WEBVTT 00:00:00.060 --> 00:00:01.770 The following content is provided 00:00:01.770 --> 00:00:04.010 under a Creative Commons license. 00:00:04.010 --> 00:00:06.860 Your support will help MIT OpenCourseWare continue 00:00:06.860 --> 00:00:10.720 to offer high-quality educational resources for free. 00:00:10.720 --> 00:00:13.330 To make a donation or view additional materials 00:00:13.330 --> 00:00:17.209 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:17.209 --> 00:00:17.834 at ocw.mit.edu. 00:00:21.180 --> 00:00:23.140 PROFESSOR: Welcome back. 00:00:23.140 --> 00:00:29.546 Today, we will do two problems involving many oscillators, 00:00:29.546 --> 00:00:34.420 or at least several oscillators, coupled to each other. 00:00:34.420 --> 00:00:40.850 Now, you will not be surprised, from past experience, 00:00:40.850 --> 00:00:45.340 that in the situation with several oscillators, 00:00:45.340 --> 00:00:49.700 we are going to end up with many equations and a lot 00:00:49.700 --> 00:00:55.270 of grindy, tough solving of equations. 00:00:55.270 --> 00:00:58.510 Already, with a single harmonic oscillator, 00:00:58.510 --> 00:01:01.820 for example, with damping, the equations 00:01:01.820 --> 00:01:04.370 have been pretty horrendous. 00:01:04.370 --> 00:01:09.020 Now, we considered a couple of oscillators 00:01:09.020 --> 00:01:13.440 with all the details, with damping, driven, et cetera, 00:01:13.440 --> 00:01:16.900 and it really becomes complicated. 00:01:16.900 --> 00:01:19.230 Now, what we are trying to do by doing 00:01:19.230 --> 00:01:23.460 problems is to get a feel, an understanding what goes on 00:01:23.460 --> 00:01:29.160 in a given situation, so one wants to focus on new features 00:01:29.160 --> 00:01:32.180 and try to simplify the mathematics as 00:01:32.180 --> 00:01:34.120 much as possible. 00:01:34.120 --> 00:01:37.330 And so in the problems are we doing now, 00:01:37.330 --> 00:01:42.710 I am intentionally making them even more ideal than before. 00:01:42.710 --> 00:01:45.640 Consider the first problem. 00:01:45.640 --> 00:01:52.660 Here is a highly idealized system 00:01:52.660 --> 00:01:56.680 which has 3 degrees of freedom. 00:01:56.680 --> 00:02:03.750 What we have is we have a single mass, 2m, two identical springs 00:02:03.750 --> 00:02:06.920 attached to two masses, each of mass m. 00:02:09.690 --> 00:02:11.680 This is completely idealized. 00:02:11.680 --> 00:02:14.550 We are assuming there's no friction, 00:02:14.550 --> 00:02:17.510 that these springs are ideal springs. 00:02:17.510 --> 00:02:19.080 In other words, they're massless, 00:02:19.080 --> 00:02:23.230 obey Hooke's laws with constant k. 00:02:23.230 --> 00:02:25.910 We are assuming the system is constrained, 00:02:25.910 --> 00:02:28.520 so it can only move in one direction, 00:02:28.520 --> 00:02:33.670 it can't move up and down or in or out from the board. 00:02:33.670 --> 00:02:40.480 So as I say, it's an idealized situation. 00:02:40.480 --> 00:02:46.970 And the question we want to answer is, for such a system, 00:02:46.970 --> 00:02:50.860 what are the normal mode frequencies? 00:02:50.860 --> 00:02:55.880 And you know from lectures given by Professor Walter Lewin 00:02:55.880 --> 00:03:00.710 that when you have coupled oscillators, in this case, 3, 00:03:00.710 --> 00:03:09.060 all right, one finds that there are very special oscillations, 00:03:09.060 --> 00:03:13.150 which we call normal mode oscillations, in which 00:03:13.150 --> 00:03:20.980 every part of the system is oscillating 00:03:20.980 --> 00:03:24.095 with the same frequency and phase. 00:03:28.720 --> 00:03:32.680 So first, we want to find out for this system, what 00:03:32.680 --> 00:03:36.505 are those three normal mode frequencies? 00:03:39.240 --> 00:03:44.100 And secondly, suppose I take this system, 00:03:44.100 --> 00:03:49.630 place these masses at some arbitrary positions, some 00:03:49.630 --> 00:03:53.870 maybe even moving, and let go, how would I 00:03:53.870 --> 00:04:00.380 predict where each mass would be at some later time? 00:04:00.380 --> 00:04:02.070 So those are the two questions I want 00:04:02.070 --> 00:04:05.730 to answer today for this problem. 00:04:10.010 --> 00:04:18.750 Now, in general, coupled oscillator problems 00:04:18.750 --> 00:04:21.850 are quite difficult. 00:04:21.850 --> 00:04:25.300 But there are situations in which 00:04:25.300 --> 00:04:29.890 one can use logic alone, well, logic 00:04:29.890 --> 00:04:33.920 and our understanding of coupled oscillators, 00:04:33.920 --> 00:04:38.280 to do most of the solution by hand-waving 00:04:38.280 --> 00:04:41.320 and not do any calculations. 00:04:41.320 --> 00:04:45.020 In this first example I'm doing in this problem, 00:04:45.020 --> 00:04:52.930 since the system has a lot of symmetry, 00:04:52.930 --> 00:04:56.670 I know from experience that under those conditions 00:04:56.670 --> 00:05:02.910 it might be possible to solve this, as I say, by logic alone 00:05:02.910 --> 00:05:06.230 and our understanding of coupled oscillators. 00:05:06.230 --> 00:05:09.380 So the issue is, can I guess what 00:05:09.380 --> 00:05:13.835 will be the motion in the normal modes? 00:05:16.480 --> 00:05:19.200 And the answer is, actually, if you stop and think 00:05:19.200 --> 00:05:23.370 for a second, at least the first two 00:05:23.370 --> 00:05:26.770 modes you can figure out rather easily. 00:05:26.770 --> 00:05:30.270 So here I'm showing you schematically 00:05:30.270 --> 00:05:34.130 what would be one of the modes of oscillation, 00:05:34.130 --> 00:05:35.780 one of the normal modes. 00:05:35.780 --> 00:05:41.120 Imagine that I first hold this mass fixed 00:05:41.120 --> 00:05:45.390 and I displace these symmetrically, one 00:05:45.390 --> 00:05:48.565 to the right, one to the left, and I let go. 00:05:51.400 --> 00:05:54.720 What you will find is, by symmetry, 00:05:54.720 --> 00:05:59.470 this spring will always be stretched by the same amount 00:05:59.470 --> 00:06:06.500 as this is compressed, and so the force on this mass 00:06:06.500 --> 00:06:07.550 will be 0. 00:06:07.550 --> 00:06:13.210 This spring will be pushing this mass exactly by the same amount 00:06:13.210 --> 00:06:16.290 as this one will be pulling it and vice versa. 00:06:16.290 --> 00:06:19.145 So there will never be any net force on this mass, 00:06:19.145 --> 00:06:22.420 and so they'll stay put. 00:06:22.420 --> 00:06:31.160 And these two will now behave as a simple harmonic oscillator 00:06:31.160 --> 00:06:38.040 consisting of a spring attached to a fixed wall and the mass m. 00:06:38.040 --> 00:06:40.420 So that's what this one will be doing, 00:06:40.420 --> 00:06:42.070 and that's what this will be doing. 00:06:42.070 --> 00:06:44.580 It will be out of phase by 180 degrees. 00:06:44.580 --> 00:06:48.520 Or what one normally says, it'll be in phase 00:06:48.520 --> 00:06:53.440 but the amplitude will be minus the other one. 00:06:53.440 --> 00:06:56.490 This, by now, you've had enough experience 00:06:56.490 --> 00:07:00.240 and I've done it here before, I can just 00:07:00.240 --> 00:07:04.460 write what will be that frequency of oscillation 00:07:04.460 --> 00:07:08.410 here, as we've seen many times. 00:07:08.410 --> 00:07:13.310 And, by the way, throughout my solving the problems here, 00:07:13.310 --> 00:07:19.150 I do not differentiate frequency and angular frequency, that you 00:07:19.150 --> 00:07:22.320 can see which I'm talking about is whether I'm using omega 00:07:22.320 --> 00:07:25.500 or F. Omega is the angular frequency, 00:07:25.500 --> 00:07:28.230 and there's just a fact of 2 pi between the two. 00:07:28.230 --> 00:07:31.970 But I'll use, interchangeably, I'll call this frequency. 00:07:31.970 --> 00:07:37.340 So this frequency of one of the normal modes is just k/m. 00:07:37.340 --> 00:07:40.710 So we found one normal mode frequency. 00:07:40.710 --> 00:07:42.850 Let's look for another one. 00:07:42.850 --> 00:07:48.376 We stare at this and we see, sure, there is another one. 00:07:48.376 --> 00:07:55.760 Another one is where I take, simply, these two masses 00:07:55.760 --> 00:07:59.540 and simultaneously pull them out, in this mass, 00:07:59.540 --> 00:08:03.220 away and let go. 00:08:03.220 --> 00:08:07.730 Effectively, these two will look like that. 00:08:07.730 --> 00:08:13.160 It's as if I had a mass of 2m attached to a mass of 2m 00:08:13.160 --> 00:08:18.790 by two springs, each of spring constant k. 00:08:18.790 --> 00:08:23.730 And as I pull this back, this will oscillate like this. 00:08:23.730 --> 00:08:26.470 Everything, both this mass and this mass, 00:08:26.470 --> 00:08:31.570 will be oscillating with the same frequency, same phase, 00:08:31.570 --> 00:08:33.789 although there's this minus sign. 00:08:33.789 --> 00:08:39.600 I can always replace the 180-degree phase shift 00:08:39.600 --> 00:08:43.240 by saying the amplitude is minus. 00:08:43.240 --> 00:08:47.170 So those are identical statements. 00:08:47.170 --> 00:08:51.308 So you have these two masses oscillating like that. 00:08:51.308 --> 00:08:52.670 OK? 00:08:52.670 --> 00:08:58.090 By the way, I've said that in the normal mode 00:08:58.090 --> 00:09:02.040 I want everything to move with the same frequency. 00:09:02.040 --> 00:09:04.170 Some of you may argue, hold on. 00:09:04.170 --> 00:09:05.680 Did I cheat on you? 00:09:05.680 --> 00:09:09.510 Here, this mass is not oscillating. 00:09:09.510 --> 00:09:12.110 So am I contradicting myself? 00:09:12.110 --> 00:09:14.490 And the answer is no. 00:09:14.490 --> 00:09:16.230 It's a diabolical answer. 00:09:16.230 --> 00:09:18.970 But I'll say, no, this is oscillating 00:09:18.970 --> 00:09:22.760 with this frequency, but with 0 amplitude. 00:09:22.760 --> 00:09:24.970 And you can't tell me that I'm wrong. 00:09:24.970 --> 00:09:27.810 It's not moving because the amplitude is 0, 00:09:27.810 --> 00:09:32.360 but it is oscillating with the same frequency. 00:09:32.360 --> 00:09:36.180 So now, what is the frequency of this oscillation? 00:09:36.180 --> 00:09:38.550 Again, I can do it in my head. 00:09:38.550 --> 00:09:46.450 I can imagine that the middle of this spring is not moving. 00:09:46.450 --> 00:09:52.650 And so this mass is a half a length of the spring 00:09:52.650 --> 00:09:55.500 and a mass 2 attached to it oscillating, 00:09:55.500 --> 00:09:57.440 and I can calculate the frequency 00:09:57.440 --> 00:09:59.150 of oscillation of that. 00:09:59.150 --> 00:10:02.700 Or if you prefer, I can do the following. 00:10:02.700 --> 00:10:09.180 I can move this mass out a distance dx and this one dx 00:10:09.180 --> 00:10:12.030 to the other side and calculate what 00:10:12.030 --> 00:10:14.450 will be the restoring force. 00:10:14.450 --> 00:10:19.890 Well, now each one has a spring constant k. 00:10:19.890 --> 00:10:21.600 There is two of them, so that's 2k. 00:10:25.140 --> 00:10:30.320 Although this mass has been moved by dx, this by dx, 00:10:30.320 --> 00:10:35.030 the springs have been extended by twice dx, so it's 2 dx. 00:10:35.030 --> 00:10:37.240 So that's the total restoring force 00:10:37.240 --> 00:10:41.440 by Hooke's law, which is equal minus 4k dx. 00:10:41.440 --> 00:10:42.610 All right? 00:10:42.610 --> 00:10:49.650 And so this will behave like a single mass, 00:10:49.650 --> 00:10:52.880 like this system, a single mass with a single spring 00:10:52.880 --> 00:10:56.770 where the spring has a spring constant 4k. 00:10:56.770 --> 00:10:58.940 The mass is 2m. 00:10:58.940 --> 00:11:05.450 And so the angular frequency, or frequency of this mode is 2k/m. 00:11:05.450 --> 00:11:07.660 So we've found two of them. 00:11:07.660 --> 00:11:12.330 Now, we know from our studies of coupled oscillators 00:11:12.330 --> 00:11:15.685 that for a system of 3 degrees of freedom, 00:11:15.685 --> 00:11:20.940 so it would be three masses, there are three normal modes. 00:11:20.940 --> 00:11:24.370 What is the third normal mode? 00:11:24.370 --> 00:11:27.210 And I intentionally did not write it 00:11:27.210 --> 00:11:30.955 down here because I wanted you for a second to think. 00:11:30.955 --> 00:11:35.920 Well, how else can this system move such 00:11:35.920 --> 00:11:45.591 that every mass is moving in phase with the same frequency? 00:11:45.591 --> 00:11:47.110 All right? 00:11:47.110 --> 00:11:50.810 And I tell you, when I first saw it, 00:11:50.810 --> 00:11:54.290 it took me a long time to figure out, 00:11:54.290 --> 00:12:00.030 and most people don't find it. 00:12:00.030 --> 00:12:03.400 And yet the answer is extremely simple. 00:12:03.400 --> 00:12:08.430 This answer is there is one more normal mode. 00:12:08.430 --> 00:12:17.600 It is one of almost infinite amplitude, but 0 frequency. 00:12:17.600 --> 00:12:24.290 Just the opposite to here, where this was 0 amplitude, OK, 00:12:24.290 --> 00:12:26.700 and the angular frequency doesn't even matter. 00:12:26.700 --> 00:12:32.480 Here it has very large amplitude, infinite in fact, 00:12:32.480 --> 00:12:34.100 but 0 frequency. 00:12:34.100 --> 00:12:38.770 What does such motion look like at any instant of time? 00:12:38.770 --> 00:12:41.530 It's moving with the same velocity. 00:12:41.530 --> 00:12:54.770 So if I take and if I consider the motion where 00:12:54.770 --> 00:13:01.100 each one of these is moving uniformly to the right 00:13:01.100 --> 00:13:04.240 with a constant velocity, that is 00:13:04.240 --> 00:13:10.200 a normal mode with 0 frequency. 00:13:10.200 --> 00:13:18.240 So the last one is omega C is equal to 0. 00:13:21.010 --> 00:13:21.920 OK? 00:13:21.920 --> 00:13:28.800 So we have now found the three normal modes. 00:13:28.800 --> 00:13:31.000 OK. 00:13:31.000 --> 00:13:31.650 All right. 00:13:31.650 --> 00:13:38.960 Now that I've got a piece of chalk, let's continue. 00:13:38.960 --> 00:13:41.980 So we've answered the first part of the question, what 00:13:41.980 --> 00:13:44.490 are the normal mode frequencies of the system? 00:13:44.490 --> 00:13:47.340 And I found you those three normal modes. 00:13:47.340 --> 00:13:49.970 Now, the next question we want to answer 00:13:49.970 --> 00:13:54.340 is, if, at any given instant of time, 00:13:54.340 --> 00:13:57.930 I know the positions and the velocities of the three masses, 00:13:57.930 --> 00:14:01.380 can I predict what will happen at the end? 00:14:01.380 --> 00:14:04.690 That's equivalent of saying, can I 00:14:04.690 --> 00:14:09.570 write equations for the positions of the masses 00:14:09.570 --> 00:14:11.400 as a function of time? 00:14:11.400 --> 00:14:15.700 For that I need to define for myself a coordinate system. 00:14:15.700 --> 00:14:18.585 So I'll say those are those three masses. 00:14:23.120 --> 00:14:26.510 All the motion is along the x direction, 00:14:26.510 --> 00:14:35.900 so there's just one variable, one variable for each mass, x. 00:14:35.900 --> 00:14:38.080 So the position of the first mass 00:14:38.080 --> 00:14:42.090 I'll call x1, position of the second mass x2, 00:14:42.090 --> 00:14:43.760 and of the third is x3. 00:14:43.760 --> 00:14:46.960 And the origin of coordinates I will 00:14:46.960 --> 00:14:51.070 take to be through the center of each mass 00:14:51.070 --> 00:14:54.580 at a time when these are in equilibrium. 00:14:54.580 --> 00:14:57.990 In other words, the spring is unstretched, et cetera. 00:14:57.990 --> 00:14:58.880 OK. 00:14:58.880 --> 00:15:07.630 So I can now, using the information we've obtained, 00:15:07.630 --> 00:15:17.060 write down the description of each mass in each mode. 00:15:17.060 --> 00:15:23.360 So in mode A, I know that the x1 is 0 all the time. 00:15:23.360 --> 00:15:26.690 It's not moving, OK? 00:15:26.690 --> 00:15:34.830 Now, in a normal mode, that's what we mean by a normal mode, 00:15:34.830 --> 00:15:38.020 each mass is moving with the same frequency 00:15:38.020 --> 00:15:42.240 and the same phase, and it's oscillating. 00:15:42.240 --> 00:15:49.900 So there will be some sinusoidal function of the mode frequency 00:15:49.900 --> 00:15:54.730 t plus this phase times some arbitrary amplitude. 00:15:57.480 --> 00:16:00.960 The other one, the last one, will also 00:16:00.960 --> 00:16:03.330 be oscillating in the normal mode, 00:16:03.330 --> 00:16:05.025 so everything will be the same. 00:16:05.025 --> 00:16:07.930 It will be cosine of omega A t plus phi 00:16:07.930 --> 00:16:11.320 A with some other amplitude. 00:16:11.320 --> 00:16:18.350 But from our analysis here, we know 00:16:18.350 --> 00:16:23.810 that these two are not both arbitrary. 00:16:23.810 --> 00:16:27.510 Because we said that in the normal mode, 00:16:27.510 --> 00:16:35.360 it was the case where the 2m mass was stationary, 00:16:35.360 --> 00:16:37.970 but these were going like this. 00:16:37.970 --> 00:16:40.340 That was the first mode. 00:16:40.340 --> 00:16:46.080 So whenever this one was going to the right a distance A, 00:16:46.080 --> 00:16:49.440 this one was going to the left a distance A. 00:16:49.440 --> 00:16:52.830 So the overall normalization is arbitrary. 00:16:52.830 --> 00:16:54.620 It can be anything. 00:16:54.620 --> 00:16:58.210 But these are not independent of the other, 00:16:58.210 --> 00:17:01.240 because we discovered that in that mode, 00:17:01.240 --> 00:17:06.589 this was the frequency, and the amplitudes were 00:17:06.589 --> 00:17:09.849 opposite to each other. 00:17:09.849 --> 00:17:17.630 So this describes the situation for this system 00:17:17.630 --> 00:17:22.650 if it's oscillating in the first normal mode. 00:17:22.650 --> 00:17:26.540 It's the most general description of that. 00:17:26.540 --> 00:17:29.490 Next, I will describe what it will 00:17:29.490 --> 00:17:32.350 do in the second normal mode. 00:17:32.350 --> 00:17:34.880 And now I can go much quicker. 00:17:34.880 --> 00:17:37.270 This time, it'll have the second normal mode 00:17:37.270 --> 00:17:40.750 frequency, different phase. 00:17:40.750 --> 00:17:44.980 But these have to be the same because all of these masses 00:17:44.980 --> 00:17:49.910 are moving in the same normal mode, same normal frequency. 00:17:49.910 --> 00:17:54.560 These amplitudes, overall, it's arbitrary. 00:17:54.560 --> 00:17:55.960 I can make it anything. 00:17:55.960 --> 00:17:58.600 It'll satisfy the equations. 00:17:58.600 --> 00:18:03.520 But I know that in the second normal mode, 00:18:03.520 --> 00:18:06.820 this one is moving in that direction and those two 00:18:06.820 --> 00:18:10.220 in the opposite direction with the same magnitude. 00:18:10.220 --> 00:18:14.220 So these two have to have the same size and magnitude, 00:18:14.220 --> 00:18:16.620 but this has to be opposite. 00:18:16.620 --> 00:18:21.460 So this describes the motion in the second mode. 00:18:21.460 --> 00:18:25.090 And finally, in the third normal mode, 00:18:25.090 --> 00:18:29.160 we said that it was 0 frequency. 00:18:29.160 --> 00:18:30.280 All right? 00:18:30.280 --> 00:18:32.990 And all that was happening, the three masses 00:18:32.990 --> 00:18:37.780 were moving with uniform velocity. 00:18:37.780 --> 00:18:41.850 So the positions will be some constant 00:18:41.850 --> 00:18:44.265 plus the velocity of that system. 00:18:51.310 --> 00:18:55.030 So this describes the system in the third mode, 00:18:55.030 --> 00:18:58.650 and these are the other two. 00:18:58.650 --> 00:19:17.870 Now, I know that, in general, each one of these masses 00:19:17.870 --> 00:19:25.990 can move in a superposition of the normal modes. 00:19:25.990 --> 00:19:37.390 So the most general expression for x1 of t 00:19:37.390 --> 00:19:45.430 has to be of the form which is the sum of its possible motion 00:19:45.430 --> 00:19:47.360 in each of the modes. 00:19:47.360 --> 00:19:51.660 So x1, so this is the mass 2m, will 00:19:51.660 --> 00:19:56.320 have possible motion in the first mode, which is 0. 00:19:56.320 --> 00:19:57.520 All right? 00:19:57.520 --> 00:20:13.570 So I'll write 0 plus B times cosine omega B t plus phi 00:20:13.570 --> 00:20:21.920 B, all right, plus C plus v. So that 00:20:21.920 --> 00:20:26.940 will be the most general description of the motion 00:20:26.940 --> 00:20:30.880 of the big mass, the 2m one. 00:20:30.880 --> 00:20:34.310 The B is arbitrary, this phase is arbitrary, 00:20:34.310 --> 00:20:37.990 this is arbitrary, and v. Those quantities 00:20:37.990 --> 00:20:42.986 will be determined by the initial conditions. 00:20:42.986 --> 00:20:44.280 Well, how about x2? 00:20:48.340 --> 00:21:08.190 Well, I know that if x1 is 0, x2 is A cosine omega A t 00:21:08.190 --> 00:21:26.250 plus phi A. When the x1 is this, then x2 is minus B cosine omega 00:21:26.250 --> 00:21:37.270 B t plus phi B, OK, and plus C plus v. OK? 00:21:37.270 --> 00:21:45.210 And finally, x3 of t, again now just following 00:21:45.210 --> 00:21:50.360 the same pattern, this is minus A cosine omega A t 00:21:50.360 --> 00:21:59.620 plus phi A. This one is minus B cosine omega B 00:21:59.620 --> 00:22:15.940 t plus phi B plus C plus v. 00:22:15.940 --> 00:22:19.780 And I'm almost home. 00:22:19.780 --> 00:22:22.650 I have described the motion of each one. 00:22:27.015 --> 00:22:31.620 For each particle, it is oscillating 00:22:31.620 --> 00:22:34.410 in a superposition of its normal modes. 00:22:38.230 --> 00:22:45.460 And the amplitudes are arbitrary within the constraint 00:22:45.460 --> 00:22:50.650 that the ratios between them is determined 00:22:50.650 --> 00:22:56.070 by the constraints of the system. 00:22:56.070 --> 00:23:00.450 So now the question is, can I predict the future? 00:23:00.450 --> 00:23:02.720 Well, in order to predict the future, 00:23:02.720 --> 00:23:06.645 say, where will particle 3 be at time t? 00:23:06.645 --> 00:23:11.876 I need to know the value of these arbitrary constants. 00:23:14.420 --> 00:23:16.640 Omega is not an arbitrary constant. 00:23:16.640 --> 00:23:17.740 We found it. 00:23:17.740 --> 00:23:19.630 Omega A is here. 00:23:19.630 --> 00:23:20.580 OK? 00:23:20.580 --> 00:23:26.370 This is arbitrary, this is arbitrary, this, and that. 00:23:26.370 --> 00:23:30.670 So there are six arbitrary constants. 00:23:30.670 --> 00:23:32.280 How do I find them? 00:23:32.280 --> 00:23:36.730 They are determined by the initial conditions. 00:23:36.730 --> 00:23:40.420 And you know, fortunately, or else it 00:23:40.420 --> 00:23:44.660 would mean we've made a mistake, we have three conditions. 00:23:47.280 --> 00:23:51.560 At the beginning, someone has to tell me 00:23:51.560 --> 00:23:58.000 where each mass was, the location, and with what 00:23:58.000 --> 00:24:01.810 velocity is it moving at that time. 00:24:01.810 --> 00:24:09.290 So I can write these three equations at, say, 00:24:09.290 --> 00:24:14.920 with t equals 0, at t equals 0, and equate this 00:24:14.920 --> 00:24:21.220 to the position of x1 of t equals 0, wherever it is. 00:24:21.220 --> 00:24:26.220 When the t is 0, I make it equal to the position of x2 at t 00:24:26.220 --> 00:24:28.170 equals 0. 00:24:28.170 --> 00:24:32.890 When t is 0, this tells me has to be equal to where x3 is. 00:24:32.890 --> 00:24:34.650 That's three equations. 00:24:34.650 --> 00:24:36.680 I can differentiate each with respect 00:24:36.680 --> 00:24:40.650 to time to get the velocities, and likewise, 00:24:40.650 --> 00:24:42.440 get three more equations. 00:24:42.440 --> 00:24:46.500 So I'm going to end up with six algebraic equations. 00:24:46.500 --> 00:24:49.210 And you, as well as I, know that if we 00:24:49.210 --> 00:24:53.540 have six algebraic equations, I can solve for six unknowns. 00:24:53.540 --> 00:24:56.550 And that will give me those six unknowns. 00:24:56.550 --> 00:25:00.010 So once I've used the initial conditions, 00:25:00.010 --> 00:25:03.880 I can then find these arbitrary constants. 00:25:03.880 --> 00:25:07.420 They're no longer arbitrary for a specific situation. 00:25:07.420 --> 00:25:11.960 And I can predict the future, what will happen in this case. 00:25:11.960 --> 00:25:12.610 OK? 00:25:12.610 --> 00:25:19.430 So that's the way one would solve a problem when 00:25:19.430 --> 00:25:21.570 you are lucky enough that there is 00:25:21.570 --> 00:25:25.340 sufficient symmetry in the original situation 00:25:25.340 --> 00:25:29.510 so you can guess the normal modes. 00:25:29.510 --> 00:25:34.730 The next problem I'll do, I will do one where you cannot guess 00:25:34.730 --> 00:25:36.110 the normal modes. 00:25:36.110 --> 00:25:38.160 What do you do under those circumstances? 00:25:38.160 --> 00:25:40.180 And you won't be surprised that it's 00:25:40.180 --> 00:25:46.350 going to be mathematically much, much harder, but conceptually 00:25:46.350 --> 00:25:47.500 no harder. 00:25:47.500 --> 00:25:50.950 So now we'll move to the next problem. 00:25:50.950 --> 00:25:51.720 Here it is. 00:25:51.720 --> 00:25:54.050 But in order to be able to do this problem, 00:25:54.050 --> 00:25:55.600 I need more board space. 00:25:55.600 --> 00:25:59.281 So we're going to erase the boards and continue from there. 00:25:59.281 --> 00:25:59.780 OK. 00:25:59.780 --> 00:26:03.130 So now let's go and do the second problem. 00:26:03.130 --> 00:26:07.190 And it's going to be a problem with no symmetry, 00:26:07.190 --> 00:26:10.760 so we can't guess the answer like we did before. 00:26:10.760 --> 00:26:12.870 And I'll tell you what I decided to do. 00:26:12.870 --> 00:26:15.770 You remember Professor Walter Lewin 00:26:15.770 --> 00:26:21.660 in his lectures discussed a double pendulum. 00:26:21.660 --> 00:26:23.400 In other words, a situation where 00:26:23.400 --> 00:26:27.990 you had a string, a mass, another string, and mass. 00:26:27.990 --> 00:26:30.200 He gave you the answer, but he didn't prove it. 00:26:30.200 --> 00:26:32.590 He said it's hard and lengthy, et cetera. 00:26:32.590 --> 00:26:35.140 And I thought that would be a perfect example 00:26:35.140 --> 00:26:39.660 to do here from first principles. 00:26:39.660 --> 00:26:44.930 So let me just describe in detail 00:26:44.930 --> 00:26:47.230 the problem we're trying to solve. 00:26:47.230 --> 00:26:51.230 What we have is, again, an idealized situation. 00:26:51.230 --> 00:26:57.250 We have a simple pendulum with a string of length L, 00:26:57.250 --> 00:26:59.960 attached to it a mass m. 00:26:59.960 --> 00:27:03.850 Attached to this mass at one end is another string of length 00:27:03.850 --> 00:27:06.930 L to another mass m. 00:27:06.930 --> 00:27:09.930 We'll again idealize the situation. 00:27:09.930 --> 00:27:12.200 So the assumptions we are making, 00:27:12.200 --> 00:27:18.460 that this string and this string is massless, which, by the way, 00:27:18.460 --> 00:27:22.530 is equivalent to saying that it's taut and straight all 00:27:22.530 --> 00:27:23.310 the time. 00:27:23.310 --> 00:27:25.970 You can think about that for yourself. 00:27:25.970 --> 00:27:31.350 Next, these masses are point masses. 00:27:31.350 --> 00:27:34.330 And we're assuming no friction. 00:27:34.330 --> 00:27:37.980 Furthermore, as we've done in the past, 00:27:37.980 --> 00:27:41.500 we'll assume that all displacements 00:27:41.500 --> 00:27:47.260 when this is oscillating, at all times the angle that the string 00:27:47.260 --> 00:27:52.220 makes with the vertical is always such that we can ignore 00:27:52.220 --> 00:27:56.520 the difference between a sine theta, theta, 00:27:56.520 --> 00:28:01.695 or [INAUDIBLE] can make the assumption that cosine theta is 00:28:01.695 --> 00:28:03.090 0. 00:28:03.090 --> 00:28:06.410 By the way, the fact that we'll make this assumption 00:28:06.410 --> 00:28:09.290 is equivalent to saying that we'll 00:28:09.290 --> 00:28:13.080 ignore vertical motion of these masses. 00:28:13.080 --> 00:28:16.780 The motion is sufficiently small that vertically the masses 00:28:16.780 --> 00:28:18.210 are not moving. 00:28:18.210 --> 00:28:22.780 We can assume the acceleration vertically is 0, et cetera. 00:28:22.780 --> 00:28:23.340 OK? 00:28:23.340 --> 00:28:24.850 This is in the vertical plane. 00:28:24.850 --> 00:28:26.420 Gravity is down. 00:28:26.420 --> 00:28:27.180 OK? 00:28:27.180 --> 00:28:32.440 And what we are trying to derive for this, 00:28:32.440 --> 00:28:37.980 predict what are the normal mode frequencies of this. 00:28:37.980 --> 00:28:40.170 And once we do that, of course, we 00:28:40.170 --> 00:28:43.610 can use the same kind of technique as we did before. 00:28:43.610 --> 00:28:45.800 Once we've managed to find the normal modes 00:28:45.800 --> 00:28:47.810 and the frequencies, we can always 00:28:47.810 --> 00:28:50.450 write the most general expression. 00:28:50.450 --> 00:28:53.920 And then using the boundary conditions, initial conditions, 00:28:53.920 --> 00:28:57.440 predict what this will do as a function of time. 00:28:57.440 --> 00:28:58.774 OK, so let's get going. 00:29:05.280 --> 00:29:05.780 OK. 00:29:05.780 --> 00:29:07.660 Now, we are not guessing. 00:29:07.660 --> 00:29:09.120 We are not using logic. 00:29:09.120 --> 00:29:11.950 We are following the kind of prescription 00:29:11.950 --> 00:29:14.450 I've told you in the past. 00:29:14.450 --> 00:29:17.020 This is our description of the situation. 00:29:17.020 --> 00:29:19.700 And all the new words and ordinary language, we 00:29:19.700 --> 00:29:23.820 must now translate it into mathematics. 00:29:23.820 --> 00:29:26.230 We've got to describe this problem 00:29:26.230 --> 00:29:29.580 in terms of mathematical equations. 00:29:29.580 --> 00:29:33.600 So step one is we redraw this and define 00:29:33.600 --> 00:29:35.380 some coordinate system. 00:29:35.380 --> 00:29:39.690 So we will say that the angle this first string makes 00:29:39.690 --> 00:29:43.200 with respect to the vertical is theta 1. 00:29:43.200 --> 00:29:45.980 This angle the second string makes with respect 00:29:45.980 --> 00:29:48.420 to the vertical is theta 2. 00:29:48.420 --> 00:29:51.420 We will take the vertical through this pivot 00:29:51.420 --> 00:29:56.140 here as our origin of coordinate x. 00:29:56.140 --> 00:30:02.870 x equals 0 here, is defined here. 00:30:02.870 --> 00:30:05.190 From the point of your motion of the masses, 00:30:05.190 --> 00:30:06.960 it's a one-dimensional problem. 00:30:06.960 --> 00:30:10.750 The masses only move along the x-axis. 00:30:10.750 --> 00:30:13.430 So we'll define this distance as x1, 00:30:13.430 --> 00:30:17.440 and we'll define this distance as x2. 00:30:17.440 --> 00:30:22.720 Now we have to use the laws of physics, Newtonian mechanics, 00:30:22.720 --> 00:30:29.390 to derive the equations of motion for the two masses. 00:30:29.390 --> 00:30:36.010 So I draw a force diagram separately for the two masses. 00:30:36.010 --> 00:30:39.930 So the mass is there, what forces act on that mass? 00:30:39.930 --> 00:30:43.790 Well, I have to look at the mass and see what's attached to it. 00:30:43.790 --> 00:30:46.670 There is a string here attached to it. 00:30:46.670 --> 00:30:49.280 It's taut, so there will be a tension in it. 00:30:49.280 --> 00:30:53.890 So along this string, there will be tension T1, 00:30:53.890 --> 00:30:58.690 which will exert a force T1 along this string. 00:30:58.690 --> 00:31:02.100 This string is attached to that mass. 00:31:02.100 --> 00:31:06.030 It has a tension, so it's pulling on this mass. 00:31:06.030 --> 00:31:08.630 That tension I call T2, so there will 00:31:08.630 --> 00:31:13.990 be a force along that direction T2. 00:31:13.990 --> 00:31:16.820 This sits in a gravitational field. 00:31:16.820 --> 00:31:19.360 Gravity acts on this mass. 00:31:19.360 --> 00:31:23.330 There is a force downwards due to gravity, Fg. 00:31:26.330 --> 00:31:29.660 The sum of these forces, by Newton's laws, 00:31:29.660 --> 00:31:33.460 must equal to the mass times the acceleration. 00:31:33.460 --> 00:31:37.810 That acceleration, in this approximation that is not 00:31:37.810 --> 00:31:40.550 moving up, is horizontally, and it's 00:31:40.550 --> 00:31:43.920 the second derivative of x1. 00:31:43.920 --> 00:31:50.160 Similarly, for the second mass, again, I'll go now faster. 00:31:50.160 --> 00:31:52.390 There is this mass m, and this tension 00:31:52.390 --> 00:31:57.636 exerts a force here of T2 and the gravity on it Fg. 00:31:57.636 --> 00:32:00.310 The only subtlety is, why did I say 00:32:00.310 --> 00:32:03.350 this force is equal to that force? 00:32:03.350 --> 00:32:04.650 Think about it for a second. 00:32:04.650 --> 00:32:08.490 Why should the two forces be equal? 00:32:08.490 --> 00:32:12.870 Why is the tension at both ends of the string equal? 00:32:12.870 --> 00:32:16.756 And the answer is, actually, a subtle one. 00:32:16.756 --> 00:32:20.740 It's equal because we made the assumption that the string has 00:32:20.740 --> 00:32:28.240 no mass, and there can be no net force on an object of 0 mass. 00:32:28.240 --> 00:32:31.190 Because if there was, that object would disappear, 00:32:31.190 --> 00:32:33.650 would have an infinite acceleration. 00:32:33.650 --> 00:32:37.620 So if you treat the string as a mass, 00:32:37.620 --> 00:32:39.930 there cannot be net force on it. 00:32:39.930 --> 00:32:44.780 And so the force of each of these masses 00:32:44.780 --> 00:32:49.460 must be pulling on this string with exactly the same force 00:32:49.460 --> 00:32:51.031 equal and opposite. 00:32:51.031 --> 00:32:51.530 OK? 00:32:51.530 --> 00:32:55.020 And we're using the third law to equate those forces. 00:32:55.020 --> 00:32:58.770 So the net result is this T2 is the same as that, 00:32:58.770 --> 00:33:02.850 but in opposite direction, and this is the mass of that. 00:33:02.850 --> 00:33:03.710 OK. 00:33:03.710 --> 00:33:07.410 So these are the force diagrams. 00:33:07.410 --> 00:33:11.020 Using now Newton's laws of motion, 00:33:11.020 --> 00:33:15.220 I can translate this into equations of motion. 00:33:15.220 --> 00:33:18.810 So let me consider the horizontal motion 00:33:18.810 --> 00:33:20.950 of each mass separately. 00:33:20.950 --> 00:33:26.050 First this mass, and so its mass times acceleration 00:33:26.050 --> 00:33:31.320 is equal to the horizontal force of T2. 00:33:31.320 --> 00:33:35.690 And remember that this angle here is theta 2. 00:33:35.690 --> 00:33:38.600 And we see that the force due to T2, 00:33:38.600 --> 00:33:40.930 and there's a T2 sine theta 2. 00:33:40.930 --> 00:33:45.150 And the force due to T1 is minus T1 sine theta 1 00:33:45.150 --> 00:33:47.530 because it's in the opposite direction. 00:33:47.530 --> 00:33:50.190 For this, the only horizontal component 00:33:50.190 --> 00:33:53.180 is the force horizontal component of T2. 00:33:53.180 --> 00:33:58.800 And x2 double dot is equal to minus T2 sine theta 2. 00:33:58.800 --> 00:34:01.350 OK, that's horizontal motion. 00:34:01.350 --> 00:34:06.190 Applying Newton's laws of motion to the vertical motion, 00:34:06.190 --> 00:34:12.170 we said that, because cosine theta is 0 or approximately 0, 00:34:12.170 --> 00:34:14.250 the masses are not moving up and down. 00:34:14.250 --> 00:34:16.739 There's no acceleration of the masses. 00:34:16.739 --> 00:34:20.800 So the vertical acceleration of the first mass 00:34:20.800 --> 00:34:28.570 is 0 must be equal to the vertical component of T1, which 00:34:28.570 --> 00:34:33.219 is T1 cosine theta 1, minus the vertical component of this, 00:34:33.219 --> 00:34:40.940 which is minus T2 cosine theta 2 minus mg. 00:34:40.940 --> 00:34:41.440 All right. 00:34:41.440 --> 00:34:47.219 Similarly for the second mass, it's 0 is equal to this. 00:34:47.219 --> 00:34:51.850 0 must equal to the vertical component 00:34:51.850 --> 00:34:58.390 of T1 minus the vertical component of T2 00:34:58.390 --> 00:35:02.170 minus the force of gravity down. 00:35:02.170 --> 00:35:06.040 And similarly for the second mass, 00:35:06.040 --> 00:35:09.170 we know that vertically the acceleration is 0. 00:35:09.170 --> 00:35:12.560 That must be equal to the vertical component of T2 00:35:12.560 --> 00:35:18.650 minus the gravitational force pulling down. 00:35:18.650 --> 00:35:26.790 Now I can make use of what we made the assumption-- oops, 00:35:26.790 --> 00:35:29.130 I see there is an error. 00:35:29.130 --> 00:35:31.540 If you were in this room, I'm sure you 00:35:31.540 --> 00:35:35.020 would have corrected me. 00:35:35.020 --> 00:35:38.540 For very small angles, the sine of an angle 00:35:38.540 --> 00:35:40.790 is approximately equal to the angle. 00:35:40.790 --> 00:35:44.740 But for small angles, the cosine is 1. 00:35:44.740 --> 00:35:48.250 This is an error, and I apologize for that. 00:35:48.250 --> 00:35:50.450 But that's the approximation we are making. 00:35:50.450 --> 00:35:54.580 And it is this approximation which 00:35:54.580 --> 00:36:00.810 is equivalent to saying that we can ignore the vertical motion. 00:36:00.810 --> 00:36:02.200 All right. 00:36:02.200 --> 00:36:07.230 So with the assumption that the cosines are all 1, 00:36:07.230 --> 00:36:12.470 I can, from these two equations, I clearly 00:36:12.470 --> 00:36:18.430 derive that T2 must equal to mg, and T1 equals twice mg. 00:36:18.430 --> 00:36:20.580 Actually, it makes sense. 00:36:20.580 --> 00:36:23.770 Imagine this is hanging completely vertically. 00:36:23.770 --> 00:36:28.210 Obviously, this string, the upper part of the string, 00:36:28.210 --> 00:36:30.875 is supporting not only this mass but also that. 00:36:30.875 --> 00:36:35.530 It's supporting 2m masses, while this string is only 00:36:35.530 --> 00:36:36.640 supporting 1. 00:36:36.640 --> 00:36:40.440 So that's consistent with what we see, 00:36:40.440 --> 00:36:45.230 that the top string has twice the tension and the lower 00:36:45.230 --> 00:36:47.780 one half the tension. 00:36:47.780 --> 00:36:50.640 Having determined T1 and T2, we can now 00:36:50.640 --> 00:36:56.170 go back into our equation, replace the T1 and T2. 00:36:56.170 --> 00:36:58.830 We also can replace-- we know what 00:36:58.830 --> 00:37:02.440 sine theta 1 and sine theta 2 are. 00:37:02.440 --> 00:37:04.440 We can replace those. 00:37:04.440 --> 00:37:06.990 And we end up-- the two equations 00:37:06.990 --> 00:37:10.750 of motions are written here. 00:37:10.750 --> 00:37:11.770 All right? 00:37:11.770 --> 00:37:15.750 So here is the equation of motion for x1. 00:37:15.750 --> 00:37:19.380 And the second one, the equation for motion of x2, 00:37:19.380 --> 00:37:21.380 is written here. 00:37:21.380 --> 00:37:22.140 OK? 00:37:22.140 --> 00:37:27.200 So these are the equations of motion, 00:37:27.200 --> 00:37:37.930 all right, which we have to solve. 00:37:37.930 --> 00:37:46.300 Now, to simplify the algebra, let me define the quantity g/l 00:37:46.300 --> 00:37:49.525 by omega squared over 2. 00:37:49.525 --> 00:37:51.140 And you'll recognize this. 00:37:51.140 --> 00:37:52.240 This is not [INAUDIBLE]. 00:37:52.240 --> 00:37:54.010 I'm using that terminology. 00:37:54.010 --> 00:37:56.620 This is the frequency of oscillation 00:37:56.620 --> 00:38:01.140 of a mass on a string of length l. 00:38:01.140 --> 00:38:02.260 All right? 00:38:02.260 --> 00:38:08.930 So here are our two differential equations, the two equations 00:38:08.930 --> 00:38:14.620 of motion, one for x1 and one for x2. 00:38:14.620 --> 00:38:17.980 When we had one mass, one harmonic oscillator, 00:38:17.980 --> 00:38:21.420 we had a single second-order differential equation. 00:38:21.420 --> 00:38:25.070 We now have two masses, so we have 00:38:25.070 --> 00:38:27.740 two second-order differential equations. 00:38:27.740 --> 00:38:28.490 These are it. 00:38:28.490 --> 00:38:30.970 They are coupled differential equations. 00:38:30.970 --> 00:38:38.860 You see, the derivatives of x1 is related both to x1 and x2. 00:38:38.860 --> 00:38:43.870 The second derivative of x2 is related both to x1 and to x2. 00:38:43.870 --> 00:38:47.210 So these are two coupled differential equations. 00:38:47.210 --> 00:38:48.220 OK? 00:38:48.220 --> 00:38:51.655 This is the end of step 1. 00:38:51.655 --> 00:38:55.370 What we succeeded in, we've translated 00:38:55.370 --> 00:39:00.610 the physical situation into mathematics. 00:39:00.610 --> 00:39:06.580 These two second-order differential equations 00:39:06.580 --> 00:39:11.285 describe exactly that idealized situation we had. 00:39:14.050 --> 00:39:17.280 So if I now want to answer the question, what 00:39:17.280 --> 00:39:19.770 will be the motion of those masses, 00:39:19.770 --> 00:39:22.240 I have to go into the world of mathematics. 00:39:22.240 --> 00:39:25.630 I have to solve these equations. 00:39:25.630 --> 00:39:28.290 And life is not as easy as it was 00:39:28.290 --> 00:39:31.600 for a single second-order differential equation. 00:39:31.600 --> 00:39:33.720 It's more complicated. 00:39:33.720 --> 00:39:39.060 But here I will use my general knowledge 00:39:39.060 --> 00:39:44.040 of what happens when you have coupled oscillators. 00:39:44.040 --> 00:39:51.380 I know that the general solution of coupled oscillations 00:39:51.380 --> 00:39:57.410 is a superposition of normal modes. 00:39:57.410 --> 00:40:00.620 Here I have 2 degrees of freedom, 00:40:00.620 --> 00:40:03.020 two second-order differential equations. 00:40:03.020 --> 00:40:05.940 There will be two normal modes. 00:40:05.940 --> 00:40:09.490 If I succeed in finding them, I will 00:40:09.490 --> 00:40:13.890 have found the most general solution to this problem, 00:40:13.890 --> 00:40:17.220 because it will be the sum of the two normal modes. 00:40:17.220 --> 00:40:24.900 So I will now do it by trial. 00:40:24.900 --> 00:40:30.630 In a normal mode, we know that x1 00:40:30.630 --> 00:40:34.230 will be oscillating with a single frequency 00:40:34.230 --> 00:40:39.500 omega, some phase phi, some amplitude A1. 00:40:39.500 --> 00:40:41.820 This will be a solution to these equations. 00:40:44.370 --> 00:40:47.250 If this is a solution to those equations, 00:40:47.250 --> 00:40:51.280 the other mass must also be oscillating 00:40:51.280 --> 00:40:55.165 in the same normal mode, meaning with the same frequency 00:40:55.165 --> 00:40:58.080 and same phase. 00:40:58.080 --> 00:41:00.260 It'll have some arbitrary amplitude. 00:41:02.860 --> 00:41:05.820 So I don't know what A1 is, and I don't know what omega is. 00:41:05.820 --> 00:41:07.030 I don't know what phi is. 00:41:07.030 --> 00:41:12.600 But I know that the solutions to these equations 00:41:12.600 --> 00:41:16.980 must be of this form. 00:41:16.980 --> 00:41:18.140 OK? 00:41:18.140 --> 00:41:29.780 So let me now try to find these various constants. 00:41:29.780 --> 00:41:37.274 Well, if this and that satisfies these equations-- 00:41:37.274 --> 00:41:39.190 and it has to because that's what it's saying, 00:41:39.190 --> 00:41:41.470 these are the solutions of those equations-- 00:41:41.470 --> 00:41:49.440 then I can take these, x1 and x2, plug it into this equation. 00:41:49.440 --> 00:41:51.800 In other words, calculate the second derivative 00:41:51.800 --> 00:41:54.780 of x1, et cetera, calculate second, 00:41:54.780 --> 00:41:58.750 and I'll end up with two equations. 00:42:02.415 --> 00:42:02.915 OK? 00:42:06.640 --> 00:42:11.270 So I won't bother to go in great detail here. 00:42:11.270 --> 00:42:17.560 But just to start off with, x1 double dot, 00:42:17.560 --> 00:42:23.010 I have to differentiate A1 cosine omega t twice. 00:42:23.010 --> 00:42:34.830 So I get minus omega squared A1 times cosine omega t plus phi. 00:42:34.830 --> 00:42:40.220 If I take the next term, I get this, and the third term that. 00:42:40.220 --> 00:42:44.470 And in each case, if you notice, since it'll 00:42:44.470 --> 00:42:47.730 be multiplied by the same cosine function, 00:42:47.730 --> 00:42:51.210 I've just canceled in my head the cosine function there. 00:42:51.210 --> 00:42:52.270 All right? 00:42:52.270 --> 00:42:53.540 Just saving time. 00:42:53.540 --> 00:42:58.360 Similarly, next equation, I end up with this. 00:42:58.360 --> 00:43:04.190 So if my guessed functions, and I'm returning to this, 00:43:04.190 --> 00:43:08.810 if these two satisfy those equations, 00:43:08.810 --> 00:43:16.390 then here these algebraic equations must be satisfied. 00:43:16.390 --> 00:43:18.125 But you'll notice something interesting. 00:43:20.930 --> 00:43:24.010 If I take the first of these equations, 00:43:24.010 --> 00:43:30.000 it boils down to A1 times this quantity plus A2 00:43:30.000 --> 00:43:33.160 to [? then ?] this quantity equal to 0. 00:43:33.160 --> 00:43:36.320 And the second equation boils down to this. 00:43:39.580 --> 00:43:44.450 If this is true, then this must be true. 00:43:44.450 --> 00:43:49.850 It's A1 divided by A2 has to be equal to that. 00:43:49.850 --> 00:43:54.940 If this is true, then A1 divided by A2 00:43:54.940 --> 00:43:56.590 has got to be equal to this. 00:43:59.300 --> 00:44:02.960 That means that this has to be equal to that, 00:44:02.960 --> 00:44:06.460 or else I have an inconsistency. 00:44:06.460 --> 00:44:13.660 So what I have found is that, in general, my guess 00:44:13.660 --> 00:44:15.020 is not a good one. 00:44:15.020 --> 00:44:19.010 In general, these equations would not 00:44:19.010 --> 00:44:24.580 satisfy my second-order differential equations. 00:44:24.580 --> 00:44:30.660 There is only under very special circumstances 00:44:30.660 --> 00:44:33.470 that they do satisfy it, and that 00:44:33.470 --> 00:44:37.470 is if this is equal to that. 00:44:40.440 --> 00:44:42.824 In general, they will not be equal. 00:44:46.460 --> 00:44:50.410 So under what conditions will these be equal? 00:44:50.410 --> 00:44:51.850 Well, let me force them. 00:44:51.850 --> 00:44:55.760 Let me say this is equal to that and see 00:44:55.760 --> 00:44:59.680 what it means for omega. 00:44:59.680 --> 00:45:03.890 I originally took omega to be an unknown quantity. 00:45:03.890 --> 00:45:08.060 What I am now seeing, that those functions 00:45:08.060 --> 00:45:13.430 would work only for specific values of omega. 00:45:13.430 --> 00:45:18.050 So I'm going to solve this equation for omega 00:45:18.050 --> 00:45:24.090 and see for what values of omega do I get a consistent solution. 00:45:26.620 --> 00:45:29.820 And this is a hard grind, unfortunately. 00:45:29.820 --> 00:45:34.880 If I take this, multiply it out, I 00:45:34.880 --> 00:45:39.730 get a quadratic equation in omega squared. 00:45:39.730 --> 00:45:43.200 If this quadratic equation is satisfied, 00:45:43.200 --> 00:45:49.790 then all these are self-consistent. 00:45:49.790 --> 00:45:50.290 All right? 00:45:50.290 --> 00:45:56.170 You know how to solve a quadratic equation, 00:45:56.170 --> 00:46:05.430 and this is a quadratic equation in omega squared. 00:46:05.430 --> 00:46:08.530 Omega square will have to be equal to minus 00:46:08.530 --> 00:46:13.350 this plus or minus the square root of this, 00:46:13.350 --> 00:46:16.840 4 times this times that, divided by 2a, right? 00:46:16.840 --> 00:46:19.960 This is the standard formula for the solution 00:46:19.960 --> 00:46:21.840 of a quadratic equation. 00:46:26.090 --> 00:46:28.280 And there's a plus or minus. 00:46:28.280 --> 00:46:32.970 If I calculate this out, I get that omega squared 00:46:32.970 --> 00:46:36.470 has to be equal to 2 plus or minus 00:46:36.470 --> 00:46:40.680 the square root of 2 times omega 0 squared. 00:46:40.680 --> 00:46:41.690 All right? 00:46:41.690 --> 00:46:49.200 Now, omega 0 squared, earlier on, I defined to be g/l. 00:46:49.200 --> 00:46:53.830 So omega squared has to be this. 00:46:53.830 --> 00:46:57.030 So let's stop for a second and think 00:46:57.030 --> 00:47:00.560 and review what we've done. 00:47:00.560 --> 00:47:06.370 We found that our physical situation 00:47:06.370 --> 00:47:13.690 can be described by these two coupled differential equations. 00:47:13.690 --> 00:47:20.666 We looked for solutions, which are normal modes where both x1 00:47:20.666 --> 00:47:25.990 and x2 is oscillating with same frequency and phase. 00:47:25.990 --> 00:47:33.700 And we found that it was possible to find solution 00:47:33.700 --> 00:47:42.730 of this form to these equations if and only if omega squared 00:47:42.730 --> 00:47:46.210 has one of two possible values. 00:47:46.210 --> 00:47:48.050 I've rewritten them here. 00:47:48.050 --> 00:47:53.020 One is 2 plus root 2 times g/l, and the other's 00:47:53.020 --> 00:47:54.890 2 minus root 2 g/l. 00:47:57.420 --> 00:48:04.900 With these values, these two equations satisfies that. 00:48:04.900 --> 00:48:10.850 So what we found is the two normal modes. 00:48:10.850 --> 00:48:17.420 Since this system consists of the coupled oscillators, 00:48:17.420 --> 00:48:21.290 in other words, two oscillators coupled with each other, 00:48:21.290 --> 00:48:23.960 we know there are two normal modes. 00:48:23.960 --> 00:48:30.180 So these are the normal mode frequencies in this situation. 00:48:30.180 --> 00:48:33.440 And by the way, I'm delighted to see 00:48:33.440 --> 00:48:38.430 that, if you remember Professor Walter Lewin's lectures, 00:48:38.430 --> 00:48:41.830 he actually quoted these numbers. 00:48:41.830 --> 00:48:42.580 So he was right. 00:48:42.580 --> 00:48:45.270 He did not make a mistake, as we've just seen. 00:48:45.270 --> 00:48:46.280 OK? 00:48:46.280 --> 00:48:48.960 These are the two normal mode frequencies. 00:48:48.960 --> 00:48:51.830 Now, how about if we want to predict 00:48:51.830 --> 00:48:54.250 everything, the amplitude, et cetera? 00:48:57.330 --> 00:49:00.380 We've got to be careful here with the logic we've applied. 00:49:03.500 --> 00:49:10.780 For any one of these normal frequencies, say this one, 00:49:10.780 --> 00:49:23.390 earlier on we found that A1 to A2 is given by this equation 00:49:23.390 --> 00:49:25.220 or by this equation. 00:49:25.220 --> 00:49:27.820 But for the given values of omega, 00:49:27.820 --> 00:49:30.940 we found these are now equivalent. 00:49:30.940 --> 00:49:37.890 And so if you know what omega is, say omega A, 00:49:37.890 --> 00:49:43.400 one of the normal modes, this we know. 00:49:43.400 --> 00:49:45.930 It's g/l. 00:49:45.930 --> 00:49:47.550 And this is known. 00:49:47.550 --> 00:49:56.170 So A1/A2, once we fix omega A, is fixed. 00:49:56.170 --> 00:50:01.370 So these are not two independent amplitudes. 00:50:01.370 --> 00:50:07.620 So if I go back to here when we said let's guess a solution, 00:50:07.620 --> 00:50:10.150 and this was an arbitrary number and this was an arbitrary 00:50:10.150 --> 00:50:15.550 number, now we see that in a normal mode 00:50:15.550 --> 00:50:18.720 the ratio between them is fixed. 00:50:18.720 --> 00:50:22.840 It depends on the normal mode frequency. 00:50:22.840 --> 00:50:25.340 But the [? overall ?] normalization 00:50:25.340 --> 00:50:26.530 is still arbitrary. 00:50:26.530 --> 00:50:29.230 I could make this 10 times this and 10 times that. 00:50:29.230 --> 00:50:31.460 [? It would still ?] work. 00:50:31.460 --> 00:50:39.670 So back to here, we found one of the normal mode's frequencies. 00:50:39.670 --> 00:50:43.630 Using the equation above, it determines 00:50:43.630 --> 00:50:48.560 the value of A1 to A2. 00:50:48.560 --> 00:50:52.400 Taking the other normal mode frequency 00:50:52.400 --> 00:50:57.000 determines the ratio of those two 00:50:57.000 --> 00:51:00.380 but not the overall amplitude. 00:51:00.380 --> 00:51:04.780 Not to confuse the two, I will call this B1 and B2. 00:51:04.780 --> 00:51:07.180 This is the amplitude of the first mass, 00:51:07.180 --> 00:51:08.870 and this is of the other one. 00:51:08.870 --> 00:51:14.840 The ratio is fixed, determined by the value of omega B. 00:51:14.840 --> 00:51:18.570 But the overall amplitude is not. 00:51:18.570 --> 00:51:24.020 So now we have ended up in the same situation 00:51:24.020 --> 00:51:29.110 we were in our three masses, where, if you remember, 00:51:29.110 --> 00:51:33.020 I started by guessing the normal modes 00:51:33.020 --> 00:51:37.690 and guessing the relative amplitudes. 00:51:37.690 --> 00:51:42.490 And with that information alone and the initial conditions, 00:51:42.490 --> 00:51:44.500 I could predict what will happen. 00:51:44.500 --> 00:51:46.650 I could now repeat that here. 00:51:46.650 --> 00:51:51.510 But to save time, I think you can do that for yourself. 00:51:51.510 --> 00:51:55.970 For each, you can write now a single equation 00:51:55.970 --> 00:52:03.830 saying what a given mass will do as the sum of two normal modes, 00:52:03.830 --> 00:52:11.280 where you know the frequency of them and the amplitudes. 00:52:11.280 --> 00:52:14.040 And you have to remember that for the two masses, 00:52:14.040 --> 00:52:16.570 the ratio of the amplitudes in each normal mode 00:52:16.570 --> 00:52:20.330 have to be fixed given by what we've done here. 00:52:20.330 --> 00:52:24.810 And therefore, you can get the single equation 00:52:24.810 --> 00:52:28.530 for each mass, one for one mass, one for the other. 00:52:28.530 --> 00:52:31.690 Which will tell you what they will do in the future 00:52:31.690 --> 00:52:36.500 if you know what they're doing initially, for example, 00:52:36.500 --> 00:52:40.700 the initial conditions, that you can solve 00:52:40.700 --> 00:52:45.710 to get the final predicted position of each mass. 00:52:45.710 --> 00:52:47.350 So that's as much as I was hoping 00:52:47.350 --> 00:52:51.700 to do today on coupled oscillators. 00:52:51.700 --> 00:52:56.010 And we'll continue next time on situations 00:52:56.010 --> 00:52:58.920 with many, many more degrees of freedom. 00:52:58.920 --> 00:53:00.770 Thank you.