20:15.010","Text":"and then we\u0027ll have some expression where"},{"Start":"20:15.010 ","End":"20:17.830","Text":"the sum of all of our forces or whatever it will be,"},{"Start":"20:17.830 ","End":"20:19.450","Text":"is equal to 0,"},{"Start":"20:19.450 ","End":"20:27.040","Text":"then we have to expand our cosine Theta to a further degree of accuracy,"},{"Start":"20:27.040 ","End":"20:28.810","Text":"and if this 2 cancels out,"},{"Start":"20:28.810 ","End":"20:31.779","Text":"then to a further degree of accuracy and so on and so forth."},{"Start":"20:31.779 ","End":"20:34.525","Text":"Here\u0027s specifically my 1 doesn\u0027t cancel out,"},{"Start":"20:34.525 ","End":"20:39.775","Text":"and my Theta^2 divided by 2 is such a small number in comparison to my 1"},{"Start":"20:39.775 ","End":"20:45.475","Text":"that it won\u0027t really make a difference or a significant difference at all to my answer,"},{"Start":"20:45.475 ","End":"20:47.230","Text":"so I don\u0027t have to include it."},{"Start":"20:47.230 ","End":"20:52.420","Text":"The only time that I have to include this term or further terms down the line,"},{"Start":"20:52.420 ","End":"20:56.095","Text":"you\u0027ll learn this in 1 of your mathematics courses, how to do this,"},{"Start":"20:56.095 ","End":"21:00.790","Text":"is only if my terms before it cancel out,"},{"Start":"21:00.790 ","End":"21:03.190","Text":"such that I get that my force is equal to"},{"Start":"21:03.190 ","End":"21:06.340","Text":"0 when my energy is equal to 0 or whatever it is."},{"Start":"21:06.340 ","End":"21:08.530","Text":"When we know that it isn\u0027t equal to 0,"},{"Start":"21:08.530 ","End":"21:11.350","Text":"it\u0027s equal to something even if it\u0027s very small."},{"Start":"21:11.350 ","End":"21:15.549","Text":"That\u0027s the only time we expand this out further and include these terms."},{"Start":"21:15.549 ","End":"21:17.665","Text":"Here specifically, my 1,"},{"Start":"21:17.665 ","End":"21:19.285","Text":"obviously doesn\u0027t cancel out."},{"Start":"21:19.285 ","End":"21:24.080","Text":"So I don\u0027t have to include the further terms down the line."},{"Start":"21:25.140 ","End":"21:27.865","Text":"Back to our question."},{"Start":"21:27.865 ","End":"21:30.790","Text":"We have this equation and now we want to get it"},{"Start":"21:30.790 ","End":"21:35.545","Text":"into the format of an equation for harmonic motion."},{"Start":"21:35.545 ","End":"21:38.335","Text":"Just to remind you what the equation looks like."},{"Start":"21:38.335 ","End":"21:41.977","Text":"It goes negative k multiplied by x minus"},{"Start":"21:41.977 ","End":"21:49.044","Text":"x_0= m x-double dot."},{"Start":"21:49.044 ","End":"21:53.875","Text":"Now, we\u0027ll notice that our variables over here are xs."},{"Start":"21:53.875 ","End":"21:56.439","Text":"That\u0027s not a problem, we can change our variables to"},{"Start":"21:56.439 ","End":"21:59.710","Text":"anything as long as we maintain the relationship."},{"Start":"21:59.710 ","End":"22:05.360","Text":"Over here, specifically our x is Theta."},{"Start":"22:05.550 ","End":"22:08.200","Text":"Let\u0027s take a look at our other term,"},{"Start":"22:08.200 ","End":"22:09.580","Text":"then we have our x_0."},{"Start":"22:09.580 ","End":"22:17.320","Text":"Our x_0 represents when we\u0027re at a point of equilibrium."},{"Start":"22:17.320 ","End":"22:21.010","Text":"Our point of equilibrium was exactly when we remember in"},{"Start":"22:21.010 ","End":"22:25.045","Text":"this position when our rod was perpendicular to the ceiling."},{"Start":"22:25.045 ","End":"22:28.299","Text":"At what angle will our Theta be at"},{"Start":"22:28.299 ","End":"22:32.130","Text":"our angle over here when we\u0027re at our point of equilibrium?"},{"Start":"22:32.130 ","End":"22:34.089","Text":"Our Theta is going to be equal to 0,"},{"Start":"22:34.089 ","End":"22:42.379","Text":"so that means that our x_0 is equal to 0 in this case."},{"Start":"22:42.379 ","End":"22:46.780","Text":"Let\u0027s scroll down a little bit."},{"Start":"22:46.780 ","End":"22:49.030","Text":"Let\u0027s take a look at what we do."},{"Start":"22:49.030 ","End":"22:51.475","Text":"We want to get this into this position."},{"Start":"22:51.475 ","End":"22:53.544","Text":"We\u0027re looking at this equation over here,"},{"Start":"22:53.544 ","End":"22:56.890","Text":"and we can see that we have negatives which are common factors."},{"Start":"22:56.890 ","End":"23:00.310","Text":"We take our negatives out because they\u0027re common factors,"},{"Start":"23:00.310 ","End":"23:07.390","Text":"and then inside our brackets we have MgL/2."},{"Start":"23:07.390 ","End":"23:09.520","Text":"We\u0027ll notice that we also have Thetas,"},{"Start":"23:09.520 ","End":"23:17.995","Text":"so we\u0027re going to leave them on the side and then plus KL^2."},{"Start":"23:17.995 ","End":"23:22.705","Text":"Then here we have our Thetas,"},{"Start":"23:22.705 ","End":"23:26.360","Text":"so we can multiply this by Theta."},{"Start":"23:26.460 ","End":"23:30.280","Text":"Now, we can see back at this equation,"},{"Start":"23:30.280 ","End":"23:32.864","Text":"we have our negative over here,"},{"Start":"23:32.864 ","End":"23:38.215","Text":"we have some value over here k which can be this inside over here."},{"Start":"23:38.215 ","End":"23:42.280","Text":"Then we have in our brackets our x minus x_0"},{"Start":"23:42.280 ","End":"23:47.765","Text":"which in our terms when we\u0027re dealing with Theta is going to be Theta minus 0."},{"Start":"23:47.765 ","End":"23:52.210","Text":"Here we have our Theta and we can include the minus 0 to represent that,"},{"Start":"23:52.210 ","End":"23:53.485","Text":"but it doesn\u0027t really matter."},{"Start":"23:53.485 ","End":"23:55.599","Text":"Let\u0027s do it anyway,"},{"Start":"23:55.599 ","End":"23:58.281","Text":"so Theta minus 0."},{"Start":"23:58.281 ","End":"24:03.066","Text":"We can see that we\u0027re really getting the same format."},{"Start":"24:03.066 ","End":"24:09.415","Text":"What I can do is I can call this whole square bracket over here."},{"Start":"24:09.415 ","End":"24:13.885","Text":"I can call this k-star instead of my k,"},{"Start":"24:13.885 ","End":"24:18.781","Text":"and then my x minus x_0 is Theta minus 0."},{"Start":"24:18.781 ","End":"24:28.315","Text":"This is going to be equal to my 1/3 ML^2 and my x-double dot"},{"Start":"24:28.315 ","End":"24:31.630","Text":"over here as we"},{"Start":"24:31.630 ","End":"24:34.750","Text":"know my linear acceleration is just"},{"Start":"24:34.750 ","End":"24:38.680","Text":"going to be my angular acceleration, my Theta-double dot."},{"Start":"24:38.680 ","End":"24:42.669","Text":"That\u0027s, that. This is going to be equal"},{"Start":"24:42.669 ","End":"24:48.624","Text":"to 1/3 ML^2 multiplied by my Theta double-dot which is my x-double dot."},{"Start":"24:48.624 ","End":"24:51.895","Text":"I have here my Theta-double dot,"},{"Start":"24:51.895 ","End":"24:56.660","Text":"and then I can call this whole section over here."},{"Start":"24:56.730 ","End":"25:02.170","Text":"What is multiplying my x-double dot or my Theta double-dot is this M,"},{"Start":"25:02.170 ","End":"25:06.230","Text":"so I\u0027m going to call this M-star."},{"Start":"25:07.170 ","End":"25:10.135","Text":"It\u0027s the coefficient of my x-double dot."},{"Start":"25:10.135 ","End":"25:15.054","Text":"We can really see that we have the exact same format."},{"Start":"25:15.054 ","End":"25:21.320","Text":"Our equation is in the exact same format as that for harmonic motion."},{"Start":"25:22.080 ","End":"25:25.123","Text":"Let\u0027s scroll 1 second to the question,"},{"Start":"25:25.123 ","End":"25:28.480","Text":"so we\u0027re being asked to prove that the motion of the rod"},{"Start":"25:28.480 ","End":"25:32.365","Text":"at small angles is harmonic and to find our Omega."},{"Start":"25:32.365 ","End":"25:37.330","Text":"When we set up that we\u0027re dealing with small angles which was this condition over"},{"Start":"25:37.330 ","End":"25:44.020","Text":"here we substituted everything into our equation for the sum of all of the torques."},{"Start":"25:44.020 ","End":"25:46.749","Text":"Then what we did is we changed"},{"Start":"25:46.749 ","End":"25:51.685","Text":"our variable from our Delta x to something to do with Theta."},{"Start":"25:51.685 ","End":"25:56.335","Text":"We can see that we can rearrange our equation as we did over here"},{"Start":"25:56.335 ","End":"26:01.594","Text":"to be exactly like this equation over here representing harmonic motion."},{"Start":"26:01.594 ","End":"26:06.069","Text":"We\u0027ve proved that this rod moves in harmonic motion because"},{"Start":"26:06.069 ","End":"26:12.640","Text":"any equation that we can get into this format means that it\u0027s moving in harmonic motion."},{"Start":"26:12.640 ","End":"26:15.310","Text":"That was the first part of our question."},{"Start":"26:15.310 ","End":"26:20.065","Text":"The second part was to find what our Omega is."},{"Start":"26:20.065 ","End":"26:21.658","Text":"As we know,"},{"Start":"26:21.658 ","End":"26:27.805","Text":"our Omega is the square root of the coefficient of x minus x_0,"},{"Start":"26:27.805 ","End":"26:29.530","Text":"so that\u0027s our k,"},{"Start":"26:29.530 ","End":"26:37.560","Text":"divided by the coefficient of x-double dot which is our m. Over here,"},{"Start":"26:37.560 ","End":"26:40.055","Text":"we can see what is our k and what is our m?"},{"Start":"26:40.055 ","End":"26:43.505","Text":"Our k is our k-star and our m is our M-star."},{"Start":"26:43.505 ","End":"26:46.959","Text":"Instead of writing everything out because it will take too long,"},{"Start":"26:46.959 ","End":"26:51.985","Text":"I can just say k-star divided by M-star."},{"Start":"26:51.985 ","End":"26:57.775","Text":"If you want you can substitute in our actual values, and that\u0027s it."},{"Start":"26:57.775 ","End":"27:01.130","Text":"We\u0027ve answered question number 1."},{"Start":"27:01.830 ","End":"27:06.952","Text":"This is the end of question 1 and these are the answers."},{"Start":"27:06.952 ","End":"27:09.984","Text":"Now let\u0027s go on to the second part of the question."},{"Start":"27:09.984 ","End":"27:13.040","Text":"I\u0027m going to rub out all of our working out."},{"Start":"27:13.920 ","End":"27:18.879","Text":"In my second question I\u0027m being asked to find the angle of the rod as a function of"},{"Start":"27:18.879 ","End":"27:23.910","Text":"time if the rod is released from rest at an angle of Theta_0,"},{"Start":"27:23.910 ","End":"27:26.875","Text":"so let\u0027s see how we do this."},{"Start":"27:26.875 ","End":"27:28.465","Text":"From question number 1,"},{"Start":"27:28.465 ","End":"27:31.780","Text":"I\u0027ve seen that the motion of the rod is harmonic."},{"Start":"27:31.780 ","End":"27:36.490","Text":"Once I know that it\u0027s harmonic and I\u0027ve gotten it into this type of format,"},{"Start":"27:36.490 ","End":"27:41.575","Text":"I have this differential equation where are my variables are Theta."},{"Start":"27:41.575 ","End":"27:44.860","Text":"That means that I can use the general solution to"},{"Start":"27:44.860 ","End":"27:48.304","Text":"this differential equation which we know according to x,"},{"Start":"27:48.304 ","End":"27:50.860","Text":"but according to Theta it\u0027s exactly the same thing."},{"Start":"27:50.860 ","End":"27:56.785","Text":"Which says that my angle as a function of time is going to be equal to"},{"Start":"27:56.785 ","End":"28:00.985","Text":"my A cosine of"},{"Start":"28:00.985 ","End":"28:07.990","Text":"my Omega t plus Phi and then plus my sighting position."},{"Start":"28:07.990 ","End":"28:14.019","Text":"Because we said that my point of equilibrium is where my angle Theta is equal to 0,"},{"Start":"28:14.019 ","End":"28:17.454","Text":"so that means it was plus 0."},{"Start":"28:17.454 ","End":"28:22.550","Text":"I can cross that out because that doesn\u0027t change my equation."},{"Start":"28:23.130 ","End":"28:28.239","Text":"I already have found my Omega in the previous question,"},{"Start":"28:28.239 ","End":"28:33.250","Text":"and now what I need to do is I have to work out what my A and what my Phi is equal to."},{"Start":"28:33.250 ","End":"28:38.395","Text":"How I\u0027m I going to do that? I\u0027m going to work that out by using initial conditions."},{"Start":"28:38.395 ","End":"28:44.485","Text":"I know that my position at time t is going to equal to 0,"},{"Start":"28:44.485 ","End":"28:47.634","Text":"is equal to my Theta_0."},{"Start":"28:47.634 ","End":"28:53.035","Text":"From the question, the rod is released from rest at angle of Theta_0."},{"Start":"28:53.035 ","End":"28:56.095","Text":"That means right at the start our t=0,"},{"Start":"28:56.095 ","End":"29:00.807","Text":"my starting position is Theta_0."},{"Start":"29:00.807 ","End":"29:03.445","Text":"My next initial condition,"},{"Start":"29:03.445 ","End":"29:06.775","Text":"I\u0027m told that the rod is released from rest."},{"Start":"29:06.775 ","End":"29:11.132","Text":"That means that my first derivative of Theta\u0027"},{"Start":"29:11.132 ","End":"29:16.495","Text":"meaning my angular velocity at my time t=0,"},{"Start":"29:16.495 ","End":"29:20.395","Text":"so right at the start is going to be equal to 0."},{"Start":"29:20.395 ","End":"29:26.350","Text":"Because it\u0027s released from rest which means that its angular velocity at t=0 is 0."},{"Start":"29:26.350 ","End":"29:33.670","Text":"This also equals to of course my angular velocity at t=0."},{"Start":"29:33.670 ","End":"29:37.489","Text":"These are my initial conditions."},{"Start":"29:37.500 ","End":"29:42.530","Text":"We just have to substitute these in to this equation."},{"Start":"29:43.680 ","End":"29:51.159","Text":"My Theta at t=0 is going to be equal"},{"Start":"29:51.159 ","End":"29:58.704","Text":"to A multiplied by cosine and then Omega times t=0,"},{"Start":"29:58.704 ","End":"30:01.270","Text":"so that\u0027s 0 plus Phi,"},{"Start":"30:01.270 ","End":"30:03.340","Text":"so cosine of Phi."},{"Start":"30:03.340 ","End":"30:08.899","Text":"My position as we know has to be equal to Theta_0."},{"Start":"30:09.210 ","End":"30:13.299","Text":"Then, I\u0027m going to derive this equation in order to"},{"Start":"30:13.299 ","End":"30:16.750","Text":"find out what my Theta-dot will be. I\u0027ve done that."},{"Start":"30:16.750 ","End":"30:24.175","Text":"My Theta-dot as a function of t is equal to negative A Omega sine of Omega t plus Phi."},{"Start":"30:24.175 ","End":"30:26.815","Text":"Then as we know from my initial conditions,"},{"Start":"30:26.815 ","End":"30:29.920","Text":"this is meant to be equal to 0."},{"Start":"30:29.920 ","End":"30:35.068","Text":"Now, once I rearrange everything I\u0027m going to skip all of the algebra,"},{"Start":"30:35.068 ","End":"30:39.309","Text":"so we\u0027ll find that my A is going to be equal to"},{"Start":"30:39.309 ","End":"30:46.190","Text":"Theta_0 and that my Phi is going to be equal to 0."},{"Start":"30:47.280 ","End":"30:50.035","Text":"We now have all of our unknowns,"},{"Start":"30:50.035 ","End":"30:52.659","Text":"we have A Omega and Phi."},{"Start":"30:52.659 ","End":"30:56.395","Text":"We can simply substitute that in to our equation,"},{"Start":"30:56.395 ","End":"30:59.110","Text":"so our Theta as a function of t because"},{"Start":"30:59.110 ","End":"31:03.235","Text":"our question was to find the angle of the rod as a function of time."},{"Start":"31:03.235 ","End":"31:08.289","Text":"That\u0027s our Theta as a function of t which is going to be equal to A which is"},{"Start":"31:08.289 ","End":"31:14.920","Text":"our Theta 0 multiplied by cosine of Omega."},{"Start":"31:14.920 ","End":"31:21.489","Text":"My Omega is the square root of k-star divided by"},{"Start":"31:21.489 ","End":"31:29.350","Text":"M-star and obviously you should substitute in the actual values of k-star and of M-star,"},{"Start":"31:29.350 ","End":"31:33.700","Text":"and that\u0027s multiplied by t plus Phi which is equal to 0,"},{"Start":"31:33.700 ","End":"31:35.890","Text":"so nothing plus our 0,"},{"Start":"31:35.890 ","End":"31:39.520","Text":"which is obviously just that."},{"Start":"31:39.520 ","End":"31:44.319","Text":"This is our answer to question 2."},{"Start":"31:44.319 ","End":"31:45.879","Text":"That\u0027s the end of this question."},{"Start":"31:45.879 ","End":"31:49.419","Text":"This question is very important and it comes up"},{"Start":"31:49.419 ","End":"31:53.155","Text":"similar sorts of questions a lot in tests,"},{"Start":"31:53.155 ","End":"31:55.539","Text":"so maybe watch this video again,"},{"Start":"31:55.539 ","End":"31:57.452","Text":"and try, and understand,"},{"Start":"31:57.452 ","End":"31:59.659","Text":"and memorize the steps."}],"ID":10629},{"Watched":false,"Name":"1.6 Mathematical Pendulum (torque)","Duration":"20m 17s","ChapterTopicVideoID":10289,"CourseChapterTopicPlaylistID":5365,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.280","Text":"Hello. In this question,"},{"Start":"00:02.280 ","End":"00:06.945","Text":"we\u0027re being told that a pendulum is attached to a ceiling and that the length of the rope"},{"Start":"00:06.945 ","End":"00:12.585","Text":"is l. We\u0027re being asked to find the frequency of small angle oscillations."},{"Start":"00:12.585 ","End":"00:18.390","Text":"Small angular oscillations are when our angle Theta is significantly smaller than Pi,"},{"Start":"00:18.390 ","End":"00:23.055","Text":"and we\u0027re being asked to find the angle as a function of time."},{"Start":"00:23.055 ","End":"00:25.920","Text":"We\u0027re being told that the pendulum begins its movement from"},{"Start":"00:25.920 ","End":"00:29.520","Text":"rest at an initial angle of Theta."},{"Start":"00:29.520 ","End":"00:34.740","Text":"The usual way to tackle these questions is to write out our equation for the sum of"},{"Start":"00:34.740 ","End":"00:41.100","Text":"the forces and or to write out our equation for the sum of the torques."},{"Start":"00:41.100 ","End":"00:43.205","Text":"Now, just as a reminder,"},{"Start":"00:43.205 ","End":"00:47.660","Text":"what we want to do when we\u0027re dealing with harmonic motion is to end up"},{"Start":"00:47.660 ","End":"00:53.090","Text":"with our force equation dealing with harmonic motion,"},{"Start":"00:53.090 ","End":"00:55.820","Text":"which is negative k,"},{"Start":"00:55.820 ","End":"00:57.485","Text":"which is our spring coefficient,"},{"Start":"00:57.485 ","End":"01:07.205","Text":"multiplied by x our position minus x_ 0 our of equilibrium."},{"Start":"01:07.205 ","End":"01:12.470","Text":"This is meant to equal our mass multiplied by acceleration,"},{"Start":"01:12.470 ","End":"01:16.645","Text":"which is x double dot, it\u0027s a differential equation."},{"Start":"01:16.645 ","End":"01:21.630","Text":"Now, I\u0027m reminding you that our k,"},{"Start":"01:21.630 ","End":"01:25.950","Text":"so the coefficient of our x minus x_ 0 and our m,"},{"Start":"01:25.950 ","End":"01:28.115","Text":"a coefficient of our x double dot,"},{"Start":"01:28.115 ","End":"01:31.880","Text":"are both positive constants."},{"Start":"01:31.880 ","End":"01:37.540","Text":"K and m both have to be positive and constant our x_ 0."},{"Start":"01:37.540 ","End":"01:41.765","Text":"This can be a positive or a negative,"},{"Start":"01:41.765 ","End":"01:43.909","Text":"and our x is our position,"},{"Start":"01:43.909 ","End":"01:46.460","Text":"and our x_ 0 is a positive or negative,"},{"Start":"01:46.460 ","End":"01:48.760","Text":"but it\u0027s also a constant."},{"Start":"01:48.760 ","End":"01:53.915","Text":"Now another reminder is that it doesn\u0027t matter because this is a mathematical equation."},{"Start":"01:53.915 ","End":"01:56.960","Text":"Our k could be an A our m could be a Z."},{"Start":"01:56.960 ","End":"01:58.850","Text":"It doesn\u0027t really matter,"},{"Start":"01:58.850 ","End":"02:03.055","Text":"we just have to have something in this format."},{"Start":"02:03.055 ","End":"02:05.665","Text":"Now, similarly, our x,"},{"Start":"02:05.665 ","End":"02:06.920","Text":"which is our variable,"},{"Start":"02:06.920 ","End":"02:08.810","Text":"can also not be an x."},{"Start":"02:08.810 ","End":"02:10.070","Text":"Specifically over here,"},{"Start":"02:10.070 ","End":"02:14.550","Text":"our x is in fact going to be Theta,"},{"Start":"02:14.550 ","End":"02:16.865","Text":"because we\u0027re dealing with angles."},{"Start":"02:16.865 ","End":"02:20.770","Text":"Our position is denoted by the angle which we are at."},{"Start":"02:20.770 ","End":"02:24.440","Text":"We\u0027re going to start by drawing out our forces,"},{"Start":"02:24.440 ","End":"02:26.750","Text":"because we\u0027re going to write equations for the sum of"},{"Start":"02:26.750 ","End":"02:29.285","Text":"all of our forces and the sum of all torques,"},{"Start":"02:29.285 ","End":"02:32.450","Text":"and for that, we need to know which forces we have."},{"Start":"02:32.450 ","End":"02:36.184","Text":"The first force that we have is our tension,"},{"Start":"02:36.184 ","End":"02:39.770","Text":"which is going in the direction of the string,"},{"Start":"02:39.770 ","End":"02:43.025","Text":"of course, and we have our force going downwards,"},{"Start":"02:43.025 ","End":"02:46.720","Text":"which our mg. Now,"},{"Start":"02:46.720 ","End":"02:50.960","Text":"another force that we have acting is over here at the axis of rotation,"},{"Start":"02:50.960 ","End":"02:53.420","Text":"right over here we have an axis of rotation."},{"Start":"02:53.420 ","End":"02:57.470","Text":"Now, it\u0027s very difficult to find out what force is acting over here,"},{"Start":"02:57.470 ","End":"03:00.320","Text":"so instead of writing out the sum of all of our forces,"},{"Start":"03:00.320 ","End":"03:03.575","Text":"we\u0027re going to write out the sum of all of our torques."},{"Start":"03:03.575 ","End":"03:07.900","Text":"The reason that this is good is because"},{"Start":"03:07.900 ","End":"03:14.135","Text":"our force over here at the axis of rotation is acting from the axis of rotation,"},{"Start":"03:14.135 ","End":"03:18.470","Text":"which means that its value for its r vector is going to be"},{"Start":"03:18.470 ","End":"03:23.990","Text":"0 because it\u0027s r vector is equal to 0 because it\u0027s coming out of the axis of rotation."},{"Start":"03:23.990 ","End":"03:27.770","Text":"Which means that its torque is going to be equal to 0,"},{"Start":"03:27.770 ","End":"03:29.780","Text":"which means that we don\u0027t have to add it into"},{"Start":"03:29.780 ","End":"03:33.785","Text":"our calculation and we don\u0027t have to get carried away with that."},{"Start":"03:33.785 ","End":"03:36.005","Text":"What we\u0027re going to do is we\u0027re going to write"},{"Start":"03:36.005 ","End":"03:38.105","Text":"an equation for the sum of all of our torques,"},{"Start":"03:38.105 ","End":"03:43.435","Text":"and the torques that we\u0027re going to use is the torque for our tension and for our mg."},{"Start":"03:43.435 ","End":"03:46.265","Text":"This is going to be equal to,"},{"Start":"03:46.265 ","End":"03:49.850","Text":"and now I\u0027ll just give you a brief reminder over here."},{"Start":"03:49.850 ","End":"03:54.200","Text":"As we remember when we\u0027re writing out an equation for the size of a torque,"},{"Start":"03:54.200 ","End":"03:57.650","Text":"it\u0027s going to be equal to this,"},{"Start":"03:57.650 ","End":"04:00.583","Text":"the size of our r vector,"},{"Start":"04:00.583 ","End":"04:07.910","Text":"so our distance from the axis of rotation multiplied by the size of our force,"},{"Start":"04:07.910 ","End":"04:14.030","Text":"then multiplied by sine of the angle between them."},{"Start":"04:14.030 ","End":"04:17.090","Text":"Given this, we can write down,"},{"Start":"04:17.090 ","End":"04:21.960","Text":"so we know that force is mg. We\u0027re going to start with"},{"Start":"04:21.960 ","End":"04:27.755","Text":"our mg. We have mg and then multiplied by our r,"},{"Start":"04:27.755 ","End":"04:32.344","Text":"which is l, the distance from our axis of rotation and"},{"Start":"04:32.344 ","End":"04:37.410","Text":"then multiplied by sine of the angle between the two, which is Theta."},{"Start":"04:37.410 ","End":"04:40.255","Text":"Now how do we know that it\u0027s Theta?"},{"Start":"04:40.255 ","End":"04:41.735","Text":"Let\u0027s just take a look."},{"Start":"04:41.735 ","End":"04:46.505","Text":"If I carry on and I imagine that my rope is carrying on,"},{"Start":"04:46.505 ","End":"04:49.325","Text":"then I can see that if this angle here is Theta,"},{"Start":"04:49.325 ","End":"04:53.435","Text":"then that means that this angle here is Theta as well,"},{"Start":"04:53.435 ","End":"04:55.025","Text":"so it\u0027s sine Theta."},{"Start":"04:55.025 ","End":"04:57.620","Text":"Or if we want to be more correct,"},{"Start":"04:57.620 ","End":"05:01.370","Text":"then what we will do is we will take the size of this angle,"},{"Start":"05:01.370 ","End":"05:04.940","Text":"which is simply going to be 180 minus Theta,"},{"Start":"05:04.940 ","End":"05:11.405","Text":"and sine of Theta is equal to sine of 180 minus Theta."},{"Start":"05:11.405 ","End":"05:14.600","Text":"This has a size of our torque for our mg."},{"Start":"05:14.600 ","End":"05:16.535","Text":"However, because our torque is a vector,"},{"Start":"05:16.535 ","End":"05:18.905","Text":"we need to see in which direction it\u0027s going."},{"Start":"05:18.905 ","End":"05:20.750","Text":"Now there\u0027s two ways to do this."},{"Start":"05:20.750 ","End":"05:26.495","Text":"The first way is to assume that your angle has been opened in the positive direction."},{"Start":"05:26.495 ","End":"05:28.895","Text":"If this angle is positive,"},{"Start":"05:28.895 ","End":"05:34.444","Text":"then that means that this clockwise direction is the positive direction."},{"Start":"05:34.444 ","End":"05:39.320","Text":"Then it will make it much easier later in the question to figure"},{"Start":"05:39.320 ","End":"05:44.675","Text":"out your positive and negative signs in your equations and not get confused."},{"Start":"05:44.675 ","End":"05:46.580","Text":"Then if we say that this is"},{"Start":"05:46.580 ","End":"05:49.895","Text":"the positive direction because this is a positive angle Theta,"},{"Start":"05:49.895 ","End":"05:54.455","Text":"then we can see that our mg force,"},{"Start":"05:54.455 ","End":"06:00.260","Text":"is trying to push the pendulum in this direction,"},{"Start":"06:00.260 ","End":"06:03.740","Text":"and If we carry this on in a circular direction,"},{"Start":"06:03.740 ","End":"06:06.950","Text":"we\u0027ll see that that is going anticlockwise."},{"Start":"06:06.950 ","End":"06:10.235","Text":"That means that it\u0027s against the positive direction,"},{"Start":"06:10.235 ","End":"06:13.370","Text":"which means that we have to have a negative over here."},{"Start":"06:13.370 ","End":"06:16.730","Text":"Now, the second way to think of this in order to get"},{"Start":"06:16.730 ","End":"06:20.915","Text":"to finding out if this is positive or negative expression,"},{"Start":"06:20.915 ","End":"06:24.110","Text":"is if we look at our mg,"},{"Start":"06:24.110 ","End":"06:27.205","Text":"the force that we\u0027re looking at in order to work out the torque."},{"Start":"06:27.205 ","End":"06:32.855","Text":"Here we see this dotted line represents a point of equilibrium."},{"Start":"06:32.855 ","End":"06:35.360","Text":"Now, if our force,"},{"Start":"06:35.360 ","End":"06:37.520","Text":"the torque of this,"},{"Start":"06:37.520 ","End":"06:41.650","Text":"is pushing our objects back to its point of equilibrium,"},{"Start":"06:41.650 ","End":"06:44.209","Text":"then there\u0027s always a negative."},{"Start":"06:44.209 ","End":"06:48.755","Text":"If our force is pushing our objects away from the point of equilibrium,"},{"Start":"06:48.755 ","End":"06:50.500","Text":"then it\u0027s a positive."},{"Start":"06:50.500 ","End":"06:54.590","Text":"That\u0027s the second way and it\u0027s a very good and easy way to take a look at it,"},{"Start":"06:54.590 ","End":"06:57.160","Text":"especially when dealing with pendulums."},{"Start":"06:57.160 ","End":"06:58.820","Text":"Now what we want to do,"},{"Start":"06:58.820 ","End":"07:01.700","Text":"is we want to work out the torque for our tension,"},{"Start":"07:01.700 ","End":"07:07.390","Text":"for a T. Going back to the size of our torque over here,"},{"Start":"07:07.390 ","End":"07:10.790","Text":"we can see that we have to do the dot product between"},{"Start":"07:10.790 ","End":"07:17.060","Text":"our r vector and our force vector and then multiplied by sine of the angle between them."},{"Start":"07:17.060 ","End":"07:22.459","Text":"Now you\u0027ll notice that your force vector is pointing in the same direction,"},{"Start":"07:22.459 ","End":"07:27.390","Text":"but opposite direction, give up parallel to our r vector."},{"Start":"07:27.390 ","End":"07:29.325","Text":"Let me just draw this,"},{"Start":"07:29.325 ","End":"07:38.140","Text":"so we can see if this is our axis of rotation and this black is our r,"},{"Start":"07:38.140 ","End":"07:43.095","Text":"so our torque is just going in this direction."},{"Start":"07:43.095 ","End":"07:46.100","Text":"In the exact opposite direction,"},{"Start":"07:46.100 ","End":"07:47.765","Text":"but they\u0027re both parallel."},{"Start":"07:47.765 ","End":"07:54.310","Text":"Which means that the angle between them is 0 and sine of 0 is equal to 0."},{"Start":"07:54.310 ","End":"07:56.795","Text":"Because our sine of 0 is equal to 0,"},{"Start":"07:56.795 ","End":"07:58.895","Text":"the size of our torque is equal to 0,"},{"Start":"07:58.895 ","End":"08:03.040","Text":"which means that we have here plus 0."},{"Start":"08:03.040 ","End":"08:06.410","Text":"We have our equation for the sum of all the torques,"},{"Start":"08:06.410 ","End":"08:07.790","Text":"and as we know,"},{"Start":"08:07.790 ","End":"08:12.100","Text":"this is equal to I Alpha."},{"Start":"08:12.100 ","End":"08:14.885","Text":"I Alpha is, as we know,"},{"Start":"08:14.885 ","End":"08:18.080","Text":"our angular acceleration and our I,"},{"Start":"08:18.080 ","End":"08:20.930","Text":"we are right now dealing with the moment of inertia,"},{"Start":"08:20.930 ","End":"08:23.270","Text":"as we can see is a point mass over here."},{"Start":"08:23.270 ","End":"08:27.215","Text":"We\u0027re dealing with the moment of inertia of a point mass,"},{"Start":"08:27.215 ","End":"08:30.860","Text":"which is simply the mass of the mass multiplied"},{"Start":"08:30.860 ","End":"08:34.430","Text":"by its distance from the axis of rotation."},{"Start":"08:34.430 ","End":"08:36.050","Text":"In this case,"},{"Start":"08:36.050 ","End":"08:38.890","Text":"it\u0027s a distance l from the axis of rotation,"},{"Start":"08:38.890 ","End":"08:41.720","Text":"so it\u0027s multiplied by l^2,"},{"Start":"08:41.720 ","End":"08:44.320","Text":"so mass times distance squared."},{"Start":"08:44.320 ","End":"08:47.300","Text":"That\u0027s the moment of inertia of a point mass."},{"Start":"08:47.300 ","End":"08:54.365","Text":"Now, we\u0027re just going to remind ourselves that our Alpha is equal to Theta double dot."},{"Start":"08:54.365 ","End":"08:58.235","Text":"Now what I\u0027m going to do is I\u0027m going to rewrite this equation,"},{"Start":"08:58.235 ","End":"09:01.360","Text":"putting in my I and my Alpha."},{"Start":"09:01.360 ","End":"09:04.125","Text":"Let\u0027s go a little bit to the side,"},{"Start":"09:04.125 ","End":"09:13.805","Text":"and I will get the equation being negative mgl sine of Theta is equal to my I,"},{"Start":"09:13.805 ","End":"09:17.585","Text":"which is my ml^2 multiplied by my Alpha,"},{"Start":"09:17.585 ","End":"09:19.790","Text":"which is my Theta double dot."},{"Start":"09:19.790 ","End":"09:22.895","Text":"Now I can cross off the ms on both sides,"},{"Start":"09:22.895 ","End":"09:25.935","Text":"and this l divide by l,"},{"Start":"09:25.935 ","End":"09:27.975","Text":"and now I\u0027m left with this."},{"Start":"09:27.975 ","End":"09:30.485","Text":"What I want to do is I want to transform"},{"Start":"09:30.485 ","End":"09:36.080","Text":"this equation into this equation over here for harmonic motion in red."},{"Start":"09:36.080 ","End":"09:40.100","Text":"Now, my problem is that I have my sine Theta over"},{"Start":"09:40.100 ","End":"09:44.815","Text":"here and I need it to be with regards to Theta."},{"Start":"09:44.815 ","End":"09:48.485","Text":"We\u0027re going to be reminded that we\u0027re dealing with small angles,"},{"Start":"09:48.485 ","End":"09:53.660","Text":"that our Theta is significantly smaller than Pi. I\u0027m going to write in red."},{"Start":"09:53.660 ","End":"09:56.005","Text":"When we\u0027re dealing with small angles,"},{"Start":"09:56.005 ","End":"10:02.088","Text":"so a sine of Theta is approximately equal to Theta itself."},{"Start":"10:02.088 ","End":"10:06.740","Text":"Given that, I can rewrite this as negative g,"},{"Start":"10:06.740 ","End":"10:08.645","Text":"and then instead of sine Theta,"},{"Start":"10:08.645 ","End":"10:15.805","Text":"I can write Theta and this is going to be equal to l multiplied by Theta double dot."},{"Start":"10:15.805 ","End":"10:18.560","Text":"Now what we can see is that my equation over"},{"Start":"10:18.560 ","End":"10:22.675","Text":"here is completely identical to this equation over here."},{"Start":"10:22.675 ","End":"10:25.445","Text":"Let\u0027s take a look at why."},{"Start":"10:25.445 ","End":"10:28.965","Text":"We can see that my k is a positive coefficient,"},{"Start":"10:28.965 ","End":"10:30.835","Text":"it\u0027s a positive constant."},{"Start":"10:30.835 ","End":"10:33.574","Text":"My g, gravity,"},{"Start":"10:33.574 ","End":"10:35.299","Text":"it\u0027s a positive constant."},{"Start":"10:35.299 ","End":"10:37.895","Text":"Then I have my l, my length,"},{"Start":"10:37.895 ","End":"10:41.900","Text":"and length is always positive and it\u0027s a constant,"},{"Start":"10:41.900 ","End":"10:44.960","Text":"so that\u0027s like my m. That\u0027s great."},{"Start":"10:44.960 ","End":"10:48.350","Text":"This is perfect. Now what we can take a look at,"},{"Start":"10:48.350 ","End":"10:50.150","Text":"is that just like we said,"},{"Start":"10:50.150 ","End":"10:53.505","Text":"my x becomes my Theta over here,"},{"Start":"10:53.505 ","End":"10:56.120","Text":"my x_ 0, let\u0027s take a look at what that is."},{"Start":"10:56.120 ","End":"11:01.305","Text":"My point of equilibrium is when my angle Theta is 0,"},{"Start":"11:01.305 ","End":"11:04.520","Text":"so when my pendulum is perpendicular to the ceiling."},{"Start":"11:04.520 ","End":"11:06.650","Text":"That means that my x_ 0,"},{"Start":"11:06.650 ","End":"11:09.595","Text":"my point of equilibrium, is 0."},{"Start":"11:09.595 ","End":"11:12.425","Text":"Then, if you look at this equation,"},{"Start":"11:12.425 ","End":"11:13.655","Text":"given all of this,"},{"Start":"11:13.655 ","End":"11:17.995","Text":"we can see that it is exactly identical to this equation over here."},{"Start":"11:17.995 ","End":"11:23.495","Text":"In the question we were being asked to find the frequency of small angle oscillations."},{"Start":"11:23.495 ","End":"11:28.760","Text":"As we know, our frequency Omega is equal to the square root"},{"Start":"11:28.760 ","End":"11:35.215","Text":"of our coefficient over here divided by this coefficient over here."},{"Start":"11:35.215 ","End":"11:40.175","Text":"In our case, that is g divided by i."},{"Start":"11:40.175 ","End":"11:43.195","Text":"That\u0027s the answer to this section."},{"Start":"11:43.195 ","End":"11:48.550","Text":"Really quickly, before we go on to the second question which is finding the angle as"},{"Start":"11:48.550 ","End":"11:50.710","Text":"a function of time let\u0027s quickly do"},{"Start":"11:50.710 ","End":"11:54.145","Text":"a little conclusion of how we answer the first question."},{"Start":"11:54.145 ","End":"11:57.000","Text":"What we want to do is we want to find the frequency of"},{"Start":"11:57.000 ","End":"12:00.470","Text":"small angle oscillations which means we want to find Omega."},{"Start":"12:00.470 ","End":"12:03.706","Text":"If we want to find Omega always, always, always,"},{"Start":"12:03.706 ","End":"12:06.935","Text":"we have to get to this equation and it\u0027s"},{"Start":"12:06.935 ","End":"12:10.935","Text":"the square root of this coefficient divided by this coefficient."},{"Start":"12:10.935 ","End":"12:12.635","Text":"That\u0027s what we need to do."},{"Start":"12:12.635 ","End":"12:14.895","Text":"Now, if we need to find this equation,"},{"Start":"12:14.895 ","End":"12:16.595","Text":"we can do it in 2 ways."},{"Start":"12:16.595 ","End":"12:18.875","Text":"We can either write the equation for the sum of"},{"Start":"12:18.875 ","End":"12:22.885","Text":"all the forces or we can write the equation for the sum of all of the torques."},{"Start":"12:22.885 ","End":"12:25.420","Text":"Now specifically here because we have an axis of"},{"Start":"12:25.420 ","End":"12:29.738","Text":"rotation and a force acting from the axis of rotation,"},{"Start":"12:29.738 ","End":"12:33.730","Text":"we\u0027re not going to do the equation for the sum of all of the forces because we don\u0027t"},{"Start":"12:33.730 ","End":"12:38.115","Text":"know what this force is over here and that will over-complicate the calculation,"},{"Start":"12:38.115 ","End":"12:41.800","Text":"so we\u0027re going to use the equation for the sum of all of the torques."},{"Start":"12:41.800 ","End":"12:44.430","Text":"We draw a free-body diagram see,"},{"Start":"12:44.430 ","End":"12:45.780","Text":"which forces are acting,"},{"Start":"12:45.780 ","End":"12:48.670","Text":"and then we write out our equation for the torques."},{"Start":"12:48.670 ","End":"12:54.085","Text":"Remember that this is the equation for the size of the torque and we have to"},{"Start":"12:54.085 ","End":"12:59.935","Text":"multiply it by sine of the angle which means that if our r and our f are parallel,"},{"Start":"12:59.935 ","End":"13:03.605","Text":"then the angle between them is going to be 0 which means that the torque is going"},{"Start":"13:03.605 ","End":"13:07.670","Text":"to be 0 which is what happened to us in the case over here with our tension."},{"Start":"13:07.670 ","End":"13:11.065","Text":"Then when we were dealing with our other force; our mg,"},{"Start":"13:11.065 ","End":"13:16.165","Text":"so you write the size of the force multiplied by its distance from the axis of rotation,"},{"Start":"13:16.165 ","End":"13:23.110","Text":"multiplied by sine of the angle between the distance and the force."},{"Start":"13:23.110 ","End":"13:28.690","Text":"Then in order to find the direction because our torque is a vector,"},{"Start":"13:28.690 ","End":"13:31.105","Text":"what we have to do is we have to look"},{"Start":"13:31.105 ","End":"13:35.085","Text":"which direction is our positive rotational direction."},{"Start":"13:35.085 ","End":"13:38.461","Text":"Here specifically we want our angle Theta to be positive,"},{"Start":"13:38.461 ","End":"13:41.215","Text":"so we said that if this is the positive direction then"},{"Start":"13:41.215 ","End":"13:46.865","Text":"our clockwise direction is the positive direction of rotation."},{"Start":"13:46.865 ","End":"13:51.420","Text":"Another way to look at it is if our force is a restorative force which means that"},{"Start":"13:51.420 ","End":"13:56.150","Text":"if it\u0027s pushing our objects back to its point of equilibrium,"},{"Start":"13:56.150 ","End":"14:00.820","Text":"and if so then the sign is going to be negative."},{"Start":"14:00.820 ","End":"14:04.410","Text":"Now we have the sum of all of our torques and then we say that,"},{"Start":"14:04.410 ","End":"14:06.910","Text":"that is equal to I Alpha."},{"Start":"14:06.910 ","End":"14:09.605","Text":"Our I is of course when dealing with this type of pendulum."},{"Start":"14:09.605 ","End":"14:13.195","Text":"It\u0027s the moment of inertia of a point mass which is simply"},{"Start":"14:13.195 ","End":"14:17.965","Text":"the mass multiplied by its distance from the axis of rotation squared."},{"Start":"14:17.965 ","End":"14:21.385","Text":"Here the distance was l, so we have ml^2."},{"Start":"14:21.385 ","End":"14:26.065","Text":"Then what we did is we use the fact that when dealing with small angles,"},{"Start":"14:26.065 ","End":"14:29.785","Text":"sine of Theta approximately is equal to Theta."},{"Start":"14:29.785 ","End":"14:34.360","Text":"Then once we rewrote the equation and simplified it we were left with"},{"Start":"14:34.360 ","End":"14:38.900","Text":"this which is exactly identical to this equation,"},{"Start":"14:38.900 ","End":"14:42.460","Text":"and then we could find our Omega because as we know that equals to"},{"Start":"14:42.460 ","End":"14:47.985","Text":"the square root of this coefficient divided by this coefficient."},{"Start":"14:47.985 ","End":"14:53.910","Text":"Now the second section of our question is to find the angle as a function of time."},{"Start":"14:53.910 ","End":"14:56.875","Text":"I\u0027m going to scroll down a little bit."},{"Start":"14:56.875 ","End":"15:03.235","Text":"This means that what we want to do is we want to find our Theta as a function of time."},{"Start":"15:03.235 ","End":"15:08.050","Text":"As we know, the general equation for the differential equation that"},{"Start":"15:08.050 ","End":"15:12.420","Text":"we get when dealing with harmonic motion is that our angle Theta as"},{"Start":"15:12.420 ","End":"15:18.515","Text":"a function of time is going to be equal to our A cosine of"},{"Start":"15:18.515 ","End":"15:25.560","Text":"Omega t plus Phi and then plus its starting position,"},{"Start":"15:25.560 ","End":"15:29.908","Text":"so an initial angle of let\u0027s say Omega Theta_0."},{"Start":"15:29.908 ","End":"15:34.119","Text":"So plus Theta_0. We have our Theta_0,"},{"Start":"15:34.119 ","End":"15:35.782","Text":"it\u0027s given to us in the question."},{"Start":"15:35.782 ","End":"15:38.650","Text":"We have this Omega over here;"},{"Start":"15:38.650 ","End":"15:44.455","Text":"this is this, and now what we want to find is what is our A and what is our Phi."},{"Start":"15:44.455 ","End":"15:49.295","Text":"The way we do this is by working out our initial conditions."},{"Start":"15:49.295 ","End":"15:51.235","Text":"Let\u0027s begin."},{"Start":"15:51.235 ","End":"15:58.685","Text":"We know from the question our position at time t is equal to 0."},{"Start":"15:58.685 ","End":"16:04.240","Text":"The pendulum begins its movement from rest at an initial angle of Theta_0."},{"Start":"16:04.240 ","End":"16:10.396","Text":"At t equals 0, our position is Theta_0."},{"Start":"16:10.396 ","End":"16:12.635","Text":"We know that our velocity,"},{"Start":"16:12.635 ","End":"16:14.375","Text":"so our first derivative;"},{"Start":"16:14.375 ","End":"16:18.010","Text":"our Theta-dot, also at the beginning at t equals"},{"Start":"16:18.010 ","End":"16:22.540","Text":"0 is going to be the pendulum begins its movement from rest."},{"Start":"16:22.540 ","End":"16:25.300","Text":"That means it equals to 0."},{"Start":"16:25.300 ","End":"16:35.340","Text":"Our Theta-dot is our angular velocity as a function of time."},{"Start":"16:35.340 ","End":"16:43.615","Text":"Now I\u0027m doing this to show you this is angular velocity,"},{"Start":"16:43.615 ","End":"16:48.965","Text":"this and this over here is our frequency of oscillation."},{"Start":"16:48.965 ","End":"16:51.335","Text":"They\u0027re both completely different things."},{"Start":"16:51.335 ","End":"16:54.079","Text":"Now, I\u0027ve written over here in arrows;"},{"Start":"16:54.079 ","End":"16:57.320","Text":"color coordinated, so that you can see the difference."},{"Start":"16:57.320 ","End":"16:59.270","Text":"It\u0027s very confusing."},{"Start":"16:59.270 ","End":"17:03.100","Text":"Please don\u0027t get confused and make this mistake."},{"Start":"17:03.100 ","End":"17:05.230","Text":"Your Theta-dot over here is"},{"Start":"17:05.230 ","End":"17:09.605","Text":"your angular velocity which as we know is generally denoted by Omega."},{"Start":"17:09.605 ","End":"17:13.320","Text":"If it helps you include the swirl over"},{"Start":"17:13.320 ","End":"17:17.420","Text":"to remind you that this is the angular velocity which is different,"},{"Start":"17:17.420 ","End":"17:21.125","Text":"this Omega over here which is the frequency of oscillation."},{"Start":"17:21.125 ","End":"17:23.470","Text":"Now, why are they completely different?Our"},{"Start":"17:23.470 ","End":"17:27.275","Text":"angular velocity as we know throughout the movement is changing."},{"Start":"17:27.275 ","End":"17:32.255","Text":"Once it gets to its maximum height it\u0027s going to have an angular velocity of 0,"},{"Start":"17:32.255 ","End":"17:35.165","Text":"and once it gets to its point of equilibrium over here"},{"Start":"17:35.165 ","End":"17:38.560","Text":"it\u0027s going to have maximum velocity and back and forth."},{"Start":"17:38.560 ","End":"17:40.910","Text":"Its velocity is changing constantly."},{"Start":"17:40.910 ","End":"17:44.450","Text":"As opposed to this the frequency oscillation is"},{"Start":"17:44.450 ","End":"17:49.205","Text":"something that\u0027s positive and it\u0027s something that is constant, it\u0027s never changing."},{"Start":"17:49.205 ","End":"17:51.160","Text":"G divided by l,"},{"Start":"17:51.160 ","End":"17:53.645","Text":"k divided by m, whatever it will be,"},{"Start":"17:53.645 ","End":"17:58.790","Text":"these are positive numbers and they\u0027re also constants."},{"Start":"17:58.830 ","End":"18:04.400","Text":"It doesn\u0027t matter where we are and from which initial angle we\u0027re starting at,"},{"Start":"18:04.400 ","End":"18:08.400","Text":"our Omega is always going to be our Omega over here."},{"Start":"18:08.400 ","End":"18:12.700","Text":"Our frequency of oscillation is constant throughout the movement and is going to"},{"Start":"18:12.700 ","End":"18:17.380","Text":"be the same and angular velocity is ever-changing."},{"Start":"18:17.380 ","End":"18:18.910","Text":"Now that we\u0027ve dealt with that,"},{"Start":"18:18.910 ","End":"18:21.170","Text":"let\u0027s substitute this in."},{"Start":"18:21.170 ","End":"18:23.750","Text":"Theta as a function of t is equal to this,"},{"Start":"18:23.750 ","End":"18:34.027","Text":"so we have that our Theta_0 is equal to our A cosine of Omega t but our t is equal to 0,"},{"Start":"18:34.027 ","End":"18:36.220","Text":"so that crosses out."},{"Start":"18:36.220 ","End":"18:41.690","Text":"So A cosine of Phi plus Theta_0."},{"Start":"18:41.690 ","End":"18:46.166","Text":"Then we have our Theta-dot which is our angular velocity,"},{"Start":"18:46.166 ","End":"18:51.240","Text":"so what we have to do is we have to take the derivative of this; the first derivative."},{"Start":"18:51.240 ","End":"18:54.730","Text":"That means that we\u0027re going to have 0 because that\u0027s what our Theta-dot is"},{"Start":"18:54.730 ","End":"18:58.120","Text":"equal to and the derivative of this is going to"},{"Start":"18:58.120 ","End":"19:06.463","Text":"be negative A Omega sine of Omega t plus Phi,"},{"Start":"19:06.463 ","End":"19:08.700","Text":"but then because our t is equal to 0,"},{"Start":"19:08.700 ","End":"19:10.490","Text":"so it\u0027s going to cross out."},{"Start":"19:10.490 ","End":"19:13.430","Text":"It\u0027s going to be a Phi solely,"},{"Start":"19:13.430 ","End":"19:17.570","Text":"and then our 0 obviously doesn\u0027t exist."},{"Start":"19:17.570 ","End":"19:22.695","Text":"Then we can start by looking at this equation over here."},{"Start":"19:22.695 ","End":"19:25.475","Text":"We know that this equation has to equal 0,"},{"Start":"19:25.475 ","End":"19:27.758","Text":"we know that our amplitude;"},{"Start":"19:27.758 ","End":"19:29.240","Text":"our A can never equal 0,"},{"Start":"19:29.240 ","End":"19:31.524","Text":"we know that our Omega;"},{"Start":"19:31.524 ","End":"19:32.960","Text":"this over here,"},{"Start":"19:32.960 ","End":"19:37.692","Text":"is also not equal to 0 which means that our sine of Phi must equal to 0."},{"Start":"19:37.692 ","End":"19:39.607","Text":"When is sine of Phi equal to 0?"},{"Start":"19:39.607 ","End":"19:43.435","Text":"When Phi is equal to 0."},{"Start":"19:43.435 ","End":"19:46.265","Text":"Then we can say that our Phi is equal to 0,"},{"Start":"19:46.265 ","End":"19:49.631","Text":"and then substituting that in we can take a look,"},{"Start":"19:49.631 ","End":"19:54.405","Text":"cosine of 0 is equal to 1 which means that"},{"Start":"19:54.405 ","End":"20:00.935","Text":"our amplitude is equal to our Theta_0."},{"Start":"20:00.935 ","End":"20:04.300","Text":"Now, all we have to do is substitute in our values for"},{"Start":"20:04.300 ","End":"20:08.825","Text":"A Phi and our Omega into our initial equation,"},{"Start":"20:08.825 ","End":"20:11.575","Text":"and we\u0027ll get that this is our answer."},{"Start":"20:11.575 ","End":"20:14.950","Text":"This is the angle as a function of time."},{"Start":"20:14.950 ","End":"20:17.630","Text":"That\u0027s the end of this lesson."}],"ID":10630},{"Watched":false,"Name":"1.7 Explnation About A Physical Pendulum","Duration":"14m 44s","ChapterTopicVideoID":10290,"CourseChapterTopicPlaylistID":5365,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:02.150 ","End":"00:04.965","Text":"Hello, in this question,"},{"Start":"00:04.965 ","End":"00:09.045","Text":"we\u0027re dealing with a compound or a physical pendulum."},{"Start":"00:09.045 ","End":"00:12.840","Text":"Now, a compound pendulum or a physical pendulum is"},{"Start":"00:12.840 ","End":"00:18.090","Text":"any rigid body which is attached to some axis of rotation."},{"Start":"00:18.090 ","End":"00:22.770","Text":"It hangs from some point and that is the axis of rotation."},{"Start":"00:22.770 ","End":"00:26.970","Text":"The shape that I\u0027ve drawn here is meant to be some random shape."},{"Start":"00:26.970 ","End":"00:30.045","Text":"Now when we\u0027re speaking about compound or physical pendulums,"},{"Start":"00:30.045 ","End":"00:31.710","Text":"we\u0027re speaking about any shape,"},{"Start":"00:31.710 ","End":"00:33.720","Text":"so that means that it can be a rod,"},{"Start":"00:33.720 ","End":"00:34.965","Text":"it can be a triangle,"},{"Start":"00:34.965 ","End":"00:37.155","Text":"it can be an oblong,"},{"Start":"00:37.155 ","End":"00:38.620","Text":"any shape that you want,"},{"Start":"00:38.620 ","End":"00:44.900","Text":"as long as it\u0027s a solid and it\u0027s attached to some axis of rotation which is fixed."},{"Start":"00:44.900 ","End":"00:47.945","Text":"Then we can consider it as"},{"Start":"00:47.945 ","End":"00:53.740","Text":"a physical pendulum and work out our question according to what we will learn now."},{"Start":"00:53.740 ","End":"00:57.380","Text":"Let\u0027s take a look right now on how"},{"Start":"00:57.380 ","End":"01:01.760","Text":"to solve these types of questions and what this actually means."},{"Start":"01:01.760 ","End":"01:04.474","Text":"If we\u0027re dealing with a physical pendulum,"},{"Start":"01:04.474 ","End":"01:08.525","Text":"the first thing that we have to know is where its center of mass is located."},{"Start":"01:08.525 ","End":"01:10.820","Text":"Now, it\u0027s very easy to find its center"},{"Start":"01:10.820 ","End":"01:13.930","Text":"of mass because that\u0027s at its point of equilibrium."},{"Start":"01:13.930 ","End":"01:15.620","Text":"I wrote over here,"},{"Start":"01:15.620 ","End":"01:19.264","Text":"CM center of mass is equal to PE point of equilibrium."},{"Start":"01:19.264 ","End":"01:22.640","Text":"The next thing that we have to know where it\u0027s hanging from,"},{"Start":"01:22.640 ","End":"01:24.790","Text":"the point, the hanging point,"},{"Start":"01:24.790 ","End":"01:30.455","Text":"and then say that we\u0027ve measured the distance between our CM and our hanging point,"},{"Start":"01:30.455 ","End":"01:33.770","Text":"and that we\u0027re going to call our r_cm."},{"Start":"01:33.770 ","End":"01:40.520","Text":"Now let\u0027s assume that if this is our point of equilibrium,"},{"Start":"01:40.520 ","End":"01:46.070","Text":"when we\u0027re perpendicular to a point of attachment to our ceiling,"},{"Start":"01:46.070 ","End":"01:49.055","Text":"let\u0027s say, when our angle Theta is equal to 0."},{"Start":"01:49.055 ","End":"01:54.979","Text":"This is the equilibrium if we release our pendulum from this position,"},{"Start":"01:54.979 ","End":"01:56.885","Text":"there won\u0027t be any movements."},{"Start":"01:56.885 ","End":"01:59.720","Text":"Say we move it by some angle,"},{"Start":"01:59.720 ","End":"02:02.905","Text":"over here, we can say that this is an angle Theta."},{"Start":"02:02.905 ","End":"02:08.600","Text":"Now we know that according to our moments our pendulum is going to"},{"Start":"02:08.600 ","End":"02:14.430","Text":"go and try and fall backwards towards its point of equilibrium."},{"Start":"02:14.430 ","End":"02:17.540","Text":"It\u0027s going to try and return to its point of equilibrium."},{"Start":"02:17.540 ","End":"02:20.330","Text":"Let\u0027s take a look at why that happens."},{"Start":"02:20.330 ","End":"02:24.845","Text":"Over here we know that from our center of mass,"},{"Start":"02:24.845 ","End":"02:28.025","Text":"we have our mg acting."},{"Start":"02:28.025 ","End":"02:32.930","Text":"This mass has some moment from the hanging point,"},{"Start":"02:32.930 ","End":"02:34.910","Text":"which is going to try and return,"},{"Start":"02:34.910 ","End":"02:36.595","Text":"it\u0027s a restorative for us."},{"Start":"02:36.595 ","End":"02:41.680","Text":"Try and return our physical pendulum to its point of equilibrium."},{"Start":"02:41.680 ","End":"02:45.170","Text":"Now, if we\u0027re dealing with a starting point or"},{"Start":"02:45.170 ","End":"02:50.840","Text":"a starting position where the physical pendulum is at its point of equilibrium,"},{"Start":"02:50.840 ","End":"02:56.315","Text":"then that means that our center of mass is going to be somewhere along this dotted line,"},{"Start":"02:56.315 ","End":"02:58.295","Text":"exactly on the dotted line."},{"Start":"02:58.295 ","End":"03:02.090","Text":"Now another important thing to remember when dealing with these types of"},{"Start":"03:02.090 ","End":"03:04.850","Text":"questions is that the only way that we can"},{"Start":"03:04.850 ","End":"03:07.850","Text":"solve them is if we\u0027re dealing with small angles,"},{"Start":"03:07.850 ","End":"03:11.480","Text":"so to remind you what the definition of a small angle is,"},{"Start":"03:11.480 ","End":"03:17.440","Text":"it means that our angle Theta is going to be significantly smaller than Pi."},{"Start":"03:17.440 ","End":"03:20.875","Text":"We\u0027re dealing with our Theta in radians over here."},{"Start":"03:20.875 ","End":"03:23.060","Text":"When we\u0027re dealing with small angles,"},{"Start":"03:23.060 ","End":"03:29.330","Text":"we can start solving this question by using our moment equations."},{"Start":"03:29.330 ","End":"03:34.560","Text":"Our equations for moments or our equation for torque, that the same thing."},{"Start":"03:34.700 ","End":"03:39.815","Text":"What we\u0027re going to do is we\u0027ll see that our equation for our torques,"},{"Start":"03:39.815 ","End":"03:42.770","Text":"so the sum of all of our torques,"},{"Start":"03:42.770 ","End":"03:45.345","Text":"which is a vector quantity."},{"Start":"03:45.345 ","End":"03:49.340","Text":"We\u0027re going to see we only have to work out the torque of this force over here,"},{"Start":"03:49.340 ","End":"03:51.965","Text":"mg. Now of course,"},{"Start":"03:51.965 ","End":"03:53.300","Text":"at this point over here,"},{"Start":"03:53.300 ","End":"03:55.460","Text":"which is our axis of rotation,"},{"Start":"03:55.460 ","End":"03:58.985","Text":"there\u0027s also going to be a force acting over here."},{"Start":"03:58.985 ","End":"04:03.935","Text":"However, because the force is acting out of the axis of rotation,"},{"Start":"04:03.935 ","End":"04:07.939","Text":"it means that its radius from the axis of rotation"},{"Start":"04:07.939 ","End":"04:12.035","Text":"is going to be zero because it\u0027s a distance zero from the axis of rotation,"},{"Start":"04:12.035 ","End":"04:13.595","Text":"because it\u0027s coming out of there,"},{"Start":"04:13.595 ","End":"04:18.560","Text":"which means that its torque will be therefore equal to 0."},{"Start":"04:18.560 ","End":"04:21.260","Text":"Let\u0027s do this. We have the sum of"},{"Start":"04:21.260 ","End":"04:25.655","Text":"all our torques is equal to the torque of this mg. As we know,"},{"Start":"04:25.655 ","End":"04:28.445","Text":"this is going to be equal to the force"},{"Start":"04:28.445 ","End":"04:33.215","Text":"mg multiplied by its distance from the axis of rotation,"},{"Start":"04:33.215 ","End":"04:35.870","Text":"which in this case is exactly our r_cm."},{"Start":"04:35.870 ","End":"04:43.265","Text":"So multiply by r_cm and then always multiply by sine of the angle between the two."},{"Start":"04:43.265 ","End":"04:46.430","Text":"Now we can see that the angle between the two is Theta,"},{"Start":"04:46.430 ","End":"04:49.984","Text":"because with vectors, as long as we keep the size and direction,"},{"Start":"04:49.984 ","End":"04:51.815","Text":"we can move them to any point."},{"Start":"04:51.815 ","End":"04:55.414","Text":"If you see, if you copy this mg to the axis of rotation,"},{"Start":"04:55.414 ","End":"05:00.470","Text":"then it goes along this dotted line and we can see easily that the angle between"},{"Start":"05:00.470 ","End":"05:06.017","Text":"the dotted line where we\u0027ve just put our mg and our r_cm is going to be Theta."},{"Start":"05:06.017 ","End":"05:09.855","Text":"It\u0027s mg r_cm sine Theta."},{"Start":"05:09.855 ","End":"05:13.865","Text":"Right now what we have is the size of our torque,"},{"Start":"05:13.865 ","End":"05:15.755","Text":"but because our torque is a vector,"},{"Start":"05:15.755 ","End":"05:17.675","Text":"we also need its direction."},{"Start":"05:17.675 ","End":"05:21.005","Text":"Our direction is going to be denoted by a sign over here,"},{"Start":"05:21.005 ","End":"05:23.090","Text":"either positive or negative."},{"Start":"05:23.090 ","End":"05:25.535","Text":"How are we going to decide the sign?"},{"Start":"05:25.535 ","End":"05:27.860","Text":"Now we\u0027ve spoken about this in previous videos,"},{"Start":"05:27.860 ","End":"05:29.765","Text":"but we\u0027ll go over it again over here."},{"Start":"05:29.765 ","End":"05:33.605","Text":"There are two ways to decide what this is going to be."},{"Start":"05:33.605 ","End":"05:36.170","Text":"We can look at our mg and as we know,"},{"Start":"05:36.170 ","End":"05:38.610","Text":"our mg is a restorative force."},{"Start":"05:38.610 ","End":"05:44.270","Text":"It\u0027s trying to push our system back to its point of equilibrium."},{"Start":"05:44.270 ","End":"05:47.750","Text":"Now, any force which is trying to bring"},{"Start":"05:47.750 ","End":"05:52.775","Text":"a system to its point of equilibrium is called a starter force,"},{"Start":"05:52.775 ","End":"05:56.080","Text":"and the sine will always be a negative."},{"Start":"05:56.080 ","End":"05:59.315","Text":"If we\u0027re being pushed to our point of equilibrium,"},{"Start":"05:59.315 ","End":"06:00.845","Text":"the size is negative,"},{"Start":"06:00.845 ","End":"06:04.175","Text":"if we\u0027re being pushed away from our point of equilibrium,"},{"Start":"06:04.175 ","End":"06:07.352","Text":"then the sign will be a positive,"},{"Start":"06:07.352 ","End":"06:12.461","Text":"so mg is a restorative force with which means that it\u0027s sign is negative,"},{"Start":"06:12.461 ","End":"06:13.535","Text":"so that\u0027s the first way."},{"Start":"06:13.535 ","End":"06:17.675","Text":"The second way is to decide or is as a rule,"},{"Start":"06:17.675 ","End":"06:23.045","Text":"that anytime you move your system out of its point of equilibrium,"},{"Start":"06:23.045 ","End":"06:26.929","Text":"the direction that you\u0027ve moved it is going to be the positive direction."},{"Start":"06:26.929 ","End":"06:31.700","Text":"That means that your angle Theta over here is going to be a positive angle."},{"Start":"06:31.700 ","End":"06:33.710","Text":"That\u0027s the easiest way."},{"Start":"06:33.710 ","End":"06:35.780","Text":"If this Theta is a positive angle,"},{"Start":"06:35.780 ","End":"06:39.205","Text":"then that means that anything going in this direction,"},{"Start":"06:39.205 ","End":"06:42.410","Text":"anticlockwise like so,"},{"Start":"06:42.410 ","End":"06:44.929","Text":"is going to be in the positive direction."},{"Start":"06:44.929 ","End":"06:46.385","Text":"Because you don\u0027t want to have"},{"Start":"06:46.385 ","End":"06:48.860","Text":"a negative angle over here because that will mean that all of"},{"Start":"06:48.860 ","End":"06:53.810","Text":"your signs will get confused and you\u0027ll make mistakes in your calculation."},{"Start":"06:53.810 ","End":"06:55.760","Text":"This is a positive angle,"},{"Start":"06:55.760 ","End":"06:57.665","Text":"which means that this is the positive direction,"},{"Start":"06:57.665 ","End":"07:02.400","Text":"which means that here specifically anticlockwise is the positive direction."},{"Start":"07:02.400 ","End":"07:05.710","Text":"Then if we take a look at our mg,"},{"Start":"07:05.710 ","End":"07:13.540","Text":"the torque or the force over here is trying to push the system in the opposite direction,"},{"Start":"07:13.540 ","End":"07:17.500","Text":"in this direction, in the clockwise direction,"},{"Start":"07:17.500 ","End":"07:20.155","Text":"which is the negative direction."},{"Start":"07:20.155 ","End":"07:26.290","Text":"Then again, we get a negative sign over here because your mg is pushing your system,"},{"Start":"07:26.290 ","End":"07:29.050","Text":"not in the positive direction."},{"Start":"07:29.050 ","End":"07:31.883","Text":"Now I\u0027ll quickly draw this in a slightly easier way."},{"Start":"07:31.883 ","End":"07:33.970","Text":"Let\u0027s say this is our origin,"},{"Start":"07:33.970 ","End":"07:38.316","Text":"we\u0027re going to put our two vectors onto the same origin."},{"Start":"07:38.316 ","End":"07:42.145","Text":"I have my r_cm going in this direction."},{"Start":"07:42.145 ","End":"07:43.705","Text":"This is my r_cm,"},{"Start":"07:43.705 ","End":"07:47.980","Text":"and then I have my mg going straight downwards."},{"Start":"07:47.980 ","End":"07:49.225","Text":"Now what I want to do,"},{"Start":"07:49.225 ","End":"07:51.925","Text":"because I\u0027m doing the dot product,"},{"Start":"07:51.925 ","End":"07:57.100","Text":"I want my r vector to go in the direction of my mg, of my force."},{"Start":"07:57.100 ","End":"07:59.260","Text":"My r vector and two, my force."},{"Start":"07:59.260 ","End":"08:00.574","Text":"I want to bring them together."},{"Start":"08:00.574 ","End":"08:04.345","Text":"Which means that I\u0027m doing this."},{"Start":"08:04.345 ","End":"08:07.075","Text":"As we can see, this is clockwise,"},{"Start":"08:07.075 ","End":"08:08.890","Text":"which is the negative direction because we"},{"Start":"08:08.890 ","End":"08:14.403","Text":"said that the positive direction is anticlockwise."},{"Start":"08:14.403 ","End":"08:16.375","Text":"This is our minus over here,"},{"Start":"08:16.375 ","End":"08:18.295","Text":"and then as we know,"},{"Start":"08:18.295 ","End":"08:26.425","Text":"the sum of all of our torques is going to equal to our I multiplied by Alpha."},{"Start":"08:26.425 ","End":"08:30.805","Text":"Moment of inertia multiplied by angular acceleration."},{"Start":"08:30.805 ","End":"08:37.180","Text":"Specifically here, because our axis of rotation is not at the center of mass,"},{"Start":"08:37.180 ","End":"08:39.565","Text":"it\u0027s at the origin, if you will,"},{"Start":"08:39.565 ","End":"08:42.355","Text":"you can say that it\u0027s here,"},{"Start":"08:42.355 ","End":"08:44.500","Text":"the 0, not at the cm."},{"Start":"08:44.500 ","End":"08:46.360","Text":"Now, if specifically,"},{"Start":"08:46.360 ","End":"08:50.770","Text":"we know the moment of inertia at the cm of the center of mass,"},{"Start":"08:50.770 ","End":"08:55.015","Text":"then we can use Steiner\u0027s theorem in order to work"},{"Start":"08:55.015 ","End":"08:59.860","Text":"out our moment of inertia around this point over here,"},{"Start":"08:59.860 ","End":"09:01.330","Text":"and it will go,"},{"Start":"09:01.330 ","End":"09:06.805","Text":"our I around the origin is going to be equal to our I center of mass,"},{"Start":"09:06.805 ","End":"09:11.830","Text":"given that we know this plus our Steiner additive,"},{"Start":"09:11.830 ","End":"09:17.830","Text":"which is mass multiplied by radius squared."},{"Start":"09:17.830 ","End":"09:21.730","Text":"The distance from the center of mass squared."},{"Start":"09:21.730 ","End":"09:26.500","Text":"Now comes in the section where we\u0027re speaking about small angles."},{"Start":"09:26.500 ","End":"09:29.800","Text":"As we know, when dealing with small angles,"},{"Start":"09:29.800 ","End":"09:32.440","Text":"when Theta is significantly smaller than Pi,"},{"Start":"09:32.440 ","End":"09:37.465","Text":"then that means that whenever we see the expression of sine Theta,"},{"Start":"09:37.465 ","End":"09:41.695","Text":"so we can say that it is approximately equal to Theta."},{"Start":"09:41.695 ","End":"09:44.065","Text":"Then going with that,"},{"Start":"09:44.065 ","End":"09:47.425","Text":"we can say that instead of our sine Theta over here,"},{"Start":"09:47.425 ","End":"09:49.990","Text":"that this is equal to Theta."},{"Start":"09:49.990 ","End":"09:52.450","Text":"Of course, our Alpha,"},{"Start":"09:52.450 ","End":"09:56.785","Text":"our angular acceleration is just the second derivative of our Theta,"},{"Start":"09:56.785 ","End":"09:59.575","Text":"so that\u0027s Theta double dots."},{"Start":"09:59.575 ","End":"10:03.940","Text":"Now we see that we have a differential equation and we obviously want to"},{"Start":"10:03.940 ","End":"10:08.394","Text":"get it into the format for a harmonic equation."},{"Start":"10:08.394 ","End":"10:10.630","Text":"If we rewrite this equation that we have"},{"Start":"10:10.630 ","End":"10:13.000","Text":"over here for the sum of all of the torques we\u0027ll have"},{"Start":"10:13.000 ","End":"10:18.880","Text":"the negative mgr_cm multiplied by"},{"Start":"10:18.880 ","End":"10:25.315","Text":"Theta is equal to our I_0 multiplied by Theta double dot."},{"Start":"10:25.315 ","End":"10:28.150","Text":"Now, if we remember our harmonic equation,"},{"Start":"10:28.150 ","End":"10:30.910","Text":"so I\u0027ve written it above over here."},{"Start":"10:30.910 ","End":"10:33.505","Text":"This is the format that we want to get."},{"Start":"10:33.505 ","End":"10:39.505","Text":"Now, remember, our K can represent anything as long it\u0027s a positive constant,"},{"Start":"10:39.505 ","End":"10:41.710","Text":"and so can our M,"},{"Start":"10:41.710 ","End":"10:45.535","Text":"it can represent anything not necessarily actual mass,"},{"Start":"10:45.535 ","End":"10:48.310","Text":"as long as it\u0027s also a positive constant."},{"Start":"10:48.310 ","End":"10:51.190","Text":"If we look at this equation over here,"},{"Start":"10:51.190 ","End":"10:55.330","Text":"we can see our mg and our r are positive constants,"},{"Start":"10:55.330 ","End":"10:58.300","Text":"a moment of inertia about the axis of"},{"Start":"10:58.300 ","End":"11:03.535","Text":"rotation is also going to be positive and constant all the time."},{"Start":"11:03.535 ","End":"11:05.755","Text":"Then over here, our Theta,"},{"Start":"11:05.755 ","End":"11:10.675","Text":"we said our x minus x_0 can also be Theta minus Theta 0,"},{"Start":"11:10.675 ","End":"11:14.005","Text":"and our x double-dot can therefore also be Theta double dot."},{"Start":"11:14.005 ","End":"11:15.669","Text":"Now here specifically,"},{"Start":"11:15.669 ","End":"11:18.820","Text":"our x_0 represents a point of equilibrium."},{"Start":"11:18.820 ","End":"11:21.475","Text":"Now when we look over here when dealing with our Theta\u0027s,"},{"Start":"11:21.475 ","End":"11:25.150","Text":"a point of equilibrium is when our Theta is equal to 0."},{"Start":"11:25.150 ","End":"11:28.105","Text":"That means that we have Theta minus 0,"},{"Start":"11:28.105 ","End":"11:29.345","Text":"which is just Theta,"},{"Start":"11:29.345 ","End":"11:31.560","Text":"which is what we have over here and our x"},{"Start":"11:31.560 ","End":"11:34.635","Text":"double dot is our Theta double dot, which we have over here."},{"Start":"11:34.635 ","End":"11:39.711","Text":"Now we have this equation in terms of the harmonic motion equation."},{"Start":"11:39.711 ","End":"11:45.198","Text":"That means that we can find through this our value for Omega."},{"Start":"11:45.198 ","End":"11:49.690","Text":"As we remember, our value for Omega is going to be the square roots of"},{"Start":"11:49.690 ","End":"11:57.040","Text":"the coefficient of our Theta divided by the coefficient of our Theta double dot."},{"Start":"11:57.040 ","End":"12:05.335","Text":"Over here it\u0027s going to be mgr_cm divided by I_0."},{"Start":"12:05.335 ","End":"12:09.265","Text":"Now what we have here is the classic equation for"},{"Start":"12:09.265 ","End":"12:13.615","Text":"Omega when dealing with a physical or a compound pendulum."},{"Start":"12:13.615 ","End":"12:16.420","Text":"It\u0027s always going to be something that looks like this."},{"Start":"12:16.420 ","End":"12:20.830","Text":"If you remember, when dealing with a mathematical pendulum,"},{"Start":"12:20.830 ","End":"12:24.040","Text":"where we have some rope with a mass attached to it,"},{"Start":"12:24.040 ","End":"12:27.625","Text":"our value for Omega, in that case,"},{"Start":"12:27.625 ","End":"12:32.920","Text":"it\u0027s the square root of g divided by L. This is for a mathematical pendulum."},{"Start":"12:32.920 ","End":"12:38.230","Text":"Now, if you substitute in these values into your Omega for the physical pendulum,"},{"Start":"12:38.230 ","End":"12:41.230","Text":"you\u0027ll see that you\u0027ll get the same Omega."},{"Start":"12:41.230 ","End":"12:43.405","Text":"You\u0027ll see that you\u0027ll get the same values."},{"Start":"12:43.405 ","End":"12:48.205","Text":"This is the general equation for an Omega of any pendulum."},{"Start":"12:48.205 ","End":"12:51.085","Text":"What we have here in our mathematical pendulum,"},{"Start":"12:51.085 ","End":"12:57.100","Text":"which I\u0027ll remind you is a mass attached to a rope which is swinging."},{"Start":"12:57.100 ","End":"13:00.460","Text":"This is one specific example,"},{"Start":"13:00.460 ","End":"13:05.350","Text":"depending on your conditions of the equation that you\u0027ll get for your Omega."},{"Start":"13:05.350 ","End":"13:09.550","Text":"This is a specific example and this is the general equation."},{"Start":"13:09.550 ","End":"13:12.520","Text":"Of course, another thing which is also important to remember,"},{"Start":"13:12.520 ","End":"13:16.630","Text":"I\u0027ll write it here in red that we can"},{"Start":"13:16.630 ","End":"13:21.700","Text":"also find through this equation for our angle as a function of time."},{"Start":"13:21.700 ","End":"13:24.850","Text":"Theta as a function of t is just going to be"},{"Start":"13:24.850 ","End":"13:29.575","Text":"the general solution to this differential equation that we have over here."},{"Start":"13:29.575 ","End":"13:38.260","Text":"It\u0027s just going to be equal to A cosine of Omega t plus Phi."},{"Start":"13:38.260 ","End":"13:41.680","Text":"Of course, we know that our Omega we\u0027ve already found,"},{"Start":"13:41.680 ","End":"13:43.750","Text":"and in order to find our A and our Phi,"},{"Start":"13:43.750 ","End":"13:48.445","Text":"we\u0027re going to need to use the initial conditions given to us in the question."},{"Start":"13:48.445 ","End":"13:50.185","Text":"A short little recap."},{"Start":"13:50.185 ","End":"13:53.980","Text":"We have to remember that a compound pendulum or a physical pendulum,"},{"Start":"13:53.980 ","End":"13:59.740","Text":"is any rigid body which is free to rotate about a fixed axis of rotation."},{"Start":"13:59.740 ","End":"14:01.855","Text":"It can be any shape."},{"Start":"14:01.855 ","End":"14:06.320","Text":"The equation that you\u0027re going to get for its Omega is always going to"},{"Start":"14:06.320 ","End":"14:11.490","Text":"be the square root of mgr_cm divided by I_0."},{"Start":"14:11.520 ","End":"14:14.180","Text":"Your mg is going to be a force,"},{"Start":"14:14.180 ","End":"14:20.434","Text":"your r_cm is your distance between the center of mass and your point of rotation,"},{"Start":"14:20.434 ","End":"14:25.250","Text":"and your I_0 is the moment of inertia around the axis of rotation."},{"Start":"14:25.250 ","End":"14:27.320","Text":"Then how did we get to this equation?"},{"Start":"14:27.320 ","End":"14:30.785","Text":"We always use the fact that we\u0027re dealing with small angles,"},{"Start":"14:30.785 ","End":"14:32.104","Text":"which means this over here,"},{"Start":"14:32.104 ","End":"14:35.210","Text":"then we write out the sum of all of"},{"Start":"14:35.210 ","End":"14:40.985","Text":"the torques and say that it\u0027s equal to I Alpha and then we get to this answer."},{"Start":"14:40.985 ","End":"14:44.010","Text":"That\u0027s the end of this lesson."}],"ID":10631},{"Watched":false,"Name":"1.8 Example- Physical Pendulum","Duration":"9m 7s","ChapterTopicVideoID":10291,"CourseChapterTopicPlaylistID":5365,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.040","Text":"Hello. In this example,"},{"Start":"00:02.040 ","End":"00:08.770","Text":"we\u0027re going to be working out how to calculate the frequencies of a physical pendulum."},{"Start":"00:08.770 ","End":"00:11.085","Text":"Now, what is a physical pendulum?"},{"Start":"00:11.085 ","End":"00:15.195","Text":"In the past, we\u0027ve looked at mathematical pendulums, or simple pendulums,"},{"Start":"00:15.195 ","End":"00:20.550","Text":"which are simply some point mass tied to a massless string,"},{"Start":"00:20.550 ","End":"00:22.605","Text":"and then we work out their frequencies."},{"Start":"00:22.605 ","End":"00:24.465","Text":"However, in real life,"},{"Start":"00:24.465 ","End":"00:26.625","Text":"a string will not be massless,"},{"Start":"00:26.625 ","End":"00:30.630","Text":"and our mass itself will have some radius,"},{"Start":"00:30.630 ","End":"00:33.225","Text":"and it will not be a simple point mass."},{"Start":"00:33.225 ","End":"00:35.930","Text":"Therefore, we\u0027re going to now find out in"},{"Start":"00:35.930 ","End":"00:39.545","Text":"this lesson how to calculate their frequencies."},{"Start":"00:39.545 ","End":"00:44.965","Text":"Now, again, we\u0027re going to be working with small oscillations with small angles."},{"Start":"00:44.965 ","End":"00:46.670","Text":"Let\u0027s see how we do this."},{"Start":"00:46.670 ","End":"00:52.460","Text":"Now the first step that I have to do is I have to locate its axis of rotation,"},{"Start":"00:52.460 ","End":"00:55.310","Text":"which I\u0027m going to label over here."},{"Start":"00:55.310 ","End":"00:58.580","Text":"It\u0027s rotating about this x point."},{"Start":"00:58.580 ","End":"01:03.365","Text":"Then what I have to do is I have to find the sum of all the moments,"},{"Start":"01:03.365 ","End":"01:06.570","Text":"this, which are acting around this point."},{"Start":"01:06.570 ","End":"01:10.780","Text":"Then I say that this is equal to I Alpha,"},{"Start":"01:10.780 ","End":"01:19.985","Text":"where Alpha equals Theta double dot the second derivative of the angle."},{"Start":"01:19.985 ","End":"01:23.875","Text":"What is the only moment acting in the system?"},{"Start":"01:23.875 ","End":"01:27.170","Text":"Mg is the only force acting in the system."},{"Start":"01:27.170 ","End":"01:33.125","Text":"The moment is the force multiplied by the distance,"},{"Start":"01:33.125 ","End":"01:36.495","Text":"multiplied by sine of the angle."},{"Start":"01:36.495 ","End":"01:38.510","Text":"Let\u0027s see how we do this."},{"Start":"01:38.510 ","End":"01:41.195","Text":"We\u0027re going to find the center of mass,"},{"Start":"01:41.195 ","End":"01:45.730","Text":"which is going to be around about here. Why over here?"},{"Start":"01:45.730 ","End":"01:50.210","Text":"Because the center of mass of the rope is going to be right in the center of the rope."},{"Start":"01:50.210 ","End":"01:53.185","Text":"However, there\u0027s then an extra mass,"},{"Start":"01:53.185 ","End":"01:55.430","Text":"this m attached to it,"},{"Start":"01:55.430 ","End":"01:58.805","Text":"which will mean that the center of mass moves down slightly."},{"Start":"01:58.805 ","End":"02:01.020","Text":"It\u0027s not symmetrical."},{"Start":"02:01.960 ","End":"02:06.080","Text":"Now, what we\u0027re going to do is we\u0027re going to call"},{"Start":"02:06.080 ","End":"02:09.620","Text":"this length of rope up until the center of mass,"},{"Start":"02:09.620 ","End":"02:12.815","Text":"we\u0027ll call it c. From the center of mass,"},{"Start":"02:12.815 ","End":"02:15.935","Text":"working directly downwards in the direction of Earth,"},{"Start":"02:15.935 ","End":"02:21.100","Text":"we have the force m\u0027g."},{"Start":"02:21.410 ","End":"02:26.210","Text":"M\u0027 represents the mass of the entire system."},{"Start":"02:26.210 ","End":"02:33.515","Text":"M\u0027 equals this mass plus the mass of the string,"},{"Start":"02:33.515 ","End":"02:39.310","Text":"which is M. What we\u0027ve done up until now is"},{"Start":"02:39.310 ","End":"02:42.850","Text":"similar to making this entire system"},{"Start":"02:42.850 ","End":"02:46.360","Text":"from a whole system of strings and disks and whatever,"},{"Start":"02:46.360 ","End":"02:52.780","Text":"into a single point mass where the force acting downwards."},{"Start":"02:52.780 ","End":"02:59.710","Text":"Now let\u0027s see what the moments are that are being applied to the system."},{"Start":"02:59.710 ","End":"03:04.215","Text":"We have m\u0027g,"},{"Start":"03:04.215 ","End":"03:09.235","Text":"the force multiplied by this distance c,"},{"Start":"03:09.235 ","End":"03:12.145","Text":"the distance to the center of mass."},{"Start":"03:12.145 ","End":"03:15.525","Text":"We\u0027re going to find out what c is in just a minute."},{"Start":"03:15.525 ","End":"03:20.045","Text":"Multiplied by sine of the angle. Now, what\u0027s the angle?"},{"Start":"03:20.045 ","End":"03:26.750","Text":"If this angle is Theta then because this and this are parallel,"},{"Start":"03:26.750 ","End":"03:29.510","Text":"then this angle is also Theta,"},{"Start":"03:29.510 ","End":"03:32.840","Text":"so multiplied by sine of Theta."},{"Start":"03:32.840 ","End":"03:36.005","Text":"Now we have to find the sine of this."},{"Start":"03:36.005 ","End":"03:39.995","Text":"Here our Theta is obviously going to be a positive number,"},{"Start":"03:39.995 ","End":"03:42.950","Text":"which means that going in the anticlockwise direction,"},{"Start":"03:42.950 ","End":"03:45.805","Text":"we\u0027re going to say is the positive direction."},{"Start":"03:45.805 ","End":"03:48.320","Text":"However, if we look,"},{"Start":"03:48.320 ","End":"03:53.660","Text":"mg is trying to force this whole system in a clockwise direction."},{"Start":"03:53.660 ","End":"03:56.150","Text":"If you look, it\u0027s being pushed like this,"},{"Start":"03:56.150 ","End":"03:58.415","Text":"which means that it\u0027s in the negative direction,"},{"Start":"03:58.415 ","End":"04:00.925","Text":"so we add a minus at the start."},{"Start":"04:00.925 ","End":"04:09.110","Text":"Now I have to say that this is equal to I multiplied by Theta double dot."},{"Start":"04:09.110 ","End":"04:11.900","Text":"Now all I need to do is I need to rearrange"},{"Start":"04:11.900 ","End":"04:17.195","Text":"this equation in order to have the relationship between Theta and Theta double dot."},{"Start":"04:17.195 ","End":"04:20.825","Text":"Now because we\u0027re speaking about small angles,"},{"Start":"04:20.825 ","End":"04:26.210","Text":"I can say that my sine Theta is equal to Theta."},{"Start":"04:26.210 ","End":"04:28.970","Text":"At small angles, sine Theta is equal to Theta,"},{"Start":"04:28.970 ","End":"04:31.355","Text":"and cosine of Theta is equal to 1."},{"Start":"04:31.355 ","End":"04:34.315","Text":"Then I can just divide both sides by I."},{"Start":"04:34.315 ","End":"04:39.860","Text":"Then I will get negative m\u0027g multiplied by c"},{"Start":"04:39.860 ","End":"04:47.285","Text":"divided by I Theta equals Theta double dot."},{"Start":"04:47.285 ","End":"04:53.990","Text":"Then what I have here is my Omega^2, which is frequency^2."},{"Start":"04:53.990 ","End":"04:59.950","Text":"Now, all I have to do is I need to figure out what my c is and what my I is."},{"Start":"04:59.950 ","End":"05:01.895","Text":"Then I have my frequency,"},{"Start":"05:01.895 ","End":"05:04.160","Text":"which is the answer to the question."},{"Start":"05:04.160 ","End":"05:06.215","Text":"Now I hear you asking me,"},{"Start":"05:06.215 ","End":"05:07.655","Text":"What is this I?"},{"Start":"05:07.655 ","End":"05:10.925","Text":"Now my I is the moment of inertia,"},{"Start":"05:10.925 ","End":"05:20.190","Text":"and the c is the distance of my point mass to its axis of rotation."},{"Start":"05:20.600 ","End":"05:25.865","Text":"Now, what we\u0027re going to do first is we\u0027re going to find out what c is."},{"Start":"05:25.865 ","End":"05:27.590","Text":"Now, c again,"},{"Start":"05:27.590 ","End":"05:31.850","Text":"represents the location of the center of mass."},{"Start":"05:31.850 ","End":"05:35.795","Text":"All we have to do is we have to find"},{"Start":"05:35.795 ","End":"05:41.930","Text":"the different masses and their distances from the center of rotation,"},{"Start":"05:41.930 ","End":"05:48.530","Text":"which we\u0027ve said is this x over here divided by the sum of all of the masses."},{"Start":"05:48.530 ","End":"05:51.980","Text":"First, we\u0027re going to have small m"},{"Start":"05:51.980 ","End":"05:55.625","Text":"multiplied by its distance from here, which we\u0027ve said is L,"},{"Start":"05:55.625 ","End":"06:01.370","Text":"because that\u0027s the length of the string plus M. Now,"},{"Start":"06:01.370 ","End":"06:05.385","Text":"M is the point mass and the center of the string."},{"Start":"06:05.385 ","End":"06:08.895","Text":"It\u0027s got to be L over 2."},{"Start":"06:08.895 ","End":"06:13.550","Text":"Then we just simply divided by m plus"},{"Start":"06:13.550 ","End":"06:20.258","Text":"M. Now we found c. Now let\u0027s find out what our I is,"},{"Start":"06:20.258 ","End":"06:22.745","Text":"what our moment of inertia is."},{"Start":"06:22.745 ","End":"06:29.115","Text":"It\u0027s going to be the moment of inertia of the string or of a rod,"},{"Start":"06:29.115 ","End":"06:33.440","Text":"it\u0027s the same thing when we\u0027re working these calculations."},{"Start":"06:33.440 ","End":"06:37.465","Text":"We\u0027re going to add to that the moment of inertia of a disk."},{"Start":"06:37.465 ","End":"06:43.720","Text":"First, we\u0027re going to work out the moment of inertia of the string or of a rod."},{"Start":"06:43.720 ","End":"06:46.120","Text":"It\u0027s just something that you have to know off by heart,"},{"Start":"06:46.120 ","End":"06:48.965","Text":"and if you can take notes in with you to the exam,"},{"Start":"06:48.965 ","End":"06:51.575","Text":"then I suggest writing this formula down."},{"Start":"06:51.575 ","End":"06:55.650","Text":"It\u0027s 1/3 multiplied by the mass."},{"Start":"06:55.650 ","End":"07:00.739","Text":"Here the mass is M multiplied by L^2,"},{"Start":"07:00.739 ","End":"07:04.290","Text":"the length of the rod squared."},{"Start":"07:04.290 ","End":"07:11.345","Text":"Then we have to add in the moment of inertia of the disk, so plus,"},{"Start":"07:11.345 ","End":"07:14.735","Text":"and now again, it\u0027s something that your mentor know off by heart,"},{"Start":"07:14.735 ","End":"07:19.280","Text":"1/2m multiplied by the radius^2."},{"Start":"07:19.280 ","End":"07:23.704","Text":"Now, I didn\u0027t write it in the question because I forgot."},{"Start":"07:23.704 ","End":"07:28.190","Text":"But imagine that the disk is of mass m and"},{"Start":"07:28.190 ","End":"07:34.475","Text":"radius R. It wouldn\u0027t need to be a given in the question."},{"Start":"07:34.475 ","End":"07:36.770","Text":"However, now, let\u0027s look at this."},{"Start":"07:36.770 ","End":"07:41.735","Text":"This moment of inertia is for the disk around its center."},{"Start":"07:41.735 ","End":"07:45.240","Text":"However, this disk isn\u0027t spinning around its center,"},{"Start":"07:45.240 ","End":"07:47.760","Text":"it\u0027s spinning around this axis of rotation."},{"Start":"07:47.760 ","End":"07:50.280","Text":"We have to add in Steiner."},{"Start":"07:50.280 ","End":"07:52.035","Text":"What is Steiner?"},{"Start":"07:52.035 ","End":"08:01.705","Text":"Steiner is the mass multiplied by its distance away from the axis of rotation squared."},{"Start":"08:01.705 ","End":"08:05.910","Text":"We\u0027re going to do plus m because that\u0027s the mass of the disk."},{"Start":"08:06.220 ","End":"08:09.815","Text":"Now, it\u0027s center is over here,"},{"Start":"08:09.815 ","End":"08:14.720","Text":"so multiplied by L because it\u0027s alloway plus R,"},{"Start":"08:14.720 ","End":"08:17.660","Text":"the distance to the center of the disk,"},{"Start":"08:17.660 ","End":"08:20.360","Text":"and then all of that squared."},{"Start":"08:20.360 ","End":"08:26.990","Text":"Now we have our I and we also have our c. We just have to plug it into this equation."},{"Start":"08:26.990 ","End":"08:29.855","Text":"Now we go, we found the frequency squared."},{"Start":"08:29.855 ","End":"08:31.775","Text":"Then all you have to do is square root this,"},{"Start":"08:31.775 ","End":"08:32.900","Text":"and that\u0027s the frequency,"},{"Start":"08:32.900 ","End":"08:35.515","Text":"and that is the answer to our question."},{"Start":"08:35.515 ","End":"08:41.450","Text":"This question is the basis for every question that we\u0027re going to see in the future."},{"Start":"08:41.450 ","End":"08:44.329","Text":"These questions always come up in tests."},{"Start":"08:44.329 ","End":"08:48.560","Text":"I suggest learning this formula for how to solve it,"},{"Start":"08:48.560 ","End":"08:53.580","Text":"because it is almost exactly like this every single question."},{"Start":"08:53.580 ","End":"08:55.399","Text":"It\u0027s really a recipe."},{"Start":"08:55.399 ","End":"08:57.155","Text":"You do this, then you do that."},{"Start":"08:57.155 ","End":"08:59.405","Text":"If you follow this exactly,"},{"Start":"08:59.405 ","End":"09:02.660","Text":"then it will be very hard to go wrong."},{"Start":"09:02.660 ","End":"09:04.880","Text":"If this lesson wasn\u0027t clear,"},{"Start":"09:04.880 ","End":"09:07.590","Text":"please go over it again."}],"ID":10632},{"Watched":false,"Name":"1.9 Energy Analysis","Duration":"8m 30s","ChapterTopicVideoID":10292,"CourseChapterTopicPlaylistID":5365,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.735","Text":"Hello. In this lesson we\u0027re going to see how to look at"},{"Start":"00:03.735 ","End":"00:08.520","Text":"harmonic motion with regards to thinking about energy."},{"Start":"00:08.520 ","End":"00:10.665","Text":"There\u0027s 2 ways to do this."},{"Start":"00:10.665 ","End":"00:12.915","Text":"Let\u0027s go through this now."},{"Start":"00:12.915 ","End":"00:16.116","Text":"A little recap from previous lessons;"},{"Start":"00:16.116 ","End":"00:19.980","Text":"up until now we\u0027ve been trying to work out"},{"Start":"00:19.980 ","End":"00:23.760","Text":"our harmonic motion of some system that we"},{"Start":"00:23.760 ","End":"00:28.620","Text":"have through using our force equations or our torque equations."},{"Start":"00:28.620 ","End":"00:31.980","Text":"What we\u0027ve been trying to get to is this equation over here,"},{"Start":"00:31.980 ","End":"00:39.860","Text":"which is negative k(x-x_0) which is equal to mx double dot."},{"Start":"00:39.860 ","End":"00:43.250","Text":"Where k and m are positive constants,"},{"Start":"00:43.250 ","End":"00:49.160","Text":"our x_0 is our point of equilibrium and it can be a positive or a negative number,"},{"Start":"00:49.160 ","End":"00:50.959","Text":"but it must be a constant as well,"},{"Start":"00:50.959 ","End":"00:54.310","Text":"and our x double dot is our acceleration,"},{"Start":"00:54.310 ","End":"00:56.390","Text":"the second derivative of our x."},{"Start":"00:56.390 ","End":"00:59.780","Text":"Then what we got is this differential equation,"},{"Start":"00:59.780 ","End":"01:04.750","Text":"and from that we could find out all information about our system."},{"Start":"01:04.750 ","End":"01:07.670","Text":"This equation was from using forces and of course,"},{"Start":"01:07.670 ","End":"01:12.050","Text":"I\u0027ll remind you that our x and our x double-dot could be any other variables."},{"Start":"01:12.050 ","End":"01:13.430","Text":"It could be our Theta,"},{"Start":"01:13.430 ","End":"01:16.655","Text":"our angles, or r, whatever it might be."},{"Start":"01:16.655 ","End":"01:18.695","Text":"But those were our variables."},{"Start":"01:18.695 ","End":"01:22.475","Text":"Now when using energy we\u0027re going to be using"},{"Start":"01:22.475 ","End":"01:25.310","Text":"a slightly different equation to this and it\u0027s also"},{"Start":"01:25.310 ","End":"01:28.885","Text":"going to describe harmonic motion but in terms of energy."},{"Start":"01:28.885 ","End":"01:31.070","Text":"Let\u0027s write this over here."},{"Start":"01:31.070 ","End":"01:39.080","Text":"E energy is equal to 1/2 mx dot squared."},{"Start":"01:39.080 ","End":"01:47.170","Text":"1 dot over here, plus 1/2 kx^2."},{"Start":"01:47.170 ","End":"01:49.550","Text":"What we have over here is in fact,"},{"Start":"01:49.550 ","End":"01:52.650","Text":"the energy of a spring."},{"Start":"01:52.730 ","End":"01:57.980","Text":"In a similar way that our k and our m have to be"},{"Start":"01:57.980 ","End":"02:05.680","Text":"positive constants so also over here our m and our k are positive constants."},{"Start":"02:05.680 ","End":"02:12.635","Text":"In the same way we have our x and our x dots that are squared. Now our variables."},{"Start":"02:12.635 ","End":"02:16.745","Text":"Obviously, instead of x they can be Theta or r,"},{"Start":"02:16.745 ","End":"02:19.770","Text":"or any other variable."},{"Start":"02:20.150 ","End":"02:26.090","Text":"Now in a similar way that we use this equation over here for our forces and once"},{"Start":"02:26.090 ","End":"02:31.310","Text":"we could show that our force equation was according to this format,"},{"Start":"02:31.310 ","End":"02:35.245","Text":"then we can say that our system was moving in harmonic motion."},{"Start":"02:35.245 ","End":"02:39.845","Text":"Similarly, if we can take our energy equations from the system and put"},{"Start":"02:39.845 ","End":"02:43.070","Text":"our energy equation into this type of format then we"},{"Start":"02:43.070 ","End":"02:46.730","Text":"can say that our system is moving in harmonic motion."},{"Start":"02:46.730 ","End":"02:48.290","Text":"This is the proof as well,"},{"Start":"02:48.290 ","End":"02:50.300","Text":"in the same way that this is."},{"Start":"02:50.300 ","End":"02:53.704","Text":"Now similarly to over here,"},{"Start":"02:53.704 ","End":"02:55.865","Text":"if we get this equation for our energy,"},{"Start":"02:55.865 ","End":"02:58.880","Text":"it\u0027s also a differential equation and you\u0027ll see that it\u0027s"},{"Start":"02:58.880 ","End":"03:03.530","Text":"solution is the exact same solution that we got for our forces."},{"Start":"03:03.530 ","End":"03:07.590","Text":"That means that its position as a function of time and again,"},{"Start":"03:07.590 ","End":"03:09.645","Text":"this can be Theta, r, whatever,"},{"Start":"03:09.645 ","End":"03:17.940","Text":"is equal to A cosine of Omega t plus Phi."},{"Start":"03:17.940 ","End":"03:21.393","Text":"Of course, our Omega,"},{"Start":"03:21.393 ","End":"03:22.625","Text":"so how do we work it out?"},{"Start":"03:22.625 ","End":"03:29.405","Text":"Exactly like over here where we had the square root of this coefficient k,"},{"Start":"03:29.405 ","End":"03:31.145","Text":"which could be any other thing."},{"Start":"03:31.145 ","End":"03:34.610","Text":"But the coefficient of our x minus x_0 divided"},{"Start":"03:34.610 ","End":"03:38.090","Text":"by the coefficient of x double-dot in the square root."},{"Start":"03:38.090 ","End":"03:42.830","Text":"Similar over here, it\u0027s going to be the coefficient of our x^2,"},{"Start":"03:42.830 ","End":"03:48.814","Text":"so that\u0027s k. In the same way divided by the coefficient of our x dot squared,"},{"Start":"03:48.814 ","End":"03:56.920","Text":"divided by m. The exact same Omega for the 2 equations."},{"Start":"03:57.320 ","End":"03:59.590","Text":"The reason that this works,"},{"Start":"03:59.590 ","End":"04:05.515","Text":"that we get this general solution and this solution for Omega is because in actual fact,"},{"Start":"04:05.515 ","End":"04:08.875","Text":"these 2 equations are the exact same thing."},{"Start":"04:08.875 ","End":"04:13.675","Text":"What does that mean? That this equation over here,"},{"Start":"04:13.675 ","End":"04:17.370","Text":"they\u0027re the derivatives of 1 of another."},{"Start":"04:17.370 ","End":"04:19.715","Text":"Let\u0027s take a look at this."},{"Start":"04:19.715 ","End":"04:22.450","Text":"Because we\u0027re dealing with simple harmonic motion,"},{"Start":"04:22.450 ","End":"04:27.170","Text":"we know that our energy is going to be a constant."},{"Start":"04:31.760 ","End":"04:34.230","Text":"Let\u0027s see how we do this."},{"Start":"04:34.230 ","End":"04:35.995","Text":"Our energy is a constant,"},{"Start":"04:35.995 ","End":"04:38.290","Text":"as we said in simple harmonic motion."},{"Start":"04:38.290 ","End":"04:45.805","Text":"Now in order to differentiate this we need to differentiate as a function of time,"},{"Start":"04:45.805 ","End":"04:54.300","Text":"because these x\u0027s represents our position as a function of t. In actual fact,"},{"Start":"04:54.300 ","End":"04:56.570","Text":"we\u0027re doing d by dt;"},{"Start":"04:56.570 ","End":"04:59.065","Text":"differentiating with respect to time."},{"Start":"04:59.065 ","End":"05:02.660","Text":"If we do d by dt of our E,"},{"Start":"05:02.770 ","End":"05:06.040","Text":"let\u0027s see what we get in the equation."},{"Start":"05:06.040 ","End":"05:10.580","Text":"Of this side when we differentiate because we\u0027re differentiating both sides,"},{"Start":"05:10.580 ","End":"05:11.860","Text":"our E is constant,"},{"Start":"05:11.860 ","End":"05:15.970","Text":"which means when we differentiate we\u0027re going to have 0 over here."},{"Start":"05:15.970 ","End":"05:17.995","Text":"Then this is going to be equal to,"},{"Start":"05:17.995 ","End":"05:19.884","Text":"and then let\u0027s differentiate this."},{"Start":"05:19.884 ","End":"05:28.290","Text":"We\u0027re going to have 1/2m and then we\u0027ll have 2 multiplied by our x dot."},{"Start":"05:28.290 ","End":"05:33.740","Text":"Then the inner differential over here which will be x double dot."},{"Start":"05:33.740 ","End":"05:38.015","Text":"Then plus, let\u0027s differentiate this,"},{"Start":"05:38.015 ","End":"05:46.580","Text":"1/2k multiplied by 2x multiplied by the inner differential over here,"},{"Start":"05:46.580 ","End":"05:49.870","Text":"which is going to be x dot."},{"Start":"05:49.870 ","End":"05:53.300","Text":"Now we can see that our 1/2 and our 2 over here"},{"Start":"05:53.300 ","End":"05:56.570","Text":"cancel out and our 1/2 and our 2 over here cancel out."},{"Start":"05:56.570 ","End":"05:59.660","Text":"Then because this side is equal to 0,"},{"Start":"05:59.660 ","End":"06:03.693","Text":"and we can see that we have an x dot over here and an x dot over here,"},{"Start":"06:03.693 ","End":"06:06.275","Text":"we can divide everything by x dot."},{"Start":"06:06.275 ","End":"06:09.590","Text":"So this will remain 0 and then our x dot over here disappears,"},{"Start":"06:09.590 ","End":"06:12.025","Text":"and our x dot over here disappears."},{"Start":"06:12.025 ","End":"06:14.570","Text":"Now what we\u0027re going to do is we\u0027re going to move"},{"Start":"06:14.570 ","End":"06:18.990","Text":"this expression to the other side of the equals sign,"},{"Start":"06:18.990 ","End":"06:27.465","Text":"and we\u0027re going to get negative kx is equal to mx double dots."},{"Start":"06:27.465 ","End":"06:32.720","Text":"As we can see, we get to an equation which is almost identical to this,"},{"Start":"06:32.720 ","End":"06:34.880","Text":"we just don\u0027t have this negative x_0."},{"Start":"06:34.880 ","End":"06:41.890","Text":"But as you saw, sometimes it doesn\u0027t even come into the equation because it can equal 0."},{"Start":"06:41.890 ","End":"06:44.255","Text":"We get the exact same equation."},{"Start":"06:44.255 ","End":"06:48.695","Text":"Later, I\u0027ll show you how you can get this exact version which is x minus x_0."},{"Start":"06:48.695 ","End":"06:50.870","Text":"But you can see basically,"},{"Start":"06:50.870 ","End":"06:54.500","Text":"that these 2 equations are the exact same equations,"},{"Start":"06:54.500 ","End":"06:57.830","Text":"which explains why we get the same solutions."},{"Start":"06:57.830 ","End":"06:59.330","Text":"That was the explanation."},{"Start":"06:59.330 ","End":"07:02.005","Text":"Let\u0027s get back to the lesson."},{"Start":"07:02.005 ","End":"07:07.594","Text":"What we want to do is with our energy equation we want to get back to this format."},{"Start":"07:07.594 ","End":"07:09.590","Text":"Now another way of writing this format,"},{"Start":"07:09.590 ","End":"07:11.660","Text":"so I\u0027m going to write this over here,"},{"Start":"07:11.660 ","End":"07:20.455","Text":"is to say that our E is equal to 1/2mx dot squared, plus 1/2k."},{"Start":"07:20.455 ","End":"07:22.115","Text":"Then instead of x^2,"},{"Start":"07:22.115 ","End":"07:26.850","Text":"we can say (x-x_0)^2."},{"Start":"07:26.850 ","End":"07:31.640","Text":"Then if we do this and we differentiate this,"},{"Start":"07:31.640 ","End":"07:36.305","Text":"then we will see that we get this exact equation with our x minus x_0."},{"Start":"07:36.305 ","End":"07:38.555","Text":"Remember here, because we had our x,"},{"Start":"07:38.555 ","End":"07:42.245","Text":"we got an equation of negative kx equals mx double dot."},{"Start":"07:42.245 ","End":"07:44.540","Text":"Here, if our variable is x minus x_0,"},{"Start":"07:44.540 ","End":"07:50.170","Text":"we\u0027ll get here x minus x_0 as well when we differentiate this."},{"Start":"07:50.360 ","End":"07:55.370","Text":"What we wrote over here is a more specific case where we say that"},{"Start":"07:55.370 ","End":"08:01.110","Text":"our x_0 is equal to 0 because of our point of equilibrium,"},{"Start":"08:01.110 ","End":"08:05.915","Text":"and this is especially useful when dealing with our angle Theta."},{"Start":"08:05.915 ","End":"08:10.340","Text":"However, this is the more general equation which takes into"},{"Start":"08:10.340 ","End":"08:15.245","Text":"account that our point of equilibrium might not be at the origin."},{"Start":"08:15.245 ","End":"08:18.365","Text":"I\u0027m just going to underline this in red,"},{"Start":"08:18.365 ","End":"08:21.515","Text":"because this is also important to know."},{"Start":"08:21.515 ","End":"08:25.580","Text":"In the next lesson I\u0027m going to go over a simple example"},{"Start":"08:25.580 ","End":"08:30.600","Text":"for how to solve these types of questions when dealing with energy and harmonic motion."}],"ID":10633},{"Watched":false,"Name":"1.10 Example- Mathematical Pendulum (Energy)","Duration":"16m 36s","ChapterTopicVideoID":10293,"CourseChapterTopicPlaylistID":5365,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.914","Text":"Hello. What we have here is the mathematical pendulum,"},{"Start":"00:03.914 ","End":"00:06.990","Text":"which means that we have some mass and soon,"},{"Start":"00:06.990 ","End":"00:09.240","Text":"we\u0027ll see that its mass is irrelevant."},{"Start":"00:09.240 ","End":"00:11.595","Text":"It doesn\u0027t play a role in the equation."},{"Start":"00:11.595 ","End":"00:14.280","Text":"It\u0027s attached to a rope of"},{"Start":"00:14.280 ","End":"00:19.620","Text":"length l. Now this dotted line represents the point of equilibrium."},{"Start":"00:19.620 ","End":"00:22.605","Text":"It\u0027s 90 degrees to the ceiling."},{"Start":"00:22.605 ","End":"00:24.185","Text":"This is the ceiling."},{"Start":"00:24.185 ","End":"00:32.145","Text":"Then we can see that our pendulum has been moved in this direction by Theta degrees."},{"Start":"00:32.145 ","End":"00:34.460","Text":"Now what we\u0027re going to do is we\u0027re going to use"},{"Start":"00:34.460 ","End":"00:37.130","Text":"the fact that we\u0027re dealing with small angles,"},{"Start":"00:37.130 ","End":"00:43.280","Text":"which means that our angle Theta is always going to be significantly smaller than Pi."},{"Start":"00:43.280 ","End":"00:46.905","Text":"These are small angles."},{"Start":"00:46.905 ","End":"00:50.810","Text":"Now what we\u0027re going to do is we\u0027re going to analyze its motion in"},{"Start":"00:50.810 ","End":"00:54.470","Text":"the harmonic way by using the ideas of energy."},{"Start":"00:54.470 ","End":"00:58.955","Text":"Now we can also do the analysis by using forces and torques."},{"Start":"00:58.955 ","End":"01:02.695","Text":"But in this example we\u0027re going to be using energy."},{"Start":"01:02.695 ","End":"01:06.110","Text":"As we saw in our previous lesson,"},{"Start":"01:06.110 ","End":"01:07.609","Text":"when we\u0027re dealing with energy,"},{"Start":"01:07.609 ","End":"01:10.880","Text":"in order to show that something is moving in harmonic motion,"},{"Start":"01:10.880 ","End":"01:13.940","Text":"we have to achieve this equation,"},{"Start":"01:13.940 ","End":"01:19.790","Text":"which is E is equal to 1/2 of mx dot"},{"Start":"01:19.790 ","End":"01:27.715","Text":"squared plus 1/2 kx squared."},{"Start":"01:27.715 ","End":"01:29.495","Text":"As we said before,"},{"Start":"01:29.495 ","End":"01:32.705","Text":"these xs simply represent our variable."},{"Start":"01:32.705 ","End":"01:34.415","Text":"Our x over here,"},{"Start":"01:34.415 ","End":"01:38.195","Text":"because we\u0027re using our position as Theta,"},{"Start":"01:38.195 ","End":"01:45.935","Text":"so our x will become Theta and the equation will still work, so let\u0027s begin."},{"Start":"01:45.935 ","End":"01:50.450","Text":"Now up until now, we\u0027ve been using the idea of forces and moments in order to solve this,"},{"Start":"01:50.450 ","End":"01:53.735","Text":"and what we would do is we would draw out our forces,"},{"Start":"01:53.735 ","End":"01:56.540","Text":"write out our force and torque equations,"},{"Start":"01:56.540 ","End":"02:00.350","Text":"and say that that\u0027s equal to MA or equal to I Alpha,"},{"Start":"02:00.350 ","End":"02:04.650","Text":"and then we\u0027d solve and try and rearrange to get our equation."},{"Start":"02:04.650 ","End":"02:06.890","Text":"Let\u0027s see how we do this with energy."},{"Start":"02:06.890 ","End":"02:09.530","Text":"With energy, what we\u0027re going to do is we\u0027re going to write"},{"Start":"02:09.530 ","End":"02:13.770","Text":"the equation for the energy of a pendulum."},{"Start":"02:13.770 ","End":"02:19.200","Text":"We\u0027re going to rise it correct for any general angle,"},{"Start":"02:19.200 ","End":"02:21.020","Text":"so not for 1 specific Theta,"},{"Start":"02:21.020 ","End":"02:22.490","Text":"let\u0027s say not over here."},{"Start":"02:22.490 ","End":"02:26.795","Text":"Because our energy is constantly shifting between"},{"Start":"02:26.795 ","End":"02:33.040","Text":"potential and kinetic because the pendulum is constantly moving."},{"Start":"02:33.110 ","End":"02:37.130","Text":"Our energy is going to be as a function of Theta."},{"Start":"02:37.130 ","End":"02:42.035","Text":"We can write our energy as a function of Theta is going to be,"},{"Start":"02:42.035 ","End":"02:43.865","Text":"so we have our kinetic energy,"},{"Start":"02:43.865 ","End":"02:48.750","Text":"so that\u0027s going to be always 1/2 mv squared."},{"Start":"02:48.750 ","End":"02:51.665","Text":"Then we have plus our potential energy."},{"Start":"02:51.665 ","End":"02:55.805","Text":"Now, potential energy U is as a function of"},{"Start":"02:55.805 ","End":"03:00.670","Text":"Theta because it depends on our height and our height is constantly changing."},{"Start":"03:00.670 ","End":"03:07.470","Text":"As you see here, the pendulum will be at a much lower height than when it\u0027s over here."},{"Start":"03:07.610 ","End":"03:09.975","Text":"We know that our U,"},{"Start":"03:09.975 ","End":"03:18.660","Text":"our potential energy is a function of Theta and that it equals to mgh."},{"Start":"03:18.660 ","End":"03:23.940","Text":"Now we\u0027re going to see that our height is dependent on Theta, like we just said before."},{"Start":"03:23.940 ","End":"03:28.805","Text":"Now we\u0027re going to write an equation for our height as a function of Theta."},{"Start":"03:28.805 ","End":"03:31.355","Text":"Now, a lot of people, what they do,"},{"Start":"03:31.355 ","End":"03:36.050","Text":"is they say that their 0 point is right over here at the bottom."},{"Start":"03:36.050 ","End":"03:39.140","Text":"Now you\u0027ll see once you play around with this,"},{"Start":"03:39.140 ","End":"03:44.405","Text":"that it\u0027s much easier to solve this if you say that your 0 is at this point over here."},{"Start":"03:44.405 ","End":"03:49.375","Text":"We\u0027re going to say that our 0 is over here at the top."},{"Start":"03:49.375 ","End":"03:57.020","Text":"Now we can write over here that our h is equal to 0 over here, so that we remember."},{"Start":"03:57.020 ","End":"04:01.055","Text":"Now we can show the height."},{"Start":"04:01.055 ","End":"04:07.900","Text":"Our height is going to be all the way from here until here."},{"Start":"04:07.900 ","End":"04:10.650","Text":"The line is meant to be straight."},{"Start":"04:10.650 ","End":"04:13.420","Text":"This is our height."},{"Start":"04:14.240 ","End":"04:19.590","Text":"Now what we can see is that if this is our angle Theta,"},{"Start":"04:19.590 ","End":"04:21.905","Text":"then because of alternate angles,"},{"Start":"04:21.905 ","End":"04:26.215","Text":"then that means that this is also our angle Theta."},{"Start":"04:26.215 ","End":"04:32.855","Text":"Now we can see that our hypotenuse over here is our l. This is obviously 90 degrees,"},{"Start":"04:32.855 ","End":"04:34.640","Text":"so we have a right angle triangle."},{"Start":"04:34.640 ","End":"04:39.890","Text":"The hypotenuse is of length l and the adjacent edge is of length"},{"Start":"04:39.890 ","End":"04:46.275","Text":"h. Then by using our trig equations,"},{"Start":"04:46.275 ","End":"04:48.820","Text":"we can see that"},{"Start":"04:50.690 ","End":"05:01.030","Text":"the adjacent side h divided by the hypotenuse l is equal to cosine of Theta."},{"Start":"05:01.250 ","End":"05:03.470","Text":"What I want is my height,"},{"Start":"05:03.470 ","End":"05:07.760","Text":"so all I have to do is isolate out my h. I get that"},{"Start":"05:07.760 ","End":"05:13.470","Text":"my height is equal to l multiplied by cosine of Theta."},{"Start":"05:13.470 ","End":"05:16.775","Text":"Now we have our h as a function of Theta,"},{"Start":"05:16.775 ","End":"05:19.310","Text":"which means that our U is as a function of Theta,"},{"Start":"05:19.310 ","End":"05:23.075","Text":"which gives us our energy equation as a function of Theta."},{"Start":"05:23.075 ","End":"05:25.700","Text":"Now something to notice,"},{"Start":"05:25.700 ","End":"05:29.330","Text":"because my 0 is over here and my height is"},{"Start":"05:29.330 ","End":"05:34.050","Text":"always going to be below the 0 line, so it\u0027s negative."},{"Start":"05:34.460 ","End":"05:41.680","Text":"I\u0027m going to add a negative over here just because I said that my 0 point is up here."},{"Start":"05:41.680 ","End":"05:45.305","Text":"Now we can substitute this into our energy equation."},{"Start":"05:45.305 ","End":"05:50.030","Text":"We can rewrite this so we have that our energy as a function of Theta is going to"},{"Start":"05:50.030 ","End":"05:55.950","Text":"equal to 1/2 mv squared plus our potential energy,"},{"Start":"05:55.950 ","End":"06:04.665","Text":"which is mgh and our h is negative l cosine of Theta."},{"Start":"06:04.665 ","End":"06:09.545","Text":"We have this equation and we want to get to this equation."},{"Start":"06:09.545 ","End":"06:12.190","Text":"We\u0027re going to have to play around over here."},{"Start":"06:12.190 ","End":"06:14.105","Text":"Let\u0027s see what I can do."},{"Start":"06:14.105 ","End":"06:18.290","Text":"Now we\u0027re going to use the fact that we\u0027re dealing with small angles over here."},{"Start":"06:18.290 ","End":"06:26.560","Text":"In small angles, we know that our cosine of Theta is approximately equal to 1."},{"Start":"06:26.560 ","End":"06:29.595","Text":"Then let\u0027s see what we get if we have that."},{"Start":"06:29.595 ","End":"06:35.360","Text":"Instead of having mg multiplied by negative l multiplied by cosine of Theta,"},{"Start":"06:35.360 ","End":"06:41.655","Text":"so with small angles we can have multiplied by 1 over here."},{"Start":"06:41.655 ","End":"06:44.840","Text":"Now, we\u0027ve struck a problem because we see"},{"Start":"06:44.840 ","End":"06:48.220","Text":"that our energy is dependent on our angle Theta."},{"Start":"06:48.220 ","End":"06:55.340","Text":"Once we change our cosine Theta to 1 when using this approximation of small angles,"},{"Start":"06:55.340 ","End":"06:57.985","Text":"we lose our dependence of Theta,"},{"Start":"06:57.985 ","End":"07:04.495","Text":"which means that our analysis of the harmonic motion is going to be incorrect."},{"Start":"07:04.495 ","End":"07:06.920","Text":"We\u0027ve seen that this is a problem."},{"Start":"07:06.920 ","End":"07:08.900","Text":"Now, in one of the previous videos,"},{"Start":"07:08.900 ","End":"07:13.520","Text":"I stated that our cosine Theta and also our sine Theta,"},{"Start":"07:13.520 ","End":"07:16.160","Text":"when we do approximations with small angles,"},{"Start":"07:16.160 ","End":"07:22.325","Text":"we say that our first order approximation is 1 with cosine Theta."},{"Start":"07:22.325 ","End":"07:28.985","Text":"However, if we see that we cancel out our variable that this is dependent on,"},{"Start":"07:28.985 ","End":"07:32.135","Text":"then this isn\u0027t a good approximation and we\u0027ll get the wrong answer."},{"Start":"07:32.135 ","End":"07:38.525","Text":"Which means that we have to go to the next order in order to get a better approximation."},{"Start":"07:38.525 ","End":"07:42.350","Text":"That will mean that we keep our dependence on Theta."},{"Start":"07:42.350 ","End":"07:44.795","Text":"When dealing with cosine of Theta,"},{"Start":"07:44.795 ","End":"07:53.580","Text":"this 2nd order is going to be negative Theta squared divided by 2."},{"Start":"07:53.580 ","End":"07:55.535","Text":"When going to the 2nd order,"},{"Start":"07:55.535 ","End":"07:59.075","Text":"we have 1 minus Theta squared divided by 2."},{"Start":"07:59.075 ","End":"08:04.475","Text":"A good note is that when you\u0027re dealing with forces and torques,"},{"Start":"08:04.475 ","End":"08:06.320","Text":"you just need the 1st order,"},{"Start":"08:06.320 ","End":"08:10.475","Text":"so I\u0027ll just write forces and torques."},{"Start":"08:10.475 ","End":"08:13.100","Text":"However, when you\u0027re dealing with energy,"},{"Start":"08:13.100 ","End":"08:17.730","Text":"you need both orders because we also need Theta squared."},{"Start":"08:17.730 ","End":"08:19.955","Text":"Because here we have our x squared,"},{"Start":"08:19.955 ","End":"08:21.875","Text":"so we want our Theta squared."},{"Start":"08:21.875 ","End":"08:24.380","Text":"Remember because our X goes to Theta."},{"Start":"08:24.380 ","End":"08:26.824","Text":"This is when dealing with energy."},{"Start":"08:26.824 ","End":"08:30.505","Text":"This is something very useful to remember."},{"Start":"08:30.505 ","End":"08:33.255","Text":"Let\u0027s rewrite this."},{"Start":"08:33.255 ","End":"08:37.445","Text":"Let\u0027s take the negative and put it over here."},{"Start":"08:37.445 ","End":"08:43.170","Text":"We have negative and then we can also move the l over here, lm."},{"Start":"08:43.680 ","End":"08:47.695","Text":"Because we had mgh, so we have our l,"},{"Start":"08:47.695 ","End":"08:54.970","Text":"and then we can have here 1 minus Theta squared divided by 2."},{"Start":"08:54.970 ","End":"08:57.250","Text":"I know this is a bit complicated."},{"Start":"08:57.250 ","End":"09:00.040","Text":"If you want to open up these brackets and take"},{"Start":"09:00.040 ","End":"09:04.295","Text":"a few steps back and see that we get this result."},{"Start":"09:04.295 ","End":"09:06.810","Text":"Now what we\u0027re going to do is we\u0027re going to take"},{"Start":"09:06.810 ","End":"09:11.115","Text":"this equation and write it with all the brackets open."},{"Start":"09:11.115 ","End":"09:15.990","Text":"We\u0027re going to get that our energy as a function of Theta is going to be"},{"Start":"09:15.990 ","End":"09:21.400","Text":"equal to half mv squared minus"},{"Start":"09:21.400 ","End":"09:28.960","Text":"lmg plus lmg Theta"},{"Start":"09:28.960 ","End":"09:32.200","Text":"squared divided by 2."},{"Start":"09:32.200 ","End":"09:37.105","Text":"Now we can see that we\u0027re getting close to this equation over here."},{"Start":"09:37.105 ","End":"09:39.070","Text":"But here we have a problem."},{"Start":"09:39.070 ","End":"09:40.975","Text":"Here, we have our v squared,"},{"Start":"09:40.975 ","End":"09:43.915","Text":"which is obviously linear velocity squared."},{"Start":"09:43.915 ","End":"09:46.690","Text":"However, we\u0027re dealing with Theta,"},{"Start":"09:46.690 ","End":"09:49.345","Text":"which is dealing with angles."},{"Start":"09:49.345 ","End":"09:54.685","Text":"What do we want to do is we want to get our angular velocity squared over here."},{"Start":"09:54.685 ","End":"09:58.190","Text":"Let\u0027s take a look at how we do this."},{"Start":"09:58.760 ","End":"10:01.335","Text":"As the pendulum moves,"},{"Start":"10:01.335 ","End":"10:03.555","Text":"it moves in circular motion,"},{"Start":"10:03.555 ","End":"10:08.940","Text":"because it\u0027s always located a distance l away from my origin."},{"Start":"10:08.940 ","End":"10:11.255","Text":"The radius of the circle is l,"},{"Start":"10:11.255 ","End":"10:15.370","Text":"which means that we can use our relationship,"},{"Start":"10:15.370 ","End":"10:21.610","Text":"which is that v is equal to Omega multiplied by the radius of the circle."},{"Start":"10:21.610 ","End":"10:25.225","Text":"Over here, it equals to Omega multiplied by"},{"Start":"10:25.225 ","End":"10:33.745","Text":"l. Then we also know that our Omega is our angular velocity,"},{"Start":"10:33.745 ","End":"10:37.795","Text":"which means that it equals to Theta dot."},{"Start":"10:37.795 ","End":"10:43.105","Text":"Now what we\u0027re going to do is we\u0027re going to substitute this in over here."},{"Start":"10:43.105 ","End":"10:46.825","Text":"We\u0027re going to have that our E as a function of Theta,"},{"Start":"10:46.825 ","End":"10:50.290","Text":"is equal to half multiplied by m,"},{"Start":"10:50.290 ","End":"10:54.010","Text":"and then v squared is going to be Omega l squared,"},{"Start":"10:54.010 ","End":"10:55.900","Text":"where Omega is r Theta dot."},{"Start":"10:55.900 ","End":"11:01.390","Text":"We\u0027re going to have Theta.l squared minus"},{"Start":"11:01.390 ","End":"11:08.860","Text":"mlg plus mlg multiplied"},{"Start":"11:08.860 ","End":"11:12.775","Text":"by Theta squared divided by 2."},{"Start":"11:12.775 ","End":"11:16.540","Text":"Now, let\u0027s open up this bracket."},{"Start":"11:16.540 ","End":"11:21.640","Text":"Now I\u0027ve opened up the brackets and let\u0027s compare our equation"},{"Start":"11:21.640 ","End":"11:26.845","Text":"now to what we want it to look like. Let\u0027s take a look over here."},{"Start":"11:26.845 ","End":"11:28.930","Text":"Here we have a half, half."},{"Start":"11:28.930 ","End":"11:32.335","Text":"Then we have our ml squared,"},{"Start":"11:32.335 ","End":"11:34.270","Text":"which is a positive constant."},{"Start":"11:34.270 ","End":"11:37.165","Text":"That\u0027s this, which is also a positive constant."},{"Start":"11:37.165 ","End":"11:38.602","Text":"Then we have Theta."},{"Start":"11:38.602 ","End":"11:41.935","Text":"squared, which is the same as x dot squared."},{"Start":"11:41.935 ","End":"11:46.000","Text":"Perfect. This section is exactly like this."},{"Start":"11:46.000 ","End":"11:49.660","Text":"Then we have this over here,"},{"Start":"11:49.660 ","End":"11:55.960","Text":"which is some kind of constant that we\u0027re going to look at in a second,"},{"Start":"11:55.960 ","End":"12:00.175","Text":"and then over here we have our half, our half,"},{"Start":"12:00.175 ","End":"12:02.590","Text":"and then our mlg,"},{"Start":"12:02.590 ","End":"12:04.015","Text":"which is our constant,"},{"Start":"12:04.015 ","End":"12:06.730","Text":"k, is a positive constant."},{"Start":"12:06.730 ","End":"12:08.920","Text":"Then instead of our x squared,"},{"Start":"12:08.920 ","End":"12:10.285","Text":"because our x goes to Theta,"},{"Start":"12:10.285 ","End":"12:14.125","Text":"we have our Theta squared. That\u0027s perfect."},{"Start":"12:14.125 ","End":"12:15.490","Text":"Now what we have over here,"},{"Start":"12:15.490 ","End":"12:18.790","Text":"this mlg is some kind of constant."},{"Start":"12:18.790 ","End":"12:25.540","Text":"Now when we\u0027re dealing specifically with energy and we have an extra constant over here,"},{"Start":"12:25.540 ","End":"12:28.945","Text":"what we can do is we can just cross it out."},{"Start":"12:28.945 ","End":"12:31.015","Text":"Now, why can we cross it out?"},{"Start":"12:31.015 ","End":"12:32.875","Text":"Because when we\u0027re dealing with energy,"},{"Start":"12:32.875 ","End":"12:36.355","Text":"it\u0027s correct until a final constant."},{"Start":"12:36.355 ","End":"12:39.880","Text":"We have to have it depending on our variables."},{"Start":"12:39.880 ","End":"12:41.860","Text":"Here it\u0027s our Theta or Theta dot,"},{"Start":"12:41.860 ","End":"12:44.260","Text":"dot and their derivatives."},{"Start":"12:44.260 ","End":"12:50.350","Text":"But when we\u0027re dealing with an added constant that has no variable Theta in it."},{"Start":"12:50.350 ","End":"12:54.370","Text":"So that means we can cross it out when dealing with energy,"},{"Start":"12:54.370 ","End":"12:59.200","Text":"because energy is accurate up until a certain constant."},{"Start":"12:59.200 ","End":"13:01.944","Text":"Now we can also see that if our height,"},{"Start":"13:01.944 ","End":"13:03.310","Text":"if our zero point,"},{"Start":"13:03.310 ","End":"13:05.230","Text":"we said was located over here,"},{"Start":"13:05.230 ","End":"13:08.170","Text":"then we wouldn\u0027t have this constant over here."},{"Start":"13:08.170 ","End":"13:10.720","Text":"That also shows us that our energy is going to be"},{"Start":"13:10.720 ","End":"13:15.340","Text":"correct even without this section over here,"},{"Start":"13:15.340 ","End":"13:18.880","Text":"and another way to look at it is if we take"},{"Start":"13:18.880 ","End":"13:23.020","Text":"the derivative of this in order to get our equation for the forces,"},{"Start":"13:23.020 ","End":"13:25.360","Text":"then by taking the derivative,"},{"Start":"13:25.360 ","End":"13:29.845","Text":"anyway, our constant is going to disappear."},{"Start":"13:29.845 ","End":"13:36.230","Text":"Then again, it doesn\u0027t count and it doesn\u0027t factor into the calculation."},{"Start":"13:36.360 ","End":"13:40.060","Text":"Now it\u0027s very clear that we have harmonic motion,"},{"Start":"13:40.060 ","End":"13:45.520","Text":"and our equation now over here is the exact equation over here."},{"Start":"13:45.520 ","End":"13:49.780","Text":"What we have over here is our m constant and what we"},{"Start":"13:49.780 ","End":"13:54.340","Text":"have over here is our k constant from this equation over here."},{"Start":"13:54.340 ","End":"13:56.440","Text":"Now that we know we have harmonic motion,"},{"Start":"13:56.440 ","End":"13:59.530","Text":"we can write down our Omega, and as we know,"},{"Start":"13:59.530 ","End":"14:04.720","Text":"our Omega is the square root of k divided by m. That\u0027s going to"},{"Start":"14:04.720 ","End":"14:11.545","Text":"be mlg divided by ml squared,"},{"Start":"14:11.545 ","End":"14:17.080","Text":"and of course, we can cross out this m, and this."},{"Start":"14:17.080 ","End":"14:21.430","Text":"What we\u0027ll be left with is the square root of"},{"Start":"14:21.430 ","End":"14:26.875","Text":"g divided by l. Two things we can say about this"},{"Start":"14:26.875 ","End":"14:31.390","Text":"is that Omega equals to the square root of g over l"},{"Start":"14:31.390 ","End":"14:36.730","Text":"is a known equation that we have when dealing with mathematical pendulums,"},{"Start":"14:36.730 ","End":"14:41.575","Text":"and the second thing that we can say is that we can see that our m is here, cancel out."},{"Start":"14:41.575 ","End":"14:46.150","Text":"Remember at the beginning when I said that there\u0027s a mass over here,"},{"Start":"14:46.150 ","End":"14:47.890","Text":"but it doesn\u0027t really matter what mass this"},{"Start":"14:47.890 ","End":"14:51.145","Text":"is because it will cancel out and we\u0027ll see later."},{"Start":"14:51.145 ","End":"14:58.280","Text":"Now we can see that our mass doesn\u0027t play a role in this calculation."},{"Start":"14:58.650 ","End":"15:03.670","Text":"What I\u0027ve done is I\u0027ve squared this result for our energy as a function of"},{"Start":"15:03.670 ","End":"15:08.050","Text":"Theta to show that it\u0027s really harmonic motion when dealing with energy,"},{"Start":"15:08.050 ","End":"15:10.285","Text":"and this is the Omega that we got."},{"Start":"15:10.285 ","End":"15:13.210","Text":"Now the next thing that we can do with this is we can"},{"Start":"15:13.210 ","End":"15:16.435","Text":"find the angle as a function of time."},{"Start":"15:16.435 ","End":"15:18.745","Text":"Theta as a function of time,"},{"Start":"15:18.745 ","End":"15:23.020","Text":"it\u0027s going to be the general solution to this differential equation,"},{"Start":"15:23.020 ","End":"15:24.460","Text":"which as we know,"},{"Start":"15:24.460 ","End":"15:31.510","Text":"is going to be equal to a cosine of Omega t plus Phi."},{"Start":"15:31.510 ","End":"15:34.825","Text":"Now our Omega we already have over here,"},{"Start":"15:34.825 ","End":"15:37.915","Text":"and we know that in order to work out our and our Phi,"},{"Start":"15:37.915 ","End":"15:41.035","Text":"we have to use our initial conditions."},{"Start":"15:41.035 ","End":"15:43.405","Text":"Now here specifically,"},{"Start":"15:43.405 ","End":"15:45.055","Text":"so scroll up a little bit."},{"Start":"15:45.055 ","End":"15:53.305","Text":"Our a, our amplitude is going to be some maximum angle that our pendulum reaches."},{"Start":"15:53.305 ","End":"15:54.970","Text":"It\u0027s not going to be a distance,"},{"Start":"15:54.970 ","End":"15:56.800","Text":"but rather the maximum angle."},{"Start":"15:56.800 ","End":"16:00.715","Text":"If it opens to this angle or to a greater angle,"},{"Start":"16:00.715 ","End":"16:03.790","Text":"and then obviously by using initial conditions,"},{"Start":"16:03.790 ","End":"16:06.325","Text":"we can find out what our a in our Phi is equal to."},{"Start":"16:06.325 ","End":"16:10.360","Text":"The initial conditions that we need to set is we have to find what"},{"Start":"16:10.360 ","End":"16:14.815","Text":"Theta is at t equals zero and what our Theta."},{"Start":"16:14.815 ","End":"16:18.475","Text":"is equal to when t equals zero."},{"Start":"16:18.475 ","End":"16:20.500","Text":"Then just like in previous questions,"},{"Start":"16:20.500 ","End":"16:26.110","Text":"we set our initial conditions and we will find out what our a and what I Phi is equal to."},{"Start":"16:26.110 ","End":"16:31.060","Text":"Here we don\u0027t have the starting position is given in the question,"},{"Start":"16:31.060 ","End":"16:33.830","Text":"so I won\u0027t do it now."},{"Start":"16:34.200 ","End":"16:37.580","Text":"That\u0027s the end of the question."}],"ID":10634}],"Thumbnail":null,"ID":5365},{"Name":"Damped Harmonic Motion","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Damped Hamonic Motion Intro","Duration":"17m 32s","ChapterTopicVideoID":9144,"CourseChapterTopicPlaylistID":5366,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.145","Text":"Hello. In this lesson,"},{"Start":"00:02.145 ","End":"00:04.540","Text":"we\u0027re going to be learning about damping."},{"Start":"00:04.540 ","End":"00:07.650","Text":"That means damped harmonic motion."},{"Start":"00:07.650 ","End":"00:10.275","Text":"We already know simple harmonic motion,"},{"Start":"00:10.275 ","End":"00:13.200","Text":"which is where we have some wall over here,"},{"Start":"00:13.200 ","End":"00:17.955","Text":"and we have a spring with constant k. Attached to it,"},{"Start":"00:17.955 ","End":"00:22.950","Text":"we have some mass m located at point x,"},{"Start":"00:22.950 ","End":"00:25.125","Text":"0, which is our point of equilibrium."},{"Start":"00:25.125 ","End":"00:31.230","Text":"Then, we know that the equation for this will be the sum of all of the forces is equal"},{"Start":"00:31.230 ","End":"00:37.875","Text":"to negative k multiplied by x minus our x_0,"},{"Start":"00:37.875 ","End":"00:44.095","Text":"and this is going to be equal to mx double dot."},{"Start":"00:44.095 ","End":"00:47.600","Text":"Now, when we\u0027re dealing with damping,"},{"Start":"00:47.600 ","End":"00:50.525","Text":"so that means that our harmonic motion is being dampened."},{"Start":"00:50.525 ","End":"00:52.040","Text":"It\u0027s being slowed down."},{"Start":"00:52.040 ","End":"00:56.190","Text":"That means that we have some added force."},{"Start":"00:57.080 ","End":"00:59.305","Text":"An added force,"},{"Start":"00:59.305 ","End":"01:04.610","Text":"which we say is equal to negative Lambda multiplied"},{"Start":"01:04.610 ","End":"01:11.255","Text":"by v. It\u0027s pointing in the opposite direction to the direction of travel."},{"Start":"01:11.255 ","End":"01:13.280","Text":"That\u0027s r minus over here."},{"Start":"01:13.280 ","End":"01:16.265","Text":"Now, usually, when we\u0027re speaking about this damping force,"},{"Start":"01:16.265 ","End":"01:20.130","Text":"we\u0027re speaking about some frictional force."},{"Start":"01:20.840 ","End":"01:26.810","Text":"It\u0027s generally describing some friction with water or air resistance,"},{"Start":"01:26.810 ","End":"01:30.240","Text":"things like that. That\u0027s this force."},{"Start":"01:31.760 ","End":"01:34.880","Text":"We know that our v is equal to x dot,"},{"Start":"01:34.880 ","End":"01:36.275","Text":"so we can write that over here."},{"Start":"01:36.275 ","End":"01:39.830","Text":"Negative Lambda x dot."},{"Start":"01:39.830 ","End":"01:45.170","Text":"Now, if we\u0027re dealing with harmonic motion being damped,"},{"Start":"01:45.170 ","End":"01:46.325","Text":"so what we\u0027re going to do,"},{"Start":"01:46.325 ","End":"01:49.525","Text":"our new equation over here is going to look like this."},{"Start":"01:49.525 ","End":"01:53.297","Text":"We\u0027re going to have the sum of all of the forces is equal to,"},{"Start":"01:53.297 ","End":"01:55.805","Text":"and then we have negative k,"},{"Start":"01:55.805 ","End":"02:00.110","Text":"x minus x_0, like over here."},{"Start":"02:00.110 ","End":"02:03.165","Text":"Then we just add in this."},{"Start":"02:03.165 ","End":"02:08.385","Text":"Add in our F. Negative Lambda x dot,"},{"Start":"02:08.385 ","End":"02:14.755","Text":"and then that\u0027s going to be equal to our mx double dot."},{"Start":"02:14.755 ","End":"02:18.785","Text":"Now, notice we always add it in here with a minus"},{"Start":"02:18.785 ","End":"02:23.137","Text":"because remember we still haven\u0027t decided which direction is positive or negative,"},{"Start":"02:23.137 ","End":"02:25.490","Text":"and because this is an oscillating system,"},{"Start":"02:25.490 ","End":"02:27.950","Text":"meaning that the sign is going to change all the time,"},{"Start":"02:27.950 ","End":"02:32.074","Text":"it works out in the end to be correct."},{"Start":"02:32.074 ","End":"02:37.110","Text":"We always have to have a negative over here and a negative over here."},{"Start":"02:38.390 ","End":"02:42.965","Text":"What we have over here is a differential equation,"},{"Start":"02:42.965 ","End":"02:44.600","Text":"just like what we had over here."},{"Start":"02:44.600 ","End":"02:47.960","Text":"However, this time it\u0027s a little bit more complicated."},{"Start":"02:47.960 ","End":"02:53.675","Text":"Now, we\u0027re going to go over how to put this in a more simple format."},{"Start":"02:53.675 ","End":"02:57.940","Text":"We see over here we have our x minus x_0."},{"Start":"02:57.940 ","End":"03:01.389","Text":"We\u0027re going to name some constant,"},{"Start":"03:01.389 ","End":"03:03.735","Text":"let\u0027s start over here, called z."},{"Start":"03:03.735 ","End":"03:08.620","Text":"Z is going to be equal to x minus x_0."},{"Start":"03:08.620 ","End":"03:12.050","Text":"Now, we can rewrite out this equation,"},{"Start":"03:12.050 ","End":"03:16.745","Text":"and we can say that the sum of all of the forces is going to be equal to"},{"Start":"03:16.745 ","End":"03:22.900","Text":"negative kz, negative Lambda."},{"Start":"03:22.900 ","End":"03:28.610","Text":"Then, we can see that our z dot is going to be equal to x dot,"},{"Start":"03:28.610 ","End":"03:32.580","Text":"so Lambda z dot."},{"Start":"03:32.580 ","End":"03:40.740","Text":"This is going to be equal to m. Then z double dot is equal to x double dot."},{"Start":"03:40.740 ","End":"03:43.775","Text":"So mz double dot."},{"Start":"03:43.775 ","End":"03:48.050","Text":"Now, what we\u0027re going to do is we\u0027re going to move all of these terms"},{"Start":"03:48.050 ","End":"03:53.485","Text":"over here to the other side of the equals sign and set it equal to 0."},{"Start":"03:53.485 ","End":"04:00.555","Text":"We\u0027re going to add to both sides kz and Lambda z dot."},{"Start":"04:00.555 ","End":"04:06.225","Text":"Then, we\u0027re going to have that 0 is equal to mz"},{"Start":"04:06.225 ","End":"04:14.430","Text":"double dot plus Lambda z dot, plus kz."},{"Start":"04:14.430 ","End":"04:17.150","Text":"I just moved them all to the other side."},{"Start":"04:17.150 ","End":"04:18.520","Text":"Now, what I\u0027m going to do,"},{"Start":"04:18.520 ","End":"04:22.430","Text":"is I\u0027m going to divide everything by m because I want to"},{"Start":"04:22.430 ","End":"04:26.360","Text":"have my z double-dot term without a coefficient."},{"Start":"04:26.360 ","End":"04:30.155","Text":"I\u0027m going to have 0 is equal to z"},{"Start":"04:30.155 ","End":"04:40.420","Text":"double dot plus Lambda divided by m z dot plus k divided by mz."},{"Start":"04:41.980 ","End":"04:47.240","Text":"Now, what I\u0027m going to do is I\u0027m going to say over here that may"},{"Start":"04:47.240 ","End":"04:52.925","Text":"Lambda divided by m is going to be equal to Gamma."},{"Start":"04:52.925 ","End":"05:02.335","Text":"This is Gamma, and that my k divided by m is going to be equal to Omega 0^2."},{"Start":"05:02.335 ","End":"05:06.435","Text":"Remember these. These are good to know what they stand for."},{"Start":"05:06.435 ","End":"05:08.575","Text":"Then if I substitute this M,"},{"Start":"05:08.575 ","End":"05:15.695","Text":"I\u0027ll be left with my equation of 0 is equal to z double dot plus"},{"Start":"05:15.695 ","End":"05:21.109","Text":"Gamma z dot"},{"Start":"05:21.109 ","End":"05:28.400","Text":"plus Omega 0^2z."},{"Start":"05:28.400 ","End":"05:32.245","Text":"This is the differential equation that we\u0027ll get to."},{"Start":"05:32.245 ","End":"05:34.750","Text":"Soon or in the next video,"},{"Start":"05:34.750 ","End":"05:38.620","Text":"I\u0027m going to show how you get from here to the general solution."},{"Start":"05:38.620 ","End":"05:42.175","Text":"Most of the time they won\u0027t expect you to know how to do this,"},{"Start":"05:42.175 ","End":"05:44.200","Text":"so you don\u0027t have to see the next video."},{"Start":"05:44.200 ","End":"05:49.240","Text":"However, if you know that you will need to know how to do this or you\u0027re interested in,"},{"Start":"05:49.240 ","End":"05:51.950","Text":"I\u0027ll explain it in the next video."},{"Start":"05:53.420 ","End":"05:57.795","Text":"Now, we\u0027re going to go to the solution."},{"Start":"05:57.795 ","End":"06:04.470","Text":"We\u0027ll show the two or three options that we have in order to solve this."},{"Start":"06:05.360 ","End":"06:08.030","Text":"The first option that we have,"},{"Start":"06:08.030 ","End":"06:14.505","Text":"is that our Gamma divided by 2 is bigger than Omega 0."},{"Start":"06:14.505 ","End":"06:16.230","Text":"Now, here comes that,"},{"Start":"06:16.230 ","End":"06:22.795","Text":"if there\u0027s two or three options bigger than or equal to Omega 0, acts the same."},{"Start":"06:22.795 ","End":"06:24.740","Text":"We can just put like this,"},{"Start":"06:24.740 ","End":"06:26.555","Text":"and now we have two options."},{"Start":"06:26.555 ","End":"06:32.580","Text":"The first is that our Gamma divided by 2 is bigger or equal to Omega 0."},{"Start":"06:33.740 ","End":"06:36.380","Text":"If this is the case,"},{"Start":"06:36.380 ","End":"06:44.150","Text":"then our solution to this differential equation is going to be something very strange."},{"Start":"06:44.150 ","End":"06:48.365","Text":"We\u0027re going to have z as a function of t,"},{"Start":"06:48.365 ","End":"06:51.410","Text":"is going to be equal to e to"},{"Start":"06:51.410 ","End":"06:58.200","Text":"the negative Gamma divided by 2 multiplied by t. Then, in brackets,"},{"Start":"06:58.200 ","End":"07:08.295","Text":"Ae to the negative Omega wave t plus Be"},{"Start":"07:08.295 ","End":"07:16.815","Text":"to the Omega wave t. Our Omega wave is equal to"},{"Start":"07:16.815 ","End":"07:26.860","Text":"the ^2 of Gamma divided by 2^2 minus Omega 0^2."},{"Start":"07:31.170 ","End":"07:34.015","Text":"Now what we\u0027re going to do is we\u0027re going to speak"},{"Start":"07:34.015 ","End":"07:36.865","Text":"about this solution, what does this mean?"},{"Start":"07:36.865 ","End":"07:41.440","Text":"Now the actual solution to the differential equation,"},{"Start":"07:41.440 ","End":"07:45.715","Text":"let\u0027s simplify it and draw a graph to explain what\u0027s happening."},{"Start":"07:45.715 ","End":"07:48.115","Text":"What we can do is,"},{"Start":"07:48.115 ","End":"07:50.620","Text":"because we can see that our Omega wave is going to be"},{"Start":"07:50.620 ","End":"07:53.350","Text":"significantly smaller than our Gamma divided by 2,"},{"Start":"07:53.350 ","End":"07:57.850","Text":"we can simplify this by saying that"},{"Start":"07:57.850 ","End":"08:03.445","Text":"our Z as a function of time is going to be equal to some constant,"},{"Start":"08:03.445 ","End":"08:09.325","Text":"let\u0027s say C multiplied by e to the negative, another constant,"},{"Start":"08:09.325 ","End":"08:15.900","Text":"let\u0027s say D multiplied by t. What that means,"},{"Start":"08:15.900 ","End":"08:17.820","Text":"if we\u0027re going to look at the graph,"},{"Start":"08:17.820 ","End":"08:20.580","Text":"so we\u0027ll have something like this."},{"Start":"08:20.580 ","End":"08:25.510","Text":"Here\u0027s our Z, and then over here this is our time,"},{"Start":"08:25.510 ","End":"08:31.670","Text":"and what\u0027s going to happen is that we\u0027re going to start at some amplitude, say over here."},{"Start":"08:32.130 ","End":"08:35.920","Text":"Then because we have a negative over here,"},{"Start":"08:35.920 ","End":"08:41.680","Text":"exponential to k. What we\u0027ll have is we\u0027ll start at this amplitude and we\u0027ll"},{"Start":"08:41.680 ","End":"08:48.850","Text":"slowly go down until we reach the 0."},{"Start":"08:48.850 ","End":"08:54.490","Text":"What does that mean? We\u0027re going to approach the 0,"},{"Start":"08:54.490 ","End":"08:56.620","Text":"where our amplitude is equal to 0,"},{"Start":"08:56.620 ","End":"08:59.305","Text":"and then that\u0027s where it\u0027s going to stop."},{"Start":"08:59.305 ","End":"09:04.420","Text":"As our time reaches to infinity,"},{"Start":"09:04.420 ","End":"09:08.125","Text":"our harmonic motion is just going to stop at the 0."},{"Start":"09:08.125 ","End":"09:10.105","Text":"Let\u0027s take a look at what that means,"},{"Start":"09:10.105 ","End":"09:12.730","Text":"scroll back up to our diagram over here."},{"Start":"09:12.730 ","End":"09:14.800","Text":"So that means that we\u0027re going to start,"},{"Start":"09:14.800 ","End":"09:16.930","Text":"let\u0027s say at this amplitude over here,"},{"Start":"09:16.930 ","End":"09:20.665","Text":"this distance d,"},{"Start":"09:20.665 ","End":"09:26.665","Text":"and then what\u0027s going to happen is our mass is going to go back to our point x_0,"},{"Start":"09:26.665 ","End":"09:28.870","Text":"to our 0 point,"},{"Start":"09:28.870 ","End":"09:30.490","Text":"our x_0 over here."},{"Start":"09:30.490 ","End":"09:33.445","Text":"It\u0027s not going to oscillate back and forth."},{"Start":"09:33.445 ","End":"09:37.149","Text":"It\u0027s just going to go back to its point of equilibrium,"},{"Start":"09:37.149 ","End":"09:41.020","Text":"so there\u0027ll be no back-and-forth oscillating."},{"Start":"09:41.020 ","End":"09:44.620","Text":"It\u0027s just going to go to its 0 point."},{"Start":"09:44.620 ","End":"09:53.420","Text":"This happens when our Gamma divided by 2 is bigger or equal to our Omega 0,"},{"Start":"09:53.420 ","End":"09:57.805","Text":"when this is our Gamma and this is our Omega 0^2."},{"Start":"09:57.805 ","End":"10:04.495","Text":"When this happens, then this is called critically damped."},{"Start":"10:04.495 ","End":"10:08.005","Text":"What that means is that the system returns to"},{"Start":"10:08.005 ","End":"10:12.550","Text":"equilibrium without oscillating, critically damped."},{"Start":"10:12.550 ","End":"10:19.270","Text":"Now critically damped oscillating systems don\u0027t happen that often and"},{"Start":"10:19.270 ","End":"10:22.600","Text":"it\u0027s slightly less interesting because it\u0027s not really speaking about"},{"Start":"10:22.600 ","End":"10:26.110","Text":"harmonic motion and so you\u0027re less likely to get this in a test,"},{"Start":"10:26.110 ","End":"10:27.970","Text":"but just so that you know."},{"Start":"10:27.970 ","End":"10:33.175","Text":"Now I\u0027m going to rub this out and I\u0027m going to speak about the second option."},{"Start":"10:33.175 ","End":"10:43.045","Text":"The second case is going to be when our Gamma divided by 2 is now smaller than Omega 0."},{"Start":"10:43.045 ","End":"10:45.190","Text":"Now what does this mean?"},{"Start":"10:45.190 ","End":"10:47.230","Text":"Our Gamma has got to do,"},{"Start":"10:47.230 ","End":"10:50.260","Text":"so we can see over here it\u0027s equal to Lambda divided by"},{"Start":"10:50.260 ","End":"10:56.575","Text":"m. What this tells us is that it\u0027s to do with frictional force,"},{"Start":"10:56.575 ","End":"10:58.720","Text":"this over here because this is our Lambda,"},{"Start":"10:58.720 ","End":"11:01.285","Text":"and our Omega 0 if you can see,"},{"Start":"11:01.285 ","End":"11:06.535","Text":"is the square root of k divided by m,"},{"Start":"11:06.535 ","End":"11:11.725","Text":"which is simply the frequency of the oscillating system."},{"Start":"11:11.725 ","End":"11:15.130","Text":"Even if we didn\u0027t have this frictional force over here,"},{"Start":"11:15.130 ","End":"11:19.045","Text":"then we would see that our Omega from this equation,"},{"Start":"11:19.045 ","End":"11:21.655","Text":"the original for a simple harmonic motion,"},{"Start":"11:21.655 ","End":"11:30.805","Text":"so it\u0027s simply the frequency which is the square root of k divided by m. Over here,"},{"Start":"11:30.805 ","End":"11:34.150","Text":"the second option that we have when"},{"Start":"11:34.150 ","End":"11:41.785","Text":"Gamma divided by 2 is smaller than Omega 0 it\u0027s called underdamped."},{"Start":"11:41.785 ","End":"11:45.040","Text":"Underdamped is when the system is oscillating,"},{"Start":"11:45.040 ","End":"11:48.895","Text":"the amplitude is gradually decreasing."},{"Start":"11:48.895 ","End":"11:54.760","Text":"Slowly, slowly the amplitude is becoming less and less until the amplitude reaches 0."},{"Start":"11:54.760 ","End":"11:57.950","Text":"Let\u0027s take a look at the solution."},{"Start":"11:57.950 ","End":"12:03.955","Text":"Our solution now is going to be Z as a function of time,"},{"Start":"12:03.955 ","End":"12:07.960","Text":"is going to be equal to some constant, A,"},{"Start":"12:07.960 ","End":"12:16.440","Text":"multiplied by e to the negative Gamma divided by 2 multiplied by t,"},{"Start":"12:16.440 ","End":"12:23.365","Text":"and then this is multiplied by cosine of Omega let\u0027s call it star,"},{"Start":"12:23.365 ","End":"12:28.735","Text":"multiplied by t plus our Phi."},{"Start":"12:28.735 ","End":"12:34.465","Text":"Then over here, our Omega star is going to be equal to"},{"Start":"12:34.465 ","End":"12:43.465","Text":"the square root of Omega 0 squared minus Gamma divided by 2^2."},{"Start":"12:43.465 ","End":"12:48.520","Text":"Notice that our Omega star and our Omega wave from"},{"Start":"12:48.520 ","End":"12:56.380","Text":"the first case are similar just we\u0027ve switched the order of these 2 variables."},{"Start":"12:56.380 ","End":"12:58.255","Text":"Why is that? Because over here,"},{"Start":"12:58.255 ","End":"13:01.945","Text":"our Gamma divided by 2 is smaller than Omega 0."},{"Start":"13:01.945 ","End":"13:04.495","Text":"When we have it in this order,"},{"Start":"13:04.495 ","End":"13:12.790","Text":"we have a positive constant inside our square root sign,"},{"Start":"13:12.790 ","End":"13:15.400","Text":"which means that we can get an answer."},{"Start":"13:15.400 ","End":"13:17.260","Text":"If we had a negative inside here,"},{"Start":"13:17.260 ","End":"13:20.155","Text":"then we would run into a problem."},{"Start":"13:20.155 ","End":"13:23.440","Text":"Now let\u0027s take a look at what we have."},{"Start":"13:23.440 ","End":"13:29.815","Text":"What we can see is that we have our cosine of some Omega multiplied by t plus our Phi."},{"Start":"13:29.815 ","End":"13:34.990","Text":"This is our general solution for harmonic motion."},{"Start":"13:34.990 ","End":"13:37.360","Text":"We have some oscillation,"},{"Start":"13:37.360 ","End":"13:41.995","Text":"and then sometimes people call this whole thing as"},{"Start":"13:41.995 ","End":"13:47.425","Text":"the amplitude but the amplitude as a function of time."},{"Start":"13:47.425 ","End":"13:53.425","Text":"Our amplitude is changing as the time goes by and we still have harmonic motion."},{"Start":"13:53.425 ","End":"13:57.250","Text":"We can already see the graph that we\u0027re going to get,"},{"Start":"13:57.250 ","End":"13:59.680","Text":"and we can start to imagine what this kind of"},{"Start":"13:59.680 ","End":"14:04.315","Text":"system with underdamped oscillations is going to look like."},{"Start":"14:04.315 ","End":"14:07.495","Text":"Let\u0027s take a look at what this looks like over here."},{"Start":"14:07.495 ","End":"14:13.075","Text":"We can imagine again that we moved our mass to over here and this is the distance"},{"Start":"14:13.075 ","End":"14:16.900","Text":"d. Then that means that we"},{"Start":"14:16.900 ","End":"14:21.400","Text":"stretched all the way to here so it\u0027s going to oscillate past the point of equilibrium,"},{"Start":"14:21.400 ","End":"14:23.800","Text":"but a bit less than this distance d,"},{"Start":"14:23.800 ","End":"14:27.460","Text":"and then it\u0027s going to come back again past the point of equilibrium,"},{"Start":"14:27.460 ","End":"14:29.680","Text":"but a bit less, bit less, bit less,"},{"Start":"14:29.680 ","End":"14:34.135","Text":"bit less, until it eventually comes to a stop."},{"Start":"14:34.135 ","End":"14:39.145","Text":"Now let\u0027s take a look at what this looks like on the graph."},{"Start":"14:39.145 ","End":"14:40.945","Text":"We have our z-axis,"},{"Start":"14:40.945 ","End":"14:44.980","Text":"and notice we have to drag it down because we are oscillating."},{"Start":"14:44.980 ","End":"14:47.020","Text":"So we\u0027re going below the 0,"},{"Start":"14:47.020 ","End":"14:49.540","Text":"we\u0027re going into the negative amplitude,"},{"Start":"14:49.540 ","End":"14:53.815","Text":"and we have our axis representing our time."},{"Start":"14:53.815 ","End":"14:57.009","Text":"How we draw this is first,"},{"Start":"14:57.009 ","End":"14:59.605","Text":"we\u0027re going to draw our amplitude."},{"Start":"14:59.605 ","End":"15:01.435","Text":"We can see that it\u0027s decaying,"},{"Start":"15:01.435 ","End":"15:03.925","Text":"it\u0027s decreasing as time goes by."},{"Start":"15:03.925 ","End":"15:08.710","Text":"We can say that it\u0027s going to look like this,"},{"Start":"15:08.710 ","End":"15:12.905","Text":"and then also from the other side because we are oscillating."},{"Start":"15:12.905 ","End":"15:19.495","Text":"Then eventually, we\u0027ll get to the 0 as our time goes by,"},{"Start":"15:19.495 ","End":"15:21.790","Text":"so we\u0027ll get to this arrow."},{"Start":"15:21.790 ","End":"15:25.610","Text":"Then in order to draw our actual graph,"},{"Start":"15:25.610 ","End":"15:27.200","Text":"we can see that we have our cosine,"},{"Start":"15:27.200 ","End":"15:31.715","Text":"which means that we\u0027re going plus amplitude minus amplitude, so we\u0027re oscillating."},{"Start":"15:31.715 ","End":"15:34.670","Text":"In the end, our graph is going to look like this."},{"Start":"15:34.670 ","End":"15:40.295","Text":"The amplitude starts off really high and then as time goes by,"},{"Start":"15:40.295 ","End":"15:49.370","Text":"the amplitude begins to decrease until it gets to the 0."},{"Start":"15:49.370 ","End":"15:56.440","Text":"These green lines are meant to be mirror images of one another on this t-axis."},{"Start":"15:56.440 ","End":"15:59.900","Text":"What\u0027s important to see over here is"},{"Start":"15:59.900 ","End":"16:03.500","Text":"that although our amplitude is decreasing with time,"},{"Start":"16:03.500 ","End":"16:06.155","Text":"our frequency remains constant."},{"Start":"16:06.155 ","End":"16:12.865","Text":"The distance between two peaks is going to be the same throughout,"},{"Start":"16:12.865 ","End":"16:17.390","Text":"the distance along the horizontal axis."},{"Start":"16:17.390 ","End":"16:23.670","Text":"However, the amplitude is decreasing eventually reaching 0."},{"Start":"16:23.670 ","End":"16:26.495","Text":"This is what you have to know."},{"Start":"16:26.495 ","End":"16:32.310","Text":"Remember that your Omega 0 is simply your normal frequency."},{"Start":"16:33.390 ","End":"16:39.220","Text":"It\u0027s the exact same even if you don\u0027t have any damping at all,"},{"Start":"16:39.220 ","End":"16:41.690","Text":"and of course, this over here,"},{"Start":"16:41.690 ","End":"16:46.925","Text":"your Gamma divided by 2 is to do with your Lambda divided by m,"},{"Start":"16:46.925 ","End":"16:50.285","Text":"which has to do with the damping force."},{"Start":"16:50.285 ","End":"16:55.610","Text":"This is the damping force or some friction."},{"Start":"16:56.190 ","End":"16:59.840","Text":"This only happens when you have damping."},{"Start":"16:59.840 ","End":"17:03.170","Text":"Then of course, you find your amplitude and your"},{"Start":"17:03.170 ","End":"17:08.480","Text":"Phi when you have initial conditions, just like before."},{"Start":"17:09.030 ","End":"17:13.715","Text":"This is the end of our lesson dealing with damping."},{"Start":"17:13.715 ","End":"17:15.365","Text":"This was just the theory."},{"Start":"17:15.365 ","End":"17:18.020","Text":"We\u0027re going to go over some questions so that we can practice"},{"Start":"17:18.020 ","End":"17:21.095","Text":"solving and you can give this a go."},{"Start":"17:21.095 ","End":"17:24.410","Text":"It looks very complicated all of this,"},{"Start":"17:24.410 ","End":"17:25.790","Text":"but once you get the hang of it,"},{"Start":"17:25.790 ","End":"17:28.430","Text":"it\u0027s really not that scary."},{"Start":"17:28.430 ","End":"17:32.520","Text":"Now you can go on to other questions."}],"ID":9414},{"Watched":false,"Name":"Solution To Differential Equation (Damping)","Duration":"13m 59s","ChapterTopicVideoID":9145,"CourseChapterTopicPlaylistID":5366,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.505","Text":"Hello. In the previous lesson,"},{"Start":"00:02.505 ","End":"00:08.490","Text":"we were speaking about damping and we got to this differential equation."},{"Start":"00:08.490 ","End":"00:14.160","Text":"Then we showed the different solutions that we can have and what they mean."},{"Start":"00:14.160 ","End":"00:18.540","Text":"They said in that video that if you wanted to look at how"},{"Start":"00:18.540 ","End":"00:23.293","Text":"you get to the solutions so that you can understand the Maths behind it,"},{"Start":"00:23.293 ","End":"00:25.660","Text":"I\u0027ll explain it here."},{"Start":"00:25.910 ","End":"00:31.185","Text":"What we have over here is our differential equation and now,"},{"Start":"00:31.185 ","End":"00:33.075","Text":"we\u0027re going to solve it."},{"Start":"00:33.075 ","End":"00:37.845","Text":"Now, this was the working out that we had before and this is the actual equation."},{"Start":"00:37.845 ","End":"00:44.060","Text":"The first thing that we\u0027re going to do is we\u0027re going to guess an equation like so."},{"Start":"00:44.060 ","End":"00:50.195","Text":"Of course our solution is going to be z as a function of t. We\u0027re guessing a solution,"},{"Start":"00:50.195 ","End":"00:59.840","Text":"so z is a function of t. This is equal to a constant A multiplied by e^p,"},{"Start":"00:59.840 ","End":"01:04.775","Text":"which is just some constant multiplied by t. Now,"},{"Start":"01:04.775 ","End":"01:07.715","Text":"I said that we\u0027re going to guess this solution."},{"Start":"01:07.715 ","End":"01:09.850","Text":"Now you\u0027re probably wondering, what do you mean guess?"},{"Start":"01:09.850 ","End":"01:14.486","Text":"I\u0027m not going to go into why this is the solution to this differential equation,"},{"Start":"01:14.486 ","End":"01:20.210","Text":"you\u0027ll hopefully be learning this in one of your math\u0027s courses, so they\u0027ll explain it."},{"Start":"01:20.210 ","End":"01:23.450","Text":"But what we need to know is that when we have an equation like this,"},{"Start":"01:23.450 ","End":"01:27.180","Text":"then we guess a solution in this form."},{"Start":"01:27.230 ","End":"01:33.080","Text":"Now what we\u0027re going to do is we\u0027re going to substitute this into our equation."},{"Start":"01:33.080 ","End":"01:39.060","Text":"We have that 0 is equal to z-double dot."},{"Start":"01:39.060 ","End":"01:41.210","Text":"We take the second derivative of this,"},{"Start":"01:41.210 ","End":"01:47.105","Text":"which is going to be simply p^2 multiplied by this,"},{"Start":"01:47.105 ","End":"01:52.190","Text":"multiplied by z as a function of t. I hope"},{"Start":"01:52.190 ","End":"01:57.230","Text":"you remember how to take the second derivative with your e. Then"},{"Start":"01:57.230 ","End":"02:01.400","Text":"we\u0027re going to have plus Gamma multiplied by"},{"Start":"02:01.400 ","End":"02:06.230","Text":"the first derivative of our z so that\u0027s going to be multiplied by p"},{"Start":"02:06.230 ","End":"02:15.900","Text":"multiplied by our z as a function of t plus Omega 0^2 multiplied by"},{"Start":"02:15.900 ","End":"02:20.900","Text":"our z as a function of t. Now we can"},{"Start":"02:20.900 ","End":"02:27.300","Text":"see that we can just divide both sides by z as a function of t and get rid of these."},{"Start":"02:27.590 ","End":"02:30.855","Text":"Then what we have is"},{"Start":"02:30.855 ","End":"02:38.645","Text":"a quadratic equation with regards to our p. In our quadratic equation,"},{"Start":"02:38.645 ","End":"02:42.225","Text":"we have over here our A,"},{"Start":"02:42.225 ","End":"02:44.865","Text":"so let\u0027s in blue."},{"Start":"02:44.865 ","End":"02:50.705","Text":"Here we have our A, which has just 1 multiplied by our p^2 then"},{"Start":"02:50.705 ","End":"02:57.345","Text":"our b is this and our c, Is this."},{"Start":"02:57.345 ","End":"03:02.015","Text":"Now I hope you all know the equation for the quadratic equation,"},{"Start":"03:02.015 ","End":"03:06.590","Text":"which in general terms will be x_1 and 2,"},{"Start":"03:06.590 ","End":"03:08.285","Text":"you always get 2 solutions,"},{"Start":"03:08.285 ","End":"03:14.285","Text":"is going to be equal to negative b plus or minus."},{"Start":"03:14.285 ","End":"03:16.445","Text":"Here are the 2 solutions."},{"Start":"03:16.445 ","End":"03:23.550","Text":"The square root of b^2 minus 4ac,"},{"Start":"03:23.550 ","End":"03:28.425","Text":"and all of this divided by 2a."},{"Start":"03:28.425 ","End":"03:32.690","Text":"This is the general solution to a quadratic equation."},{"Start":"03:32.690 ","End":"03:38.500","Text":"In our case, what we\u0027re going to have is that our p _1 and 2,"},{"Start":"03:38.500 ","End":"03:45.201","Text":"so the 2 solutions that we have for p is going to be equal to negative Gamma"},{"Start":"03:45.201 ","End":"03:53.060","Text":"plus or minus the square root of Gamma^2 minus 4 times a,"},{"Start":"03:53.060 ","End":"03:55.325","Text":"which is 1, times c,"},{"Start":"03:55.325 ","End":"04:03.435","Text":"which is Omega 0^2 and then all of this divided by 2a and our a is 1,"},{"Start":"04:03.435 ","End":"04:07.020","Text":"so it\u0027s just going to be divided by 2."},{"Start":"04:07.700 ","End":"04:11.375","Text":"Now what we\u0027re going to do is we\u0027re going to separate this"},{"Start":"04:11.375 ","End":"04:14.240","Text":"out so we\u0027re going to have that this is equal to"},{"Start":"04:14.240 ","End":"04:16.370","Text":"negative Gamma divided by"},{"Start":"04:16.370 ","End":"04:23.655","Text":"2 plus minus and then we\u0027ll add in our 2 into our square root sign."},{"Start":"04:23.655 ","End":"04:31.785","Text":"It\u0027s going to be the square root of Gamma divided by 2^2,"},{"Start":"04:31.785 ","End":"04:33.900","Text":"because it\u0027s going to be divided by 4."},{"Start":"04:33.900 ","End":"04:35.925","Text":"Then we have Gamma^2 divided by 4,"},{"Start":"04:35.925 ","End":"04:44.080","Text":"which is this negative and then 4 divided by 4 is just 1 Omega 0^2."},{"Start":"04:44.360 ","End":"04:52.460","Text":"In actual fact, what we get over here is this is our Omega wave or Omega star;"},{"Start":"04:52.460 ","End":"04:55.140","Text":"however, we call this."},{"Start":"04:55.810 ","End":"04:59.915","Text":"From this, we can see that we have 2 options."},{"Start":"04:59.915 ","End":"05:06.500","Text":"Either our Gamma divided by 2^2 is bigger than our Omega 0, in which case,"},{"Start":"05:06.500 ","End":"05:10.789","Text":"the expression inside the square root sign is positive"},{"Start":"05:10.789 ","End":"05:16.625","Text":"or our Gamma divided by 2^2 is smaller than our Omega 0^2."},{"Start":"05:16.625 ","End":"05:22.800","Text":"In which case, our expression inside the square root sign is going to be a negative."},{"Start":"05:23.290 ","End":"05:27.050","Text":"Let\u0027s look at our first option,"},{"Start":"05:27.050 ","End":"05:35.101","Text":"and our first option is that Gamma divided by 2 is bigger or equal to our Omega 0."},{"Start":"05:35.101 ","End":"05:40.160","Text":"Which means that its squared is also going to be bigger than our Omega 0^2,"},{"Start":"05:40.160 ","End":"05:42.190","Text":"so it\u0027s the same thing."},{"Start":"05:42.190 ","End":"05:48.690","Text":"In that case, let\u0027s substitute this in to our answer over here, to our solution."},{"Start":"05:48.690 ","End":"05:51.640","Text":"Then we\u0027ll have that as z as a function of t,"},{"Start":"05:51.640 ","End":"05:57.340","Text":"is going to be equal to some constant multiplied by"},{"Start":"05:57.340 ","End":"06:05.870","Text":"e^pt so p is going to be negative Gamma divided by 2."},{"Start":"06:05.870 ","End":"06:07.745","Text":"Then we have our first option,"},{"Start":"06:07.745 ","End":"06:13.830","Text":"which is plus Omega wave and they\u0027ll all of this multiplied by"},{"Start":"06:13.830 ","End":"06:20.245","Text":"t. Then our next option is with the minus Omega wave over here,"},{"Start":"06:20.245 ","End":"06:27.015","Text":"plus some other constant b multiplied by e, and then we have 2ipt."},{"Start":"06:27.015 ","End":"06:35.630","Text":"Again, negative Gamma divided by 2 but this time minus our Omega wave,"},{"Start":"06:35.630 ","End":"06:40.110","Text":"then this multiplied by t."},{"Start":"06:41.060 ","End":"06:48.815","Text":"Now because we know that our Omega 0 is smaller than our Gamma divided by 2,"},{"Start":"06:48.815 ","End":"06:51.665","Text":"we also know that therefore"},{"Start":"06:51.665 ","End":"06:57.260","Text":"our Omega wave is always going to be smaller than our Gamma divided by 2,"},{"Start":"06:57.260 ","End":"07:02.200","Text":"because our Omega wave is this."},{"Start":"07:03.330 ","End":"07:09.565","Text":"That means that what we\u0027re going to have here is our e is always going to"},{"Start":"07:09.565 ","End":"07:15.895","Text":"be to the power of some negative constant multiplied by t,"},{"Start":"07:15.895 ","End":"07:18.430","Text":"in both these cases."},{"Start":"07:18.430 ","End":"07:20.770","Text":"Over here we\u0027re going to have a negative,"},{"Start":"07:20.770 ","End":"07:22.315","Text":"which means that we have decayed."},{"Start":"07:22.315 ","End":"07:26.580","Text":"In that case, we can rewrite our solution"},{"Start":"07:26.580 ","End":"07:31.690","Text":"z(t) is always going to be equal to some constant."},{"Start":"07:31.690 ","End":"07:38.765","Text":"To not confuse you, let\u0027s say C for constant multiplied by e to the negative,"},{"Start":"07:38.765 ","End":"07:45.225","Text":"and then some expression multiplied by t. This is what interests us."},{"Start":"07:45.225 ","End":"07:46.485","Text":"This form over here,"},{"Start":"07:46.485 ","End":"07:49.290","Text":"because this we can understand and we can take"},{"Start":"07:49.290 ","End":"07:52.930","Text":"a look at what this means and we\u0027ll look at it a little bit later."},{"Start":"07:52.930 ","End":"07:56.170","Text":"In the meantime, before we look at what this means,"},{"Start":"07:56.170 ","End":"07:59.350","Text":"let\u0027s look at our second option."},{"Start":"07:59.350 ","End":"08:06.220","Text":"Our second option is that our Gamma divided by 2 is smaller than our Omega 0."},{"Start":"08:06.220 ","End":"08:13.130","Text":"Which means that the expression that we have inside the square root sign is a negative."},{"Start":"08:13.340 ","End":"08:18.285","Text":"What I\u0027m going to do, when we\u0027re speaking about this in mathematical terms,"},{"Start":"08:18.285 ","End":"08:22.120","Text":"I can say this is an expression that we"},{"Start":"08:22.120 ","End":"08:26.560","Text":"can understand properly to have a square root of a negative expression."},{"Start":"08:26.560 ","End":"08:30.310","Text":"Then what we do in order to solve this is that we switch"},{"Start":"08:30.310 ","End":"08:34.390","Text":"the order of these 2 terms and we define something else."},{"Start":"08:34.390 ","End":"08:35.845","Text":"I\u0027m going to rub this out."},{"Start":"08:35.845 ","End":"08:38.860","Text":"This is no longer going to be our Omega wave."},{"Start":"08:38.860 ","End":"08:41.650","Text":"We\u0027re going to say therefore that our Omega wave"},{"Start":"08:41.650 ","End":"08:44.830","Text":"is the opposite of this, something positive."},{"Start":"08:44.830 ","End":"08:48.055","Text":"It\u0027s going to be Omega 0^2."},{"Start":"08:48.055 ","End":"08:50.770","Text":"We\u0027re multiplying it by negative 1,"},{"Start":"08:50.770 ","End":"08:56.065","Text":"negative Gamma divided by 2^2."},{"Start":"08:56.065 ","End":"08:58.930","Text":"What we\u0027ve done is we\u0027ve multiplied what\u0027s inside"},{"Start":"08:58.930 ","End":"09:03.940","Text":"the square root sign by negative 1 in order to get a positive expression."},{"Start":"09:03.940 ","End":"09:07.135","Text":"Now this is our Omega wave,"},{"Start":"09:07.135 ","End":"09:10.465","Text":"which is the negative of what we have over here."},{"Start":"09:10.465 ","End":"09:15.700","Text":"That means that when we\u0027re dealing with, here specifically,"},{"Start":"09:15.700 ","End":"09:19.066","Text":"our Gamma divided by 2 being smaller than our Omega is 0,"},{"Start":"09:19.066 ","End":"09:22.630","Text":"then we\u0027ll have that our P1,2 is going to be equal"},{"Start":"09:22.630 ","End":"09:27.055","Text":"to negative Gamma divided by 2 plus or minus."},{"Start":"09:27.055 ","End":"09:34.900","Text":"This is going to be the square root of a negative Omega wave."},{"Start":"09:34.900 ","End":"09:36.610","Text":"What does that mean?"},{"Start":"09:36.610 ","End":"09:43.375","Text":"We have to write it via i Omega wave."},{"Start":"09:43.375 ","End":"09:47.200","Text":"When we take the negative out of the square root sign,"},{"Start":"09:47.200 ","End":"09:49.765","Text":"we multiply it by i."},{"Start":"09:49.765 ","End":"09:52.855","Text":"If before this was our Omega wave,"},{"Start":"09:52.855 ","End":"09:54.205","Text":"but this is now a negative,"},{"Start":"09:54.205 ","End":"09:56.320","Text":"to get the negative out of the square root sign,"},{"Start":"09:56.320 ","End":"10:00.115","Text":"we have i and then we have our normal Omega wave."},{"Start":"10:00.115 ","End":"10:03.985","Text":"This is what we are left with over here."},{"Start":"10:03.985 ","End":"10:06.924","Text":"Once we have this,"},{"Start":"10:06.924 ","End":"10:08.980","Text":"and this equation,"},{"Start":"10:08.980 ","End":"10:13.570","Text":"so we can substitute it again into this over here, into our z."},{"Start":"10:13.570 ","End":"10:20.035","Text":"That means that we\u0027ll have our z as a function of t is going to be equal to."},{"Start":"10:20.035 ","End":"10:24.970","Text":"We\u0027re going to have some constant A multiplied by e,"},{"Start":"10:24.970 ","End":"10:29.065","Text":"multiplied by our P. Now our P is this over here."},{"Start":"10:29.065 ","End":"10:36.190","Text":"We\u0027re going to have negative Gamma divided by 2 multiplied by"},{"Start":"10:36.190 ","End":"10:42.475","Text":"t plus i Omega wave"},{"Start":"10:42.475 ","End":"10:46.345","Text":"multiplied by t. Because we have PT, so our P is this."},{"Start":"10:46.345 ","End":"10:49.165","Text":"Just already open the brackets."},{"Start":"10:49.165 ","End":"10:53.860","Text":"Then what I\u0027m going to have next is my second option,"},{"Start":"10:53.860 ","End":"10:56.290","Text":"the option with the negative over here."},{"Start":"10:56.290 ","End":"11:01.360","Text":"Plus some different constant and then e to"},{"Start":"11:01.360 ","End":"11:06.580","Text":"our p multiplied by t. Now my P is going to be negative Gamma divided by 2."},{"Start":"11:06.580 ","End":"11:08.650","Text":"Then already opening the brackets with a t,"},{"Start":"11:08.650 ","End":"11:12.085","Text":"so t and then this time negative"},{"Start":"11:12.085 ","End":"11:20.155","Text":"i Omega wave t. Now we can see that in both terms,"},{"Start":"11:20.155 ","End":"11:24.280","Text":"I have e to the power of negative Gamma divided by 2t."},{"Start":"11:24.280 ","End":"11:27.415","Text":"I\u0027m going to take that out as a common factor."},{"Start":"11:27.415 ","End":"11:31.985","Text":"I\u0027ll have e to the negative Gamma divided by 2t."},{"Start":"11:31.985 ","End":"11:42.205","Text":"Then in brackets, I\u0027ll have my A multiplied by e to the i Omega wave t"},{"Start":"11:42.205 ","End":"11:47.305","Text":"plus my B multiplied by e to the negative"},{"Start":"11:47.305 ","End":"11:55.840","Text":"i Omega wave t. I hope you see what I\u0027ve done here using this exponent,"},{"Start":"11:55.840 ","End":"11:58.495","Text":"and rearranging it like so."},{"Start":"11:58.495 ","End":"12:00.505","Text":"Now according to Euler,"},{"Start":"12:00.505 ","End":"12:04.555","Text":"we can rewrite this over here as cosine."},{"Start":"12:04.555 ","End":"12:07.465","Text":"I\u0027m not going to go into the explanation of that."},{"Start":"12:07.465 ","End":"12:09.955","Text":"You don\u0027t need to know for this course the explanation."},{"Start":"12:09.955 ","End":"12:12.160","Text":"If you\u0027re interested, go and ask one of"},{"Start":"12:12.160 ","End":"12:15.340","Text":"your maths professors or you can look it up on the Internet."},{"Start":"12:15.340 ","End":"12:20.440","Text":"But basically, you just need to know that this can be rewritten as cosine."},{"Start":"12:20.440 ","End":"12:22.450","Text":"Then if we do that,"},{"Start":"12:22.450 ","End":"12:25.690","Text":"we\u0027re going to have some constant."},{"Start":"12:25.690 ","End":"12:31.660","Text":"Let\u0027s call it C. C has a connection to your A and B,"},{"Start":"12:31.660 ","End":"12:41.995","Text":"and this is equal to some constant multiplied by e to the negative Gamma divided by 2t,"},{"Start":"12:41.995 ","End":"12:43.990","Text":"what you have over here."},{"Start":"12:43.990 ","End":"12:50.695","Text":"Then instead of this e to the i Omega t plus e to the negative i Omega t,"},{"Start":"12:50.695 ","End":"12:59.480","Text":"this turns to cosine of Omega wave t plus Phi."},{"Start":"12:59.850 ","End":"13:09.250","Text":"What we have over here is we have 2 new constants, this and this."},{"Start":"13:09.250 ","End":"13:13.660","Text":"These are new and you can find their connections to your A and your B."},{"Start":"13:13.660 ","End":"13:15.399","Text":"They\u0027re connected."},{"Start":"13:15.399 ","End":"13:21.280","Text":"But this, what we have over here on this line is equal to this."},{"Start":"13:21.280 ","End":"13:24.805","Text":"This is our second solution."},{"Start":"13:24.805 ","End":"13:29.770","Text":"This is z as a function of t. We\u0027ve gotten our 2 solutions"},{"Start":"13:29.770 ","End":"13:35.590","Text":"depending on the relationship between Gamma divided by 2 and Omega 0."},{"Start":"13:35.590 ","End":"13:37.360","Text":"We have our 2 solutions."},{"Start":"13:37.360 ","End":"13:42.280","Text":"I said I\u0027ll speak about what these formats mean."},{"Start":"13:42.280 ","End":"13:45.265","Text":"What these solutions mean in a graphical way."},{"Start":"13:45.265 ","End":"13:47.530","Text":"I described it in the previous video."},{"Start":"13:47.530 ","End":"13:52.180","Text":"You can go back there and watch it to get a bit of an explanation of what these look"},{"Start":"13:52.180 ","End":"13:57.460","Text":"like graphically and what this means when we\u0027re speaking about harmonic motion."},{"Start":"13:57.460 ","End":"14:00.380","Text":"That\u0027s the end of this video."}],"ID":9415},{"Watched":false,"Name":"Example - Ball in Water Tank","Duration":"9m 41s","ChapterTopicVideoID":10424,"CourseChapterTopicPlaylistID":5366,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.400","Text":"Hello. In this question,"},{"Start":"00:02.400 ","End":"00:06.690","Text":"we\u0027re being given a ball of mass M and radius I,"},{"Start":"00:06.690 ","End":"00:12.090","Text":"which is placed inside a water tank and attached to the wall via a spring of"},{"Start":"00:12.090 ","End":"00:18.000","Text":"spring constant k. The water in the tank applies a resistive force on the ball."},{"Start":"00:18.000 ","End":"00:20.595","Text":"This force is called Stokes law,"},{"Start":"00:20.595 ","End":"00:22.950","Text":"and it\u0027s given by F,"},{"Start":"00:22.950 ","End":"00:29.535","Text":"the resistive force is equal to negative 6Pi R Eta V,"},{"Start":"00:29.535 ","End":"00:34.785","Text":"where Eta is the viscosity of the water and I is the radius of the ball."},{"Start":"00:34.785 ","End":"00:36.380","Text":"Assume that I,"},{"Start":"00:36.380 ","End":"00:38.684","Text":"Eta, k and m are given,"},{"Start":"00:38.684 ","End":"00:42.890","Text":"and find the ball\u0027s frequency of oscillations given that I is"},{"Start":"00:42.890 ","End":"00:48.040","Text":"smaller than the square root of m k divided by 3Pi Eta."},{"Start":"00:48.040 ","End":"00:50.135","Text":"We\u0027re being told that in this question,"},{"Start":"00:50.135 ","End":"00:55.590","Text":"there is no friction between the ball and the bottom of the tank."},{"Start":"00:56.060 ","End":"01:02.615","Text":"In this question, we\u0027re trying to find the frequency of oscillations."},{"Start":"01:02.615 ","End":"01:05.570","Text":"The first thing that we\u0027re going to do is we\u0027re going to write out"},{"Start":"01:05.570 ","End":"01:11.010","Text":"the force equations for the forces acting on the ball."},{"Start":"01:11.230 ","End":"01:19.970","Text":"Let\u0027s begin by writing out the sum of all of the forces in the x-direction."},{"Start":"01:19.970 ","End":"01:23.869","Text":"First of all, we have the force of the spring,"},{"Start":"01:23.869 ","End":"01:27.770","Text":"so that\u0027s equal to negative kx."},{"Start":"01:27.770 ","End":"01:31.520","Text":"Then we have our resistive force,"},{"Start":"01:31.520 ","End":"01:36.215","Text":"or Stokes law, which is negative 6Pi."},{"Start":"01:36.215 ","End":"01:40.325","Text":"Now R, we\u0027re being told that the radius of the ball is lowercase I."},{"Start":"01:40.325 ","End":"01:49.010","Text":"Let\u0027s substitute that n multiplied by Eta multiplied by v. Now I,"},{"Start":"01:49.010 ","End":"01:53.270","Text":"V is the velocity and we don\u0027t have to add in a vector sum over"},{"Start":"01:53.270 ","End":"01:58.325","Text":"here because we know that we are taking the velocity in the x-direction."},{"Start":"01:58.325 ","End":"02:03.105","Text":"That\u0027s why I\u0027m emitting the vector arrow at the top."},{"Start":"02:03.105 ","End":"02:08.300","Text":"That is of course equals to mass times acceleration."},{"Start":"02:08.300 ","End":"02:11.855","Text":"Now what I\u0027m going to do is I know that my v,"},{"Start":"02:11.855 ","End":"02:15.785","Text":"my velocity is equal x dot the first derivative of my x,"},{"Start":"02:15.785 ","End":"02:18.740","Text":"and that my acceleration is equal"},{"Start":"02:18.740 ","End":"02:22.790","Text":"the second derivative of my x. I\u0027m going to rewrite that."},{"Start":"02:22.790 ","End":"02:26.510","Text":"I\u0027ll have that the sum of all of my forces in the x-direction is equal"},{"Start":"02:26.510 ","End":"02:34.665","Text":"negative kx minus 6Pi r Eta x dot,"},{"Start":"02:34.665 ","End":"02:39.760","Text":"and that is equal to mx double dot."},{"Start":"02:40.940 ","End":"02:45.454","Text":"Now what you\u0027ll notice is that we\u0027ve got into an equation"},{"Start":"02:45.454 ","End":"02:50.390","Text":"describing harmonic motion when there is a damping process."},{"Start":"02:50.390 ","End":"02:55.385","Text":"Where obviously the damping process is due to Stokes law,"},{"Start":"02:55.385 ","End":"03:00.570","Text":"due to the viscosity of the water applying a resistive force."},{"Start":"03:00.570 ","End":"03:07.065","Text":"We already know that 6Pi r and Eta are constants."},{"Start":"03:07.065 ","End":"03:12.455","Text":"What I\u0027m going to do, instead of to continue copying this all out,"},{"Start":"03:12.455 ","End":"03:15.739","Text":"I\u0027m just going to call 6Pi r Eta,"},{"Start":"03:15.739 ","End":"03:19.564","Text":"some constant, let\u0027s call it Lambda."},{"Start":"03:19.564 ","End":"03:23.070","Text":"Then let\u0027s carry on from here."},{"Start":"03:23.630 ","End":"03:26.805","Text":"Now, like in previous questions,"},{"Start":"03:26.805 ","End":"03:29.400","Text":"what we want to do is,"},{"Start":"03:29.400 ","End":"03:33.710","Text":"we want all of the variables or"},{"Start":"03:33.710 ","End":"03:37.955","Text":"expressions in this equation to move to one side of the equals sign."},{"Start":"03:37.955 ","End":"03:39.575","Text":"Let\u0027s do that."},{"Start":"03:39.575 ","End":"03:44.280","Text":"We\u0027ll have that mx double dot plus"},{"Start":"03:44.440 ","End":"03:52.875","Text":"our Lambda x dot plus our kx,"},{"Start":"03:52.875 ","End":"03:58.920","Text":"because we\u0027ve moved everything to the other side so they become positives is equal to 0."},{"Start":"03:58.920 ","End":"04:03.455","Text":"This is exactly the equation for dampened harmonic motion."},{"Start":"04:03.455 ","End":"04:07.355","Text":"Now what we deal with all of these questions is we divide everything by"},{"Start":"04:07.355 ","End":"04:12.245","Text":"m. What do we want to do is we want to get our x double dot with no coefficients,"},{"Start":"04:12.245 ","End":"04:15.190","Text":"or that it\u0027s coefficient is equal to 1."},{"Start":"04:15.190 ","End":"04:18.405","Text":"We\u0027ll divide both sides by m,"},{"Start":"04:18.405 ","End":"04:24.850","Text":"we\u0027ll get that x double dot plus lambda divided"},{"Start":"04:24.850 ","End":"04:34.530","Text":"m x dot plus k divided by mx = 0."},{"Start":"04:34.530 ","End":"04:38.180","Text":"Now what we\u0027re going to do is we\u0027re going to say that this coefficient"},{"Start":"04:38.180 ","End":"04:43.120","Text":"Lambda divided by m is going to be called Gamma,"},{"Start":"04:43.120 ","End":"04:45.440","Text":"and that this coefficient,"},{"Start":"04:45.440 ","End":"04:47.165","Text":"k divided by m,"},{"Start":"04:47.165 ","End":"04:55.860","Text":"as we know is equal to Omega 0^2."},{"Start":"04:55.860 ","End":"05:00.385","Text":"Now let\u0027s remember that our question is to find the frequency of oscillations,"},{"Start":"05:00.385 ","End":"05:04.810","Text":"which means that our system does oscillate."},{"Start":"05:04.810 ","End":"05:10.420","Text":"In which case we have to take the only case where we can have oscillations,"},{"Start":"05:10.420 ","End":"05:13.135","Text":"when we\u0027re located in some fluid,"},{"Start":"05:13.135 ","End":"05:18.590","Text":"is when we have under damping."},{"Start":"05:18.600 ","End":"05:28.720","Text":"For this situation, we have to be in the region of under damping,"},{"Start":"05:28.720 ","End":"05:32.949","Text":"because only then we will have oscillations."},{"Start":"05:32.949 ","End":"05:38.640","Text":"What we need in order to have under damping is that"},{"Start":"05:38.640 ","End":"05:45.880","Text":"our Gamma |divided by 2 must be smaller than Omega 0."},{"Start":"05:46.760 ","End":"05:50.520","Text":"Now we know that this is our condition,"},{"Start":"05:50.520 ","End":"05:52.110","Text":"this is what we need to fulfill."},{"Start":"05:52.110 ","End":"05:56.305","Text":"I\u0027m just going to scroll down a little bit to give us some extra space."},{"Start":"05:56.305 ","End":"06:01.925","Text":"Let\u0027s substitute in what we have for Gamma and Omega 0."},{"Start":"06:01.925 ","End":"06:05.450","Text":"Our Gamma is this over here,"},{"Start":"06:05.450 ","End":"06:13.520","Text":"Lambda divided by m. Gamma divided by 2 will be equal to Lambda divided by 2m,"},{"Start":"06:13.520 ","End":"06:16.390","Text":"and this is equal to."},{"Start":"06:16.390 ","End":"06:18.420","Text":"What is our Lambda?"},{"Start":"06:18.420 ","End":"06:25.320","Text":"We said that our Lambda = 6 Pi Eta,"},{"Start":"06:25.320 ","End":"06:27.045","Text":"so that\u0027s our Lambda."},{"Start":"06:27.045 ","End":"06:29.895","Text":"Then divided by 2m."},{"Start":"06:29.895 ","End":"06:32.330","Text":"Then we can cross that out,"},{"Start":"06:32.330 ","End":"06:34.220","Text":"cancel out with the 2s."},{"Start":"06:34.220 ","End":"06:35.960","Text":"This will be equal to"},{"Start":"06:35.960 ","End":"06:46.320","Text":"3Pi Eta divided by m. This has to be smaller than Omega_0."},{"Start":"06:46.320 ","End":"06:50.610","Text":"We know that Omega_0 ^2 is k divide by m,"},{"Start":"06:50.610 ","End":"06:52.200","Text":"but we just want Omega_0,"},{"Start":"06:52.200 ","End":"06:57.120","Text":"so it\u0027s going to be the square root of k divided by"},{"Start":"06:57.120 ","End":"07:05.325","Text":"m. This is our condition."},{"Start":"07:05.325 ","End":"07:12.085","Text":"Now what we want to do is we want to isolate out our r. We want to see"},{"Start":"07:12.085 ","End":"07:21.050","Text":"that our equation for I in this inequality is the same as this over here."},{"Start":"07:21.870 ","End":"07:27.990","Text":"Let\u0027s do that. We\u0027ll get that I is equal to."},{"Start":"07:27.990 ","End":"07:35.358","Text":"Once we multiply both sides by m will have inside the square root sign m^2,"},{"Start":"07:35.358 ","End":"07:38.170","Text":"and then one m from the numerator will"},{"Start":"07:38.170 ","End":"07:41.268","Text":"cancel out with this m over here in the denominator,"},{"Start":"07:41.268 ","End":"07:52.100","Text":"and we\u0027ll be left with square root of km divided by 3Pi Eta."},{"Start":"07:52.940 ","End":"07:58.025","Text":"This is the exact equation that we wanted. That\u0027s great."},{"Start":"07:58.025 ","End":"08:01.070","Text":"Which means that we are fulfilling"},{"Start":"08:01.070 ","End":"08:07.770","Text":"this need for our system to be in the under damped region."},{"Start":"08:08.810 ","End":"08:13.160","Text":"Now we know that when we\u0027re dealing with under damping,"},{"Start":"08:13.160 ","End":"08:16.490","Text":"we don\u0027t have to solve this because we already know"},{"Start":"08:16.490 ","End":"08:22.520","Text":"the equation for the frequency of oscillations when we\u0027re dealing with under damping."},{"Start":"08:22.520 ","End":"08:29.585","Text":"That is that the frequency of oscillation when dealing with under damping is equal to"},{"Start":"08:29.585 ","End":"08:36.650","Text":"the square root of"},{"Start":"08:36.650 ","End":"08:46.915","Text":"(Lambda divided by 2)^2 minus Omega_0^2."},{"Start":"08:46.915 ","End":"08:49.580","Text":"This is always the equation for"},{"Start":"08:49.580 ","End":"08:53.125","Text":"the frequency of oscillations when dealing with under damping."},{"Start":"08:53.125 ","End":"08:56.240","Text":"Now all we have to do is we have to substitute in"},{"Start":"08:56.240 ","End":"09:00.245","Text":"what we know for our Lambda and our Omega_0^2."},{"Start":"09:00.245 ","End":"09:05.940","Text":"Lambda divided by 2^2 is going to be equal to"},{"Start":"09:05.940 ","End":"09:14.015","Text":"(Gamma divide by by 2m)^2 minus Omega_0^2,"},{"Start":"09:14.015 ","End":"09:23.520","Text":"which is simply k divided by m. Then we take the square root of all of this."},{"Start":"09:24.410 ","End":"09:30.470","Text":"This is the answer to this question for finding the frequency of oscillations."},{"Start":"09:30.470 ","End":"09:37.760","Text":"Now of course, this Lambda over here represents this over here,"},{"Start":"09:37.760 ","End":"09:39.855","Text":"these constants over here."},{"Start":"09:39.855 ","End":"09:42.670","Text":"That\u0027s the end of this question."}],"ID":10780}],"Thumbnail":null,"ID":5366},{"Name":"Driven Harmonic Motion","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Explanation And The Resonance Effect","Duration":"18m 38s","ChapterTopicVideoID":9146,"CourseChapterTopicPlaylistID":5367,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.205","Text":"Hello. We\u0027re going to be speaking about driven harmonic oscillators and resonance."},{"Start":"00:05.205 ","End":"00:07.875","Text":"What exactly is resonance?"},{"Start":"00:07.875 ","End":"00:14.175","Text":"As per usual, we have a simple harmonic setup."},{"Start":"00:14.175 ","End":"00:22.485","Text":"We have our wall, we have our spring of constant k attached to our mass m. Now,"},{"Start":"00:22.485 ","End":"00:28.250","Text":"what\u0027s different from simple harmonic motion is that we have some driving force."},{"Start":"00:28.250 ","End":"00:35.400","Text":"Now, this is a force which is dependent on time and that equals to F_0,"},{"Start":"00:35.400 ","End":"00:37.885","Text":"in case of some constant,"},{"Start":"00:37.885 ","End":"00:43.505","Text":"multiplied by cosine of Omega t. Now of course,"},{"Start":"00:43.505 ","End":"00:45.050","Text":"sine can be also written here,"},{"Start":"00:45.050 ","End":"00:47.165","Text":"but we tend to write it with cosine."},{"Start":"00:47.165 ","End":"00:49.430","Text":"Now our constant over here,"},{"Start":"00:49.430 ","End":"00:52.910","Text":"this is when our cosine Omega t is equal to 1,"},{"Start":"00:52.910 ","End":"00:55.050","Text":"so at the maximum."},{"Start":"00:56.030 ","End":"01:02.180","Text":"The maximum driving force that we can reach is our F_0 over here."},{"Start":"01:02.180 ","End":"01:04.925","Text":"In order to solve these types of questions,"},{"Start":"01:04.925 ","End":"01:12.330","Text":"we solve it with our simple harmonic motion plus our driving force; adding the sine."},{"Start":"01:12.330 ","End":"01:17.255","Text":"We can either do it like that or we can also add in a damping force,"},{"Start":"01:17.255 ","End":"01:20.630","Text":"which as you remember from the previous lessons,"},{"Start":"01:20.630 ","End":"01:26.489","Text":"is going to be F is equal to negative Lambda multiplied"},{"Start":"01:26.489 ","End":"01:34.970","Text":"by v. Then what we do at the end is that we say that our Lambda over here is equal to 0."},{"Start":"01:34.970 ","End":"01:38.260","Text":"We can solve it both ways."},{"Start":"01:38.260 ","End":"01:40.440","Text":"The first thing I want to do,"},{"Start":"01:40.440 ","End":"01:43.530","Text":"is I want to get my equation of motion."},{"Start":"01:43.530 ","End":"01:48.290","Text":"I say that I have the sum of all of the forces is equal to."},{"Start":"01:48.290 ","End":"01:50.600","Text":"I have the force from the spring,"},{"Start":"01:50.600 ","End":"01:54.780","Text":"which is negative k(x-x_0)."},{"Start":"01:54.780 ","End":"01:57.360","Text":"Then I have my damping for us,"},{"Start":"01:57.360 ","End":"02:00.090","Text":"so negative Lambda v,"},{"Start":"02:00.090 ","End":"02:03.210","Text":"plus I have my driving force,"},{"Start":"02:03.210 ","End":"02:08.550","Text":"so plus F_0 cosine of"},{"Start":"02:08.550 ","End":"02:15.630","Text":"Omega t. All of this is equal to mx double dot."},{"Start":"02:15.630 ","End":"02:18.140","Text":"I said it exactly like this,"},{"Start":"02:18.140 ","End":"02:20.675","Text":"where I have my negative k here,"},{"Start":"02:20.675 ","End":"02:22.865","Text":"my negative Lambda v,"},{"Start":"02:22.865 ","End":"02:29.105","Text":"and here I have my positive F as a function of t. Positive driving force."},{"Start":"02:29.105 ","End":"02:30.545","Text":"If I wanted it to be negative,"},{"Start":"02:30.545 ","End":"02:35.720","Text":"then I would add in some phase difference inside the brackets over here."},{"Start":"02:35.720 ","End":"02:38.075","Text":"This is the equation that is good to know."},{"Start":"02:38.075 ","End":"02:43.490","Text":"Now we\u0027re going to write out the general solution to this,"},{"Start":"02:43.490 ","End":"02:47.245","Text":"which is x as a function of t. Now I\u0027m going to write out the solution."},{"Start":"02:47.245 ","End":"02:50.030","Text":"Again, just like when we were speaking about damping,"},{"Start":"02:50.030 ","End":"02:52.805","Text":"if you want to see how I get to the solution,"},{"Start":"02:52.805 ","End":"02:55.520","Text":"which you don\u0027t have to do most of the time,"},{"Start":"02:55.520 ","End":"02:58.405","Text":"then you can look at the next video."},{"Start":"02:58.405 ","End":"03:04.745","Text":"Our general solution is going to be our amplitude as a function of Omega,"},{"Start":"03:04.745 ","End":"03:13.000","Text":"multiplied by cosine of Omega t plus Phi,"},{"Start":"03:13.000 ","End":"03:17.525","Text":"and then plus our x_0 point of equilibrium."},{"Start":"03:17.525 ","End":"03:21.920","Text":"Now the biggest difference between this general solution and"},{"Start":"03:21.920 ","End":"03:24.200","Text":"the general solution that we got when dealing"},{"Start":"03:24.200 ","End":"03:26.870","Text":"with simple harmonic motion and also a damping,"},{"Start":"03:26.870 ","End":"03:28.556","Text":"is that over here,"},{"Start":"03:28.556 ","End":"03:34.455","Text":"the entire equation is independent of our initial conditions."},{"Start":"03:34.455 ","End":"03:37.625","Text":"Our A is already given."},{"Start":"03:37.625 ","End":"03:40.534","Text":"We\u0027re going to see what that is in a second."},{"Start":"03:40.534 ","End":"03:42.830","Text":"Our Omega over here,"},{"Start":"03:42.830 ","End":"03:44.443","Text":"let\u0027s take a look,"},{"Start":"03:44.443 ","End":"03:48.215","Text":"our Omega over here is the same Omega over here,"},{"Start":"03:48.215 ","End":"03:50.225","Text":"which is the same Omega over here."},{"Start":"03:50.225 ","End":"03:54.060","Text":"It\u0027s dependent on the driving force."},{"Start":"03:54.890 ","End":"04:02.270","Text":"Our Phi is also given and we\u0027re also in a second going to see how we find it."},{"Start":"04:02.270 ","End":"04:04.385","Text":"Also another note on our Omega,"},{"Start":"04:04.385 ","End":"04:07.970","Text":"it\u0027s independent of our k and our m. It\u0027s"},{"Start":"04:07.970 ","End":"04:12.410","Text":"just what we\u0027re given over here in our driving force."},{"Start":"04:12.410 ","End":"04:15.605","Text":"That means that once we have this,"},{"Start":"04:15.605 ","End":"04:18.390","Text":"then we already know the answer."},{"Start":"04:18.390 ","End":"04:22.805","Text":"We don\u0027t have to start doing initial conditions and working out and everything."},{"Start":"04:22.805 ","End":"04:26.839","Text":"This is because our driving force is forcing"},{"Start":"04:26.839 ","End":"04:32.820","Text":"our oscillating system to move at the frequency that it dictates."},{"Start":"04:32.820 ","End":"04:35.590","Text":"That\u0027s the whole point of this force."},{"Start":"04:35.590 ","End":"04:41.630","Text":"Notice this driving force is also deciding on the amplitude,"},{"Start":"04:41.630 ","End":"04:45.339","Text":"and is also deciding on the phase."},{"Start":"04:45.339 ","End":"04:51.230","Text":"This is what decides what the harmonic motion of our body is going to look like,"},{"Start":"04:51.230 ","End":"04:53.480","Text":"not the body itself."},{"Start":"04:53.480 ","End":"04:54.995","Text":"Now, a little side note,"},{"Start":"04:54.995 ","End":"04:57.590","Text":"which is important but not critical to know,"},{"Start":"04:57.590 ","End":"05:01.250","Text":"is that when we have our simple harmonic motion,"},{"Start":"05:01.250 ","End":"05:04.175","Text":"and then we add in a driving force,"},{"Start":"05:04.175 ","End":"05:08.580","Text":"our simple harmonic motion doesn\u0027t automatically straightaway in"},{"Start":"05:08.580 ","End":"05:12.664","Text":"the first millisecond move over to this type of motion,"},{"Start":"05:12.664 ","End":"05:14.600","Text":"exactly how the driving force dictates."},{"Start":"05:14.600 ","End":"05:17.525","Text":"It takes a certain amount of time,"},{"Start":"05:17.525 ","End":"05:21.175","Text":"be it 1 millisecond or a minute, whatever it might be."},{"Start":"05:21.175 ","End":"05:24.020","Text":"If you want to be super accurate,"},{"Start":"05:24.020 ","End":"05:27.455","Text":"so you have to write in this equation,"},{"Start":"05:27.455 ","End":"05:30.020","Text":"which is the harmonic motion of the system"},{"Start":"05:30.020 ","End":"05:33.410","Text":"once it\u0027s going according to the driving force,"},{"Start":"05:33.410 ","End":"05:37.035","Text":"plus how it started initially."},{"Start":"05:37.035 ","End":"05:42.810","Text":"Now, this extra added plus how it started initially,"},{"Start":"05:42.810 ","End":"05:46.330","Text":"I will speak about in a different lesson, not over here."},{"Start":"05:46.330 ","End":"05:48.380","Text":"But in the meantime, this is all you have to know."},{"Start":"05:48.380 ","End":"05:52.835","Text":"This is the most important because this is the motion that it will be"},{"Start":"05:52.835 ","End":"05:59.200","Text":"moving in throughout its movement aside from the initial beginning period."},{"Start":"05:59.200 ","End":"06:04.710","Text":"Now let\u0027s take a look at what our constant over here are."},{"Start":"06:04.710 ","End":"06:10.875","Text":"We have our A Omega, this over here."},{"Start":"06:10.875 ","End":"06:16.670","Text":"This is simply equal to our F_0 over here or over here,"},{"Start":"06:16.670 ","End":"06:20.685","Text":"divided by our m, this."},{"Start":"06:20.685 ","End":"06:25.535","Text":"Then all of this is going to be divided by the square root"},{"Start":"06:25.535 ","End":"06:31.415","Text":"of Omega_0^2 minus Omega^2."},{"Start":"06:31.415 ","End":"06:33.910","Text":"This Omega is this,"},{"Start":"06:33.910 ","End":"06:41.595","Text":"then all of this squared plus Gamma Omega^2,"},{"Start":"06:41.595 ","End":"06:45.585","Text":"and then the square root of this."},{"Start":"06:45.585 ","End":"06:50.220","Text":"This is our A as a function of Omega, this over here."},{"Start":"06:50.220 ","End":"06:52.920","Text":"Then our Phi, how we solve this,"},{"Start":"06:52.920 ","End":"06:56.220","Text":"is we say that our sine Phi,"},{"Start":"06:56.220 ","End":"06:59.360","Text":"and then we just do arc sine to both sides and we get our Phi."},{"Start":"06:59.360 ","End":"07:02.825","Text":"That\u0027s going to be equal to our Gamma Omega,"},{"Start":"07:02.825 ","End":"07:05.840","Text":"divided by the square root."},{"Start":"07:05.840 ","End":"07:10.560","Text":"The square root of Omega_0^2 minus Omega^2"},{"Start":"07:10.560 ","End":"07:18.975","Text":"plus our Gamma Omega^2."},{"Start":"07:18.975 ","End":"07:21.870","Text":"Then all of this, the square root."},{"Start":"07:21.870 ","End":"07:26.100","Text":"It\u0027s the same square root, the same denominator."},{"Start":"07:26.100 ","End":"07:31.060","Text":"All of this is if there is damping."},{"Start":"07:33.950 ","End":"07:38.130","Text":"Our negative Lambda v is happening,"},{"Start":"07:38.130 ","End":"07:39.980","Text":"so this is if there\u0027s damping."},{"Start":"07:39.980 ","End":"07:42.410","Text":"However, if there isn\u0027t damping,"},{"Start":"07:42.410 ","End":"07:46.755","Text":"so our A as a function of Omega,"},{"Start":"07:46.755 ","End":"07:48.375","Text":"is going to be equal to."},{"Start":"07:48.375 ","End":"07:53.885","Text":"All we have to do is, we have to substitute in over here that our Gamma is equal to 0."},{"Start":"07:53.885 ","End":"07:55.250","Text":"Again also here,"},{"Start":"07:55.250 ","End":"07:57.150","Text":"our Gamma and also over here,"},{"Start":"07:57.150 ","End":"07:58.815","Text":"our Gamma is equal to 0."},{"Start":"07:58.815 ","End":"08:01.470","Text":"What does that mean? We\u0027ll get that our A as a function of"},{"Start":"08:01.470 ","End":"08:04.905","Text":"Omega is equal to F_0 divided by"},{"Start":"08:04.905 ","End":"08:11.300","Text":"m. This over here will be equal to 0 because our Gamma is equal to 0."},{"Start":"08:11.300 ","End":"08:15.735","Text":"Then we have the square root of what\u0027s inside the brackets squared."},{"Start":"08:15.735 ","End":"08:17.370","Text":"Just what\u0027s inside the brackets,"},{"Start":"08:17.370 ","End":"08:22.425","Text":"so Omega_0^2, negative Omega^2."},{"Start":"08:22.425 ","End":"08:26.525","Text":"Then our sine Phi is going to be equal to,"},{"Start":"08:26.525 ","End":"08:28.513","Text":"because our Gamma over here is equal to 0,"},{"Start":"08:28.513 ","End":"08:30.905","Text":"so it\u0027s going to be equal to 0."},{"Start":"08:30.905 ","End":"08:33.830","Text":"Then that means that our Phi,"},{"Start":"08:33.830 ","End":"08:36.245","Text":"if we do arc sine to both sides,"},{"Start":"08:36.245 ","End":"08:40.185","Text":"we\u0027ll see that our Phi is equal to 0."},{"Start":"08:40.185 ","End":"08:44.370","Text":"This is when we have no damping,"},{"Start":"08:44.370 ","End":"08:48.700","Text":"when our Lambda over here is equal to 0."},{"Start":"08:48.700 ","End":"08:51.665","Text":"I\u0027ve squared all of this in red,"},{"Start":"08:51.665 ","End":"08:53.764","Text":"these 3 are boxes,"},{"Start":"08:53.764 ","End":"08:59.540","Text":"and please copy them out and try to remember this or write this in your notes."},{"Start":"09:00.140 ","End":"09:06.335","Text":"Now let\u0027s take a look at this over here when we do have damping."},{"Start":"09:06.335 ","End":"09:11.075","Text":"Let\u0027s try and draw out our graph of what our A,"},{"Start":"09:11.075 ","End":"09:12.488","Text":"as a function of Omega,"},{"Start":"09:12.488 ","End":"09:16.820","Text":"what our amplitude as a function of Omega is going to look like."},{"Start":"09:16.820 ","End":"09:20.315","Text":"Now, remember it\u0027s independent of our initial conditions."},{"Start":"09:20.315 ","End":"09:23.090","Text":"Let\u0027s draw this out as a graph."},{"Start":"09:23.090 ","End":"09:27.080","Text":"Over here, we have our A as a function of Omega,"},{"Start":"09:27.080 ","End":"09:29.095","Text":"and over here,"},{"Start":"09:29.095 ","End":"09:31.815","Text":"we have our Omega."},{"Start":"09:31.815 ","End":"09:37.550","Text":"What we can see is that our maximum amplitude that we\u0027re going to get,"},{"Start":"09:37.550 ","End":"09:40.020","Text":"our maximum A as a function of Omega,"},{"Start":"09:40.020 ","End":"09:44.990","Text":"is going to be when our denominator is as small as possible."},{"Start":"09:44.990 ","End":"09:47.105","Text":"When does this happen?"},{"Start":"09:47.105 ","End":"09:51.985","Text":"As a rule, it\u0027s when our Omega_0 is equal to our Omega."},{"Start":"09:51.985 ","End":"09:57.730","Text":"Then that means that this expression over here is going to be equal to 0,"},{"Start":"09:57.730 ","End":"10:01.310","Text":"and then our denominator will just consist of Gamma Omega,"},{"Start":"10:01.310 ","End":"10:03.605","Text":"which is the smallest it can be."},{"Start":"10:03.605 ","End":"10:05.945","Text":"Then if we draw that out,"},{"Start":"10:05.945 ","End":"10:09.680","Text":"we\u0027ll see that our amplitude increases to"},{"Start":"10:09.680 ","End":"10:15.260","Text":"some peak and then decreases back with this peak,"},{"Start":"10:15.260 ","End":"10:20.325","Text":"so our maximum point is at Omega_0."},{"Start":"10:20.325 ","End":"10:24.485","Text":"Now let\u0027s take a look. We also have our Gamma over here."},{"Start":"10:24.485 ","End":"10:27.980","Text":"Our Gamma is this over here."},{"Start":"10:27.980 ","End":"10:30.275","Text":"It\u0027s to do with our damping for us."},{"Start":"10:30.275 ","End":"10:37.970","Text":"Let\u0027s take a look. If I take a larger value for my Gamma over here,"},{"Start":"10:37.970 ","End":"10:41.370","Text":"then my denominator is going to increase."},{"Start":"10:41.650 ","End":"10:48.280","Text":"A larger Gamma will result in a lower peak."},{"Start":"10:48.280 ","End":"10:51.370","Text":"What I\u0027ve done over here is that this Lambda"},{"Start":"10:51.370 ","End":"10:56.270","Text":"represents some arbitrary Lambda over here, which goes over here."},{"Start":"10:56.270 ","End":"11:00.880","Text":"My lambda wave represents if I take a bigger,"},{"Start":"11:00.880 ","End":"11:03.465","Text":"if I change the sizes of my Lambdas."},{"Start":"11:03.465 ","End":"11:06.565","Text":"If I go larger, my peak will reduce."},{"Start":"11:06.565 ","End":"11:13.825","Text":"My peak is still going to be at Omega_0 on my axis for Omega,"},{"Start":"11:13.825 ","End":"11:18.689","Text":"however, my amplitude is going to be lower."},{"Start":"11:18.689 ","End":"11:23.475","Text":"Now, let\u0027s speak about if I take"},{"Start":"11:23.475 ","End":"11:30.465","Text":"a Lambda which is smaller than my arbitrary Lambda."},{"Start":"11:30.465 ","End":"11:35.625","Text":"So if I go smaller then my peak is still going to be at my Omega 0."},{"Start":"11:35.625 ","End":"11:39.735","Text":"However, this time it\u0027s going to be higher."},{"Start":"11:39.735 ","End":"11:42.720","Text":"My amplitude is going to be higher."},{"Start":"11:42.720 ","End":"11:44.310","Text":"This is still at Omega 0."},{"Start":"11:44.310 ","End":"11:46.230","Text":"The peak is here."},{"Start":"11:46.230 ","End":"11:50.130","Text":"Now my amplitude is higher and then,"},{"Start":"11:50.130 ","End":"11:55.662","Text":"let\u0027s say I\u0027m going to take my Lambda and I\u0027m going to send it off to 0,"},{"Start":"11:55.662 ","End":"11:58.050","Text":"so make my Lambda as small as possible."},{"Start":"11:58.050 ","End":"12:01.710","Text":"What\u0027s going to happen is that my amplitude is going to explode,"},{"Start":"12:01.710 ","End":"12:06.930","Text":"which means that it\u0027s going to go off to infinity."},{"Start":"12:06.930 ","End":"12:11.985","Text":"We\u0027re going to have something like this."},{"Start":"12:11.985 ","End":"12:18.075","Text":"So its limit is infinity on the amplitude axis."},{"Start":"12:18.075 ","End":"12:20.805","Text":"Before I explain what this is,"},{"Start":"12:20.805 ","End":"12:23.430","Text":"I forgot to mention what my Omega 0 is."},{"Start":"12:23.430 ","End":"12:28.710","Text":"My Omega 0 is my resonant frequency,"},{"Start":"12:28.710 ","End":"12:33.195","Text":"and so my Omega 0 squared is equal to just like before,"},{"Start":"12:33.195 ","End":"12:43.650","Text":"which is k divided by m. This is my frequency without my driving force and of"},{"Start":"12:43.650 ","End":"12:48.870","Text":"course my Gamma is equal"},{"Start":"12:48.870 ","End":"12:55.545","Text":"to Lambda divided by m. I forgot to say that."},{"Start":"12:55.545 ","End":"13:02.988","Text":"Now what we can see is that when my omega is equal to my Omega 0,"},{"Start":"13:02.988 ","End":"13:06.390","Text":"so that means the frequency of"},{"Start":"13:06.390 ","End":"13:11.730","Text":"my driving force is similar to that of my resonant frequency."},{"Start":"13:11.730 ","End":"13:14.175","Text":"So as they become equal,"},{"Start":"13:14.175 ","End":"13:16.005","Text":"then I can get resonance,"},{"Start":"13:16.005 ","End":"13:18.419","Text":"which means that my amplitude can explode."},{"Start":"13:18.419 ","End":"13:21.150","Text":"It can go to infinity."},{"Start":"13:21.150 ","End":"13:23.115","Text":"Now what I\u0027m going to do is I\u0027m going to give"},{"Start":"13:23.115 ","End":"13:29.130","Text":"a few examples for what this means with a driving force and everything."},{"Start":"13:29.130 ","End":"13:32.445","Text":"You get a little bit more of an idea what this means."},{"Start":"13:32.445 ","End":"13:38.655","Text":"Let\u0027s say that we have some swing with a little boy sitting on it."},{"Start":"13:38.655 ","End":"13:43.650","Text":"He\u0027s sitting on the swing and let\u0027s say he\u0027s swinging so we can work it out."},{"Start":"13:43.650 ","End":"13:46.695","Text":"It\u0027s like simple harmonic motion of a normal pendulum."},{"Start":"13:46.695 ","End":"13:50.160","Text":"Its frequency is going to be the square root of g"},{"Start":"13:50.160 ","End":"13:54.135","Text":"divided by l. He\u0027s swinging back and forth, back and forth."},{"Start":"13:54.135 ","End":"14:02.550","Text":"If however his mother or father suddenly comes and starts pushing the swing,"},{"Start":"14:02.550 ","End":"14:05.385","Text":"then that is the driving force."},{"Start":"14:05.385 ","End":"14:10.500","Text":"In that case, the swing is going to swing with a frequency and"},{"Start":"14:10.500 ","End":"14:16.660","Text":"amplitude as defined by the father who\u0027s pushing the swing."},{"Start":"14:16.880 ","End":"14:21.810","Text":"Now, what happens is that if the father pushes"},{"Start":"14:21.810 ","End":"14:27.255","Text":"the swing at the exact same frequency as the resonant frequency."},{"Start":"14:27.255 ","End":"14:30.330","Text":"At the exact same frequency of the swing acting as"},{"Start":"14:30.330 ","End":"14:33.960","Text":"a pendulum itself so the square root of g divided by l,"},{"Start":"14:33.960 ","End":"14:42.299","Text":"then that means that slowly the swing is going to go to a higher amplitude,"},{"Start":"14:42.299 ","End":"14:44.055","Text":"higher and higher and higher and higher,"},{"Start":"14:44.055 ","End":"14:47.910","Text":"until eventually it gets to infinity."},{"Start":"14:47.910 ","End":"14:51.660","Text":"We\u0027ve all seen that when we were little or maybe even now still,"},{"Start":"14:51.660 ","End":"14:55.020","Text":"when we sit on a swing and our friends swings us and if they"},{"Start":"14:55.020 ","End":"14:58.665","Text":"push the swing at the exact right moment,"},{"Start":"14:58.665 ","End":"15:03.420","Text":"you notice that you go a lot higher than what you were in last time."},{"Start":"15:03.420 ","End":"15:06.465","Text":"You can get a bit of a rush."},{"Start":"15:06.465 ","End":"15:11.070","Text":"That\u0027s because your friend or whoever it is pushed you at the resonant frequency,"},{"Start":"15:11.070 ","End":"15:14.144","Text":"at the same frequency of the swing itself."},{"Start":"15:14.144 ","End":"15:17.340","Text":"If however, the father would push the swing when it"},{"Start":"15:17.340 ","End":"15:20.790","Text":"didn\u0027t exactly return and complete a whole period,"},{"Start":"15:20.790 ","End":"15:27.540","Text":"for instance, then the swing won\u0027t go to such a high amplitude."},{"Start":"15:27.540 ","End":"15:29.460","Text":"Now, of course,"},{"Start":"15:29.460 ","End":"15:31.515","Text":"in order to get your swing to go really high,"},{"Start":"15:31.515 ","End":"15:34.500","Text":"it doesn\u0027t depend on your F_0 over here."},{"Start":"15:34.500 ","End":"15:41.655","Text":"It just depends on this Omega that you\u0027re being pushed at the resonant frequency."},{"Start":"15:41.655 ","End":"15:48.990","Text":"Now, this is called resonance and resonance occurs in almost everything that you know."},{"Start":"15:48.990 ","End":"15:52.980","Text":"It\u0027s very important to know this, to understand it."},{"Start":"15:52.980 ","End":"15:55.800","Text":"Because it will come up in a test most probably and"},{"Start":"15:55.800 ","End":"15:58.305","Text":"it\u0027s very useful to know when it comes up everywhere"},{"Start":"15:58.305 ","End":"16:04.050","Text":"and then subjects that you\u0027ll do later on in your degree."},{"Start":"16:04.050 ","End":"16:08.145","Text":"It\u0027s very important in industry, for example."},{"Start":"16:08.145 ","End":"16:11.775","Text":"You can look it up on Google or YouTube."},{"Start":"16:11.775 ","End":"16:16.230","Text":"There\u0027s a video of the bridge in Canada which you might have heard of."},{"Start":"16:16.230 ","End":"16:22.215","Text":"It was built and 1 day the wind was blowing at the exact frequency,"},{"Start":"16:22.215 ","End":"16:27.750","Text":"the resonant frequency of the bridge so of the material that the bridge was made out of,"},{"Start":"16:27.750 ","End":"16:30.210","Text":"which cause the bridge you can really see it"},{"Start":"16:30.210 ","End":"16:33.735","Text":"wobbling as if it isn\u0027t made out of solid material,"},{"Start":"16:33.735 ","End":"16:35.730","Text":"and eventually it collapsed."},{"Start":"16:35.730 ","End":"16:41.835","Text":"In industry and engineering this is very important to know and take into account."},{"Start":"16:41.835 ","End":"16:46.050","Text":"The bridge is 1 example where it\u0027s useful and another example"},{"Start":"16:46.050 ","End":"16:50.040","Text":"that you probably use every single day is the microwave."},{"Start":"16:50.040 ","End":"16:53.953","Text":"If you have your microwave over here,"},{"Start":"16:53.953 ","End":"17:02.430","Text":"you have some box and inside you have your little plates,"},{"Start":"17:02.430 ","End":"17:04.620","Text":"with your food inside."},{"Start":"17:04.620 ","End":"17:10.605","Text":"What happens, the microwave is sending electromagnetic waves onto the field."},{"Start":"17:10.605 ","End":"17:13.050","Text":"Now what happens is that the frequency of"},{"Start":"17:13.050 ","End":"17:20.590","Text":"the electromagnetic waves is the exact same frequency of the water inside your field."},{"Start":"17:20.900 ","End":"17:24.570","Text":"Because they\u0027re moving at the same frequency,"},{"Start":"17:24.570 ","End":"17:27.120","Text":"that means that the water molecules inside"},{"Start":"17:27.120 ","End":"17:30.675","Text":"your field are going to move up and down a lot faster."},{"Start":"17:30.675 ","End":"17:37.575","Text":"They\u0027re going to vibrate much faster and then your food heats up because of the friction"},{"Start":"17:37.575 ","End":"17:46.125","Text":"occurring due to the water molecules vibrating so quickly and at such large amplitudes."},{"Start":"17:46.125 ","End":"17:48.090","Text":"This friction heats up your food,"},{"Start":"17:48.090 ","End":"17:52.620","Text":"and that\u0027s how you can heat up food using this idea of resonance."},{"Start":"17:52.620 ","End":"17:58.245","Text":"Of course, the example that you must have seen in your cartoons"},{"Start":"17:58.245 ","End":"18:03.945","Text":"is that you have the opera singer and she\u0027s singing at very high note,"},{"Start":"18:03.945 ","End":"18:07.680","Text":"and there\u0027s a glass and the room and it shatters."},{"Start":"18:07.680 ","End":"18:13.860","Text":"That\u0027s because supposedly her voice reaches such a high frequency,"},{"Start":"18:13.860 ","End":"18:17.190","Text":"a high tone, high note that it\u0027s at"},{"Start":"18:17.190 ","End":"18:21.210","Text":"the resonant frequency of the glass and the glass smashes."},{"Start":"18:21.210 ","End":"18:24.209","Text":"Resonance is really something very interesting."},{"Start":"18:24.209 ","End":"18:28.125","Text":"It\u0027s very relevant to everything"},{"Start":"18:28.125 ","End":"18:33.585","Text":"in your everyday life so it\u0027s very good to understand this."},{"Start":"18:33.585 ","End":"18:35.385","Text":"That\u0027s the end of this lesson."},{"Start":"18:35.385 ","End":"18:39.220","Text":"Let\u0027s go on to solving some questions."}],"ID":9416},{"Watched":false,"Name":"Complete Solution To The Differential Equation of Driven Harmonic Motion","Duration":"14m 12s","ChapterTopicVideoID":9147,"CourseChapterTopicPlaylistID":5367,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.690","Text":"Hello. In the previous video we were dealing with"},{"Start":"00:03.690 ","End":"00:07.230","Text":"our equation for our harmonic motion which is"},{"Start":"00:07.230 ","End":"00:15.610","Text":"dealing with a case where there\u0027s both potentially damping and also a driving force."},{"Start":"00:16.370 ","End":"00:19.350","Text":"Then we got to a few solutions,"},{"Start":"00:19.350 ","End":"00:24.060","Text":"and I said that in this video I\u0027ll explain to you how we got to these solutions."},{"Start":"00:24.060 ","End":"00:25.812","Text":"You don\u0027t have to watch this video,"},{"Start":"00:25.812 ","End":"00:29.160","Text":"this is just for some extra understanding."},{"Start":"00:29.160 ","End":"00:35.245","Text":"In your maths lessons you should be learning how to solve differential equations,"},{"Start":"00:35.245 ","End":"00:38.000","Text":"but in the meantime if you haven\u0027t done that yet then you"},{"Start":"00:38.000 ","End":"00:41.255","Text":"still won\u0027t need to understand it at this stage."},{"Start":"00:41.255 ","End":"00:46.490","Text":"What we have to do is we have to guess a solution of this type."},{"Start":"00:46.490 ","End":"00:50.360","Text":"We\u0027ll have x as a function of time which is going to be"},{"Start":"00:50.360 ","End":"00:57.060","Text":"equal to A cosine of some Omega;"},{"Start":"00:57.060 ","End":"01:04.450","Text":"so Omega-tag, multiplied by t plus our Phi."},{"Start":"01:04.630 ","End":"01:10.340","Text":"Then we add in our point of equilibrium, x_0 over here."},{"Start":"01:10.340 ","End":"01:13.160","Text":"Now what we\u0027re going to do is we\u0027re going to substitute in"},{"Start":"01:13.160 ","End":"01:16.369","Text":"this solution into this equation."},{"Start":"01:16.369 ","End":"01:22.640","Text":"We\u0027re going to get negative k and then we substitute in our x."},{"Start":"01:22.640 ","End":"01:27.380","Text":"We\u0027re going to have multiplied by A cosine of"},{"Start":"01:27.380 ","End":"01:33.405","Text":"Omega-tag t plus Phi plus x_0 minus x_0,"},{"Start":"01:33.405 ","End":"01:40.740","Text":"so our x_0s cancel out and then minus Lambda multiplied by v. Our v is our x-dot,"},{"Start":"01:40.740 ","End":"01:42.890","Text":"it\u0027s our first derivative of this."},{"Start":"01:42.890 ","End":"01:45.140","Text":"We\u0027re going to derive this."},{"Start":"01:45.140 ","End":"01:49.418","Text":"We\u0027re going to have Omega-tag"},{"Start":"01:49.418 ","End":"01:57.500","Text":"A sine of Omega-tag t plus Phi,"},{"Start":"01:57.500 ","End":"02:03.590","Text":"and when we take the first derivative our x_0 crosses out and then plus."},{"Start":"02:03.590 ","End":"02:05.030","Text":"Here we have no x,"},{"Start":"02:05.030 ","End":"02:12.575","Text":"so F_0 cosine of our Omega t and this is going to be equal to;"},{"Start":"02:12.575 ","End":"02:14.810","Text":"let\u0027s just scroll a little bit,"},{"Start":"02:14.810 ","End":"02:19.450","Text":"so it\u0027s going to be equal to m and then x-double dot."},{"Start":"02:19.450 ","End":"02:21.380","Text":"We\u0027re going to take the second derivative,"},{"Start":"02:21.380 ","End":"02:23.125","Text":"so it\u0027s going to be"},{"Start":"02:23.125 ","End":"02:29.850","Text":"negative Omega-tag^2 cosine of"},{"Start":"02:29.850 ","End":"02:34.300","Text":"Omega-tag t plus Phi."},{"Start":"02:34.760 ","End":"02:42.360","Text":"This is really complicated and it\u0027s going to get slightly more complicated now."},{"Start":"02:42.520 ","End":"02:51.270","Text":"Something that you should remember is that cosine of Alpha plus Beta;"},{"Start":"02:51.270 ","End":"02:53.990","Text":"so these are very useful trig identities for solving"},{"Start":"02:53.990 ","End":"02:57.274","Text":"these types of complicated situations,"},{"Start":"02:57.274 ","End":"03:02.810","Text":"is equal to cosine of Alpha cosine of"},{"Start":"03:02.810 ","End":"03:10.369","Text":"Beta minus sine of Alpha sine of Beta,"},{"Start":"03:10.369 ","End":"03:20.000","Text":"and sine of Alpha plus Beta is equal to sine of"},{"Start":"03:20.000 ","End":"03:24.095","Text":"Alpha cosine of Beta"},{"Start":"03:24.095 ","End":"03:32.260","Text":"plus sine of Beta cosine of Alpha."},{"Start":"03:32.270 ","End":"03:35.855","Text":"Using now these trig identities,"},{"Start":"03:35.855 ","End":"03:41.895","Text":"we\u0027re going to separate this expression out. Let\u0027s do that."},{"Start":"03:41.895 ","End":"03:47.070","Text":"We\u0027re going to have negative kA and then we have cosine Alpha plus Beta,"},{"Start":"03:47.070 ","End":"03:56.535","Text":"so it\u0027s going to be cosine of Omega-tag t cosine of"},{"Start":"03:56.535 ","End":"04:02.985","Text":"Phi plus kA sine"},{"Start":"04:02.985 ","End":"04:09.955","Text":"of Omega-tag t sine of Phi."},{"Start":"04:09.955 ","End":"04:20.610","Text":"Then we\u0027re going to have negative Lambda Omega-tag A sine"},{"Start":"04:20.610 ","End":"04:29.260","Text":"of Omega-tag t cosine of Phi"},{"Start":"04:30.050 ","End":"04:36.630","Text":"plus sine of Phi cosine"},{"Start":"04:36.630 ","End":"04:43.834","Text":"of Omega-tag t and close this bracket,"},{"Start":"04:43.834 ","End":"04:48.450","Text":"and we\u0027ll multiply this all by this."},{"Start":"04:49.180 ","End":"04:52.640","Text":"Try and see how I did this."},{"Start":"04:52.640 ","End":"04:58.745","Text":"I\u0027m using these identities every time I see a cosine or a sine I\u0027m substituting in this."},{"Start":"04:58.745 ","End":"05:07.860","Text":"Then we\u0027re going to have plus our F_0 cosine of Omega t,"},{"Start":"05:07.860 ","End":"05:17.720","Text":"and then that\u0027s going to be equal to negative m Omega-tag^2A and then we\u0027ll"},{"Start":"05:17.720 ","End":"05:27.260","Text":"multiply this all by cosine of Omega-tag t cosine of Phi"},{"Start":"05:27.260 ","End":"05:37.830","Text":"minus sine of Omega-tag t sine of Phi."},{"Start":"05:37.830 ","End":"05:42.160","Text":"We have over here this really long expression."},{"Start":"05:42.160 ","End":"05:46.550","Text":"Now what we have to do is we have to compare all of"},{"Start":"05:46.550 ","End":"05:48.260","Text":"the coefficients of our cosine"},{"Start":"05:48.260 ","End":"05:51.944","Text":"Omega t with all of the coefficients of our sine of Omega t,"},{"Start":"05:51.944 ","End":"05:54.830","Text":"and what we want to get is that everything cancels"},{"Start":"05:54.830 ","End":"06:00.290","Text":"out because this side of the equation is equal to this side of the equation."},{"Start":"06:00.290 ","End":"06:03.395","Text":"Everything is meant to cancel out,"},{"Start":"06:03.395 ","End":"06:08.010","Text":"and also more than that everything is meant to cancel out at every time."},{"Start":"06:09.350 ","End":"06:11.525","Text":"How can that happen?"},{"Start":"06:11.525 ","End":"06:16.745","Text":"The only way that this can happen is if we say that our Omega-tag is equal to"},{"Start":"06:16.745 ","End":"06:22.275","Text":"our Omega because otherwise this term won\u0027t cancel out,"},{"Start":"06:22.275 ","End":"06:24.875","Text":"and then we\u0027re going to have a bit of a problem."},{"Start":"06:24.875 ","End":"06:28.580","Text":"We say that our Omega-tag is equal to our Omega so that everything can"},{"Start":"06:28.580 ","End":"06:32.640","Text":"cancel out at every single time,"},{"Start":"06:32.640 ","End":"06:34.910","Text":"and then we will see that both sides of the equation are"},{"Start":"06:34.910 ","End":"06:38.240","Text":"equal which means that this really is the solution."},{"Start":"06:38.240 ","End":"06:42.060","Text":"Let\u0027s see how we do this. Now we\u0027ve"},{"Start":"06:42.060 ","End":"06:46.065","Text":"said that the Omega of the force is equal to the Omega of the body."},{"Start":"06:46.065 ","End":"06:49.955","Text":"Now what we\u0027re going to do is we\u0027re going to compare coefficients."},{"Start":"06:49.955 ","End":"06:55.160","Text":"I\u0027m going to start by writing an equation with only the coefficients of our cosine of"},{"Start":"06:55.160 ","End":"07:05.905","Text":"our Omega t. It\u0027s going to be that with this."},{"Start":"07:05.905 ","End":"07:12.410","Text":"Then our next is going to be this with this,"},{"Start":"07:12.870 ","End":"07:19.315","Text":"and our F_0, and then what\u0027s over here and this."},{"Start":"07:19.315 ","End":"07:22.615","Text":"That\u0027s going to be our first equation."},{"Start":"07:22.615 ","End":"07:27.280","Text":"This is the equation, negative kA cosine of Phi is this."},{"Start":"07:27.280 ","End":"07:33.310","Text":"Negative Lambda Omega A sine of Phi is this,"},{"Start":"07:33.310 ","End":"07:35.230","Text":"plus our F_0 was this,"},{"Start":"07:35.230 ","End":"07:41.620","Text":"and then negative m Omega^2 A multiplied by cosine of Phi."},{"Start":"07:41.620 ","End":"07:44.890","Text":"Now we\u0027re going to do the exact same thing,"},{"Start":"07:44.890 ","End":"07:52.240","Text":"but with the coefficients of A sine Omega t. So we\u0027re going to have kA sine Phi."},{"Start":"07:52.240 ","End":"07:59.485","Text":"Then we\u0027re going to take this as well with our sine Omega t and this over here."},{"Start":"07:59.485 ","End":"08:03.100","Text":"Here there\u0027s no sine Omega t. Here we have our sine Omega t,"},{"Start":"08:03.100 ","End":"08:08.200","Text":"so we\u0027re also going to take this again with this,"},{"Start":"08:08.200 ","End":"08:10.880","Text":"and this negative as well."},{"Start":"08:11.580 ","End":"08:17.290","Text":"Now we can see over here that my make A sine Phi over here,"},{"Start":"08:17.290 ","End":"08:22.960","Text":"my negative Lambda Omega A cosine of Phi over here,"},{"Start":"08:22.960 ","End":"08:30.430","Text":"and that is going to be equal to my m Omega^2 A multiplied by sine Phi."},{"Start":"08:30.430 ","End":"08:36.595","Text":"Now what I have is I have 2 equations with 2 unknowns,"},{"Start":"08:36.595 ","End":"08:38.350","Text":"my A and my Phi."},{"Start":"08:38.350 ","End":"08:41.410","Text":"Now all I have to do is I have to work out"},{"Start":"08:41.410 ","End":"08:45.350","Text":"these equations in order to find out my variables."},{"Start":"08:45.870 ","End":"08:48.400","Text":"Let\u0027s begin simplifying this."},{"Start":"08:48.400 ","End":"08:50.905","Text":"We\u0027re going to take the second equation first,"},{"Start":"08:50.905 ","End":"08:55.690","Text":"and I\u0027m going to put all of my sine Phi on one side and my cosine Phi on the other side."},{"Start":"08:55.690 ","End":"09:00.145","Text":"What I\u0027m going to have is I can divide everything by A,"},{"Start":"09:00.145 ","End":"09:03.055","Text":"so cross this, cross this, and cross this."},{"Start":"09:03.055 ","End":"09:09.340","Text":"Then I\u0027ll have k, and then here I have a sine Phi,"},{"Start":"09:09.340 ","End":"09:17.860","Text":"so negative m Omega^2 multiplied by sine Phi is equal to."},{"Start":"09:17.860 ","End":"09:21.040","Text":"Then I\u0027m going to move this expression to the other side of the equal sign,"},{"Start":"09:21.040 ","End":"09:28.720","Text":"so this becomes a positive Lambda Omega cosine of Phi."},{"Start":"09:28.720 ","End":"09:34.795","Text":"Now an easy trig identity in order to turn my cosine Phi into my sine."},{"Start":"09:34.795 ","End":"09:38.253","Text":"I can say that this is equal to Lambda Omega A,"},{"Start":"09:38.253 ","End":"09:39.655","Text":"and then instead of cosine Phi,"},{"Start":"09:39.655 ","End":"09:47.455","Text":"I have 1 minus sine^2 Phi^1/2."},{"Start":"09:47.455 ","End":"09:53.150","Text":"Cosine Phi is equal to the square root of 1 minus sine^2 Phi."},{"Start":"09:53.520 ","End":"09:59.769","Text":"Now what I can do is I want to get rid of this square root sign,"},{"Start":"09:59.769 ","End":"10:02.240","Text":"so I\u0027m going to raise everything."},{"Start":"10:02.240 ","End":"10:04.060","Text":"So I\u0027m going to square everything,"},{"Start":"10:04.060 ","End":"10:06.265","Text":"raise everything to the power of 2."},{"Start":"10:06.265 ","End":"10:08.215","Text":"So this is going to be^2,"},{"Start":"10:08.215 ","End":"10:10.270","Text":"this is going to be^2,"},{"Start":"10:10.270 ","End":"10:12.400","Text":"this is squared,"},{"Start":"10:12.400 ","End":"10:16.600","Text":"and then this is squared,"},{"Start":"10:16.600 ","End":"10:19.915","Text":"and I can get rid of my 1/2 over here."},{"Start":"10:19.915 ","End":"10:25.735","Text":"Now what I\u0027m going to do is I want to get my sine Phi."},{"Start":"10:25.735 ","End":"10:28.060","Text":"I want to isolate out my sine Phi."},{"Start":"10:28.060 ","End":"10:30.580","Text":"First, I\u0027m going to isolate out my sine^2 Phi,"},{"Start":"10:30.580 ","End":"10:33.100","Text":"and I\u0027m going to just not look at"},{"Start":"10:33.100 ","End":"10:37.510","Text":"this middle over here because that has cosine Phi and I don\u0027t want that."},{"Start":"10:37.510 ","End":"10:40.015","Text":"Let\u0027s isolate out my sine Phi."},{"Start":"10:40.015 ","End":"10:50.755","Text":"I\u0027ll get that my sine^2 Phi is going to be equal to Lambda Omega^2 divided by"},{"Start":"10:50.755 ","End":"10:53.200","Text":"(K minus m"},{"Start":"10:53.200 ","End":"11:03.020","Text":"Omega^2)^2^2 plus Lambda Omega^2."},{"Start":"11:04.260 ","End":"11:08.380","Text":"Now I want the square root because I want just my sine."},{"Start":"11:08.380 ","End":"11:10.600","Text":"I\u0027m going to square root everything."},{"Start":"11:10.600 ","End":"11:13.240","Text":"I\u0027m going to lose this square, this square,"},{"Start":"11:13.240 ","End":"11:17.270","Text":"and here I\u0027m going to get a square root."},{"Start":"11:18.090 ","End":"11:23.335","Text":"Now I\u0027m going to find my relationship between my sine Phi and my cosine."},{"Start":"11:23.335 ","End":"11:27.535","Text":"Now I can look at this section over here,"},{"Start":"11:27.535 ","End":"11:32.590","Text":"and we can see that the relationship between my sine Phi and cosine Phi,"},{"Start":"11:32.590 ","End":"11:36.895","Text":"so I\u0027ll isolate out my cosine of Phi from over here,"},{"Start":"11:36.895 ","End":"11:40.855","Text":"is going to be equal to k minus"},{"Start":"11:40.855 ","End":"11:49.458","Text":"m Omega^2 multiplied by sine Phi,"},{"Start":"11:49.458 ","End":"11:54.505","Text":"and this is going to be divided by Lambda Omega."},{"Start":"11:54.505 ","End":"12:01.040","Text":"Now I can substitute this in to this equation over here."},{"Start":"12:01.500 ","End":"12:07.405","Text":"Then I will get that my cosine Phi is therefore"},{"Start":"12:07.405 ","End":"12:14.360","Text":"equal to k minus m Omega^2,"},{"Start":"12:17.670 ","End":"12:28.060","Text":"and then it\u0027s going to be divided by k minus m Omega^2,"},{"Start":"12:28.060 ","End":"12:30.640","Text":"and this is squared just like here,"},{"Start":"12:30.640 ","End":"12:35.275","Text":"plus my Lambda Omega^2."},{"Start":"12:35.275 ","End":"12:38.200","Text":"The same denominator, exactly,"},{"Start":"12:38.200 ","End":"12:40.675","Text":"just the numerator is a bit different."},{"Start":"12:40.675 ","End":"12:43.690","Text":"I use this over here to substitute in here to"},{"Start":"12:43.690 ","End":"12:47.260","Text":"get this expression here with the same numerator."},{"Start":"12:47.260 ","End":"12:51.610","Text":"Then if I substitute this cosine PHI and"},{"Start":"12:51.610 ","End":"12:56.200","Text":"sine Phi into this equation over here at the top,"},{"Start":"12:56.200 ","End":"12:57.940","Text":"the equation that I\u0027m working with,"},{"Start":"12:57.940 ","End":"13:06.580","Text":"so I\u0027ll get therefore that my A is going to be equal to my F_0 divided by"},{"Start":"13:06.580 ","End":"13:11.245","Text":"my m divided by the square root"},{"Start":"13:11.245 ","End":"13:17.754","Text":"of Omega 0^2 minus Omega^2,"},{"Start":"13:17.754 ","End":"13:23.170","Text":"all of this squared, plus Omega^2,"},{"Start":"13:23.170 ","End":"13:28.270","Text":"Gamma^2, all of this squared,"},{"Start":"13:28.270 ","End":"13:32.964","Text":"and then this square root."},{"Start":"13:32.964 ","End":"13:35.845","Text":"Where of course, again,"},{"Start":"13:35.845 ","End":"13:40.105","Text":"my Gamma is equal to my Lambda divided by m,"},{"Start":"13:40.105 ","End":"13:45.340","Text":"and my Omega 0^2 is equal to my k divided by"},{"Start":"13:45.340 ","End":"13:51.850","Text":"m. You\u0027ll recognize this answer from the previous lesson where I said what my A was."},{"Start":"13:51.850 ","End":"13:56.005","Text":"This is how I get this funny expression for my A."},{"Start":"13:56.005 ","End":"13:59.190","Text":"Then the Phi which was another unknown,"},{"Start":"13:59.190 ","End":"14:01.690","Text":"I can just undo it from over here."},{"Start":"14:01.690 ","End":"14:07.540","Text":"I do arc cos and I get this over here, or arc sine."},{"Start":"14:07.540 ","End":"14:10.265","Text":"That\u0027s the end of this explanation."},{"Start":"14:10.265 ","End":"14:12.750","Text":"Well done for having a look."}],"ID":9417},{"Watched":false,"Name":"Initial Conditions Explanation","Duration":"10m 8s","ChapterTopicVideoID":9148,"CourseChapterTopicPlaylistID":5367,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.665","Text":"Hello. In this video,"},{"Start":"00:01.665 ","End":"00:05.520","Text":"I\u0027m going to be describing what happens with our initial condition."},{"Start":"00:05.520 ","End":"00:09.435","Text":"We\u0027re going to also be focusing on when we spoke about"},{"Start":"00:09.435 ","End":"00:14.820","Text":"our resonance and we didn\u0027t at all pay any attention to our initial conditions,"},{"Start":"00:14.820 ","End":"00:17.730","Text":"and what happens if we anyway want to"},{"Start":"00:17.730 ","End":"00:21.615","Text":"look at our initial conditions when dealing with resonance."},{"Start":"00:21.615 ","End":"00:24.675","Text":"Let\u0027s start doing a little recap."},{"Start":"00:24.675 ","End":"00:30.465","Text":"First, let\u0027s speak about resonance and damping with no initial conditions."},{"Start":"00:30.465 ","End":"00:33.880","Text":"We have to remember that we have 2 forces acting,"},{"Start":"00:33.880 ","End":"00:36.240","Text":"we have our driving force."},{"Start":"00:36.240 ","End":"00:38.530","Text":"This is driving force,"},{"Start":"00:38.530 ","End":"00:45.260","Text":"and that is equal to F_0 multiplied by cosine"},{"Start":"00:45.260 ","End":"00:53.340","Text":"of Omega t. This is as a function of time,"},{"Start":"00:53.340 ","End":"00:55.600","Text":"and then we have our damping force,"},{"Start":"00:55.600 ","End":"01:00.785","Text":"which is simply negative Lambda multiplied by v. Now,"},{"Start":"01:00.785 ","End":"01:03.620","Text":"we can work with both of these forces or we can"},{"Start":"01:03.620 ","End":"01:06.860","Text":"leave out our damping force which is usually a frictional force,"},{"Start":"01:06.860 ","End":"01:11.020","Text":"by simply saying that our Lambda is equal to 0."},{"Start":"01:11.020 ","End":"01:14.255","Text":"Then we\u0027re left with a differential equation,"},{"Start":"01:14.255 ","End":"01:24.050","Text":"which is x.. plus Gamma x. plus Omega"},{"Start":"01:24.630 ","End":"01:31.670","Text":"0^2x is equal to F_0 divided by m multiplied"},{"Start":"01:31.670 ","End":"01:38.878","Text":"by cosine of Omega t. Gone a bit out of the line over here."},{"Start":"01:38.878 ","End":"01:44.250","Text":"To remind you, our Gamma over here is to do with our frictional force or damping force,"},{"Start":"01:44.250 ","End":"01:48.440","Text":"and our Gamma is set to be equal"},{"Start":"01:48.440 ","End":"01:53.394","Text":"to our Lambda divided by m. Our Lambda comes from our frictional force,"},{"Start":"01:53.394 ","End":"02:00.270","Text":"and our Omega 0^2 is our resonance frequency."},{"Start":"02:00.270 ","End":"02:04.570","Text":"We have Omega 0^2 and this has to do with the spring,"},{"Start":"02:04.570 ","End":"02:09.830","Text":"and it\u0027s equal to k divided by m. Then we saw in"},{"Start":"02:09.830 ","End":"02:15.735","Text":"that lesson that our solution is x as a function of t,"},{"Start":"02:15.735 ","End":"02:19.475","Text":"and it\u0027s equal to an amplitude which is dependent"},{"Start":"02:19.475 ","End":"02:23.255","Text":"on our Omega which is from our driving force,"},{"Start":"02:23.255 ","End":"02:28.685","Text":"the same Omega multiplied by cosine of Omega,"},{"Start":"02:28.685 ","End":"02:33.200","Text":"the same Omega multiplied by t plus Phi."},{"Start":"02:33.200 ","End":"02:39.690","Text":"We saw that our A as a function of Omega is given from the question,"},{"Start":"02:39.690 ","End":"02:44.155","Text":"and it equals to F_0 divided by m,"},{"Start":"02:44.155 ","End":"02:54.640","Text":"and then all of this is divided by the square root of Omega^2 minus Omega 0^2,"},{"Start":"02:55.280 ","End":"03:03.480","Text":"all this squared, plus Gamma Omega and all of this squared,"},{"Start":"03:03.480 ","End":"03:07.710","Text":"and the square root of all of this."},{"Start":"03:07.710 ","End":"03:13.175","Text":"We saw this, so we can see that our amplitude over here is dependent"},{"Start":"03:13.175 ","End":"03:18.920","Text":"on our driving force and the Gamma which is given over here,"},{"Start":"03:18.920 ","End":"03:21.200","Text":"which is our Lambda divided by m,"},{"Start":"03:21.200 ","End":"03:22.895","Text":"which is also given in the question."},{"Start":"03:22.895 ","End":"03:26.690","Text":"We saw that our Phi is we just have to arc sine."},{"Start":"03:26.690 ","End":"03:30.950","Text":"We can say that our sign Phi is equal"},{"Start":"03:30.950 ","End":"03:37.520","Text":"to our Gamma Omega divided by the same denominator over here."},{"Start":"03:37.520 ","End":"03:43.195","Text":"The square root of Omega^2 minus Omega 0^2,"},{"Start":"03:43.195 ","End":"03:44.820","Text":"all of this squared,"},{"Start":"03:44.820 ","End":"03:54.660","Text":"plus Gamma Omega^2 and then the square root of all of this."},{"Start":"03:55.010 ","End":"03:59.370","Text":"This is the same denominator in both."},{"Start":"03:59.630 ","End":"04:04.520","Text":"Because we saw that we can find the amplitude and the Phis, our Phi,"},{"Start":"04:04.520 ","End":"04:07.589","Text":"from just what is given to us in the question,"},{"Start":"04:07.589 ","End":"04:11.155","Text":"we don\u0027t even need to look at the initial conditions at all."},{"Start":"04:11.155 ","End":"04:15.675","Text":"As I highlighted, over here are no initial conditions."},{"Start":"04:15.675 ","End":"04:17.675","Text":"Next, we\u0027re going to speak about damping."},{"Start":"04:17.675 ","End":"04:19.640","Text":"Sorry before it said here resonance,"},{"Start":"04:19.640 ","End":"04:21.820","Text":"but I rubbed it out, it was a mistake."},{"Start":"04:21.820 ","End":"04:23.690","Text":"Now we\u0027re speaking about damping."},{"Start":"04:23.690 ","End":"04:26.755","Text":"When we only have our frictional force,"},{"Start":"04:26.755 ","End":"04:28.330","Text":"our damping force,"},{"Start":"04:28.330 ","End":"04:36.560","Text":"so that is going to be our F is equal to negative Lambda v. This is a frictional force."},{"Start":"04:36.560 ","End":"04:40.720","Text":"Then we saw that our differential equation,"},{"Start":"04:40.720 ","End":"04:43.460","Text":"so I\u0027ll write it here so that everything\u0027s in the same line,"},{"Start":"04:43.460 ","End":"04:46.010","Text":"is going to be very similar to this 1."},{"Start":"04:46.010 ","End":"04:47.990","Text":"We\u0027re going to have again x double dot,"},{"Start":"04:47.990 ","End":"04:53.805","Text":"and then we\u0027re going to have plus Gamma multiplied by x dot,"},{"Start":"04:53.805 ","End":"04:59.595","Text":"so just like here, plus our Omega 0^2x."},{"Start":"04:59.595 ","End":"05:01.970","Text":"Again, just like here is equal to,"},{"Start":"05:01.970 ","End":"05:04.460","Text":"and then because we don\u0027t have our driving force,"},{"Start":"05:04.460 ","End":"05:07.145","Text":"so it\u0027s going to be equal to 0."},{"Start":"05:07.145 ","End":"05:13.130","Text":"It\u0027s exact same differential equation aside from this over here,"},{"Start":"05:13.130 ","End":"05:14.763","Text":"it\u0027s equal to 0."},{"Start":"05:14.763 ","End":"05:23.195","Text":"Then we saw that our solution where is equal to x as a function of t,"},{"Start":"05:23.195 ","End":"05:25.670","Text":"which is equal to A tag."},{"Start":"05:25.670 ","End":"05:27.080","Text":"Now I\u0027m doing a tag over here,"},{"Start":"05:27.080 ","End":"05:29.654","Text":"so that we don\u0027t get confused with this A over here,"},{"Start":"05:29.654 ","End":"05:31.115","Text":"it\u0027s a different A,"},{"Start":"05:31.115 ","End":"05:38.240","Text":"multiplied by e to the power of negative Gamma divided by 2t."},{"Start":"05:38.240 ","End":"05:43.580","Text":"This is multiplied by cosine of Omega wave,"},{"Start":"05:43.580 ","End":"05:46.955","Text":"which soon we\u0027ll have a look at what that is and it\u0027s different to this Omega,"},{"Start":"05:46.955 ","End":"05:51.180","Text":"multiplied by t plus our Phi tag."},{"Start":"05:51.180 ","End":"05:55.355","Text":"Again, Phi tag to differentiate it from this,"},{"Start":"05:55.355 ","End":"06:00.310","Text":"so that you don\u0027t think that this Phi tag is the same as this Phi over here."},{"Start":"06:00.310 ","End":"06:05.750","Text":"Our Omega wave because we don\u0027t have any driving force over here,"},{"Start":"06:05.750 ","End":"06:08.480","Text":"we only have damping,"},{"Start":"06:08.480 ","End":"06:12.950","Text":"so it\u0027s going to be the square root of our resonant frequency squared,"},{"Start":"06:12.950 ","End":"06:17.951","Text":"so Omega 0^2 plus our gamma squared."},{"Start":"06:17.951 ","End":"06:20.210","Text":"It\u0027s going to be that."},{"Start":"06:20.210 ","End":"06:24.185","Text":"Then we also saw in the lesson that in order to find our A tag in our Phi tag,"},{"Start":"06:24.185 ","End":"06:27.610","Text":"then what we had to do is we have to use initial conditions."},{"Start":"06:27.610 ","End":"06:30.440","Text":"We would set our x,"},{"Start":"06:30.440 ","End":"06:34.910","Text":"so this at time t equals 0 and the first derivative of it,"},{"Start":"06:34.910 ","End":"06:37.760","Text":"so its velocity also at t equals 0,"},{"Start":"06:37.760 ","End":"06:40.880","Text":"and then through these initial conditions,"},{"Start":"06:40.880 ","End":"06:45.840","Text":"we would get our A tag and our Phi tag."},{"Start":"06:45.840 ","End":"06:50.750","Text":"Now what we\u0027re going to take a look at is why we don\u0027t usually use this,"},{"Start":"06:50.750 ","End":"06:56.419","Text":"the working out for the initial conditions when dealing with resonance and damping."},{"Start":"06:56.419 ","End":"06:59.655","Text":"Let\u0027s see why we don\u0027t use this."},{"Start":"06:59.655 ","End":"07:01.830","Text":"Now we\u0027re taking a look over here,"},{"Start":"07:01.830 ","End":"07:06.680","Text":"so we\u0027re dealing with resonance and damping with initial conditions."},{"Start":"07:06.680 ","End":"07:09.170","Text":"Let\u0027s take a look at what that will look like."},{"Start":"07:09.170 ","End":"07:11.144","Text":"Our differential equations,"},{"Start":"07:11.144 ","End":"07:12.900","Text":"let\u0027s leave that for a second,"},{"Start":"07:12.900 ","End":"07:18.095","Text":"but our solution is going to be the solution for"},{"Start":"07:18.095 ","End":"07:21.350","Text":"our resonance and damping with no initial conditions"},{"Start":"07:21.350 ","End":"07:25.430","Text":"plus the solution over here for damping."},{"Start":"07:25.430 ","End":"07:31.995","Text":"Our x as a function of t is going to be equal to our A as a function of"},{"Start":"07:31.995 ","End":"07:38.610","Text":"Omega multiplied by cosine of our Omega t plus Phi,"},{"Start":"07:38.610 ","End":"07:43.865","Text":"and then we\u0027re going to have plus this over here."},{"Start":"07:43.865 ","End":"07:51.200","Text":"Our A tag e to the negative Gamma divided by 2 multiplied by"},{"Start":"07:51.200 ","End":"07:59.120","Text":"t cosine of our Omega wave t plus our Phi tag."},{"Start":"07:59.120 ","End":"08:03.430","Text":"The answer here is a full answer,"},{"Start":"08:03.430 ","End":"08:05.870","Text":"so that we can know the position of"},{"Start":"08:05.870 ","End":"08:10.115","Text":"our pendulum or our body that\u0027s moving in harmonic motion."},{"Start":"08:10.115 ","End":"08:12.905","Text":"We can know at every single moment from"},{"Start":"08:12.905 ","End":"08:17.215","Text":"the first moment of our harmonic motion experiment."},{"Start":"08:17.215 ","End":"08:20.690","Text":"This is a really full and concise answer."},{"Start":"08:20.690 ","End":"08:25.925","Text":"How come we never use this and we usually when dealing with resonance and damping,"},{"Start":"08:25.925 ","End":"08:28.667","Text":"only use this section of the solution,"},{"Start":"08:28.667 ","End":"08:31.590","Text":"only this expression over here."},{"Start":"08:32.030 ","End":"08:36.365","Text":"The reason is because we can see that over here,"},{"Start":"08:36.365 ","End":"08:39.320","Text":"we have this exponent e to the power of"},{"Start":"08:39.320 ","End":"08:44.390","Text":"negative Gamma divided by 2 multiplied by t. We can"},{"Start":"08:44.390 ","End":"08:48.410","Text":"see that this is a decaying exponent which"},{"Start":"08:48.410 ","End":"08:52.980","Text":"means that after a very short time this is going to go to 0,"},{"Start":"08:52.980 ","End":"08:59.780","Text":"which means that this whole expression over here will go to 0 so will be insignificant."},{"Start":"08:59.780 ","End":"09:06.575","Text":"Very quickly we can see that this expression goes to 0,"},{"Start":"09:06.575 ","End":"09:11.900","Text":"which means that we don\u0027t have to add it into our calculation after a certainty,"},{"Start":"09:11.900 ","End":"09:18.080","Text":"a very small t. This expression just tells us after how"},{"Start":"09:18.080 ","End":"09:24.560","Text":"long our harmonic motion system starts moving according to the driving force,"},{"Start":"09:24.560 ","End":"09:26.437","Text":"to what the driving force dictates,"},{"Start":"09:26.437 ","End":"09:30.440","Text":"and it takes a very short period of time for this to happen."},{"Start":"09:30.440 ","End":"09:32.210","Text":"That\u0027s why we generally just"},{"Start":"09:32.210 ","End":"09:36.620","Text":"ignore the section which has to do with the initial conditions."},{"Start":"09:36.620 ","End":"09:40.060","Text":"However, if you are interested to add this and to add this"},{"Start":"09:40.060 ","End":"09:43.565","Text":"into your calculations or to look at this,"},{"Start":"09:43.565 ","End":"09:47.335","Text":"then you can add this in, it\u0027s not incorrect."},{"Start":"09:47.335 ","End":"09:55.150","Text":"However, it is rendered arbitrary within a very short period of time."},{"Start":"09:55.150 ","End":"09:58.670","Text":"I hope that this little explanation has given you"},{"Start":"09:58.670 ","End":"10:03.395","Text":"a bit more insight into when we use our initial conditions,"},{"Start":"10:03.395 ","End":"10:06.658","Text":"and why not and why yes when we do."},{"Start":"10:06.658 ","End":"10:09.690","Text":"This is the end of the lesson."}],"ID":9418},{"Watched":false,"Name":"Exercise 1","Duration":"11m 7s","ChapterTopicVideoID":9149,"CourseChapterTopicPlaylistID":5367,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.620","Text":"Hello. In this lesson,"},{"Start":"00:01.620 ","End":"00:06.465","Text":"we\u0027re going to practice tackling the question of harmonic motion with resonance."},{"Start":"00:06.465 ","End":"00:10.260","Text":"Let\u0027s imagine that we have some plank,"},{"Start":"00:10.260 ","End":"00:13.199","Text":"and attached to it, we have our spring."},{"Start":"00:13.199 ","End":"00:15.765","Text":"Into the other end of the spring,"},{"Start":"00:15.765 ","End":"00:18.195","Text":"we have a mass attached."},{"Start":"00:18.195 ","End":"00:21.435","Text":"Now, you could imagine that someone is grabbing"},{"Start":"00:21.435 ","End":"00:25.185","Text":"this plank over here and is bouncing it up and down,"},{"Start":"00:25.185 ","End":"00:27.165","Text":"like this, up and down,"},{"Start":"00:27.165 ","End":"00:36.989","Text":"up and down and the bouncing is being done by y_1 is equal to y_0 multiplied"},{"Start":"00:36.989 ","End":"00:41.810","Text":"by cosine of Omega t. Someone is"},{"Start":"00:41.810 ","End":"00:48.050","Text":"forcing this plank to move up and down according to this type of movement."},{"Start":"00:48.050 ","End":"00:50.435","Text":"But notice this isn\u0027t a force,"},{"Start":"00:50.435 ","End":"00:52.450","Text":"so this isn\u0027t a driving force."},{"Start":"00:52.450 ","End":"01:00.530","Text":"The only forces acting on this mass is mg and the force from the spring."},{"Start":"01:00.530 ","End":"01:04.855","Text":"It\u0027s just moving because our plank of wood up here is moving."},{"Start":"01:04.855 ","End":"01:09.080","Text":"Now what we\u0027re going to do is we\u0027re going to analyze this,"},{"Start":"01:09.080 ","End":"01:10.640","Text":"the movement of this system."},{"Start":"01:10.640 ","End":"01:15.290","Text":"We\u0027re going to start with trying to find its position as a function of time."},{"Start":"01:15.290 ","End":"01:18.635","Text":"Let\u0027s call this mass, mass number 2."},{"Start":"01:18.635 ","End":"01:21.890","Text":"Because its position is going to be in the y-direction,"},{"Start":"01:21.890 ","End":"01:23.450","Text":"because it\u0027s moving up and down,"},{"Start":"01:23.450 ","End":"01:26.318","Text":"so in order to find its position as a function of time,"},{"Start":"01:26.318 ","End":"01:31.195","Text":"it\u0027s going to be y_2(t) is equal to what?"},{"Start":"01:31.195 ","End":"01:39.145","Text":"Now, let\u0027s say that the length of the spring when it\u0027s slack is going to be l_0."},{"Start":"01:39.145 ","End":"01:42.795","Text":"Now, let\u0027s start labeling our axes."},{"Start":"01:42.795 ","End":"01:48.440","Text":"Let\u0027s start by saying that our y-axis is going in this downwards direction."},{"Start":"01:48.440 ","End":"01:54.055","Text":"This is our y-direction and the origin is somewhere over here."},{"Start":"01:54.055 ","End":"01:58.730","Text":"Then what we can do is we can say that the position of the plank,"},{"Start":"01:58.730 ","End":"02:06.235","Text":"which is given to us by this equation over here is going to be this, this is y_1."},{"Start":"02:06.235 ","End":"02:09.875","Text":"Then that means that the position of our mass,"},{"Start":"02:09.875 ","End":"02:14.590","Text":"mass number 2 is going to be this, y_2."},{"Start":"02:14.590 ","End":"02:18.830","Text":"Now, in order to get our harmonic motion and then to find our position,"},{"Start":"02:18.830 ","End":"02:23.404","Text":"what we\u0027re going to do is we\u0027re going to work with our force equations."},{"Start":"02:23.404 ","End":"02:27.815","Text":"Right here that we have the sum of all of our forces is equal to,"},{"Start":"02:27.815 ","End":"02:29.405","Text":"which forces do we have?"},{"Start":"02:29.405 ","End":"02:31.850","Text":"We have our mg working over here,"},{"Start":"02:31.850 ","End":"02:36.830","Text":"so we can write here mg and notice even though it\u0027s pointing down,"},{"Start":"02:36.830 ","End":"02:38.479","Text":"it\u0027s in the positive y direction."},{"Start":"02:38.479 ","End":"02:42.395","Text":"Then we have our force from our spring."},{"Start":"02:42.395 ","End":"02:48.060","Text":"That\u0027s going to be negative k. Now,"},{"Start":"02:48.060 ","End":"02:50.835","Text":"we have to write x minus x_0,"},{"Start":"02:50.835 ","End":"02:58.490","Text":"so we have to include in here how much it has been elongated from this slack length."},{"Start":"02:58.490 ","End":"03:03.560","Text":"That means that we have to find this distance here."},{"Start":"03:03.560 ","End":"03:06.320","Text":"This in green is our Delta y."},{"Start":"03:06.320 ","End":"03:08.615","Text":"What is our Delta y?"},{"Start":"03:08.615 ","End":"03:10.640","Text":"Let\u0027s write over here,"},{"Start":"03:10.640 ","End":"03:18.385","Text":"our Delta y is going to be equal to out y_2 minus out l_0,"},{"Start":"03:18.385 ","End":"03:21.780","Text":"then minus our y_1."},{"Start":"03:21.780 ","End":"03:27.500","Text":"If you look at your y_1 plus l_0 plus Delta y,"},{"Start":"03:27.500 ","End":"03:32.690","Text":"we\u0027ll get to the full length of our y_2. Take a look at that."},{"Start":"03:32.690 ","End":"03:36.200","Text":"Now what we\u0027re going to do is we\u0027re going to write that in."},{"Start":"03:36.200 ","End":"03:39.980","Text":"This is our x, if we remember in the equation it\u0027s"},{"Start":"03:39.980 ","End":"03:43.505","Text":"usually negative k multiplied by x minus x_0,"},{"Start":"03:43.505 ","End":"03:46.910","Text":"which represents the Delta x."},{"Start":"03:46.910 ","End":"03:48.725","Text":"Here we\u0027re doing Delta y,"},{"Start":"03:48.725 ","End":"03:51.925","Text":"which we\u0027ve worked out through this."},{"Start":"03:51.925 ","End":"03:59.335","Text":"We have y_2 minus y_1 minus l_0."},{"Start":"03:59.335 ","End":"04:01.285","Text":"Now, this is going to be equal 2,"},{"Start":"04:01.285 ","End":"04:04.025","Text":"of course m multiplied by,"},{"Start":"04:04.025 ","End":"04:08.495","Text":"we\u0027re trying to find the movement of this pendulum,"},{"Start":"04:08.495 ","End":"04:12.295","Text":"so of y_2 double dot."},{"Start":"04:12.295 ","End":"04:14.805","Text":"Let\u0027s open this up."},{"Start":"04:14.805 ","End":"04:19.050","Text":"We\u0027ll have mg minus and"},{"Start":"04:19.050 ","End":"04:23.279","Text":"then I\u0027m going to keep my y_2 and my l_0 together in the brackets."},{"Start":"04:23.279 ","End":"04:33.350","Text":"I\u0027ll have y_2 minus l_0 and then negative k multiplied by negative y_1."},{"Start":"04:33.350 ","End":"04:42.810","Text":"That\u0027s plus ky_1 and this is going to be equal to our my_2 double dot."},{"Start":"04:42.810 ","End":"04:45.810","Text":"Now, let\u0027s substitute in our y_1,"},{"Start":"04:45.810 ","End":"04:47.115","Text":"so we\u0027re going to have"},{"Start":"04:47.115 ","End":"04:56.390","Text":"mg negative k(y_2 minus l_0) plus k,"},{"Start":"04:56.390 ","End":"05:06.000","Text":"and then y_1 is equal to y_0 cosine of Omega t my double dot."},{"Start":"05:06.000 ","End":"05:11.840","Text":"We have this equation and what we want to do is we want to move it or shift it and"},{"Start":"05:11.840 ","End":"05:14.420","Text":"play around with it in order for it to look like"},{"Start":"05:14.420 ","End":"05:18.275","Text":"the equation for harmonic motion with resonance."},{"Start":"05:18.275 ","End":"05:21.035","Text":"Just to remind you what the equation is,"},{"Start":"05:21.035 ","End":"05:22.460","Text":"it\u0027s going to be negative k,"},{"Start":"05:22.460 ","End":"05:25.460","Text":"now in this case, because we\u0027re working in the y direction."},{"Start":"05:25.460 ","End":"05:32.450","Text":"It\u0027s y minus y_0 and then we have plus our driving force,"},{"Start":"05:32.450 ","End":"05:38.820","Text":"which is going to be f_0 multiplied by cosine of Omega t,"},{"Start":"05:38.820 ","End":"05:46.110","Text":"and that is supposed to equal my double dot."},{"Start":"05:46.110 ","End":"05:49.835","Text":"We want this equation over here to look like this."},{"Start":"05:49.835 ","End":"05:53.540","Text":"Now, of course, here it\u0027s without using a damping force,"},{"Start":"05:53.540 ","End":"05:55.400","Text":"which we could have added into this equation,"},{"Start":"05:55.400 ","End":"05:58.235","Text":"and then we would have just said that Lambda"},{"Start":"05:58.235 ","End":"06:01.705","Text":"is equal to 0 and we would be left with this equation."},{"Start":"06:01.705 ","End":"06:03.525","Text":"Let\u0027s see how we do this."},{"Start":"06:03.525 ","End":"06:06.290","Text":"I\u0027m going to do my little trick,"},{"Start":"06:06.290 ","End":"06:08.300","Text":"which isn\u0027t very nice, but it doesn\u0027t really matter."},{"Start":"06:08.300 ","End":"06:15.120","Text":"I\u0027m going to try and get rid of this mg from being some weird term over here."},{"Start":"06:15.120 ","End":"06:18.830","Text":"I\u0027m going to take my negative k and"},{"Start":"06:18.830 ","End":"06:23.400","Text":"put it as a common factor with my mg. Let\u0027s see how this works."},{"Start":"06:23.650 ","End":"06:28.700","Text":"Let\u0027s move this more. I\u0027ll have negative k as"},{"Start":"06:28.700 ","End":"06:36.795","Text":"my common factor and then I have here my y_2 minus my l_0,"},{"Start":"06:36.795 ","End":"06:39.890","Text":"and then to add in my mg into these brackets."},{"Start":"06:39.890 ","End":"06:47.425","Text":"I have to multiply by negative mg divided by k. You see this,"},{"Start":"06:47.425 ","End":"06:51.410","Text":"then, if you try and open the brackets now over here,"},{"Start":"06:51.410 ","End":"06:53.300","Text":"you\u0027ll see that you get this."},{"Start":"06:53.300 ","End":"06:56.300","Text":"Then plus over here,"},{"Start":"06:56.300 ","End":"07:00.890","Text":"so ky_0 cosine of"},{"Start":"07:00.890 ","End":"07:09.185","Text":"Omega t and this is equal to my_2 double dot."},{"Start":"07:09.185 ","End":"07:15.840","Text":"Now, I\u0027m going to say that I\u0027m going to call all of"},{"Start":"07:15.840 ","End":"07:23.945","Text":"this my y_0 tag to differentiate it from this y_0 tag."},{"Start":"07:23.945 ","End":"07:29.975","Text":"But still, sorry, y_0 tag to differentiate this y_0 from this y_0."},{"Start":"07:29.975 ","End":"07:34.440","Text":"But to have the y_0 to remind you that it\u0027s this term,"},{"Start":"07:34.440 ","End":"07:39.604","Text":"to connect it to this so that you understand how it links into this equation."},{"Start":"07:39.604 ","End":"07:43.205","Text":"But to differentiate it from this y_0 because it\u0027s not the same."},{"Start":"07:43.205 ","End":"07:48.390","Text":"Then I\u0027m going to call this over here my f_0,"},{"Start":"07:48.390 ","End":"07:50.265","Text":"and my y_2,"},{"Start":"07:50.265 ","End":"07:54.710","Text":"I\u0027m just going to name as y simultaneously here."},{"Start":"07:54.710 ","End":"08:01.570","Text":"This will be y double-dot to link it into my equation in red so that it makes more sense."},{"Start":"08:01.570 ","End":"08:03.480","Text":"Now I can rewrite this,"},{"Start":"08:03.480 ","End":"08:10.805","Text":"so I\u0027ll have negative k multiplied by y minus all of this,"},{"Start":"08:10.805 ","End":"08:14.765","Text":"which is my y_0 tag, because it\u0027s not this y_0,"},{"Start":"08:14.765 ","End":"08:21.855","Text":"close my brackets plus my f_0 multiplied by cosine"},{"Start":"08:21.855 ","End":"08:29.430","Text":"of Omega t. This is equal to my my double dot."},{"Start":"08:29.430 ","End":"08:32.914","Text":"Good, so we\u0027ve come to this equation,"},{"Start":"08:32.914 ","End":"08:36.949","Text":"which represents harmonic motion with resonance."},{"Start":"08:36.949 ","End":"08:41.420","Text":"That\u0027s perfect and we have that our y_0 tag is of course a point of"},{"Start":"08:41.420 ","End":"08:48.285","Text":"equilibrium and that is at l_0 minus mg divided by k,"},{"Start":"08:48.285 ","End":"08:49.735","Text":"and we have our f_0,"},{"Start":"08:49.735 ","End":"08:53.360","Text":"which is our driving force amplitude,"},{"Start":"08:53.360 ","End":"08:59.830","Text":"which is equal to k multiplied by the amplitude of this over here."},{"Start":"08:59.830 ","End":"09:03.095","Text":"Now let\u0027s find our solution."},{"Start":"09:03.095 ","End":"09:06.715","Text":"Our position as a function of time."},{"Start":"09:06.715 ","End":"09:12.935","Text":"Our equation for y as a function of time is as we know,"},{"Start":"09:12.935 ","End":"09:19.935","Text":"going to be a as a function of Omega multiplied by cosine of"},{"Start":"09:19.935 ","End":"09:27.645","Text":"Omega t plus Phi plus y_0 from over here."},{"Start":"09:27.645 ","End":"09:31.910","Text":"This is the general equation and as we remember,"},{"Start":"09:31.910 ","End":"09:33.620","Text":"because we\u0027re dealing with resonance,"},{"Start":"09:33.620 ","End":"09:37.120","Text":"so we don\u0027t have to look at our initial conditions."},{"Start":"09:37.120 ","End":"09:39.560","Text":"In our example specifically,"},{"Start":"09:39.560 ","End":"09:43.055","Text":"because we said that Gamma is equal to 0,"},{"Start":"09:43.055 ","End":"09:45.410","Text":"because we don\u0027t have any damping."},{"Start":"09:45.410 ","End":"09:47.060","Text":"That means that this,"},{"Start":"09:47.060 ","End":"09:52.485","Text":"our Phi is going to be equal to 0,"},{"Start":"09:52.485 ","End":"09:56.165","Text":"and that\u0027s because our Gamma is equal to 0."},{"Start":"09:56.165 ","End":"09:57.815","Text":"If you go back to the lecture,"},{"Start":"09:57.815 ","End":"09:59.585","Text":"then you can see that."},{"Start":"09:59.585 ","End":"10:04.370","Text":"Then our y_0 over here is going to be our y_0 tag."},{"Start":"10:04.370 ","End":"10:06.155","Text":"We can add a tag over here."},{"Start":"10:06.155 ","End":"10:08.960","Text":"Then our a as a function of Omega,"},{"Start":"10:08.960 ","End":"10:11.075","Text":"as we know, it\u0027s our usual equation,"},{"Start":"10:11.075 ","End":"10:13.205","Text":"so you should remember this."},{"Start":"10:13.205 ","End":"10:20.150","Text":"A as a function of omega is equal to f_0 divided by m,"},{"Start":"10:20.150 ","End":"10:23.750","Text":"so here f_0 is k multiplied by y_0,"},{"Start":"10:23.750 ","End":"10:30.185","Text":"and notice y_0 without a tag and all of this divided by because our gamma is equal to 0,"},{"Start":"10:30.185 ","End":"10:31.655","Text":"because we have no damping."},{"Start":"10:31.655 ","End":"10:36.985","Text":"It\u0027s going to be omega 0 squared minus Omega squared,"},{"Start":"10:36.985 ","End":"10:40.610","Text":"where this Omega and this Omega are the same"},{"Start":"10:40.610 ","End":"10:43.750","Text":"as this Omega over here or this omega over here,"},{"Start":"10:43.750 ","End":"10:50.000","Text":"so it\u0027s given an or question and our Omega_0^2 is to do with our spring,"},{"Start":"10:50.000 ","End":"10:53.900","Text":"which is going to be equal to k divided by m. Again,"},{"Start":"10:53.900 ","End":"10:56.610","Text":"these are constants in the question."},{"Start":"10:56.610 ","End":"10:58.980","Text":"Now I wrote out the final answer,"},{"Start":"10:58.980 ","End":"11:04.390","Text":"substituting in all of our variables and this is the final answer that you should get."},{"Start":"11:04.390 ","End":"11:07.360","Text":"That\u0027s the end of our lesson."}],"ID":9419}],"Thumbnail":null,"ID":5367},{"Name":"Double Masses","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Double Masses","Duration":"18m 24s","ChapterTopicVideoID":10483,"CourseChapterTopicPlaylistID":5368,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.770","Text":"Hello. In this question,"},{"Start":"00:01.770 ","End":"00:04.425","Text":"we\u0027re going to still be dealing with harmonic motion,"},{"Start":"00:04.425 ","End":"00:07.895","Text":"however, with a slightly more complicated case."},{"Start":"00:07.895 ","End":"00:10.950","Text":"We\u0027re going to start at an easy example for"},{"Start":"00:10.950 ","End":"00:14.895","Text":"this complicated thing and then move harder later on."},{"Start":"00:14.895 ","End":"00:18.630","Text":"Over here we have 2 masses which are attached by"},{"Start":"00:18.630 ","End":"00:22.470","Text":"a spring of constant k and for the meantime,"},{"Start":"00:22.470 ","End":"00:26.610","Text":"they\u0027re just going to be moving along the x-axis to make it a bit easier."},{"Start":"00:26.610 ","End":"00:31.545","Text":"What we want to do is to find the frequency of oscillations."},{"Start":"00:31.545 ","End":"00:34.140","Text":"Here we have our x and y axis,"},{"Start":"00:34.140 ","End":"00:36.915","Text":"we have our mass number 1 and its position,"},{"Start":"00:36.915 ","End":"00:39.075","Text":"X_1, and because it\u0027s moving,"},{"Start":"00:39.075 ","End":"00:40.555","Text":"it\u0027s dependent on time."},{"Start":"00:40.555 ","End":"00:41.990","Text":"We have a spring of constant k,"},{"Start":"00:41.990 ","End":"00:44.195","Text":"and then we have our mass number 2,"},{"Start":"00:44.195 ","End":"00:46.670","Text":"at a position of X_2,"},{"Start":"00:46.670 ","End":"00:48.740","Text":"which is also is a function of time because it\u0027s"},{"Start":"00:48.740 ","End":"00:51.590","Text":"also going to be moving along the x-axis."},{"Start":"00:51.590 ","End":"00:55.025","Text":"How we\u0027re going to begin is we\u0027re going to write out"},{"Start":"00:55.025 ","End":"00:58.880","Text":"the r equation for the energy of this system,"},{"Start":"00:58.880 ","End":"01:02.495","Text":"so we\u0027re going to say that our energy is equal to."},{"Start":"01:02.495 ","End":"01:05.640","Text":"We have the kinetic energy of our first mass,"},{"Start":"01:05.640 ","End":"01:09.930","Text":"so that\u0027s going to be 1/2 of m_1 multiplied"},{"Start":"01:09.930 ","End":"01:15.080","Text":"by V_1^2 plus the kinetic energy of our mass number 2,"},{"Start":"01:15.080 ","End":"01:18.935","Text":"so that\u0027s going to be 1/2m_2,"},{"Start":"01:18.935 ","End":"01:26.825","Text":"V_2^2 and then plus our energy from our spring potential,"},{"Start":"01:26.825 ","End":"01:35.540","Text":"so we\u0027re going to have plus 1/2k and then we have to add in the length of the spring,"},{"Start":"01:35.540 ","End":"01:37.265","Text":"so multiply it in over here."},{"Start":"01:37.265 ","End":"01:41.660","Text":"The length of the spring is going to be our X_2,"},{"Start":"01:41.660 ","End":"01:42.904","Text":"which is as a function of time,"},{"Start":"01:42.904 ","End":"01:46.470","Text":"minus our X_1, and this squared."},{"Start":"01:46.470 ","End":"01:48.814","Text":"That\u0027s the length of our spring."},{"Start":"01:48.814 ","End":"01:51.770","Text":"We can also add in over here plus L_0,"},{"Start":"01:51.770 ","End":"01:56.675","Text":"which would represent the length of the spring when it\u0027s limp, when it\u0027s slack."},{"Start":"01:56.675 ","End":"02:00.350","Text":"However, we can say that the length of the spring when it\u0027s slack,"},{"Start":"02:00.350 ","End":"02:03.245","Text":"is really, really tiny,"},{"Start":"02:03.245 ","End":"02:06.185","Text":"in which case, we don\u0027t have to look at its length."},{"Start":"02:06.185 ","End":"02:07.670","Text":"Also, even if it isn\u0027t,"},{"Start":"02:07.670 ","End":"02:11.480","Text":"it doesn\u0027t really play much of a role in this exercise,"},{"Start":"02:11.480 ","End":"02:12.895","Text":"so it doesn\u0027t matter."},{"Start":"02:12.895 ","End":"02:15.650","Text":"This is the total energy of the system over"},{"Start":"02:15.650 ","End":"02:19.190","Text":"here and our X_2 minus X_1 is our Delta X of the spring,"},{"Start":"02:19.190 ","End":"02:21.595","Text":"so the length of it."},{"Start":"02:21.595 ","End":"02:23.820","Text":"I changed it from Delta X_2L,"},{"Start":"02:23.820 ","End":"02:26.055","Text":"the length of the spring."},{"Start":"02:26.055 ","End":"02:29.840","Text":"Now what we want to do is we want to put"},{"Start":"02:29.840 ","End":"02:35.390","Text":"this equation into our equation for energy when dealing with harmonic motion."},{"Start":"02:35.390 ","End":"02:37.490","Text":"If you remember back to the lesson,"},{"Start":"02:37.490 ","End":"02:39.350","Text":"when we\u0027re dealing with harmonic motion,"},{"Start":"02:39.350 ","End":"02:43.565","Text":"we want our equation for energy of the system to look something like this."},{"Start":"02:43.565 ","End":"02:48.390","Text":"It\u0027s going to be 1/2 of M X dot squared"},{"Start":"02:48.390 ","End":"02:54.970","Text":"plus 1/2 of kx^2."},{"Start":"02:55.220 ","End":"02:57.420","Text":"Over here we can see that we have"},{"Start":"02:57.420 ","End":"03:00.545","Text":"our 1 variable X and we have the first derivative of it."},{"Start":"03:00.545 ","End":"03:03.440","Text":"Now here, we don\u0027t have anything that"},{"Start":"03:03.440 ","End":"03:06.500","Text":"looks remotely like this because we have 2 variables,"},{"Start":"03:06.500 ","End":"03:09.620","Text":"we have our X_2 and our X_1 and then we also have"},{"Start":"03:09.620 ","End":"03:13.240","Text":"our first derivatives of each one and this is being squared,"},{"Start":"03:13.240 ","End":"03:15.290","Text":"so when we open out the brackets,"},{"Start":"03:15.290 ","End":"03:18.830","Text":"we\u0027re going to have our X_2 multiplied by our X_1 as one of the terms."},{"Start":"03:18.830 ","End":"03:21.815","Text":"It\u0027s a really big mess and it\u0027s really hard"},{"Start":"03:21.815 ","End":"03:25.930","Text":"to look at this and see this equation over here."},{"Start":"03:25.930 ","End":"03:30.790","Text":"Let\u0027s see how we can manipulate this in order to get something like this."},{"Start":"03:30.790 ","End":"03:33.305","Text":"Up until now in dealing with harmonic motion,"},{"Start":"03:33.305 ","End":"03:36.575","Text":"we\u0027ve always been dealing with 1 variable, with 1 mass."},{"Start":"03:36.575 ","End":"03:41.060","Text":"Here we have 2 masses and we\u0027ve never seen how to deal with this,"},{"Start":"03:41.060 ","End":"03:42.880","Text":"so let\u0027s take a look."},{"Start":"03:42.880 ","End":"03:47.880","Text":"The first thing that we\u0027re going to do is variable substitution."},{"Start":"03:48.610 ","End":"03:51.380","Text":"Instead of using 2 masses,"},{"Start":"03:51.380 ","End":"03:55.070","Text":"we\u0027re going to use the center of mass in order to show our position."},{"Start":"03:55.070 ","End":"03:59.255","Text":"We\u0027re going to have our X center of mass,"},{"Start":"03:59.255 ","End":"04:00.740","Text":"and that is, as we know,"},{"Start":"04:00.740 ","End":"04:04.245","Text":"going to be equal to M_1, X_1,"},{"Start":"04:04.245 ","End":"04:06.845","Text":"so Mass 1 multiplied by its position,"},{"Start":"04:06.845 ","End":"04:14.060","Text":"plus Mass 2 multiplied by its position and this divided by the total mass of the system,"},{"Start":"04:14.060 ","End":"04:17.700","Text":"which is M_1 plus M_2."},{"Start":"04:17.700 ","End":"04:20.495","Text":"Then we\u0027re going to have our X relative,"},{"Start":"04:20.495 ","End":"04:23.975","Text":"which is in actual fact going to be the length of the spring,"},{"Start":"04:23.975 ","End":"04:30.680","Text":"so we can see that this over here is going to be our X relative."},{"Start":"04:30.680 ","End":"04:35.000","Text":"That is going to simply be equal to this over here."},{"Start":"04:35.000 ","End":"04:39.230","Text":"Our X_2 minus our X_1,"},{"Start":"04:39.230 ","End":"04:40.930","Text":"the length of the spring."},{"Start":"04:40.930 ","End":"04:43.695","Text":"Now I have my new variables,"},{"Start":"04:43.695 ","End":"04:48.365","Text":"and I can also write my equations for my V_CM."},{"Start":"04:48.365 ","End":"04:53.505","Text":"That\u0027s just simply going to be my first derivative of my X_CM,"},{"Start":"04:53.505 ","End":"04:57.005","Text":"which is simply going to be equal to,"},{"Start":"04:57.005 ","End":"04:58.625","Text":"let\u0027s just do this quickly,"},{"Start":"04:58.625 ","End":"05:05.700","Text":"M_1V_1 plus M_2V2 divided"},{"Start":"05:05.700 ","End":"05:11.005","Text":"by M_1 plus M_2."},{"Start":"05:11.005 ","End":"05:15.810","Text":"Then similarly with my V relative,"},{"Start":"05:15.810 ","End":"05:21.619","Text":"so that\u0027s simply going to be the first derivative of my X relative,"},{"Start":"05:21.619 ","End":"05:26.955","Text":"which is simply going to be V_2 minus V_1."},{"Start":"05:26.955 ","End":"05:31.894","Text":"Now what I have over here is using what I have in my question,"},{"Start":"05:31.894 ","End":"05:34.430","Text":"I can get to my new variables,"},{"Start":"05:34.430 ","End":"05:39.130","Text":"so I can map from what I have in the question to my new variables."},{"Start":"05:39.130 ","End":"05:40.530","Text":"This is this mapping,"},{"Start":"05:40.530 ","End":"05:44.240","Text":"now I\u0027m just going to show you what we do just in"},{"Start":"05:44.240 ","End":"05:49.460","Text":"case we have our V_CM and our V relative and our X_CM and our X relative."},{"Start":"05:49.460 ","End":"05:54.330","Text":"How we get from this to our M_1 and X_1 and M_2 and X_2,"},{"Start":"05:54.330 ","End":"05:56.610","Text":"sorry our X_1 and our X_2,"},{"Start":"05:56.610 ","End":"05:58.800","Text":"so let\u0027s take a look."},{"Start":"05:58.800 ","End":"06:02.800","Text":"Our opposite transform is going to look like this,"},{"Start":"06:02.800 ","End":"06:07.745","Text":"so if we have our X_CM and our X relative and our V_CM and our V relative,"},{"Start":"06:07.745 ","End":"06:09.155","Text":"so we can get to our X_1,"},{"Start":"06:09.155 ","End":"06:11.040","Text":"X_2, V_1, and V_2."},{"Start":"06:11.040 ","End":"06:12.540","Text":"This is just the opposite,"},{"Start":"06:12.540 ","End":"06:14.990","Text":"you can take a look at these equations."},{"Start":"06:14.990 ","End":"06:18.155","Text":"Also I suggest just to also practice"},{"Start":"06:18.155 ","End":"06:21.980","Text":"your algebra and also to get it faster for if you\u0027re in an exam to solve"},{"Start":"06:21.980 ","End":"06:26.060","Text":"things quicker then I suggest you do the algebra"},{"Start":"06:26.060 ","End":"06:31.590","Text":"to transform and isolate out your X_1 and your X_2 and your V_1 and V_2."},{"Start":"06:32.900 ","End":"06:39.274","Text":"Now, what we\u0027re going to do is we\u0027re going to rewrite our energy equation,"},{"Start":"06:39.274 ","End":"06:41.015","Text":"this one over here,"},{"Start":"06:41.015 ","End":"06:44.990","Text":"except instead of adding in our V_1,"},{"Start":"06:44.990 ","End":"06:46.935","Text":"so we\u0027ll put in our V_1 over here,"},{"Start":"06:46.935 ","End":"06:48.735","Text":"we\u0027ll substitute in all of this,"},{"Start":"06:48.735 ","End":"06:51.135","Text":"instead of having our V_2,"},{"Start":"06:51.135 ","End":"06:53.265","Text":"we\u0027ll substitute in our V_2 over here,"},{"Start":"06:53.265 ","End":"06:56.495","Text":"and our X_2 minus X_1 is going to be the easiest thing."},{"Start":"06:56.495 ","End":"07:00.220","Text":"It\u0027s simply our X relative."},{"Start":"07:00.230 ","End":"07:03.590","Text":"Let\u0027s see what this is going to look like."},{"Start":"07:03.590 ","End":"07:07.049","Text":"Now, I\u0027m just going to write out the equation now."},{"Start":"07:07.049 ","End":"07:11.920","Text":"After I\u0027ve already sorted it out and made it look a little bit"},{"Start":"07:11.920 ","End":"07:16.980","Text":"neater because you have to open some brackets and move stuff around."},{"Start":"07:16.980 ","End":"07:23.685","Text":"We\u0027re going to have 1/2 of m_1 plus m_2,"},{"Start":"07:23.685 ","End":"07:32.785","Text":"and all of this is going to be multiplied by V_CM^2 and then plus 1/2 of Mu."},{"Start":"07:32.785 ","End":"07:39.250","Text":"Soon we\u0027ll see what this is multiplied by V relative squared,"},{"Start":"07:39.250 ","End":"07:48.400","Text":"and then plus 1/2 of k X relative squared."},{"Start":"07:48.400 ","End":"07:50.050","Text":"So let\u0027s write this out."},{"Start":"07:50.050 ","End":"07:59.440","Text":"So 1 divided by Mu is going to be equal to 1 over m_1 plus 1 over m_2."},{"Start":"07:59.440 ","End":"08:01.120","Text":"It\u0027s the inverse of this,"},{"Start":"08:01.120 ","End":"08:02.845","Text":"that\u0027s our Mu over here."},{"Start":"08:02.845 ","End":"08:06.460","Text":"This is our equation where I\u0027ve substituted in my v_1,"},{"Start":"08:06.460 ","End":"08:11.410","Text":"v_2 and my X relative into my original equation for energy,"},{"Start":"08:11.410 ","End":"08:14.920","Text":"and I\u0027ve done some algebra to make it look slightly neater."},{"Start":"08:14.920 ","End":"08:19.270","Text":"Now, what I have over here is I have 1/2 of my total mass"},{"Start":"08:19.270 ","End":"08:24.445","Text":"multiplied by my velocity of the center of mass squared plus 1/2 of my Mu,"},{"Start":"08:24.445 ","End":"08:25.570","Text":"which we said it\u0027s over here,"},{"Start":"08:25.570 ","End":"08:28.120","Text":"which is also equal to this,"},{"Start":"08:28.120 ","End":"08:37.255","Text":"multiplied by the V relative squared plus 1/2 of my k multiplied by X relative squared."},{"Start":"08:37.255 ","End":"08:43.390","Text":"What I have here is everything is in addition and I don\u0027t have a mix up between terms,"},{"Start":"08:43.390 ","End":"08:47.635","Text":"so I don\u0027t have my V_CM multiplied by my V relative, or anything."},{"Start":"08:47.635 ","End":"08:50.755","Text":"Everything is plus, everything is separated."},{"Start":"08:50.755 ","End":"08:52.840","Text":"Now when things unseparated,"},{"Start":"08:52.840 ","End":"08:56.410","Text":"that\u0027s when things get confusing and very difficult to work out."},{"Start":"08:56.410 ","End":"08:58.855","Text":"So here I don\u0027t have this problem."},{"Start":"08:58.855 ","End":"09:02.650","Text":"Now I need to carry on solving."},{"Start":"09:02.650 ","End":"09:09.955","Text":"What I can say is that if no external forces are acting on this system over here,"},{"Start":"09:09.955 ","End":"09:18.580","Text":"then that will mean that my sum of my external forces is equal to 0."},{"Start":"09:18.580 ","End":"09:20.350","Text":"No force is pushing this,"},{"Start":"09:20.350 ","End":"09:22.360","Text":"my system is just moving, so yes,"},{"Start":"09:22.360 ","End":"09:25.180","Text":"my m_2 is going to stretch the spring a bit more to here,"},{"Start":"09:25.180 ","End":"09:27.520","Text":"my m_1 to there and back and whatever."},{"Start":"09:27.520 ","End":"09:32.185","Text":"But those are the inner forces happening within the system."},{"Start":"09:32.185 ","End":"09:38.050","Text":"Because notice we\u0027ve written our V_CM and we have our X_CM and everything,"},{"Start":"09:38.050 ","End":"09:40.405","Text":"so they\u0027re all internal forces."},{"Start":"09:40.405 ","End":"09:43.150","Text":"My external forces which is what I\u0027m looking at,"},{"Start":"09:43.150 ","End":"09:45.850","Text":"is going to be equal to 0."},{"Start":"09:45.850 ","End":"09:49.660","Text":"Now, if my external forces are equal to 0,"},{"Start":"09:49.660 ","End":"09:51.910","Text":"so as we know from Newton,"},{"Start":"09:51.910 ","End":"09:56.320","Text":"the sum of all of the forces is equal to mass times acceleration."},{"Start":"09:56.320 ","End":"09:59.650","Text":"If the sum of all my external forces is equal to 0,"},{"Start":"09:59.650 ","End":"10:06.100","Text":"then that means that my V center of mass is going to be,"},{"Start":"10:06.100 ","End":"10:09.970","Text":"so that means that I have mass multiplied by my acceleration."},{"Start":"10:09.970 ","End":"10:12.760","Text":"My acceleration is going to have to be equal to 0,"},{"Start":"10:12.760 ","End":"10:15.025","Text":"so if my acceleration is equal to 0,"},{"Start":"10:15.025 ","End":"10:19.825","Text":"that means that my velocity is going to be equal to some constant."},{"Start":"10:19.825 ","End":"10:26.320","Text":"Because I\u0027m reminding you, the first derivative of velocity is acceleration."},{"Start":"10:26.320 ","End":"10:28.315","Text":"If my velocity is a constant,"},{"Start":"10:28.315 ","End":"10:31.660","Text":"then my first derivative which has the acceleration will be equal to 0,"},{"Start":"10:31.660 ","End":"10:35.860","Text":"if that\u0027s equal to 0, the sum of all my external forces is equal to 0."},{"Start":"10:35.860 ","End":"10:40.885","Text":"That\u0027s perfect. We also learned in lesson that our V_CM,"},{"Start":"10:40.885 ","End":"10:46.750","Text":"or velocity of our center of mass moves only because of external forces."},{"Start":"10:46.750 ","End":"10:48.790","Text":"If there aren\u0027t external forces,"},{"Start":"10:48.790 ","End":"10:52.840","Text":"then our center of mass moves at a constant velocity."},{"Start":"10:52.840 ","End":"10:55.135","Text":"Now let\u0027s take a look."},{"Start":"10:55.135 ","End":"10:57.850","Text":"If we say that our V_CM is a constant,"},{"Start":"10:57.850 ","End":"11:04.240","Text":"then let\u0027s go back to our energy equation after we did our transforms with our variables."},{"Start":"11:04.240 ","End":"11:06.460","Text":"Then our V_CM is over here,"},{"Start":"11:06.460 ","End":"11:07.780","Text":"so a 1/2 is a constant,"},{"Start":"11:07.780 ","End":"11:09.070","Text":"m_1 and m_2 is a constant,"},{"Start":"11:09.070 ","End":"11:16.525","Text":"and our V_CM is a constant which means that this entire term over here is a constant."},{"Start":"11:16.525 ","End":"11:20.545","Text":"Now, we can look at our V relative."},{"Start":"11:20.545 ","End":"11:22.465","Text":"Now what is our V relative?"},{"Start":"11:22.465 ","End":"11:24.800","Text":"It\u0027s our X dot relative."},{"Start":"11:25.520 ","End":"11:28.860","Text":"In energy, we already saw that if we have plus"},{"Start":"11:28.860 ","End":"11:32.805","Text":"constant because then we take the derivative to get to our force,"},{"Start":"11:32.805 ","End":"11:36.045","Text":"so this constant doesn\u0027t really play a role."},{"Start":"11:36.045 ","End":"11:40.780","Text":"The important trick here is to show that"},{"Start":"11:40.780 ","End":"11:45.790","Text":"this extra term inside our equation is a constant,"},{"Start":"11:45.790 ","End":"11:49.075","Text":"in which case we can just ignore it in our calculations."},{"Start":"11:49.075 ","End":"11:53.545","Text":"It\u0027s very important we have to show that it\u0027s constant and it has to be a constant."},{"Start":"11:53.545 ","End":"11:56.650","Text":"Then going back to our equation that we wanted it to look like,"},{"Start":"11:56.650 ","End":"12:01.135","Text":"so we have to have a variable dot squared."},{"Start":"12:01.135 ","End":"12:04.735","Text":"Here we have a variable dot squared plus"},{"Start":"12:04.735 ","End":"12:08.950","Text":"our variable squared and then here we have our variable squared over here."},{"Start":"12:08.950 ","End":"12:10.870","Text":"X dot relative squared,"},{"Start":"12:10.870 ","End":"12:14.275","Text":"and now our X relative squared, so perfect."},{"Start":"12:14.275 ","End":"12:20.315","Text":"Now we have our energy equation in the exact form that we need for harmonic motion."},{"Start":"12:20.315 ","End":"12:22.485","Text":"Then as we know,"},{"Start":"12:22.485 ","End":"12:24.480","Text":"we can just write that"},{"Start":"12:24.480 ","End":"12:31.060","Text":"our Omega which is our frequency of oscillation which is what our aim is to find."},{"Start":"12:31.060 ","End":"12:35.065","Text":"As we know, it\u0027s going to be the square root of"},{"Start":"12:35.065 ","End":"12:41.080","Text":"our coefficient of our X relative over here,"},{"Start":"12:41.080 ","End":"12:42.370","Text":"so our X relative squared,"},{"Start":"12:42.370 ","End":"12:46.645","Text":"its coefficient is k divided by"},{"Start":"12:46.645 ","End":"12:52.870","Text":"our coefficient of our X relative dot which over here is Mu."},{"Start":"12:52.870 ","End":"12:56.395","Text":"This coefficient divided by this coefficient."},{"Start":"12:56.395 ","End":"13:00.070","Text":"Yes, we also have the halves here but they will cancel out."},{"Start":"13:00.070 ","End":"13:03.025","Text":"Then we just take the square root of that."},{"Start":"13:03.025 ","End":"13:06.610","Text":"Our Mu again is something that we know. It\u0027s over here."},{"Start":"13:06.610 ","End":"13:12.100","Text":"Before we finish, this Mu is not something that I just made up for ease of use,"},{"Start":"13:12.100 ","End":"13:15.295","Text":"it\u0027s actually something that we use a lot in physics,"},{"Start":"13:15.295 ","End":"13:18.400","Text":"and it\u0027s very useful to know,"},{"Start":"13:18.400 ","End":"13:20.260","Text":"and it actually has a name."},{"Start":"13:20.260 ","End":"13:24.310","Text":"It\u0027s called the reduced mass."},{"Start":"13:24.310 ","End":"13:26.455","Text":"This will come a lot."},{"Start":"13:26.455 ","End":"13:30.025","Text":"It\u0027s on the Internet, you can look it up and it"},{"Start":"13:30.025 ","End":"13:33.520","Text":"just means that when we have our 2 variables, or 2 masses,"},{"Start":"13:33.520 ","End":"13:38.080","Text":"2 bodies, then what we can do is find the reduced mass of the 2 of them,"},{"Start":"13:38.080 ","End":"13:40.870","Text":"and then it\u0027s like we\u0027ve moved to 1 variable."},{"Start":"13:40.870 ","End":"13:43.330","Text":"Then instead of having our m over here,"},{"Start":"13:43.330 ","End":"13:44.845","Text":"we have our Mu."},{"Start":"13:44.845 ","End":"13:46.435","Text":"It\u0027s still a mass,"},{"Start":"13:46.435 ","End":"13:48.235","Text":"but it\u0027s a reduced 1."},{"Start":"13:48.235 ","End":"13:50.800","Text":"Now we have a constant."},{"Start":"13:50.800 ","End":"13:55.105","Text":"Again, I\u0027m reminding you this is very important that this term is a constant,"},{"Start":"13:55.105 ","End":"13:56.890","Text":"and to show that it\u0027s constant."},{"Start":"13:56.890 ","End":"14:00.655","Text":"Then we have the exact same terms that we wanted."},{"Start":"14:00.655 ","End":"14:07.390","Text":"The trick for working out these types of questions is to take a look at your system."},{"Start":"14:07.390 ","End":"14:10.945","Text":"Notice that that\u0027s a bit more complicated because there\u0027s 2 variables,"},{"Start":"14:10.945 ","End":"14:15.595","Text":"and then to just start off by writing the energy of the system."},{"Start":"14:15.595 ","End":"14:19.820","Text":"Then when you see that really there are 2 variables and it\u0027s complicated,"},{"Start":"14:19.820 ","End":"14:23.225","Text":"remind yourself of what you want the equation to look like."},{"Start":"14:23.225 ","End":"14:27.050","Text":"Then all you have to do is you have to write out your X_CM,"},{"Start":"14:27.050 ","End":"14:29.300","Text":"which you should know already how to do."},{"Start":"14:29.300 ","End":"14:32.420","Text":"Your X relative, which is simply going to be the length of the spring,"},{"Start":"14:32.420 ","End":"14:38.630","Text":"and then take the first derivative of each term to find your V_CM and your V relative."},{"Start":"14:38.630 ","End":"14:43.205","Text":"Then what you have to do is either write out"},{"Start":"14:43.205 ","End":"14:48.605","Text":"this opposite transform to get from your X_CM and your X relative to your x_1 and x_2,"},{"Start":"14:48.605 ","End":"14:51.770","Text":"so mapping backwards, or you can just do it in"},{"Start":"14:51.770 ","End":"14:55.640","Text":"the exam via algebra if you\u0027re super quick at it anyway."},{"Start":"14:55.640 ","End":"14:57.560","Text":"But you can also write this if you have space in"},{"Start":"14:57.560 ","End":"15:01.550","Text":"your paper that you can take with you into the exam,"},{"Start":"15:01.550 ","End":"15:03.560","Text":"and then you just want to write this out."},{"Start":"15:03.560 ","End":"15:06.530","Text":"Then all you want to do is you want to switch in your v_1,"},{"Start":"15:06.530 ","End":"15:08.600","Text":"v_2 in your X relative into"},{"Start":"15:08.600 ","End":"15:13.100","Text":"your original equation and then do a little bit of algebra in order to get this."},{"Start":"15:13.100 ","End":"15:16.805","Text":"Then you\u0027ll see that you have your Mu, your reduce mass."},{"Start":"15:16.805 ","End":"15:19.580","Text":"Remember that because it might make it a little bit"},{"Start":"15:19.580 ","End":"15:22.715","Text":"easier to sort out this equation over here."},{"Start":"15:22.715 ","End":"15:25.280","Text":"Then what you want to do is you want to say that the sum"},{"Start":"15:25.280 ","End":"15:27.875","Text":"of all of the external forces is equal to 0."},{"Start":"15:27.875 ","End":"15:30.889","Text":"If so, then your V_CM is constant,"},{"Start":"15:30.889 ","End":"15:33.140","Text":"which is super important."},{"Start":"15:33.140 ","End":"15:38.785","Text":"Please do not forget, do remember this."},{"Start":"15:38.785 ","End":"15:42.230","Text":"This over here it\u0027s very important. Do remember it."},{"Start":"15:42.230 ","End":"15:44.360","Text":"Then if your V_CM is constant,"},{"Start":"15:44.360 ","End":"15:49.010","Text":"that means that this term over here is going to be a constant,"},{"Start":"15:49.010 ","End":"15:54.330","Text":"which means that perfect you\u0027ve arrived at your destination and"},{"Start":"15:54.330 ","End":"15:59.870","Text":"then your Omega is just the coefficients divided by each other and the square root."},{"Start":"15:59.870 ","End":"16:02.030","Text":"Now, obviously if I had"},{"Start":"16:02.030 ","End":"16:07.030","Text":"some other questions under this asking for my position as a function of time,"},{"Start":"16:07.030 ","End":"16:10.795","Text":"so it works like any harmonic equation,"},{"Start":"16:10.795 ","End":"16:16.259","Text":"so I can see that my variable is my X relative,"},{"Start":"16:16.259 ","End":"16:23.835","Text":"so my X relative as a function of time is simply going to be the general solution,"},{"Start":"16:23.835 ","End":"16:29.470","Text":"A cosine of Omega t plus Phi,"},{"Start":"16:29.470 ","End":"16:31.525","Text":"where Omega we\u0027ve already found,"},{"Start":"16:31.525 ","End":"16:33.665","Text":"and my A and my Phi as per usual,"},{"Start":"16:33.665 ","End":"16:36.410","Text":"I find from my initial conditions."},{"Start":"16:36.410 ","End":"16:42.935","Text":"If they ask for my X_CM as a function of time so my center of mass as a function of time,"},{"Start":"16:42.935 ","End":"16:45.395","Text":"because we saw that my V_CM is constant,"},{"Start":"16:45.395 ","End":"16:52.460","Text":"so I can just write that my X_CM as a function of time is going to be equal to my V_CM"},{"Start":"16:52.460 ","End":"16:54.830","Text":"multiplied by t. You\u0027re on"},{"Start":"16:54.830 ","End":"16:59.630","Text":"your normal equation for your position and depending on where we started,"},{"Start":"16:59.630 ","End":"17:05.125","Text":"I can have plus X0, it\u0027s starting position."},{"Start":"17:05.125 ","End":"17:09.695","Text":"Now that I have my X relative and my X_CM as a function of time,"},{"Start":"17:09.695 ","End":"17:11.585","Text":"I can know at any time where they are."},{"Start":"17:11.585 ","End":"17:15.110","Text":"If they go back to the original question and"},{"Start":"17:15.110 ","End":"17:18.400","Text":"they want to know where my m_1 is at any period of time,"},{"Start":"17:18.400 ","End":"17:20.320","Text":"or my m_2 at any period of time,"},{"Start":"17:20.320 ","End":"17:24.480","Text":"so I substitute in my X relative over here,"},{"Start":"17:24.480 ","End":"17:30.620","Text":"and I can find my x_2 and x_1 from my m_1 and m_2 and same with my V_CM,"},{"Start":"17:30.620 ","End":"17:33.145","Text":"I can also substitute it in here."},{"Start":"17:33.145 ","End":"17:38.770","Text":"Then I can find these positions as a function of time of each individual mass."},{"Start":"17:38.770 ","End":"17:41.450","Text":"Sorry, you wouldn\u0027t substitute it into here."},{"Start":"17:41.450 ","End":"17:46.249","Text":"You would substitute our X relative and our X_CM into"},{"Start":"17:46.249 ","End":"17:51.690","Text":"here and then you\u0027ll get your x_1 and x_2 for each mass."},{"Start":"17:51.690 ","End":"17:53.645","Text":"That\u0027s the end of this example."},{"Start":"17:53.645 ","End":"17:55.760","Text":"I hope it\u0027s been clear what to do,"},{"Start":"17:55.760 ","End":"17:58.370","Text":"and further questions that they can ask,"},{"Start":"17:58.370 ","End":"18:01.520","Text":"and how to solve it by what you\u0027ve already done."},{"Start":"18:01.520 ","End":"18:04.475","Text":"But up until now, we\u0027ve only been working"},{"Start":"18:04.475 ","End":"18:09.890","Text":"in 1 dimension where our masses are moving up and only on the x-axis."},{"Start":"18:09.890 ","End":"18:13.220","Text":"In the next question, I\u0027m going to speak about the masses"},{"Start":"18:13.220 ","End":"18:17.345","Text":"moving not just on the x-axis but in 2D."},{"Start":"18:17.345 ","End":"18:19.340","Text":"But you\u0027ll see it\u0027s not very complicated."},{"Start":"18:19.340 ","End":"18:22.205","Text":"It\u0027s pretty similar to what we have over here."},{"Start":"18:22.205 ","End":"18:24.930","Text":"That\u0027s the end of this question."}],"ID":10845},{"Watched":false,"Name":"Double Masses Multiple Dimensions","Duration":"7m 27s","ChapterTopicVideoID":9150,"CourseChapterTopicPlaylistID":5368,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.830","Text":"Hello. In the last lesson,"},{"Start":"00:01.830 ","End":"00:05.175","Text":"we were speaking about two masses that were attached by a string,"},{"Start":"00:05.175 ","End":"00:08.759","Text":"which we\u0027re only moving along the x-axis in one dimension."},{"Start":"00:08.759 ","End":"00:10.200","Text":"Now, in this lesson,"},{"Start":"00:10.200 ","End":"00:12.840","Text":"what we\u0027re going to be speaking about is what happens if we"},{"Start":"00:12.840 ","End":"00:16.260","Text":"have these two masses and they\u0027re moving in space,"},{"Start":"00:16.260 ","End":"00:18.075","Text":"in 2-dimensional space,"},{"Start":"00:18.075 ","End":"00:23.025","Text":"with some x coordinate and also some y coordinate."},{"Start":"00:23.025 ","End":"00:26.293","Text":"What we do is we describe their position,"},{"Start":"00:26.293 ","End":"00:28.275","Text":"instead of by x_1 and x_2,"},{"Start":"00:28.275 ","End":"00:32.520","Text":"we describe them by r vectors, r_1,"},{"Start":"00:32.520 ","End":"00:37.395","Text":"and r_2 going from the origin until each mass."},{"Start":"00:37.395 ","End":"00:45.195","Text":"Then, all that we\u0027ll have to change is our v_1 and our v_2 will now be vectors."},{"Start":"00:45.195 ","End":"00:47.055","Text":"Then now vector quantities,"},{"Start":"00:47.055 ","End":"00:48.750","Text":"because we\u0027re moving in space,"},{"Start":"00:48.750 ","End":"00:50.970","Text":"and then defined by these,"},{"Start":"00:50.970 ","End":"00:53.685","Text":"and instead of x_2 minus x_1,"},{"Start":"00:53.685 ","End":"01:01.380","Text":"we will have r_2 vector minus r_1 vector."},{"Start":"01:01.380 ","End":"01:04.950","Text":"When we take a look at r_2 minus r_1,"},{"Start":"01:04.950 ","End":"01:08.165","Text":"then if we go over here,"},{"Start":"01:08.165 ","End":"01:12.065","Text":"we\u0027ll see that this is exactly in terms of vectors."},{"Start":"01:12.065 ","End":"01:14.135","Text":"This distance over here,"},{"Start":"01:14.135 ","End":"01:16.700","Text":"which is our r relative,"},{"Start":"01:16.700 ","End":"01:18.565","Text":"and this is also a vector."},{"Start":"01:18.565 ","End":"01:22.535","Text":"Now looking at our harmonic equation when dealing with energy,"},{"Start":"01:22.535 ","End":"01:28.025","Text":"we still want it to look like this and not to be a vector with a certain dimensions."},{"Start":"01:28.025 ","End":"01:30.620","Text":"Soon we\u0027re going to take a look at this,"},{"Start":"01:30.620 ","End":"01:31.960","Text":"so we\u0027ll come back here."},{"Start":"01:31.960 ","End":"01:35.200","Text":"In the meantime, let\u0027s look at our variable substitution."},{"Start":"01:35.200 ","End":"01:39.485","Text":"All I have to do is change over here everything into vectors."},{"Start":"01:39.485 ","End":"01:42.860","Text":"I\u0027m looking at every single axis alone."},{"Start":"01:42.860 ","End":"01:45.865","Text":"Instead of having my x_cm,"},{"Start":"01:45.865 ","End":"01:49.920","Text":"I\u0027m going to have my r_cm,"},{"Start":"01:49.920 ","End":"01:51.885","Text":"and this is of course can be a vector."},{"Start":"01:51.885 ","End":"01:54.255","Text":"Then instead of my x_1,"},{"Start":"01:54.255 ","End":"01:56.220","Text":"I\u0027ll have my r_1,"},{"Start":"01:56.220 ","End":"01:57.930","Text":"instead of my x_2,"},{"Start":"01:57.930 ","End":"02:01.245","Text":"I\u0027ll have my r_2 vector."},{"Start":"02:01.245 ","End":"02:04.170","Text":"The same here instead of x relative,"},{"Start":"02:04.170 ","End":"02:07.245","Text":"I\u0027ll have my r_relative,"},{"Start":"02:07.245 ","End":"02:11.250","Text":"and then instead of my x_2 minus x_1, we\u0027ll have what I had there."},{"Start":"02:11.250 ","End":"02:16.315","Text":"I\u0027ll have my r_2 minus my r_1 vectors."},{"Start":"02:16.315 ","End":"02:18.410","Text":"Then to do with our velocity,"},{"Start":"02:18.410 ","End":"02:20.584","Text":"so they just changed to vectors."},{"Start":"02:20.584 ","End":"02:25.360","Text":"All I have to add on my arrows on top over here."},{"Start":"02:25.360 ","End":"02:27.345","Text":"It\u0027s as simple as that."},{"Start":"02:27.345 ","End":"02:32.305","Text":"Now also when we\u0027re looking at our opposite transformation,"},{"Start":"02:32.305 ","End":"02:38.060","Text":"so everything over here changes over to vectors similarly like so."},{"Start":"02:38.060 ","End":"02:40.340","Text":"Let\u0027s get to our actual equation."},{"Start":"02:40.340 ","End":"02:42.070","Text":"Let\u0027s take a look at this."},{"Start":"02:42.070 ","End":"02:44.700","Text":"Looking at this equation over here,"},{"Start":"02:44.700 ","End":"02:48.300","Text":"my v_cm is just going to have a vector,"},{"Start":"02:48.300 ","End":"02:50.325","Text":"this is just going be a vector,"},{"Start":"02:50.325 ","End":"02:52.740","Text":"same with my v_relative,"},{"Start":"02:52.740 ","End":"02:55.110","Text":"is also going to be a vector quantity."},{"Start":"02:55.110 ","End":"02:58.875","Text":"Now instead of my x^2_relative,"},{"Start":"02:58.875 ","End":"03:02.910","Text":"I\u0027m going to have my r_relative."},{"Start":"03:02.910 ","End":"03:05.235","Text":"This is also a vector."},{"Start":"03:05.235 ","End":"03:09.855","Text":"Now what we want to do, is we want to look at my r_relative."},{"Start":"03:09.855 ","End":"03:15.605","Text":"It\u0027s a vector quantity and it\u0027s squared. What does that mean?"},{"Start":"03:15.605 ","End":"03:20.265","Text":"That\u0027s going to mean that this whole expression over here,"},{"Start":"03:20.265 ","End":"03:24.990","Text":"1/2 kr_relative squared is going to turn into the size."},{"Start":"03:24.990 ","End":"03:31.250","Text":"That\u0027s going to be the size of this distance between my m_1 and"},{"Start":"03:31.250 ","End":"03:40.170","Text":"my m_2 multiplied by 1/2 k. That means that we can call this distance over here,"},{"Start":"03:40.170 ","End":"03:44.895","Text":"we can just call it r_relative without the vector sign,"},{"Start":"03:44.895 ","End":"03:46.385","Text":"and in that case,"},{"Start":"03:46.385 ","End":"03:49.744","Text":"what we can do is we can simply erase"},{"Start":"03:49.744 ","End":"03:56.495","Text":"a vector sign over there and just leave it as r_relative because it\u0027s a size."},{"Start":"03:56.495 ","End":"04:00.400","Text":"It\u0027s just the distance between these two."},{"Start":"04:00.400 ","End":"04:03.850","Text":"Now we can look at our v_relative and we said that"},{"Start":"04:03.850 ","End":"04:07.075","Text":"it\u0027s a vector and we can see that it\u0027s a vector squared."},{"Start":"04:07.075 ","End":"04:08.905","Text":"That means that again,"},{"Start":"04:08.905 ","End":"04:10.810","Text":"we\u0027re dealing with size."},{"Start":"04:10.810 ","End":"04:13.390","Text":"Let\u0027s write it here so that it\u0027s more clear."},{"Start":"04:13.390 ","End":"04:17.965","Text":"A v_relative squared is the same"},{"Start":"04:17.965 ","End":"04:23.995","Text":"as the size of Iv_relative vector."},{"Start":"04:23.995 ","End":"04:25.690","Text":"Now, what does that mean?"},{"Start":"04:25.690 ","End":"04:30.740","Text":"It\u0027s the change between these two."},{"Start":"04:30.740 ","End":"04:33.940","Text":"The change of our r_relative,"},{"Start":"04:33.940 ","End":"04:36.610","Text":"sorry, this size over here,"},{"Start":"04:36.610 ","End":"04:45.525","Text":"so that means that we can rewrite this simply as r_relative dot."},{"Start":"04:45.525 ","End":"04:49.640","Text":"Now instead of writing v vector relative squared,"},{"Start":"04:49.640 ","End":"04:54.440","Text":"so I can change it over here to simply being,"},{"Start":"04:54.440 ","End":"05:01.213","Text":"so we\u0027ll have 1/2 Mu multiplied by r_relative dot,"},{"Start":"05:01.213 ","End":"05:04.360","Text":"get the first derivative, squared."},{"Start":"05:04.360 ","End":"05:06.805","Text":"Then finally we have our v_cm."},{"Start":"05:06.805 ","End":"05:10.145","Text":"Now, our v_cm is still going to be some kind of constant,"},{"Start":"05:10.145 ","End":"05:13.040","Text":"but because it\u0027s a vector,"},{"Start":"05:13.040 ","End":"05:15.145","Text":"it\u0027s telling us in which direction."},{"Start":"05:15.145 ","End":"05:17.645","Text":"But it\u0027s still a constant value,"},{"Start":"05:17.645 ","End":"05:22.025","Text":"which is what we said was very important and when we\u0027re dealing with energy anyway,"},{"Start":"05:22.025 ","End":"05:25.385","Text":"usually the direction doesn\u0027t really matter."},{"Start":"05:25.385 ","End":"05:27.605","Text":"Here this is a constant,"},{"Start":"05:27.605 ","End":"05:30.680","Text":"and now we\u0027re left with the exact same equation that we had before."},{"Start":"05:30.680 ","End":"05:37.820","Text":"Except this time instead of having our x_relative as our variable,"},{"Start":"05:37.820 ","End":"05:39.840","Text":"we have still 1 variable,"},{"Start":"05:39.840 ","End":"05:42.370","Text":"but it\u0027s now r_relative."},{"Start":"05:42.620 ","End":"05:45.995","Text":"r_relative and r_relative dot."},{"Start":"05:45.995 ","End":"05:48.950","Text":"That is exactly like this equation here,"},{"Start":"05:48.950 ","End":"05:51.445","Text":"which is what we wanted in the first place."},{"Start":"05:51.445 ","End":"05:58.220","Text":"Now, even though our r_relative is a vector and here specifically we wrote it in the x,"},{"Start":"05:58.220 ","End":"06:01.550","Text":"y plane, but it can also be an x, y, and z."},{"Start":"06:01.550 ","End":"06:04.380","Text":"Although it\u0027s potentially x,"},{"Start":"06:04.380 ","End":"06:06.375","Text":"y, and z in 3-dimensions,"},{"Start":"06:06.375 ","End":"06:09.270","Text":"we can still write it as just 1 single variable,"},{"Start":"06:09.270 ","End":"06:11.160","Text":"r_rel and r_rel dot."},{"Start":"06:11.160 ","End":"06:14.080","Text":"All of the energy is, in actual fact,"},{"Start":"06:14.080 ","End":"06:18.185","Text":"dependent on the distance between our 2 masses,"},{"Start":"06:18.185 ","End":"06:25.780","Text":"r_relative, and how this distance changes with regards to time."},{"Start":"06:25.780 ","End":"06:32.155","Text":"This is correct, even if the direction of this change is constantly changing."},{"Start":"06:32.155 ","End":"06:36.890","Text":"It\u0027s the exact same thing as what we had when we were working in 1-dimension."},{"Start":"06:36.890 ","End":"06:42.530","Text":"Just our x\u0027s have changed to our distance between the bodies."},{"Start":"06:42.530 ","End":"06:49.115","Text":"Then of course, instead of having our position with x_relative as a function of time,"},{"Start":"06:49.115 ","End":"06:52.460","Text":"now it\u0027s going to be with r_relative and the exact same thing,"},{"Start":"06:52.460 ","End":"06:57.200","Text":"so we have our r_relative with respect to time,"},{"Start":"06:57.200 ","End":"06:59.105","Text":"and it\u0027s going to be the exact same thing."},{"Start":"06:59.105 ","End":"07:05.730","Text":"A cosine of Omega t plus Phi,"},{"Start":"07:05.730 ","End":"07:08.550","Text":"where r Omega is the same Omega."},{"Start":"07:08.550 ","End":"07:14.395","Text":"We just still are taking the coefficients and our Phi and our A,"},{"Start":"07:14.395 ","End":"07:18.140","Text":"just like before, we\u0027ll work it out via initial conditions."},{"Start":"07:18.140 ","End":"07:21.740","Text":"As you can see, when working with multiple dimensions,"},{"Start":"07:21.740 ","End":"07:25.730","Text":"it\u0027s going to be pretty much the exact same deal."},{"Start":"07:25.730 ","End":"07:28.500","Text":"That\u0027s the end of the lesson."}],"ID":9420}],"Thumbnail":null,"ID":5368},{"Name":"Exercises","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Two Masses and a Spring","Duration":"17m 22s","ChapterTopicVideoID":9164,"CourseChapterTopicPlaylistID":5369,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.860 ","End":"00:04.304","Text":"Hello. In this question,"},{"Start":"00:04.304 ","End":"00:07.320","Text":"we are given a system below,"},{"Start":"00:07.320 ","End":"00:11.070","Text":"and we\u0027re being asked what is the frequency of the system."},{"Start":"00:11.070 ","End":"00:14.385","Text":"Now, we know what m is,"},{"Start":"00:14.385 ","End":"00:15.990","Text":"we know what M is,"},{"Start":"00:15.990 ","End":"00:19.770","Text":"and we know what the spring constant here is, it\u0027s."},{"Start":"00:19.770 ","End":"00:23.705","Text":"Let\u0027s see when we know all of this information,"},{"Start":"00:23.705 ","End":"00:28.865","Text":"given that this mass is attached to a spring of constant k,"},{"Start":"00:28.865 ","End":"00:33.125","Text":"and it\u0027s also attached by a string to this M,"},{"Start":"00:33.125 ","End":"00:37.070","Text":"let\u0027s find out what the frequency is of this system."},{"Start":"00:37.070 ","End":"00:43.670","Text":"Now the first thing I\u0027m going to do is I\u0027m going to mark where my spring is at rest,"},{"Start":"00:43.670 ","End":"00:46.805","Text":"where it\u0027s not being compressed and it\u0027s not being stretched,"},{"Start":"00:46.805 ","End":"00:50.525","Text":"and that\u0027s going to be at this point, at 0."},{"Start":"00:50.525 ","End":"00:54.965","Text":"My next point that I\u0027m going to mark is this x_0."},{"Start":"00:54.965 ","End":"00:59.885","Text":"Now this x_0 arrow marks my point of equilibrium. Now what does that mean?"},{"Start":"00:59.885 ","End":"01:02.555","Text":"It means that the force of the spring,"},{"Start":"01:02.555 ","End":"01:08.405","Text":"which is pulling this whole system backup is equal and opposite to"},{"Start":"01:08.405 ","End":"01:11.360","Text":"the forces that these masses or"},{"Start":"01:11.360 ","End":"01:16.565","Text":"this massive least is providing to stretch the spring more,"},{"Start":"01:16.565 ","End":"01:19.060","Text":"so that\u0027s the point of equilibrium."},{"Start":"01:19.060 ","End":"01:24.205","Text":"Then the third thing I mark is my x as a function of t,"},{"Start":"01:24.205 ","End":"01:26.075","Text":"x is a function of time,"},{"Start":"01:26.075 ","End":"01:32.540","Text":"and that represents the location that my mass is at any point,"},{"Start":"01:32.540 ","End":"01:35.910","Text":"so this is my variable that I\u0027m trying to find."},{"Start":"01:35.910 ","End":"01:39.515","Text":"Now I have to choose my axis."},{"Start":"01:39.515 ","End":"01:45.715","Text":"As we can see, this system is going in this sideways L-shape,"},{"Start":"01:45.715 ","End":"01:48.220","Text":"so it\u0027s convenient for me to draw"},{"Start":"01:48.220 ","End":"01:53.734","Text":"my axis going in this direction being the positive direction."},{"Start":"01:53.734 ","End":"01:55.360","Text":"Now what can I do this?"},{"Start":"01:55.360 ","End":"01:57.670","Text":"Because all of my buddies will be moving in"},{"Start":"01:57.670 ","End":"02:02.600","Text":"the same direction if this mass goes down and this will go in that way."},{"Start":"02:03.450 ","End":"02:11.260","Text":"Now what I need to do is imagine that I stretch my spring,"},{"Start":"02:11.260 ","End":"02:16.825","Text":"so I pull this mass some distance x_t away,"},{"Start":"02:16.825 ","End":"02:18.830","Text":"and then I release it."},{"Start":"02:18.830 ","End":"02:22.535","Text":"Then I want to see what my frequency is,"},{"Start":"02:22.535 ","End":"02:24.290","Text":"what is going on in the system."},{"Start":"02:24.290 ","End":"02:27.230","Text":"The first thing I\u0027m going to have to do now is"},{"Start":"02:27.230 ","End":"02:30.660","Text":"to figure out which forces are acting on my system,"},{"Start":"02:30.660 ","End":"02:34.115","Text":"and then once we\u0027ve found all the forces acting on the system,"},{"Start":"02:34.115 ","End":"02:39.170","Text":"then we can rearrange our equation into the format of a harmonic oscillator."},{"Start":"02:39.170 ","End":"02:44.760","Text":"Let\u0027s draw a diagram of small m,"},{"Start":"02:44.760 ","End":"02:50.240","Text":"and then of capital M. Which forces are acting on my small m. From this rope,"},{"Start":"02:50.240 ","End":"02:58.859","Text":"I have T going in that direction and aside from that, I have the spring."},{"Start":"02:58.859 ","End":"03:02.140","Text":"Now the spring is a restorative for us."},{"Start":"03:02.140 ","End":"03:05.660","Text":"If I stretch the spring this way,"},{"Start":"03:05.660 ","End":"03:10.775","Text":"it\u0027s going to be trying to restore and pull my mass in this direction."},{"Start":"03:10.775 ","End":"03:17.734","Text":"In that case, I\u0027m going to draw my force of the spring in this direction,"},{"Start":"03:17.734 ","End":"03:21.635","Text":"but add a negative sign in front of it,"},{"Start":"03:21.635 ","End":"03:25.400","Text":"like we\u0027ve spoken about in previous exercises."},{"Start":"03:25.400 ","End":"03:28.730","Text":"I always draw the arrow in"},{"Start":"03:28.730 ","End":"03:34.025","Text":"the opposite direction to where the spring will be trying to restore itself,"},{"Start":"03:34.025 ","End":"03:37.990","Text":"but with a negative coefficient."},{"Start":"03:37.990 ","End":"03:49.160","Text":"Then I can write that the sum of the forces on m is going to equal negative kx plus T,"},{"Start":"03:49.160 ","End":"03:52.655","Text":"because we\u0027re going in the direction of the arrows,"},{"Start":"03:52.655 ","End":"03:58.730","Text":"which will equal mx double dot mass times acceleration."},{"Start":"03:58.730 ","End":"04:03.025","Text":"Now let\u0027s take a look at our mass capital M,"},{"Start":"04:03.025 ","End":"04:08.615","Text":"so there we go and then going in this direction on this rope,"},{"Start":"04:08.615 ","End":"04:11.900","Text":"we have T again,"},{"Start":"04:11.900 ","End":"04:18.810","Text":"and then going downwards we have mg."},{"Start":"04:19.370 ","End":"04:22.890","Text":"They\u0027re both pointing in the opposite direction,"},{"Start":"04:22.890 ","End":"04:25.174","Text":"so I can say that the sum of my forces,"},{"Start":"04:25.174 ","End":"04:28.370","Text":"because we\u0027ve said that this direction,"},{"Start":"04:28.370 ","End":"04:32.975","Text":"the positive direction, so my mg is in the positive meaning my T is in the negative."},{"Start":"04:32.975 ","End":"04:37.835","Text":"I\u0027m going to have mg negative T is"},{"Start":"04:37.835 ","End":"04:44.030","Text":"equal to mx double dot mass times acceleration."},{"Start":"04:44.030 ","End":"04:45.905","Text":"Now a note here,"},{"Start":"04:45.905 ","End":"04:48.995","Text":"I know that these 2 accelerations are the same."},{"Start":"04:48.995 ","End":"04:55.485","Text":"Why do I know that? Because my whole system is on the same axes."},{"Start":"04:55.485 ","End":"05:00.920","Text":"If this small m moves 1 centimeter to the right,"},{"Start":"05:00.920 ","End":"05:05.630","Text":"then this will move 1 centimeter on the same axis,"},{"Start":"05:05.630 ","End":"05:08.960","Text":"so downwards and that\u0027s why"},{"Start":"05:08.960 ","End":"05:12.675","Text":"I can say that the accelerations are the same and I can mark them as the same."},{"Start":"05:12.675 ","End":"05:17.970","Text":"Now what I can do is I know that my T is unknown,"},{"Start":"05:17.970 ","End":"05:19.455","Text":"so I want to cancel them out."},{"Start":"05:19.455 ","End":"05:23.570","Text":"If I do this equation plus this equation,"},{"Start":"05:23.570 ","End":"05:27.320","Text":"it will cancel out my T\u0027 s because then I\u0027ll have plus T and then I\u0027ll have minus"},{"Start":"05:27.320 ","End":"05:32.195","Text":"T. Let\u0027s do that so I\u0027ll have this plus this."},{"Start":"05:32.195 ","End":"05:41.995","Text":"I have mg minus kx, sorry,"},{"Start":"05:41.995 ","End":"05:44.820","Text":"and then I have plus T minus T,"},{"Start":"05:44.820 ","End":"05:54.540","Text":"so that cancels out equals small m plus capital M multiplied by x double-dot."},{"Start":"05:54.540 ","End":"05:57.735","Text":"Now I want to get rid of this section."},{"Start":"05:57.735 ","End":"05:59.390","Text":"Sorry, this is an x,"},{"Start":"05:59.390 ","End":"06:02.570","Text":"I want to get rid of this mass on"},{"Start":"06:02.570 ","End":"06:06.090","Text":"this side to find the relationship between x and x double dot."},{"Start":"06:06.090 ","End":"06:09.305","Text":"I\u0027ll divide both sides by m and capital M,"},{"Start":"06:09.305 ","End":"06:14.430","Text":"so I get mg over plus M"},{"Start":"06:14.430 ","End":"06:19.205","Text":"negative k over m"},{"Start":"06:19.205 ","End":"06:25.175","Text":"plus M x equals x double dot."},{"Start":"06:25.175 ","End":"06:28.550","Text":"Now, this is great for getting there,"},{"Start":"06:28.550 ","End":"06:30.590","Text":"but this isn\u0027t the format that we want."},{"Start":"06:30.590 ","End":"06:34.790","Text":"I\u0027m reminding you in blue that"},{"Start":"06:34.790 ","End":"06:39.605","Text":"the format that we want is we want x double dot is equal to"},{"Start":"06:39.605 ","End":"06:49.635","Text":"negative something multiplied by x minus something else."},{"Start":"06:49.635 ","End":"06:57.935","Text":"We need this to be in this format where what we have over here is our Omega squared,"},{"Start":"06:57.935 ","End":"06:59.500","Text":"our frequency squared,"},{"Start":"06:59.500 ","End":"07:06.390","Text":"and what we have over here is is I point of equilibrium,"},{"Start":"07:06.390 ","End":"07:10.305","Text":"so I\u0027ll label this x_0."},{"Start":"07:10.305 ","End":"07:13.355","Text":"Now let\u0027s take a look at how we can do this,"},{"Start":"07:13.355 ","End":"07:17.170","Text":"how we can rearrange this into this format."},{"Start":"07:17.170 ","End":"07:21.660","Text":"This bit is actually super easy we always know,"},{"Start":"07:21.660 ","End":"07:26.085","Text":"let\u0027s start this x double dot is equal to negative,"},{"Start":"07:26.085 ","End":"07:28.925","Text":"and here we put in the frequency squared."},{"Start":"07:28.925 ","End":"07:30.590","Text":"What is the frequency squared?"},{"Start":"07:30.590 ","End":"07:33.800","Text":"It\u0027s always going to be my coefficient of the x,"},{"Start":"07:33.800 ","End":"07:36.110","Text":"now I\u0027ve already put the minus outside,"},{"Start":"07:36.110 ","End":"07:37.820","Text":"remember this minus always has to be"},{"Start":"07:37.820 ","End":"07:43.175","Text":"outside just because that\u0027s the accepted way to write it."},{"Start":"07:43.175 ","End":"07:49.575","Text":"Then I just put in k divided by m plus m,"},{"Start":"07:49.575 ","End":"07:54.755","Text":"so the frequency squared is always my negative coefficient of the x,"},{"Start":"07:54.755 ","End":"07:58.180","Text":"and I just put my negative sign on the outside,"},{"Start":"07:58.180 ","End":"08:03.997","Text":"next brackets, x minus something."},{"Start":"08:03.997 ","End":"08:07.210","Text":"What is this something? Let\u0027s see."},{"Start":"08:07.210 ","End":"08:13.255","Text":"This needs to multiply by something in order to equal this."},{"Start":"08:13.255 ","End":"08:17.050","Text":"What can this be? Let\u0027s see."},{"Start":"08:17.050 ","End":"08:18.625","Text":"I need to have an mg,"},{"Start":"08:18.625 ","End":"08:21.715","Text":"so add my mg onto the top."},{"Start":"08:21.715 ","End":"08:23.725","Text":"When I multiply this by this,"},{"Start":"08:23.725 ","End":"08:27.145","Text":"I\u0027ll get some mg multiplied by k,"},{"Start":"08:27.145 ","End":"08:35.080","Text":"over m plus M. But I don\u0027t have a k in this expression,"},{"Start":"08:35.080 ","End":"08:38.770","Text":"so I\u0027m just going to divide by k over here,"},{"Start":"08:38.770 ","End":"08:42.310","Text":"and then I\u0027ll have mgk divided by k,"},{"Start":"08:42.310 ","End":"08:45.505","Text":"so it will just be mg,"},{"Start":"08:45.505 ","End":"08:47.680","Text":"divided by m plus M,"},{"Start":"08:47.680 ","End":"08:49.510","Text":"and there we have it."},{"Start":"08:49.510 ","End":"08:51.535","Text":"This is our frequency squared,"},{"Start":"08:51.535 ","End":"08:54.145","Text":"and this is our point of equilibrium."},{"Start":"08:54.145 ","End":"08:56.110","Text":"Now, another way that I could have found my point of"},{"Start":"08:56.110 ","End":"08:59.590","Text":"equilibrium if we were going to be asked that,"},{"Start":"08:59.590 ","End":"09:03.175","Text":"is we can go back to this equation here."},{"Start":"09:03.175 ","End":"09:05.635","Text":"At point of equilibrium,"},{"Start":"09:05.635 ","End":"09:08.176","Text":"there\u0027s no acceleration,"},{"Start":"09:08.176 ","End":"09:12.160","Text":"so my mx double dot will equal 0,"},{"Start":"09:12.160 ","End":"09:17.290","Text":"and my t will equal mg. Why?"},{"Start":"09:17.290 ","End":"09:20.845","Text":"Because it will just have mg pulling down at it,"},{"Start":"09:20.845 ","End":"09:22.750","Text":"so my t will equal mg."},{"Start":"09:22.750 ","End":"09:29.755","Text":"Therefore, I could just write that mg minus kx is equal to 0,"},{"Start":"09:29.755 ","End":"09:36.850","Text":"and then I could have just written that mg equals kx,"},{"Start":"09:36.850 ","End":"09:40.660","Text":"and therefore, that my x equals mg"},{"Start":"09:40.660 ","End":"09:44.860","Text":"over k. Which is the exact result that we got over here."},{"Start":"09:44.860 ","End":"09:50.030","Text":"This is a good way to check just to make sure that you\u0027ve done this whole calculation,"},{"Start":"09:50.030 ","End":"09:52.630","Text":"because sometimes you might get some complicated numbers."},{"Start":"09:52.630 ","End":"09:56.130","Text":"That you\u0027ve done this whole calculation correctly,"},{"Start":"09:56.130 ","End":"09:58.240","Text":"you can just do this little thing and see that you get"},{"Start":"09:58.240 ","End":"10:01.690","Text":"the exact same answer. Now question B."},{"Start":"10:01.690 ","End":"10:05.230","Text":"Sorry, I forgot to write it before,"},{"Start":"10:05.230 ","End":"10:06.625","Text":"but I\u0027ve added it now."},{"Start":"10:06.625 ","End":"10:10.540","Text":"At t=0, the mass was released to rest after being"},{"Start":"10:10.540 ","End":"10:14.920","Text":"pulled a distance D. The spring was elongated a distance D,"},{"Start":"10:14.920 ","End":"10:18.265","Text":"find the solution to the equation of motion."},{"Start":"10:18.265 ","End":"10:21.970","Text":"We\u0027ve already learned that what we need to do is we have to"},{"Start":"10:21.970 ","End":"10:25.869","Text":"find an equation in this format."},{"Start":"10:25.869 ","End":"10:32.650","Text":"We\u0027re meant to find that x(t) is equal to A cosine"},{"Start":"10:32.650 ","End":"10:40.180","Text":"of Omega t plus Phi plus x_0,"},{"Start":"10:40.180 ","End":"10:42.835","Text":"which is my point of equilibrium."},{"Start":"10:42.835 ","End":"10:45.280","Text":"Whenever we\u0027re asked this question,"},{"Start":"10:45.280 ","End":"10:50.970","Text":"this is the equation that we need to immediately pop into our head,"},{"Start":"10:50.970 ","End":"10:54.375","Text":"and then we have to find what A is, what Omega is,"},{"Start":"10:54.375 ","End":"10:57.935","Text":"what Phi is, and what x_0 is."},{"Start":"10:57.935 ","End":"10:59.815","Text":"From our previous section,"},{"Start":"10:59.815 ","End":"11:03.790","Text":"we already found out what Omega and what x_0 is,"},{"Start":"11:03.790 ","End":"11:14.180","Text":"because we have over here that Omega is equal to the square root of k over m plus M,"},{"Start":"11:14.340 ","End":"11:19.705","Text":"and then x_0 is equal to mg"},{"Start":"11:19.705 ","End":"11:25.510","Text":"over k. Now all we need to do is we need to find what A is,"},{"Start":"11:25.510 ","End":"11:27.220","Text":"and what Phi is."},{"Start":"11:27.220 ","End":"11:28.615","Text":"How do we do this?"},{"Start":"11:28.615 ","End":"11:32.155","Text":"We have to use initial conditions."},{"Start":"11:32.155 ","End":"11:34.120","Text":"Now, we haven\u0027t been given them in the question."},{"Start":"11:34.120 ","End":"11:37.795","Text":"Well, we have, not in a mathematical way,"},{"Start":"11:37.795 ","End":"11:42.700","Text":"so now we have to rewrite what we have here in words into a mathematical way,"},{"Start":"11:42.700 ","End":"11:44.605","Text":"plug it into the equation,"},{"Start":"11:44.605 ","End":"11:48.430","Text":"and through that work out what A and what Phi is."},{"Start":"11:48.430 ","End":"11:50.110","Text":"Let\u0027s see what we do. Now,"},{"Start":"11:50.110 ","End":"11:53.395","Text":"they\u0027ve already told us that at t=0,"},{"Start":"11:53.395 ","End":"12:01.480","Text":"so we can already write that the position at t=0 is going to be equal to,"},{"Start":"12:01.480 ","End":"12:03.265","Text":"stretched a distance,"},{"Start":"12:03.265 ","End":"12:09.910","Text":"D. Then we always write plus x_0."},{"Start":"12:09.910 ","End":"12:11.755","Text":"Why do we write plus x_ 0?"},{"Start":"12:11.755 ","End":"12:18.160","Text":"Because it\u0027s always stretched a distance D past its point of equilibrium."},{"Start":"12:18.160 ","End":"12:23.475","Text":"If this was the point of equilibrium and this distance was D,"},{"Start":"12:23.475 ","End":"12:25.335","Text":"so now it makes a bit more sense."},{"Start":"12:25.335 ","End":"12:28.170","Text":"When the forces were balanced,"},{"Start":"12:28.170 ","End":"12:31.145","Text":"the mass was resting at this point of equilibrium,"},{"Start":"12:31.145 ","End":"12:32.830","Text":"and then I stretched it."},{"Start":"12:32.830 ","End":"12:34.810","Text":"I overcame the forces,"},{"Start":"12:34.810 ","End":"12:38.455","Text":"and I stretched the D. It\u0027s actual distance from the 0,"},{"Start":"12:38.455 ","End":"12:41.755","Text":"from it\u0027s resting length of the spring,"},{"Start":"12:41.755 ","End":"12:44.695","Text":"when it\u0027s not being compressed or stretched,"},{"Start":"12:44.695 ","End":"12:51.130","Text":"it\u0027s going to be a distance of x_0 plus D in order to get to this point."},{"Start":"12:51.130 ","End":"12:56.185","Text":"That\u0027s that. Then its velocity. Let\u0027s see."},{"Start":"12:56.185 ","End":"12:57.910","Text":"Velocity, I\u0027m reminding you, is x dot at t=0,"},{"Start":"12:57.910 ","End":"13:07.165","Text":"and then we\u0027re told in the question that it is released from rest,"},{"Start":"13:07.165 ","End":"13:09.355","Text":"which means that its velocity at the beginning,"},{"Start":"13:09.355 ","End":"13:11.575","Text":"starting velocity is 0."},{"Start":"13:11.575 ","End":"13:13.345","Text":"Now let\u0027s see what we do."},{"Start":"13:13.345 ","End":"13:16.105","Text":"Now, it\u0027s usually easier to start off with"},{"Start":"13:16.105 ","End":"13:20.980","Text":"the initial condition of velocity. Let\u0027s do that."},{"Start":"13:20.980 ","End":"13:23.155","Text":"Now, because its velocity, its x dot,"},{"Start":"13:23.155 ","End":"13:26.214","Text":"so we have to derive this equation."},{"Start":"13:26.214 ","End":"13:30.330","Text":"Then we\u0027re going to get, so x dot is equal to 0,"},{"Start":"13:30.330 ","End":"13:33.720","Text":"so 0 equals, now let\u0027s derive this,"},{"Start":"13:33.720 ","End":"13:42.355","Text":"so it\u0027s going to be A multiplied by the derivative of cosine is negative sine,"},{"Start":"13:42.355 ","End":"13:51.430","Text":"so negative sine of Omega t plus Phi,"},{"Start":"13:51.430 ","End":"13:54.130","Text":"and then I enter derivative,"},{"Start":"13:54.130 ","End":"13:57.380","Text":"which is going to be Omega."},{"Start":"13:57.660 ","End":"14:00.700","Text":"We say that this is equal to 0."},{"Start":"14:00.700 ","End":"14:02.680","Text":"Now, we know that our A,"},{"Start":"14:02.680 ","End":"14:04.450","Text":"our amplitude is not equal to 0,"},{"Start":"14:04.450 ","End":"14:07.150","Text":"and we know that our Omega is not equal to 0,"},{"Start":"14:07.150 ","End":"14:10.000","Text":"and we know that t is equal to 0,"},{"Start":"14:10.000 ","End":"14:13.300","Text":"so this crosses out."},{"Start":"14:13.300 ","End":"14:21.160","Text":"The only thing that can cause this expression to equal 0 is if sine of Phi is equal to 0."},{"Start":"14:21.160 ","End":"14:23.530","Text":"When is sine of v equal to 0?"},{"Start":"14:23.530 ","End":"14:27.640","Text":"When Phi is equal to 0."},{"Start":"14:27.640 ","End":"14:31.525","Text":"Great. Now we found what Phi is."},{"Start":"14:31.525 ","End":"14:35.020","Text":"Now we do the exact same thing with this equation."},{"Start":"14:35.020 ","End":"14:40.630","Text":"We say that x at t=0 is equal to D plus x_0."},{"Start":"14:40.630 ","End":"14:46.120","Text":"We can write D plus x_0 is equal to"},{"Start":"14:46.120 ","End":"14:52.674","Text":"A cosine of Omega t plus 0."},{"Start":"14:52.674 ","End":"14:55.045","Text":"We don\u0027t have to do that."},{"Start":"14:55.045 ","End":"14:59.185","Text":"Plus 0, plus x(0)."},{"Start":"14:59.185 ","End":"15:00.940","Text":"Now, in both sides we have an x(0),"},{"Start":"15:00.940 ","End":"15:02.275","Text":"so we can cross those off."},{"Start":"15:02.275 ","End":"15:03.880","Text":"We know that t=0,"},{"Start":"15:03.880 ","End":"15:05.860","Text":"so this is equal to 0."},{"Start":"15:05.860 ","End":"15:08.800","Text":"Then we have what is cosine of 0?"},{"Start":"15:08.800 ","End":"15:10.510","Text":"Cosine of 0 is 1,"},{"Start":"15:10.510 ","End":"15:14.290","Text":"so then we have D equals A multiplied by 1,"},{"Start":"15:14.290 ","End":"15:20.395","Text":"so we know that my amplitude is equal to D. Therefore,"},{"Start":"15:20.395 ","End":"15:23.380","Text":"my equation, back to here,"},{"Start":"15:23.380 ","End":"15:33.070","Text":"is going to be x(t) is equal to D cosine of Omega t plus x_0,"},{"Start":"15:33.070 ","End":"15:36.955","Text":"and we can plug in this x_0."},{"Start":"15:36.955 ","End":"15:39.610","Text":"Let\u0027s actually do that to make it a bit clear."},{"Start":"15:39.610 ","End":"15:43.165","Text":"Plus mg over k,"},{"Start":"15:43.165 ","End":"15:47.575","Text":"and we can plug in what our Omega is,"},{"Start":"15:47.575 ","End":"15:53.334","Text":"which is the square root of k over m plus"},{"Start":"15:53.334 ","End":"16:02.590","Text":"M. I just subbed in my values for Omega and for x_0 into the equation,"},{"Start":"16:02.590 ","End":"16:07.945","Text":"and here there\u0027s a t. There you go."},{"Start":"16:07.945 ","End":"16:09.640","Text":"Now, a handy trick."},{"Start":"16:09.640 ","End":"16:11.620","Text":"In most of the questions,"},{"Start":"16:11.620 ","End":"16:13.370","Text":"90 percent of the questions,"},{"Start":"16:13.370 ","End":"16:16.819","Text":"we\u0027re going to be given similar starting conditions,"},{"Start":"16:16.819 ","End":"16:19.430","Text":"initial conditions, as to here."},{"Start":"16:19.430 ","End":"16:29.065","Text":"That our x at t=0 will equal to some stretch in a certain direction,"},{"Start":"16:29.065 ","End":"16:32.210","Text":"and that my starting velocity will be 0,"},{"Start":"16:32.210 ","End":"16:34.790","Text":"and it will be released from rest."},{"Start":"16:34.790 ","End":"16:38.835","Text":"Now, if I have initial conditions of this form,"},{"Start":"16:38.835 ","End":"16:44.150","Text":"with this being the important defining factor here,"},{"Start":"16:44.150 ","End":"16:47.025","Text":"that my velocity is 0 at the beginning,"},{"Start":"16:47.025 ","End":"16:50.585","Text":"then I know straight away that my amplitude will equal"},{"Start":"16:50.585 ","End":"16:58.460","Text":"just the stretch D without the point of equilibrium,"},{"Start":"16:58.460 ","End":"17:01.190","Text":"and that my Phi will equal 0."},{"Start":"17:01.190 ","End":"17:04.580","Text":"Then without doing all of these steps,"},{"Start":"17:04.580 ","End":"17:08.600","Text":"I can straight away will just write that my amplitude is D, this,"},{"Start":"17:08.600 ","End":"17:12.606","Text":"the stretch itself, not including the point of equilibrium,"},{"Start":"17:12.606 ","End":"17:16.550","Text":"cosine of Omega t plus x_0,"},{"Start":"17:16.550 ","End":"17:18.440","Text":"and cross out the Phi."},{"Start":"17:18.440 ","End":"17:22.770","Text":"That\u0027s a quick tip. That is the end of our lesson."}],"ID":9433},{"Watched":false,"Name":"Pulley Mass And Spring","Duration":"33m 37s","ChapterTopicVideoID":9165,"CourseChapterTopicPlaylistID":5369,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.065","Text":"Hello. Here I\u0027m going to try answering this question."},{"Start":"00:04.065 ","End":"00:06.480","Text":"We\u0027re told that we have a mass m_1,"},{"Start":"00:06.480 ","End":"00:12.734","Text":"which is tied via string over an unideal pulley system,"},{"Start":"00:12.734 ","End":"00:17.490","Text":"and we\u0027re told that it\u0027s attached on the other end to a spring of constant k,"},{"Start":"00:17.490 ","End":"00:20.730","Text":"and that is also attached to the ground."},{"Start":"00:20.730 ","End":"00:26.249","Text":"We\u0027re being told that the radius and the mass of the pulley is r and m_2,"},{"Start":"00:26.249 ","End":"00:30.015","Text":"and we\u0027re being told that the string does not slip on the pulley."},{"Start":"00:30.015 ","End":"00:32.120","Text":"Our first question is,"},{"Start":"00:32.120 ","End":"00:34.160","Text":"where is the point of equilibrium?"},{"Start":"00:34.160 ","End":"00:37.084","Text":"We have to find that, and second is,"},{"Start":"00:37.084 ","End":"00:39.530","Text":"what is the frequency of oscillations?"},{"Start":"00:39.530 ","End":"00:42.860","Text":"We\u0027re going to answer these 2 questions together,"},{"Start":"00:42.860 ","End":"00:49.000","Text":"and how we\u0027re going to do that is by first drawing all of our forces in the system."},{"Start":"00:49.000 ","End":"00:53.375","Text":"Let\u0027s take a look at which forces we have acting over here."},{"Start":"00:53.375 ","End":"00:56.449","Text":"Now because we\u0027re being told that the pulley is not ideal,"},{"Start":"00:56.449 ","End":"00:59.839","Text":"it means that the tension in the string over here is"},{"Start":"00:59.839 ","End":"01:03.665","Text":"going to be different to the tension on this side."},{"Start":"01:03.665 ","End":"01:11.140","Text":"Let\u0027s take a look. Here we can draw that we have T_1 and also over here we have our T_1."},{"Start":"01:11.140 ","End":"01:15.479","Text":"It\u0027s always going in the direction of into the string."},{"Start":"01:15.479 ","End":"01:18.310","Text":"Then going down here,"},{"Start":"01:18.310 ","End":"01:21.069","Text":"we have our mg,"},{"Start":"01:21.069 ","End":"01:25.110","Text":"and then here we have our T_2,"},{"Start":"01:25.110 ","End":"01:26.580","Text":"the different tension,"},{"Start":"01:26.580 ","End":"01:30.460","Text":"and here again we have our T_2."},{"Start":"01:30.460 ","End":"01:34.669","Text":"Now we also have our force of the spring."},{"Start":"01:34.669 ","End":"01:37.115","Text":"But I\u0027m not going to draw this in,"},{"Start":"01:37.115 ","End":"01:40.250","Text":"so that we don\u0027t get confused with a negative sign."},{"Start":"01:40.250 ","End":"01:42.005","Text":"However, if I was going to draw it in,"},{"Start":"01:42.005 ","End":"01:46.350","Text":"I would draw it in whichever direction I choose to be the positive direction."},{"Start":"01:46.350 ","End":"01:50.390","Text":"Then when I put it into my force equations,"},{"Start":"01:50.390 ","End":"01:52.610","Text":"so I just add a negative."},{"Start":"01:52.610 ","End":"01:54.829","Text":"That will stop me from getting confused."},{"Start":"01:54.829 ","End":"01:59.780","Text":"Now, what we should do is we should choose which is our positive direction."},{"Start":"01:59.780 ","End":"02:03.410","Text":"Because in my third question,"},{"Start":"02:03.410 ","End":"02:07.475","Text":"I\u0027m being told that the mass is pulled down a distance d,"},{"Start":"02:07.475 ","End":"02:14.374","Text":"so I\u0027m going to say that this is the positive x direction,"},{"Start":"02:14.374 ","End":"02:16.744","Text":"which means that I can say that this,"},{"Start":"02:16.744 ","End":"02:19.708","Text":"on this side is the positive x direction,"},{"Start":"02:19.708 ","End":"02:22.550","Text":"and then I\u0027ll say that my positive direction for"},{"Start":"02:22.550 ","End":"02:26.859","Text":"rotating is therefore in this direction,"},{"Start":"02:26.859 ","End":"02:30.215","Text":"so that it\u0027s going with the direction of travel."},{"Start":"02:30.215 ","End":"02:34.020","Text":"We start off here, and it goes all the way like that."},{"Start":"02:34.550 ","End":"02:40.044","Text":"Now we\u0027re going to see that because I\u0027ve written my axis like this,"},{"Start":"02:40.044 ","End":"02:44.080","Text":"that my axis is just 1 log axis going in the direction of travel,"},{"Start":"02:44.080 ","End":"02:48.695","Text":"so I have to write separate equations for the 2 sides."},{"Start":"02:48.695 ","End":"02:52.215","Text":"My first equation, switch colors,"},{"Start":"02:52.215 ","End":"02:57.680","Text":"it\u0027s going to be the sum of all of my forces on body number 1."},{"Start":"02:57.680 ","End":"02:59.750","Text":"That\u0027s my mass 1 over here,"},{"Start":"02:59.750 ","End":"03:05.620","Text":"so the sum of all the forces is going to be;"},{"Start":"03:05.620 ","End":"03:08.145","Text":"so I have my mg, It\u0027s just add in a 1 over here."},{"Start":"03:08.145 ","End":"03:12.260","Text":"I have my m_1g going down here,"},{"Start":"03:12.260 ","End":"03:14.495","Text":"which is in my positive x direction,"},{"Start":"03:14.495 ","End":"03:16.674","Text":"so It\u0027s positive m_1g."},{"Start":"03:16.674 ","End":"03:20.925","Text":"Then I have my T_1 going in the negative direction,"},{"Start":"03:20.925 ","End":"03:24.584","Text":"so I\u0027ll have negative T_1,"},{"Start":"03:24.584 ","End":"03:28.208","Text":"and that is going to be equal to m_1a,"},{"Start":"03:28.208 ","End":"03:32.560","Text":"and we\u0027ll write 1 over here to show that it\u0027s for this body."},{"Start":"03:33.020 ","End":"03:37.279","Text":"This is my first equation."},{"Start":"03:37.279 ","End":"03:39.430","Text":"Now let\u0027s go on to my second equation."},{"Start":"03:39.430 ","End":"03:42.349","Text":"Now we\u0027re going to be looking at our pulley system over here."},{"Start":"03:42.349 ","End":"03:44.510","Text":"Now because there\u0027s an axis and"},{"Start":"03:44.510 ","End":"03:48.950","Text":"my pulley system and I know that there\u0027s a force acting in here,"},{"Start":"03:48.950 ","End":"03:50.990","Text":"but I don\u0027t know which force it is,"},{"Start":"03:50.990 ","End":"03:52.415","Text":"so what I\u0027m going to do,"},{"Start":"03:52.415 ","End":"03:55.939","Text":"instead of writing an equation for the sum of all of my forces,"},{"Start":"03:55.939 ","End":"03:57.725","Text":"which will get a little bit complicated,"},{"Start":"03:57.725 ","End":"04:03.154","Text":"I\u0027m going to write my second equation as the sum of all of my torques."},{"Start":"04:03.154 ","End":"04:04.760","Text":"Why can I do that?"},{"Start":"04:04.760 ","End":"04:07.969","Text":"Because then I can write out my forces acting"},{"Start":"04:07.969 ","End":"04:13.180","Text":"here and my force acting from my axis of rotation,"},{"Start":"04:13.180 ","End":"04:18.290","Text":"its radius is going to be equal to 0 because it\u0027s acting at the axis of"},{"Start":"04:18.290 ","End":"04:23.195","Text":"rotation at the origin which means that I don\u0027t have to take it into account,"},{"Start":"04:23.195 ","End":"04:26.824","Text":"and then I\u0027ve bypassed all of the confusion."},{"Start":"04:26.824 ","End":"04:32.439","Text":"Let\u0027s do this. I have the sum of all of my torques is going to be equal to."},{"Start":"04:32.439 ","End":"04:37.254","Text":"I know that my anticlockwise direction as my positive direction."},{"Start":"04:37.254 ","End":"04:44.360","Text":"I\u0027m going to have positive T_1 multiplied by its distance from the axis of rotation,"},{"Start":"04:44.360 ","End":"04:45.735","Text":"which is R,"},{"Start":"04:45.735 ","End":"04:48.960","Text":"the radius and then minus,"},{"Start":"04:48.960 ","End":"04:52.249","Text":"because my T_2 is acting in the opposite direction,"},{"Start":"04:52.249 ","End":"04:55.490","Text":"my negative direction of rotation,"},{"Start":"04:55.490 ","End":"04:57.980","Text":"so negative T_2,"},{"Start":"04:57.980 ","End":"05:03.005","Text":"and its distance from the axis of rotation is also R. This is of course,"},{"Start":"05:03.005 ","End":"05:04.699","Text":"equal to I,"},{"Start":"05:04.699 ","End":"05:09.520","Text":"our moment of inertia multiplied by Alpha."},{"Start":"05:10.190 ","End":"05:17.030","Text":"My I_2 comes from the fact that my mass of the pulley system is m_2."},{"Start":"05:17.030 ","End":"05:22.415","Text":"Then let\u0027s just write a note over here for what my I_2 is going to be equal to."},{"Start":"05:22.415 ","End":"05:26.369","Text":"My moment of inertia of this is going to just"},{"Start":"05:26.369 ","End":"05:30.300","Text":"be the moment of inertia of a disc rotating about its center,"},{"Start":"05:30.300 ","End":"05:34.095","Text":"you should write this in your notes,"},{"Start":"05:34.095 ","End":"05:36.600","Text":"which is going to be 1/2 the mass,"},{"Start":"05:36.600 ","End":"05:37.770","Text":"which is m_2,"},{"Start":"05:37.770 ","End":"05:42.700","Text":"multiplied by its radius squared."},{"Start":"05:43.220 ","End":"05:50.310","Text":"Now notice from where we said that our pulley system is an unideal pulley system."},{"Start":"05:50.310 ","End":"05:56.370","Text":"If it was ideal then what we would have is that our mass would be equal to 0."},{"Start":"05:56.370 ","End":"05:57.690","Text":"An ideal pulley system,"},{"Start":"05:57.690 ","End":"05:59.340","Text":"its mass is equal to 0,"},{"Start":"05:59.340 ","End":"06:04.189","Text":"in which case our I_2 would be equal to 0 and if our I_2 is equal to 0,"},{"Start":"06:04.189 ","End":"06:07.790","Text":"then that means that this side of the equation is equal to 0,"},{"Start":"06:07.790 ","End":"06:11.209","Text":"which means that we\u0027ll have that our T_1 is equal to T_2,"},{"Start":"06:11.209 ","End":"06:15.439","Text":"and then we would see that both of our tensions are the same and that\u0027s"},{"Start":"06:15.439 ","End":"06:20.550","Text":"why I wrote them differently here because our mass is not equal to 0."},{"Start":"06:21.160 ","End":"06:27.335","Text":"Now let\u0027s go on to our third equation for what\u0027s happening over here."},{"Start":"06:27.335 ","End":"06:30.429","Text":"This is my third equation and here again,"},{"Start":"06:30.429 ","End":"06:33.954","Text":"I can go back to writing the sum of all of my forces,"},{"Start":"06:33.954 ","End":"06:35.875","Text":"and I\u0027m just going to call this area,"},{"Start":"06:35.875 ","End":"06:39.205","Text":"area 3, sum of forces 3."},{"Start":"06:39.205 ","End":"06:44.920","Text":"Now because I said that my positive x direction was going up in the direction of travel,"},{"Start":"06:44.920 ","End":"06:49.450","Text":"so I can say that my T_2 is going in the positive direction,"},{"Start":"06:49.450 ","End":"06:50.965","Text":"so I have positive T_2."},{"Start":"06:50.965 ","End":"06:55.900","Text":"Then my force for the spring is going to be kx,"},{"Start":"06:55.900 ","End":"06:58.673","Text":"but it always has a minus sign before it."},{"Start":"06:58.673 ","End":"07:01.165","Text":"It\u0027s going to be negative kx."},{"Start":"07:01.165 ","End":"07:04.149","Text":"Now this is always correct and even if"},{"Start":"07:04.149 ","End":"07:07.520","Text":"I draw my positive x direction in the opposite direction,"},{"Start":"07:07.520 ","End":"07:09.475","Text":"say going this way,"},{"Start":"07:09.475 ","End":"07:14.030","Text":"then all I would do is I would write negative T_2,"},{"Start":"07:14.030 ","End":"07:16.640","Text":"but it would still be negative kx."},{"Start":"07:16.640 ","End":"07:20.645","Text":"It\u0027s always and always negative kx."},{"Start":"07:20.645 ","End":"07:24.370","Text":"Doesn\u0027t matter in which direction my axis is going."},{"Start":"07:24.370 ","End":"07:32.999","Text":"Now, this is obviously going to be equal to some mass multiplied by acceleration."},{"Start":"07:33.210 ","End":"07:35.815","Text":"Let\u0027s take a look at what this means."},{"Start":"07:35.815 ","End":"07:39.114","Text":"When I only have a spring and a string,"},{"Start":"07:39.114 ","End":"07:44.905","Text":"then the sum of all of my forces is always going to be equal to 0."},{"Start":"07:44.905 ","End":"07:47.560","Text":"What\u0027s a way of looking at this?"},{"Start":"07:47.560 ","End":"07:51.640","Text":"If I imagine that I have some tiny little mass over here,"},{"Start":"07:51.640 ","End":"07:54.490","Text":"let\u0027s call it m tag,"},{"Start":"07:54.490 ","End":"07:58.869","Text":"then I can write over here that the sum of all of my forces is"},{"Start":"07:58.869 ","End":"08:04.014","Text":"equal to m tag multiplied by it\u0027s acceleration, a tag."},{"Start":"08:04.014 ","End":"08:08.170","Text":"Now, if my m tag is extremely small,"},{"Start":"08:08.170 ","End":"08:10.419","Text":"if I imagine it as for instance,"},{"Start":"08:10.419 ","End":"08:12.850","Text":"a grain of dust or sand,"},{"Start":"08:12.850 ","End":"08:16.194","Text":"then I can say that it\u0027s mass is so tiny"},{"Start":"08:16.194 ","End":"08:23.035","Text":"that it\u0027s really not substantial enough to put into my calculations."},{"Start":"08:23.035 ","End":"08:25.465","Text":"In that case, I can just cancel it out,"},{"Start":"08:25.465 ","End":"08:27.579","Text":"I can just rub this out right now,"},{"Start":"08:27.579 ","End":"08:31.059","Text":"and I can just say that this whole equation is equal to"},{"Start":"08:31.059 ","End":"08:35.005","Text":"0 because it\u0027s mass is practically equal to 0."},{"Start":"08:35.005 ","End":"08:39.070","Text":"This is only correct when we have a spring and a string."},{"Start":"08:39.070 ","End":"08:41.770","Text":"Now, let\u0027s go back to taking a look at"},{"Start":"08:41.770 ","End":"08:47.214","Text":"my x-axis and how I have described where my positive direction is."},{"Start":"08:47.214 ","End":"08:50.005","Text":"Now, it\u0027s okay to do what we\u0027ve done over here,"},{"Start":"08:50.005 ","End":"08:54.730","Text":"which is showing that the positive direction is in 1 axes going like this."},{"Start":"08:54.730 ","End":"08:58.119","Text":"All we have to do is remember that both sides are"},{"Start":"08:58.119 ","End":"09:02.125","Text":"pointing in different directions,even though it\u0027s going with the direction of travel."},{"Start":"09:02.125 ","End":"09:06.160","Text":"Here, the positive direction is up and here the positive direction is down."},{"Start":"09:06.160 ","End":"09:10.435","Text":"We just have to know that there\u0027s a difference here and how to work with that."},{"Start":"09:10.435 ","End":"09:12.954","Text":"Specifically, in this question,"},{"Start":"09:12.954 ","End":"09:16.884","Text":"how we\u0027ve designed our axis to be is very useful,"},{"Start":"09:16.884 ","End":"09:20.020","Text":"because it\u0027s going to be easy when we\u0027re looking at the movements"},{"Start":"09:20.020 ","End":"09:23.365","Text":"of m1 and our imaginary mass over here."},{"Start":"09:23.365 ","End":"09:26.020","Text":"When this imaginary mass moves up"},{"Start":"09:26.020 ","End":"09:31.675","Text":"10 centimeters our m1 is going to move down also 10 centimeters."},{"Start":"09:31.675 ","End":"09:36.099","Text":"But in fact, they\u0027re still both moving in the positive direction and it will"},{"Start":"09:36.099 ","End":"09:41.545","Text":"make this x link in over here very nicely."},{"Start":"09:41.545 ","End":"09:43.119","Text":"How we\u0027ve done it like this,"},{"Start":"09:43.119 ","End":"09:46.105","Text":"we don\u0027t have to play around with our signs."},{"Start":"09:46.105 ","End":"09:51.565","Text":"We can say that this x is going to be equal to the displacement of our m1,"},{"Start":"09:51.565 ","End":"09:54.130","Text":"which here we would have called maybe x1."},{"Start":"09:54.130 ","End":"09:58.930","Text":"If on the other hand, I would have chosen here my positive x-direction to"},{"Start":"09:58.930 ","End":"10:03.249","Text":"be down and also here my positive x-direction to be down,"},{"Start":"10:03.249 ","End":"10:07.720","Text":"then I would have had to have remembered to put a negative each time,"},{"Start":"10:07.720 ","End":"10:12.280","Text":"which will make it much more complicated and it\u0027s a shame."},{"Start":"10:12.280 ","End":"10:15.009","Text":"That\u0027s why in all of these types of questions,"},{"Start":"10:15.009 ","End":"10:21.310","Text":"I\u0027m always going to choose my axis to be going in the direction of travel on both sides,"},{"Start":"10:21.310 ","End":"10:22.989","Text":"so it will look like this."},{"Start":"10:22.989 ","End":"10:26.799","Text":"If however you really want to do it the other way, then you can,"},{"Start":"10:26.799 ","End":"10:31.449","Text":"but you have to remember to add in negatives where it\u0027s applicable in"},{"Start":"10:31.449 ","End":"10:36.910","Text":"order to maintain the correct relationship between the movements of each mass."},{"Start":"10:36.910 ","End":"10:40.090","Text":"I hope that that was understandable."},{"Start":"10:40.090 ","End":"10:42.729","Text":"I\u0027ll just rewrite this,"},{"Start":"10:42.729 ","End":"10:46.735","Text":"that our x is going to be equal to our x1."},{"Start":"10:46.735 ","End":"10:51.040","Text":"Let\u0027s move on. What we have now is"},{"Start":"10:51.040 ","End":"10:58.810","Text":"3 equations and we have 4 unknowns."},{"Start":"10:58.810 ","End":"11:01.915","Text":"Our unknowns are our T1,"},{"Start":"11:01.915 ","End":"11:05.410","Text":"our T2, our a1, and our Alpha."},{"Start":"11:05.410 ","End":"11:08.995","Text":"That means that we need another equation."},{"Start":"11:08.995 ","End":"11:15.069","Text":"What am I going to do? Whenever I\u0027m working with a rigid body and harmonic motion is"},{"Start":"11:15.069 ","End":"11:21.130","Text":"that I\u0027m going to write my fourth equation as the relationship between the accelerations."},{"Start":"11:21.130 ","End":"11:22.989","Text":"What does that mean? The relationship between"},{"Start":"11:22.989 ","End":"11:26.785","Text":"my linear acceleration and my angular acceleration."},{"Start":"11:26.785 ","End":"11:28.840","Text":"Just like with my velocities,"},{"Start":"11:28.840 ","End":"11:33.010","Text":"we just take the derivative and we get that our a1 is going to be equal"},{"Start":"11:33.010 ","End":"11:38.845","Text":"to our Alpha multiplied by the radius of the pulley system."},{"Start":"11:38.845 ","End":"11:45.820","Text":"Now, I have 4 unknowns and 4 equations, perfect."},{"Start":"11:45.820 ","End":"11:51.530","Text":"A final thing, this equation is only because there\u0027s no slipping."},{"Start":"11:52.560 ","End":"11:55.407","Text":"This is very important."},{"Start":"11:55.407 ","End":"11:59.140","Text":"Otherwise, this equation can be used."},{"Start":"11:59.180 ","End":"12:04.735","Text":"We can do this because we\u0027re being told that the string does not slip on the pulley."},{"Start":"12:04.735 ","End":"12:07.495","Text":"That\u0027s where this equation comes from."},{"Start":"12:07.495 ","End":"12:10.374","Text":"Now, with these 4 equations,"},{"Start":"12:10.374 ","End":"12:13.119","Text":"I\u0027m going to simply rearrange"},{"Start":"12:13.119 ","End":"12:16.945","Text":"everything in order to get a more clear and concise equation."},{"Start":"12:16.945 ","End":"12:20.995","Text":"First thing I\u0027m going to do from equation Number 1,"},{"Start":"12:20.995 ","End":"12:23.630","Text":"I\u0027m going to isolate out my T1."},{"Start":"12:23.970 ","End":"12:33.430","Text":"I\u0027m just going to get that my T1 = m1g negative m1a1."},{"Start":"12:33.430 ","End":"12:37.145","Text":"Then from 3, I\u0027m going to isolate out my T2,"},{"Start":"12:37.145 ","End":"12:43.000","Text":"so I\u0027m going to get that my T2 is simply equal to k multiplied by x."},{"Start":"12:45.120 ","End":"12:47.485","Text":"This is from 1,"},{"Start":"12:47.485 ","End":"12:49.300","Text":"this is from 3."},{"Start":"12:49.300 ","End":"12:52.600","Text":"Now, I\u0027m going to substitute in 1, 3,"},{"Start":"12:52.600 ","End":"12:58.005","Text":"and 4 into my equation Number 2."},{"Start":"12:58.005 ","End":"13:02.324","Text":"My equation Number 2 is simply going to be equal to,"},{"Start":"13:02.324 ","End":"13:08.180","Text":"so I have T1 multiplied by R. I\u0027m going to have m1g minus"},{"Start":"13:08.180 ","End":"13:15.250","Text":"m1a multiplied by R minus T2 multiplied by R,"},{"Start":"13:15.250 ","End":"13:24.460","Text":"so minus kx multiplied by R. This is going to be equal to my I2 multiplied by my Alpha."},{"Start":"13:24.460 ","End":"13:31.750","Text":"That\u0027s going to be equal to my 1/2 m_2 R^2 multiplied by my Alpha,"},{"Start":"13:31.750 ","End":"13:37.930","Text":"and my Alpha is simply a1 divided by R. If I isolate out my Alpha,"},{"Start":"13:37.930 ","End":"13:42.310","Text":"so it\u0027s a1 divided by R. Now,"},{"Start":"13:42.310 ","End":"13:43.839","Text":"the next thing that I want to do,"},{"Start":"13:43.839 ","End":"13:45.324","Text":"now that I have this equation,"},{"Start":"13:45.324 ","End":"13:46.930","Text":"is to tidy it up a little."},{"Start":"13:46.930 ","End":"13:50.358","Text":"First of all, this R and this R can cancel out,"},{"Start":"13:50.358 ","End":"13:54.020","Text":"and then I can in fact cancel out all my Rs."},{"Start":"13:54.990 ","End":"13:58.210","Text":"Sorry, this is not going to be there."},{"Start":"13:58.210 ","End":"14:01.870","Text":"I can divide both sides by R,"},{"Start":"14:01.870 ","End":"14:06.610","Text":"so divided by R divided by R divided by R. Perfect."},{"Start":"14:06.610 ","End":"14:10.419","Text":"Now, I can see that I have my acceleration over here,"},{"Start":"14:10.419 ","End":"14:13.555","Text":"my a1, and I also have an a1 over here."},{"Start":"14:13.555 ","End":"14:18.940","Text":"What I want to do is I want to get my accelerations to be on 1 side."},{"Start":"14:18.940 ","End":"14:22.150","Text":"Now, I\u0027m just going to play around with the algebra."},{"Start":"14:22.150 ","End":"14:23.965","Text":"You can do this on a piece of paper,"},{"Start":"14:23.965 ","End":"14:31.180","Text":"but I\u0027ll get negative k multiplied by x plus m1g."},{"Start":"14:31.180 ","End":"14:38.780","Text":"This is going to be equal to m1 plus 1/2m2."},{"Start":"14:38.780 ","End":"14:43.590","Text":"2 and all of this multiplied by a_1."},{"Start":"14:43.590 ","End":"14:45.869","Text":"All I did was some algebra,"},{"Start":"14:45.869 ","End":"14:49.620","Text":"just rearranging this, you\u0027re welcome to pause the video and do it yourself."},{"Start":"14:49.620 ","End":"14:52.349","Text":"Now, what I\u0027m going to do is I\u0027m going to do"},{"Start":"14:52.349 ","End":"14:56.100","Text":"my sneaky trick in order to get this equation over"},{"Start":"14:56.100 ","End":"15:04.725","Text":"here to be more similar to or equal to an equation representing harmonic motion."},{"Start":"15:04.725 ","End":"15:10.664","Text":"My stinky and sneaky trick is to take out my k as a common factor."},{"Start":"15:10.664 ","End":"15:14.033","Text":"I\u0027m going to have my negative k over here,"},{"Start":"15:14.033 ","End":"15:16.364","Text":"and this is going to be equal to x,"},{"Start":"15:16.364 ","End":"15:22.560","Text":"and then minus m_1g divided by k."},{"Start":"15:22.560 ","End":"15:33.675","Text":"Then this is going to be equal to m_1 plus 1/2 m_2 multiplied by a_1."},{"Start":"15:33.675 ","End":"15:41.805","Text":"What we actually have over here is that this over here is my x_0."},{"Start":"15:41.805 ","End":"15:46.410","Text":"Notice, my first question is where is the point of equilibrium?"},{"Start":"15:46.410 ","End":"15:49.485","Text":"X_0 is my point of equilibrium, that\u0027s that."},{"Start":"15:49.485 ","End":"15:52.050","Text":"Then what I have over here"},{"Start":"15:52.050 ","End":"16:00.225","Text":"is what switches my mass."},{"Start":"16:00.225 ","End":"16:04.619","Text":"Remember I have negative k multiplied by x minus x_0 is equal"},{"Start":"16:04.619 ","End":"16:10.200","Text":"to mass times x double dot so I multiply by acceleration."},{"Start":"16:10.200 ","End":"16:13.395","Text":"This is my m over there, my coefficient."},{"Start":"16:13.395 ","End":"16:17.145","Text":"Here, we\u0027ve answered our first 2 questions."},{"Start":"16:17.145 ","End":"16:19.964","Text":"I have that, where\u0027s my points of equilibrium?"},{"Start":"16:19.964 ","End":"16:25.215","Text":"It\u0027s at x_0. My point of equilibrium is at x_0,"},{"Start":"16:25.215 ","End":"16:28.283","Text":"which is equal to here,"},{"Start":"16:28.283 ","End":"16:34.185","Text":"and m_1g divided by k. Then my second question is,"},{"Start":"16:34.185 ","End":"16:36.405","Text":"what is the frequency of oscillation?"},{"Start":"16:36.405 ","End":"16:38.219","Text":"My frequency of oscillation,"},{"Start":"16:38.219 ","End":"16:40.560","Text":"as we know, it\u0027s equal to Omega."},{"Start":"16:40.560 ","End":"16:46.365","Text":"The square root of the coefficient of my x minus x_0,"},{"Start":"16:46.365 ","End":"16:48.419","Text":"which is k,"},{"Start":"16:48.419 ","End":"16:51.704","Text":"and then divided by my coefficient of my x double dot,"},{"Start":"16:51.704 ","End":"16:53.174","Text":"or of my acceleration,"},{"Start":"16:53.174 ","End":"16:54.885","Text":"which is over here,"},{"Start":"16:54.885 ","End":"17:03.180","Text":"equal to m_1 plus 1/2(m_2)."},{"Start":"17:03.180 ","End":"17:07.395","Text":"Now, another way that I could have found my point of equilibrium, my x_0,"},{"Start":"17:07.395 ","End":"17:12.915","Text":"is that I could have said that the sum of all of my forces was equal to 0."},{"Start":"17:12.915 ","End":"17:16.155","Text":"This would have equal to 0 and this would have been equal to 0."},{"Start":"17:16.155 ","End":"17:20.895","Text":"Then I would have just had to isolate out my x."},{"Start":"17:20.895 ","End":"17:24.300","Text":"Then we would have found our x_0."},{"Start":"17:24.300 ","End":"17:27.825","Text":"Then a lot of times people use this in order to"},{"Start":"17:27.825 ","End":"17:31.125","Text":"move the origin to a point of equilibrium."},{"Start":"17:31.125 ","End":"17:34.440","Text":"Then sometimes it makes the question a little bit easier to solve."},{"Start":"17:34.440 ","End":"17:37.980","Text":"Here\u0027s specifically, and also in general,"},{"Start":"17:37.980 ","End":"17:40.905","Text":"I prefer to work it out like this because then I"},{"Start":"17:40.905 ","End":"17:44.190","Text":"get to see that really this is working in harmonic motion"},{"Start":"17:44.190 ","End":"17:46.680","Text":"and I get to find my point of equilibrium"},{"Start":"17:46.680 ","End":"17:51.705","Text":"and simultaneously I can find what my frequency is."},{"Start":"17:51.705 ","End":"17:54.810","Text":"We finished with questions 1 and 2."},{"Start":"17:54.810 ","End":"17:57.555","Text":"Let\u0027s move on to Question number 3."},{"Start":"17:57.555 ","End":"18:01.830","Text":"Question number 3, we\u0027re being told that the mass number"},{"Start":"18:01.830 ","End":"18:06.270","Text":"1 is pulled a distance d from the point of equilibrium."},{"Start":"18:06.270 ","End":"18:10.470","Text":"That means if we take our pen over here,"},{"Start":"18:10.470 ","End":"18:13.800","Text":"that if this was our point of equilibrium over here,"},{"Start":"18:13.800 ","End":"18:17.970","Text":"so a mass is pulled some distance d,"},{"Start":"18:17.970 ","End":"18:23.010","Text":"and then it\u0027s released and it moves up and down in harmonic motion."},{"Start":"18:23.010 ","End":"18:27.854","Text":"We\u0027re being asked, what is the maximum distance,"},{"Start":"18:27.854 ","End":"18:30.179","Text":"or because we\u0027re moving in harmonic motion,"},{"Start":"18:30.179 ","End":"18:37.035","Text":"the maximum amplitude which the mass can be pulled such that the tension does not reset,"},{"Start":"18:37.035 ","End":"18:41.280","Text":"does not go to 0 or equals 0 throughout the motion?"},{"Start":"18:41.280 ","End":"18:44.040","Text":"That means that neither tension,"},{"Start":"18:44.040 ","End":"18:49.020","Text":"neither T_1 nor T_2 can be equal to 0."},{"Start":"18:49.020 ","End":"18:56.310","Text":"What is the maximum distance which our m_1 can be pulled down such that our Ts are never,"},{"Start":"18:56.310 ","End":"18:57.765","Text":"ever equal to 0?"},{"Start":"18:57.765 ","End":"19:00.675","Text":"Now, the reason this is important to me,"},{"Start":"19:00.675 ","End":"19:05.459","Text":"the reason that I have to have that my tension is never equal to 0."},{"Start":"19:05.459 ","End":"19:07.739","Text":"Because if my tension is equal to 0,"},{"Start":"19:07.739 ","End":"19:11.309","Text":"that means that one of the sides of the string is Slack."},{"Start":"19:11.309 ","End":"19:16.769","Text":"If that happens, then I\u0027m no longer moving in harmonic motion and there\u0027s a little bit of"},{"Start":"19:16.769 ","End":"19:23.174","Text":"a mess between the motions and how everything is working inside the system."},{"Start":"19:23.174 ","End":"19:27.540","Text":"I need to assume that all the time there is some tension in"},{"Start":"19:27.540 ","End":"19:31.604","Text":"my string and that my system is working in harmonic motion."},{"Start":"19:31.604 ","End":"19:36.390","Text":"I need to find the maximum distance so that my calculations are still correct."},{"Start":"19:36.390 ","End":"19:39.366","Text":"If one of my Ts is ever equal to 0,"},{"Start":"19:39.366 ","End":"19:42.510","Text":"then everything we\u0027ve done up until now is incorrect."},{"Start":"19:42.510 ","End":"19:45.239","Text":"What I\u0027m actually doing in this question is I\u0027m finding"},{"Start":"19:45.239 ","End":"19:49.830","Text":"the limit for which all my calculations are correct."},{"Start":"19:49.830 ","End":"19:52.770","Text":"The maximum amplitude such that"},{"Start":"19:52.770 ","End":"19:57.960","Text":"harmonic motion is still occurring and all my calculations are correct."},{"Start":"19:57.960 ","End":"19:59.864","Text":"Let\u0027s begin doing this."},{"Start":"19:59.864 ","End":"20:01.410","Text":"In our Question 1,"},{"Start":"20:01.410 ","End":"20:04.949","Text":"we found out our equation for harmonic motion,"},{"Start":"20:04.949 ","End":"20:07.200","Text":"which was our differential equation."},{"Start":"20:07.200 ","End":"20:10.800","Text":"That means that we know that the general solution to"},{"Start":"20:10.800 ","End":"20:15.435","Text":"a differential equation is going to be that our x as a function of t is going to be"},{"Start":"20:15.435 ","End":"20:18.705","Text":"equal to A multiplied by"},{"Start":"20:18.705 ","End":"20:25.739","Text":"Cosine of Omega t plus Phi and then plus our starting position,"},{"Start":"20:25.739 ","End":"20:28.455","Text":"which was a point of equilibrium x_0."},{"Start":"20:28.455 ","End":"20:31.845","Text":"Now, obviously, we know what our Omega is,"},{"Start":"20:31.845 ","End":"20:35.295","Text":"we worked it out and we know what our x_0 is, we worked it out."},{"Start":"20:35.295 ","End":"20:38.955","Text":"Now, all we have to do is we have to find our A and Phi."},{"Start":"20:38.955 ","End":"20:42.910","Text":"This is obviously from our initial conditions."},{"Start":"20:43.220 ","End":"20:46.035","Text":"What are my initial conditions?"},{"Start":"20:46.035 ","End":"20:50.880","Text":"I know that my position at T=0 when I start."},{"Start":"20:50.880 ","End":"20:58.050","Text":"I\u0027m being told that the mass is pulled down a distance d from the point of equilibrium,"},{"Start":"20:58.050 ","End":"21:02.040","Text":"which means that it\u0027s d plus x_0."},{"Start":"21:02.040 ","End":"21:04.785","Text":"That\u0027s its starting position."},{"Start":"21:04.785 ","End":"21:07.938","Text":"Then it\u0027s movement,"},{"Start":"21:07.938 ","End":"21:11.264","Text":"it\u0027s x dot T is equal to 0,"},{"Start":"21:11.264 ","End":"21:14.748","Text":"is equal to, let\u0027s take a look,"},{"Start":"21:14.748 ","End":"21:16.874","Text":"pull down from a distance,"},{"Start":"21:16.874 ","End":"21:18.435","Text":"and then it\u0027s released."},{"Start":"21:18.435 ","End":"21:21.255","Text":"That means that it\u0027s released from rest."},{"Start":"21:21.255 ","End":"21:24.495","Text":"Its starting velocity is 0."},{"Start":"21:24.495 ","End":"21:26.880","Text":"Without doing any algebra,"},{"Start":"21:26.880 ","End":"21:28.590","Text":"I\u0027m just going to write out,"},{"Start":"21:28.590 ","End":"21:30.383","Text":"you can solve this if you want,"},{"Start":"21:30.383 ","End":"21:40.305","Text":"that my x(t) is going to be equal to d multiplied by Cosine of my Omega,"},{"Start":"21:40.305 ","End":"21:43.650","Text":"I have up here, so I\u0027ll just write Omega t. My Phi is going"},{"Start":"21:43.650 ","End":"21:47.775","Text":"to be equal to 0 once you workout from your initial conditions."},{"Start":"21:47.775 ","End":"21:55.155","Text":"My x_0, I\u0027ll just substitute it in m_1g divided by k. Now,"},{"Start":"21:55.155 ","End":"21:58.994","Text":"a little tip to save you some time on working out this algebra."},{"Start":"21:58.994 ","End":"22:02.969","Text":"If ever, you ever get a question in harmonic motion,"},{"Start":"22:02.969 ","End":"22:07.409","Text":"where you\u0027re starting distance is some distance"},{"Start":"22:07.409 ","End":"22:11.970","Text":"from the point of equilibrium or from your x_0,"},{"Start":"22:11.970 ","End":"22:15.495","Text":"that this value over here is the same value over her."},{"Start":"22:15.495 ","End":"22:18.240","Text":"This is 0 that your x_0 will be over here,"},{"Start":"22:18.240 ","End":"22:20.048","Text":"your value will be 0,"},{"Start":"22:20.048 ","End":"22:21.885","Text":"and that you\u0027re starting from rest."},{"Start":"22:21.885 ","End":"22:25.064","Text":"That means that your x dot at t=0 is 0."},{"Start":"22:25.064 ","End":"22:29.520","Text":"Then you can always know in these cases that your Phi will equal 0"},{"Start":"22:29.520 ","End":"22:34.680","Text":"and your A will be the distance that you are pulled down."},{"Start":"22:34.680 ","End":"22:36.915","Text":"I have d plus x_0,"},{"Start":"22:36.915 ","End":"22:38.430","Text":"but because I have an x_0 here,"},{"Start":"22:38.430 ","End":"22:40.680","Text":"so it will cancel out in the calculation."},{"Start":"22:40.680 ","End":"22:43.440","Text":"You\u0027ll see if you do this on a piece of paper."},{"Start":"22:43.440 ","End":"22:44.969","Text":"My starting velocity is 0,"},{"Start":"22:44.969 ","End":"22:50.369","Text":"which means that always my A will equal this value and my Phi will equal 0."},{"Start":"22:50.369 ","End":"22:52.064","Text":"Now, if that wasn\u0027t clear to you,"},{"Start":"22:52.064 ","End":"22:53.685","Text":"please pause the video,"},{"Start":"22:53.685 ","End":"22:57.929","Text":"take time to substitute in these initial conditions into"},{"Start":"22:57.929 ","End":"23:02.940","Text":"the equation and to really see what you get for A and what you get for Phi."},{"Start":"23:02.940 ","End":"23:07.335","Text":"Now, the next thing that I have to find is my acceleration."},{"Start":"23:07.335 ","End":"23:11.760","Text":"Because I\u0027m going to be using this equation such that my t doesn\u0027t equal 0."},{"Start":"23:11.760 ","End":"23:16.560","Text":"Then also subsequently this equation. Let\u0027s see."},{"Start":"23:16.560 ","End":"23:21.840","Text":"My acceleration is simply going to be my second derivative of"},{"Start":"23:21.840 ","End":"23:27.915","Text":"my x as a function of t. If I take the derivative of this twice,"},{"Start":"23:27.915 ","End":"23:36.640","Text":"I will get negative Omega^2 d multiplied by Cosine(Omega t)."},{"Start":"23:38.210 ","End":"23:42.870","Text":"Now if I substitute in my Omega^2,"},{"Start":"23:42.870 ","End":"23:48.240","Text":"it\u0027s going to be equal to negative and my Omega^2 is just without the square root,"},{"Start":"23:48.240 ","End":"23:55.530","Text":"negative k divided by m_1 plus1/2 m_2,"},{"Start":"23:55.530 ","End":"24:00.950","Text":"then multiplied by d Cosine(Omega t)."},{"Start":"24:00.950 ","End":"24:02.830","Text":"I won\u0027t write it in now."},{"Start":"24:02.830 ","End":"24:07.559","Text":"Now, what I\u0027m going to do is I have my x and I have my acceleration,"},{"Start":"24:07.559 ","End":"24:10.815","Text":"so I can substitute them in over here and over here,"},{"Start":"24:10.815 ","End":"24:14.490","Text":"and then I can see what my T_1 and T_2 is equal to."},{"Start":"24:14.490 ","End":"24:17.799","Text":"Then I can set a condition,"},{"Start":"24:18.060 ","End":"24:25.975","Text":"and enforce that my T_1 and T_2 will always be bigger than 0."},{"Start":"24:25.975 ","End":"24:33.370","Text":"Let\u0027s begin by substituting my acceleration into my equation for T_1. Let\u0027s see."},{"Start":"24:33.370 ","End":"24:42.400","Text":"I have my T_1 is equal to m_1g negative m_1 multiplied by my acceleration."},{"Start":"24:42.400 ","End":"24:47.649","Text":"Negative and negative, this will be a plus m_1 multiplied by k"},{"Start":"24:47.649 ","End":"24:53.440","Text":"divided by m_1 plus 1/2 of m_2,"},{"Start":"24:53.440 ","End":"25:00.725","Text":"and this is multiplied by dcosine(omega t)."},{"Start":"25:00.725 ","End":"25:06.669","Text":"Now I\u0027m putting in that it has to be bigger or equal to 0."},{"Start":"25:06.669 ","End":"25:11.920","Text":"Notice that just means that throughout the whole time I\u0027m doing this experiment,"},{"Start":"25:11.920 ","End":"25:14.695","Text":"my T_1 has to be bigger, or equal to 0."},{"Start":"25:14.695 ","End":"25:16.510","Text":"That means any t,"},{"Start":"25:16.510 ","End":"25:19.209","Text":"I put in for time,"},{"Start":"25:19.209 ","End":"25:22.254","Text":"anytime value I substitute in,"},{"Start":"25:22.254 ","End":"25:25.465","Text":"my T_1 must be bigger or equal to 0."},{"Start":"25:25.465 ","End":"25:31.375","Text":"Now what I\u0027m going to find is the smallest value I can have for T_1,"},{"Start":"25:31.375 ","End":"25:34.675","Text":"and make sure that that\u0027s always bigger than 0,"},{"Start":"25:34.675 ","End":"25:38.155","Text":"and if my smallest value for tension is always bigger than 0,"},{"Start":"25:38.155 ","End":"25:44.170","Text":"then obviously bigger values for my T_1 are also going to be bigger than 0."},{"Start":"25:44.170 ","End":"25:46.554","Text":"There\u0027s 2 ways."},{"Start":"25:46.554 ","End":"25:54.610","Text":"1 of the ways in order to find our minimum tension is to take the derivative of this."},{"Start":"25:54.610 ","End":"25:57.895","Text":"Then with the derivative I say that that is equal to 0,"},{"Start":"25:57.895 ","End":"26:00.444","Text":"and then through that I can find the minimum."},{"Start":"26:00.444 ","End":"26:03.400","Text":"That\u0027s 1 way. However, it\u0027s a little bit complicated."},{"Start":"26:03.400 ","End":"26:06.139","Text":"Sometimes it can be taking the derivative,"},{"Start":"26:06.139 ","End":"26:07.899","Text":"so instead of messing around with that,"},{"Start":"26:07.899 ","End":"26:09.355","Text":"here it\u0027s pretty simple."},{"Start":"26:09.355 ","End":"26:11.965","Text":"My m_1 g is a constant,"},{"Start":"26:11.965 ","End":"26:14.770","Text":"and what I have here is all constants."},{"Start":"26:14.770 ","End":"26:20.950","Text":"The only thing that is changing or they can change the value is my cosine(Omega t)."},{"Start":"26:20.950 ","End":"26:26.440","Text":"What I have to do is I have to find the lowest value that my cosine(Omega t) can"},{"Start":"26:26.440 ","End":"26:32.080","Text":"be and make sure that that is bigger than 0 because everything else is a constant."},{"Start":"26:32.080 ","End":"26:35.034","Text":"It\u0027s going to remain the same all the time."},{"Start":"26:35.034 ","End":"26:42.070","Text":"The lowest value that my cosine(Omega t) can be is negative 1."},{"Start":"26:42.070 ","End":"26:46.360","Text":"The lowest value that cosine or sine can ever be is negative 1."},{"Start":"26:46.360 ","End":"26:48.670","Text":"Cosine and sine,"},{"Start":"26:48.670 ","End":"26:52.315","Text":"its value is always between negative 1 and 1,"},{"Start":"26:52.315 ","End":"26:57.055","Text":"which means that its lowest value is negative 1."},{"Start":"26:57.055 ","End":"27:01.300","Text":"Now what I\u0027m going to do is I\u0027m going to rewrite this equation."},{"Start":"27:01.300 ","End":"27:03.130","Text":"But instead of writing cosine(Omega t),"},{"Start":"27:03.130 ","End":"27:05.005","Text":"I\u0027m going to write negative 1."},{"Start":"27:05.005 ","End":"27:09.535","Text":"Then that is the minimum value that my T_1 can be."},{"Start":"27:09.535 ","End":"27:13.869","Text":"Then, because it always has to be bigger or equal to 0,"},{"Start":"27:13.869 ","End":"27:22.060","Text":"then I can rearrange it and isolate out my d and find the range that my d can be in."},{"Start":"27:22.060 ","End":"27:23.890","Text":"Let\u0027s rewrite this."},{"Start":"27:23.890 ","End":"27:32.170","Text":"I\u0027m going to have m_1g plus m_1k divided"},{"Start":"27:32.170 ","End":"27:41.814","Text":"by m_1 plus 1/2(m_2) multiplied by d multiplied by negative 1,"},{"Start":"27:41.814 ","End":"27:45.849","Text":"so I\u0027m just going to take this and put a negative over here,"},{"Start":"27:45.849 ","End":"27:49.479","Text":"and this is always bigger or equal to 0."},{"Start":"27:49.479 ","End":"27:55.883","Text":"Then if I just rearrange this equation,"},{"Start":"27:55.883 ","End":"28:00.535","Text":"then you can do the algebra on a piece of paper just to cut time over here."},{"Start":"28:00.535 ","End":"28:06.098","Text":"We should get our d is always smaller or equal to g divided by"},{"Start":"28:06.098 ","End":"28:15.730","Text":"k multiplied by m_1 plus 1/2(m_2)."},{"Start":"28:15.730 ","End":"28:18.339","Text":"That means that the maximum amplitude,"},{"Start":"28:18.339 ","End":"28:20.950","Text":"I can have my maximum d,"},{"Start":"28:20.950 ","End":"28:27.520","Text":"my d_max, so long as it\u0027s smaller than this value over here."},{"Start":"28:27.520 ","End":"28:32.230","Text":"The maximum that my d can be is if it\u0027s equal to this,"},{"Start":"28:32.230 ","End":"28:38.275","Text":"and that just promises me that my T_1 will not be equal to 0."},{"Start":"28:38.275 ","End":"28:42.115","Text":"But now I have to do the exact same thing for my T_2."},{"Start":"28:42.115 ","End":"28:46.225","Text":"I\u0027m going to rewrite my equations but for T_2 now."},{"Start":"28:46.225 ","End":"28:48.100","Text":"Let\u0027s rewrite this."},{"Start":"28:48.100 ","End":"28:53.455","Text":"I have that my T_2 is equal to kx."},{"Start":"28:53.455 ","End":"28:59.109","Text":"That means that it\u0027s equal to k multiplied by my x over here,"},{"Start":"28:59.109 ","End":"29:05.695","Text":"so It\u0027s going to be k multiplied by d cosine(Omega t),"},{"Start":"29:05.695 ","End":"29:08.064","Text":"and then k multiplied by this."},{"Start":"29:08.064 ","End":"29:09.570","Text":"The ks\u0027 will cancel out,"},{"Start":"29:09.570 ","End":"29:12.145","Text":"so I\u0027ll have m_1g."},{"Start":"29:12.145 ","End":"29:17.380","Text":"Then again, I put my condition that it\u0027s greater or equal to 0."},{"Start":"29:17.380 ","End":"29:19.615","Text":"Then the same trick as before,"},{"Start":"29:19.615 ","End":"29:22.000","Text":"I can see that my m_1g is a constant,"},{"Start":"29:22.000 ","End":"29:24.970","Text":"my k and my d is a constant,"},{"Start":"29:24.970 ","End":"29:27.025","Text":"but it\u0027s an unknown and my k is a constant."},{"Start":"29:27.025 ","End":"29:30.700","Text":"In order to find the minimum value that my T_2 can be such that"},{"Start":"29:30.700 ","End":"29:34.735","Text":"it\u0027s bigger or equal to 0,"},{"Start":"29:34.735 ","End":"29:40.404","Text":"is when my cosine of Omega t is at its smallest value,"},{"Start":"29:40.404 ","End":"29:42.355","Text":"which is negative 1."},{"Start":"29:42.355 ","End":"29:45.970","Text":"If the minimum value for my T_2 is bigger or equal to 0,"},{"Start":"29:45.970 ","End":"29:52.309","Text":"then obviously any value bigger than that is also going to be bigger or equal than 0."},{"Start":"29:52.830 ","End":"29:56.094","Text":"Now I can rewrite that."},{"Start":"29:56.094 ","End":"30:00.940","Text":"I have that my kd multiplied by negative"},{"Start":"30:00.940 ","End":"30:08.125","Text":"1 plus my m_1g is bigger or equal to 0."},{"Start":"30:08.125 ","End":"30:12.295","Text":"I\u0027ve just written instead of my cosine of substituted in here, negative 1."},{"Start":"30:12.295 ","End":"30:16.795","Text":"Then via some rearranging and some algebra,"},{"Start":"30:16.795 ","End":"30:24.444","Text":"I\u0027ll get that my d has to be smaller or equal to m_1g divided by k."},{"Start":"30:24.444 ","End":"30:33.535","Text":"Then I have to add in and my condition that my T_1 won\u0027t also equal to 0,"},{"Start":"30:33.535 ","End":"30:37.044","Text":"that it also won\u0027t reset."},{"Start":"30:37.044 ","End":"30:42.325","Text":"That was that my d must be smaller or equal to g"},{"Start":"30:42.325 ","End":"30:48.650","Text":"divided by k multiplied by m_1 plus 1/2(m_2)."},{"Start":"30:50.820 ","End":"30:55.599","Text":"I have to have that both of these conditions that"},{"Start":"30:55.599 ","End":"31:00.249","Text":"my d has to be smaller than this and smaller than this,"},{"Start":"31:00.249 ","End":"31:02.859","Text":"then none of my tensions,"},{"Start":"31:02.859 ","End":"31:05.845","Text":"not my T_1 or T_2 will reset."},{"Start":"31:05.845 ","End":"31:07.120","Text":"They\u0027ll never be equal to 0,"},{"Start":"31:07.120 ","End":"31:08.979","Text":"and then that answers my question."},{"Start":"31:08.979 ","End":"31:10.629","Text":"In order to do that,"},{"Start":"31:10.629 ","End":"31:15.130","Text":"I have to take the more extreme value."},{"Start":"31:15.130 ","End":"31:18.085","Text":"That means the smallest value between the 2."},{"Start":"31:18.085 ","End":"31:23.634","Text":"Then if my d is smaller than the smallest maximum value,"},{"Start":"31:23.634 ","End":"31:28.480","Text":"then I know that I\u0027m within the limits for the question."},{"Start":"31:28.480 ","End":"31:33.399","Text":"That will be the maximum distance. Let\u0027s take a look."},{"Start":"31:33.399 ","End":"31:35.184","Text":"How do I know which is smaller?"},{"Start":"31:35.184 ","End":"31:38.619","Text":"I can see my masses are always going to be positive."},{"Start":"31:38.619 ","End":"31:39.940","Text":"Here I have m_1,"},{"Start":"31:39.940 ","End":"31:42.310","Text":"in both places I have g divided by k,"},{"Start":"31:42.310 ","End":"31:44.530","Text":"g divided by k. Here I have my m_1,"},{"Start":"31:44.530 ","End":"31:51.025","Text":"and here I have m_1 plus another positive constant."},{"Start":"31:51.025 ","End":"31:53.319","Text":"Obviously, this number over here,"},{"Start":"31:53.319 ","End":"31:58.420","Text":"m_1 plus another positive constant is going to be bigger than my m_1,"},{"Start":"31:58.420 ","End":"32:05.484","Text":"which means that m_1g divided by k is going to be smaller than this over here."},{"Start":"32:05.484 ","End":"32:09.355","Text":"That means therefore that my d_max,"},{"Start":"32:09.355 ","End":"32:14.440","Text":"I\u0027m reminding you the maximum distance which the mass can be pulled such that"},{"Start":"32:14.440 ","End":"32:20.979","Text":"the tension both T_1 and T_2 don\u0027t reset throughout the motion."},{"Start":"32:20.979 ","End":"32:25.910","Text":"That means that my d_max is simply going to be equal to this,"},{"Start":"32:28.620 ","End":"32:31.705","Text":"m_1g divided by k. Because my m_1g,"},{"Start":"32:31.705 ","End":"32:33.445","Text":"the d that I can be pulled."},{"Start":"32:33.445 ","End":"32:35.500","Text":"If this is my point of equilibrium,"},{"Start":"32:35.500 ","End":"32:39.475","Text":"my m_1g divided by k is up until here."},{"Start":"32:39.475 ","End":"32:48.295","Text":"This is m_1g divided by k. Then this value over here is lower."},{"Start":"32:48.295 ","End":"32:51.350","Text":"It\u0027s this value of g divided by k,"},{"Start":"32:51.350 ","End":"32:57.085","Text":"and m_1 plus 1/2(m_2)."},{"Start":"32:57.085 ","End":"33:02.215","Text":"I need the maximum distance such that my tensions won\u0027t become 0."},{"Start":"33:02.215 ","End":"33:05.574","Text":"If I pull more than this value,"},{"Start":"33:05.574 ","End":"33:07.989","Text":"so if my d is larger,"},{"Start":"33:07.989 ","End":"33:11.410","Text":"then that means that my T_1 wouldn\u0027t be equal to 0,"},{"Start":"33:11.410 ","End":"33:13.465","Text":"but my T_2 will be equal to 0,"},{"Start":"33:13.465 ","End":"33:18.490","Text":"and then I\u0027m not meeting the conditions set to me from the question."},{"Start":"33:18.490 ","End":"33:21.520","Text":"That means I have to take the smallest value because,"},{"Start":"33:21.520 ","End":"33:22.915","Text":"obviously, at this value,"},{"Start":"33:22.915 ","End":"33:28.585","Text":"I\u0027m at a point where neither my T_2 nor my T_1 will be equal to 0."},{"Start":"33:28.585 ","End":"33:33.025","Text":"That is the end of the question,"},{"Start":"33:33.025 ","End":"33:38.060","Text":"and this is the answer to my question 3."}],"ID":9434},{"Watched":false,"Name":"Rod On Two Wheels","Duration":"11m 36s","ChapterTopicVideoID":9166,"CourseChapterTopicPlaylistID":5369,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.235","Text":"Hello. In this question,"},{"Start":"00:02.235 ","End":"00:06.750","Text":"we have a plank which is resting on 2 wheels."},{"Start":"00:06.750 ","End":"00:13.750","Text":"Each wheel is spinning with an angular velocity of Omega 0."},{"Start":"00:15.140 ","End":"00:24.120","Text":"The distance from the center of mass of a plank or of a rod,"},{"Start":"00:24.120 ","End":"00:27.195","Text":"until the center of one of the wheels,"},{"Start":"00:27.195 ","End":"00:30.705","Text":"is a distance d. Of course,"},{"Start":"00:30.705 ","End":"00:32.265","Text":"we have over here,"},{"Start":"00:32.265 ","End":"00:41.170","Text":"kinetic friction with a coefficient Mu_k and the same over here, Mu_k."},{"Start":"00:43.220 ","End":"00:47.625","Text":"The rod is of mass m and now,"},{"Start":"00:47.625 ","End":"00:50.205","Text":"we give a small little knock to the rod,"},{"Start":"00:50.205 ","End":"00:54.760","Text":"and we want to show that this rod is moving in harmonic motion."},{"Start":"00:54.800 ","End":"00:58.245","Text":"When we\u0027re working with kinetic friction,"},{"Start":"00:58.245 ","End":"01:02.340","Text":"so the best way to work this out is to"},{"Start":"01:02.340 ","End":"01:07.690","Text":"use the equation for the sum of forces and the sum of all of the torques."},{"Start":"01:07.880 ","End":"01:12.135","Text":"Let\u0027s imagine that our center of mass,"},{"Start":"01:12.135 ","End":"01:16.815","Text":"this was our center of mass at the point where it\u0027s even,"},{"Start":"01:16.815 ","End":"01:19.080","Text":"an equal distance between each wheel."},{"Start":"01:19.080 ","End":"01:22.440","Text":"Now let\u0027s say that we\u0027ve started our emotion and our center of mass is over here"},{"Start":"01:22.440 ","End":"01:26.400","Text":"and the whole rod has moved a bit forward."},{"Start":"01:26.400 ","End":"01:28.197","Text":"Let\u0027s draw in our forces."},{"Start":"01:28.197 ","End":"01:31.980","Text":"We have our mg pointing downwards."},{"Start":"01:31.980 ","End":"01:34.020","Text":"Then we have normal forces,"},{"Start":"01:34.020 ","End":"01:41.490","Text":"so we have our N_1 and over here we\u0027ll have our N_2."},{"Start":"01:41.490 ","End":"01:51.330","Text":"Our normal forces are different because it depends to which side our rod is leaning to."},{"Start":"01:51.330 ","End":"01:54.195","Text":"Because for instance, now if our center of mass is moved here,"},{"Start":"01:54.195 ","End":"01:57.120","Text":"then more of the rod is on this wheel."},{"Start":"01:57.120 ","End":"02:02.595","Text":"This normal force will be bigger than this one because it has to hold more weight."},{"Start":"02:02.595 ","End":"02:07.770","Text":"Then the next forces that we have to draw are our frictional forces."},{"Start":"02:07.770 ","End":"02:10.500","Text":"Now, of course, our frictional forces are also going to be"},{"Start":"02:10.500 ","End":"02:14.070","Text":"different because our normal forces are different."},{"Start":"02:14.070 ","End":"02:16.755","Text":"This is going to be f_k1,"},{"Start":"02:16.755 ","End":"02:18.255","Text":"our first frictional force,"},{"Start":"02:18.255 ","End":"02:23.160","Text":"and then in this direction will have our f_k2."},{"Start":"02:23.160 ","End":"02:25.845","Text":"Now they\u0027re pointing in this direction because if we look,"},{"Start":"02:25.845 ","End":"02:28.260","Text":"when the rod is moving in this direction,"},{"Start":"02:28.260 ","End":"02:32.190","Text":"the friction f_k1 is going to try to oppose the force,"},{"Start":"02:32.190 ","End":"02:34.455","Text":"so it\u0027s going to be pushing in the opposite direction."},{"Start":"02:34.455 ","End":"02:36.000","Text":"A similar story here,"},{"Start":"02:36.000 ","End":"02:38.850","Text":"when the rod is moving in the leftward direction,"},{"Start":"02:38.850 ","End":"02:43.330","Text":"so f_k2 is going to try and oppose that motion."},{"Start":"02:43.760 ","End":"02:49.245","Text":"Now let\u0027s write out our equation for the sum of all of our forces."},{"Start":"02:49.245 ","End":"02:54.525","Text":"We have the sum of all of our forces in our y-axis is going to be."},{"Start":"02:54.525 ","End":"02:59.370","Text":"Let\u0027s say that this is the positive x-direction."},{"Start":"02:59.370 ","End":"03:02.235","Text":"We\u0027re going to have mg,"},{"Start":"03:02.235 ","End":"03:05.910","Text":"and then we have minus our N_1 and minus"},{"Start":"03:05.910 ","End":"03:10.810","Text":"our N_2 because they\u0027re pointing in the negative y-direction."},{"Start":"03:12.020 ","End":"03:14.655","Text":"We\u0027re working in the y-axis now."},{"Start":"03:14.655 ","End":"03:20.625","Text":"Then this is going to be equal to our mass multiplied by acceleration in the y-direction."},{"Start":"03:20.625 ","End":"03:23.205","Text":"Now acceleration in the y-direction is equal to 0,"},{"Start":"03:23.205 ","End":"03:24.855","Text":"so it\u0027s going to be equal to 0."},{"Start":"03:24.855 ","End":"03:29.535","Text":"Then we can write out the sum of all of the forces in the x-direction."},{"Start":"03:29.535 ","End":"03:34.200","Text":"Let\u0027s say that this is the positive x-direction."},{"Start":"03:34.200 ","End":"03:37.484","Text":"We\u0027re going to have our frictional force f_k2"},{"Start":"03:37.484 ","End":"03:40.785","Text":"going in the positive x-direction and then negative,"},{"Start":"03:40.785 ","End":"03:45.555","Text":"our f_k1 because it\u0027s going in the negative direction,"},{"Start":"03:45.555 ","End":"03:48.165","Text":"x-direction, and that\u0027s going to be a mass"},{"Start":"03:48.165 ","End":"03:52.185","Text":"multiplied by acceleration in the x-direction."},{"Start":"03:52.185 ","End":"03:57.165","Text":"Then our last equation is going to be the sum of all of the torques."},{"Start":"03:57.165 ","End":"03:59.070","Text":"Let\u0027s take a look."},{"Start":"03:59.070 ","End":"04:04.080","Text":"The first thing we have to do is we have to choose our axis of rotation."},{"Start":"04:04.080 ","End":"04:10.860","Text":"Let\u0027s say that our axis of rotation is at this corner over here,"},{"Start":"04:10.860 ","End":"04:16.035","Text":"where our plank is leaning on one of the wheels."},{"Start":"04:16.035 ","End":"04:19.395","Text":"Now let\u0027s do according to forces."},{"Start":"04:19.395 ","End":"04:22.990","Text":"We\u0027re going to have our mg working from there."},{"Start":"04:22.990 ","End":"04:25.415","Text":"Why have I done this also?"},{"Start":"04:25.415 ","End":"04:27.230","Text":"Because then I don\u0027t have to take into account"},{"Start":"04:27.230 ","End":"04:30.395","Text":"these forces because they\u0027re coming out of the axis of rotation."},{"Start":"04:30.395 ","End":"04:32.840","Text":"It\u0027s slightly less complicated equation,"},{"Start":"04:32.840 ","End":"04:35.950","Text":"it\u0027s still correct and less working out."},{"Start":"04:35.950 ","End":"04:42.975","Text":"We have mg multiplied by its distance away from this point. Let\u0027s take a look."},{"Start":"04:42.975 ","End":"04:47.145","Text":"We said before that at a starting position where"},{"Start":"04:47.145 ","End":"04:51.570","Text":"the rod is completely symmetrical around these wheels."},{"Start":"04:51.570 ","End":"04:55.230","Text":"The distance from the center of the wheel until our center of mass is"},{"Start":"04:55.230 ","End":"04:59.220","Text":"a distance of d. Then we said that we shifted our rod a little bit,"},{"Start":"04:59.220 ","End":"05:01.425","Text":"so its center of mass also shifted."},{"Start":"05:01.425 ","End":"05:05.190","Text":"Let\u0027s call this distance that it will shift,"},{"Start":"05:05.190 ","End":"05:11.700","Text":"let\u0027s call it x because"},{"Start":"05:11.700 ","End":"05:18.705","Text":"this is d. Then we can say that this distance therefore over here,"},{"Start":"05:18.705 ","End":"05:23.470","Text":"is going to be equal to d minus x."},{"Start":"05:23.630 ","End":"05:27.210","Text":"Because this distance originally was d,"},{"Start":"05:27.210 ","End":"05:30.675","Text":"and then we shifted backwards x, so minus x."},{"Start":"05:30.675 ","End":"05:35.580","Text":"It\u0027s going to be mg multiplied by d minus x,"},{"Start":"05:35.580 ","End":"05:39.090","Text":"which is our distance away from the axis of rotation."},{"Start":"05:39.090 ","End":"05:44.800","Text":"Then our next force that we\u0027re dealing with is our N_2."},{"Start":"05:44.870 ","End":"05:51.195","Text":"Because we\u0027re working with forces that are acting in the y-direction."},{"Start":"05:51.195 ","End":"05:53.055","Text":"We\u0027re going to have"},{"Start":"05:53.055 ","End":"05:59.370","Text":"negative N_2 because it\u0027s working in the negative y-direction, it\u0027s pointing upwards,"},{"Start":"05:59.370 ","End":"06:04.875","Text":"multiplied by its distance from the axis of rotation,"},{"Start":"06:04.875 ","End":"06:13.500","Text":"which is going to be d plus x plus d minus x."},{"Start":"06:13.500 ","End":"06:14.970","Text":"Our x\u0027s will cancel out."},{"Start":"06:14.970 ","End":"06:18.765","Text":"It\u0027s going to be multiplied by 2d,"},{"Start":"06:18.765 ","End":"06:22.950","Text":"the distance from the axis of rotation and this is going to be equal"},{"Start":"06:22.950 ","End":"06:27.405","Text":"to our Alpha and because our angular acceleration is equal to 0,"},{"Start":"06:27.405 ","End":"06:31.480","Text":"so this is going to be equal to 0."},{"Start":"06:31.700 ","End":"06:34.065","Text":"For my third equation,"},{"Start":"06:34.065 ","End":"06:36.195","Text":"which is this one, over here,"},{"Start":"06:36.195 ","End":"06:38.880","Text":"I can isolate out my N_2."},{"Start":"06:38.880 ","End":"06:42.615","Text":"I\u0027ll get that my N_2 is equal to,"},{"Start":"06:42.615 ","End":"06:44.550","Text":"by doing some algebra,"},{"Start":"06:44.550 ","End":"06:53.590","Text":"it\u0027s equal to mg divided by 2d multiplied by d minus x."},{"Start":"06:53.900 ","End":"07:00.130","Text":"Then if I substitute this into my first equation,"},{"Start":"07:00.500 ","End":"07:08.085","Text":"then I will get mg minus N_1 minus my N_2."},{"Start":"07:08.085 ","End":"07:17.175","Text":"I\u0027ll substitute in this mg divided by 2d multiplied by d minus x is equal to 0."},{"Start":"07:17.175 ","End":"07:20.955","Text":"Then I can simply isolate out my N_1."},{"Start":"07:20.955 ","End":"07:28.740","Text":"I\u0027ll get that my N_1 is equal to mg multiplied by 1 minus 1 divided by"},{"Start":"07:28.740 ","End":"07:36.465","Text":"2d multiplied by d minus x and that\u0027s it."},{"Start":"07:36.465 ","End":"07:38.560","Text":"That\u0027s our N_1."},{"Start":"07:38.570 ","End":"07:42.690","Text":"Then from our second equation,"},{"Start":"07:42.690 ","End":"07:46.770","Text":"what we\u0027re going to do is we\u0027re going to substitute instead of our f_k,"},{"Start":"07:46.770 ","End":"07:51.845","Text":"we\u0027re going to use Mu_k multiplied by the normal forces."},{"Start":"07:51.845 ","End":"07:53.955","Text":"Let\u0027s do that."},{"Start":"07:53.955 ","End":"07:55.710","Text":"Instead of f_k2,"},{"Start":"07:55.710 ","End":"07:59.060","Text":"we\u0027re going to have Mu_k and then we can take that as"},{"Start":"07:59.060 ","End":"08:04.540","Text":"a common factor multiplied by N_2 minus N_1."},{"Start":"08:04.540 ","End":"08:05.880","Text":"If you open out the brackets,"},{"Start":"08:05.880 ","End":"08:08.870","Text":"you\u0027ll get exactly this and this is going to be equal"},{"Start":"08:08.870 ","End":"08:13.145","Text":"to our m multiplied by acceleration in the x-axis."},{"Start":"08:13.145 ","End":"08:15.710","Text":"Now let\u0027s substitute in our N_2 and N_1."},{"Start":"08:15.710 ","End":"08:20.715","Text":"We\u0027ll have that Mu_k multiplied by our N_2,"},{"Start":"08:20.715 ","End":"08:27.450","Text":"which is mg divided by 2d multiplied by d minus x,"},{"Start":"08:27.450 ","End":"08:29.325","Text":"minus our N_1,"},{"Start":"08:29.325 ","End":"08:39.840","Text":"which is mg multiplied by 1 minus 1 divided by 2d multiplied by d minus x."},{"Start":"08:39.840 ","End":"08:49.740","Text":"Then we know that this is equal to our m a_x."},{"Start":"08:49.940 ","End":"08:54.590","Text":"Now what we want to do is we want to move this equation"},{"Start":"08:54.590 ","End":"08:58.830","Text":"into that of the form of harmonic motion."},{"Start":"08:58.830 ","End":"09:02.670","Text":"Let\u0027s start by taking out common terms."},{"Start":"09:02.670 ","End":"09:05.480","Text":"We have in both cases,"},{"Start":"09:05.480 ","End":"09:13.940","Text":"obviously our Mu_k and then we also have multiplied by mg. Then inside the brackets,"},{"Start":"09:13.940 ","End":"09:17.240","Text":"we have d minus x divided by"},{"Start":"09:17.240 ","End":"09:23.520","Text":"2d minus 1 plus"},{"Start":"09:23.520 ","End":"09:28.685","Text":"d minus x and again divided by 2d."},{"Start":"09:28.685 ","End":"09:32.930","Text":"This is equal to our m a_x."},{"Start":"09:32.930 ","End":"09:35.660","Text":"Now I can divide both sides by m,"},{"Start":"09:35.660 ","End":"09:39.010","Text":"and I can add these on to each other."},{"Start":"09:39.010 ","End":"09:45.735","Text":"I have Mu_k is equal to d minus x divided by d,"},{"Start":"09:45.735 ","End":"09:47.265","Text":"because I\u0027ll have two of these,"},{"Start":"09:47.265 ","End":"09:52.570","Text":"minus 1 is equal to my a_x."},{"Start":"09:52.760 ","End":"09:56.355","Text":"Now I can expand this over here."},{"Start":"09:56.355 ","End":"09:59.775","Text":"Let\u0027s see what this will look like."},{"Start":"09:59.775 ","End":"10:09.060","Text":"My d minus x divided by d is the same as 1 minus x over d."},{"Start":"10:09.060 ","End":"10:18.800","Text":"Now I can rewrite this as Mu_k multiplied by 1 minus x over d minus 1,"},{"Start":"10:18.800 ","End":"10:22.020","Text":"so 1 and minus 1 will cancel out."},{"Start":"10:22.020 ","End":"10:25.480","Text":"Negative x divided by d,"},{"Start":"10:25.480 ","End":"10:29.300","Text":"which is equal to our a,"},{"Start":"10:29.300 ","End":"10:32.210","Text":"because this and this cross out together."},{"Start":"10:32.210 ","End":"10:36.360","Text":"Now let\u0027s open this up a little bit more."},{"Start":"10:36.410 ","End":"10:39.765","Text":"Now we can just take out these brackets."},{"Start":"10:39.765 ","End":"10:45.950","Text":"We\u0027ll get negative Mu_k divided by,"},{"Start":"10:45.950 ","End":"10:50.460","Text":"I\u0027m missing a g over here and also over here."},{"Start":"10:50.460 ","End":"10:57.950","Text":"Mu_kg divided by d multiplied by x is equal to our a,"},{"Start":"10:57.950 ","End":"11:00.290","Text":"as we know, is equal to x double dot."},{"Start":"11:00.290 ","End":"11:02.405","Text":"Because it\u0027s in x-direction,"},{"Start":"11:02.405 ","End":"11:04.715","Text":"so this is equal to x double dot."},{"Start":"11:04.715 ","End":"11:09.980","Text":"Here we\u0027ve gotten to our equation that this is harmonic motion."},{"Start":"11:09.980 ","End":"11:13.510","Text":"Then in order to find the frequency of oscillations, as we know,"},{"Start":"11:13.510 ","End":"11:16.670","Text":"it\u0027s going to be the square root of"},{"Start":"11:16.670 ","End":"11:21.080","Text":"our coefficient of our x divided by our coefficient over x double dot."},{"Start":"11:21.080 ","End":"11:23.630","Text":"Here the coefficient of my x double dot is 1."},{"Start":"11:23.630 ","End":"11:33.670","Text":"It\u0027s simply going to be Mu_kg divided by d. This is our frequency of oscillations."},{"Start":"11:34.100 ","End":"11:37.930","Text":"That\u0027s the end of this lesson."}],"ID":9435},{"Watched":false,"Name":"Mass On Surface On Vertical Spring","Duration":"22m 58s","ChapterTopicVideoID":9167,"CourseChapterTopicPlaylistID":5369,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.710","Text":"Hello. The question in front of us is a more difficult example."},{"Start":"00:04.710 ","End":"00:10.500","Text":"Because what we\u0027re going to be using is energy and springs and harmonic motion,"},{"Start":"00:10.500 ","End":"00:14.970","Text":"and it\u0027s going to be a slightly more complicated way to solve these types of questions"},{"Start":"00:14.970 ","End":"00:19.695","Text":"so it\u0027s very good to look at this video and really understand the tricks."},{"Start":"00:19.695 ","End":"00:22.230","Text":"Let\u0027s start by looking at the question,"},{"Start":"00:22.230 ","End":"00:24.735","Text":"we\u0027re told that on a spring of constant k,"},{"Start":"00:24.735 ","End":"00:26.280","Text":"let\u0027s try it over here,"},{"Start":"00:26.280 ","End":"00:31.980","Text":"our k rests a surface of mass 1 of m_1."},{"Start":"00:31.980 ","End":"00:36.270","Text":"A mass m_2 is placed on the surface so that\u0027s this box,"},{"Start":"00:36.270 ","End":"00:38.810","Text":"and then we\u0027re being told that the spring is pressed down"},{"Start":"00:38.810 ","End":"00:41.660","Text":"to a height of Delta y and is released."},{"Start":"00:41.660 ","End":"00:44.555","Text":"Let\u0027s just explain this section,"},{"Start":"00:44.555 ","End":"00:52.580","Text":"here they\u0027re saying that right now the spring is already and it\u0027s point of equilibrium,"},{"Start":"00:52.580 ","End":"00:56.030","Text":"our system is at this point of equilibrium and it\u0027s being pressed"},{"Start":"00:56.030 ","End":"01:00.985","Text":"down by a height of Delta y,"},{"Start":"01:00.985 ","End":"01:02.380","Text":"to a height of Delta y,"},{"Start":"01:02.380 ","End":"01:07.320","Text":"from the top, that means let\u0027s say this is our Delta y,"},{"Start":"01:07.320 ","End":"01:13.934","Text":"it\u0027s being pressed down till here away from its point of equilibrium,"},{"Start":"01:13.934 ","End":"01:17.675","Text":"and then the whole system is released."},{"Start":"01:17.675 ","End":"01:19.250","Text":"In our first question,"},{"Start":"01:19.250 ","End":"01:22.610","Text":"we\u0027re being asked at what value of Delta y minimum,"},{"Start":"01:22.610 ","End":"01:25.295","Text":"the minimum Delta y."},{"Start":"01:25.295 ","End":"01:29.600","Text":"What\u0027s the minimum value of Delta y that will have to have in order for"},{"Start":"01:29.600 ","End":"01:33.350","Text":"a mass at the top over here m_2 to break away from the surface?"},{"Start":"01:33.350 ","End":"01:36.004","Text":"That means that it will be moving"},{"Start":"01:36.004 ","End":"01:41.829","Text":"independently even for a moment from this surface over here."},{"Start":"01:41.829 ","End":"01:46.940","Text":"I\u0027m pushing down my spring with the surface and the mass on top of it,"},{"Start":"01:46.940 ","End":"01:48.350","Text":"some Delta y,"},{"Start":"01:48.350 ","End":"01:50.495","Text":"and I\u0027m looking for the minimum Delta y,"},{"Start":"01:50.495 ","End":"01:53.120","Text":"such that when it flies back up,"},{"Start":"01:53.120 ","End":"01:55.100","Text":"my surface will now be here,"},{"Start":"01:55.100 ","End":"01:59.060","Text":"which is a Delta y away."},{"Start":"01:59.060 ","End":"02:01.865","Text":"The box, my mass 2,"},{"Start":"02:01.865 ","End":"02:07.905","Text":"will go flying up and there will be a gap between these 2."},{"Start":"02:07.905 ","End":"02:12.190","Text":"Then I\u0027m looking for the minimum Delta y because obviously,"},{"Start":"02:12.190 ","End":"02:15.130","Text":"at Delta y\u0027s that are greater,"},{"Start":"02:15.130 ","End":"02:17.150","Text":"if I compress the spring more,"},{"Start":"02:17.150 ","End":"02:19.820","Text":"then obviously my mass will also fly off."},{"Start":"02:19.820 ","End":"02:26.225","Text":"What is the boundary or what is the limit that I can have?"},{"Start":"02:26.225 ","End":"02:28.320","Text":"We\u0027re answering question 1,"},{"Start":"02:28.320 ","End":"02:32.524","Text":"we have to find conditions so that we understand when"},{"Start":"02:32.524 ","End":"02:37.550","Text":"our mass 2 has broken away from a surface."},{"Start":"02:37.550 ","End":"02:39.905","Text":"Now, let\u0027s take a look."},{"Start":"02:39.905 ","End":"02:43.789","Text":"In order to know if our mass 2 is broken away from my surface,"},{"Start":"02:43.789 ","End":"02:46.210","Text":"that means that the normal force,"},{"Start":"02:46.210 ","End":"02:53.225","Text":"what the surface is pushing back on our m_2 if that\u0027s equal to 0,"},{"Start":"02:53.225 ","End":"02:57.560","Text":"then that means that our m_2 and our surface are not connected."},{"Start":"02:57.560 ","End":"03:00.505","Text":"Potentially, there will be touching,"},{"Start":"03:00.505 ","End":"03:06.560","Text":"but it\u0027s exactly the limit between our m_2 resting somewhat"},{"Start":"03:06.560 ","End":"03:12.880","Text":"or some of its weights on our surface and it being in the air completely disconnected."},{"Start":"03:12.880 ","End":"03:15.634","Text":"Let\u0027s draw a free force diagram,"},{"Start":"03:15.634 ","End":"03:23.360","Text":"join our forces and then I\u0027ll set this condition that might normal force is equal to 0."},{"Start":"03:23.360 ","End":"03:28.085","Text":"Let\u0027s see, we have our surface of mass m_1,"},{"Start":"03:28.085 ","End":"03:32.390","Text":"then on top of it we have our box of mass m_2,"},{"Start":"03:32.390 ","End":"03:37.335","Text":"and then we have our spring and it\u0027s connected to the floor."},{"Start":"03:37.335 ","End":"03:43.805","Text":"The forces that we have acting are N going up"},{"Start":"03:43.805 ","End":"03:50.670","Text":"and our m_2g pointing downwards."},{"Start":"03:50.670 ","End":"03:54.675","Text":"These are the forces on our body, our m_2."},{"Start":"03:54.675 ","End":"03:57.765","Text":"Now, let\u0027s say that"},{"Start":"03:57.765 ","End":"04:04.250","Text":"the positive y direction is this direction and then we can write our force equation."},{"Start":"04:04.250 ","End":"04:08.270","Text":"The sum of all of the forces on the y-axis is going to be,"},{"Start":"04:08.270 ","End":"04:10.669","Text":"so we have N in the positive direction,"},{"Start":"04:10.669 ","End":"04:15.570","Text":"negative m_2g, because our m_2g is in the negative direction,"},{"Start":"04:15.570 ","End":"04:18.335","Text":"and then this is going to be equal to."},{"Start":"04:18.335 ","End":"04:23.810","Text":"Now notice, when we\u0027re at our beginning stage before the compression,"},{"Start":"04:23.810 ","End":"04:26.120","Text":"where I point of equilibrium,"},{"Start":"04:26.120 ","End":"04:27.785","Text":"which means that there\u0027s no movement,"},{"Start":"04:27.785 ","End":"04:32.105","Text":"which means that our acceleration is equal to 0 over here."},{"Start":"04:32.105 ","End":"04:35.465","Text":"However, once we\u0027ve had some compression,"},{"Start":"04:35.465 ","End":"04:38.810","Text":"we\u0027ve moved away from a point of equilibrium,"},{"Start":"04:38.810 ","End":"04:43.550","Text":"that means that our mass over here is going to have some acceleration."},{"Start":"04:43.550 ","End":"04:47.315","Text":"The sum of all of the forces over here is not going to be equal to 0,"},{"Start":"04:47.315 ","End":"04:52.285","Text":"but it\u0027s going to be equal to m_2 multiplied by its acceleration."},{"Start":"04:52.285 ","End":"04:57.380","Text":"Now I can set in that I need my normal force to be equal to 0 in order"},{"Start":"04:57.380 ","End":"05:02.960","Text":"to represent that my m_2 has broken away from my surface."},{"Start":"05:02.960 ","End":"05:07.970","Text":"Here, I\u0027ve said that this is equal to 0 and then we can cross out our m_2"},{"Start":"05:07.970 ","End":"05:13.580","Text":"from both sides and we\u0027ll get that when our acceleration of body number 2 is equal to g,"},{"Start":"05:14.520 ","End":"05:19.785","Text":"then we know that our mass is broken away."},{"Start":"05:19.785 ","End":"05:21.930","Text":"This is the first stage."},{"Start":"05:21.930 ","End":"05:25.430","Text":"Now the next stage is to know that our acceleration"},{"Start":"05:25.430 ","End":"05:29.915","Text":"is dependent on the force coming from the spring."},{"Start":"05:29.915 ","End":"05:34.005","Text":"Let\u0027s say that we\u0027re at some Delta y,"},{"Start":"05:34.005 ","End":"05:38.430","Text":"let\u0027s say that over here is our point of equilibrium,"},{"Start":"05:38.430 ","End":"05:40.535","Text":"y is 0 tag,"},{"Start":"05:40.535 ","End":"05:47.575","Text":"and this is our Delta y which is of course exactly what we compressed,"},{"Start":"05:47.575 ","End":"05:49.790","Text":"that\u0027s exactly what we compressed over here."},{"Start":"05:49.790 ","End":"05:54.669","Text":"We went from a point of equilibrium compressed down to here,"},{"Start":"05:54.669 ","End":"05:55.850","Text":"our Delta y,"},{"Start":"05:55.850 ","End":"05:57.620","Text":"and then because we compressed down to here,"},{"Start":"05:57.620 ","End":"06:00.500","Text":"it\u0027s going to rise up the exact same amplitude,"},{"Start":"06:00.500 ","End":"06:02.260","Text":"the delta y up here."},{"Start":"06:02.260 ","End":"06:06.905","Text":"Now, our goal is to get some force equation where"},{"Start":"06:06.905 ","End":"06:12.080","Text":"our acceleration in that equation is equal to our g. From over here."},{"Start":"06:12.080 ","End":"06:13.460","Text":"Let\u0027s write this down,"},{"Start":"06:13.460 ","End":"06:17.000","Text":"the force is coming from the spring, I\u0027m reminding you again."},{"Start":"06:17.000 ","End":"06:23.270","Text":"That\u0027s going to be equal to negative k multiplied by the distance it was compressed,"},{"Start":"06:23.270 ","End":"06:25.375","Text":"which is Delta y,"},{"Start":"06:25.375 ","End":"06:27.515","Text":"and this is going to be equal to,"},{"Start":"06:27.515 ","End":"06:29.150","Text":"so up until this point,"},{"Start":"06:29.150 ","End":"06:32.920","Text":"our m_1 and m_2 are still joined."},{"Start":"06:32.920 ","End":"06:35.750","Text":"Our m_2 is still resting on our surface,"},{"Start":"06:35.750 ","End":"06:43.705","Text":"it\u0027s going to be m_1 plus m_2 multiplied by acceleration."},{"Start":"06:43.705 ","End":"06:49.385","Text":"Now that we can see the relationship between our a and our Delta y,"},{"Start":"06:49.385 ","End":"06:55.610","Text":"we can see that our a is going to be equal to k divided"},{"Start":"06:55.610 ","End":"07:03.820","Text":"by m_1 plus m_2 and then multiplied by our Delta y."},{"Start":"07:03.820 ","End":"07:08.060","Text":"Now, I haven\u0027t included in the negative sign over here,"},{"Start":"07:08.060 ","End":"07:11.600","Text":"because obviously the sign of our acceleration,"},{"Start":"07:11.600 ","End":"07:12.830","Text":"whether it\u0027s positive or negative,"},{"Start":"07:12.830 ","End":"07:16.960","Text":"depends on if we\u0027re above or below a point of equilibrium."},{"Start":"07:16.960 ","End":"07:21.680","Text":"Our sign is constantly changing so I could just ignore the sign over here."},{"Start":"07:21.680 ","End":"07:24.710","Text":"Another thing to note is our Delta y,"},{"Start":"07:24.710 ","End":"07:28.050","Text":"I didn\u0027t have to include in my forces,"},{"Start":"07:28.050 ","End":"07:30.290","Text":"this is obviously the force of the spring,"},{"Start":"07:30.290 ","End":"07:35.340","Text":"but they didn\u0027t have to include into this also my mg going downwards,"},{"Start":"07:35.340 ","End":"07:38.885","Text":"because that\u0027s already included in my Delta y."},{"Start":"07:38.885 ","End":"07:41.945","Text":"This Delta y is representing"},{"Start":"07:41.945 ","End":"07:45.830","Text":"something to do with our point of equilibrium and because of that,"},{"Start":"07:45.830 ","End":"07:48.505","Text":"because it\u0027s got something to do with a point of equilibrium,"},{"Start":"07:48.505 ","End":"07:53.085","Text":"it takes into account our mg. Now,"},{"Start":"07:53.085 ","End":"07:55.035","Text":"this is going to be my acceleration,"},{"Start":"07:55.035 ","End":"07:57.530","Text":"and I want my acceleration to be equal to"},{"Start":"07:57.530 ","End":"08:01.160","Text":"g in order for my normal force to be equal to 0,"},{"Start":"08:01.160 ","End":"08:06.814","Text":"in order to represent that my m_2 has broken away from my surface."},{"Start":"08:06.814 ","End":"08:11.780","Text":"I say that this is equal to g. Now,"},{"Start":"08:11.780 ","End":"08:14.915","Text":"because I\u0027m being asked something about my Delta y,"},{"Start":"08:14.915 ","End":"08:20.050","Text":"I\u0027m now going to isolate out my Delta y."},{"Start":"08:20.050 ","End":"08:29.740","Text":"We\u0027ll get this over here so my Delta y is going to be equal to m_1 plus m_2,"},{"Start":"08:29.740 ","End":"08:31.264","Text":"I\u0027m just rearranging this,"},{"Start":"08:31.264 ","End":"08:39.035","Text":"multiplied by g divided by k. This is in fact my Delta y min."},{"Start":"08:39.035 ","End":"08:43.340","Text":"This is the minimum compression that I need to have in"},{"Start":"08:43.340 ","End":"08:47.915","Text":"order for my mass number 2 to disconnect from the surface."},{"Start":"08:47.915 ","End":"08:52.490","Text":"Because when I have this and any compression which is larger,"},{"Start":"08:52.490 ","End":"08:54.020","Text":"if I push this down more,"},{"Start":"08:54.020 ","End":"08:56.420","Text":"obviously my mass will still disconnect,"},{"Start":"08:56.420 ","End":"08:58.115","Text":"but any smaller compression,"},{"Start":"08:58.115 ","End":"09:01.205","Text":"my mass will still stay on the surface."},{"Start":"09:01.205 ","End":"09:05.980","Text":"This came from the fact that my acceleration was equal to g,"},{"Start":"09:05.980 ","End":"09:10.190","Text":"and that came from the fact that my normal force have to be"},{"Start":"09:10.190 ","End":"09:14.705","Text":"equal to 0 to represent the breaking away."},{"Start":"09:14.705 ","End":"09:16.570","Text":"Now question 2,"},{"Start":"09:16.570 ","End":"09:20.540","Text":"we\u0027re being told what our Delta y is going to be equal to,"},{"Start":"09:20.540 ","End":"09:22.010","Text":"what are k, what are m_1,"},{"Start":"09:22.010 ","End":"09:23.795","Text":"and what are m_2 are equal to?"},{"Start":"09:23.795 ","End":"09:29.005","Text":"And then we\u0027re being asked at what moment does our m_2 breakaway from the surface?"},{"Start":"09:29.005 ","End":"09:32.855","Text":"So what are they actually asking? Here\u0027s our question 2."},{"Start":"09:32.855 ","End":"09:37.610","Text":"They\u0027re asking us to find the time,"},{"Start":"09:37.610 ","End":"09:42.440","Text":"at which our mass breaks away from the surface."},{"Start":"09:42.440 ","End":"09:45.080","Text":"What does that mean? We\u0027re going to have to find"},{"Start":"09:45.080 ","End":"09:48.890","Text":"some equation for the position of our mass."},{"Start":"09:48.890 ","End":"09:52.805","Text":"Our x as a function of t. In order to find that,"},{"Start":"09:52.805 ","End":"09:56.755","Text":"we\u0027re going to have to use our equations for harmonic motion."},{"Start":"09:56.755 ","End":"09:59.385","Text":"Our t is what we\u0027re looking for."},{"Start":"09:59.385 ","End":"10:01.700","Text":"Let\u0027s draw our diagram out again."},{"Start":"10:01.700 ","End":"10:05.390","Text":"We have our block which is m_2,"},{"Start":"10:05.390 ","End":"10:09.410","Text":"and it\u0027s resting on our surface over here."},{"Start":"10:09.410 ","End":"10:12.580","Text":"Here we have the spring attached to the ground."},{"Start":"10:12.580 ","End":"10:18.870","Text":"Again, this is our positive y direction and let\u0027s say that our 0 point is over"},{"Start":"10:18.870 ","End":"10:25.240","Text":"here and this is of course m_1."},{"Start":"10:25.240 ","End":"10:32.885","Text":"We\u0027re going to have negative m_1 plus m_2g,"},{"Start":"10:32.885 ","End":"10:35.150","Text":"so the total mass of this multiplied by"},{"Start":"10:35.150 ","End":"10:37.970","Text":"gravity and it\u0027s going in the negative y direction,"},{"Start":"10:37.970 ","End":"10:40.414","Text":"and then as always,"},{"Start":"10:40.414 ","End":"10:46.765","Text":"the force from the spring is going to be negative k multiplied by our Delta y,"},{"Start":"10:46.765 ","End":"10:48.850","Text":"instead of our y minus y_0,"},{"Start":"10:48.850 ","End":"10:53.335","Text":"we\u0027re using our delta y to represent that and that is equal to a mass,"},{"Start":"10:53.335 ","End":"10:59.990","Text":"that\u0027s m_1 plus m_2 multiplied by acceleration."},{"Start":"10:59.990 ","End":"11:05.780","Text":"Now, because I said that my point of equilibrium is at this point where this is our 0,"},{"Start":"11:05.780 ","End":"11:09.545","Text":"I can say that when it\u0027s being compressed Delta y,"},{"Start":"11:09.545 ","End":"11:12.440","Text":"I can just say that it\u0027s equal to y,"},{"Start":"11:12.440 ","End":"11:15.530","Text":"because it\u0027s going to be Delta y minus 0,"},{"Start":"11:15.530 ","End":"11:18.905","Text":"it\u0027s going to be y minus 0 is our Delta y, so that\u0027s why."},{"Start":"11:18.905 ","End":"11:22.235","Text":"Now, let\u0027s rewrite that."},{"Start":"11:22.235 ","End":"11:26.660","Text":"Now, I want to put this into the format for harmonic motion."},{"Start":"11:26.660 ","End":"11:32.900","Text":"I\u0027m going to do my little sneaky trick where I take my k out as a common factor,"},{"Start":"11:32.900 ","End":"11:37.167","Text":"so I have my k and then I have my y."},{"Start":"11:37.167 ","End":"11:44.815","Text":"Then plus my m_1 plus m_2 g"},{"Start":"11:44.815 ","End":"11:48.130","Text":"divided by k. Now,"},{"Start":"11:48.130 ","End":"11:49.330","Text":"if you open up the brackets,"},{"Start":"11:49.330 ","End":"11:51.160","Text":"you\u0027ll see you get that exact same thing."},{"Start":"11:51.160 ","End":"11:58.560","Text":"This is going to be equal to my m_1 plus m_2 and it\u0027s of course,"},{"Start":"11:58.560 ","End":"12:03.135","Text":"going to be multiplied by instead of a y double dot."},{"Start":"12:03.135 ","End":"12:06.080","Text":"Now we have a differential equation."},{"Start":"12:06.080 ","End":"12:08.680","Text":"Now, we\u0027ve gotten to the stage where we have"},{"Start":"12:08.680 ","End":"12:12.100","Text":"a differential equation and it\u0027s in the harmonic motion form."},{"Start":"12:12.100 ","End":"12:14.815","Text":"Let\u0027s just take a look at that."},{"Start":"12:14.815 ","End":"12:21.895","Text":"What we have over here is we have negative k multiplied by our y minus y_0."},{"Start":"12:21.895 ","End":"12:24.370","Text":"This is in fact our y_0."},{"Start":"12:24.370 ","End":"12:27.459","Text":"Now, notice I\u0027ve done this incorrectly."},{"Start":"12:27.459 ","End":"12:32.560","Text":"We have to include the positive here because the usual format is inside the brackets,"},{"Start":"12:32.560 ","End":"12:34.690","Text":"we have y minus y_0."},{"Start":"12:34.690 ","End":"12:37.825","Text":"That means that our y_0 tag,"},{"Start":"12:37.825 ","End":"12:40.135","Text":"our point of equilibrium,"},{"Start":"12:40.135 ","End":"12:43.750","Text":"is in fact equal to negative of"},{"Start":"12:43.750 ","End":"12:52.050","Text":"m_1 plus m_2 g divided by k because this is meant to be a negative sign,"},{"Start":"12:52.050 ","End":"12:54.750","Text":"which means that our point of equilibrium, our y tag 0,"},{"Start":"12:54.750 ","End":"12:58.975","Text":"which is over here, so it\u0027s going to be negative this."},{"Start":"12:58.975 ","End":"13:05.455","Text":"Then what we have over here is our effective mass."},{"Start":"13:05.455 ","End":"13:07.600","Text":"We have this equation."},{"Start":"13:07.600 ","End":"13:11.470","Text":"Now we can write our position equation."},{"Start":"13:11.470 ","End":"13:14.200","Text":"This will be actually y as a function of t."},{"Start":"13:14.200 ","End":"13:18.280","Text":"We know that the general solution for this differential equation,"},{"Start":"13:18.280 ","End":"13:21.010","Text":"for harmonic motion, is going to be y(t),"},{"Start":"13:21.010 ","End":"13:25.900","Text":"which is equal to A cosine of Omega t,"},{"Start":"13:25.900 ","End":"13:35.305","Text":"where our Omega is simply the square root of our k divided by our m_1 plus m_2,"},{"Start":"13:35.305 ","End":"13:42.850","Text":"plus Phi and then plus here our y_0,"},{"Start":"13:42.850 ","End":"13:45.865","Text":"our point of equilibrium, our y_0 tag."},{"Start":"13:45.865 ","End":"13:48.220","Text":"But here our y_0 tag is a negative,"},{"Start":"13:48.220 ","End":"13:51.550","Text":"so this will be negative and then here"},{"Start":"13:51.550 ","End":"14:00.070","Text":"m_1 plus m_2 multiplied by g divided by k. Now,"},{"Start":"14:00.070 ","End":"14:02.350","Text":"it\u0027s going to be a little bit harder for us to solve"},{"Start":"14:02.350 ","End":"14:05.680","Text":"things with this over here, this extra."},{"Start":"14:05.680 ","End":"14:08.050","Text":"What is a very common thing to do,"},{"Start":"14:08.050 ","End":"14:12.550","Text":"is to just say that where our y_0 tag was,"},{"Start":"14:12.550 ","End":"14:14.680","Text":"our point of equilibrium."},{"Start":"14:14.680 ","End":"14:20.065","Text":"We can call some new variable"},{"Start":"14:20.065 ","End":"14:23.950","Text":"our y Tilde and say that is at 0"},{"Start":"14:23.950 ","End":"14:28.765","Text":"so that we can move our origin over here to our point of equilibrium."},{"Start":"14:28.765 ","End":"14:31.120","Text":"Then when I do that,"},{"Start":"14:31.120 ","End":"14:33.400","Text":"when I move my origin to my point of equilibrium,"},{"Start":"14:33.400 ","End":"14:37.930","Text":"then I can say that my y Tilde as a function of t is simply equal"},{"Start":"14:37.930 ","End":"14:44.380","Text":"to A cosine of Omegat plus Phi."},{"Start":"14:44.380 ","End":"14:49.285","Text":"I don\u0027t have to take in this constant into my calculations."},{"Start":"14:49.285 ","End":"14:54.310","Text":"Now, we\u0027re going to find our A and our Phi from our initial conditions."},{"Start":"14:54.310 ","End":"14:57.280","Text":"Our A we know because in the question we\u0027re being told"},{"Start":"14:57.280 ","End":"15:01.495","Text":"that this whole system is being compressed and then released,"},{"Start":"15:01.495 ","End":"15:03.115","Text":"so it\u0027s released from rest."},{"Start":"15:03.115 ","End":"15:05.140","Text":"Whenever something is released from rest,"},{"Start":"15:05.140 ","End":"15:13.345","Text":"we know that our A is simply going to be equal to its starting amplitude."},{"Start":"15:13.345 ","End":"15:18.745","Text":"We know that our starting amplitude is over here from the question."},{"Start":"15:18.745 ","End":"15:20.425","Text":"Let\u0027s go back 1 second."},{"Start":"15:20.425 ","End":"15:22.510","Text":"Our starting amplitude,"},{"Start":"15:22.510 ","End":"15:25.555","Text":"our Delta y is 2 of Delta y_min,"},{"Start":"15:25.555 ","End":"15:29.320","Text":"which we worked out over here. Let\u0027s go back."},{"Start":"15:29.320 ","End":"15:35.150","Text":"This is going to be 2 Delta y_min."},{"Start":"15:35.730 ","End":"15:43.630","Text":"We know that our amplitude is simply going to be 2 Delta y_min"},{"Start":"15:43.630 ","End":"15:47.500","Text":"because we have started from rest and because we\u0027re"},{"Start":"15:47.500 ","End":"15:51.925","Text":"starting from a compression which is below our point of equilibrium,"},{"Start":"15:51.925 ","End":"15:54.283","Text":"we\u0027re going to have a negative over here"},{"Start":"15:54.283 ","End":"15:56.635","Text":"because we said that our positive direction is going"},{"Start":"15:56.635 ","End":"16:02.080","Text":"upwards and we\u0027re below our 0 point, so it\u0027s negative."},{"Start":"16:02.080 ","End":"16:07.630","Text":"Of course, our Phi when we\u0027re being released from rest is going to be equal to 0."},{"Start":"16:07.630 ","End":"16:09.340","Text":"Of course, our Omega,"},{"Start":"16:09.340 ","End":"16:11.050","Text":"which I already said what it was before,"},{"Start":"16:11.050 ","End":"16:16.810","Text":"it\u0027s going to be the square root of our coefficient for our y minus y_0,"},{"Start":"16:16.810 ","End":"16:19.000","Text":"which is going to be k,"},{"Start":"16:19.000 ","End":"16:22.510","Text":"divided by the coefficient of our y double dot,"},{"Start":"16:22.510 ","End":"16:26.530","Text":"which is going to be m_1 plus m_2,"},{"Start":"16:26.530 ","End":"16:29.875","Text":"k could have also written m star, and that\u0027s it."},{"Start":"16:29.875 ","End":"16:32.575","Text":"Let\u0027s rewrite our equation."},{"Start":"16:32.575 ","End":"16:36.655","Text":"I\u0027m reminding you, we have our y Tilde as a function"},{"Start":"16:36.655 ","End":"16:41.650","Text":"of t because we\u0027re taking it from our point of equilibrium."},{"Start":"16:41.650 ","End":"16:44.140","Text":"We\u0027ve said that that\u0027s where our origin is."},{"Start":"16:44.140 ","End":"16:46.720","Text":"That\u0027s going to be equal to our A which is"},{"Start":"16:46.720 ","End":"16:54.640","Text":"negative 2 Delta y_min multiplied by cosine of Omega,"},{"Start":"16:54.640 ","End":"16:58.750","Text":"which is this multiplied by t. Now,"},{"Start":"16:58.750 ","End":"17:00.040","Text":"what we\u0027re trying to find,"},{"Start":"17:00.040 ","End":"17:02.064","Text":"let\u0027s look at the question again."},{"Start":"17:02.064 ","End":"17:07.696","Text":"We\u0027re trying to find at what moment does m_2 break away from the surface?"},{"Start":"17:07.696 ","End":"17:12.430","Text":"At what time do I get to this distance?"},{"Start":"17:12.430 ","End":"17:17.245","Text":"Let\u0027s say it\u0027s over here at this distance,"},{"Start":"17:17.245 ","End":"17:21.880","Text":"which I know is at Delta y_min."},{"Start":"17:21.880 ","End":"17:24.070","Text":"When do I get to this distance,"},{"Start":"17:24.070 ","End":"17:31.449","Text":"because that\u0027s when I know that my mass_2 is going to fly off from the surface."},{"Start":"17:31.449 ","End":"17:33.265","Text":"What time does it happen?"},{"Start":"17:33.265 ","End":"17:36.010","Text":"I need that my y Tilde,"},{"Start":"17:36.010 ","End":"17:39.940","Text":"a t_1, I\u0027m trying to find what my t_1 is."},{"Start":"17:39.940 ","End":"17:48.175","Text":"Well, my y Tilde equal to 2 Delta y_min."},{"Start":"17:48.175 ","End":"17:52.900","Text":"I\u0027m reminding you this is going to be a positive Delta y_min because I\u0027m going in"},{"Start":"17:52.900 ","End":"18:00.175","Text":"the positive y-direction and I\u0027m above my_0 line,"},{"Start":"18:00.175 ","End":"18:02.755","Text":"above my point of equilibrium."},{"Start":"18:02.755 ","End":"18:10.150","Text":"My Delta y_min isn\u0027t changing even though I\u0027ve compressed my spring more."},{"Start":"18:10.150 ","End":"18:12.820","Text":"The force remains the same in spring,"},{"Start":"18:12.820 ","End":"18:16.405","Text":"which means that my acceleration remains the same in the spring,"},{"Start":"18:16.405 ","End":"18:21.835","Text":"which means that everything I did in my question number 1, is still relevant."},{"Start":"18:21.835 ","End":"18:24.985","Text":"I\u0027m trying to find this t_1."},{"Start":"18:24.985 ","End":"18:31.990","Text":"Also, I know that my equation for my y Tilde is equal to"},{"Start":"18:31.990 ","End":"18:40.195","Text":"negative 2 Delta y_min multiplied by cosine of"},{"Start":"18:40.195 ","End":"18:45.580","Text":"Omega t. Now what I can do is I can cross out"},{"Start":"18:45.580 ","End":"18:52.420","Text":"my Delta y_min from both sides and I will get that when I rearrange,"},{"Start":"18:52.420 ","End":"18:53.964","Text":"and here there\u0027s a minus,"},{"Start":"18:53.964 ","End":"18:57.790","Text":"I will get that my cosine of"},{"Start":"18:57.790 ","End":"19:05.149","Text":"Omega t is equal to negative a half."},{"Start":"19:05.149 ","End":"19:11.560","Text":"This is obviously our t_1 and also over here, t_1."},{"Start":"19:11.560 ","End":"19:17.380","Text":"Then we\u0027ll get that our t_1 is going to be equal to 1 divided by our Omega,"},{"Start":"19:17.380 ","End":"19:20.905","Text":"where our Omega is this over here,"},{"Start":"19:20.905 ","End":"19:24.955","Text":"multiplied by cosine to the minus 1,"},{"Start":"19:24.955 ","End":"19:29.485","Text":"so i cos of negative a half."},{"Start":"19:29.485 ","End":"19:35.960","Text":"This is the time at which our mass will disconnect."},{"Start":"19:36.690 ","End":"19:39.955","Text":"Take a note that when you\u0027re doing this i cos,"},{"Start":"19:39.955 ","End":"19:45.560","Text":"we\u0027re working in radians, not degrees."},{"Start":"19:45.840 ","End":"19:49.120","Text":"Final thing, I\u0027m not going to substitute it in"},{"Start":"19:49.120 ","End":"19:52.030","Text":"because at this stage it\u0027s just a waste of time."},{"Start":"19:52.030 ","End":"19:54.670","Text":"If you want, you can plug it into a calculator."},{"Start":"19:54.670 ","End":"19:59.269","Text":"But we were given values for everything,"},{"Start":"19:59.269 ","End":"20:02.215","Text":"for our k, for our m_1, for our m_2."},{"Start":"20:02.215 ","End":"20:10.525","Text":"You can substitute if you want the values into here and you\u0027ll get the actual answer."},{"Start":"20:10.525 ","End":"20:12.805","Text":"Now, to question 3."},{"Start":"20:12.805 ","End":"20:14.245","Text":"Given the values above,"},{"Start":"20:14.245 ","End":"20:15.865","Text":"so that\u0027s these values,"},{"Start":"20:15.865 ","End":"20:20.170","Text":"what is the position and velocity of the surface,"},{"Start":"20:20.170 ","End":"20:24.280","Text":"so this plot over here,"},{"Start":"20:24.280 ","End":"20:28.585","Text":"at the moment that m_2 breaks away?"},{"Start":"20:28.585 ","End":"20:34.300","Text":"The position is going to be exactly what we found in question number 1."},{"Start":"20:34.300 ","End":"20:39.220","Text":"It\u0027s going to be Delta y_min because that\u0027s the position that"},{"Start":"20:39.220 ","End":"20:44.335","Text":"our platter has to be in for a mass to break away from it."},{"Start":"20:44.335 ","End":"20:46.765","Text":"That was the exact same question that we were asking."},{"Start":"20:46.765 ","End":"20:52.535","Text":"Now what we have to do is we have to find the velocity of the surface at that moment."},{"Start":"20:52.535 ","End":"20:55.770","Text":"As for the velocity, its position,"},{"Start":"20:55.770 ","End":"21:02.785","Text":"its y Tilde is simply going to be equal to Delta y_min."},{"Start":"21:02.785 ","End":"21:04.870","Text":"Now, as for our velocity,"},{"Start":"21:04.870 ","End":"21:08.380","Text":"it\u0027s going to be as a function of time."},{"Start":"21:08.380 ","End":"21:14.215","Text":"All we have to do is we have to take the derivative of this over here,"},{"Start":"21:14.215 ","End":"21:17.860","Text":"and then once we\u0027ve taken the derivative, that is our velocity."},{"Start":"21:17.860 ","End":"21:23.290","Text":"It\u0027s going to be equal to our Delta Tilde dot, with the derivative,"},{"Start":"21:23.290 ","End":"21:25.090","Text":"and that\u0027s of course,"},{"Start":"21:25.090 ","End":"21:28.720","Text":"as a function of t. Let\u0027s see,"},{"Start":"21:28.720 ","End":"21:30.250","Text":"it\u0027s going to be equal to."},{"Start":"21:30.250 ","End":"21:32.860","Text":"I\u0027m going to carry it on over here because I don\u0027t have space."},{"Start":"21:32.860 ","End":"21:43.065","Text":"We\u0027ll have negative 2 Delta y_min multiplied by the derivative of a cosine Omegat."},{"Start":"21:43.065 ","End":"21:45.360","Text":"It\u0027s going to be negative sign."},{"Start":"21:45.360 ","End":"21:50.295","Text":"This will become a negative Omega sine"},{"Start":"21:50.295 ","End":"21:55.705","Text":"of Omega t. Then negative and negative becomes a positive."},{"Start":"21:55.705 ","End":"22:06.700","Text":"Now, we want to find the velocity when our platter is at this position."},{"Start":"22:06.700 ","End":"22:10.375","Text":"All we have to do is we have to substitute in our t_1."},{"Start":"22:10.375 ","End":"22:12.160","Text":"Our velocity at t_1,"},{"Start":"22:12.160 ","End":"22:17.080","Text":"which is the time at which our mass breaks away from a platter,"},{"Start":"22:17.080 ","End":"22:25.410","Text":"is simply going to be equal to 2 multiplied by Delta y_min multiplied by our Omega,"},{"Start":"22:25.410 ","End":"22:26.820","Text":"which I won\u0027t substitute in here."},{"Start":"22:26.820 ","End":"22:29.205","Text":"I\u0027ll just write, Omega it\u0027s this over here,"},{"Start":"22:29.205 ","End":"22:32.545","Text":"multiplied by sine, again,"},{"Start":"22:32.545 ","End":"22:36.850","Text":"of Omega multiplied by t_1."},{"Start":"22:36.850 ","End":"22:39.580","Text":"Let\u0027s see. It\u0027s going to be 1 over Omega."},{"Start":"22:39.580 ","End":"22:41.650","Text":"The Omega will cancel out,"},{"Start":"22:41.650 ","End":"22:46.585","Text":"sine of arc cos of a negative half."},{"Start":"22:46.585 ","End":"22:47.755","Text":"Of course, again,"},{"Start":"22:47.755 ","End":"22:49.855","Text":"this is in radians."},{"Start":"22:49.855 ","End":"22:55.090","Text":"That\u0027s the answer for our question 3 which is also our final question."},{"Start":"22:55.090 ","End":"22:58.700","Text":"That is the end of the lesson."}],"ID":9436},{"Watched":false,"Name":"Disk Rolling in Pipe","Duration":"26m 25s","ChapterTopicVideoID":10283,"CourseChapterTopicPlaylistID":5369,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.280","Text":"Hello. In this question,"},{"Start":"00:02.280 ","End":"00:05.190","Text":"we\u0027re given a disk here of radius"},{"Start":"00:05.190 ","End":"00:08.790","Text":"r. We\u0027re told that it\u0027s rolling up and down the pipes in a wall."},{"Start":"00:08.790 ","End":"00:12.435","Text":"Going like this, up and down, up and down."},{"Start":"00:12.435 ","End":"00:15.400","Text":"Pipes in a wall of radius R."},{"Start":"00:15.400 ","End":"00:19.725","Text":"The pipe has a radius R and the pipe is attached to the ground,"},{"Start":"00:19.725 ","End":"00:25.575","Text":"meaning that the pipe isn\u0027t rotating or rolling or anything, it\u0027s still."},{"Start":"00:25.575 ","End":"00:27.975","Text":"In the first question we\u0027re being asked,"},{"Start":"00:27.975 ","End":"00:30.090","Text":"what will be the frequency of the oscillations of"},{"Start":"00:30.090 ","End":"00:33.100","Text":"the disk given that there isn\u0027t friction?"},{"Start":"00:33.100 ","End":"00:35.645","Text":"In the second question we\u0027re being asked,"},{"Start":"00:35.645 ","End":"00:37.760","Text":"what will be the answer to number 1,"},{"Start":"00:37.760 ","End":"00:43.055","Text":"if friction was added to the pipe\u0027s walls and in brackets that there was no slipping?"},{"Start":"00:43.055 ","End":"00:44.615","Text":"Now what does this mean,"},{"Start":"00:44.615 ","End":"00:50.750","Text":"when there\u0027s no friction in the equation then number 1,"},{"Start":"00:50.750 ","End":"00:52.865","Text":"the disk will remain like this."},{"Start":"00:52.865 ","End":"00:57.800","Text":"Let\u0027s say if I was to draw an arrow going in this direction on the disk,"},{"Start":"00:57.800 ","End":"01:01.835","Text":"as the disk would oscillate up and down the wall of the pipe,"},{"Start":"01:01.835 ","End":"01:07.690","Text":"the arrow will always be pointing in this exact direction as the disk goes up and down."},{"Start":"01:07.690 ","End":"01:09.620","Text":"That would be for Question 1."},{"Start":"01:09.620 ","End":"01:12.080","Text":"But now we\u0027re being asked what will be the answer to"},{"Start":"01:12.080 ","End":"01:15.710","Text":"1 friction was added to the pipe\u0027s walls and there would be no slipping,"},{"Start":"01:15.710 ","End":"01:21.290","Text":"which would mean that if I had that same arrow pointing in a direction as this disk."},{"Start":"01:21.290 ","End":"01:24.410","Text":"Because there\u0027s friction and there\u0027s no slipping,"},{"Start":"01:24.410 ","End":"01:26.255","Text":"the disk is rolling without slipping."},{"Start":"01:26.255 ","End":"01:30.830","Text":"As well as the disk going in this formation up and down the wars,"},{"Start":"01:30.830 ","End":"01:34.475","Text":"the disk itself will be rotating in on itself."},{"Start":"01:34.475 ","End":"01:37.815","Text":"Been going round and round and round itself."},{"Start":"01:37.815 ","End":"01:39.860","Text":"Then Question 3,"},{"Start":"01:39.860 ","End":"01:42.005","Text":"what will be the answer to number 1, if,"},{"Start":"01:42.005 ","End":"01:44.135","Text":"as well as friction on the floor being added,"},{"Start":"01:44.135 ","End":"01:45.950","Text":"a further frictional force,"},{"Start":"01:45.950 ","End":"01:49.385","Text":"F equals negative bv would be added?"},{"Start":"01:49.385 ","End":"01:53.000","Text":"Here we\u0027re being asked what would be the frequency of"},{"Start":"01:53.000 ","End":"01:57.020","Text":"oscillation when we also have friction like in Question 2."},{"Start":"01:57.020 ","End":"02:01.715","Text":"But aside from that, we also have a further frictional force this,"},{"Start":"02:01.715 ","End":"02:12.275","Text":"which is called in other words a drag speaking about an oscillatory system."},{"Start":"02:12.275 ","End":"02:15.635","Text":"Damping is an influence within or on"},{"Start":"02:15.635 ","End":"02:19.130","Text":"an oscillatory system that has the effect of reducing,"},{"Start":"02:19.130 ","End":"02:22.250","Text":"restricting, or preventing its oscillations."},{"Start":"02:22.250 ","End":"02:27.125","Text":"When we get to this question, question number 3,"},{"Start":"02:27.125 ","End":"02:32.620","Text":"we\u0027re going to see how we work out this question when we\u0027re speaking about damping."},{"Start":"02:32.620 ","End":"02:36.410","Text":"Let\u0027s begin by answering question number 1."},{"Start":"02:36.410 ","End":"02:38.450","Text":"The first thing I\u0027m going to do is,"},{"Start":"02:38.450 ","End":"02:40.685","Text":"I\u0027m going to draw a free body diagram."},{"Start":"02:40.685 ","End":"02:42.695","Text":"If this is my small disk,"},{"Start":"02:42.695 ","End":"02:51.110","Text":"it has the normal going in this direction and it has mg going downwards."},{"Start":"02:51.110 ","End":"02:58.025","Text":"Now this means that my only moment acting on the small disk is mg. Why do I know this?"},{"Start":"02:58.025 ","End":"03:01.445","Text":"Because the normal force will always be pointing into"},{"Start":"03:01.445 ","End":"03:06.755","Text":"the center of my pipe into the center of my circle."},{"Start":"03:06.755 ","End":"03:12.695","Text":"Which means that only mg is the only moment working on the system."},{"Start":"03:12.695 ","End":"03:16.805","Text":"Now why do I know that there are moments involved at all?"},{"Start":"03:16.805 ","End":"03:23.255","Text":"Because I know that the disk is doing some rolling, some circular motion."},{"Start":"03:23.255 ","End":"03:26.805","Text":"I know that there are moments acting in the system."},{"Start":"03:26.805 ","End":"03:29.675","Text":"Now my center of rotation will be here,"},{"Start":"03:29.675 ","End":"03:31.370","Text":"the center of the pipe."},{"Start":"03:31.370 ","End":"03:41.490","Text":"Now I can draw some axes which points to the center of my pipe."},{"Start":"03:41.490 ","End":"03:45.815","Text":"I can say that because the normal is pointing towards the center,"},{"Start":"03:45.815 ","End":"03:47.525","Text":"it\u0027s equal to 0."},{"Start":"03:47.525 ","End":"03:51.130","Text":"How do I find out what my sum of moments is?"},{"Start":"03:51.130 ","End":"03:54.010","Text":"Let\u0027s see, so a moment is"},{"Start":"03:54.010 ","End":"04:00.520","Text":"force times distance to the center multiplied by sine of the angle."},{"Start":"04:00.520 ","End":"04:02.590","Text":"Mg is my force."},{"Start":"04:02.590 ","End":"04:07.540","Text":"Distance from here to here is"},{"Start":"04:07.710 ","End":"04:15.180","Text":"R and my angle is this angle here."},{"Start":"04:15.180 ","End":"04:17.185","Text":"How do we find this angle?"},{"Start":"04:17.185 ","End":"04:22.330","Text":"Because we can see that this shape is like 1.5 of a parallelogram."},{"Start":"04:22.330 ","End":"04:24.730","Text":"We know that this angle here is Theta."},{"Start":"04:24.730 ","End":"04:33.570","Text":"We know from geometry that this angle must be 180 or Pi minus Theta."},{"Start":"04:33.570 ","End":"04:42.140","Text":"Then sine of Pi minus Theta or sine of 180 minus Theta is just the same as sine of Theta."},{"Start":"04:42.140 ","End":"04:48.755","Text":"Then we can write down our equation in the meantime as mg, the force,"},{"Start":"04:48.755 ","End":"04:53.840","Text":"multiplied by the distance which is R,"},{"Start":"04:53.840 ","End":"04:57.575","Text":"multiplied by sine of the angle,"},{"Start":"04:57.575 ","End":"05:00.690","Text":"which is sine of Theta."},{"Start":"05:01.070 ","End":"05:04.175","Text":"Now let\u0027s see what the sign of this is."},{"Start":"05:04.175 ","End":"05:06.410","Text":"Now it\u0027s a negative."},{"Start":"05:06.410 ","End":"05:08.255","Text":"Why is it a negative?"},{"Start":"05:08.255 ","End":"05:13.565","Text":"Because here we\u0027ve defined that Theta is a positive angle,"},{"Start":"05:13.565 ","End":"05:22.234","Text":"which means that going in the anticlockwise direction is the positive direction."},{"Start":"05:22.234 ","End":"05:24.560","Text":"If this is the positive direction,"},{"Start":"05:24.560 ","End":"05:30.800","Text":"it means that if mg is over here and what\u0027s it trying to do is trying to go down,"},{"Start":"05:30.800 ","End":"05:32.375","Text":"go in this direction,"},{"Start":"05:32.375 ","End":"05:35.885","Text":"in the anticlockwise direction."},{"Start":"05:35.885 ","End":"05:39.080","Text":"Because it\u0027s trying to go in an anticlockwise direction,"},{"Start":"05:39.080 ","End":"05:42.275","Text":"which we\u0027ve defined as the negative direction."},{"Start":"05:42.275 ","End":"05:45.095","Text":"It\u0027s trying to reduce this angle,"},{"Start":"05:45.095 ","End":"05:46.550","Text":"not increase the angle,"},{"Start":"05:46.550 ","End":"05:49.840","Text":"which means that we put a minus over here."},{"Start":"05:49.840 ","End":"05:51.515","Text":"Now as we remember,"},{"Start":"05:51.515 ","End":"05:57.350","Text":"we\u0027re trying to write that the sum of all of the moments is equal"},{"Start":"05:57.350 ","End":"06:04.590","Text":"to I Alpha or I Theta double dot."},{"Start":"06:04.590 ","End":"06:06.920","Text":"This is equal to,"},{"Start":"06:06.920 ","End":"06:10.550","Text":"now we\u0027re going to try and find out what I is,"},{"Start":"06:10.550 ","End":"06:14.350","Text":"but I\u0027m already going to write in the Theta double dot from here."},{"Start":"06:14.350 ","End":"06:20.330","Text":"Remember it\u0027s always easier to work with Theta double dot then working with Alpha,"},{"Start":"06:20.330 ","End":"06:22.880","Text":"because it\u0027s easier to do all of our work things out when we\u0027re"},{"Start":"06:22.880 ","End":"06:26.970","Text":"working with the derivatives of angles."},{"Start":"06:27.770 ","End":"06:32.765","Text":"What\u0027s happening here is that we have this disk which is rotating."},{"Start":"06:32.765 ","End":"06:36.960","Text":"We have to find the I of this disk."},{"Start":"06:36.960 ","End":"06:40.775","Text":"Then we also have to add the deviation"},{"Start":"06:40.775 ","End":"06:44.940","Text":"from the center from the axis of rotation, which is this."},{"Start":"06:44.940 ","End":"06:49.530","Text":"Reminding you that this addition is called Steiner."},{"Start":"06:49.700 ","End":"06:51.845","Text":"The I of a disk,"},{"Start":"06:51.845 ","End":"06:54.065","Text":"I\u0027m just going to open up the brackets."},{"Start":"06:54.065 ","End":"07:02.335","Text":"The I of a disk is half m and the radius of it, r squared."},{"Start":"07:02.335 ","End":"07:06.870","Text":"I\u0027m going to write half m r squared."},{"Start":"07:06.870 ","End":"07:09.260","Text":"This is the cost,"},{"Start":"07:09.260 ","End":"07:13.445","Text":"if you will, for this disk to rotate around itself."},{"Start":"07:13.445 ","End":"07:20.140","Text":"But then, because it\u0027s this R distance away from the center,"},{"Start":"07:20.140 ","End":"07:30.133","Text":"so we have to add m the mass times the distance squared."},{"Start":"07:30.133 ","End":"07:35.550","Text":"Again, this is equal to the I of the disk,"},{"Start":"07:35.550 ","End":"07:39.570","Text":"and this is Steiner,"},{"Start":"07:39.570 ","End":"07:45.750","Text":"which we have to add each time we have some rotating body,"},{"Start":"07:45.750 ","End":"07:47.640","Text":"here, in this case,"},{"Start":"07:47.640 ","End":"07:52.845","Text":"it\u0027s a disk that is rotating at a distance,"},{"Start":"07:52.845 ","End":"07:54.224","Text":"this is the distance,"},{"Start":"07:54.224 ","End":"07:56.160","Text":"away from the center of rotation."},{"Start":"07:56.160 ","End":"07:58.810","Text":"We always do the distance squared."},{"Start":"07:59.180 ","End":"08:03.990","Text":"Now all I have to do is find the relationship between my Theta and"},{"Start":"08:03.990 ","End":"08:08.315","Text":"my Theta double dot in order to find what my frequencies are."},{"Start":"08:08.315 ","End":"08:11.975","Text":"If you don\u0027t remember this and you don\u0027t really know what I\u0027m talking about,"},{"Start":"08:11.975 ","End":"08:15.900","Text":"please go back to the earlier lessons while I go over this."},{"Start":"08:16.310 ","End":"08:19.560","Text":"Now we have to get,"},{"Start":"08:19.560 ","End":"08:23.250","Text":"as I just said, the relationship between Theta and Theta double dot."},{"Start":"08:23.250 ","End":"08:25.770","Text":"Now, you\u0027ll notice because we\u0027re working with"},{"Start":"08:25.770 ","End":"08:30.164","Text":"small angles that we can say that the sine of Theta,"},{"Start":"08:30.164 ","End":"08:32.110","Text":"if Theta is a small number,"},{"Start":"08:32.110 ","End":"08:34.200","Text":"is around about equal to just Theta."},{"Start":"08:34.200 ","End":"08:43.005","Text":"We can rewrite this as negative mgR Theta equals Theta double dot."},{"Start":"08:43.005 ","End":"08:47.310","Text":"Then because we want to just get Theta double dot alone,"},{"Start":"08:47.310 ","End":"08:50.925","Text":"so we can divide by all of this on both sides."},{"Start":"08:50.925 ","End":"09:00.435","Text":"Then we\u0027ll have divided by 1/2 mr squared plus m capital R squared."},{"Start":"09:00.435 ","End":"09:02.340","Text":"This is what we get."},{"Start":"09:02.340 ","End":"09:06.370","Text":"Now, in the question,"},{"Start":"09:07.310 ","End":"09:10.140","Text":"the coefficient of this Theta,"},{"Start":"09:10.140 ","End":"09:14.100","Text":"so let me write this,"},{"Start":"09:14.100 ","End":"09:20.530","Text":"this is the frequency squared, without the minus."},{"Start":"09:22.070 ","End":"09:25.050","Text":"This is the frequency squared,"},{"Start":"09:25.050 ","End":"09:29.115","Text":"so this is omega squared."},{"Start":"09:29.115 ","End":"09:31.125","Text":"However, in the question,"},{"Start":"09:31.125 ","End":"09:34.000","Text":"it wasn\u0027t written, I forgot to write it."},{"Start":"09:35.060 ","End":"09:40.980","Text":"But because the small radius is so small compared to the large radius,"},{"Start":"09:40.980 ","End":"09:44.745","Text":"we can ignore this."},{"Start":"09:44.745 ","End":"09:49.470","Text":"We can cross it out. Because if this r is such a small number,"},{"Start":"09:49.470 ","End":"09:55.530","Text":"then this whole expression of 1/2 m small r squared will be also a really tiny number,"},{"Start":"09:55.530 ","End":"09:57.030","Text":"which means that we can just do this."},{"Start":"09:57.030 ","End":"10:02.205","Text":"Then we can estimate that the omega, the frequency,"},{"Start":"10:02.205 ","End":"10:07.238","Text":"will be root of g over capital R."},{"Start":"10:07.238 ","End":"10:11.280","Text":"This is the frequency if we"},{"Start":"10:11.280 ","End":"10:15.825","Text":"cancel out this because it\u0027s such a small number compared to this."},{"Start":"10:15.825 ","End":"10:21.480","Text":"Now notice that this frequency is the exact same frequency as that of a pendulum."},{"Start":"10:21.480 ","End":"10:26.925","Text":"But instead with a pendulum because there\u0027s a string of a certain length L,"},{"Start":"10:26.925 ","End":"10:29.310","Text":"so instead of writing here R,"},{"Start":"10:29.310 ","End":"10:35.385","Text":"you would write L. If you had a question of a pendulum attached to a rope of length L,"},{"Start":"10:35.385 ","End":"10:37.609","Text":"this would just be the frequency,"},{"Start":"10:37.609 ","End":"10:43.095","Text":"square root of g divided by L. Why is it like this here?"},{"Start":"10:43.095 ","End":"10:48.105","Text":"Because we don\u0027t have the disk rotating around itself."},{"Start":"10:48.105 ","End":"10:51.405","Text":"But now that\u0027s it. This is the answer to Question Number 1."},{"Start":"10:51.405 ","End":"10:53.220","Text":"But now in Question Number 2,"},{"Start":"10:53.220 ","End":"10:58.110","Text":"and we say that because friction is added and there\u0027s no slipping,"},{"Start":"10:58.110 ","End":"11:01.860","Text":"that now this disk is going to start rotating around itself,"},{"Start":"11:01.860 ","End":"11:04.395","Text":"which means that it\u0027s going to stop acting like a pendulum."},{"Start":"11:04.395 ","End":"11:07.230","Text":"Now let\u0027s see how we can solve this."},{"Start":"11:07.230 ","End":"11:11.700","Text":"In this question, we\u0027re adding in friction."},{"Start":"11:11.700 ","End":"11:15.420","Text":"I can draw this on my free body diagram."},{"Start":"11:15.420 ","End":"11:19.814","Text":"Here we have the addition of friction."},{"Start":"11:19.814 ","End":"11:21.735","Text":"Let\u0027s see how I calculate it."},{"Start":"11:21.735 ","End":"11:25.920","Text":"Again, I have to use this equation."},{"Start":"11:25.920 ","End":"11:32.400","Text":"The sum of all of the moments is equal to I multiplied by Theta double dot."},{"Start":"11:32.400 ","End":"11:35.505","Text":"Let\u0027s do the sum of all the moments."},{"Start":"11:35.505 ","End":"11:44.955","Text":"Just like before, I have negative mgR sine of Theta."},{"Start":"11:44.955 ","End":"11:48.750","Text":"But then, now I have a moment which is trying to,"},{"Start":"11:48.750 ","End":"11:52.869","Text":"if the force mg,"},{"Start":"11:52.869 ","End":"11:55.890","Text":"wasn\u0027t being applied on this disk,"},{"Start":"11:55.890 ","End":"11:59.370","Text":"then this frictional force would be acting"},{"Start":"11:59.370 ","End":"12:03.917","Text":"in a clockwise direction so as to increase this angle,"},{"Start":"12:03.917 ","End":"12:07.140","Text":"so it\u0027s positive,"},{"Start":"12:07.140 ","End":"12:12.825","Text":"positive f multiplied by the length."},{"Start":"12:12.825 ","End":"12:19.260","Text":"Remember, just like here we did force times distance times sine of the angle,"},{"Start":"12:19.260 ","End":"12:22.200","Text":"here we\u0027re doing force times distance."},{"Start":"12:22.200 ","End":"12:23.580","Text":"What is the distance?"},{"Start":"12:23.580 ","End":"12:31.050","Text":"It\u0027s capital R plus small r because we\u0027re having a capital R over here"},{"Start":"12:31.050 ","End":"12:38.360","Text":"and then plus the small r. Then this force is acting in 90 degrees."},{"Start":"12:38.360 ","End":"12:40.925","Text":"I didn\u0027t draw it very well in this diagram,"},{"Start":"12:40.925 ","End":"12:44.510","Text":"but it\u0027s at 90 degrees to the normal force."},{"Start":"12:44.510 ","End":"12:50.625","Text":"Frictional force is always at 90 degrees to the normal force."},{"Start":"12:50.625 ","End":"12:55.530","Text":"Sine of 90 degrees is just 1."},{"Start":"12:55.530 ","End":"12:59.655","Text":"I can write here sine of,"},{"Start":"12:59.655 ","End":"13:01.290","Text":"I\u0027ll write Pi over 2,"},{"Start":"13:01.290 ","End":"13:06.360","Text":"because that\u0027s 90, but this whole thing is equal to 1."},{"Start":"13:06.360 ","End":"13:10.515","Text":"Then equals I times Theta double dot."},{"Start":"13:10.515 ","End":"13:15.510","Text":"Now the I is going to be exactly the same because it\u0027s still a disk and it\u0027s"},{"Start":"13:15.510 ","End":"13:21.929","Text":"still a distance of capital R away from the center of rotation."},{"Start":"13:21.929 ","End":"13:24.750","Text":"Again, I\u0027m going to do plus 1/2 mr squared,"},{"Start":"13:24.750 ","End":"13:27.720","Text":"the I of the disk,"},{"Start":"13:27.720 ","End":"13:30.240","Text":"plus m capital R squared,"},{"Start":"13:30.240 ","End":"13:35.790","Text":"the Steiner, multiplied by Theta double dot."},{"Start":"13:35.790 ","End":"13:38.175","Text":"Let\u0027s take a look at this."},{"Start":"13:38.175 ","End":"13:42.315","Text":"Now, it looks exactly the same as this equation over here."},{"Start":"13:42.315 ","End":"13:49.770","Text":"However, because I have this f here and this frictional force is an unknown variable,"},{"Start":"13:49.770 ","End":"13:52.815","Text":"I don\u0027t know what the value of this is."},{"Start":"13:52.815 ","End":"13:54.660","Text":"Which means that I have to find"},{"Start":"13:54.660 ","End":"13:59.070","Text":"some other equation in order to find out what this unknown is so that"},{"Start":"13:59.070 ","End":"14:06.134","Text":"I can make this equation into this format and find out the frequency."},{"Start":"14:06.134 ","End":"14:13.035","Text":"Again, like in here where I crossed out the expression with the small r in it"},{"Start":"14:13.035 ","End":"14:16.155","Text":"because it\u0027s such a small number compared to"},{"Start":"14:16.155 ","End":"14:20.490","Text":"this capital R radius that it\u0027s insignificant and I can just erase it,"},{"Start":"14:20.490 ","End":"14:25.215","Text":"I\u0027m going to do the same thing here and the same thing here."},{"Start":"14:25.215 ","End":"14:29.685","Text":"It will also make the mathematics and all the algebra a lot easier."},{"Start":"14:29.685 ","End":"14:33.270","Text":"Now, how do I find the value of this frictional force?"},{"Start":"14:33.270 ","End":"14:35.790","Text":"What I\u0027m going to do is I\u0027m going to try and find out"},{"Start":"14:35.790 ","End":"14:40.320","Text":"how many times this disk rotates around itself."},{"Start":"14:40.320 ","End":"14:44.145","Text":"I\u0027m going to try and find out this rotation around itself,"},{"Start":"14:44.145 ","End":"14:45.840","Text":"which will give me hopefully,"},{"Start":"14:45.840 ","End":"14:50.340","Text":"the friction in variables that are known to me from the question,"},{"Start":"14:50.340 ","End":"14:56.430","Text":"and then we can begin to reshape this equation into this format."},{"Start":"14:56.430 ","End":"15:01.560","Text":"Let\u0027s see which moments are working on the disk when it\u0027s rotating around itself."},{"Start":"15:01.560 ","End":"15:05.410","Text":"It\u0027s rotating around itself from the center."},{"Start":"15:05.410 ","End":"15:12.660","Text":"We have mg and the normal force working out of the center of the disk,"},{"Start":"15:12.660 ","End":"15:17.610","Text":"which means that I can cancel each 1 of those forces out."},{"Start":"15:18.140 ","End":"15:21.060","Text":"My N and mg cancel out,"},{"Start":"15:21.060 ","End":"15:25.245","Text":"and then I have the frictional force from the center,"},{"Start":"15:25.245 ","End":"15:34.656","Text":"it\u0027s working from its point of contact with this outer pipe."},{"Start":"15:34.656 ","End":"15:40.840","Text":"Now let\u0027s write the sum of moments on the disk rotating about itself."},{"Start":"15:40.840 ","End":"15:45.640","Text":"Again, it\u0027s the force multiplied by the distance,"},{"Start":"15:45.640 ","End":"15:48.655","Text":"multiplied by the sine of the angle."},{"Start":"15:48.655 ","End":"15:51.700","Text":"We have the force which is f. Again,"},{"Start":"15:51.700 ","End":"15:54.055","Text":"remember it\u0027s going in the positive direction because it\u0027s"},{"Start":"15:54.055 ","End":"15:57.595","Text":"acting in a way to increase this angle Theta."},{"Start":"15:57.595 ","End":"16:02.740","Text":"It\u0027s positive f multiplied by the distance from the center,"},{"Start":"16:02.740 ","End":"16:08.905","Text":"which is r multiplied by sine of Pi over 2,"},{"Start":"16:08.905 ","End":"16:10.900","Text":"which as we\u0027ve said, is 1."},{"Start":"16:10.900 ","End":"16:12.940","Text":"I\u0027m just going to leave it as that,"},{"Start":"16:12.940 ","End":"16:18.010","Text":"which equals I multiplied by Theta double dot."},{"Start":"16:18.010 ","End":"16:19.525","Text":"Now what\u0027s my I?"},{"Start":"16:19.525 ","End":"16:28.150","Text":"My I is again the I of a disk 1/2 mr^2 squared multiplied by Theta double dot."},{"Start":"16:28.150 ","End":"16:32.140","Text":"But this Theta double dot isn\u0027t the same as this Theta double dot."},{"Start":"16:32.140 ","End":"16:35.170","Text":"I\u0027m going to write Theta double dot prime."},{"Start":"16:35.170 ","End":"16:38.170","Text":"It\u0027s a different Theta double dot because"},{"Start":"16:38.170 ","End":"16:41.500","Text":"we\u0027re working out the sum of all of the moments,"},{"Start":"16:41.500 ","End":"16:43.765","Text":"not on this system,"},{"Start":"16:43.765 ","End":"16:46.690","Text":"but on just this system."},{"Start":"16:46.690 ","End":"16:49.390","Text":"It\u0027s Theta double dot prime."},{"Start":"16:49.390 ","End":"16:51.745","Text":"Now what\u0027s the problem here?"},{"Start":"16:51.745 ","End":"16:55.330","Text":"Of course, I can do things to isolate out this f,"},{"Start":"16:55.330 ","End":"16:59.755","Text":"and then substitute in that into this equation."},{"Start":"16:59.755 ","End":"17:01.480","Text":"However, what\u0027s the problem with that?"},{"Start":"17:01.480 ","End":"17:04.975","Text":"I don\u0027t know what my Theta double dot prime is equal to."},{"Start":"17:04.975 ","End":"17:08.740","Text":"I\u0027m not doing anything by substituting 1 unknown in,"},{"Start":"17:08.740 ","End":"17:10.225","Text":"with a different unknown."},{"Start":"17:10.225 ","End":"17:15.350","Text":"I have to find out what this Theta double dot prime is."},{"Start":"17:15.630 ","End":"17:18.790","Text":"What can I do in this situation?"},{"Start":"17:18.790 ","End":"17:22.495","Text":"Because in the question I\u0027m told that there\u0027s no slipping,"},{"Start":"17:22.495 ","End":"17:25.990","Text":"I know that there\u0027s a relationship between how many times it turns"},{"Start":"17:25.990 ","End":"17:30.505","Text":"this way and how much it turns this way."},{"Start":"17:30.505 ","End":"17:36.820","Text":"Let\u0027s assume that the radius of the small disk is 1/10 of the radius of the large disk."},{"Start":"17:36.820 ","End":"17:38.440","Text":"What does this say to me?"},{"Start":"17:38.440 ","End":"17:45.100","Text":"It tells me that if this disk does 1 turn around this large radius,"},{"Start":"17:45.100 ","End":"17:48.280","Text":"it will turn in on itself 10 times."},{"Start":"17:48.280 ","End":"17:50.530","Text":"Then from that logic,"},{"Start":"17:50.530 ","End":"17:58.420","Text":"I can say that my Theta double dot prime multiplied by small r is equal to"},{"Start":"17:58.420 ","End":"18:02.835","Text":"my Theta double dot multiplied by my"},{"Start":"18:02.835 ","End":"18:08.430","Text":"R. This is the idea of spinning,"},{"Start":"18:08.430 ","End":"18:10.425","Text":"of turning without slipping."},{"Start":"18:10.425 ","End":"18:16.150","Text":"Now we have to think of the sign of this equation. Let\u0027s see."},{"Start":"18:16.150 ","End":"18:20.620","Text":"As the small disk rotates in a clockwise direction,"},{"Start":"18:20.620 ","End":"18:23.755","Text":"we can see that here it\u0027s going in a positive direction."},{"Start":"18:23.755 ","End":"18:27.955","Text":"Remember we defined the clockwise direction as the positive direction."},{"Start":"18:27.955 ","End":"18:31.360","Text":"But as this is rotating in the positive direction,"},{"Start":"18:31.360 ","End":"18:33.955","Text":"it\u0027s rolling in this direction,"},{"Start":"18:33.955 ","End":"18:36.940","Text":"which is anticlockwise, which is the negative direction."},{"Start":"18:36.940 ","End":"18:39.865","Text":"We have to add in a negative sign."},{"Start":"18:39.865 ","End":"18:42.550","Text":"Because in and around itself, it\u0027s positive,"},{"Start":"18:42.550 ","End":"18:46.850","Text":"but it\u0027s going in a negative direction, which is this."},{"Start":"18:47.010 ","End":"18:50.650","Text":"Now I have all of my equations,"},{"Start":"18:50.650 ","End":"18:53.905","Text":"and now I just have to sub in all of my equations,"},{"Start":"18:53.905 ","End":"18:58.195","Text":"this equation into my Theta double dot prime,"},{"Start":"18:58.195 ","End":"19:00.835","Text":"and then I have to isolate out my f,"},{"Start":"19:00.835 ","End":"19:03.475","Text":"and then substitute it into here."},{"Start":"19:03.475 ","End":"19:05.995","Text":"Let\u0027s do this working out together."},{"Start":"19:05.995 ","End":"19:12.023","Text":"I have that my f will equal 1/2"},{"Start":"19:12.023 ","End":"19:20.200","Text":"mr^2 divided by r multiplied by Theta double dot prime,"},{"Start":"19:20.200 ","End":"19:22.563","Text":"which will then equal,"},{"Start":"19:22.563 ","End":"19:24.850","Text":"my Theta double dot prime is equal"},{"Start":"19:24.850 ","End":"19:32.560","Text":"to Theta double dot R"},{"Start":"19:32.560 ","End":"19:36.560","Text":"divided by r with a negative."},{"Start":"19:36.600 ","End":"19:44.065","Text":"Then I\u0027ll have that my f is equal to 1/2mr^2 squared"},{"Start":"19:44.065 ","End":"19:54.690","Text":"over r. Negative Theta double dot R"},{"Start":"19:54.690 ","End":"19:58.800","Text":"over r. Then we can just do this."},{"Start":"19:58.800 ","End":"20:08.565","Text":"Then we get that my f is equal to negative a 1/2 mR Theta double dot."},{"Start":"20:08.565 ","End":"20:13.230","Text":"Now let substitute everything into our equation."},{"Start":"20:13.230 ","End":"20:16.965","Text":"I\u0027ll have negative mgR."},{"Start":"20:16.965 ","End":"20:20.985","Text":"Now, my sine Theta, I can say,"},{"Start":"20:20.985 ","End":"20:25.740","Text":"is around about equal to Theta because we\u0027re working in the small angles."},{"Start":"20:25.740 ","End":"20:33.625","Text":"Theta plus my f, which is negative."},{"Start":"20:33.625 ","End":"20:40.135","Text":"Negative a 1/2 mR Theta double dot."},{"Start":"20:40.135 ","End":"20:45.565","Text":"Then, we have the brackets of R plus r. But remember again,"},{"Start":"20:45.565 ","End":"20:49.855","Text":"we crossed out our expressions with our r\u0027s."},{"Start":"20:49.855 ","End":"20:53.470","Text":"With our r\u0027s because that number is so"},{"Start":"20:53.470 ","End":"20:56.995","Text":"small that it\u0027s insignificant compared to the value"},{"Start":"20:56.995 ","End":"21:04.180","Text":"of the R. Then we multiply it by this R. This R becomes squared."},{"Start":"21:04.180 ","End":"21:07.195","Text":"Then we said sine Pi over 2=1."},{"Start":"21:07.195 ","End":"21:10.795","Text":"Then this equals just this expression here."},{"Start":"21:10.795 ","End":"21:12.925","Text":"Remember again, we cancel that out."},{"Start":"21:12.925 ","End":"21:19.670","Text":"Because r is an insignificant number compared to R,"},{"Start":"21:19.890 ","End":"21:23.575","Text":"mR^2 Theta double dot."},{"Start":"21:23.575 ","End":"21:28.990","Text":"Will notice that here we have this expression of Theta double dot."},{"Start":"21:28.990 ","End":"21:32.290","Text":"We can move that to the other side of the equation."},{"Start":"21:32.290 ","End":"21:36.830","Text":"Meaning that we\u0027re going to have negative mgR Theta,"},{"Start":"21:37.200 ","End":"21:48.565","Text":"which equals 3/2 mR^2 Theta double dot."},{"Start":"21:48.565 ","End":"21:54.865","Text":"Then, I can divide both sides by m. Then,"},{"Start":"21:54.865 ","End":"22:00.655","Text":"I will just divide both sides by 3/2 and R^2."},{"Start":"22:00.655 ","End":"22:05.215","Text":"I will be left with negative g over r,"},{"Start":"22:05.215 ","End":"22:12.670","Text":"because r divided by R^2 will just be 1/R Theta equals,"},{"Start":"22:12.670 ","End":"22:18.940","Text":"then 2/3 will equal Theta double dot."},{"Start":"22:18.940 ","End":"22:26.365","Text":"Which means that this is the frequency squared."},{"Start":"22:26.365 ","End":"22:29.950","Text":"Which means that if we go over here,"},{"Start":"22:29.950 ","End":"22:36.190","Text":"the frequency equals the square root of 2g over 3 times"},{"Start":"22:36.190 ","End":"22:42.925","Text":"R. Now what\u0027s interesting is that we can see that our frequency here,"},{"Start":"22:42.925 ","End":"22:46.360","Text":"when we have this frictional force,"},{"Start":"22:46.360 ","End":"22:52.375","Text":"is less than our frequency when there isn\u0027t friction."},{"Start":"22:52.375 ","End":"22:56.020","Text":"Now what happens when we have the friction force?"},{"Start":"22:56.020 ","End":"23:01.400","Text":"The friction force caused a little disk to rotate."},{"Start":"23:01.770 ","End":"23:06.155","Text":"This rotating little disk is what changed our frequency,"},{"Start":"23:06.155 ","End":"23:07.885","Text":"is what reduced our frequency."},{"Start":"23:07.885 ","End":"23:10.045","Text":"Which is, if you think about it,"},{"Start":"23:10.045 ","End":"23:14.080","Text":"different because even though I canceled out of the equation,"},{"Start":"23:14.080 ","End":"23:17.755","Text":"these r\u0027s, the radius of the small disk,"},{"Start":"23:17.755 ","End":"23:19.855","Text":"because I said that the small disk was"},{"Start":"23:19.855 ","End":"23:24.970","Text":"such a small and insignificant size compared to the large disk so we can cancel out"},{"Start":"23:24.970 ","End":"23:28.960","Text":"these values even so we can see that it"},{"Start":"23:28.960 ","End":"23:33.520","Text":"reduced our frequency by 1/3 which is pretty significant."},{"Start":"23:33.520 ","End":"23:36.505","Text":"It\u0027s quite interesting that this is what happened."},{"Start":"23:36.505 ","End":"23:40.495","Text":"If you want to look at this from an energy perspective,"},{"Start":"23:40.495 ","End":"23:43.015","Text":"the total energy of this disk,"},{"Start":"23:43.015 ","End":"23:48.160","Text":"when it was oscillating without friction was just going on, oscillating like this."},{"Start":"23:48.160 ","End":"23:51.985","Text":"However, because there was the friction and the disk now had to spin,"},{"Start":"23:51.985 ","End":"23:56.785","Text":"some of the energy would go on the disk spinning around itself,"},{"Start":"23:56.785 ","End":"24:02.785","Text":"and then only what was left would go on allowing the disk to oscillate up and down,"},{"Start":"24:02.785 ","End":"24:08.530","Text":"which is what reduced this rate of oscillation, the frequency."},{"Start":"24:08.530 ","End":"24:12.250","Text":"This is the onset to the second question."},{"Start":"24:12.250 ","End":"24:14.395","Text":"Moving on to the third question."},{"Start":"24:14.395 ","End":"24:16.525","Text":"What will be the onset to 1,"},{"Start":"24:16.525 ","End":"24:19.375","Text":"if, as well as friction on the floor being added."},{"Start":"24:19.375 ","End":"24:21.970","Text":"That\u0027s what we did in question number 2."},{"Start":"24:21.970 ","End":"24:24.170","Text":"A further frictional force,"},{"Start":"24:24.170 ","End":"24:28.605","Text":"drag force, F equals negative b_v would be added."},{"Start":"24:28.605 ","End":"24:33.700","Text":"This type of question, I meant to know how to answer in a second."},{"Start":"24:33.700 ","End":"24:35.635","Text":"Let\u0027s see how we deal with this."},{"Start":"24:35.635 ","End":"24:41.380","Text":"There\u0027s a very easy equation that you have to remember and you have to learn,"},{"Start":"24:41.380 ","End":"24:45.100","Text":"which is the damping frequency."},{"Start":"24:45.100 ","End":"24:48.010","Text":"Omega prime, Omega tag,"},{"Start":"24:48.010 ","End":"24:50.710","Text":"however you want to call it, the damping frequency."},{"Start":"24:50.710 ","End":"24:55.390","Text":"The new frequency that you get after a dampening effect is"},{"Start":"24:55.390 ","End":"25:00.605","Text":"equal to the square root of Omega 0^2."},{"Start":"25:00.605 ","End":"25:02.320","Text":"Which is, what is this?"},{"Start":"25:02.320 ","End":"25:06.440","Text":"This is the frequency before the dampening."},{"Start":"25:06.720 ","End":"25:12.830","Text":"Negative b divided by 2m^2."},{"Start":"25:13.440 ","End":"25:18.900","Text":"Therefore, for us, our Omega after this,"},{"Start":"25:18.900 ","End":"25:22.095","Text":"our new Omega after the dampening effect,"},{"Start":"25:22.095 ","End":"25:28.380","Text":"will be equal to the square root of Omega 0^2 which is this."},{"Start":"25:28.380 ","End":"25:31.270","Text":"It will just be 2g over 3R."},{"Start":"25:31.680 ","End":"25:34.450","Text":"Because this is a square root squared,"},{"Start":"25:34.450 ","End":"25:38.120","Text":"negative b over 2m^2."},{"Start":"25:41.420 ","End":"25:44.535","Text":"This is the end of our lesson."},{"Start":"25:44.535 ","End":"25:51.855","Text":"The third question is a great and easy question that if you get in an exam,"},{"Start":"25:51.855 ","End":"25:56.550","Text":"you literally just plug in your values into this equation."},{"Start":"25:56.550 ","End":"26:00.365","Text":"Important to write this and memorize it."},{"Start":"26:00.365 ","End":"26:03.594","Text":"The first 2 questions are harder,"},{"Start":"26:03.594 ","End":"26:07.465","Text":"but this is a type of a question that you can get in your exam."},{"Start":"26:07.465 ","End":"26:09.415","Text":"If you know how to do it,"},{"Start":"26:09.415 ","End":"26:12.790","Text":"then you\u0027ll know how to do all the questions which are similar."},{"Start":"26:12.790 ","End":"26:15.675","Text":"It\u0027s usually worth a lot of points."},{"Start":"26:15.675 ","End":"26:19.380","Text":"It\u0027s worth, if you didn\u0027t understand this 100 percent,"},{"Start":"26:19.380 ","End":"26:25.720","Text":"then to go over this again until you do understand it. That\u0027s it."}],"ID":10627},{"Watched":false,"Name":"Double Disk, Mass and Spring","Duration":"31m 16s","ChapterTopicVideoID":10284,"CourseChapterTopicPlaylistID":5369,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.565","Text":"Hello. In this question,"},{"Start":"00:02.565 ","End":"00:03.840","Text":"we have 1 disk,"},{"Start":"00:03.840 ","End":"00:08.355","Text":"which is actually made of 2 different size disks."},{"Start":"00:08.355 ","End":"00:11.820","Text":"We have this small disk of radius small r,"},{"Start":"00:11.820 ","End":"00:17.295","Text":"and we have the larger disk of radius capital R. Now,"},{"Start":"00:17.295 ","End":"00:20.730","Text":"these 2 are nailed at the center to the wall,"},{"Start":"00:20.730 ","End":"00:25.200","Text":"meaning that they can rotate around in any direction,"},{"Start":"00:25.200 ","End":"00:26.684","Text":"but they can\u0027t move up and down,"},{"Start":"00:26.684 ","End":"00:28.259","Text":"or left and right."},{"Start":"00:28.259 ","End":"00:31.139","Text":"There are strings wrapped around the disks and"},{"Start":"00:31.139 ","End":"00:34.185","Text":"the strings don\u0027t slip. Where are the strings?"},{"Start":"00:34.185 ","End":"00:37.275","Text":"This is 1 string wrapped around the outer disk,"},{"Start":"00:37.275 ","End":"00:42.630","Text":"and it\u0027s moving the disks in this direction,"},{"Start":"00:42.630 ","End":"00:45.964","Text":"here or here, depending on how the spring will work."},{"Start":"00:45.964 ","End":"00:48.904","Text":"Then there\u0027s this string attached to the inner disk,"},{"Start":"00:48.904 ","End":"00:51.349","Text":"which has a mass attached to the end of it,"},{"Start":"00:51.349 ","End":"00:56.945","Text":"which is pulling the disk in anticlockwise direction."},{"Start":"00:56.945 ","End":"00:59.726","Text":"We\u0027re being asked in the first question to find,"},{"Start":"00:59.726 ","End":"01:02.060","Text":"what is the frequency of oscillation?"},{"Start":"01:02.060 ","End":"01:06.190","Text":"What frequency is this system oscillating at?"},{"Start":"01:06.190 ","End":"01:07.655","Text":"Then our second question,"},{"Start":"01:07.655 ","End":"01:10.415","Text":"what is the energy of this system?"},{"Start":"01:10.415 ","End":"01:12.775","Text":"Let\u0027s answer the first question first."},{"Start":"01:12.775 ","End":"01:18.710","Text":"Now, we have a very rigid way of solving frequency questions."},{"Start":"01:18.710 ","End":"01:22.310","Text":"The first thing that we do is we want to find an area in"},{"Start":"01:22.310 ","End":"01:28.429","Text":"the spring where the starting point that we\u0027re going to label 0,"},{"Start":"01:28.429 ","End":"01:33.979","Text":"where we take the spring as being not extended."},{"Start":"01:33.979 ","End":"01:36.965","Text":"Like a spring would look at rest."},{"Start":"01:36.965 ","End":"01:42.995","Text":"Here we can say that is a roundabout at this point."},{"Start":"01:42.995 ","End":"01:44.465","Text":"Now, why is it here?"},{"Start":"01:44.465 ","End":"01:46.729","Text":"Because at a balance,"},{"Start":"01:46.729 ","End":"01:49.535","Text":"when the systems at balance, at equilibrium,"},{"Start":"01:49.535 ","End":"01:53.180","Text":"then here because this mass is pulling this downwards,"},{"Start":"01:53.180 ","End":"01:57.040","Text":"so the spring will be slightly extended over here."},{"Start":"01:57.040 ","End":"02:00.650","Text":"The balance comes when the spring is slightly"},{"Start":"02:00.650 ","End":"02:04.850","Text":"extended due to the mass and"},{"Start":"02:04.850 ","End":"02:10.500","Text":"the system isn\u0027t moving or rotating in any direction, it\u0027s stationary."},{"Start":"02:11.020 ","End":"02:14.960","Text":"Now, the second stage is to find"},{"Start":"02:14.960 ","End":"02:19.474","Text":"my axes and to see which direction is positive and which direction is negative."},{"Start":"02:19.474 ","End":"02:25.984","Text":"Now, if I do the classic way that the upwards direction is positive,"},{"Start":"02:25.984 ","End":"02:29.479","Text":"then when this side moves downwards,"},{"Start":"02:29.479 ","End":"02:31.389","Text":"this will be in the negative direction."},{"Start":"02:31.389 ","End":"02:36.814","Text":"Then I would have to write another equation in order to connect this side to this side,"},{"Start":"02:36.814 ","End":"02:38.914","Text":"the accelerations in both sides,"},{"Start":"02:38.914 ","End":"02:42.050","Text":"which makes it a little bit more complicated."},{"Start":"02:42.050 ","End":"02:49.830","Text":"Instead, what I can do is I can choose my axis going in this direction."},{"Start":"02:49.830 ","End":"02:52.294","Text":"Here it will be the positive,"},{"Start":"02:52.294 ","End":"02:55.018","Text":"and here it will be the position,"},{"Start":"02:55.018 ","End":"02:57.154","Text":"so when on this side, on this string,"},{"Start":"02:57.154 ","End":"02:59.690","Text":"it goes upwards, it\u0027s in the positive direction."},{"Start":"02:59.690 ","End":"03:02.805","Text":"Here, when this string goes downwards,"},{"Start":"03:02.805 ","End":"03:04.959","Text":"it\u0027s also in the positive direction."},{"Start":"03:04.959 ","End":"03:09.984","Text":"Then I don\u0027t have to write lots of equations in order to connect the 2."},{"Start":"03:09.984 ","End":"03:15.034","Text":"Now, because the disks can rotate in a clockwise or anticlockwise direction,"},{"Start":"03:15.034 ","End":"03:19.463","Text":"we have to decide which direction"},{"Start":"03:19.463 ","End":"03:24.510","Text":"be it anticlockwise or clockwise will be the positive direction."},{"Start":"03:24.510 ","End":"03:32.819","Text":"Because this is our axis and this is how the positive direction looks,"},{"Start":"03:32.819 ","End":"03:41.149","Text":"so then we can say that the anticlockwise direction will be the positive direction."},{"Start":"03:41.149 ","End":"03:42.980","Text":"Now, why is it like that?"},{"Start":"03:42.980 ","End":"03:45.180","Text":"Because as this goes up,"},{"Start":"03:45.180 ","End":"03:49.963","Text":"and this goes down and both are in the positive direction,"},{"Start":"03:49.963 ","End":"03:55.029","Text":"then they will rotate the disk in an anticlockwise direction."},{"Start":"03:55.029 ","End":"04:01.230","Text":"That means that we\u0027ll define that the anticlockwise direction is the positive direction."},{"Start":"04:01.880 ","End":"04:04.144","Text":"Now, what we have to do,"},{"Start":"04:04.144 ","End":"04:07.595","Text":"step number 3, is write the sum of all of the forces."},{"Start":"04:07.595 ","End":"04:16.429","Text":"Let\u0027s start on the forces on the box of mass M. We have mg pointing downwards,"},{"Start":"04:16.429 ","End":"04:19.159","Text":"which we\u0027ve said is in the positive direction,"},{"Start":"04:19.159 ","End":"04:26.375","Text":"so we\u0027ll write positive mg. Then remember that there\u0027s also a T in this system."},{"Start":"04:26.375 ","End":"04:31.775","Text":"Now, remember that the T will always be pointing in the direction of the string."},{"Start":"04:31.775 ","End":"04:35.745","Text":"Here the T will be pointing upwards."},{"Start":"04:35.745 ","End":"04:39.695","Text":"But if we\u0027re looking over here around this disk,"},{"Start":"04:39.695 ","End":"04:42.739","Text":"so if we were doing the forces on the disk,"},{"Start":"04:42.739 ","End":"04:46.710","Text":"then the T would be pointing downwards."},{"Start":"04:46.810 ","End":"04:51.050","Text":"Now, because the T is pointing upwards and on this side,"},{"Start":"04:51.050 ","End":"04:52.940","Text":"that\u0027s a negative direction,"},{"Start":"04:52.940 ","End":"04:57.080","Text":"so I\u0027ll do minus T and that\u0027s it."},{"Start":"04:57.080 ","End":"05:02.210","Text":"Those are all the forces working on box of mass small m,"},{"Start":"05:02.210 ","End":"05:06.905","Text":"so we can do equals m multiplied by the acceleration."},{"Start":"05:06.905 ","End":"05:10.404","Text":"Now, let\u0027s take a look at this body."},{"Start":"05:10.404 ","End":"05:12.664","Text":"Now, in my opinion,"},{"Start":"05:12.664 ","End":"05:17.180","Text":"the easiest way to solve this is to imagine that there\u0027s a mass here."},{"Start":"05:17.180 ","End":"05:18.920","Text":"Let\u0027s call it m_2,"},{"Start":"05:18.920 ","End":"05:23.670","Text":"and we can see that m_2 weighs 0."},{"Start":"05:25.880 ","End":"05:29.735","Text":"Now, because we\u0027re using a spring,"},{"Start":"05:29.735 ","End":"05:35.249","Text":"so we can say that the spring is going upwards in the positive direction with"},{"Start":"05:35.249 ","End":"05:43.550","Text":"a spring coefficient of negative k. Because we\u0027ve said that this is our 0 point,"},{"Start":"05:43.550 ","End":"05:48.334","Text":"then we can say that this distance here is x."},{"Start":"05:48.334 ","End":"05:51.605","Text":"Now, why am I labeling the masses"},{"Start":"05:51.605 ","End":"05:55.984","Text":"m_2\u0027s position over here but I didn\u0027t label the position here?"},{"Start":"05:55.984 ","End":"05:59.449","Text":"Because when I\u0027m looking at the mass,"},{"Start":"05:59.449 ","End":"06:06.665","Text":"this box, then its position isn\u0027t really important to me when working out the forces."},{"Start":"06:06.665 ","End":"06:11.629","Text":"But here, because it\u0027s a spring and it\u0027s working on harmonic motion,"},{"Start":"06:11.629 ","End":"06:14.179","Text":"so that\u0027s the whole idea about harmonic motion."},{"Start":"06:14.179 ","End":"06:21.950","Text":"It\u0027s relating the position as a function of time and trying to see where the position is."},{"Start":"06:21.950 ","End":"06:23.665","Text":"That\u0027s why here with a spring,"},{"Start":"06:23.665 ","End":"06:27.890","Text":"I need to work out the position of mass number 2."},{"Start":"06:27.890 ","End":"06:30.874","Text":"I\u0027m sorry, not the position relative to time."},{"Start":"06:30.874 ","End":"06:34.100","Text":"I meant to say the position relative to the force,"},{"Start":"06:34.100 ","End":"06:37.495","Text":"how the force affects the position."},{"Start":"06:37.495 ","End":"06:40.890","Text":"The x is constantly changing,"},{"Start":"06:40.890 ","End":"06:42.260","Text":"and so is the force."},{"Start":"06:42.260 ","End":"06:44.419","Text":"Depending on how the x changes,"},{"Start":"06:44.419 ","End":"06:47.630","Text":"the force is always changing as well."},{"Start":"06:47.630 ","End":"06:56.434","Text":"I advise to always take the direction of mass number 2 or the spring to be in the;"},{"Start":"06:56.434 ","End":"06:59.450","Text":"here upwards but always in the positive direction,"},{"Start":"06:59.450 ","End":"07:06.706","Text":"and then to write k with a negative sign in front of it,"},{"Start":"07:06.706 ","End":"07:09.159","Text":"with a negative coefficient."},{"Start":"07:09.159 ","End":"07:12.123","Text":"Then because here there is a minus,"},{"Start":"07:12.123 ","End":"07:13.975","Text":"when the spring moves upwards,"},{"Start":"07:13.975 ","End":"07:17.229","Text":"then of course the force of the spring is going to want to push it downwards,"},{"Start":"07:17.229 ","End":"07:18.550","Text":"which is in the opposite direction,"},{"Start":"07:18.550 ","End":"07:20.775","Text":"so that\u0027s a minus."},{"Start":"07:20.775 ","End":"07:23.234","Text":"If you push the spring down,"},{"Start":"07:23.234 ","End":"07:27.699","Text":"then the k coefficient will be pushing it upwards in"},{"Start":"07:27.699 ","End":"07:30.190","Text":"the positive direction to get this spring"},{"Start":"07:30.190 ","End":"07:33.650","Text":"back to its resting position, so it makes sense."},{"Start":"07:33.650 ","End":"07:38.020","Text":"Let\u0027s continue. We\u0027ll write this."},{"Start":"07:38.020 ","End":"07:43.900","Text":"We have negative k multiplied by x,"},{"Start":"07:43.900 ","End":"07:48.779","Text":"the spring constant, and this is in the positive direction."},{"Start":"07:48.779 ","End":"07:53.584","Text":"Then here you\u0027ll notice because it\u0027s a string, sorry,"},{"Start":"07:53.584 ","End":"07:56.720","Text":"we have T, but it\u0027s a different T to this,"},{"Start":"07:56.720 ","End":"07:59.450","Text":"so we\u0027ll label this T_2 and this one T_1."},{"Start":"07:59.450 ","End":"08:01.830","Text":"Let\u0027s just correct everything."},{"Start":"08:02.020 ","End":"08:04.310","Text":"Because this is a different string,"},{"Start":"08:04.310 ","End":"08:05.705","Text":"it\u0027s a different tension."},{"Start":"08:05.705 ","End":"08:10.601","Text":"Now, notice also even if it was the same string,"},{"Start":"08:10.601 ","End":"08:15.455","Text":"only if this was going over an ideal pulley system,"},{"Start":"08:15.455 ","End":"08:18.155","Text":"which means a frictionless pulley system,"},{"Start":"08:18.155 ","End":"08:22.190","Text":"would we be able to say that the 2 T\u0027s are the same."},{"Start":"08:22.190 ","End":"08:26.876","Text":"If this was a pulley system with friction,"},{"Start":"08:26.876 ","End":"08:29.120","Text":"then it would still, even if the strings were the same,"},{"Start":"08:29.120 ","End":"08:33.060","Text":"it would still be T_1 and T_2 because they aren\u0027t equal."},{"Start":"08:35.010 ","End":"08:41.065","Text":"Negative kx is going in the positive direction and so is T_2,"},{"Start":"08:41.065 ","End":"08:44.125","Text":"so we can add T_2."},{"Start":"08:44.125 ","End":"08:50.030","Text":"Those are the forces acting on it, which equals m_2a_2."},{"Start":"08:50.040 ","End":"08:56.800","Text":"However, we said at the beginning because we made up mass number 2 it equals to 0."},{"Start":"08:56.800 ","End":"08:59.034","Text":"This crosses out and equal to 0,"},{"Start":"08:59.034 ","End":"09:01.270","Text":"which means that we can say,"},{"Start":"09:01.270 ","End":"09:08.000","Text":"this is a positive, so we can say that T_2 = kx."},{"Start":"09:11.640 ","End":"09:17.470","Text":"Now, of course, some of you will be able to see immediately that T_2 ="},{"Start":"09:17.470 ","End":"09:20.379","Text":"kx without having to make up"},{"Start":"09:20.379 ","End":"09:25.179","Text":"this imaginary mass number 2 and work out all of the forces,"},{"Start":"09:25.179 ","End":"09:29.290","Text":"because you know that you have the force of the spring pulling"},{"Start":"09:29.290 ","End":"09:34.059","Text":"it and you have T_2 pulling it in the other direction,"},{"Start":"09:34.059 ","End":"09:40.930","Text":"so you could easily just say that T_2 = kx and finish."},{"Start":"09:40.930 ","End":"09:43.480","Text":"But if you don\u0027t understand it properly,"},{"Start":"09:43.480 ","End":"09:45.970","Text":"sometimes it\u0027s better to go the long way and then be"},{"Start":"09:45.970 ","End":"09:48.925","Text":"certain that you\u0027re getting the correct answer."},{"Start":"09:48.925 ","End":"09:50.785","Text":"As for the disks now."},{"Start":"09:50.785 ","End":"09:53.859","Text":"On the disks, even though they\u0027re not moving up, down,"},{"Start":"09:53.859 ","End":"09:56.154","Text":"left or right, they can rotate,"},{"Start":"09:56.154 ","End":"09:59.350","Text":"which means that they have moments."},{"Start":"09:59.350 ","End":"10:02.320","Text":"Let\u0027s write down all of their moments."},{"Start":"10:02.320 ","End":"10:05.214","Text":"This arrow here is T_1 as well,"},{"Start":"10:05.214 ","End":"10:10.120","Text":"and this arrow going in this direction is T_2."},{"Start":"10:10.120 ","End":"10:12.745","Text":"Let\u0027s write down the forces."},{"Start":"10:12.745 ","End":"10:18.340","Text":"On this string, we have T_2 pointing downwards,"},{"Start":"10:18.340 ","End":"10:25.630","Text":"and then we multiply it by the radius from the center until where the string is,"},{"Start":"10:25.630 ","End":"10:28.059","Text":"which is the larger radius,"},{"Start":"10:28.059 ","End":"10:29.920","Text":"so it\u0027s R,"},{"Start":"10:29.920 ","End":"10:35.650","Text":"and then when you have"},{"Start":"10:35.650 ","End":"10:41.590","Text":"a string attached to some pulley or disk,"},{"Start":"10:41.590 ","End":"10:48.519","Text":"we always have to multiply it by sine of Pi over 2 or sine of 90 degrees."},{"Start":"10:48.519 ","End":"10:50.589","Text":"Now, sine of 90 degrees,"},{"Start":"10:50.589 ","End":"10:53.410","Text":"sine of Pi over 2 is always equal to 1,"},{"Start":"10:53.410 ","End":"10:59.619","Text":"but you always have to remember to multiply it by that, so that\u0027s that."},{"Start":"10:59.619 ","End":"11:03.174","Text":"Now have to decide if it\u0027s positive or negative."},{"Start":"11:03.174 ","End":"11:04.914","Text":"Let\u0027s look at the diagram."},{"Start":"11:04.914 ","End":"11:12.489","Text":"This will be pulling the pulley in this direction, this string,"},{"Start":"11:12.489 ","End":"11:17.560","Text":"which would be pulling the disks in a clockwise direction,"},{"Start":"11:17.560 ","End":"11:19.569","Text":"which is in the negative direction,"},{"Start":"11:19.569 ","End":"11:22.630","Text":"because we said that the anticlockwise direction was the positive,"},{"Start":"11:22.630 ","End":"11:25.330","Text":"so we add a minus."},{"Start":"11:25.330 ","End":"11:29.965","Text":"Now, let\u0027s take a look at the next string with the mass attached to it."},{"Start":"11:29.965 ","End":"11:38.590","Text":"Here we have T_1 multiplied by from the center until the outer disk,"},{"Start":"11:38.590 ","End":"11:42.984","Text":"which is T_1 and r, and, again,"},{"Start":"11:42.984 ","End":"11:46.990","Text":"multiplied by sine of Pi over 2 or sine of 90,"},{"Start":"11:46.990 ","End":"11:48.220","Text":"which again is 1,"},{"Start":"11:48.220 ","End":"11:50.050","Text":"so we can just leave it like that."},{"Start":"11:50.050 ","End":"11:54.700","Text":"Now, let\u0027s decide if it\u0027s going in the positive direction or not."},{"Start":"11:54.700 ","End":"11:57.729","Text":"It\u0027s being pulled down on this side,"},{"Start":"11:57.729 ","End":"12:02.575","Text":"which means that the disk would rotate in the anticlockwise direction,"},{"Start":"12:02.575 ","End":"12:05.335","Text":"which is what we said was the positive direction,"},{"Start":"12:05.335 ","End":"12:07.585","Text":"so we can add a plus."},{"Start":"12:07.585 ","End":"12:14.118","Text":"Then, of course, we know that the sum of all the moments is equal to I Alpha,"},{"Start":"12:14.118 ","End":"12:19.100","Text":"and that Alpha is equal to Theta double dot."},{"Start":"12:20.100 ","End":"12:25.600","Text":"Now, we\u0027ve written down all of our force and moment equations,"},{"Start":"12:25.600 ","End":"12:30.250","Text":"so now we have to make a connection between all of the accelerations."},{"Start":"12:30.250 ","End":"12:37.480","Text":"Between a_1, between a_2 and between the angular acceleration, which is Alpha."},{"Start":"12:37.480 ","End":"12:39.478","Text":"Now, I\u0027m reminding, again,"},{"Start":"12:39.478 ","End":"12:40.795","Text":"that for each body,"},{"Start":"12:40.795 ","End":"12:47.829","Text":"it\u0027s always good to write down that body\u0027s acceleration and like here with this,"},{"Start":"12:47.829 ","End":"12:51.550","Text":"it\u0027s a_1 and here it\u0027s a_2."},{"Start":"12:51.550 ","End":"12:57.475","Text":"Why is this? Because here we can see that it won\u0027t be necessarily the same acceleration,"},{"Start":"12:57.475 ","End":"13:00.185","Text":"because here it\u0027s moving upwards,"},{"Start":"13:00.185 ","End":"13:03.505","Text":"but here the radius is much larger."},{"Start":"13:03.505 ","End":"13:06.850","Text":"Here is moving downwards and the radius is much smaller,"},{"Start":"13:06.850 ","End":"13:09.610","Text":"the accelerations will be different."},{"Start":"13:09.610 ","End":"13:14.875","Text":"Now, let\u0027s write down the relationship between all of the accelerations."},{"Start":"13:14.875 ","End":"13:18.040","Text":"Let\u0027s start with a_2."},{"Start":"13:18.040 ","End":"13:22.345","Text":"When a_2 is going in the positive direction,"},{"Start":"13:22.345 ","End":"13:27.100","Text":"so say this point is going up 1 centimeter,"},{"Start":"13:27.100 ","End":"13:33.649","Text":"because there isn\u0027t slipping it means that this also goes up 1 centimeter."},{"Start":"13:33.649 ","End":"13:37.245","Text":"Then this is going a_2 upwards,"},{"Start":"13:37.245 ","End":"13:41.955","Text":"and then this one is going Alpha r upwards,"},{"Start":"13:41.955 ","End":"13:48.850","Text":"because it\u0027s the tangential acceleration multiplied by the radius."},{"Start":"13:48.930 ","End":"13:56.140","Text":"Because of that, a_2 is equal to Alpha multiplied by R,"},{"Start":"13:56.140 ","End":"14:01.510","Text":"because the acceleration in a straight line here is a_2, which is equal to,"},{"Start":"14:01.510 ","End":"14:09.835","Text":"if we convert the Alpha into a straight line acceleration,"},{"Start":"14:09.835 ","End":"14:14.170","Text":"then it\u0027s Alpha multiplied by R. Now,"},{"Start":"14:14.170 ","End":"14:17.665","Text":"let\u0027s take a look at this box mass."},{"Start":"14:17.665 ","End":"14:20.020","Text":"Here we have a_1,"},{"Start":"14:20.020 ","End":"14:22.660","Text":"because that was the acceleration here."},{"Start":"14:22.660 ","End":"14:25.224","Text":"When this is accelerating,"},{"Start":"14:25.224 ","End":"14:30.565","Text":"then we have equals Alpha r,"},{"Start":"14:30.565 ","End":"14:32.110","Text":"exactly like over here,"},{"Start":"14:32.110 ","End":"14:34.210","Text":"just with a different radius."},{"Start":"14:34.210 ","End":"14:36.910","Text":"Now, we have to see what our I is."},{"Start":"14:36.910 ","End":"14:40.209","Text":"Now our I is just the I of a disk,"},{"Start":"14:40.209 ","End":"14:44.709","Text":"which is equal to 1/2 the mass,"},{"Start":"14:44.709 ","End":"14:46.285","Text":"which is M,"},{"Start":"14:46.285 ","End":"14:50.544","Text":"multiplied by the outer radius,"},{"Start":"14:50.544 ","End":"14:54.650","Text":"so the larger radius, R^2."},{"Start":"14:55.260 ","End":"14:58.269","Text":"Now, let me just explain this 1 second."},{"Start":"14:58.269 ","End":"14:59.769","Text":"Here we have a disk,"},{"Start":"14:59.769 ","End":"15:06.250","Text":"and in the question they stated that the whole disk,"},{"Start":"15:06.250 ","End":"15:08.980","Text":"this whole thing is of mass M,"},{"Start":"15:08.980 ","End":"15:13.090","Text":"and it has the outer radius from the center is equal to 3,"},{"Start":"15:13.090 ","End":"15:18.530","Text":"and then there\u0027s a small disk of radius r."},{"Start":"15:20.100 ","End":"15:23.964","Text":"Here, because they only gave us 1 mass,"},{"Start":"15:23.964 ","End":"15:25.479","Text":"it means that it\u0027s referring to"},{"Start":"15:25.479 ","End":"15:29.275","Text":"the entire disk including the small one and the large one,"},{"Start":"15:29.275 ","End":"15:31.344","Text":"which means that we can really refer to"},{"Start":"15:31.344 ","End":"15:34.420","Text":"this small disk as being massless and just with a radius."},{"Start":"15:34.420 ","End":"15:40.330","Text":"It\u0027s is if just a circle was drawn on the desk and it\u0027s not actually a separate disk."},{"Start":"15:40.330 ","End":"15:44.800","Text":"If they would\u0027ve said there\u0027s the outer disk of mass M_1 and of"},{"Start":"15:44.800 ","End":"15:49.825","Text":"radius R and the smaller disk of mass M_2 and radius r,"},{"Start":"15:49.825 ","End":"15:54.160","Text":"then we would have had to have written 2 different I\u0027s for each disk."},{"Start":"15:54.160 ","End":"15:56.620","Text":"But because here there\u0027s only 1 mass,"},{"Start":"15:56.620 ","End":"16:00.580","Text":"so we can refer to it like this and then we can just write"},{"Start":"16:00.580 ","End":"16:05.950","Text":"that I just referring to this large disk."},{"Start":"16:05.950 ","End":"16:09.775","Text":"Now, all that\u0027s left for me to do is to"},{"Start":"16:09.775 ","End":"16:14.109","Text":"connect all of the equations into a harmonic equation."},{"Start":"16:14.109 ","End":"16:17.139","Text":"Now in a harmonic equation,"},{"Start":"16:17.139 ","End":"16:22.510","Text":"I have to connect between position and acceleration."},{"Start":"16:22.510 ","End":"16:26.634","Text":"Let me remind you what a harmonic equation is meant to look like."},{"Start":"16:26.634 ","End":"16:34.100","Text":"It\u0027s x double dot is equal to negative something multiplied by;"},{"Start":"16:35.430 ","End":"16:38.935","Text":"Now it\u0027s meant to be in exactly this form."},{"Start":"16:38.935 ","End":"16:40.299","Text":"Whoever doesn\u0027t remember this,"},{"Start":"16:40.299 ","End":"16:45.565","Text":"I suggest you go back to the beginning lessons of harmonic motion."},{"Start":"16:45.565 ","End":"16:54.145","Text":"Now we have to connect between x and a_2 because that is the acceleration of this point."},{"Start":"16:54.145 ","End":"16:57.115","Text":"In order for me to find this connection,"},{"Start":"16:57.115 ","End":"17:01.840","Text":"I have to have either the same amount of unknowns as"},{"Start":"17:01.840 ","End":"17:07.090","Text":"I do equations or more equations than amount of unknowns."},{"Start":"17:07.090 ","End":"17:08.980","Text":"Let\u0027s see over here."},{"Start":"17:08.980 ","End":"17:11.110","Text":"Now, we can cross out"},{"Start":"17:11.110 ","End":"17:16.585","Text":"this equation because this equation is just this equation that was rearranged."},{"Start":"17:16.585 ","End":"17:19.450","Text":"What are our unknowns? T_1 is unknown,"},{"Start":"17:19.450 ","End":"17:20.755","Text":"a_1 is unknown,"},{"Start":"17:20.755 ","End":"17:22.255","Text":"T_2 is unknown,"},{"Start":"17:22.255 ","End":"17:25.119","Text":"x is unknown, Alpha is unknown,"},{"Start":"17:25.119 ","End":"17:28.390","Text":"and a_2 is unknown."},{"Start":"17:28.390 ","End":"17:30.310","Text":"We have 1, 2,"},{"Start":"17:30.310 ","End":"17:31.645","Text":"3, 4, 5,"},{"Start":"17:31.645 ","End":"17:34.210","Text":"6 unknowns and we have 1, 2,"},{"Start":"17:34.210 ","End":"17:35.500","Text":"3, 4,"},{"Start":"17:35.500 ","End":"17:37.855","Text":"5, 6 equations."},{"Start":"17:37.855 ","End":"17:40.450","Text":"Perfect so we can work everything out."},{"Start":"17:40.450 ","End":"17:44.515","Text":"Now because we\u0027re working out an equation in harmonic motion,"},{"Start":"17:44.515 ","End":"17:48.190","Text":"and we\u0027re referring to the acceleration here because"},{"Start":"17:48.190 ","End":"17:53.245","Text":"the spring is the thing that can work in harmonic motion here as a_2."},{"Start":"17:53.245 ","End":"17:57.565","Text":"We\u0027re going to say that x double dot is just a_2."},{"Start":"17:57.565 ","End":"18:02.500","Text":"That means that in order to get all of my equations into this format,"},{"Start":"18:02.500 ","End":"18:07.569","Text":"I have to just sub in all of these equations 1 into another in order to be"},{"Start":"18:07.569 ","End":"18:13.240","Text":"left with an equation with just a_2 and x."},{"Start":"18:13.240 ","End":"18:15.115","Text":"Let\u0027s begin our algebra."},{"Start":"18:15.115 ","End":"18:16.795","Text":"Let\u0027s start with this equation."},{"Start":"18:16.795 ","End":"18:19.015","Text":"I have negative T_2,"},{"Start":"18:19.015 ","End":"18:21.085","Text":"but here I have an equation for T_2."},{"Start":"18:21.085 ","End":"18:24.650","Text":"I can write negative kx"},{"Start":"18:28.770 ","End":"18:38.680","Text":"multiplied by R plus T_1 multiplied by r equals i."},{"Start":"18:38.680 ","End":"18:46.510","Text":"Instead of r am going to do 1/2MR squared multiplied by Alpha."},{"Start":"18:46.510 ","End":"18:49.345","Text":"Now, I have Alpha in these 2 equations."},{"Start":"18:49.345 ","End":"18:53.341","Text":"Let\u0027s see which one I want to use now because I\u0027m trying to find a_2,"},{"Start":"18:53.341 ","End":"18:57.970","Text":"I\u0027m going to write Alpha in terms of a_2 and not in terms of a_1,"},{"Start":"18:57.970 ","End":"19:05.620","Text":"so I\u0027m going to multiply it by a_2 divided by R. Now,"},{"Start":"19:05.620 ","End":"19:09.309","Text":"we can already see our equation taking shape because here,"},{"Start":"19:09.309 ","End":"19:13.630","Text":"I have already an equation between x and a_2."},{"Start":"19:13.630 ","End":"19:16.704","Text":"But now I have to get rid of"},{"Start":"19:16.704 ","End":"19:23.470","Text":"anything that is either an unknown or that can be changing in time."},{"Start":"19:23.470 ","End":"19:27.204","Text":"Because into this section and into this section,"},{"Start":"19:27.204 ","End":"19:30.040","Text":"I can only add in constants."},{"Start":"19:30.040 ","End":"19:32.455","Text":"Let\u0027s see how we do this."},{"Start":"19:32.455 ","End":"19:37.029","Text":"Let\u0027s go and delete all of these unknowns."},{"Start":"19:37.029 ","End":"19:40.345","Text":"T_1, I don\u0027t really know what this is,"},{"Start":"19:40.345 ","End":"19:41.875","Text":"and it\u0027s changing in time."},{"Start":"19:41.875 ","End":"19:45.385","Text":"However, here I have an equation with T_1."},{"Start":"19:45.385 ","End":"19:50.586","Text":"If I move T_1 to this side and I move this Ma_1 to this side,"},{"Start":"19:50.586 ","End":"19:59.470","Text":"this will become m(g) minus a_1."},{"Start":"19:59.470 ","End":"20:02.905","Text":"Good. But now a_1 is also an unknown,"},{"Start":"20:02.905 ","End":"20:04.209","Text":"and it\u0027s also changing in time."},{"Start":"20:04.209 ","End":"20:05.815","Text":"I have to get rid of this."},{"Start":"20:05.815 ","End":"20:08.890","Text":"Now here, we can see that a_1 equals"},{"Start":"20:08.890 ","End":"20:15.520","Text":"Alpha r. We can say that this equals Alpha r. However,"},{"Start":"20:15.520 ","End":"20:17.620","Text":"Alpha, we also need to change,"},{"Start":"20:17.620 ","End":"20:27.235","Text":"but we know that Alpha is a_2 divided by R. We can change this as well."},{"Start":"20:27.235 ","End":"20:31.210","Text":"Now we have to rearrange this in order to"},{"Start":"20:31.210 ","End":"20:35.125","Text":"just be with these constants that we\u0027ve now written."},{"Start":"20:35.125 ","End":"20:39.640","Text":"Let\u0027s rewrite this equation."},{"Start":"20:39.640 ","End":"20:48.760","Text":"We have negative kx multiplied by R plus r,"},{"Start":"20:48.760 ","End":"20:50.950","Text":"and then let\u0027s open brackets."},{"Start":"20:50.950 ","End":"20:59.740","Text":"T_1 is m multiplied by g minus a_1."},{"Start":"20:59.740 ","End":"21:02.215","Text":"But a_1 is Alpha r,"},{"Start":"21:02.215 ","End":"21:11.320","Text":"but Alpha is a_2 divided by R minus a_2 divided by R,"},{"Start":"21:11.320 ","End":"21:14.110","Text":"and then backup multiplied by"},{"Start":"21:14.110 ","End":"21:21.977","Text":"r. Then we can close these brackets."},{"Start":"21:21.977 ","End":"21:26.059","Text":"Then it equals the same as here,"},{"Start":"21:26.059 ","End":"21:28.389","Text":"1/2 multiplied by m. Now,"},{"Start":"21:28.389 ","End":"21:32.395","Text":"this r squared can cross out R,"},{"Start":"21:32.395 ","End":"21:38.934","Text":"and then I\u0027m going to rewrite a_2 as x double dot."},{"Start":"21:38.934 ","End":"21:42.760","Text":"Because I called this a_2 x double dot."},{"Start":"21:42.760 ","End":"21:45.415","Text":"This a_2 can also be x double dot."},{"Start":"21:45.415 ","End":"21:52.975","Text":"I\u0027m just going to rub this out and replace it with x double-dot."},{"Start":"21:52.975 ","End":"22:00.550","Text":"Now I\u0027m left with an equation with everything being known,"},{"Start":"22:00.550 ","End":"22:03.445","Text":"all my variables being known and being constant,"},{"Start":"22:03.445 ","End":"22:08.935","Text":"and I have my non-constant variables which are x and x double dot,"},{"Start":"22:08.935 ","End":"22:12.055","Text":"which is exactly what I wanted to be left with."},{"Start":"22:12.055 ","End":"22:19.300","Text":"Now what we can do is we can open the brackets and start to simplify this."},{"Start":"22:19.300 ","End":"22:22.600","Text":"Here I\u0027ll write negative kr"},{"Start":"22:22.600 ","End":"22:26.710","Text":"multiplied by x. I\u0027ve just changed the order in order to make it"},{"Start":"22:26.710 ","End":"22:28.600","Text":"more obvious and clear to myself that"},{"Start":"22:28.600 ","End":"22:34.570","Text":"the negative kr refers to the coefficient of my x value."},{"Start":"22:34.570 ","End":"22:38.530","Text":"Plus, I\u0027ll open up my brackets here,"},{"Start":"22:38.530 ","End":"22:43.330","Text":"so I have rmg."},{"Start":"22:43.330 ","End":"22:52.495","Text":"Then I\u0027ll have rm multiplied again by this r divided by R times x double dot."},{"Start":"22:52.495 ","End":"22:56.619","Text":"But I want to get all of my x-double dots onto the same side."},{"Start":"22:56.619 ","End":"22:58.269","Text":"I\u0027ll say that this is equal,"},{"Start":"22:58.269 ","End":"23:05.365","Text":"so I\u0027ll just move this multiplied by this to the other side of the equation of 1/2."},{"Start":"23:05.365 ","End":"23:10.329","Text":"Then I\u0027ll put the x double"},{"Start":"23:10.329 ","End":"23:15.190","Text":"dot on the outside of the brackets because I want to isolate that."},{"Start":"23:15.190 ","End":"23:20.080","Text":"Then I\u0027ll put it in the 1/2mr multiplied by x double dot,"},{"Start":"23:20.080 ","End":"23:29.150","Text":"and then here I have plus mr squared divided by R."},{"Start":"23:29.150 ","End":"23:34.090","Text":"So 1/2mr squared divided"},{"Start":"23:34.090 ","End":"23:41.470","Text":"by R multiplied by x double-dot. Perfect."},{"Start":"23:41.470 ","End":"23:45.810","Text":"Now I can rearrange this by dividing,"},{"Start":"23:45.810 ","End":"23:50.334","Text":"let\u0027s call all of this A, because we\u0027re lazy."},{"Start":"23:50.334 ","End":"23:53.109","Text":"I\u0027m just going to move this down a little bit."},{"Start":"23:53.109 ","End":"23:55.000","Text":"All of this will be A."},{"Start":"23:55.000 ","End":"24:01.915","Text":"We can say that negative KRx divided by A."},{"Start":"24:01.915 ","End":"24:04.030","Text":"We\u0027re going to divide both sides by A,"},{"Start":"24:04.030 ","End":"24:12.430","Text":"plus rmg divided by A is equal to x double dot."},{"Start":"24:12.430 ","End":"24:14.889","Text":"Now, all that we have to do is,"},{"Start":"24:14.889 ","End":"24:19.075","Text":"we have to put this into this format."},{"Start":"24:19.075 ","End":"24:46.284","Text":"Let\u0027s write this x double dot is equal to minus"},{"Start":"24:46.284 ","End":"24:49.540","Text":"KR divided by A multiplied by x."},{"Start":"24:49.540 ","End":"24:57.674","Text":"Then we have x minus something and that\u0027s it."},{"Start":"24:57.674 ","End":"24:59.459","Text":"Now what is this minus something?"},{"Start":"24:59.459 ","End":"25:03.315","Text":"As you can see here,"},{"Start":"25:03.315 ","End":"25:06.505","Text":"when this kR divided by A,"},{"Start":"25:06.505 ","End":"25:09.325","Text":"this whole expression being A remember,"},{"Start":"25:09.325 ","End":"25:12.400","Text":"multiplied by x, it will equal this."},{"Start":"25:12.400 ","End":"25:14.248","Text":"Perfect."},{"Start":"25:14.248 ","End":"25:15.879","Text":"But I also have to have,"},{"Start":"25:15.879 ","End":"25:21.865","Text":"when this multiplied by whatever this is will equal this. How do I do it?"},{"Start":"25:21.865 ","End":"25:25.015","Text":"Notice they both have the same denominator."},{"Start":"25:25.015 ","End":"25:26.649","Text":"K here it\u0027s also being divided by A,"},{"Start":"25:26.649 ","End":"25:28.015","Text":"here it\u0027s being divided by A."},{"Start":"25:28.015 ","End":"25:30.265","Text":"That I don\u0027t need to touch."},{"Start":"25:30.265 ","End":"25:35.485","Text":"But here I have a K and a capital R,"},{"Start":"25:35.485 ","End":"25:40.914","Text":"and here I don\u0027t at all have a K or a capital R. In order to cancel them out,"},{"Start":"25:40.914 ","End":"25:47.200","Text":"I\u0027m going to divide by K and capital R. When this is multiplied by this,"},{"Start":"25:47.200 ","End":"25:49.750","Text":"this A in the denominator, will stay,"},{"Start":"25:49.750 ","End":"25:55.060","Text":"but this K capital R in the numerator will be canceled out by"},{"Start":"25:55.060 ","End":"26:01.239","Text":"this K and capital R in the denominator in this section."},{"Start":"26:01.239 ","End":"26:09.240","Text":"Here, I have rmg in the numerator, which I want to keep."},{"Start":"26:09.240 ","End":"26:19.315","Text":"I\u0027m just going to multiply it by rm and g. Now when this is multiplied by this,"},{"Start":"26:19.315 ","End":"26:22.780","Text":"the K and the R will cancel out because of here,"},{"Start":"26:22.780 ","End":"26:25.030","Text":"the A will remain,"},{"Start":"26:25.030 ","End":"26:29.335","Text":"and then it will be multiplied by rmg and we will end up"},{"Start":"26:29.335 ","End":"26:35.275","Text":"with this exact number or expression."},{"Start":"26:35.275 ","End":"26:37.494","Text":"That is it. That\u0027s how you get it."},{"Start":"26:37.494 ","End":"26:40.690","Text":"Now, this way, in this type of format,"},{"Start":"26:40.690 ","End":"26:43.300","Text":"we can actually learn about what\u0027s happening in the system,"},{"Start":"26:43.300 ","End":"26:48.200","Text":"and it\u0027s a lot easier to understand what\u0027s going on in this motion,"},{"Start":"26:48.240 ","End":"26:53.020","Text":"which we probably wouldn\u0027t understand or would have a very difficult time"},{"Start":"26:53.020 ","End":"26:57.910","Text":"understanding if it was in this format or even worse in this format."},{"Start":"26:57.910 ","End":"27:00.775","Text":"Let\u0027s look at what we can learn from this."},{"Start":"27:00.775 ","End":"27:08.155","Text":"We can learn that this is equal to the frequency squared,"},{"Start":"27:08.155 ","End":"27:10.660","Text":"which answers our question,"},{"Start":"27:10.660 ","End":"27:12.040","Text":"question number 1,"},{"Start":"27:12.040 ","End":"27:16.479","Text":"which was, what is the frequency of oscillation?"},{"Start":"27:16.479 ","End":"27:18.354","Text":"So here we go."},{"Start":"27:18.354 ","End":"27:22.480","Text":"It\u0027s the square root of this is the frequency because this is the frequency squared."},{"Start":"27:22.480 ","End":"27:27.980","Text":"This gives us the point of equilibrium. Equilibrium point."},{"Start":"27:34.920 ","End":"27:39.324","Text":"Let\u0027s see what this equilibrium point actually means."},{"Start":"27:39.324 ","End":"27:43.000","Text":"It\u0027s the point at which the system is in balance."},{"Start":"27:43.000 ","End":"27:46.014","Text":"If we look back over here."},{"Start":"27:46.014 ","End":"27:48.520","Text":"Here we had our 0 point."},{"Start":"27:48.520 ","End":"27:52.780","Text":"This height, the amount that this string has,"},{"Start":"27:52.780 ","End":"27:56.064","Text":"so this is EP, equilibrium point."},{"Start":"27:56.064 ","End":"28:00.595","Text":"This length that the spring had to stretch from the 0,"},{"Start":"28:00.595 ","End":"28:07.690","Text":"From the point where the spring is at rest and not being stretched or compressed,"},{"Start":"28:07.690 ","End":"28:12.579","Text":"this is the equilibrium point which denotes when the system is at rest,"},{"Start":"28:12.579 ","End":"28:14.994","Text":"when all the forces are balanced,"},{"Start":"28:14.994 ","End":"28:17.049","Text":"and nothing is moving or rotating."},{"Start":"28:17.049 ","End":"28:23.740","Text":"There\u0027s a complete balance of forces."},{"Start":"28:23.740 ","End":"28:26.035","Text":"This is the equilibrium point."},{"Start":"28:26.035 ","End":"28:29.229","Text":"Now I\u0027m going to minimize all of the writing on the screen to"},{"Start":"28:29.229 ","End":"28:32.830","Text":"make some space for this question,"},{"Start":"28:32.830 ","End":"28:35.650","Text":"what is the energy of the system?"},{"Start":"28:35.650 ","End":"28:40.150","Text":"The second question, what is the energy of the system?"},{"Start":"28:40.150 ","End":"28:43.240","Text":"The total energy of this system,"},{"Start":"28:43.240 ","End":"28:46.465","Text":"E tot is equal to,"},{"Start":"28:46.465 ","End":"28:48.069","Text":"let\u0027s start with our spring."},{"Start":"28:48.069 ","End":"28:54.710","Text":"The energy of a spring is 1/2 times Kx^2."},{"Start":"28:55.860 ","End":"29:00.580","Text":"Then my next thing will be here."},{"Start":"29:00.580 ","End":"29:04.869","Text":"Here because this box is in the air,"},{"Start":"29:04.869 ","End":"29:06.744","Text":"so it will have potential energy,"},{"Start":"29:06.744 ","End":"29:08.860","Text":"which will be mgh."},{"Start":"29:08.860 ","End":"29:10.074","Text":"Now what is my h here?"},{"Start":"29:10.074 ","End":"29:14.155","Text":"It\u0027s x because as x here changes,"},{"Start":"29:14.155 ","End":"29:19.825","Text":"it will be the exact same x over here, so it\u0027s mgx."},{"Start":"29:19.825 ","End":"29:23.230","Text":"Now, is it in the positive or negative direction?"},{"Start":"29:23.230 ","End":"29:24.999","Text":"Here it\u0027s in the positive direction."},{"Start":"29:24.999 ","End":"29:28.809","Text":"Now when this moves down,"},{"Start":"29:28.809 ","End":"29:30.775","Text":"this is according to our axes,"},{"Start":"29:30.775 ","End":"29:31.990","Text":"also in the positive direction."},{"Start":"29:31.990 ","End":"29:35.950","Text":"However, with energy, these axes"},{"Start":"29:35.950 ","End":"29:40.300","Text":"don\u0027t apply because if the energy here is going up and the energy here is going down,"},{"Start":"29:40.300 ","End":"29:41.619","Text":"they\u0027re going in different directions."},{"Start":"29:41.619 ","End":"29:43.075","Text":"It\u0027s not like where the force is."},{"Start":"29:43.075 ","End":"29:46.915","Text":"We have to say that it\u0027s negative mgx."},{"Start":"29:46.915 ","End":"29:53.184","Text":"Now, we also have to add in the energy from the rotation of the disk,"},{"Start":"29:53.184 ","End":"30:00.280","Text":"which is plus 1/2I omega squared."},{"Start":"30:00.280 ","End":"30:03.535","Text":"This is the energy of the rotation of the disk."},{"Start":"30:03.535 ","End":"30:10.165","Text":"Now remembering that this I is the I that we found over here,1/2 mI^2."},{"Start":"30:10.165 ","End":"30:16.240","Text":"Then we also have to add to it 1/2mv^2."},{"Start":"30:16.240 ","End":"30:19.944","Text":"Now what is v? It\u0027s x dot squared."},{"Start":"30:19.944 ","End":"30:28.900","Text":"Remember, the derivative of x=v."},{"Start":"30:28.900 ","End":"30:32.110","Text":"This will be 1/2mv^2,"},{"Start":"30:32.110 ","End":"30:35.875","Text":"which is the kinetic energy of this dropping."},{"Start":"30:35.875 ","End":"30:37.810","Text":"Let\u0027s go over this again,"},{"Start":"30:37.810 ","End":"30:40.075","Text":"the total energy of the system."},{"Start":"30:40.075 ","End":"30:43.720","Text":"This is the answer to question 2,"},{"Start":"30:43.720 ","End":"30:46.630","Text":"is equal to the energy, this,"},{"Start":"30:46.630 ","End":"30:50.784","Text":"the energy of the spring minus,"},{"Start":"30:50.784 ","End":"30:55.164","Text":"because it\u0027s in the opposite direction to the energy of the spring,"},{"Start":"30:55.164 ","End":"31:00.970","Text":"the gravitational potential energy, plus, sorry,"},{"Start":"31:00.970 ","End":"31:05.979","Text":"the rotational energy of the disk with the I that we already found,"},{"Start":"31:05.979 ","End":"31:12.940","Text":"and plus the kinetic energy because of the movement of the system."},{"Start":"31:12.940 ","End":"31:16.550","Text":"That is the end of our lesson."}],"ID":10628},{"Watched":false,"Name":"Disk With A Hole","Duration":"9m 11s","ChapterTopicVideoID":10285,"CourseChapterTopicPlaylistID":5369,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.044","Text":"Hello, in the following question,"},{"Start":"00:03.044 ","End":"00:05.925","Text":"we have a disk which is this,"},{"Start":"00:05.925 ","End":"00:08.460","Text":"which has had a hole drilled into it."},{"Start":"00:08.460 ","End":"00:13.305","Text":"We\u0027re told that the radius of the disk is R,"},{"Start":"00:13.305 ","End":"00:21.105","Text":"that the radius of the hole drilled into it is R divided by 4."},{"Start":"00:21.105 ","End":"00:29.415","Text":"We\u0027re told that the distance from here to here is R/2."},{"Start":"00:29.415 ","End":"00:33.045","Text":"We\u0027re told that prior to the hole being drilled into the disk,"},{"Start":"00:33.045 ","End":"00:36.270","Text":"the mass of the disk was M."},{"Start":"00:36.270 ","End":"00:42.715","Text":"We\u0027re being asked to find the frequency of the small oscillations of the disk."},{"Start":"00:42.715 ","End":"00:47.630","Text":"Let\u0027s begin. We\u0027ve already done some examples,"},{"Start":"00:47.630 ","End":"00:50.555","Text":"so we know how to work with this question,"},{"Start":"00:50.555 ","End":"00:53.625","Text":"but let\u0027s go through it anyway."},{"Start":"00:53.625 ","End":"00:57.800","Text":"First of all, we work out the sum of all the moments."},{"Start":"00:57.800 ","End":"01:00.635","Text":"Then we use this equation,"},{"Start":"01:00.635 ","End":"01:07.610","Text":"which is the moments equal I multiplied by Alpha."},{"Start":"01:07.610 ","End":"01:11.825","Text":"But we know that Alpha is equal to Theta double dot."},{"Start":"01:11.825 ","End":"01:16.670","Text":"We say that it\u0027s equal to I Theta double dot."},{"Start":"01:16.670 ","End":"01:20.285","Text":"Then I work out what this is."},{"Start":"01:20.285 ","End":"01:23.720","Text":"Then I divide this by I"},{"Start":"01:23.720 ","End":"01:27.755","Text":"and then I find the relationship between Theta and Theta double dot."},{"Start":"01:27.755 ","End":"01:32.300","Text":"Now let\u0027s go step-by-step to see how we do this exactly."},{"Start":"01:32.300 ","End":"01:37.520","Text":"Let\u0027s start off with some details that we have written in the question."},{"Start":"01:37.520 ","End":"01:40.910","Text":"This hole has been removed,"},{"Start":"01:40.910 ","End":"01:43.340","Text":"section of mass m has been removed."},{"Start":"01:43.340 ","End":"01:46.160","Text":"We\u0027re going to call this section m,"},{"Start":"01:46.160 ","End":"01:48.650","Text":"and we\u0027re going to find out what its mass is and"},{"Start":"01:48.650 ","End":"01:51.290","Text":"then we\u0027re going to subtract it from the M. Because we"},{"Start":"01:51.290 ","End":"01:56.840","Text":"know that the radius of this hole is R divided by 4 so it\u0027s a quarter of the radius,"},{"Start":"01:56.840 ","End":"02:04.235","Text":"which means that its mass is 1/16 of the mass over the entire disk."},{"Start":"02:04.235 ","End":"02:10.865","Text":"We can just write 1/16 M. Why is it like this?"},{"Start":"02:10.865 ","End":"02:16.200","Text":"Because if the radius is 1/4 because the equation is,"},{"Start":"02:16.200 ","End":"02:18.747","Text":"we take the radius is I squared,"},{"Start":"02:18.747 ","End":"02:26.154","Text":"we know that the relation is the square which 1/4 squared is 1/16."},{"Start":"02:26.154 ","End":"02:28.550","Text":"Now, for the center of mass,"},{"Start":"02:28.550 ","End":"02:35.750","Text":"we found already that it\u0027s minus 1/30R."},{"Start":"02:35.750 ","End":"02:39.680","Text":"Now we already worked this out in a question that\u0027s called a disk with"},{"Start":"02:39.680 ","End":"02:44.110","Text":"a hole under the subject title of center of mass."},{"Start":"02:44.110 ","End":"02:46.595","Text":"We also found its I value,"},{"Start":"02:46.595 ","End":"02:54.020","Text":"which is equal to 247 divided by"},{"Start":"02:54.020 ","End":"03:02.105","Text":"512 MR squared in the lesson called disk with a hole."},{"Start":"03:02.105 ","End":"03:03.830","Text":"But under the title,"},{"Start":"03:03.830 ","End":"03:07.470","Text":"the subject title of moment of inertia."},{"Start":"03:07.490 ","End":"03:10.235","Text":"Now that we have all of this data,"},{"Start":"03:10.235 ","End":"03:12.635","Text":"we can move on to the question."},{"Start":"03:12.635 ","End":"03:15.365","Text":"Let\u0027s draw the center of mass."},{"Start":"03:15.365 ","End":"03:21.420","Text":"Can say that the center of mass is say, around about here."},{"Start":"03:22.370 ","End":"03:26.855","Text":"Now, if I draw a free body diagram for this,"},{"Start":"03:26.855 ","End":"03:30.755","Text":"so I can say if this is the center of mass,"},{"Start":"03:30.755 ","End":"03:35.390","Text":"then mg is pulling downwards from the center of mass,"},{"Start":"03:35.390 ","End":"03:42.540","Text":"not from the center of the disk where the screw is but from the center of mass."},{"Start":"03:42.540 ","End":"03:49.760","Text":"Then here we have a line going to where the screw is and we"},{"Start":"03:49.760 ","End":"03:57.215","Text":"have an angle of Theta in the vertical direction over here."},{"Start":"03:57.215 ","End":"04:00.230","Text":"Now notice I\u0027ve written something wrong here."},{"Start":"04:00.230 ","End":"04:03.845","Text":"This m, I\u0027ll call it m Tilda,"},{"Start":"04:03.845 ","End":"04:08.450","Text":"isn\u0027t the same as this m. What I\u0027m speaking about is the mass of the disk."},{"Start":"04:08.450 ","End":"04:14.870","Text":"Now, of course, the mass of the disk is M minus m."},{"Start":"04:14.870 ","End":"04:24.125","Text":"Then all of this is multiplied by g. But I now have to multiply this by another thing,"},{"Start":"04:24.125 ","End":"04:31.700","Text":"which is by the distance of the center of the disk here to the center of mass."},{"Start":"04:31.700 ","End":"04:37.610","Text":"That\u0027s multiplied by 1/30R."},{"Start":"04:37.610 ","End":"04:43.325","Text":"Now the minus, right now it doesn\u0027t matter to me because I\u0027m looking for the distance,"},{"Start":"04:43.325 ","End":"04:46.795","Text":"which is just a number."},{"Start":"04:46.795 ","End":"04:49.020","Text":"It\u0027s going to be positive."},{"Start":"04:49.020 ","End":"04:51.530","Text":"Then because you\u0027ll notice an angle here,"},{"Start":"04:51.530 ","End":"04:57.530","Text":"I have to write because I want it in the vertical direction sine of the angle."},{"Start":"04:57.530 ","End":"04:58.850","Text":"Now, what is the angle?"},{"Start":"04:58.850 ","End":"05:02.655","Text":"Technically, it\u0027s 180 minus Theta."},{"Start":"05:02.655 ","End":"05:07.105","Text":"Because if we look at this, for instance,"},{"Start":"05:07.105 ","End":"05:10.490","Text":"as some kind of square or rectangle or whatever,"},{"Start":"05:10.490 ","End":"05:15.480","Text":"then we know that these 2 angles together must equal 180."},{"Start":"05:15.480 ","End":"05:17.240","Text":"If here this Theta,"},{"Start":"05:17.240 ","End":"05:19.445","Text":"and this is a 180 minus Theta."},{"Start":"05:19.445 ","End":"05:21.770","Text":"However, when we\u0027re dealing with the sine,"},{"Start":"05:21.770 ","End":"05:25.130","Text":"then sine of 180 minus Theta or sine of"},{"Start":"05:25.130 ","End":"05:29.425","Text":"Theta is the same thing so we can just write sine of Theta."},{"Start":"05:29.425 ","End":"05:33.605","Text":"Then we have to add a minus sign over here."},{"Start":"05:33.605 ","End":"05:34.895","Text":"Why the minus sign?"},{"Start":"05:34.895 ","End":"05:40.415","Text":"Because we\u0027re going to define that Theta is in the positive direction,"},{"Start":"05:40.415 ","End":"05:42.620","Text":"in the clockwise direction."},{"Start":"05:42.620 ","End":"05:45.110","Text":"Then if you look at this mg,"},{"Start":"05:45.110 ","End":"05:49.760","Text":"it\u0027s trying to push this whole thing downwards and into this direction,"},{"Start":"05:49.760 ","End":"05:51.485","Text":"into the anticlockwise direction."},{"Start":"05:51.485 ","End":"05:54.790","Text":"Therefore, we put a minus over there."},{"Start":"05:54.790 ","End":"05:57.825","Text":"That\u0027s the sum of all of the moments,"},{"Start":"05:57.825 ","End":"06:01.685","Text":"and then we know that it\u0027s equal to I Theta double dot."},{"Start":"06:01.685 ","End":"06:05.600","Text":"Now our I is 247 divided by"},{"Start":"06:05.600 ","End":"06:14.085","Text":"512 MR squared multiplied by Theta double dot."},{"Start":"06:14.085 ","End":"06:18.080","Text":"Here we have it. We found the equation of motion,"},{"Start":"06:18.080 ","End":"06:21.125","Text":"which shows the relationship between"},{"Start":"06:21.125 ","End":"06:26.560","Text":"the position Theta and the acceleration Theta double dot."},{"Start":"06:26.560 ","End":"06:32.375","Text":"Now we\u0027re going to rearrange this equation to sort it out a little bit."},{"Start":"06:32.375 ","End":"06:36.560","Text":"We\u0027re going to leave the Theta double dot on 1 side."},{"Start":"06:36.560 ","End":"06:39.410","Text":"Theta double dot because we\u0027re trying to"},{"Start":"06:39.410 ","End":"06:42.320","Text":"find the acceleration in relation to the position."},{"Start":"06:42.320 ","End":"06:49.685","Text":"Then we\u0027re going to multiply this side by 512 and divide it by 247,"},{"Start":"06:49.685 ","End":"06:51.620","Text":"and by MR squared."},{"Start":"06:51.620 ","End":"06:54.710","Text":"Now, because we\u0027re dividing by this M,"},{"Start":"06:54.710 ","End":"06:57.155","Text":"this M and this M will cross out."},{"Start":"06:57.155 ","End":"07:00.275","Text":"There\u0027s an R^2 here and an R here."},{"Start":"07:00.275 ","End":"07:09.630","Text":"We\u0027re going to be left with something that looks like minus M multiplied by"},{"Start":"07:09.630 ","End":"07:15.120","Text":"15/16 because that is this M minus"},{"Start":"07:15.120 ","End":"07:22.950","Text":"this multiplied by 1/30gR."},{"Start":"07:22.950 ","End":"07:26.230","Text":"Then here I\u0027m going to write Theta because, as you remember,"},{"Start":"07:26.230 ","End":"07:28.465","Text":"when dealing with small angles,"},{"Start":"07:28.465 ","End":"07:32.525","Text":"sine of Theta equals Theta."},{"Start":"07:32.525 ","End":"07:39.970","Text":"Then multiply it by 512 divided by 247,"},{"Start":"07:39.970 ","End":"07:45.020","Text":"and divided by MR squared."},{"Start":"07:45.020 ","End":"07:48.540","Text":"As I said, this M and this M cross out,"},{"Start":"07:48.540 ","End":"07:53.030","Text":"this R squared crosses out with this."},{"Start":"07:53.030 ","End":"07:58.180","Text":"This can cross here and make it a half."},{"Start":"07:58.180 ","End":"08:07.175","Text":"Then we\u0027re going to get something along the lines of Theta double dot equaling"},{"Start":"08:07.175 ","End":"08:15.665","Text":"negative 16/247 multiplied by"},{"Start":"08:15.665 ","End":"08:26.075","Text":"g divided by R and then multiplied by Theta minus 0,"},{"Start":"08:26.075 ","End":"08:30.390","Text":"which is going to equal Theta double dot."},{"Start":"08:30.730 ","End":"08:34.505","Text":"Now, what is this 0, I hear you asking?"},{"Start":"08:34.505 ","End":"08:37.100","Text":"This is its point of equilibrium."},{"Start":"08:37.100 ","End":"08:41.255","Text":"Now, of course, equilibrium is going to equal 0. That\u0027s fine."},{"Start":"08:41.255 ","End":"08:45.990","Text":"This is called Omega squared,"},{"Start":"08:45.990 ","End":"08:52.300","Text":"which is in fact Omega here is the frequency."},{"Start":"08:53.050 ","End":"09:00.815","Text":"Do not get confused between this frequency and angular velocity here."},{"Start":"09:00.815 ","End":"09:03.715","Text":"It\u0027s the same letter, but its frequency squared."},{"Start":"09:03.715 ","End":"09:05.075","Text":"Now you have it."},{"Start":"09:05.075 ","End":"09:10.950","Text":"We found the frequency of the small oscillations in the disk with the hole."}],"ID":10635},{"Watched":false,"Name":"Half Hoop And Two Masses","Duration":"5m 34s","ChapterTopicVideoID":10286,"CourseChapterTopicPlaylistID":5369,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.294","Text":"Hello. In this question,"},{"Start":"00:02.294 ","End":"00:07.095","Text":"we\u0027re given a hoop which is being halved, a halved hoop."},{"Start":"00:07.095 ","End":"00:10.875","Text":"We\u0027re asked to find the frequency of this halved hoop."},{"Start":"00:10.875 ","End":"00:16.845","Text":"It has a radius of R and the mass of M. At each end of the hoop,"},{"Start":"00:16.845 ","End":"00:21.660","Text":"there is a mass of mass m. This is M,"},{"Start":"00:21.660 ","End":"00:26.460","Text":"and this is m. The hoop is hung via screw at its center."},{"Start":"00:26.460 ","End":"00:29.310","Text":"This is here and the half hoop,"},{"Start":"00:29.310 ","End":"00:31.620","Text":"from now on, if I refer to it as a hoop,"},{"Start":"00:31.620 ","End":"00:37.250","Text":"I\u0027m still referring to this shape acts as a compound or a physical pendulum,"},{"Start":"00:37.250 ","End":"00:38.905","Text":"however you want to call it."},{"Start":"00:38.905 ","End":"00:41.960","Text":"We\u0027ve already seen how to answer these types of questions,"},{"Start":"00:41.960 ","End":"00:44.915","Text":"but let\u0027s go through it again in practice."},{"Start":"00:44.915 ","End":"00:48.855","Text":"Now, we already know when working with a compound pendulum,"},{"Start":"00:48.855 ","End":"00:51.200","Text":"that you have to find its center of mass,"},{"Start":"00:51.200 ","End":"00:56.135","Text":"which in this case will be something around about here."},{"Start":"00:56.135 ","End":"01:00.360","Text":"Then we have to find this distance."},{"Start":"01:00.360 ","End":"01:04.310","Text":"Then when we move the entire body,"},{"Start":"01:04.310 ","End":"01:07.070","Text":"then the center of mass will move as well."},{"Start":"01:07.070 ","End":"01:08.750","Text":"If we move the entire body,"},{"Start":"01:08.750 ","End":"01:13.435","Text":"say slightly anticlockwise, the center of mass will be here."},{"Start":"01:13.435 ","End":"01:17.865","Text":"Then we\u0027ll have a line here."},{"Start":"01:17.865 ","End":"01:23.090","Text":"Then we can work out what the moment is,"},{"Start":"01:23.090 ","End":"01:24.575","Text":"which as we know,"},{"Start":"01:24.575 ","End":"01:30.320","Text":"is I Alpha or I Theta double dot."},{"Start":"01:30.320 ","End":"01:31.775","Text":"Now, we have to do,"},{"Start":"01:31.775 ","End":"01:33.845","Text":"is we have to find what the moment is."},{"Start":"01:33.845 ","End":"01:37.220","Text":"Now the moment, as we know right now,"},{"Start":"01:37.220 ","End":"01:40.235","Text":"is of course m g,"},{"Start":"01:40.235 ","End":"01:44.585","Text":"but not this m. It\u0027s going to be m with a tilde on top."},{"Start":"01:44.585 ","End":"01:53.010","Text":"Now, this m tilde is referring to the masses in the entire system."},{"Start":"01:53.010 ","End":"01:59.770","Text":"That\u0027s too small, ms plus the mass of the halved hoop."},{"Start":"01:59.770 ","End":"02:02.029","Text":"Now according to the equation,"},{"Start":"02:02.029 ","End":"02:09.705","Text":"we have to do m tilde g multiplied by this length over here."},{"Start":"02:09.705 ","End":"02:12.300","Text":"Let\u0027s call the length for now b,"},{"Start":"02:12.300 ","End":"02:13.935","Text":"multiplied by b,"},{"Start":"02:13.935 ","End":"02:21.600","Text":"multiplied by the angle between the force and the axis of rotation."},{"Start":"02:24.860 ","End":"02:27.080","Text":"This is the force,"},{"Start":"02:27.080 ","End":"02:28.895","Text":"this is axis of rotation."},{"Start":"02:28.895 ","End":"02:33.700","Text":"We\u0027re looking for this angle and let\u0027s call it Theta."},{"Start":"02:33.700 ","End":"02:39.080","Text":"Then we\u0027ll multiply it by sine of Theta."},{"Start":"02:39.080 ","End":"02:49.155","Text":"Notice that this Theta could be referred to as 180 minus the angle."},{"Start":"02:49.155 ","End":"02:51.780","Text":"This is the actual Theta."},{"Start":"02:51.780 ","End":"03:00.630","Text":"I\u0027m actually going to change it to that so that it will be slightly more understandable."},{"Start":"03:00.630 ","End":"03:02.595","Text":"This is going to be Theta."},{"Start":"03:02.595 ","End":"03:09.980","Text":"It\u0027s the exact same thing because in sine 180 minus Theta is the same as sine Theta."},{"Start":"03:09.980 ","End":"03:15.460","Text":"Then we\u0027re going to add a minus at the beginning. Why the minus?"},{"Start":"03:15.460 ","End":"03:20.905","Text":"Because we\u0027re going to define Theta as being positive in the anticlockwise direction."},{"Start":"03:20.905 ","End":"03:24.700","Text":"In which case, because this mg is pointing downwards,"},{"Start":"03:24.700 ","End":"03:28.240","Text":"it\u0027s trying to move the system in a clockwise direction,"},{"Start":"03:28.240 ","End":"03:32.155","Text":"which is the opposite direction to our positive."},{"Start":"03:32.155 ","End":"03:33.850","Text":"It\u0027s going to be a negative."},{"Start":"03:33.850 ","End":"03:41.685","Text":"Then this entire expression is equal to i multiplied by Theta double dot."},{"Start":"03:41.685 ","End":"03:44.225","Text":"Now because we\u0027re dealing with small angles,"},{"Start":"03:44.225 ","End":"03:50.660","Text":"we can say that sine Theta is approximately equal to Theta."},{"Start":"03:50.660 ","End":"04:01.565","Text":"Then we can rearrange this expression and make it negative of m tilde g multiplied by b"},{"Start":"04:01.565 ","End":"04:09.960","Text":"divided by Theta equals"},{"Start":"04:09.960 ","End":"04:12.630","Text":"to Theta double dot."},{"Start":"04:12.630 ","End":"04:19.765","Text":"Then here, what I have in between the brackets is the frequency squared."},{"Start":"04:19.765 ","End":"04:23.590","Text":"Remember here, when we\u0027re speaking about Omega, its frequency."},{"Start":"04:23.590 ","End":"04:25.510","Text":"This is frequency squared."},{"Start":"04:25.510 ","End":"04:27.815","Text":"Now in order to find out what this is,"},{"Start":"04:27.815 ","End":"04:29.975","Text":"I have to find what B is,"},{"Start":"04:29.975 ","End":"04:32.770","Text":"which is this length."},{"Start":"04:32.770 ","End":"04:35.660","Text":"I have to find out what I is,"},{"Start":"04:35.660 ","End":"04:37.850","Text":"which will be about this point."},{"Start":"04:37.850 ","End":"04:42.380","Text":"Now, the i we\u0027ve worked out in a question called half a desk and 2"},{"Start":"04:42.380 ","End":"04:49.160","Text":"masses under the subject name of moments of inertia."},{"Start":"04:49.160 ","End":"04:51.020","Text":"Now as for the center of mass,"},{"Start":"04:51.020 ","End":"04:55.670","Text":"it\u0027s actually this length if we\u0027re going to be accurate."},{"Start":"04:55.670 ","End":"04:59.105","Text":"But then because we know that this entire length is i,"},{"Start":"04:59.105 ","End":"05:00.860","Text":"then we can find the center of mass,"},{"Start":"05:00.860 ","End":"05:03.275","Text":"do i minus center of mass,"},{"Start":"05:03.275 ","End":"05:05.480","Text":"and then we\u0027ll get this length b."},{"Start":"05:05.480 ","End":"05:08.855","Text":"Now, we worked out how to find this length,"},{"Start":"05:08.855 ","End":"05:12.575","Text":"the center of mass in the subject of the center of mass."},{"Start":"05:12.575 ","End":"05:14.205","Text":"I, we\u0027ve already found."},{"Start":"05:14.205 ","End":"05:20.705","Text":"Then very easily we can find what the frequency squared is equal to."},{"Start":"05:20.705 ","End":"05:24.470","Text":"The trick with compound or physical pendulums is"},{"Start":"05:24.470 ","End":"05:28.100","Text":"to always find the center of mass to find what i is."},{"Start":"05:28.100 ","End":"05:32.665","Text":"Then it\u0027s very quick and easy to find out what the frequency of the system is."},{"Start":"05:32.665 ","End":"05:35.590","Text":"Here we finish the question."}],"ID":10636},{"Watched":false,"Name":"Pendulum On Moving Trolley","Duration":"34m 1s","ChapterTopicVideoID":10287,"CourseChapterTopicPlaylistID":5369,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.495","Text":"Hello, let\u0027s take a look at this question."},{"Start":"00:03.495 ","End":"00:06.390","Text":"Here I have some trolley with wheels,"},{"Start":"00:06.390 ","End":"00:10.905","Text":"and it\u0027s moving with some velocity and its mass is m_2."},{"Start":"00:10.905 ","End":"00:14.310","Text":"Attached to it is a very, very tall rod,"},{"Start":"00:14.310 ","End":"00:17.849","Text":"and it can\u0027t move and attached to the rod is"},{"Start":"00:17.849 ","End":"00:23.759","Text":"this pendulum over here with a point mass m_1 attached to its end."},{"Start":"00:23.759 ","End":"00:26.699","Text":"The length of the pendulum is a,"},{"Start":"00:26.699 ","End":"00:31.710","Text":"and the pendulum is released from some angle Theta 0."},{"Start":"00:31.710 ","End":"00:36.210","Text":"The pendulum is released from rest at this starting angle of Theta 0,"},{"Start":"00:36.210 ","End":"00:40.140","Text":"and it\u0027s led to move harmonically."},{"Start":"00:40.140 ","End":"00:44.020","Text":"We\u0027re being told that our m_1,"},{"Start":"00:44.020 ","End":"00:45.859","Text":"the mass of our pendulum,"},{"Start":"00:45.859 ","End":"00:50.575","Text":"is much smaller than our m_2, the mass of the trolley."},{"Start":"00:50.575 ","End":"00:52.955","Text":"In our first question, we\u0027re being asked,"},{"Start":"00:52.955 ","End":"00:58.865","Text":"what is the velocity of the pendulum as a function of Theta and of Theta dot?"},{"Start":"00:58.865 ","End":"01:00.605","Text":"I forgot to add this section."},{"Start":"01:00.605 ","End":"01:02.735","Text":"This looks a little bit complicated,"},{"Start":"01:02.735 ","End":"01:07.790","Text":"but once we realized that our Theta dot is in fact just I Omega,"},{"Start":"01:07.790 ","End":"01:12.349","Text":"if I imagine that my Omega and my Theta are given to me in the question,"},{"Start":"01:12.349 ","End":"01:18.755","Text":"because I need to get to some expression with my Omega and my Theta."},{"Start":"01:18.755 ","End":"01:20.869","Text":"If I take it that they\u0027re given to me,"},{"Start":"01:20.869 ","End":"01:24.529","Text":"then suddenly the question becomes a lot simpler."},{"Start":"01:24.529 ","End":"01:29.910","Text":"Now we need to answer this question relative to the trolley."},{"Start":"01:32.120 ","End":"01:35.520","Text":"This is very, very important."},{"Start":"01:35.520 ","End":"01:39.350","Text":"Now what we can do is we can imagine that we\u0027re standing on the trolley,"},{"Start":"01:39.350 ","End":"01:43.430","Text":"we\u0027re on a system that\u0027s relative to the trolley and what we can"},{"Start":"01:43.430 ","End":"01:48.250","Text":"do is we can see a pendulum moving back and forth."},{"Start":"01:48.250 ","End":"01:52.610","Text":"I\u0027m standing on the trolley and because I\u0027m on the trolley,"},{"Start":"01:52.610 ","End":"01:55.429","Text":"I can ignore the motion of the trolley."},{"Start":"01:55.429 ","End":"02:00.319","Text":"I\u0027m just standing here and looking at my pendulum moving,"},{"Start":"02:00.319 ","End":"02:03.100","Text":"and now I have to find its velocity."},{"Start":"02:03.100 ","End":"02:06.410","Text":"Of course this pendulum is moving in circular motion,"},{"Start":"02:06.410 ","End":"02:12.289","Text":"because its radius or its distance from the axis of rotation is constant."},{"Start":"02:12.289 ","End":"02:15.034","Text":"I can write the relationship between"},{"Start":"02:15.034 ","End":"02:19.639","Text":"my velocity and my circular motion or my angular velocity."},{"Start":"02:19.639 ","End":"02:26.914","Text":"I can say that my V is equal to Omega multiplied by the radius of the circle,"},{"Start":"02:26.914 ","End":"02:29.795","Text":"which here the radius of the circle is a."},{"Start":"02:29.795 ","End":"02:33.650","Text":"Wrote over here that our a over here is this a which is the radius of"},{"Start":"02:33.650 ","End":"02:38.060","Text":"the circle not to be confused with acceleration,"},{"Start":"02:38.060 ","End":"02:40.495","Text":"so don\u0027t get confused in this."},{"Start":"02:40.495 ","End":"02:44.210","Text":"Then we know that our Omega is equal to our Theta dot."},{"Start":"02:44.210 ","End":"02:47.330","Text":"We can just write this in with our Theta dot multiplied by a,"},{"Start":"02:47.330 ","End":"02:50.425","Text":"again, the radius of the circle."},{"Start":"02:50.425 ","End":"02:52.924","Text":"In this question it isn\u0027t very clear,"},{"Start":"02:52.924 ","End":"02:56.725","Text":"but they\u0027re asking us to find our velocity as a vector."},{"Start":"02:56.725 ","End":"03:00.740","Text":"We need a vector quantity in this question wasn\u0027t very clear,"},{"Start":"03:00.740 ","End":"03:02.180","Text":"but when we get to question number 2,"},{"Start":"03:02.180 ","End":"03:05.950","Text":"we\u0027ll see that it makes sense that they want a vector."},{"Start":"03:05.950 ","End":"03:08.610","Text":"Let\u0027s see, the direction,"},{"Start":"03:08.610 ","End":"03:14.195","Text":"our velocity is always moving tangential to a circular motion,"},{"Start":"03:14.195 ","End":"03:17.705","Text":"so it\u0027s going to be in this direction."},{"Start":"03:17.705 ","End":"03:19.915","Text":"This is our V over here."},{"Start":"03:19.915 ","End":"03:26.554","Text":"Now what I want to do is I want to break up my velocity into its separate components,"},{"Start":"03:26.554 ","End":"03:28.775","Text":"to its x and its y component."},{"Start":"03:28.775 ","End":"03:30.980","Text":"Let\u0027s draw our axes."},{"Start":"03:30.980 ","End":"03:32.420","Text":"Let\u0027s say that this is"},{"Start":"03:32.420 ","End":"03:37.025","Text":"our positive y direction and that this is our positive x direction."},{"Start":"03:37.025 ","End":"03:41.825","Text":"Now I want to break up my velocity into its x and y components."},{"Start":"03:41.825 ","End":"03:46.630","Text":"Let\u0027s draw this diagram a little bit bigger over here so that it\u0027s a bit more clear."},{"Start":"03:46.630 ","End":"03:49.759","Text":"Here we\u0027ll have our axes in our blue,"},{"Start":"03:49.759 ","End":"03:52.220","Text":"so we have our x and our y,"},{"Start":"03:52.220 ","End":"03:55.129","Text":"this being our x-axis and this being our y."},{"Start":"03:55.129 ","End":"04:00.930","Text":"Then we have our V if we cut V it\u0027s going something like this,"},{"Start":"04:00.930 ","End":"04:05.825","Text":"so this is our V. Then we have perpendicular to our velocity,"},{"Start":"04:05.825 ","End":"04:14.370","Text":"we have our string over here and this angle going down here to 90 degrees."},{"Start":"04:14.370 ","End":"04:17.355","Text":"This over here is our Theta."},{"Start":"04:17.355 ","End":"04:22.475","Text":"Right now I\u0027m just going to rub out my 0 because we started off at an angle of Theta 0."},{"Start":"04:22.475 ","End":"04:27.500","Text":"However, now we\u0027re trying to find the velocity at any given time at any angle."},{"Start":"04:27.500 ","End":"04:29.345","Text":"A more general expression."},{"Start":"04:29.345 ","End":"04:32.075","Text":"It\u0027s just Theta, general angle Theta."},{"Start":"04:32.075 ","End":"04:38.955","Text":"As we\u0027ve said, our string and I\u0027m velocity are at 90 degrees to each other."},{"Start":"04:38.955 ","End":"04:42.980","Text":"If this is Theta and this is 90,"},{"Start":"04:42.980 ","End":"04:48.410","Text":"so this angle over here is going to be 90 minus Theta,"},{"Start":"04:48.410 ","End":"04:54.340","Text":"which means that this angle over here is again going to be Theta."},{"Start":"04:54.340 ","End":"04:56.915","Text":"In that case, if this is our Theta,"},{"Start":"04:56.915 ","End":"05:04.354","Text":"then our x component of our velocity our V_x is going to be equal to a V"},{"Start":"05:04.354 ","End":"05:07.430","Text":"multiplied by cosine of Theta and"},{"Start":"05:07.430 ","End":"05:12.680","Text":"y component V_y is going to be equal to v sine of Theta."},{"Start":"05:12.680 ","End":"05:18.065","Text":"Now we can write this in by substituting in what our V is equal to."},{"Start":"05:18.065 ","End":"05:22.370","Text":"We saw that our V is equal to Theta dot multiply by a,"},{"Start":"05:22.370 ","End":"05:26.090","Text":"a again being the radius of the circle not acceleration."},{"Start":"05:26.090 ","End":"05:32.079","Text":"It\u0027s just going to be Theta dot a multiplied by cosine of"},{"Start":"05:32.079 ","End":"05:38.755","Text":"Theta and this will be Theta dot a multiplied by sine of Theta."},{"Start":"05:38.755 ","End":"05:41.990","Text":"This is our answer for question number 1."},{"Start":"05:41.990 ","End":"05:47.160","Text":"We have our velocity in terms of r Theta dot and r Theta."},{"Start":"05:47.570 ","End":"05:50.024","Text":"We have it in vector form."},{"Start":"05:50.024 ","End":"05:52.535","Text":"We have our x component and our y components."},{"Start":"05:52.535 ","End":"05:55.150","Text":"Let\u0027s go on to the second question."},{"Start":"05:55.150 ","End":"06:00.920","Text":"Question number 2 is asking us what is the velocity of the trolley and the pendulum"},{"Start":"06:00.920 ","End":"06:07.624","Text":"relative to the lab as a function of Theta and Theta dot?"},{"Start":"06:07.624 ","End":"06:09.364","Text":"In the previous question,"},{"Start":"06:09.364 ","End":"06:13.385","Text":"I found my velocity of the pendulum relative to the trolley."},{"Start":"06:13.385 ","End":"06:15.410","Text":"Relative to this person over here."},{"Start":"06:15.410 ","End":"06:18.035","Text":"It\u0027s relative to a moving system."},{"Start":"06:18.035 ","End":"06:21.739","Text":"Whenever we have something relative to a moving system,"},{"Start":"06:21.739 ","End":"06:24.259","Text":"I\u0027m going to label the velocities with"},{"Start":"06:24.259 ","End":"06:27.830","Text":"a tag to show that they\u0027re relative to a moving system."},{"Start":"06:27.830 ","End":"06:34.944","Text":"Now I have to find the velocity of my pendulum out of my trolley relative to the lab."},{"Start":"06:34.944 ","End":"06:39.854","Text":"I\u0027m going to start by labeling some velocity for my trolley."},{"Start":"06:39.854 ","End":"06:43.505","Text":"I\u0027m going to call it V_2 because its mass is m_2,"},{"Start":"06:43.505 ","End":"06:48.049","Text":"and of course it only has an x component no y component,"},{"Start":"06:48.049 ","End":"06:50.405","Text":"the trolley isn\u0027t moving up and down."},{"Start":"06:50.405 ","End":"06:53.780","Text":"Now I want to find the relationship between the velocities."},{"Start":"06:53.780 ","End":"06:58.100","Text":"I know that through the Galilee transform that"},{"Start":"06:58.100 ","End":"07:04.065","Text":"the velocity of my mass over here, my V_1."},{"Start":"07:04.065 ","End":"07:08.659","Text":"Let\u0027s just write that over here relative to this person over here,"},{"Start":"07:08.659 ","End":"07:11.420","Text":"which is moving on the trolley at a velocity of V_2,"},{"Start":"07:11.420 ","End":"07:12.860","Text":"so let\u0027s write that."},{"Start":"07:12.860 ","End":"07:17.960","Text":"V_1 is going to be equal to my V_1,"},{"Start":"07:17.960 ","End":"07:23.370","Text":"which is relative to my trolley."},{"Start":"07:23.370 ","End":"07:26.940","Text":"That\u0027s going to be V_x1."},{"Start":"07:26.940 ","End":"07:31.375","Text":"Also here, I\u0027ll write, V_x1,"},{"Start":"07:31.375 ","End":"07:36.215","Text":"because we\u0027re only going in the x direction right now, because it\u0027s vector."},{"Start":"07:36.215 ","End":"07:40.760","Text":"The x component, so my V_x1 relative to the lab is equal to"},{"Start":"07:40.760 ","End":"07:46.929","Text":"my V_x1 relative to the trolley plus my velocity of the trolley."},{"Start":"07:46.929 ","End":"07:50.960","Text":"I\u0027ve written out labels over here so that you don\u0027t get confused."},{"Start":"07:50.960 ","End":"07:55.264","Text":"If you don\u0027t remember how to do this or this equation,"},{"Start":"07:55.264 ","End":"07:58.400","Text":"go back to the lessons on relativity,"},{"Start":"07:58.400 ","End":"08:00.605","Text":"relative velocity between bodies,"},{"Start":"08:00.605 ","End":"08:02.900","Text":"and there\u0027s a much better explanation."},{"Start":"08:02.900 ","End":"08:07.760","Text":"Of course, all of this is using the x components in the x direction."},{"Start":"08:07.760 ","End":"08:10.010","Text":"So that\u0027s why we have xs over here."},{"Start":"08:10.010 ","End":"08:16.205","Text":"Now we have the relationship linking the velocities to be relative to the lab."},{"Start":"08:16.205 ","End":"08:20.720","Text":"My V_x1 tag is this over here."},{"Start":"08:20.720 ","End":"08:23.000","Text":"I can just substitute that in over here."},{"Start":"08:23.000 ","End":"08:25.910","Text":"Now what is left to find is my V_2."},{"Start":"08:25.910 ","End":"08:29.705","Text":"I\u0027ve written an x over here because my V_2 is going in the x direction."},{"Start":"08:29.705 ","End":"08:32.095","Text":"It has no y component."},{"Start":"08:32.095 ","End":"08:35.945","Text":"How am I going to find my V_x2 over here?"},{"Start":"08:35.945 ","End":"08:41.630","Text":"I\u0027m going to use another physical traits in order to find it."},{"Start":"08:41.630 ","End":"08:44.209","Text":"What trick am I going to use?"},{"Start":"08:44.209 ","End":"08:46.325","Text":"What trick and we\u0027re going to use to find it?"},{"Start":"08:46.325 ","End":"08:50.045","Text":"Take a moment to think if you can do it on your own."},{"Start":"08:50.045 ","End":"08:55.390","Text":"I\u0027m going to be using the idea of conservation of momentum."},{"Start":"08:55.390 ","End":"08:58.735","Text":"How come I can use this idea of conservation of momentum?"},{"Start":"08:58.735 ","End":"09:00.999","Text":"Let\u0027s go back to our original diagram."},{"Start":"09:00.999 ","End":"09:05.980","Text":"I can see that whenever those wheels here generally represents that there\u0027s no friction."},{"Start":"09:05.980 ","End":"09:07.945","Text":"If there\u0027s no friction,"},{"Start":"09:07.945 ","End":"09:11.619","Text":"looking at this diagram, there\u0027s no external forces."},{"Start":"09:11.619 ","End":"09:15.190","Text":"Everything is acting internally within our system."},{"Start":"09:15.190 ","End":"09:17.785","Text":"A relationship between 2 bodies."},{"Start":"09:17.785 ","End":"09:22.240","Text":"Whenever you see some system where there\u0027s a relationship between 2 bodies or"},{"Start":"09:22.240 ","End":"09:27.204","Text":"a question where those relationship between 2 bodies and no external forces acting,"},{"Start":"09:27.204 ","End":"09:31.620","Text":"then that means that momentum must be conserved."},{"Start":"09:31.620 ","End":"09:35.585","Text":"Then you can use that to solve a great deal of questions."},{"Start":"09:35.585 ","End":"09:38.785","Text":"Now, of course, in this question, sorry,"},{"Start":"09:38.785 ","End":"09:43.090","Text":"because our movement is only in the x-axis and that\u0027s what interests us."},{"Start":"09:43.090 ","End":"09:46.450","Text":"Because we have no external forces acting on the x-axis."},{"Start":"09:46.450 ","End":"09:48.549","Text":"No friction. We can use"},{"Start":"09:48.549 ","End":"09:52.449","Text":"conservation of momentum here because we\u0027re working on the x-axis."},{"Start":"09:52.449 ","End":"09:57.220","Text":"We do have mg in the y-axis pointing downwards."},{"Start":"09:57.220 ","End":"10:00.055","Text":"In the y-axis we don\u0027t have conservation of momentum."},{"Start":"10:00.055 ","End":"10:01.509","Text":"But anyway, right now,"},{"Start":"10:01.509 ","End":"10:03.520","Text":"that\u0027s not interesting for us."},{"Start":"10:03.520 ","End":"10:07.900","Text":"In the x-axis we have conservation of momentum and so we can use that."},{"Start":"10:07.900 ","End":"10:12.549","Text":"Let\u0027s begin so we have our momentum in the x-axis."},{"Start":"10:12.549 ","End":"10:14.530","Text":"Again, no external force is acting,"},{"Start":"10:14.530 ","End":"10:16.225","Text":"so there\u0027s conservation of momentum."},{"Start":"10:16.225 ","End":"10:17.994","Text":"The momentum at the beginning,"},{"Start":"10:17.994 ","End":"10:21.474","Text":"we\u0027re told that our pendulum is released from rest,"},{"Start":"10:21.474 ","End":"10:29.440","Text":"so it has no momentum and we were also told that our trolley also started from rest."},{"Start":"10:29.440 ","End":"10:32.920","Text":"Our momentum right at the beginning is going to be equal to 0."},{"Start":"10:32.920 ","End":"10:37.240","Text":"That for conservation of momentum has to be equal to our momentum at the end."},{"Start":"10:37.240 ","End":"10:39.700","Text":"What is our momentum at the end going to be?"},{"Start":"10:39.700 ","End":"10:46.345","Text":"It\u0027s going to be our m_1 multiplied by V_1 in the x-direction, of course."},{"Start":"10:46.345 ","End":"10:50.245","Text":"Multiplied by this plus rm_2,"},{"Start":"10:50.245 ","End":"10:51.654","Text":"the momentum of this trolley,"},{"Start":"10:51.654 ","End":"10:58.839","Text":"multiplied by rv_2x, this."},{"Start":"10:58.839 ","End":"11:04.015","Text":"Now take a note. I said over here that my m_1 is significantly smaller than my m_2."},{"Start":"11:04.015 ","End":"11:06.880","Text":"However, here specifically, I cannot cross this out"},{"Start":"11:06.880 ","End":"11:10.360","Text":"yet because potentially my V_1x is going to be"},{"Start":"11:10.360 ","End":"11:14.080","Text":"significantly smaller than my V_2x in"},{"Start":"11:14.080 ","End":"11:20.034","Text":"which case my m_1 or this expression over here is an arbitrary,"},{"Start":"11:20.034 ","End":"11:21.535","Text":"and then I can cross it out."},{"Start":"11:21.535 ","End":"11:24.939","Text":"It\u0027s not so small that it\u0027s insignificant."},{"Start":"11:24.939 ","End":"11:30.279","Text":"The only time that I can cancel out my m_1 is if I have a situation where I have in"},{"Start":"11:30.279 ","End":"11:36.295","Text":"brackets m_1 plus m_2 multiplied by whatever."},{"Start":"11:36.295 ","End":"11:40.825","Text":"Only then if my m_1 is significantly smaller than my m_2,"},{"Start":"11:40.825 ","End":"11:42.610","Text":"then I can cross my m_1 out."},{"Start":"11:42.610 ","End":"11:45.730","Text":"But here because it\u0027s being multiplied by different things and"},{"Start":"11:45.730 ","End":"11:48.790","Text":"they don\u0027t know the sizes of my different things relative to each other."},{"Start":"11:48.790 ","End":"11:50.995","Text":"I cannot cross out this term."},{"Start":"11:50.995 ","End":"11:53.440","Text":"Now I have another equation."},{"Start":"11:53.440 ","End":"11:55.660","Text":"What I\u0027m left with is 2 equations."},{"Start":"11:55.660 ","End":"12:01.585","Text":"My equation for my V_x1 and my equation for my momentum or so in the x-direction."},{"Start":"12:01.585 ","End":"12:04.314","Text":"I have 2 equations and 2 unknowns."},{"Start":"12:04.314 ","End":"12:08.455","Text":"I don\u0027t know what this is and I don\u0027t know what this is."},{"Start":"12:08.455 ","End":"12:12.385","Text":"Now let\u0027s quickly do the algebra."},{"Start":"12:12.385 ","End":"12:14.365","Text":"From my equation for momentum,"},{"Start":"12:14.365 ","End":"12:18.745","Text":"I can isolate out my V_2 in the x-direction."},{"Start":"12:18.745 ","End":"12:23.590","Text":"Let\u0027s just say that it\u0027s V_2 without the x to make it a bit easier."},{"Start":"12:23.590 ","End":"12:28.464","Text":"That is going to be equal to, let\u0027s see,"},{"Start":"12:28.464 ","End":"12:35.740","Text":"negative m_1 divided by m_2 multiplied by my V_1."},{"Start":"12:35.740 ","End":"12:37.690","Text":"I\u0027m also not going to write my V_1 in"},{"Start":"12:37.690 ","End":"12:42.415","Text":"the x-direction because we know that we\u0027re working only on the x axis."},{"Start":"12:42.415 ","End":"12:45.490","Text":"Now going back to our first equation."},{"Start":"12:45.490 ","End":"12:52.150","Text":"I can say that my V_1 is going to be equal to my V_1 tag."},{"Start":"12:52.150 ","End":"12:54.715","Text":"That\u0027s going to be my Theta dot,"},{"Start":"12:54.715 ","End":"12:58.494","Text":"a cosine of Theta over here,"},{"Start":"12:58.494 ","End":"13:04.525","Text":"my V_1 tag plus my V_2 plus my V_2."},{"Start":"13:04.525 ","End":"13:13.120","Text":"That\u0027s going to be negative m_1 divided by m_2 multiplied by my V_1."},{"Start":"13:13.120 ","End":"13:17.860","Text":"Now I can see that I can move this term over to the other side."},{"Start":"13:17.860 ","End":"13:20.110","Text":"Then I\u0027m going to have my V_1,"},{"Start":"13:20.110 ","End":"13:25.705","Text":"which is going to be 1 plus m_1 divided by m_2."},{"Start":"13:25.705 ","End":"13:32.085","Text":"That\u0027s going to be equal to Theta.A cosine of Theta."},{"Start":"13:32.085 ","End":"13:35.640","Text":"Now all I have to do is I have to isolate out my V_1."},{"Start":"13:35.640 ","End":"13:37.064","Text":"That\u0027s the velocity."},{"Start":"13:37.064 ","End":"13:38.294","Text":"Let\u0027s go back 1 second."},{"Start":"13:38.294 ","End":"13:43.565","Text":"That\u0027s the velocity of my pendulum over here relative to the lab."},{"Start":"13:43.565 ","End":"13:49.026","Text":"My V_1 is simply going to be equal to my Theta."},{"Start":"13:49.026 ","End":"13:57.970","Text":"A cosine of Theta divided by 1 plus m_1 divided by m_2."},{"Start":"13:57.970 ","End":"14:00.760","Text":"Let\u0027s go back. Our question was,"},{"Start":"14:00.760 ","End":"14:02.560","Text":"what is the velocity of the trolley and of"},{"Start":"14:02.560 ","End":"14:06.460","Text":"the pendulum relative to the lab as a function of Theta and Theta dot."},{"Start":"14:06.460 ","End":"14:11.290","Text":"We found the velocity of our trolley is V_2."},{"Start":"14:11.290 ","End":"14:13.344","Text":"If we go and look over here,"},{"Start":"14:13.344 ","End":"14:17.200","Text":"we have our value for V_2 and we just have to substitute in our V_1,"},{"Start":"14:17.200 ","End":"14:19.180","Text":"which is the velocity of our pendulum,"},{"Start":"14:19.180 ","End":"14:20.875","Text":"which we\u0027ve also found."},{"Start":"14:20.875 ","End":"14:22.539","Text":"This is the velocity of a trolley."},{"Start":"14:22.539 ","End":"14:27.655","Text":"This is the velocity of a pendulum relative to the lab and in the x directions."},{"Start":"14:27.655 ","End":"14:31.480","Text":"Of course, we also have to have the velocity of"},{"Start":"14:31.480 ","End":"14:36.340","Text":"the trolley and of the pendulum relative to the lab in the y direction."},{"Start":"14:36.340 ","End":"14:39.249","Text":"The velocity of the trolley in"},{"Start":"14:39.249 ","End":"14:43.300","Text":"the y-direction is going to be equal to 0 because it\u0027s not moving up and down."},{"Start":"14:43.300 ","End":"14:48.355","Text":"That means that the velocity of the pendulum in the y-direction,"},{"Start":"14:48.355 ","End":"14:51.280","Text":"so it\u0027s simply going to be equal to this."},{"Start":"14:51.280 ","End":"14:52.659","Text":"It\u0027s going to be equal to"},{"Start":"14:52.659 ","End":"14:59.215","Text":"our V_1 the velocity of the pendulum and the y-direction tag relative to the trolley."},{"Start":"14:59.215 ","End":"15:01.719","Text":"In the third question we\u0027re being asked,"},{"Start":"15:01.719 ","End":"15:06.115","Text":"what is the equation for the conservation of mechanical energy of the system?"},{"Start":"15:06.115 ","End":"15:08.349","Text":"Notice in the first 2 questions,"},{"Start":"15:08.349 ","End":"15:11.245","Text":"we were asked to find the velocity of"},{"Start":"15:11.245 ","End":"15:14.680","Text":"both the pendulum and the trolley relative to the lab."},{"Start":"15:14.680 ","End":"15:16.540","Text":"It was quite a lot of work."},{"Start":"15:16.540 ","End":"15:22.210","Text":"Now all that\u0027s left for me to do is to find the equation for my potential energy."},{"Start":"15:22.210 ","End":"15:27.594","Text":"Let\u0027s begin by writing our equation for our initial energy,"},{"Start":"15:27.594 ","End":"15:29.020","Text":"or energy at the side,"},{"Start":"15:29.020 ","End":"15:32.275","Text":"because I pendulum and our trolleys start from rest."},{"Start":"15:32.275 ","End":"15:36.610","Text":"They\u0027ll be no kinetic energy at the beginning, only potential energy."},{"Start":"15:36.610 ","End":"15:38.755","Text":"That\u0027s going to be mgh."},{"Start":"15:38.755 ","End":"15:41.245","Text":"Let\u0027s take a look at what our h will be."},{"Start":"15:41.245 ","End":"15:48.220","Text":"It\u0027s always easiest to say that right at the tip at our point of rotation or axis."},{"Start":"15:48.220 ","End":"15:49.930","Text":"I sent her a rotation, sorry,"},{"Start":"15:49.930 ","End":"15:52.210","Text":"that our height over here is equal to 0."},{"Start":"15:52.210 ","End":"15:55.284","Text":"It just makes the calculations significantly easier."},{"Start":"15:55.284 ","End":"15:58.614","Text":"Then that means that this length here,"},{"Start":"15:58.614 ","End":"16:01.600","Text":"something like this, is going to be our height."},{"Start":"16:01.600 ","End":"16:05.170","Text":"Let\u0027s take a look because we know that the length of our pendulum"},{"Start":"16:05.170 ","End":"16:07.855","Text":"is a and this is our angle Theta."},{"Start":"16:07.855 ","End":"16:13.675","Text":"We know that our height is going to be equal to a cosine of Theta."},{"Start":"16:13.675 ","End":"16:16.705","Text":"Now notice that we have to put a negative over here."},{"Start":"16:16.705 ","End":"16:21.114","Text":"Because our height is below our line for h is equal to 0."},{"Start":"16:21.114 ","End":"16:26.980","Text":"In general, we always take our height as if it\u0027s on ground level."},{"Start":"16:26.980 ","End":"16:30.010","Text":"Here, this is our ground level, so as we go down,"},{"Start":"16:30.010 ","End":"16:32.395","Text":"it\u0027s always going to be a negative height,"},{"Start":"16:32.395 ","End":"16:33.909","Text":"as we go below."},{"Start":"16:33.909 ","End":"16:36.519","Text":"Because we always do it relative"},{"Start":"16:36.519 ","End":"16:39.730","Text":"to Earth and it doesn\u0027t matter in which direction our axis is,"},{"Start":"16:39.730 ","End":"16:42.595","Text":"if we\u0027re going down, it\u0027s always negative."},{"Start":"16:42.595 ","End":"16:45.370","Text":"Our height is negative a cosine of Theta,"},{"Start":"16:45.370 ","End":"16:50.620","Text":"so this is going to be equal to our mass times our gravity and then multiplied"},{"Start":"16:50.620 ","End":"16:56.755","Text":"by our negative a and cosine of Theta."},{"Start":"16:56.755 ","End":"17:01.254","Text":"Also, this is going to be our Theta 0 because we\u0027re starting,"},{"Start":"17:01.254 ","End":"17:04.855","Text":"we\u0027re talking about our initial energy and as we know,"},{"Start":"17:04.855 ","End":"17:06.175","Text":"as we saw over here,"},{"Start":"17:06.175 ","End":"17:10.345","Text":"we\u0027re starting right at the beginning from an angle of Theta 0,"},{"Start":"17:10.345 ","End":"17:12.115","Text":"only at the starts."},{"Start":"17:12.115 ","End":"17:16.810","Text":"Because we\u0027re doing our initial energy then we\u0027re using our beginning angle as well,"},{"Start":"17:16.810 ","End":"17:19.359","Text":"which is Theta 0, and of course,"},{"Start":"17:19.359 ","End":"17:23.305","Text":"because the trolley is at a constant height,"},{"Start":"17:23.305 ","End":"17:26.589","Text":"its potential energy is just going to"},{"Start":"17:26.589 ","End":"17:32.049","Text":"be not applicable here because its heights or its potential energy is never changing,"},{"Start":"17:32.049 ","End":"17:35.920","Text":"it\u0027s constant, because it\u0027s not moving on the y-axis."},{"Start":"17:35.920 ","End":"17:39.205","Text":"Now we\u0027re going to find our energy,"},{"Start":"17:39.205 ","End":"17:40.990","Text":"not our final energy right now,"},{"Start":"17:40.990 ","End":"17:43.675","Text":"but our energy at some Theta,"},{"Start":"17:43.675 ","End":"17:47.440","Text":"the general energy throughout the motion."},{"Start":"17:47.440 ","End":"17:51.400","Text":"First of all, we\u0027ll speak about a pendulum, our first body."},{"Start":"17:51.400 ","End":"17:53.785","Text":"Let\u0027s do its potential energy."},{"Start":"17:53.785 ","End":"17:57.490","Text":"We have its mass m_1 multiplied by g,"},{"Start":"17:57.490 ","End":"17:58.900","Text":"multiplied by its height,"},{"Start":"17:58.900 ","End":"18:02.710","Text":"which is again negative a multiplied by cosine,"},{"Start":"18:02.710 ","End":"18:04.765","Text":"but this time it\u0027s a general Theta."},{"Start":"18:04.765 ","End":"18:06.910","Text":"Here was our initial angle,"},{"Start":"18:06.910 ","End":"18:09.594","Text":"and here\u0027s a general Theta throughout its motion,"},{"Start":"18:09.594 ","End":"18:11.979","Text":"and then plus its kinetic energy."},{"Start":"18:11.979 ","End":"18:14.305","Text":"Because we\u0027re working through vectors,"},{"Start":"18:14.305 ","End":"18:16.630","Text":"let\u0027s do in the x-direction."},{"Start":"18:16.630 ","End":"18:20.440","Text":"We\u0027re going to have 1/2 multiplied its mass,"},{"Start":"18:20.440 ","End":"18:28.090","Text":"multiplied by its velocity in the x-direction squared and then in the y-direction,"},{"Start":"18:28.090 ","End":"18:32.364","Text":"so then we\u0027re going to have plus 1/2 of its mass"},{"Start":"18:32.364 ","End":"18:38.110","Text":"multiplied by its velocity in the y-direction squared."},{"Start":"18:38.110 ","End":"18:41.619","Text":"Then again for our trolley, our second body,"},{"Start":"18:41.619 ","End":"18:44.784","Text":"it has no potential energy because it\u0027s constant,"},{"Start":"18:44.784 ","End":"18:47.050","Text":"the height is never changing,"},{"Start":"18:47.050 ","End":"18:50.979","Text":"but it has kinetic energy and kinetic energy only in"},{"Start":"18:50.979 ","End":"18:55.690","Text":"the x-axis because it\u0027s only moving in the x direction,"},{"Start":"18:55.690 ","End":"18:59.575","Text":"so it\u0027s going to be 1/2 of its mass, m_2,"},{"Start":"18:59.575 ","End":"19:04.915","Text":"multiplied by its velocity in the x-direction squared."},{"Start":"19:04.915 ","End":"19:09.310","Text":"Now, let\u0027s take a look just to make this a little bit clear,"},{"Start":"19:09.310 ","End":"19:13.840","Text":"we have over here our V_1x,"},{"Start":"19:13.840 ","End":"19:17.049","Text":"which is simply this over here."},{"Start":"19:17.049 ","End":"19:18.655","Text":"This is the same V_1,"},{"Start":"19:18.655 ","End":"19:20.109","Text":"it\u0027s over here, this,"},{"Start":"19:20.109 ","End":"19:25.899","Text":"and let me just add in an x so that we know that it\u0027s more clear that it\u0027s the same one."},{"Start":"19:25.899 ","End":"19:28.855","Text":"Then let\u0027s take a look."},{"Start":"19:28.855 ","End":"19:31.435","Text":"Then we have our V_2x,"},{"Start":"19:31.435 ","End":"19:33.729","Text":"which is this,"},{"Start":"19:33.729 ","End":"19:37.000","Text":"and again, I\u0027ll add because that is an x."},{"Start":"19:37.000 ","End":"19:44.185","Text":"Then the next thing that we have is our V_1y, it\u0027s this one."},{"Start":"19:44.185 ","End":"19:47.830","Text":"This is our V_1y and we also found that,"},{"Start":"19:47.830 ","End":"19:49.990","Text":"it\u0027s our V_1y tag,"},{"Start":"19:49.990 ","End":"19:54.324","Text":"which is over here, this is our V_1y."},{"Start":"19:54.324 ","End":"19:57.445","Text":"We just have to substitute these values in over here."},{"Start":"19:57.445 ","End":"20:01.045","Text":"Now all I have to do is I have to"},{"Start":"20:01.045 ","End":"20:05.094","Text":"equate them to each other because we\u0027re using conservation of energy."},{"Start":"20:05.094 ","End":"20:06.909","Text":"I\u0027m not going to write this out again,"},{"Start":"20:06.909 ","End":"20:11.005","Text":"but we\u0027re just going to have negative mgacosine of Theta 0,"},{"Start":"20:11.005 ","End":"20:19.660","Text":"it\u0027s going to be equal to negative m_1ga cosine of Theta plus."},{"Start":"20:19.660 ","End":"20:23.080","Text":"That\u0027s the end of the answer,"},{"Start":"20:23.080 ","End":"20:24.835","Text":"we just have to equate them."},{"Start":"20:24.835 ","End":"20:32.830","Text":"Of course, remembering that our V is simply our Theta.. Let\u0027s go on to question number 4."},{"Start":"20:32.830 ","End":"20:36.340","Text":"In question number 4, we\u0027re being asked to write the equation for"},{"Start":"20:36.340 ","End":"20:40.789","Text":"energy conservation when dealing with small angles."},{"Start":"20:40.789 ","End":"20:46.015","Text":"We\u0027re being told to remember that when dealing with small angles,"},{"Start":"20:46.015 ","End":"20:47.781","Text":"we can use these equations,"},{"Start":"20:47.781 ","End":"20:55.060","Text":"that cosine^2 of Theta is around about equal to 1 minus Theta^2."},{"Start":"20:55.060 ","End":"20:57.865","Text":"That sine^2 of Theta,"},{"Start":"20:57.865 ","End":"20:59.380","Text":"again with small angles,"},{"Start":"20:59.380 ","End":"21:03.654","Text":"is around about equal to Theta^2 and that"},{"Start":"21:03.654 ","End":"21:08.890","Text":"our cosineTheta not squared is"},{"Start":"21:08.890 ","End":"21:15.145","Text":"around about equal to 1 minus Theta^2 divided by 2."},{"Start":"21:15.145 ","End":"21:19.944","Text":"I have my equation already from the previous question for my conservation of energy,"},{"Start":"21:19.944 ","End":"21:26.555","Text":"and now all I have to do is I have to substitute in these values into this equation."},{"Start":"21:26.555 ","End":"21:29.280","Text":"Long, tedious, never mind."},{"Start":"21:29.280 ","End":"21:34.370","Text":"Let\u0027s rewrite this equation. Let\u0027s see."},{"Start":"21:34.370 ","End":"21:39.595","Text":"Let\u0027s rewrite my V_2x,"},{"Start":"21:39.595 ","End":"21:44.155","Text":"because we see that we have our V_1x also over here."},{"Start":"21:44.155 ","End":"21:51.250","Text":"We can say that my energy as a function of Theta is going to be equal to,"},{"Start":"21:51.250 ","End":"21:52.945","Text":"let\u0027s take our common factors,"},{"Start":"21:52.945 ","End":"21:58.100","Text":"both have halves and m_1s."},{"Start":"21:58.380 ","End":"22:04.179","Text":"Then we can see that we have 1 plus m_1"},{"Start":"22:04.179 ","End":"22:10.790","Text":"divided by m_2 multiplied by V_1x^2,"},{"Start":"22:11.310 ","End":"22:15.175","Text":"that\u0027s these two together."},{"Start":"22:15.175 ","End":"22:18.955","Text":"How did I get this? I substituted in my V_2x^2."},{"Start":"22:18.955 ","End":"22:25.370","Text":"My V_2x^2 is negative m_1 divided by m_2 multiplied by my V_1x."},{"Start":"22:25.470 ","End":"22:28.854","Text":"I\u0027ll have a V_1x over here and I\u0027ll have"},{"Start":"22:28.854 ","End":"22:33.055","Text":"this m_2 and this division by m_2 will cancel out,"},{"Start":"22:33.055 ","End":"22:40.075","Text":"so I\u0027ll have m_1 in both and 1/2 in both and my V_1x^2 over here."},{"Start":"22:40.075 ","End":"22:45.520","Text":"Then the next term is going to be this,"},{"Start":"22:45.520 ","End":"22:54.969","Text":"plus my 1/2 of m_1V_1 in the y-axis squared and then my final term,"},{"Start":"22:54.969 ","End":"22:56.410","Text":"which is negative,"},{"Start":"22:56.410 ","End":"23:04.135","Text":"and then it\u0027s going to be m_1gacosine of Theta."},{"Start":"23:04.135 ","End":"23:06.625","Text":"Now let\u0027s write this out,"},{"Start":"23:06.625 ","End":"23:12.745","Text":"including my V_1y and my V_1x and etc."},{"Start":"23:12.745 ","End":"23:16.990","Text":"We can see that there\u0027s 1/2m_1,"},{"Start":"23:16.990 ","End":"23:19.180","Text":"which is a common factor aside from here,"},{"Start":"23:19.180 ","End":"23:21.775","Text":"we\u0027ll just multiply it by 2, so I\u0027ll take that out."},{"Start":"23:21.775 ","End":"23:25.989","Text":"I\u0027m going to have that my E as a function of Theta is going to be equal"},{"Start":"23:25.989 ","End":"23:31.030","Text":"to 1/2 of m_1 and then open up my big brackets,"},{"Start":"23:31.030 ","End":"23:40.900","Text":"and then I\u0027ll have here my 1 plus m_1 divided by m_2 multiplied by my V_1x."},{"Start":"23:40.900 ","End":"23:42.490","Text":"This is my V_1x,"},{"Start":"23:42.490 ","End":"23:45.625","Text":"so it\u0027s going to be multiplied by V_1x^2."},{"Start":"23:45.625 ","End":"23:53.680","Text":"Sorry, so Theta.acosine of Theta divided by my 1"},{"Start":"23:53.680 ","End":"24:03.595","Text":"plus m_1 divided by m_2 and then this is squared and this is squared over here as well."},{"Start":"24:03.595 ","End":"24:09.430","Text":"That\u0027s that section. Then I\u0027m going to have plus my 1/2 and my m_1 already here,"},{"Start":"24:09.430 ","End":"24:14.155","Text":"so my V_1y^2, that\u0027s over here."},{"Start":"24:14.155 ","End":"24:16.480","Text":"Let\u0027s take a look at what this is."},{"Start":"24:16.480 ","End":"24:20.559","Text":"It\u0027s going to be Theta.asine of Theta"},{"Start":"24:20.559 ","End":"24:29.439","Text":"plus Theta.a sine of Theta and of course,"},{"Start":"24:29.439 ","End":"24:33.085","Text":"it\u0027s going to be all of this squared as well because of here."},{"Start":"24:33.085 ","End":"24:39.116","Text":"Then we\u0027re going to have negative."},{"Start":"24:39.116 ","End":"24:43.038","Text":"Then my m_1 is already here,"},{"Start":"24:43.038 ","End":"24:46.855","Text":"but here there\u0027s a half and here it\u0027s 1 so I have to multiply by 2,"},{"Start":"24:46.855 ","End":"24:56.215","Text":"multiplied by g. Then I have my a cosine of Theta."},{"Start":"24:56.215 ","End":"24:58.690","Text":"That is the equation."},{"Start":"24:58.690 ","End":"25:02.739","Text":"Now, what I have to do is because I\u0027m dealing with small angles,"},{"Start":"25:02.739 ","End":"25:05.560","Text":"I have to substitute all of this in."},{"Start":"25:05.560 ","End":"25:07.780","Text":"Let\u0027s rewrite this."},{"Start":"25:07.780 ","End":"25:10.389","Text":"That means that my a is a function of theta"},{"Start":"25:10.389 ","End":"25:13.779","Text":"because I\u0027m doing approximations now to small angles,"},{"Start":"25:13.779 ","End":"25:15.550","Text":"so it\u0027s not an equal,"},{"Start":"25:15.550 ","End":"25:19.434","Text":"it\u0027s this sign saying it\u0027s approximately equal to."},{"Start":"25:19.434 ","End":"25:23.259","Text":"Then I\u0027m going to have my half multiplied by m_1,"},{"Start":"25:23.259 ","End":"25:24.955","Text":"open my big brackets."},{"Start":"25:24.955 ","End":"25:30.804","Text":"Then we can see that I have this term over here and divide it by this time squared."},{"Start":"25:30.804 ","End":"25:35.065","Text":"I can take that out and cross this out."},{"Start":"25:35.065 ","End":"25:39.669","Text":"I\u0027m going to have over here my Theta dot squared,"},{"Start":"25:39.669 ","End":"25:42.550","Text":"which remains the same multiplied by a and"},{"Start":"25:42.550 ","End":"25:48.099","Text":"my cosine squared of Theta when dealing with small angles,"},{"Start":"25:48.099 ","End":"25:54.280","Text":"is going to be equal to 1 minus Theta squared."},{"Start":"25:54.280 ","End":"26:01.524","Text":"All of this divided by 1 plus m_1 divided by m_2."},{"Start":"26:01.524 ","End":"26:03.100","Text":"That\u0027s that section."},{"Start":"26:03.100 ","End":"26:08.919","Text":"Then plus, then I\u0027m going to have my Theta dot and this is also going to be squared and"},{"Start":"26:08.919 ","End":"26:10.915","Text":"my a here is also multiplied by"},{"Start":"26:10.915 ","End":"26:15.009","Text":"a squared multiplied by my sine squared Theta in small angles,"},{"Start":"26:15.009 ","End":"26:17.185","Text":"which is approximately theta squared."},{"Start":"26:17.185 ","End":"26:25.090","Text":"Theta squared. Then I\u0027m going to have over here negative my 2ga and"},{"Start":"26:25.090 ","End":"26:29.589","Text":"then my cosine of Theta in these small angles multiply by 1"},{"Start":"26:29.589 ","End":"26:34.855","Text":"minus Theta squared divided by 2."},{"Start":"26:34.855 ","End":"26:42.610","Text":"Now again, we\u0027re dealing with small angles so some of these expressions"},{"Start":"26:42.610 ","End":"26:46.441","Text":"over here are going to be extremely small numbers"},{"Start":"26:46.441 ","End":"26:50.979","Text":"which means that we can just disregard them because they won\u0027t really change anything."},{"Start":"26:50.979 ","End":"26:54.099","Text":"Their accuracy is too high for us to even measure."},{"Start":"26:54.099 ","End":"26:57.640","Text":"Probably if this was a lab situation and"},{"Start":"26:57.640 ","End":"27:01.720","Text":"its number is so small that we can\u0027t really understand it in this context."},{"Start":"27:01.720 ","End":"27:04.795","Text":"We\u0027re going to rewrite this,"},{"Start":"27:04.795 ","End":"27:06.910","Text":"not taking that into account."},{"Start":"27:06.910 ","End":"27:09.613","Text":"If we\u0027re dealing with small angles,"},{"Start":"27:09.613 ","End":"27:14.605","Text":"that means that our Theta is going to be smaller than Pi."},{"Start":"27:14.605 ","End":"27:16.479","Text":"Then if our Theta is smaller than Pi,"},{"Start":"27:16.479 ","End":"27:21.185","Text":"then our Theta dot is also going to be smaller than Pi."},{"Start":"27:21.185 ","End":"27:23.880","Text":"Now we\u0027re going to see places where we have"},{"Start":"27:23.880 ","End":"27:27.195","Text":"our Theta dot squared multiplied by Theta squared."},{"Start":"27:27.195 ","End":"27:30.740","Text":"A small number squared is going to be really small and"},{"Start":"27:30.740 ","End":"27:33.865","Text":"a really small number multiplied by"},{"Start":"27:33.865 ","End":"27:38.320","Text":"another really small number is going to be tiny and completely insignificant."},{"Start":"27:38.320 ","End":"27:44.379","Text":"We\u0027re going to write that out by keeping only the equations or expressions where"},{"Start":"27:44.379 ","End":"27:50.485","Text":"we have Theta dot squared or our Theta squared and our squared is the highest power."},{"Start":"27:50.485 ","End":"27:54.880","Text":"Because over here we\u0027ll have a tiny number to the power of 4,"},{"Start":"27:54.880 ","End":"27:59.125","Text":"which is just absolutely ridiculous to write it."},{"Start":"27:59.125 ","End":"28:03.445","Text":"Then again, our energy is a function of Theta,"},{"Start":"28:03.445 ","End":"28:08.830","Text":"is going to be around about equal to and then again half of m_1,"},{"Start":"28:08.830 ","End":"28:10.614","Text":"and then open up our brackets."},{"Start":"28:10.614 ","End":"28:20.739","Text":"Then we\u0027re going to have our Theta dot squared multiplied"},{"Start":"28:20.739 ","End":"28:25.390","Text":"by a squared and then we\u0027re going to just multiply it by 1 and then divided"},{"Start":"28:25.390 ","End":"28:31.749","Text":"here by 1 plus our m_1 divided by m_2, or denominator."},{"Start":"28:31.749 ","End":"28:34.479","Text":"Then we\u0027re going to multiply it by this,"},{"Start":"28:34.479 ","End":"28:35.926","Text":"by our Theta squared,"},{"Start":"28:35.926 ","End":"28:38.169","Text":"but then we see that it\u0027s a small number to"},{"Start":"28:38.169 ","End":"28:41.575","Text":"the power 4 so we\u0027re just going to disregard that."},{"Start":"28:41.575 ","End":"28:46.209","Text":"Then over here, this is also going to disappear because again,"},{"Start":"28:46.209 ","End":"28:48.910","Text":"we have a small number squared multiplied by"},{"Start":"28:48.910 ","End":"28:52.420","Text":"a small number squared so again, we can disregard that."},{"Start":"28:52.420 ","End":"28:55.420","Text":"Now we\u0027re going to take a look at this."},{"Start":"28:55.420 ","End":"29:01.135","Text":"We\u0027re going to have a negative 2ga multiplied by our 1."},{"Start":"29:01.135 ","End":"29:02.785","Text":"This is a constant,"},{"Start":"29:02.785 ","End":"29:05.320","Text":"it\u0027s less interesting, but we can include it anyway."},{"Start":"29:05.320 ","End":"29:09.489","Text":"Then negative multiplied by a negative so it\u0027s going to be plus."},{"Start":"29:09.489 ","End":"29:17.155","Text":"Our 2 and our 2 will cancel out so it\u0027s going to be ga multiplied by Theta squared."},{"Start":"29:17.155 ","End":"29:21.385","Text":"This is our final answer for question Number 4."},{"Start":"29:21.385 ","End":"29:27.985","Text":"This is the equation for energy conservation when dealing with small angles."},{"Start":"29:27.985 ","End":"29:33.999","Text":"Just remember that this term over here is less important because"},{"Start":"29:33.999 ","End":"29:35.799","Text":"we\u0027re trying to find our movement as a function of"},{"Start":"29:35.799 ","End":"29:39.309","Text":"Theta and here we don\u0027t have our theta variable, it\u0027s just a constant."},{"Start":"29:39.309 ","End":"29:40.539","Text":"It\u0027s less interesting,"},{"Start":"29:40.539 ","End":"29:43.270","Text":"I could have even taken it out of the brackets."},{"Start":"29:43.270 ","End":"29:48.280","Text":"Now let\u0027s go on to question Number 5 and see why this is interesting to us."},{"Start":"29:48.280 ","End":"29:50.155","Text":"Question 5 is asking,"},{"Start":"29:50.155 ","End":"29:53.380","Text":"what is the frequency of oscillations of m_1?."},{"Start":"29:53.380 ","End":"29:56.960","Text":"I\u0027m reminding you m_1 is our pendulum."},{"Start":"29:57.270 ","End":"30:02.800","Text":"What we\u0027ve actually gotten to in our previous question is an equation for energy,"},{"Start":"30:02.800 ","End":"30:06.562","Text":"which is equal to in the most general sense,"},{"Start":"30:06.562 ","End":"30:11.290","Text":"so some constant multiplied by Theta dot squared"},{"Start":"30:11.290 ","End":"30:17.110","Text":"plus another type of constant multiplied by Theta squared."},{"Start":"30:17.110 ","End":"30:19.885","Text":"This is the most general equation."},{"Start":"30:19.885 ","End":"30:23.169","Text":"It\u0027s very similar to our equation for"},{"Start":"30:23.169 ","End":"30:26.649","Text":"our harmonic motion just when using forces which as a reminder,"},{"Start":"30:26.649 ","End":"30:28.299","Text":"is negative kx,"},{"Start":"30:28.299 ","End":"30:31.074","Text":"which is equal to mx double dot."},{"Start":"30:31.074 ","End":"30:37.045","Text":"Here this equation for energy is representing the energy in harmonic motion."},{"Start":"30:37.045 ","End":"30:41.695","Text":"This is our constant m and this is our constant k."},{"Start":"30:41.695 ","End":"30:44.845","Text":"These two equations are also the same thing"},{"Start":"30:44.845 ","End":"30:48.925","Text":"because if I differentiate this equation, I will get this."},{"Start":"30:48.925 ","End":"30:52.224","Text":"Now in our previous question,"},{"Start":"30:52.224 ","End":"30:55.270","Text":"we got this equation and we got some constant over here,"},{"Start":"30:55.270 ","End":"30:58.540","Text":"which I said, and I should have moved it out of the brackets."},{"Start":"30:58.540 ","End":"31:00.100","Text":"If it\u0027s moved out of the brackets,"},{"Start":"31:00.100 ","End":"31:01.419","Text":"it will be of the same form of,"},{"Start":"31:01.419 ","End":"31:04.749","Text":"for instance, having here plus c some constant."},{"Start":"31:04.749 ","End":"31:07.120","Text":"Now because we know our energy is constant,"},{"Start":"31:07.120 ","End":"31:11.409","Text":"the reason it works out is because I can move my constant over here and"},{"Start":"31:11.409 ","End":"31:15.909","Text":"then still what\u0027s on this side of the equation is equal to a constant and again,"},{"Start":"31:15.909 ","End":"31:16.975","Text":"when I differentiate it,"},{"Start":"31:16.975 ","End":"31:19.180","Text":"I\u0027ll get this exact equation."},{"Start":"31:19.180 ","End":"31:24.580","Text":"The same thing over here because my constants will just cancel out in differentiation."},{"Start":"31:24.580 ","End":"31:29.064","Text":"In that case, we know that our frequency of oscillation, or omega,"},{"Start":"31:29.064 ","End":"31:35.680","Text":"is simply going to be the square root of our k divided by our m. Where"},{"Start":"31:35.680 ","End":"31:39.114","Text":"the k is the coefficient of our Theta squared"},{"Start":"31:39.114 ","End":"31:43.885","Text":"and the m is the coefficient of our Theta dot squared."},{"Start":"31:43.885 ","End":"31:48.669","Text":"Whenever you\u0027re dealing with some pendulum and energy and what have you,"},{"Start":"31:48.669 ","End":"31:51.865","Text":"you always have to get into this format."},{"Start":"31:51.865 ","End":"31:57.790","Text":"Some coefficient of theta dot squared plus another coefficient of Theta squared."},{"Start":"31:57.790 ","End":"32:01.165","Text":"Then this is our frequency of oscillation."},{"Start":"32:01.165 ","End":"32:07.240","Text":"In our example, our coefficient of our Theta dot squared."},{"Start":"32:07.240 ","End":"32:09.940","Text":"Let\u0027s just highlight it."},{"Start":"32:09.940 ","End":"32:14.049","Text":"That is going to be simply over here,"},{"Start":"32:14.049 ","End":"32:17.868","Text":"this with this and also this,"},{"Start":"32:17.868 ","End":"32:19.540","Text":"not to forget that."},{"Start":"32:19.540 ","End":"32:25.420","Text":"Then our coefficient of our Theta squared so our k. Wait,"},{"Start":"32:25.420 ","End":"32:27.535","Text":"let\u0027s just write this in over here."},{"Start":"32:27.535 ","End":"32:31.209","Text":"The yellow is our k, this yellow over here."},{"Start":"32:31.209 ","End":"32:35.950","Text":"Then the coefficient of our Theta squared without the dot,"},{"Start":"32:35.950 ","End":"32:40.240","Text":"simply going to be ga and also this."},{"Start":"32:40.240 ","End":"32:47.095","Text":"This is going to be our m. This is the opposite."},{"Start":"32:47.095 ","End":"32:55.240","Text":"This is going to be our m in the yellow and our k is in the green. Let\u0027s write it."},{"Start":"32:55.240 ","End":"32:56.919","Text":"We\u0027re going to get that our omega,"},{"Start":"32:56.919 ","End":"32:58.990","Text":"which is the frequency of our oscillations,"},{"Start":"32:58.990 ","End":"33:06.009","Text":"is going to be equal to the square root of k divided by m. Our k is going to be"},{"Start":"33:06.009 ","End":"33:12.410","Text":"our m_1ga divided by"},{"Start":"33:12.410 ","End":"33:18.955","Text":"2 and then all of this divided by our m,"},{"Start":"33:18.955 ","End":"33:21.565","Text":"which is this over here."},{"Start":"33:21.565 ","End":"33:27.449","Text":"It\u0027s going to be m_1a^2 divided"},{"Start":"33:27.449 ","End":"33:33.870","Text":"by 2 multiplied by 1 plus m_1 divided by m_2."},{"Start":"33:33.870 ","End":"33:38.109","Text":"Then of course, the square root of all of that."},{"Start":"33:38.109 ","End":"33:41.679","Text":"Then by doing some simple algebra and rearranging,"},{"Start":"33:41.679 ","End":"33:45.370","Text":"you should get that this is your final answer."},{"Start":"33:45.370 ","End":"33:48.455","Text":"What is the frequency of oscillation of m_1?"},{"Start":"33:48.455 ","End":"33:51.330","Text":"This is our frequency."},{"Start":"33:51.330 ","End":"33:53.204","Text":"That\u0027s the end of the lesson."},{"Start":"33:53.204 ","End":"33:57.240","Text":"It was a bit of a long and complicated question to solve."},{"Start":"33:57.240 ","End":"33:59.609","Text":"I hope you learned a few things and if you haven\u0027t,"},{"Start":"33:59.609 ","End":"34:02.259","Text":"please go over the video again."}],"ID":10637}],"Thumbnail":null,"ID":5369},{"Name":"Exercises for Advanced","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"13m 44s","ChapterTopicVideoID":9179,"CourseChapterTopicPlaylistID":5371,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/9179.jpeg","UploadDate":"2017-03-24T11:38:25.5270000","DurationForVideoObject":"PT13M44S","Description":null,"MetaTitle":"Exercise 1 - Exercises for Advanced: Practice Makes Perfect | Proprep","MetaDescription":"Studied the topic name and want to practice? Here are some exercises on Exercises for Advanced practice questions for you to maximize your understanding.","Canonical":"https://www.proprep.uk/general-modules/all/physics-1-mechanics-waves-and-thermodynamics/harmonic-motion_oscillations/exercises-for-advanced/vid9449","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.775","Text":"Hello. In this question,"},{"Start":"00:02.775 ","End":"00:05.625","Text":"we\u0027re being told that between 2 walls,"},{"Start":"00:05.625 ","End":"00:08.175","Text":"a distance of 2L from one another,"},{"Start":"00:08.175 ","End":"00:11.610","Text":"there is a mass m. This is m,"},{"Start":"00:11.610 ","End":"00:14.760","Text":"which is attached to the walls by springs with"},{"Start":"00:14.760 ","End":"00:18.825","Text":"spring constants k and a resting length of L_0."},{"Start":"00:18.825 ","End":"00:22.305","Text":"Find the frequency on the y-axis."},{"Start":"00:22.305 ","End":"00:28.890","Text":"We\u0027re being told that this distance here is L and that"},{"Start":"00:28.890 ","End":"00:36.060","Text":"something like this distance is our resting distance,"},{"Start":"00:36.060 ","End":"00:39.470","Text":"when the spring isn\u0027t being stretched or compressed,"},{"Start":"00:39.470 ","End":"00:41.585","Text":"its length is L_0."},{"Start":"00:41.585 ","End":"00:46.445","Text":"Right now we have 2 springs of the same spring constant,"},{"Start":"00:46.445 ","End":"00:49.905","Text":"which are stretched to a length of"},{"Start":"00:49.905 ","End":"00:55.190","Text":"L. Now we\u0027re being asked to find the frequency on the y-axis."},{"Start":"00:55.190 ","End":"00:56.660","Text":"Now, on the x-axis,"},{"Start":"00:56.660 ","End":"00:59.675","Text":"it would be really easy because when you\u0027re"},{"Start":"00:59.675 ","End":"01:03.830","Text":"working out the spring constant of 2 parallel springs,"},{"Start":"01:03.830 ","End":"01:06.265","Text":"you just add the constants to one another."},{"Start":"01:06.265 ","End":"01:10.240","Text":"Our frequency would just be the square root of 1 spring"},{"Start":"01:10.240 ","End":"01:14.270","Text":"it would be k over L. But because we have 2 springs of constant k,"},{"Start":"01:14.270 ","End":"01:21.335","Text":"it would be 2k over L. This is in the x-direction,"},{"Start":"01:21.335 ","End":"01:23.945","Text":"but we\u0027re not being asked to find that."},{"Start":"01:23.945 ","End":"01:29.270","Text":"We\u0027re being asked what is the frequency in the y-direction?"},{"Start":"01:29.270 ","End":"01:31.740","Text":"This is our question."},{"Start":"01:31.840 ","End":"01:34.520","Text":"Now, just to clarify,"},{"Start":"01:34.520 ","End":"01:38.570","Text":"the x-direction would be the mass oscillating"},{"Start":"01:38.570 ","End":"01:42.590","Text":"like this but in the y-direction were asking like this."},{"Start":"01:42.590 ","End":"01:45.289","Text":"However, if I stretch my mass upwards,"},{"Start":"01:45.289 ","End":"01:50.355","Text":"then the springs are working in a diagonal type of way."},{"Start":"01:50.355 ","End":"01:56.345","Text":"I\u0027m trying to find this in the form of the sum of my forces in"},{"Start":"01:56.345 ","End":"02:03.305","Text":"the y-direction is equal to m and then my acceleration in the y-direction."},{"Start":"02:03.305 ","End":"02:07.430","Text":"Now a previous question we solved by stating that"},{"Start":"02:07.430 ","End":"02:12.230","Text":"L_0 was such a small value that we could just disregard it."},{"Start":"02:12.230 ","End":"02:13.910","Text":"However, here in this question,"},{"Start":"02:13.910 ","End":"02:17.630","Text":"it isn\u0027t such a smallest value and therefore we cannot disregard it."},{"Start":"02:17.630 ","End":"02:20.140","Text":"Let\u0027s see how we can solve this question."},{"Start":"02:20.140 ","End":"02:26.135","Text":"Let\u0027s start by imagining that I\u0027ve pulled this mass to a new position."},{"Start":"02:26.135 ","End":"02:28.460","Text":"Let\u0027s say that the new position is here."},{"Start":"02:28.460 ","End":"02:34.340","Text":"I\u0027m calling this distance that I pulled it to be y."},{"Start":"02:34.340 ","End":"02:39.380","Text":"Now what I have to do is I have to find which forces are acting in"},{"Start":"02:39.380 ","End":"02:45.350","Text":"the y-direction and also figure out which forces are acting in general."},{"Start":"02:45.350 ","End":"02:52.370","Text":"Now from symmetry, the forces in the x-direction will cancel out."},{"Start":"02:52.370 ","End":"02:54.050","Text":"Because when I pull this here,"},{"Start":"02:54.050 ","End":"02:56.390","Text":"this will be pulling in the right direction."},{"Start":"02:56.390 ","End":"02:59.045","Text":"This will be pulling the mass in the left direction,"},{"Start":"02:59.045 ","End":"03:01.940","Text":"and because they\u0027re springs of equal constants,"},{"Start":"03:01.940 ","End":"03:03.650","Text":"so they will cancel out."},{"Start":"03:03.650 ","End":"03:06.365","Text":"However, when I pull this downwards,"},{"Start":"03:06.365 ","End":"03:08.575","Text":"then it\u0027s the opposite."},{"Start":"03:08.575 ","End":"03:15.560","Text":"They\u0027ll join, their forces will join together to fling this mass up to here."},{"Start":"03:15.560 ","End":"03:17.540","Text":"Let\u0027s figure out what this force is."},{"Start":"03:17.540 ","End":"03:22.240","Text":"I\u0027m going to redraw this spring."},{"Start":"03:22.240 ","End":"03:25.000","Text":"Now, imagine that this is a straight line,"},{"Start":"03:25.000 ","End":"03:28.060","Text":"and this is the spring attaching to the mass."},{"Start":"03:28.060 ","End":"03:32.950","Text":"Now the first thing I have to do is I have to find by how much the spring"},{"Start":"03:32.950 ","End":"03:40.734","Text":"has elongated for me to get from this position to this position."},{"Start":"03:40.734 ","End":"03:47.515","Text":"Now we can work out through Pythagoras what the length of this spring is."},{"Start":"03:47.515 ","End":"03:53.080","Text":"Now first, we\u0027re going to start with when the spring is at rest,"},{"Start":"03:53.080 ","End":"03:56.920","Text":"which is here and this length is L_0."},{"Start":"03:56.920 ","End":"04:02.785","Text":"Then from this point up until here is my change in length,"},{"Start":"04:02.785 ","End":"04:05.615","Text":"which I\u0027m going to call my Delta."},{"Start":"04:05.615 ","End":"04:09.425","Text":"Now if I call this hypotenuse side,"},{"Start":"04:09.425 ","End":"04:11.225","Text":"my L_0 plus Delta,"},{"Start":"04:11.225 ","End":"04:16.175","Text":"it will make it a lot easier for me to work out the relationship between"},{"Start":"04:16.175 ","End":"04:22.260","Text":"its displacement in the y-direction to the elongation of the spring."},{"Start":"04:22.260 ","End":"04:24.980","Text":"Therefore, it will be easier for me to find out"},{"Start":"04:24.980 ","End":"04:29.810","Text":"the force and then to find out the frequency."},{"Start":"04:29.810 ","End":"04:32.675","Text":"Let\u0027s do this Pythagoras quickly."},{"Start":"04:32.675 ","End":"04:38.330","Text":"I have that L^2 plus y^2 is equal"},{"Start":"04:38.330 ","End":"04:44.460","Text":"to L_0 plus Delta^2."},{"Start":"04:44.460 ","End":"04:48.400","Text":"Now I have the relationship between y and Delta."},{"Start":"04:48.400 ","End":"04:52.550","Text":"Now what I want to do is I want to isolate out this Delta in"},{"Start":"04:52.550 ","End":"04:56.750","Text":"order to incorporate it in to my force equation."},{"Start":"04:56.750 ","End":"04:58.895","Text":"I\u0027m going to square root both sides,"},{"Start":"04:58.895 ","End":"05:03.170","Text":"and then I\u0027m going to get that Delta is equal to the square root"},{"Start":"05:03.170 ","End":"05:10.200","Text":"of L^2 plus y^2 minus L_0."},{"Start":"05:11.330 ","End":"05:14.965","Text":"Now we\u0027re working out the force of a spring."},{"Start":"05:14.965 ","End":"05:24.130","Text":"The force of a spring is equal to negative k multiplied by Delta."},{"Start":"05:24.130 ","End":"05:28.105","Text":"Now because I have 2 springs,"},{"Start":"05:28.105 ","End":"05:30.985","Text":"then I have to multiply this equation by 2,"},{"Start":"05:30.985 ","End":"05:40.060","Text":"so it\u0027ll become negative 2k multiplied by Delta."},{"Start":"05:40.110 ","End":"05:45.595","Text":"Now you\u0027ll notice that this is the force in the diagonal direction,"},{"Start":"05:45.595 ","End":"05:47.615","Text":"not in the y-direction."},{"Start":"05:47.615 ","End":"05:50.245","Text":"We\u0027ll label this as the diagonal force."},{"Start":"05:50.245 ","End":"05:54.100","Text":"But now I\u0027m trying to find the y component of this force."},{"Start":"05:54.100 ","End":"05:57.100","Text":"The force in the y-direction."},{"Start":"05:57.100 ","End":"06:03.405","Text":"This will equal to negative 2k Delta,"},{"Start":"06:03.405 ","End":"06:06.015","Text":"the force multiplied by,"},{"Start":"06:06.015 ","End":"06:09.644","Text":"I\u0027m going to call this angle Theta."},{"Start":"06:09.644 ","End":"06:12.195","Text":"Now my Theta is unknown,"},{"Start":"06:12.195 ","End":"06:13.790","Text":"in a second we\u0027re going to deal with that."},{"Start":"06:13.790 ","End":"06:21.120","Text":"But for now, I want to find this value which is opposite to my angle."},{"Start":"06:21.120 ","End":"06:25.590","Text":"I multiply this by sine of Theta,"},{"Start":"06:25.590 ","End":"06:27.210","Text":"but as I just said,"},{"Start":"06:27.210 ","End":"06:29.325","Text":"we don\u0027t know what our Theta is."},{"Start":"06:29.325 ","End":"06:31.710","Text":"But as we remember from,"},{"Start":"06:31.710 ","End":"06:36.330","Text":"SOH CAH TOA,"},{"Start":"06:36.330 ","End":"06:41.930","Text":"we know that sine is opposite over hypotenuse,"},{"Start":"06:41.930 ","End":"06:44.270","Text":"so we know what our opposite is,"},{"Start":"06:44.270 ","End":"06:49.730","Text":"it\u0027s y divided by our hypotenuse,"},{"Start":"06:49.730 ","End":"06:55.665","Text":"which we also know is L_0 plus Delta."},{"Start":"06:55.665 ","End":"07:01.010","Text":"Now we can say over here that my force in"},{"Start":"07:01.010 ","End":"07:08.205","Text":"the y-direction is equal to negative 2k multiplied by Delta."},{"Start":"07:08.205 ","End":"07:09.725","Text":"What was my Delta?"},{"Start":"07:09.725 ","End":"07:18.090","Text":"It\u0027s the square root of L^2 plus y^2 minus L_0,"},{"Start":"07:20.450 ","End":"07:29.520","Text":"multiplied by y divided by L_0 plus Delta,"},{"Start":"07:29.520 ","End":"07:37.695","Text":"which is again the square root of L^2 plus y^2 minus L_0."},{"Start":"07:37.695 ","End":"07:41.025","Text":"Then this and this can cross off."},{"Start":"07:41.025 ","End":"07:51.745","Text":"Then I can say that my F in the y-direction is equal to negative 2k multiplied"},{"Start":"07:51.745 ","End":"07:59.855","Text":"by the square root of L^2 plus y^2 minus"},{"Start":"07:59.855 ","End":"08:10.815","Text":"L_0 divided by square root of L^2 plus y^2 multiplied by y,"},{"Start":"08:10.815 ","End":"08:16.475","Text":"which is equal to m. Instead of a_y,"},{"Start":"08:16.475 ","End":"08:20.660","Text":"I\u0027m going to say that it\u0027s equal to y double dot, y double dot."},{"Start":"08:20.660 ","End":"08:25.205","Text":"Now, this is already looking more like the equation of harmonic motion."},{"Start":"08:25.205 ","End":"08:30.635","Text":"Now if I put brackets around here,"},{"Start":"08:30.635 ","End":"08:34.828","Text":"I can say that what is inside the brackets,"},{"Start":"08:34.828 ","End":"08:40.610","Text":"and then I have to divide both sides by m. Imagine I\u0027ve"},{"Start":"08:40.610 ","End":"08:47.240","Text":"just divided both sides by m to get the relationship between y double dot and y."},{"Start":"08:47.240 ","End":"08:49.340","Text":"Now when the equation is in this format,"},{"Start":"08:49.340 ","End":"08:56.285","Text":"I can say that what is inside the brackets is equal to the frequency squared."},{"Start":"08:56.285 ","End":"08:58.715","Text":"However, there\u0027s a problem,"},{"Start":"08:58.715 ","End":"09:03.815","Text":"and that is that I have ys inside this."},{"Start":"09:03.815 ","End":"09:08.360","Text":"Now my frequency can be a function of the length or"},{"Start":"09:08.360 ","End":"09:13.670","Text":"of the spring constant but it cannot be a function of my y,"},{"Start":"09:13.670 ","End":"09:15.665","Text":"it has to be independent of this."},{"Start":"09:15.665 ","End":"09:17.600","Text":"I also don\u0027t know what my y is,"},{"Start":"09:17.600 ","End":"09:19.180","Text":"so I can have it in here."},{"Start":"09:19.180 ","End":"09:21.705","Text":"How do I get out of this conundrum?"},{"Start":"09:21.705 ","End":"09:28.550","Text":"Well, I\u0027m saved because when we\u0027re speaking about these types of harmonic motions,"},{"Start":"09:28.550 ","End":"09:30.995","Text":"we\u0027re always unless stated otherwise,"},{"Start":"09:30.995 ","End":"09:34.455","Text":"going to be speaking about small angles."},{"Start":"09:34.455 ","End":"09:38.260","Text":"This idea is going to come to my rescue right now."},{"Start":"09:38.260 ","End":"09:42.320","Text":"I can write in mathematical terms that my y is"},{"Start":"09:42.320 ","End":"09:46.715","Text":"significantly smaller than my L than my L_0,"},{"Start":"09:46.715 ","End":"09:50.720","Text":"than my mass than my spring constant,"},{"Start":"09:50.720 ","End":"09:55.010","Text":"my y is significantly smaller than everything else that\u0027s given in my question."},{"Start":"09:55.010 ","End":"09:57.155","Text":"If my y is a very small value,"},{"Start":"09:57.155 ","End":"10:01.150","Text":"a very small value to the power of 2 is going to be an even smaller value."},{"Start":"10:01.150 ","End":"10:03.385","Text":"I can say that my y^2,"},{"Start":"10:03.385 ","End":"10:07.670","Text":"the value of this is so small that it becomes insignificant."},{"Start":"10:07.670 ","End":"10:10.280","Text":"I can cross it off here and cross it off here."},{"Start":"10:10.280 ","End":"10:12.920","Text":"But my y over here is still at"},{"Start":"10:12.920 ","End":"10:16.535","Text":"a size that I can say that it\u0027s significant just here, it\u0027s very small."},{"Start":"10:16.535 ","End":"10:24.135","Text":"Then I can rewrite this as my force in the y-direction is equal to negative 2k,"},{"Start":"10:24.135 ","End":"10:29.030","Text":"the square root of L^2 is just L"},{"Start":"10:29.030 ","End":"10:35.160","Text":"minus L_0 divided by m. Then again,"},{"Start":"10:35.160 ","End":"10:43.100","Text":"the square root of L^2 is L multiplied by y equals y double dot."},{"Start":"10:43.100 ","End":"10:48.570","Text":"Then this is my frequency squared."},{"Start":"10:48.570 ","End":"10:50.300","Text":"This is the end of the question."},{"Start":"10:50.300 ","End":"10:51.500","Text":"This is my onset."},{"Start":"10:51.500 ","End":"10:54.665","Text":"Now let\u0027s quickly go over what I did."},{"Start":"10:54.665 ","End":"11:00.410","Text":"I said that if I displace my mass by some distance y,"},{"Start":"11:00.410 ","End":"11:05.540","Text":"then through Pythagoras, I can find out what this horizontal force is."},{"Start":"11:05.540 ","End":"11:11.869","Text":"Now, I said that my resting length of the spring is going to be L_0,"},{"Start":"11:11.869 ","End":"11:14.435","Text":"which was given to me in the question."},{"Start":"11:14.435 ","End":"11:19.550","Text":"Then the amount that the spring was extended will be"},{"Start":"11:19.550 ","End":"11:24.305","Text":"called Delta and this will have some relationship to my force."},{"Start":"11:24.305 ","End":"11:26.045","Text":"Then through Pythagoras,"},{"Start":"11:26.045 ","End":"11:31.580","Text":"I worked out what my force is in"},{"Start":"11:31.580 ","End":"11:34.460","Text":"the diagonal direction but then because in"},{"Start":"11:34.460 ","End":"11:38.450","Text":"the question I was being asked to find the frequency on the y-axis,"},{"Start":"11:38.450 ","End":"11:45.605","Text":"I found its y component by multiplying it by sine of the angle."},{"Start":"11:45.605 ","End":"11:47.855","Text":"But because I didn\u0027t have sine of the angle,"},{"Start":"11:47.855 ","End":"11:51.350","Text":"I use my trig over here,"},{"Start":"11:51.350 ","End":"11:52.795","Text":"which was SOH CAH TOA,"},{"Start":"11:52.795 ","End":"11:59.600","Text":"which says that sine of an angle is equal to the opposite side over the hypotenuse side,"},{"Start":"11:59.600 ","End":"12:04.715","Text":"which is what I did and I worked out via Pythagoras,"},{"Start":"12:04.715 ","End":"12:06.195","Text":"what my Delta was."},{"Start":"12:06.195 ","End":"12:10.910","Text":"I isolated out my Delta then I subbed all of these values,"},{"Start":"12:10.910 ","End":"12:15.110","Text":"my Delta value, my force in the horizontal,"},{"Start":"12:15.110 ","End":"12:19.790","Text":"my force in the y component into this equation and I"},{"Start":"12:19.790 ","End":"12:25.165","Text":"said that it was equal to my double dot."},{"Start":"12:25.165 ","End":"12:29.420","Text":"Then what I did was I divided both sides by m"},{"Start":"12:29.420 ","End":"12:33.380","Text":"in order to get the relationship between y and y double dot."},{"Start":"12:33.380 ","End":"12:36.590","Text":"Then I noticed that the equation was looking very"},{"Start":"12:36.590 ","End":"12:39.860","Text":"similar to the equation for harmonic motion."},{"Start":"12:39.860 ","End":"12:46.474","Text":"Then what I did was I noticed that because I have y inside the brackets,"},{"Start":"12:46.474 ","End":"12:49.040","Text":"which means that it\u0027s part of the frequency and I know that that"},{"Start":"12:49.040 ","End":"12:51.950","Text":"isn\u0027t allowed because my y is unknown and"},{"Start":"12:51.950 ","End":"12:58.090","Text":"I know that my frequency is not dependent on the displacement of the mass."},{"Start":"12:58.090 ","End":"13:04.220","Text":"I knew because we\u0027re always working in small angles that that idea could come to save me."},{"Start":"13:04.220 ","End":"13:07.505","Text":"Because y is such a small number, when it\u0027s squared,"},{"Start":"13:07.505 ","End":"13:10.940","Text":"it becomes an even smaller and less significant number,"},{"Start":"13:10.940 ","End":"13:15.080","Text":"which allowed me to cross it out of the equation."},{"Start":"13:15.080 ","End":"13:20.830","Text":"Then I just rearranged the equation now without the y^2s."},{"Start":"13:20.830 ","End":"13:25.115","Text":"Then I got what my frequency squared was."},{"Start":"13:25.115 ","End":"13:26.900","Text":"That\u0027s the end of the question."},{"Start":"13:26.900 ","End":"13:29.660","Text":"It\u0027s a good idea to go over this again so that you"},{"Start":"13:29.660 ","End":"13:33.200","Text":"understand how to do it because it\u0027s a little bit tricky but"},{"Start":"13:33.200 ","End":"13:36.680","Text":"once you understand how to get all of your values by"},{"Start":"13:36.680 ","End":"13:40.910","Text":"using geometry and Pythagoras and very easy algebra,"},{"Start":"13:40.910 ","End":"13:44.790","Text":"it\u0027s a great question for if you get it in a test."}],"ID":9449},{"Watched":false,"Name":"Exercise 2","Duration":"19m 4s","ChapterTopicVideoID":9180,"CourseChapterTopicPlaylistID":5371,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.325","Text":"Hello. In this question,"},{"Start":"00:02.325 ","End":"00:08.040","Text":"we\u0027re given a rod of length l and of mass m and it\u0027s resting on"},{"Start":"00:08.040 ","End":"00:11.205","Text":"a ball of radius r. We\u0027re being"},{"Start":"00:11.205 ","End":"00:15.090","Text":"asked in the first question to find the frequency of the rod."},{"Start":"00:15.090 ","End":"00:22.350","Text":"What does this mean? If you imagine that I push 1 end of the rod down slightly and then"},{"Start":"00:22.350 ","End":"00:25.875","Text":"the rod would come back up and back down and"},{"Start":"00:25.875 ","End":"00:29.910","Text":"up and down and perform some harmonic motion."},{"Start":"00:29.910 ","End":"00:33.790","Text":"We\u0027re being asked to find the frequency of this harmonic motion."},{"Start":"00:33.790 ","End":"00:35.710","Text":"Let\u0027s see how we do with this."},{"Start":"00:35.710 ","End":"00:39.350","Text":"We\u0027re going to find the sum of all of the moments and make"},{"Start":"00:39.350 ","End":"00:43.210","Text":"that equal to I Theta double dot."},{"Start":"00:43.210 ","End":"00:47.225","Text":"Let\u0027s see what moments are working on the system."},{"Start":"00:47.225 ","End":"00:51.890","Text":"Here\u0027s our rod, and let\u0027s imagine that I press down on the rod slightly."},{"Start":"00:51.890 ","End":"00:53.570","Text":"Look at this diagram."},{"Start":"00:53.570 ","End":"00:55.480","Text":"This is what it looks like."},{"Start":"00:55.480 ","End":"01:03.495","Text":"I\u0027m going to call this angle Theta, my shifting angle."},{"Start":"01:03.495 ","End":"01:08.750","Text":"It\u0027s shifted by angle Theta and this is of length R"},{"Start":"01:08.750 ","End":"01:14.280","Text":"because it\u0027s the radius and this is also the same length r,"},{"Start":"01:14.280 ","End":"01:16.365","Text":"the radius of the circle."},{"Start":"01:16.365 ","End":"01:22.855","Text":"Now I know that the length of this arc is R multiplied by Theta."},{"Start":"01:22.855 ","End":"01:26.740","Text":"That\u0027s how you work out the length of an arc in a circle."},{"Start":"01:26.740 ","End":"01:29.320","Text":"If you have an angle and you know your radius,"},{"Start":"01:29.320 ","End":"01:33.175","Text":"then the length of the arc is R times Theta."},{"Start":"01:33.175 ","End":"01:35.184","Text":"Now, in every question,"},{"Start":"01:35.184 ","End":"01:37.959","Text":"unless stated that there is slipping,"},{"Start":"01:37.959 ","End":"01:42.710","Text":"you have to assume that there is no slipping in the question."},{"Start":"01:43.460 ","End":"01:49.875","Text":"Let\u0027s see. If here is my center of mass so I\u0027m going to draw it also here."},{"Start":"01:49.875 ","End":"01:51.675","Text":"This is my center of mass."},{"Start":"01:51.675 ","End":"01:54.400","Text":"What is this arc length actually mean?"},{"Start":"01:54.400 ","End":"02:01.310","Text":"It means that if my center of mass was over here when it was touching over here,"},{"Start":"02:01.310 ","End":"02:11.640","Text":"the distance R Theta that was traveled by pushing my rod by some force downwards,"},{"Start":"02:11.640 ","End":"02:14.660","Text":"it means my next point of contact with"},{"Start":"02:14.660 ","End":"02:20.420","Text":"the sphere will also be R Theta away from the original point of contact,"},{"Start":"02:20.420 ","End":"02:22.490","Text":"which was the center of mass."},{"Start":"02:22.490 ","End":"02:30.815","Text":"What I\u0027ve actually done here is I\u0027ve found where my center of mass is in relation to"},{"Start":"02:30.815 ","End":"02:34.940","Text":"my next point of contact after being applied"},{"Start":"02:34.940 ","End":"02:41.090","Text":"this force as a function of the angle that it has moved along the sphere."},{"Start":"02:41.090 ","End":"02:47.105","Text":"Now I just want to make a note that my diagram here is a little bit inaccurate"},{"Start":"02:47.105 ","End":"02:55.495","Text":"because the rod is always meant to be tangential to this sphere."},{"Start":"02:55.495 ","End":"02:57.980","Text":"What does that mean? That there\u0027s always meant to be"},{"Start":"02:57.980 ","End":"03:02.885","Text":"90 degrees between the rod and the sphere."},{"Start":"03:02.885 ","End":"03:09.320","Text":"You can see here, it\u0027s a little easier to see that there\u0027s 90 degrees over there."},{"Start":"03:09.320 ","End":"03:12.110","Text":"The reason it wasn\u0027t drawn like that is"},{"Start":"03:12.110 ","End":"03:15.710","Text":"because it would have been such a small angle here,"},{"Start":"03:15.710 ","End":"03:20.435","Text":"the movement that it would have been really difficult to see in the diagram what I meant,"},{"Start":"03:20.435 ","End":"03:25.775","Text":"but that\u0027s the point of this so the diagram is slightly inaccurate."},{"Start":"03:25.775 ","End":"03:28.430","Text":"Back to the question at hand."},{"Start":"03:28.430 ","End":"03:34.995","Text":"This new point over here is our center of rotation."},{"Start":"03:34.995 ","End":"03:38.000","Text":"Now of course, as the rod oscillates,"},{"Start":"03:38.000 ","End":"03:40.985","Text":"as it rises and drops again,"},{"Start":"03:40.985 ","End":"03:44.525","Text":"the center of rotation will move slightly to either side."},{"Start":"03:44.525 ","End":"03:46.565","Text":"For the question, it doesn\u0027t really matter."},{"Start":"03:46.565 ","End":"03:48.710","Text":"This is our center of rotation and now we have to"},{"Start":"03:48.710 ","End":"03:53.045","Text":"find the moments working around this point."},{"Start":"03:53.045 ","End":"03:57.080","Text":"Now let\u0027s work out the moments."},{"Start":"03:57.080 ","End":"04:00.980","Text":"We know that we have mg pointing downwards,"},{"Start":"04:00.980 ","End":"04:06.680","Text":"so we can already write mg and then we have to multiply it by the distance,"},{"Start":"04:06.680 ","End":"04:08.620","Text":"which here is R Theta,"},{"Start":"04:08.620 ","End":"04:14.405","Text":"multiplied by R Theta and then multiplied by sine of the angle."},{"Start":"04:14.405 ","End":"04:16.400","Text":"What is the angle?"},{"Start":"04:16.400 ","End":"04:20.269","Text":"This is going to be a little bit complicated to explain."},{"Start":"04:20.269 ","End":"04:23.450","Text":"If you don\u0027t understand it, re-watch this section."},{"Start":"04:23.450 ","End":"04:24.560","Text":"How I explain this angle,"},{"Start":"04:24.560 ","End":"04:29.840","Text":"it\u0027s just bit of some confusing geometry and you\u0027ll be fine."},{"Start":"04:29.840 ","End":"04:31.895","Text":"We know this angle is Theta."},{"Start":"04:31.895 ","End":"04:33.350","Text":"Now let\u0027s look at this diagram."},{"Start":"04:33.350 ","End":"04:37.055","Text":"This is Theta and this is 90."},{"Start":"04:37.055 ","End":"04:40.895","Text":"Now because we said that the rod is tangential to the sphere,"},{"Start":"04:40.895 ","End":"04:46.780","Text":"we can say that this angle is also 90 degrees."},{"Start":"04:46.780 ","End":"04:51.050","Text":"Now, in order to find out what this angle is,"},{"Start":"04:51.050 ","End":"04:54.485","Text":"we have to do 180,"},{"Start":"04:54.485 ","End":"04:57.155","Text":"because the sum of the angles in a triangle,"},{"Start":"04:57.155 ","End":"05:00.090","Text":"minus 90 minus Theta,"},{"Start":"05:01.310 ","End":"05:07.695","Text":"which gives an angle of 90 minus Theta."},{"Start":"05:07.695 ","End":"05:10.240","Text":"That\u0027s this angle."},{"Start":"05:10.360 ","End":"05:13.175","Text":"But now we have to find out,"},{"Start":"05:13.175 ","End":"05:16.230","Text":"imagine that you don\u0027t see the Theta here. It\u0027s a bit of a giveaway."},{"Start":"05:16.230 ","End":"05:19.695","Text":"Imagine we don\u0027t see this Theta here."},{"Start":"05:19.695 ","End":"05:22.070","Text":"We want to find out what this angle is."},{"Start":"05:22.070 ","End":"05:27.380","Text":"Then we have to do 90 because this is the angle of this section,"},{"Start":"05:27.380 ","End":"05:30.950","Text":"minus this angle over here,"},{"Start":"05:30.950 ","End":"05:32.585","Text":"which is 90 minus Theta."},{"Start":"05:32.585 ","End":"05:41.520","Text":"Then this angle will be 90 minus this angle,"},{"Start":"05:41.520 ","End":"05:44.485","Text":"90 minus Theta,"},{"Start":"05:44.485 ","End":"05:47.075","Text":"which will give an angle here of Theta."},{"Start":"05:47.075 ","End":"05:49.850","Text":"Now you understand why this angle is Theta."},{"Start":"05:49.850 ","End":"05:51.635","Text":"But I don\u0027t need this angle."},{"Start":"05:51.635 ","End":"05:57.819","Text":"I need the angle between the force and the axis of rotation."},{"Start":"05:57.819 ","End":"06:06.435","Text":"Here\u0027s my axis of rotation and my force is my mg going downwards."},{"Start":"06:06.435 ","End":"06:11.165","Text":"I need the angle between here."},{"Start":"06:11.165 ","End":"06:12.860","Text":"I need this angle."},{"Start":"06:12.860 ","End":"06:14.390","Text":"What is this angle?"},{"Start":"06:14.390 ","End":"06:16.070","Text":"Let\u0027s continue this triangle."},{"Start":"06:16.070 ","End":"06:18.145","Text":"This is also 90 degrees."},{"Start":"06:18.145 ","End":"06:21.889","Text":"If this angle is Theta and this angle is 90,"},{"Start":"06:21.889 ","End":"06:25.565","Text":"so it\u0027s exactly like what we did here. It\u0027s this angle."},{"Start":"06:25.565 ","End":"06:27.785","Text":"It\u0027s the same working out."},{"Start":"06:27.785 ","End":"06:34.690","Text":"This angle will be 180 minus 90 minus Theta,"},{"Start":"06:34.690 ","End":"06:37.070","Text":"exactly the same as what we did here,"},{"Start":"06:37.070 ","End":"06:41.850","Text":"which will equal an angle of 90 minus Theta."},{"Start":"06:41.850 ","End":"06:44.825","Text":"Because remember whenever we\u0027re working with moments,"},{"Start":"06:44.825 ","End":"06:50.240","Text":"we have to multiply the force multiplied by the distance,"},{"Start":"06:50.240 ","End":"06:55.415","Text":"the distance to the axis of rotation,"},{"Start":"06:55.415 ","End":"06:57.950","Text":"multiplied by sine of"},{"Start":"06:57.950 ","End":"07:08.090","Text":"the angle between the force and the axis of rotation."},{"Start":"07:08.090 ","End":"07:14.130","Text":"This equals I Theta double dot."},{"Start":"07:14.480 ","End":"07:18.230","Text":"Now we have to find out what the sign is over here."},{"Start":"07:18.230 ","End":"07:19.910","Text":"As we can see,"},{"Start":"07:19.910 ","End":"07:22.985","Text":"Theta is a positive number."},{"Start":"07:22.985 ","End":"07:28.355","Text":"We\u0027re moving from this point of contact to this point of contact,"},{"Start":"07:28.355 ","End":"07:31.145","Text":"which means that if we\u0027re going in a clockwise direction,"},{"Start":"07:31.145 ","End":"07:33.064","Text":"it\u0027s a positive direction."},{"Start":"07:33.064 ","End":"07:37.220","Text":"However, if we look at the rod, when it returns,"},{"Start":"07:37.220 ","End":"07:39.725","Text":"this side will be going upwards,"},{"Start":"07:39.725 ","End":"07:45.120","Text":"and then it will continue in an anticlockwise direction so it\u0027s a negative."},{"Start":"07:45.120 ","End":"07:48.470","Text":"Now what I\u0027ll have to do is to find the relationship"},{"Start":"07:48.470 ","End":"07:52.375","Text":"between this Theta and this Theta double dot."},{"Start":"07:52.375 ","End":"07:56.885","Text":"First of all, let\u0027s see what our I is equal to."},{"Start":"07:56.885 ","End":"08:00.260","Text":"We\u0027re working out the I of the rod, however,"},{"Start":"08:00.260 ","End":"08:03.965","Text":"notice that this is the center of mass however,"},{"Start":"08:03.965 ","End":"08:06.260","Text":"this is not the center of rotation,"},{"Start":"08:06.260 ","End":"08:08.830","Text":"this is not the axis of rotation."},{"Start":"08:08.830 ","End":"08:12.240","Text":"I\u0027m going to label this 0."},{"Start":"08:12.240 ","End":"08:16.790","Text":"Then what I\u0027m going to do is I\u0027m going to find out what my I_0 is,"},{"Start":"08:16.790 ","End":"08:18.035","Text":"because this point is called,"},{"Start":"08:18.035 ","End":"08:24.465","Text":"so my I_0 is going to be equal to my I of center of mass,"},{"Start":"08:24.465 ","End":"08:29.090","Text":"the I around here plus the distance which has plus Steiner."},{"Start":"08:29.090 ","End":"08:30.679","Text":"What is Steiner?"},{"Start":"08:30.679 ","End":"08:34.580","Text":"It is the mass times the distance squared."},{"Start":"08:34.580 ","End":"08:44.733","Text":"It\u0027s plus mass multiplied by the distance which is R Theta^2."},{"Start":"08:44.733 ","End":"08:47.185","Text":"When we\u0027re referring to this I,"},{"Start":"08:47.185 ","End":"08:50.000","Text":"this I_0, it\u0027s this."},{"Start":"08:50.220 ","End":"08:53.305","Text":"What is my I of center of mass?"},{"Start":"08:53.305 ","End":"08:57.400","Text":"It\u0027s my I of a rod,"},{"Start":"08:57.400 ","End":"08:59.245","Text":"which is equal to,"},{"Start":"08:59.245 ","End":"09:01.210","Text":"this you have to know off by heart,"},{"Start":"09:01.210 ","End":"09:05.905","Text":"it\u0027s equal to 1/12ml^2."},{"Start":"09:05.905 ","End":"09:08.185","Text":"l is the length of the rod."},{"Start":"09:08.185 ","End":"09:14.455","Text":"I want to get this Theta double dot to be alone, to be isolated."},{"Start":"09:14.455 ","End":"09:18.115","Text":"I\u0027m going to divide both sides by this I_0."},{"Start":"09:18.115 ","End":"09:25.990","Text":"I\u0027m going to have negative mgR Theta sin(90"},{"Start":"09:25.990 ","End":"09:30.715","Text":"minus Theta) equals Theta double dot."},{"Start":"09:30.715 ","End":"09:33.775","Text":"Then divided by"},{"Start":"09:33.775 ","End":"09:41.212","Text":"1/12ml^2 plus mR^2 Theta^2."},{"Start":"09:41.212 ","End":"09:44.275","Text":"Let\u0027s simplify this a little bit more."},{"Start":"09:44.275 ","End":"09:54.700","Text":"My sine of 90 minus Theta is actually the same as cosine of Theta."},{"Start":"09:54.700 ","End":"09:57.850","Text":"This is a trigonometric identity."},{"Start":"09:57.850 ","End":"10:01.840","Text":"That will simplify this expression quite some bit."},{"Start":"10:01.840 ","End":"10:04.270","Text":"Now what I have to do is I have to get"},{"Start":"10:04.270 ","End":"10:09.010","Text":"this Theta outside some brackets of this expression."},{"Start":"10:09.010 ","End":"10:11.080","Text":"Then whatever is in that brackets,"},{"Start":"10:11.080 ","End":"10:14.963","Text":"I know is the frequency of the rod."},{"Start":"10:14.963 ","End":"10:20.575","Text":"This section is going to be a little more complicated to explain."},{"Start":"10:20.575 ","End":"10:24.445","Text":"Because we\u0027re using such small angles, always,"},{"Start":"10:24.445 ","End":"10:29.155","Text":"whenever you are asked to find the frequency of a rod or the frequency of anything,"},{"Start":"10:29.155 ","End":"10:31.945","Text":"assume that the angles are always going to be pretty small."},{"Start":"10:31.945 ","End":"10:38.110","Text":"At small angles, cosine of Theta is equal to 1,"},{"Start":"10:38.110 ","End":"10:39.460","Text":"at small angles,"},{"Start":"10:39.460 ","End":"10:42.190","Text":"sine of Theta is equal to Theta,"},{"Start":"10:42.190 ","End":"10:45.700","Text":"and at small angles, cosine of Theta is equal to 1."},{"Start":"10:45.700 ","End":"10:47.500","Text":"That solves all of that."},{"Start":"10:47.500 ","End":"10:50.540","Text":"Then I\u0027m going to have negative mgR"},{"Start":"10:52.980 ","End":"11:00.070","Text":"multiplied by Theta multiplied by 1 divided by,"},{"Start":"11:00.070 ","End":"11:02.380","Text":"now this bit, again,"},{"Start":"11:02.380 ","End":"11:04.690","Text":"remember I said we\u0027re working with small angles,"},{"Start":"11:04.690 ","End":"11:07.840","Text":"which means that this Theta is a very small number and"},{"Start":"11:07.840 ","End":"11:11.200","Text":"a small number squared is a really tiny number."},{"Start":"11:11.200 ","End":"11:14.998","Text":"That\u0027s this. That means that, this expression,"},{"Start":"11:14.998 ","End":"11:19.390","Text":"this mR^2, is being multiplied by a very small number,"},{"Start":"11:19.390 ","End":"11:24.190","Text":"which makes this whole expression insignificant because it will"},{"Start":"11:24.190 ","End":"11:28.975","Text":"be so small in comparison to the rest of the numbers in this."},{"Start":"11:28.975 ","End":"11:36.490","Text":"Then we can just only keep this expression at the bottom, 1/12ml^2."},{"Start":"11:36.490 ","End":"11:39.520","Text":"Then I can cross out the m\u0027s."},{"Start":"11:39.520 ","End":"11:44.140","Text":"Because here I\u0027m dividing by 1/12,"},{"Start":"11:44.140 ","End":"11:46.225","Text":"I will multiply it by 12."},{"Start":"11:46.225 ","End":"11:51.205","Text":"Therefore, I get an equation saying that Theta double dot is equal to"},{"Start":"11:51.205 ","End":"12:00.910","Text":"negative 12gR divided by l^2,"},{"Start":"12:00.910 ","End":"12:05.860","Text":"all this in brackets, multiplied by Theta."},{"Start":"12:05.860 ","End":"12:07.570","Text":"Now we can say that this,"},{"Start":"12:07.570 ","End":"12:08.950","Text":"whatever is inside the brackets,"},{"Start":"12:08.950 ","End":"12:11.110","Text":"is equal to Omega^2,"},{"Start":"12:11.110 ","End":"12:13.810","Text":"which is the frequency^2."},{"Start":"12:13.810 ","End":"12:19.585","Text":"All we have to do is my frequency is the square root of this expression."},{"Start":"12:19.585 ","End":"12:22.000","Text":"If I want to be super correct,"},{"Start":"12:22.000 ","End":"12:25.870","Text":"I can make sure that this expression is always positive."},{"Start":"12:25.870 ","End":"12:28.375","Text":"I can see that 12 is a positive number,"},{"Start":"12:28.375 ","End":"12:29.665","Text":"g is always positive."},{"Start":"12:29.665 ","End":"12:31.750","Text":"My distance, R, my radius,"},{"Start":"12:31.750 ","End":"12:35.155","Text":"is always positive and the length of the rod is always positive."},{"Start":"12:35.155 ","End":"12:37.870","Text":"We are correct, that is the frequency."},{"Start":"12:37.870 ","End":"12:40.495","Text":"Now we\u0027re going to look at the next question."},{"Start":"12:40.495 ","End":"12:45.100","Text":"Find the height of the rod\u0027s center of mass as a function of its angle."},{"Start":"12:45.100 ","End":"12:46.630","Text":"Let\u0027s see how we do this."},{"Start":"12:46.630 ","End":"12:47.830","Text":"Just to be clear,"},{"Start":"12:47.830 ","End":"12:53.200","Text":"what we\u0027re being asked is what is the height of this as the function as to how"},{"Start":"12:53.200 ","End":"12:59.140","Text":"much I\u0027ve pushed down and how much I\u0027ve altered the position of the rod?"},{"Start":"12:59.140 ","End":"13:01.015","Text":"Let\u0027s write this out."},{"Start":"13:01.015 ","End":"13:06.550","Text":"My height, I\u0027m going to label it as y center of mass,"},{"Start":"13:06.550 ","End":"13:09.625","Text":"because y represents the height."},{"Start":"13:09.625 ","End":"13:11.080","Text":"That\u0027s how I\u0027ve decided to call it."},{"Start":"13:11.080 ","End":"13:13.690","Text":"It\u0027s starting height at this point,"},{"Start":"13:13.690 ","End":"13:15.355","Text":"from the center,"},{"Start":"13:15.355 ","End":"13:19.660","Text":"is R because the radius is R. It\u0027s R distance from the center,"},{"Start":"13:19.660 ","End":"13:25.200","Text":"so it\u0027s R. Then we want to add to it,"},{"Start":"13:25.200 ","End":"13:27.600","Text":"because when we push this rod down,"},{"Start":"13:27.600 ","End":"13:31.125","Text":"the center of mass goes from this point to this point,"},{"Start":"13:31.125 ","End":"13:35.760","Text":"so it\u0027s moved up this distance."},{"Start":"13:35.760 ","End":"13:37.545","Text":"What is this distance?"},{"Start":"13:37.545 ","End":"13:39.930","Text":"As we\u0027ve seen from this diagram?"},{"Start":"13:39.930 ","End":"13:43.395","Text":"This angle here is Theta."},{"Start":"13:43.395 ","End":"13:45.470","Text":"Let me label it."},{"Start":"13:45.470 ","End":"13:47.395","Text":"That angle there is Theta."},{"Start":"13:47.395 ","End":"13:50.476","Text":"In order to find what this distance, we have to do,"},{"Start":"13:50.476 ","End":"13:53.965","Text":"R Theta sine Theta."},{"Start":"13:53.965 ","End":"13:56.935","Text":"This distance multiplied by,"},{"Start":"13:56.935 ","End":"13:58.540","Text":"because we\u0027re trying to find the opposite side,"},{"Start":"13:58.540 ","End":"14:00.325","Text":"so it\u0027s sine Theta."},{"Start":"14:00.325 ","End":"14:08.020","Text":"Then we have to do plus R Theta sine of Theta."},{"Start":"14:08.020 ","End":"14:11.515","Text":"Great. But now notice a little thing."},{"Start":"14:11.515 ","End":"14:13.300","Text":"When it\u0027s at this point,"},{"Start":"14:13.300 ","End":"14:15.565","Text":"it\u0027s slightly lower than this point."},{"Start":"14:15.565 ","End":"14:19.195","Text":"If we do a line, we can see that."},{"Start":"14:19.195 ","End":"14:22.675","Text":"Now we have to find out what this little distance is."},{"Start":"14:22.675 ","End":"14:25.540","Text":"Let\u0027s go to this diagram over here."},{"Start":"14:25.540 ","End":"14:28.795","Text":"This is Theta, which is this Theta,"},{"Start":"14:28.795 ","End":"14:31.060","Text":"and this is R and this as R."},{"Start":"14:31.060 ","End":"14:38.035","Text":"This entire length is R but we want to find out what up until here,"},{"Start":"14:38.035 ","End":"14:40.585","Text":"just this section is."},{"Start":"14:40.585 ","End":"14:45.385","Text":"All we have to do is we have to do R_cosine Theta"},{"Start":"14:45.385 ","End":"14:50.005","Text":"because then we\u0027re creating a right-angled triangle."},{"Start":"14:50.005 ","End":"14:52.000","Text":"Then to find this adjacent side,"},{"Start":"14:52.000 ","End":"14:54.058","Text":"it\u0027s cosine of the angle,"},{"Start":"14:54.058 ","End":"14:58.015","Text":"Hypotenuse over adjacent, so it\u0027s R_cosine Theta."},{"Start":"14:58.015 ","End":"15:02.230","Text":"Then to find this little distance because we know that the full length is R,"},{"Start":"15:02.230 ","End":"15:05.499","Text":"so it\u0027s just R minus R_cosine Theta."},{"Start":"15:05.499 ","End":"15:08.980","Text":"Then we have to subtract it from here."},{"Start":"15:08.980 ","End":"15:11.846","Text":"We do negative and then in brackets,"},{"Start":"15:11.846 ","End":"15:16.885","Text":"R minus R_cosine of Theta."},{"Start":"15:16.885 ","End":"15:19.165","Text":"Let\u0027s quickly go over this again."},{"Start":"15:19.165 ","End":"15:26.470","Text":"What happened? My starting height was R. Then when I push the rod down,"},{"Start":"15:26.470 ","End":"15:31.856","Text":"my center of mass lifted up by a factor of R Theta sine of Theta"},{"Start":"15:31.856 ","End":"15:33.535","Text":"because we make this into"},{"Start":"15:33.535 ","End":"15:37.660","Text":"a right angle triangle and then we\u0027re trying to find this opposite."},{"Start":"15:37.660 ","End":"15:39.940","Text":"It\u0027s R Theta, the distance,"},{"Start":"15:39.940 ","End":"15:44.215","Text":"multiplied by sine of the angle but minus because"},{"Start":"15:44.215 ","End":"15:52.000","Text":"this new point from which we\u0027re measuring is slightly lower than our original point."},{"Start":"15:52.000 ","End":"15:57.010","Text":"We have to minus R minus R_cosine of Theta, again,"},{"Start":"15:57.010 ","End":"15:59.320","Text":"because this length over here is going be"},{"Start":"15:59.320 ","End":"16:03.550","Text":"cosine Theta because it\u0027s hypotenuse over adjacent."},{"Start":"16:03.550 ","End":"16:05.875","Text":"Now we can simplify this."},{"Start":"16:05.875 ","End":"16:07.570","Text":"Then I have to write."},{"Start":"16:07.570 ","End":"16:09.550","Text":"This and this is crossed out."},{"Start":"16:09.550 ","End":"16:12.055","Text":"This plus this becomes a positive."},{"Start":"16:12.055 ","End":"16:16.810","Text":"Then they both have a factor of R. Then in brackets,"},{"Start":"16:16.810 ","End":"16:22.630","Text":"cosine of Theta plus Theta sine of Theta."},{"Start":"16:22.630 ","End":"16:24.190","Text":"This is exactly like this,"},{"Start":"16:24.190 ","End":"16:26.185","Text":"just rearranged in a neater way."},{"Start":"16:26.185 ","End":"16:28.225","Text":"Here we\u0027ve basically finished,"},{"Start":"16:28.225 ","End":"16:30.115","Text":"this is our answer to Question number 2."},{"Start":"16:30.115 ","End":"16:36.015","Text":"If I want to express this in a form without my cosine Theta and my Theta sine Theta,"},{"Start":"16:36.015 ","End":"16:38.889","Text":"because it\u0027s a little bit harder to understand this,"},{"Start":"16:38.889 ","End":"16:40.450","Text":"I can write it like here,"},{"Start":"16:40.450 ","End":"16:42.430","Text":"when we wrote that in small angles,"},{"Start":"16:42.430 ","End":"16:46.840","Text":"we can estimate the cosine of Theta is equal to 1 and sine of Theta is equal to Theta."},{"Start":"16:46.840 ","End":"16:52.765","Text":"I can say instead that my y center of mass is equal to R,"},{"Start":"16:52.765 ","End":"16:56.200","Text":"that stays the same, and then cosine Theta is 1."},{"Start":"16:56.200 ","End":"17:00.145","Text":"However, because I want to be slightly more accurate."},{"Start":"17:00.145 ","End":"17:04.450","Text":"It\u0027s 1 minus Theta^2 over 2."},{"Start":"17:04.450 ","End":"17:07.690","Text":"You always write, if you want to be accurate,"},{"Start":"17:07.690 ","End":"17:08.980","Text":"1 minus Theta over 2."},{"Start":"17:08.980 ","End":"17:12.115","Text":"That\u0027s the estimate actually of cosine of Theta."},{"Start":"17:12.115 ","End":"17:17.500","Text":"However, because Theta^2 is such an insignificant number as we said over here,"},{"Start":"17:17.500 ","End":"17:22.875","Text":"it\u0027s customary and allowed to just write 1."},{"Start":"17:22.875 ","End":"17:25.110","Text":"Then sine Theta, we said,"},{"Start":"17:25.110 ","End":"17:30.820","Text":"is estimated to Theta multiplied by Theta plus Theta^2."},{"Start":"17:30.820 ","End":"17:32.240","Text":"I hear you asking,"},{"Start":"17:32.240 ","End":"17:36.290","Text":"how come here I\u0027m accepting this Theta^2 and taking this"},{"Start":"17:36.290 ","End":"17:40.400","Text":"into account to my calculations but here when I saw the Theta^2,"},{"Start":"17:40.400 ","End":"17:42.860","Text":"I said that it was such as small and"},{"Start":"17:42.860 ","End":"17:46.185","Text":"insignificant number that we can just cross out that entire expression?"},{"Start":"17:46.185 ","End":"17:48.620","Text":"Let\u0027s see why here it\u0027s important."},{"Start":"17:48.620 ","End":"17:51.380","Text":"In my question, I\u0027m being asked to find the height of"},{"Start":"17:51.380 ","End":"17:54.272","Text":"the rod center of mass as a function of its angle,"},{"Start":"17:54.272 ","End":"17:57.530","Text":"meaning that as my angle changes,"},{"Start":"17:57.530 ","End":"17:59.530","Text":"then my center of height changes."},{"Start":"17:59.530 ","End":"18:01.745","Text":"I\u0027m trying to find this change in center of height."},{"Start":"18:01.745 ","End":"18:05.570","Text":"If I look over here and I discard these Theta^2"},{"Start":"18:05.570 ","End":"18:09.879","Text":"because I say that it\u0027s such a small and insignificant number,"},{"Start":"18:09.879 ","End":"18:12.260","Text":"then I\u0027m going to get that my y center of"},{"Start":"18:12.260 ","End":"18:15.440","Text":"mass is just constantly going to be equal to R,"},{"Start":"18:15.440 ","End":"18:19.970","Text":"which cannot be because the whole point of my question was to find this change in height."},{"Start":"18:19.970 ","End":"18:24.485","Text":"In here, it is important to take into account these small changes."},{"Start":"18:24.485 ","End":"18:29.990","Text":"Then we can simplify out this expression slightly more by R(1 plus"},{"Start":"18:29.990 ","End":"18:37.385","Text":"Theta^2 over 2) because it\u0027s negative 1/2 plus 1."},{"Start":"18:37.385 ","End":"18:40.670","Text":"That is the end of our exercise."},{"Start":"18:40.670 ","End":"18:43.250","Text":"It\u0027s a slightly more complicated exercise"},{"Start":"18:43.250 ","End":"18:46.220","Text":"probably because there\u0027s a lot of factors that you have to think about."},{"Start":"18:46.220 ","End":"18:48.710","Text":"This height, Steiner, that height,"},{"Start":"18:48.710 ","End":"18:52.430","Text":"the change in when it\u0027s lowered slightly,"},{"Start":"18:52.430 ","End":"18:55.790","Text":"the angles, this little complicated bit."},{"Start":"18:55.790 ","End":"18:58.205","Text":"But it\u0027s a good question."},{"Start":"18:58.205 ","End":"19:05.100","Text":"It\u0027s not too hard once you get your head around the little finicky bits. That is it."}],"ID":9450}],"Thumbnail":null,"ID":5371}]

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