1 00:00:00,060 --> 00:00:01,770 The following content is provided 2 00:00:01,770 --> 00:00:04,010 under a Creative Commons license. 3 00:00:04,010 --> 00:00:06,860 Your support will help MIT OpenCourseWare continue 4 00:00:06,860 --> 00:00:10,720 to offer high-quality educational resources for free. 5 00:00:10,720 --> 00:00:13,330 To make a donation or view additional materials 6 00:00:13,330 --> 00:00:17,209 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,209 --> 00:00:17,834 at ocw.mit.edu. 8 00:00:21,180 --> 00:00:23,140 PROFESSOR: Welcome back. 9 00:00:23,140 --> 00:00:29,546 Today, we will do two problems involving many oscillators, 10 00:00:29,546 --> 00:00:34,420 or at least several oscillators, coupled to each other. 11 00:00:34,420 --> 00:00:40,850 Now, you will not be surprised, from past experience, 12 00:00:40,850 --> 00:00:45,340 that in the situation with several oscillators, 13 00:00:45,340 --> 00:00:49,700 we are going to end up with many equations and a lot 14 00:00:49,700 --> 00:00:55,270 of grindy, tough solving of equations. 15 00:00:55,270 --> 00:00:58,510 Already, with a single harmonic oscillator, 16 00:00:58,510 --> 00:01:01,820 for example, with damping, the equations 17 00:01:01,820 --> 00:01:04,370 have been pretty horrendous. 18 00:01:04,370 --> 00:01:09,020 Now, we considered a couple of oscillators 19 00:01:09,020 --> 00:01:13,440 with all the details, with damping, driven, et cetera, 20 00:01:13,440 --> 00:01:16,900 and it really becomes complicated. 21 00:01:16,900 --> 00:01:19,230 Now, what we are trying to do by doing 22 00:01:19,230 --> 00:01:23,460 problems is to get a feel, an understanding what goes on 23 00:01:23,460 --> 00:01:29,160 in a given situation, so one wants to focus on new features 24 00:01:29,160 --> 00:01:32,180 and try to simplify the mathematics as 25 00:01:32,180 --> 00:01:34,120 much as possible. 26 00:01:34,120 --> 00:01:37,330 And so in the problems are we doing now, 27 00:01:37,330 --> 00:01:42,710 I am intentionally making them even more ideal than before. 28 00:01:42,710 --> 00:01:45,640 Consider the first problem. 29 00:01:45,640 --> 00:01:52,660 Here is a highly idealized system 30 00:01:52,660 --> 00:01:56,680 which has 3 degrees of freedom. 31 00:01:56,680 --> 00:02:03,750 What we have is we have a single mass, 2m, two identical springs 32 00:02:03,750 --> 00:02:06,920 attached to two masses, each of mass m. 33 00:02:09,690 --> 00:02:11,680 This is completely idealized. 34 00:02:11,680 --> 00:02:14,550 We are assuming there's no friction, 35 00:02:14,550 --> 00:02:17,510 that these springs are ideal springs. 36 00:02:17,510 --> 00:02:19,080 In other words, they're massless, 37 00:02:19,080 --> 00:02:23,230 obey Hooke's laws with constant k. 38 00:02:23,230 --> 00:02:25,910 We are assuming the system is constrained, 39 00:02:25,910 --> 00:02:28,520 so it can only move in one direction, 40 00:02:28,520 --> 00:02:33,670 it can't move up and down or in or out from the board. 41 00:02:33,670 --> 00:02:40,480 So as I say, it's an idealized situation. 42 00:02:40,480 --> 00:02:46,970 And the question we want to answer is, for such a system, 43 00:02:46,970 --> 00:02:50,860 what are the normal mode frequencies? 44 00:02:50,860 --> 00:02:55,880 And you know from lectures given by Professor Walter Lewin 45 00:02:55,880 --> 00:03:00,710 that when you have coupled oscillators, in this case, 3, 46 00:03:00,710 --> 00:03:09,060 all right, one finds that there are very special oscillations, 47 00:03:09,060 --> 00:03:13,150 which we call normal mode oscillations, in which 48 00:03:13,150 --> 00:03:20,980 every part of the system is oscillating 49 00:03:20,980 --> 00:03:24,095 with the same frequency and phase. 50 00:03:28,720 --> 00:03:32,680 So first, we want to find out for this system, what 51 00:03:32,680 --> 00:03:36,505 are those three normal mode frequencies? 52 00:03:39,240 --> 00:03:44,100 And secondly, suppose I take this system, 53 00:03:44,100 --> 00:03:49,630 place these masses at some arbitrary positions, some 54 00:03:49,630 --> 00:03:53,870 maybe even moving, and let go, how would I 55 00:03:53,870 --> 00:04:00,380 predict where each mass would be at some later time? 56 00:04:00,380 --> 00:04:02,070 So those are the two questions I want 57 00:04:02,070 --> 00:04:05,730 to answer today for this problem. 58 00:04:10,010 --> 00:04:18,750 Now, in general, coupled oscillator problems 59 00:04:18,750 --> 00:04:21,850 are quite difficult. 60 00:04:21,850 --> 00:04:25,300 But there are situations in which 61 00:04:25,300 --> 00:04:29,890 one can use logic alone, well, logic 62 00:04:29,890 --> 00:04:33,920 and our understanding of coupled oscillators, 63 00:04:33,920 --> 00:04:38,280 to do most of the solution by hand-waving 64 00:04:38,280 --> 00:04:41,320 and not do any calculations. 65 00:04:41,320 --> 00:04:45,020 In this first example I'm doing in this problem, 66 00:04:45,020 --> 00:04:52,930 since the system has a lot of symmetry, 67 00:04:52,930 --> 00:04:56,670 I know from experience that under those conditions 68 00:04:56,670 --> 00:05:02,910 it might be possible to solve this, as I say, by logic alone 69 00:05:02,910 --> 00:05:06,230 and our understanding of coupled oscillators. 70 00:05:06,230 --> 00:05:09,380 So the issue is, can I guess what 71 00:05:09,380 --> 00:05:13,835 will be the motion in the normal modes? 72 00:05:16,480 --> 00:05:19,200 And the answer is, actually, if you stop and think 73 00:05:19,200 --> 00:05:23,370 for a second, at least the first two 74 00:05:23,370 --> 00:05:26,770 modes you can figure out rather easily. 75 00:05:26,770 --> 00:05:30,270 So here I'm showing you schematically 76 00:05:30,270 --> 00:05:34,130 what would be one of the modes of oscillation, 77 00:05:34,130 --> 00:05:35,780 one of the normal modes. 78 00:05:35,780 --> 00:05:41,120 Imagine that I first hold this mass fixed 79 00:05:41,120 --> 00:05:45,390 and I displace these symmetrically, one 80 00:05:45,390 --> 00:05:48,565 to the right, one to the left, and I let go. 81 00:05:51,400 --> 00:05:54,720 What you will find is, by symmetry, 82 00:05:54,720 --> 00:05:59,470 this spring will always be stretched by the same amount 83 00:05:59,470 --> 00:06:06,500 as this is compressed, and so the force on this mass 84 00:06:06,500 --> 00:06:07,550 will be 0. 85 00:06:07,550 --> 00:06:13,210 This spring will be pushing this mass exactly by the same amount 86 00:06:13,210 --> 00:06:16,290 as this one will be pulling it and vice versa. 87 00:06:16,290 --> 00:06:19,145 So there will never be any net force on this mass, 88 00:06:19,145 --> 00:06:22,420 and so they'll stay put. 89 00:06:22,420 --> 00:06:31,160 And these two will now behave as a simple harmonic oscillator 90 00:06:31,160 --> 00:06:38,040 consisting of a spring attached to a fixed wall and the mass m. 91 00:06:38,040 --> 00:06:40,420 So that's what this one will be doing, 92 00:06:40,420 --> 00:06:42,070 and that's what this will be doing. 93 00:06:42,070 --> 00:06:44,580 It will be out of phase by 180 degrees. 94 00:06:44,580 --> 00:06:48,520 Or what one normally says, it'll be in phase 95 00:06:48,520 --> 00:06:53,440 but the amplitude will be minus the other one. 96 00:06:53,440 --> 00:06:56,490 This, by now, you've had enough experience 97 00:06:56,490 --> 00:07:00,240 and I've done it here before, I can just 98 00:07:00,240 --> 00:07:04,460 write what will be that frequency of oscillation 99 00:07:04,460 --> 00:07:08,410 here, as we've seen many times. 100 00:07:08,410 --> 00:07:13,310 And, by the way, throughout my solving the problems here, 101 00:07:13,310 --> 00:07:19,150 I do not differentiate frequency and angular frequency, that you 102 00:07:19,150 --> 00:07:22,320 can see which I'm talking about is whether I'm using omega 103 00:07:22,320 --> 00:07:25,500 or F. Omega is the angular frequency, 104 00:07:25,500 --> 00:07:28,230 and there's just a fact of 2 pi between the two. 105 00:07:28,230 --> 00:07:31,970 But I'll use, interchangeably, I'll call this frequency. 106 00:07:31,970 --> 00:07:37,340 So this frequency of one of the normal modes is just k/m. 107 00:07:37,340 --> 00:07:40,710 So we found one normal mode frequency. 108 00:07:40,710 --> 00:07:42,850 Let's look for another one. 109 00:07:42,850 --> 00:07:48,376 We stare at this and we see, sure, there is another one. 110 00:07:48,376 --> 00:07:55,760 Another one is where I take, simply, these two masses 111 00:07:55,760 --> 00:07:59,540 and simultaneously pull them out, in this mass, 112 00:07:59,540 --> 00:08:03,220 away and let go. 113 00:08:03,220 --> 00:08:07,730 Effectively, these two will look like that. 114 00:08:07,730 --> 00:08:13,160 It's as if I had a mass of 2m attached to a mass of 2m 115 00:08:13,160 --> 00:08:18,790 by two springs, each of spring constant k. 116 00:08:18,790 --> 00:08:23,730 And as I pull this back, this will oscillate like this. 117 00:08:23,730 --> 00:08:26,470 Everything, both this mass and this mass, 118 00:08:26,470 --> 00:08:31,570 will be oscillating with the same frequency, same phase, 119 00:08:31,570 --> 00:08:33,789 although there's this minus sign. 120 00:08:33,789 --> 00:08:39,600 I can always replace the 180-degree phase shift 121 00:08:39,600 --> 00:08:43,240 by saying the amplitude is minus. 122 00:08:43,240 --> 00:08:47,170 So those are identical statements. 123 00:08:47,170 --> 00:08:51,308 So you have these two masses oscillating like that. 124 00:08:51,308 --> 00:08:52,670 OK? 125 00:08:52,670 --> 00:08:58,090 By the way, I've said that in the normal mode 126 00:08:58,090 --> 00:09:02,040 I want everything to move with the same frequency. 127 00:09:02,040 --> 00:09:04,170 Some of you may argue, hold on. 128 00:09:04,170 --> 00:09:05,680 Did I cheat on you? 129 00:09:05,680 --> 00:09:09,510 Here, this mass is not oscillating. 130 00:09:09,510 --> 00:09:12,110 So am I contradicting myself? 131 00:09:12,110 --> 00:09:14,490 And the answer is no. 132 00:09:14,490 --> 00:09:16,230 It's a diabolical answer. 133 00:09:16,230 --> 00:09:18,970 But I'll say, no, this is oscillating 134 00:09:18,970 --> 00:09:22,760 with this frequency, but with 0 amplitude. 135 00:09:22,760 --> 00:09:24,970 And you can't tell me that I'm wrong. 136 00:09:24,970 --> 00:09:27,810 It's not moving because the amplitude is 0, 137 00:09:27,810 --> 00:09:32,360 but it is oscillating with the same frequency. 138 00:09:32,360 --> 00:09:36,180 So now, what is the frequency of this oscillation? 139 00:09:36,180 --> 00:09:38,550 Again, I can do it in my head. 140 00:09:38,550 --> 00:09:46,450 I can imagine that the middle of this spring is not moving. 141 00:09:46,450 --> 00:09:52,650 And so this mass is a half a length of the spring 142 00:09:52,650 --> 00:09:55,500 and a mass 2 attached to it oscillating, 143 00:09:55,500 --> 00:09:57,440 and I can calculate the frequency 144 00:09:57,440 --> 00:09:59,150 of oscillation of that. 145 00:09:59,150 --> 00:10:02,700 Or if you prefer, I can do the following. 146 00:10:02,700 --> 00:10:09,180 I can move this mass out a distance dx and this one dx 147 00:10:09,180 --> 00:10:12,030 to the other side and calculate what 148 00:10:12,030 --> 00:10:14,450 will be the restoring force. 149 00:10:14,450 --> 00:10:19,890 Well, now each one has a spring constant k. 150 00:10:19,890 --> 00:10:21,600 There is two of them, so that's 2k. 151 00:10:25,140 --> 00:10:30,320 Although this mass has been moved by dx, this by dx, 152 00:10:30,320 --> 00:10:35,030 the springs have been extended by twice dx, so it's 2 dx. 153 00:10:35,030 --> 00:10:37,240 So that's the total restoring force 154 00:10:37,240 --> 00:10:41,440 by Hooke's law, which is equal minus 4k dx. 155 00:10:41,440 --> 00:10:42,610 All right? 156 00:10:42,610 --> 00:10:49,650 And so this will behave like a single mass, 157 00:10:49,650 --> 00:10:52,880 like this system, a single mass with a single spring 158 00:10:52,880 --> 00:10:56,770 where the spring has a spring constant 4k. 159 00:10:56,770 --> 00:10:58,940 The mass is 2m. 160 00:10:58,940 --> 00:11:05,450 And so the angular frequency, or frequency of this mode is 2k/m. 161 00:11:05,450 --> 00:11:07,660 So we've found two of them. 162 00:11:07,660 --> 00:11:12,330 Now, we know from our studies of coupled oscillators 163 00:11:12,330 --> 00:11:15,685 that for a system of 3 degrees of freedom, 164 00:11:15,685 --> 00:11:20,940 so it would be three masses, there are three normal modes. 165 00:11:20,940 --> 00:11:24,370 What is the third normal mode? 166 00:11:24,370 --> 00:11:27,210 And I intentionally did not write it 167 00:11:27,210 --> 00:11:30,955 down here because I wanted you for a second to think. 168 00:11:30,955 --> 00:11:35,920 Well, how else can this system move such 169 00:11:35,920 --> 00:11:45,591 that every mass is moving in phase with the same frequency? 170 00:11:45,591 --> 00:11:47,110 All right? 171 00:11:47,110 --> 00:11:50,810 And I tell you, when I first saw it, 172 00:11:50,810 --> 00:11:54,290 it took me a long time to figure out, 173 00:11:54,290 --> 00:12:00,030 and most people don't find it. 174 00:12:00,030 --> 00:12:03,400 And yet the answer is extremely simple. 175 00:12:03,400 --> 00:12:08,430 This answer is there is one more normal mode. 176 00:12:08,430 --> 00:12:17,600 It is one of almost infinite amplitude, but 0 frequency. 177 00:12:17,600 --> 00:12:24,290 Just the opposite to here, where this was 0 amplitude, OK, 178 00:12:24,290 --> 00:12:26,700 and the angular frequency doesn't even matter. 179 00:12:26,700 --> 00:12:32,480 Here it has very large amplitude, infinite in fact, 180 00:12:32,480 --> 00:12:34,100 but 0 frequency. 181 00:12:34,100 --> 00:12:38,770 What does such motion look like at any instant of time? 182 00:12:38,770 --> 00:12:41,530 It's moving with the same velocity. 183 00:12:41,530 --> 00:12:54,770 So if I take and if I consider the motion where 184 00:12:54,770 --> 00:13:01,100 each one of these is moving uniformly to the right 185 00:13:01,100 --> 00:13:04,240 with a constant velocity, that is 186 00:13:04,240 --> 00:13:10,200 a normal mode with 0 frequency. 187 00:13:10,200 --> 00:13:18,240 So the last one is omega C is equal to 0. 188 00:13:21,010 --> 00:13:21,920 OK? 189 00:13:21,920 --> 00:13:28,800 So we have now found the three normal modes. 190 00:13:28,800 --> 00:13:31,000 OK. 191 00:13:31,000 --> 00:13:31,650 All right. 192 00:13:31,650 --> 00:13:38,960 Now that I've got a piece of chalk, let's continue. 193 00:13:38,960 --> 00:13:41,980 So we've answered the first part of the question, what 194 00:13:41,980 --> 00:13:44,490 are the normal mode frequencies of the system? 195 00:13:44,490 --> 00:13:47,340 And I found you those three normal modes. 196 00:13:47,340 --> 00:13:49,970 Now, the next question we want to answer 197 00:13:49,970 --> 00:13:54,340 is, if, at any given instant of time, 198 00:13:54,340 --> 00:13:57,930 I know the positions and the velocities of the three masses, 199 00:13:57,930 --> 00:14:01,380 can I predict what will happen at the end? 200 00:14:01,380 --> 00:14:04,690 That's equivalent of saying, can I 201 00:14:04,690 --> 00:14:09,570 write equations for the positions of the masses 202 00:14:09,570 --> 00:14:11,400 as a function of time? 203 00:14:11,400 --> 00:14:15,700 For that I need to define for myself a coordinate system. 204 00:14:15,700 --> 00:14:18,585 So I'll say those are those three masses. 205 00:14:23,120 --> 00:14:26,510 All the motion is along the x direction, 206 00:14:26,510 --> 00:14:35,900 so there's just one variable, one variable for each mass, x. 207 00:14:35,900 --> 00:14:38,080 So the position of the first mass 208 00:14:38,080 --> 00:14:42,090 I'll call x1, position of the second mass x2, 209 00:14:42,090 --> 00:14:43,760 and of the third is x3. 210 00:14:43,760 --> 00:14:46,960 And the origin of coordinates I will 211 00:14:46,960 --> 00:14:51,070 take to be through the center of each mass 212 00:14:51,070 --> 00:14:54,580 at a time when these are in equilibrium. 213 00:14:54,580 --> 00:14:57,990 In other words, the spring is unstretched, et cetera. 214 00:14:57,990 --> 00:14:58,880 OK. 215 00:14:58,880 --> 00:15:07,630 So I can now, using the information we've obtained, 216 00:15:07,630 --> 00:15:17,060 write down the description of each mass in each mode. 217 00:15:17,060 --> 00:15:23,360 So in mode A, I know that the x1 is 0 all the time. 218 00:15:23,360 --> 00:15:26,690 It's not moving, OK? 219 00:15:26,690 --> 00:15:34,830 Now, in a normal mode, that's what we mean by a normal mode, 220 00:15:34,830 --> 00:15:38,020 each mass is moving with the same frequency 221 00:15:38,020 --> 00:15:42,240 and the same phase, and it's oscillating. 222 00:15:42,240 --> 00:15:49,900 So there will be some sinusoidal function of the mode frequency 223 00:15:49,900 --> 00:15:54,730 t plus this phase times some arbitrary amplitude. 224 00:15:57,480 --> 00:16:00,960 The other one, the last one, will also 225 00:16:00,960 --> 00:16:03,330 be oscillating in the normal mode, 226 00:16:03,330 --> 00:16:05,025 so everything will be the same. 227 00:16:05,025 --> 00:16:07,930 It will be cosine of omega A t plus phi 228 00:16:07,930 --> 00:16:11,320 A with some other amplitude. 229 00:16:11,320 --> 00:16:18,350 But from our analysis here, we know 230 00:16:18,350 --> 00:16:23,810 that these two are not both arbitrary. 231 00:16:23,810 --> 00:16:27,510 Because we said that in the normal mode, 232 00:16:27,510 --> 00:16:35,360 it was the case where the 2m mass was stationary, 233 00:16:35,360 --> 00:16:37,970 but these were going like this. 234 00:16:37,970 --> 00:16:40,340 That was the first mode. 235 00:16:40,340 --> 00:16:46,080 So whenever this one was going to the right a distance A, 236 00:16:46,080 --> 00:16:49,440 this one was going to the left a distance A. 237 00:16:49,440 --> 00:16:52,830 So the overall normalization is arbitrary. 238 00:16:52,830 --> 00:16:54,620 It can be anything. 239 00:16:54,620 --> 00:16:58,210 But these are not independent of the other, 240 00:16:58,210 --> 00:17:01,240 because we discovered that in that mode, 241 00:17:01,240 --> 00:17:06,589 this was the frequency, and the amplitudes were 242 00:17:06,589 --> 00:17:09,849 opposite to each other. 243 00:17:09,849 --> 00:17:17,630 So this describes the situation for this system 244 00:17:17,630 --> 00:17:22,650 if it's oscillating in the first normal mode. 245 00:17:22,650 --> 00:17:26,540 It's the most general description of that. 246 00:17:26,540 --> 00:17:29,490 Next, I will describe what it will 247 00:17:29,490 --> 00:17:32,350 do in the second normal mode. 248 00:17:32,350 --> 00:17:34,880 And now I can go much quicker. 249 00:17:34,880 --> 00:17:37,270 This time, it'll have the second normal mode 250 00:17:37,270 --> 00:17:40,750 frequency, different phase. 251 00:17:40,750 --> 00:17:44,980 But these have to be the same because all of these masses 252 00:17:44,980 --> 00:17:49,910 are moving in the same normal mode, same normal frequency. 253 00:17:49,910 --> 00:17:54,560 These amplitudes, overall, it's arbitrary. 254 00:17:54,560 --> 00:17:55,960 I can make it anything. 255 00:17:55,960 --> 00:17:58,600 It'll satisfy the equations. 256 00:17:58,600 --> 00:18:03,520 But I know that in the second normal mode, 257 00:18:03,520 --> 00:18:06,820 this one is moving in that direction and those two 258 00:18:06,820 --> 00:18:10,220 in the opposite direction with the same magnitude. 259 00:18:10,220 --> 00:18:14,220 So these two have to have the same size and magnitude, 260 00:18:14,220 --> 00:18:16,620 but this has to be opposite. 261 00:18:16,620 --> 00:18:21,460 So this describes the motion in the second mode. 262 00:18:21,460 --> 00:18:25,090 And finally, in the third normal mode, 263 00:18:25,090 --> 00:18:29,160 we said that it was 0 frequency. 264 00:18:29,160 --> 00:18:30,280 All right? 265 00:18:30,280 --> 00:18:32,990 And all that was happening, the three masses 266 00:18:32,990 --> 00:18:37,780 were moving with uniform velocity. 267 00:18:37,780 --> 00:18:41,850 So the positions will be some constant 268 00:18:41,850 --> 00:18:44,265 plus the velocity of that system. 269 00:18:51,310 --> 00:18:55,030 So this describes the system in the third mode, 270 00:18:55,030 --> 00:18:58,650 and these are the other two. 271 00:18:58,650 --> 00:19:17,870 Now, I know that, in general, each one of these masses 272 00:19:17,870 --> 00:19:25,990 can move in a superposition of the normal modes. 273 00:19:25,990 --> 00:19:37,390 So the most general expression for x1 of t 274 00:19:37,390 --> 00:19:45,430 has to be of the form which is the sum of its possible motion 275 00:19:45,430 --> 00:19:47,360 in each of the modes. 276 00:19:47,360 --> 00:19:51,660 So x1, so this is the mass 2m, will 277 00:19:51,660 --> 00:19:56,320 have possible motion in the first mode, which is 0. 278 00:19:56,320 --> 00:19:57,520 All right? 279 00:19:57,520 --> 00:20:13,570 So I'll write 0 plus B times cosine omega B t plus phi 280 00:20:13,570 --> 00:20:21,920 B, all right, plus C plus v. So that 281 00:20:21,920 --> 00:20:26,940 will be the most general description of the motion 282 00:20:26,940 --> 00:20:30,880 of the big mass, the 2m one. 283 00:20:30,880 --> 00:20:34,310 The B is arbitrary, this phase is arbitrary, 284 00:20:34,310 --> 00:20:37,990 this is arbitrary, and v. Those quantities 285 00:20:37,990 --> 00:20:42,986 will be determined by the initial conditions. 286 00:20:42,986 --> 00:20:44,280 Well, how about x2? 287 00:20:48,340 --> 00:21:08,190 Well, I know that if x1 is 0, x2 is A cosine omega A t 288 00:21:08,190 --> 00:21:26,250 plus phi A. When the x1 is this, then x2 is minus B cosine omega 289 00:21:26,250 --> 00:21:37,270 B t plus phi B, OK, and plus C plus v. OK? 290 00:21:37,270 --> 00:21:45,210 And finally, x3 of t, again now just following 291 00:21:45,210 --> 00:21:50,360 the same pattern, this is minus A cosine omega A t 292 00:21:50,360 --> 00:21:59,620 plus phi A. This one is minus B cosine omega B 293 00:21:59,620 --> 00:22:15,940 t plus phi B plus C plus v. 294 00:22:15,940 --> 00:22:19,780 And I'm almost home. 295 00:22:19,780 --> 00:22:22,650 I have described the motion of each one. 296 00:22:27,015 --> 00:22:31,620 For each particle, it is oscillating 297 00:22:31,620 --> 00:22:34,410 in a superposition of its normal modes. 298 00:22:38,230 --> 00:22:45,460 And the amplitudes are arbitrary within the constraint 299 00:22:45,460 --> 00:22:50,650 that the ratios between them is determined 300 00:22:50,650 --> 00:22:56,070 by the constraints of the system. 301 00:22:56,070 --> 00:23:00,450 So now the question is, can I predict the future? 302 00:23:00,450 --> 00:23:02,720 Well, in order to predict the future, 303 00:23:02,720 --> 00:23:06,645 say, where will particle 3 be at time t? 304 00:23:06,645 --> 00:23:11,876 I need to know the value of these arbitrary constants. 305 00:23:14,420 --> 00:23:16,640 Omega is not an arbitrary constant. 306 00:23:16,640 --> 00:23:17,740 We found it. 307 00:23:17,740 --> 00:23:19,630 Omega A is here. 308 00:23:19,630 --> 00:23:20,580 OK? 309 00:23:20,580 --> 00:23:26,370 This is arbitrary, this is arbitrary, this, and that. 310 00:23:26,370 --> 00:23:30,670 So there are six arbitrary constants. 311 00:23:30,670 --> 00:23:32,280 How do I find them? 312 00:23:32,280 --> 00:23:36,730 They are determined by the initial conditions. 313 00:23:36,730 --> 00:23:40,420 And you know, fortunately, or else it 314 00:23:40,420 --> 00:23:44,660 would mean we've made a mistake, we have three conditions. 315 00:23:47,280 --> 00:23:51,560 At the beginning, someone has to tell me 316 00:23:51,560 --> 00:23:58,000 where each mass was, the location, and with what 317 00:23:58,000 --> 00:24:01,810 velocity is it moving at that time. 318 00:24:01,810 --> 00:24:09,290 So I can write these three equations at, say, 319 00:24:09,290 --> 00:24:14,920 with t equals 0, at t equals 0, and equate this 320 00:24:14,920 --> 00:24:21,220 to the position of x1 of t equals 0, wherever it is. 321 00:24:21,220 --> 00:24:26,220 When the t is 0, I make it equal to the position of x2 at t 322 00:24:26,220 --> 00:24:28,170 equals 0. 323 00:24:28,170 --> 00:24:32,890 When t is 0, this tells me has to be equal to where x3 is. 324 00:24:32,890 --> 00:24:34,650 That's three equations. 325 00:24:34,650 --> 00:24:36,680 I can differentiate each with respect 326 00:24:36,680 --> 00:24:40,650 to time to get the velocities, and likewise, 327 00:24:40,650 --> 00:24:42,440 get three more equations. 328 00:24:42,440 --> 00:24:46,500 So I'm going to end up with six algebraic equations. 329 00:24:46,500 --> 00:24:49,210 And you, as well as I, know that if we 330 00:24:49,210 --> 00:24:53,540 have six algebraic equations, I can solve for six unknowns. 331 00:24:53,540 --> 00:24:56,550 And that will give me those six unknowns. 332 00:24:56,550 --> 00:25:00,010 So once I've used the initial conditions, 333 00:25:00,010 --> 00:25:03,880 I can then find these arbitrary constants. 334 00:25:03,880 --> 00:25:07,420 They're no longer arbitrary for a specific situation. 335 00:25:07,420 --> 00:25:11,960 And I can predict the future, what will happen in this case. 336 00:25:11,960 --> 00:25:12,610 OK? 337 00:25:12,610 --> 00:25:19,430 So that's the way one would solve a problem when 338 00:25:19,430 --> 00:25:21,570 you are lucky enough that there is 339 00:25:21,570 --> 00:25:25,340 sufficient symmetry in the original situation 340 00:25:25,340 --> 00:25:29,510 so you can guess the normal modes. 341 00:25:29,510 --> 00:25:34,730 The next problem I'll do, I will do one where you cannot guess 342 00:25:34,730 --> 00:25:36,110 the normal modes. 343 00:25:36,110 --> 00:25:38,160 What do you do under those circumstances? 344 00:25:38,160 --> 00:25:40,180 And you won't be surprised that it's 345 00:25:40,180 --> 00:25:46,350 going to be mathematically much, much harder, but conceptually 346 00:25:46,350 --> 00:25:47,500 no harder. 347 00:25:47,500 --> 00:25:50,950 So now we'll move to the next problem. 348 00:25:50,950 --> 00:25:51,720 Here it is. 349 00:25:51,720 --> 00:25:54,050 But in order to be able to do this problem, 350 00:25:54,050 --> 00:25:55,600 I need more board space. 351 00:25:55,600 --> 00:25:59,281 So we're going to erase the boards and continue from there. 352 00:25:59,281 --> 00:25:59,780 OK. 353 00:25:59,780 --> 00:26:03,130 So now let's go and do the second problem. 354 00:26:03,130 --> 00:26:07,190 And it's going to be a problem with no symmetry, 355 00:26:07,190 --> 00:26:10,760 so we can't guess the answer like we did before. 356 00:26:10,760 --> 00:26:12,870 And I'll tell you what I decided to do. 357 00:26:12,870 --> 00:26:15,770 You remember Professor Walter Lewin 358 00:26:15,770 --> 00:26:21,660 in his lectures discussed a double pendulum. 359 00:26:21,660 --> 00:26:23,400 In other words, a situation where 360 00:26:23,400 --> 00:26:27,990 you had a string, a mass, another string, and mass. 361 00:26:27,990 --> 00:26:30,200 He gave you the answer, but he didn't prove it. 362 00:26:30,200 --> 00:26:32,590 He said it's hard and lengthy, et cetera. 363 00:26:32,590 --> 00:26:35,140 And I thought that would be a perfect example 364 00:26:35,140 --> 00:26:39,660 to do here from first principles. 365 00:26:39,660 --> 00:26:44,930 So let me just describe in detail 366 00:26:44,930 --> 00:26:47,230 the problem we're trying to solve. 367 00:26:47,230 --> 00:26:51,230 What we have is, again, an idealized situation. 368 00:26:51,230 --> 00:26:57,250 We have a simple pendulum with a string of length L, 369 00:26:57,250 --> 00:26:59,960 attached to it a mass m. 370 00:26:59,960 --> 00:27:03,850 Attached to this mass at one end is another string of length 371 00:27:03,850 --> 00:27:06,930 L to another mass m. 372 00:27:06,930 --> 00:27:09,930 We'll again idealize the situation. 373 00:27:09,930 --> 00:27:12,200 So the assumptions we are making, 374 00:27:12,200 --> 00:27:18,460 that this string and this string is massless, which, by the way, 375 00:27:18,460 --> 00:27:22,530 is equivalent to saying that it's taut and straight all 376 00:27:22,530 --> 00:27:23,310 the time. 377 00:27:23,310 --> 00:27:25,970 You can think about that for yourself. 378 00:27:25,970 --> 00:27:31,350 Next, these masses are point masses. 379 00:27:31,350 --> 00:27:34,330 And we're assuming no friction. 380 00:27:34,330 --> 00:27:37,980 Furthermore, as we've done in the past, 381 00:27:37,980 --> 00:27:41,500 we'll assume that all displacements 382 00:27:41,500 --> 00:27:47,260 when this is oscillating, at all times the angle that the string 383 00:27:47,260 --> 00:27:52,220 makes with the vertical is always such that we can ignore 384 00:27:52,220 --> 00:27:56,520 the difference between a sine theta, theta, 385 00:27:56,520 --> 00:28:01,695 or [INAUDIBLE] can make the assumption that cosine theta is 386 00:28:01,695 --> 00:28:03,090 0. 387 00:28:03,090 --> 00:28:06,410 By the way, the fact that we'll make this assumption 388 00:28:06,410 --> 00:28:09,290 is equivalent to saying that we'll 389 00:28:09,290 --> 00:28:13,080 ignore vertical motion of these masses. 390 00:28:13,080 --> 00:28:16,780 The motion is sufficiently small that vertically the masses 391 00:28:16,780 --> 00:28:18,210 are not moving. 392 00:28:18,210 --> 00:28:22,780 We can assume the acceleration vertically is 0, et cetera. 393 00:28:22,780 --> 00:28:23,340 OK? 394 00:28:23,340 --> 00:28:24,850 This is in the vertical plane. 395 00:28:24,850 --> 00:28:26,420 Gravity is down. 396 00:28:26,420 --> 00:28:27,180 OK? 397 00:28:27,180 --> 00:28:32,440 And what we are trying to derive for this, 398 00:28:32,440 --> 00:28:37,980 predict what are the normal mode frequencies of this. 399 00:28:37,980 --> 00:28:40,170 And once we do that, of course, we 400 00:28:40,170 --> 00:28:43,610 can use the same kind of technique as we did before. 401 00:28:43,610 --> 00:28:45,800 Once we've managed to find the normal modes 402 00:28:45,800 --> 00:28:47,810 and the frequencies, we can always 403 00:28:47,810 --> 00:28:50,450 write the most general expression. 404 00:28:50,450 --> 00:28:53,920 And then using the boundary conditions, initial conditions, 405 00:28:53,920 --> 00:28:57,440 predict what this will do as a function of time. 406 00:28:57,440 --> 00:28:58,774 OK, so let's get going. 407 00:29:05,280 --> 00:29:05,780 OK. 408 00:29:05,780 --> 00:29:07,660 Now, we are not guessing. 409 00:29:07,660 --> 00:29:09,120 We are not using logic. 410 00:29:09,120 --> 00:29:11,950 We are following the kind of prescription 411 00:29:11,950 --> 00:29:14,450 I've told you in the past. 412 00:29:14,450 --> 00:29:17,020 This is our description of the situation. 413 00:29:17,020 --> 00:29:19,700 And all the new words and ordinary language, we 414 00:29:19,700 --> 00:29:23,820 must now translate it into mathematics. 415 00:29:23,820 --> 00:29:26,230 We've got to describe this problem 416 00:29:26,230 --> 00:29:29,580 in terms of mathematical equations. 417 00:29:29,580 --> 00:29:33,600 So step one is we redraw this and define 418 00:29:33,600 --> 00:29:35,380 some coordinate system. 419 00:29:35,380 --> 00:29:39,690 So we will say that the angle this first string makes 420 00:29:39,690 --> 00:29:43,200 with respect to the vertical is theta 1. 421 00:29:43,200 --> 00:29:45,980 This angle the second string makes with respect 422 00:29:45,980 --> 00:29:48,420 to the vertical is theta 2. 423 00:29:48,420 --> 00:29:51,420 We will take the vertical through this pivot 424 00:29:51,420 --> 00:29:56,140 here as our origin of coordinate x. 425 00:29:56,140 --> 00:30:02,870 x equals 0 here, is defined here. 426 00:30:02,870 --> 00:30:05,190 From the point of your motion of the masses, 427 00:30:05,190 --> 00:30:06,960 it's a one-dimensional problem. 428 00:30:06,960 --> 00:30:10,750 The masses only move along the x-axis. 429 00:30:10,750 --> 00:30:13,430 So we'll define this distance as x1, 430 00:30:13,430 --> 00:30:17,440 and we'll define this distance as x2. 431 00:30:17,440 --> 00:30:22,720 Now we have to use the laws of physics, Newtonian mechanics, 432 00:30:22,720 --> 00:30:29,390 to derive the equations of motion for the two masses. 433 00:30:29,390 --> 00:30:36,010 So I draw a force diagram separately for the two masses. 434 00:30:36,010 --> 00:30:39,930 So the mass is there, what forces act on that mass? 435 00:30:39,930 --> 00:30:43,790 Well, I have to look at the mass and see what's attached to it. 436 00:30:43,790 --> 00:30:46,670 There is a string here attached to it. 437 00:30:46,670 --> 00:30:49,280 It's taut, so there will be a tension in it. 438 00:30:49,280 --> 00:30:53,890 So along this string, there will be tension T1, 439 00:30:53,890 --> 00:30:58,690 which will exert a force T1 along this string. 440 00:30:58,690 --> 00:31:02,100 This string is attached to that mass. 441 00:31:02,100 --> 00:31:06,030 It has a tension, so it's pulling on this mass. 442 00:31:06,030 --> 00:31:08,630 That tension I call T2, so there will 443 00:31:08,630 --> 00:31:13,990 be a force along that direction T2. 444 00:31:13,990 --> 00:31:16,820 This sits in a gravitational field. 445 00:31:16,820 --> 00:31:19,360 Gravity acts on this mass. 446 00:31:19,360 --> 00:31:23,330 There is a force downwards due to gravity, Fg. 447 00:31:26,330 --> 00:31:29,660 The sum of these forces, by Newton's laws, 448 00:31:29,660 --> 00:31:33,460 must equal to the mass times the acceleration. 449 00:31:33,460 --> 00:31:37,810 That acceleration, in this approximation that is not 450 00:31:37,810 --> 00:31:40,550 moving up, is horizontally, and it's 451 00:31:40,550 --> 00:31:43,920 the second derivative of x1. 452 00:31:43,920 --> 00:31:50,160 Similarly, for the second mass, again, I'll go now faster. 453 00:31:50,160 --> 00:31:52,390 There is this mass m, and this tension 454 00:31:52,390 --> 00:31:57,636 exerts a force here of T2 and the gravity on it Fg. 455 00:31:57,636 --> 00:32:00,310 The only subtlety is, why did I say 456 00:32:00,310 --> 00:32:03,350 this force is equal to that force? 457 00:32:03,350 --> 00:32:04,650 Think about it for a second. 458 00:32:04,650 --> 00:32:08,490 Why should the two forces be equal? 459 00:32:08,490 --> 00:32:12,870 Why is the tension at both ends of the string equal? 460 00:32:12,870 --> 00:32:16,756 And the answer is, actually, a subtle one. 461 00:32:16,756 --> 00:32:20,740 It's equal because we made the assumption that the string has 462 00:32:20,740 --> 00:32:28,240 no mass, and there can be no net force on an object of 0 mass. 463 00:32:28,240 --> 00:32:31,190 Because if there was, that object would disappear, 464 00:32:31,190 --> 00:32:33,650 would have an infinite acceleration. 465 00:32:33,650 --> 00:32:37,620 So if you treat the string as a mass, 466 00:32:37,620 --> 00:32:39,930 there cannot be net force on it. 467 00:32:39,930 --> 00:32:44,780 And so the force of each of these masses 468 00:32:44,780 --> 00:32:49,460 must be pulling on this string with exactly the same force 469 00:32:49,460 --> 00:32:51,031 equal and opposite. 470 00:32:51,031 --> 00:32:51,530 OK? 471 00:32:51,530 --> 00:32:55,020 And we're using the third law to equate those forces. 472 00:32:55,020 --> 00:32:58,770 So the net result is this T2 is the same as that, 473 00:32:58,770 --> 00:33:02,850 but in opposite direction, and this is the mass of that. 474 00:33:02,850 --> 00:33:03,710 OK. 475 00:33:03,710 --> 00:33:07,410 So these are the force diagrams. 476 00:33:07,410 --> 00:33:11,020 Using now Newton's laws of motion, 477 00:33:11,020 --> 00:33:15,220 I can translate this into equations of motion. 478 00:33:15,220 --> 00:33:18,810 So let me consider the horizontal motion 479 00:33:18,810 --> 00:33:20,950 of each mass separately. 480 00:33:20,950 --> 00:33:26,050 First this mass, and so its mass times acceleration 481 00:33:26,050 --> 00:33:31,320 is equal to the horizontal force of T2. 482 00:33:31,320 --> 00:33:35,690 And remember that this angle here is theta 2. 483 00:33:35,690 --> 00:33:38,600 And we see that the force due to T2, 484 00:33:38,600 --> 00:33:40,930 and there's a T2 sine theta 2. 485 00:33:40,930 --> 00:33:45,150 And the force due to T1 is minus T1 sine theta 1 486 00:33:45,150 --> 00:33:47,530 because it's in the opposite direction. 487 00:33:47,530 --> 00:33:50,190 For this, the only horizontal component 488 00:33:50,190 --> 00:33:53,180 is the force horizontal component of T2. 489 00:33:53,180 --> 00:33:58,800 And x2 double dot is equal to minus T2 sine theta 2. 490 00:33:58,800 --> 00:34:01,350 OK, that's horizontal motion. 491 00:34:01,350 --> 00:34:06,190 Applying Newton's laws of motion to the vertical motion, 492 00:34:06,190 --> 00:34:12,170 we said that, because cosine theta is 0 or approximately 0, 493 00:34:12,170 --> 00:34:14,250 the masses are not moving up and down. 494 00:34:14,250 --> 00:34:16,739 There's no acceleration of the masses. 495 00:34:16,739 --> 00:34:20,800 So the vertical acceleration of the first mass 496 00:34:20,800 --> 00:34:28,570 is 0 must be equal to the vertical component of T1, which 497 00:34:28,570 --> 00:34:33,219 is T1 cosine theta 1, minus the vertical component of this, 498 00:34:33,219 --> 00:34:40,940 which is minus T2 cosine theta 2 minus mg. 499 00:34:40,940 --> 00:34:41,440 All right. 500 00:34:41,440 --> 00:34:47,219 Similarly for the second mass, it's 0 is equal to this. 501 00:34:47,219 --> 00:34:51,850 0 must equal to the vertical component 502 00:34:51,850 --> 00:34:58,390 of T1 minus the vertical component of T2 503 00:34:58,390 --> 00:35:02,170 minus the force of gravity down. 504 00:35:02,170 --> 00:35:06,040 And similarly for the second mass, 505 00:35:06,040 --> 00:35:09,170 we know that vertically the acceleration is 0. 506 00:35:09,170 --> 00:35:12,560 That must be equal to the vertical component of T2 507 00:35:12,560 --> 00:35:18,650 minus the gravitational force pulling down. 508 00:35:18,650 --> 00:35:26,790 Now I can make use of what we made the assumption-- oops, 509 00:35:26,790 --> 00:35:29,130 I see there is an error. 510 00:35:29,130 --> 00:35:31,540 If you were in this room, I'm sure you 511 00:35:31,540 --> 00:35:35,020 would have corrected me. 512 00:35:35,020 --> 00:35:38,540 For very small angles, the sine of an angle 513 00:35:38,540 --> 00:35:40,790 is approximately equal to the angle. 514 00:35:40,790 --> 00:35:44,740 But for small angles, the cosine is 1. 515 00:35:44,740 --> 00:35:48,250 This is an error, and I apologize for that. 516 00:35:48,250 --> 00:35:50,450 But that's the approximation we are making. 517 00:35:50,450 --> 00:35:54,580 And it is this approximation which 518 00:35:54,580 --> 00:36:00,810 is equivalent to saying that we can ignore the vertical motion. 519 00:36:00,810 --> 00:36:02,200 All right. 520 00:36:02,200 --> 00:36:07,230 So with the assumption that the cosines are all 1, 521 00:36:07,230 --> 00:36:12,470 I can, from these two equations, I clearly 522 00:36:12,470 --> 00:36:18,430 derive that T2 must equal to mg, and T1 equals twice mg. 523 00:36:18,430 --> 00:36:20,580 Actually, it makes sense. 524 00:36:20,580 --> 00:36:23,770 Imagine this is hanging completely vertically. 525 00:36:23,770 --> 00:36:28,210 Obviously, this string, the upper part of the string, 526 00:36:28,210 --> 00:36:30,875 is supporting not only this mass but also that. 527 00:36:30,875 --> 00:36:35,530 It's supporting 2m masses, while this string is only 528 00:36:35,530 --> 00:36:36,640 supporting 1. 529 00:36:36,640 --> 00:36:40,440 So that's consistent with what we see, 530 00:36:40,440 --> 00:36:45,230 that the top string has twice the tension and the lower 531 00:36:45,230 --> 00:36:47,780 one half the tension. 532 00:36:47,780 --> 00:36:50,640 Having determined T1 and T2, we can now 533 00:36:50,640 --> 00:36:56,170 go back into our equation, replace the T1 and T2. 534 00:36:56,170 --> 00:36:58,830 We also can replace-- we know what 535 00:36:58,830 --> 00:37:02,440 sine theta 1 and sine theta 2 are. 536 00:37:02,440 --> 00:37:04,440 We can replace those. 537 00:37:04,440 --> 00:37:06,990 And we end up-- the two equations 538 00:37:06,990 --> 00:37:10,750 of motions are written here. 539 00:37:10,750 --> 00:37:11,770 All right? 540 00:37:11,770 --> 00:37:15,750 So here is the equation of motion for x1. 541 00:37:15,750 --> 00:37:19,380 And the second one, the equation for motion of x2, 542 00:37:19,380 --> 00:37:21,380 is written here. 543 00:37:21,380 --> 00:37:22,140 OK? 544 00:37:22,140 --> 00:37:27,200 So these are the equations of motion, 545 00:37:27,200 --> 00:37:37,930 all right, which we have to solve. 546 00:37:37,930 --> 00:37:46,300 Now, to simplify the algebra, let me define the quantity g/l 547 00:37:46,300 --> 00:37:49,525 by omega squared over 2. 548 00:37:49,525 --> 00:37:51,140 And you'll recognize this. 549 00:37:51,140 --> 00:37:52,240 This is not [INAUDIBLE]. 550 00:37:52,240 --> 00:37:54,010 I'm using that terminology. 551 00:37:54,010 --> 00:37:56,620 This is the frequency of oscillation 552 00:37:56,620 --> 00:38:01,140 of a mass on a string of length l. 553 00:38:01,140 --> 00:38:02,260 All right? 554 00:38:02,260 --> 00:38:08,930 So here are our two differential equations, the two equations 555 00:38:08,930 --> 00:38:14,620 of motion, one for x1 and one for x2. 556 00:38:14,620 --> 00:38:17,980 When we had one mass, one harmonic oscillator, 557 00:38:17,980 --> 00:38:21,420 we had a single second-order differential equation. 558 00:38:21,420 --> 00:38:25,070 We now have two masses, so we have 559 00:38:25,070 --> 00:38:27,740 two second-order differential equations. 560 00:38:27,740 --> 00:38:28,490 These are it. 561 00:38:28,490 --> 00:38:30,970 They are coupled differential equations. 562 00:38:30,970 --> 00:38:38,860 You see, the derivatives of x1 is related both to x1 and x2. 563 00:38:38,860 --> 00:38:43,870 The second derivative of x2 is related both to x1 and to x2. 564 00:38:43,870 --> 00:38:47,210 So these are two coupled differential equations. 565 00:38:47,210 --> 00:38:48,220 OK? 566 00:38:48,220 --> 00:38:51,655 This is the end of step 1. 567 00:38:51,655 --> 00:38:55,370 What we succeeded in, we've translated 568 00:38:55,370 --> 00:39:00,610 the physical situation into mathematics. 569 00:39:00,610 --> 00:39:06,580 These two second-order differential equations 570 00:39:06,580 --> 00:39:11,285 describe exactly that idealized situation we had. 571 00:39:14,050 --> 00:39:17,280 So if I now want to answer the question, what 572 00:39:17,280 --> 00:39:19,770 will be the motion of those masses, 573 00:39:19,770 --> 00:39:22,240 I have to go into the world of mathematics. 574 00:39:22,240 --> 00:39:25,630 I have to solve these equations. 575 00:39:25,630 --> 00:39:28,290 And life is not as easy as it was 576 00:39:28,290 --> 00:39:31,600 for a single second-order differential equation. 577 00:39:31,600 --> 00:39:33,720 It's more complicated. 578 00:39:33,720 --> 00:39:39,060 But here I will use my general knowledge 579 00:39:39,060 --> 00:39:44,040 of what happens when you have coupled oscillators. 580 00:39:44,040 --> 00:39:51,380 I know that the general solution of coupled oscillations 581 00:39:51,380 --> 00:39:57,410 is a superposition of normal modes. 582 00:39:57,410 --> 00:40:00,620 Here I have 2 degrees of freedom, 583 00:40:00,620 --> 00:40:03,020 two second-order differential equations. 584 00:40:03,020 --> 00:40:05,940 There will be two normal modes. 585 00:40:05,940 --> 00:40:09,490 If I succeed in finding them, I will 586 00:40:09,490 --> 00:40:13,890 have found the most general solution to this problem, 587 00:40:13,890 --> 00:40:17,220 because it will be the sum of the two normal modes. 588 00:40:17,220 --> 00:40:24,900 So I will now do it by trial. 589 00:40:24,900 --> 00:40:30,630 In a normal mode, we know that x1 590 00:40:30,630 --> 00:40:34,230 will be oscillating with a single frequency 591 00:40:34,230 --> 00:40:39,500 omega, some phase phi, some amplitude A1. 592 00:40:39,500 --> 00:40:41,820 This will be a solution to these equations. 593 00:40:44,370 --> 00:40:47,250 If this is a solution to those equations, 594 00:40:47,250 --> 00:40:51,280 the other mass must also be oscillating 595 00:40:51,280 --> 00:40:55,165 in the same normal mode, meaning with the same frequency 596 00:40:55,165 --> 00:40:58,080 and same phase. 597 00:40:58,080 --> 00:41:00,260 It'll have some arbitrary amplitude. 598 00:41:02,860 --> 00:41:05,820 So I don't know what A1 is, and I don't know what omega is. 599 00:41:05,820 --> 00:41:07,030 I don't know what phi is. 600 00:41:07,030 --> 00:41:12,600 But I know that the solutions to these equations 601 00:41:12,600 --> 00:41:16,980 must be of this form. 602 00:41:16,980 --> 00:41:18,140 OK? 603 00:41:18,140 --> 00:41:29,780 So let me now try to find these various constants. 604 00:41:29,780 --> 00:41:37,274 Well, if this and that satisfies these equations-- 605 00:41:37,274 --> 00:41:39,190 and it has to because that's what it's saying, 606 00:41:39,190 --> 00:41:41,470 these are the solutions of those equations-- 607 00:41:41,470 --> 00:41:49,440 then I can take these, x1 and x2, plug it into this equation. 608 00:41:49,440 --> 00:41:51,800 In other words, calculate the second derivative 609 00:41:51,800 --> 00:41:54,780 of x1, et cetera, calculate second, 610 00:41:54,780 --> 00:41:58,750 and I'll end up with two equations. 611 00:42:02,415 --> 00:42:02,915 OK? 612 00:42:06,640 --> 00:42:11,270 So I won't bother to go in great detail here. 613 00:42:11,270 --> 00:42:17,560 But just to start off with, x1 double dot, 614 00:42:17,560 --> 00:42:23,010 I have to differentiate A1 cosine omega t twice. 615 00:42:23,010 --> 00:42:34,830 So I get minus omega squared A1 times cosine omega t plus phi. 616 00:42:34,830 --> 00:42:40,220 If I take the next term, I get this, and the third term that. 617 00:42:40,220 --> 00:42:44,470 And in each case, if you notice, since it'll 618 00:42:44,470 --> 00:42:47,730 be multiplied by the same cosine function, 619 00:42:47,730 --> 00:42:51,210 I've just canceled in my head the cosine function there. 620 00:42:51,210 --> 00:42:52,270 All right? 621 00:42:52,270 --> 00:42:53,540 Just saving time. 622 00:42:53,540 --> 00:42:58,360 Similarly, next equation, I end up with this. 623 00:42:58,360 --> 00:43:04,190 So if my guessed functions, and I'm returning to this, 624 00:43:04,190 --> 00:43:08,810 if these two satisfy those equations, 625 00:43:08,810 --> 00:43:16,390 then here these algebraic equations must be satisfied. 626 00:43:16,390 --> 00:43:18,125 But you'll notice something interesting. 627 00:43:20,930 --> 00:43:24,010 If I take the first of these equations, 628 00:43:24,010 --> 00:43:30,000 it boils down to A1 times this quantity plus A2 629 00:43:30,000 --> 00:43:33,160 to [? then ?] this quantity equal to 0. 630 00:43:33,160 --> 00:43:36,320 And the second equation boils down to this. 631 00:43:39,580 --> 00:43:44,450 If this is true, then this must be true. 632 00:43:44,450 --> 00:43:49,850 It's A1 divided by A2 has to be equal to that. 633 00:43:49,850 --> 00:43:54,940 If this is true, then A1 divided by A2 634 00:43:54,940 --> 00:43:56,590 has got to be equal to this. 635 00:43:59,300 --> 00:44:02,960 That means that this has to be equal to that, 636 00:44:02,960 --> 00:44:06,460 or else I have an inconsistency. 637 00:44:06,460 --> 00:44:13,660 So what I have found is that, in general, my guess 638 00:44:13,660 --> 00:44:15,020 is not a good one. 639 00:44:15,020 --> 00:44:19,010 In general, these equations would not 640 00:44:19,010 --> 00:44:24,580 satisfy my second-order differential equations. 641 00:44:24,580 --> 00:44:30,660 There is only under very special circumstances 642 00:44:30,660 --> 00:44:33,470 that they do satisfy it, and that 643 00:44:33,470 --> 00:44:37,470 is if this is equal to that. 644 00:44:40,440 --> 00:44:42,824 In general, they will not be equal. 645 00:44:46,460 --> 00:44:50,410 So under what conditions will these be equal? 646 00:44:50,410 --> 00:44:51,850 Well, let me force them. 647 00:44:51,850 --> 00:44:55,760 Let me say this is equal to that and see 648 00:44:55,760 --> 00:44:59,680 what it means for omega. 649 00:44:59,680 --> 00:45:03,890 I originally took omega to be an unknown quantity. 650 00:45:03,890 --> 00:45:08,060 What I am now seeing, that those functions 651 00:45:08,060 --> 00:45:13,430 would work only for specific values of omega. 652 00:45:13,430 --> 00:45:18,050 So I'm going to solve this equation for omega 653 00:45:18,050 --> 00:45:24,090 and see for what values of omega do I get a consistent solution. 654 00:45:26,620 --> 00:45:29,820 And this is a hard grind, unfortunately. 655 00:45:29,820 --> 00:45:34,880 If I take this, multiply it out, I 656 00:45:34,880 --> 00:45:39,730 get a quadratic equation in omega squared. 657 00:45:39,730 --> 00:45:43,200 If this quadratic equation is satisfied, 658 00:45:43,200 --> 00:45:49,790 then all these are self-consistent. 659 00:45:49,790 --> 00:45:50,290 All right? 660 00:45:50,290 --> 00:45:56,170 You know how to solve a quadratic equation, 661 00:45:56,170 --> 00:46:05,430 and this is a quadratic equation in omega squared. 662 00:46:05,430 --> 00:46:08,530 Omega square will have to be equal to minus 663 00:46:08,530 --> 00:46:13,350 this plus or minus the square root of this, 664 00:46:13,350 --> 00:46:16,840 4 times this times that, divided by 2a, right? 665 00:46:16,840 --> 00:46:19,960 This is the standard formula for the solution 666 00:46:19,960 --> 00:46:21,840 of a quadratic equation. 667 00:46:26,090 --> 00:46:28,280 And there's a plus or minus. 668 00:46:28,280 --> 00:46:32,970 If I calculate this out, I get that omega squared 669 00:46:32,970 --> 00:46:36,470 has to be equal to 2 plus or minus 670 00:46:36,470 --> 00:46:40,680 the square root of 2 times omega 0 squared. 671 00:46:40,680 --> 00:46:41,690 All right? 672 00:46:41,690 --> 00:46:49,200 Now, omega 0 squared, earlier on, I defined to be g/l. 673 00:46:49,200 --> 00:46:53,830 So omega squared has to be this. 674 00:46:53,830 --> 00:46:57,030 So let's stop for a second and think 675 00:46:57,030 --> 00:47:00,560 and review what we've done. 676 00:47:00,560 --> 00:47:06,370 We found that our physical situation 677 00:47:06,370 --> 00:47:13,690 can be described by these two coupled differential equations. 678 00:47:13,690 --> 00:47:20,666 We looked for solutions, which are normal modes where both x1 679 00:47:20,666 --> 00:47:25,990 and x2 is oscillating with same frequency and phase. 680 00:47:25,990 --> 00:47:33,700 And we found that it was possible to find solution 681 00:47:33,700 --> 00:47:42,730 of this form to these equations if and only if omega squared 682 00:47:42,730 --> 00:47:46,210 has one of two possible values. 683 00:47:46,210 --> 00:47:48,050 I've rewritten them here. 684 00:47:48,050 --> 00:47:53,020 One is 2 plus root 2 times g/l, and the other's 685 00:47:53,020 --> 00:47:54,890 2 minus root 2 g/l. 686 00:47:57,420 --> 00:48:04,900 With these values, these two equations satisfies that. 687 00:48:04,900 --> 00:48:10,850 So what we found is the two normal modes. 688 00:48:10,850 --> 00:48:17,420 Since this system consists of the coupled oscillators, 689 00:48:17,420 --> 00:48:21,290 in other words, two oscillators coupled with each other, 690 00:48:21,290 --> 00:48:23,960 we know there are two normal modes. 691 00:48:23,960 --> 00:48:30,180 So these are the normal mode frequencies in this situation. 692 00:48:30,180 --> 00:48:33,440 And by the way, I'm delighted to see 693 00:48:33,440 --> 00:48:38,430 that, if you remember Professor Walter Lewin's lectures, 694 00:48:38,430 --> 00:48:41,830 he actually quoted these numbers. 695 00:48:41,830 --> 00:48:42,580 So he was right. 696 00:48:42,580 --> 00:48:45,270 He did not make a mistake, as we've just seen. 697 00:48:45,270 --> 00:48:46,280 OK? 698 00:48:46,280 --> 00:48:48,960 These are the two normal mode frequencies. 699 00:48:48,960 --> 00:48:51,830 Now, how about if we want to predict 700 00:48:51,830 --> 00:48:54,250 everything, the amplitude, et cetera? 701 00:48:57,330 --> 00:49:00,380 We've got to be careful here with the logic we've applied. 702 00:49:03,500 --> 00:49:10,780 For any one of these normal frequencies, say this one, 703 00:49:10,780 --> 00:49:23,390 earlier on we found that A1 to A2 is given by this equation 704 00:49:23,390 --> 00:49:25,220 or by this equation. 705 00:49:25,220 --> 00:49:27,820 But for the given values of omega, 706 00:49:27,820 --> 00:49:30,940 we found these are now equivalent. 707 00:49:30,940 --> 00:49:37,890 And so if you know what omega is, say omega A, 708 00:49:37,890 --> 00:49:43,400 one of the normal modes, this we know. 709 00:49:43,400 --> 00:49:45,930 It's g/l. 710 00:49:45,930 --> 00:49:47,550 And this is known. 711 00:49:47,550 --> 00:49:56,170 So A1/A2, once we fix omega A, is fixed. 712 00:49:56,170 --> 00:50:01,370 So these are not two independent amplitudes. 713 00:50:01,370 --> 00:50:07,620 So if I go back to here when we said let's guess a solution, 714 00:50:07,620 --> 00:50:10,150 and this was an arbitrary number and this was an arbitrary 715 00:50:10,150 --> 00:50:15,550 number, now we see that in a normal mode 716 00:50:15,550 --> 00:50:18,720 the ratio between them is fixed. 717 00:50:18,720 --> 00:50:22,840 It depends on the normal mode frequency. 718 00:50:22,840 --> 00:50:25,340 But the [? overall ?] normalization 719 00:50:25,340 --> 00:50:26,530 is still arbitrary. 720 00:50:26,530 --> 00:50:29,230 I could make this 10 times this and 10 times that. 721 00:50:29,230 --> 00:50:31,460 [? It would still ?] work. 722 00:50:31,460 --> 00:50:39,670 So back to here, we found one of the normal mode's frequencies. 723 00:50:39,670 --> 00:50:43,630 Using the equation above, it determines 724 00:50:43,630 --> 00:50:48,560 the value of A1 to A2. 725 00:50:48,560 --> 00:50:52,400 Taking the other normal mode frequency 726 00:50:52,400 --> 00:50:57,000 determines the ratio of those two 727 00:50:57,000 --> 00:51:00,380 but not the overall amplitude. 728 00:51:00,380 --> 00:51:04,780 Not to confuse the two, I will call this B1 and B2. 729 00:51:04,780 --> 00:51:07,180 This is the amplitude of the first mass, 730 00:51:07,180 --> 00:51:08,870 and this is of the other one. 731 00:51:08,870 --> 00:51:14,840 The ratio is fixed, determined by the value of omega B. 732 00:51:14,840 --> 00:51:18,570 But the overall amplitude is not. 733 00:51:18,570 --> 00:51:24,020 So now we have ended up in the same situation 734 00:51:24,020 --> 00:51:29,110 we were in our three masses, where, if you remember, 735 00:51:29,110 --> 00:51:33,020 I started by guessing the normal modes 736 00:51:33,020 --> 00:51:37,690 and guessing the relative amplitudes. 737 00:51:37,690 --> 00:51:42,490 And with that information alone and the initial conditions, 738 00:51:42,490 --> 00:51:44,500 I could predict what will happen. 739 00:51:44,500 --> 00:51:46,650 I could now repeat that here. 740 00:51:46,650 --> 00:51:51,510 But to save time, I think you can do that for yourself. 741 00:51:51,510 --> 00:51:55,970 For each, you can write now a single equation 742 00:51:55,970 --> 00:52:03,830 saying what a given mass will do as the sum of two normal modes, 743 00:52:03,830 --> 00:52:11,280 where you know the frequency of them and the amplitudes. 744 00:52:11,280 --> 00:52:14,040 And you have to remember that for the two masses, 745 00:52:14,040 --> 00:52:16,570 the ratio of the amplitudes in each normal mode 746 00:52:16,570 --> 00:52:20,330 have to be fixed given by what we've done here. 747 00:52:20,330 --> 00:52:24,810 And therefore, you can get the single equation 748 00:52:24,810 --> 00:52:28,530 for each mass, one for one mass, one for the other. 749 00:52:28,530 --> 00:52:31,690 Which will tell you what they will do in the future 750 00:52:31,690 --> 00:52:36,500 if you know what they're doing initially, for example, 751 00:52:36,500 --> 00:52:40,700 the initial conditions, that you can solve 752 00:52:40,700 --> 00:52:45,710 to get the final predicted position of each mass. 753 00:52:45,710 --> 00:52:47,350 So that's as much as I was hoping 754 00:52:47,350 --> 00:52:51,700 to do today on coupled oscillators. 755 00:52:51,700 --> 00:52:56,010 And we'll continue next time on situations 756 00:52:56,010 --> 00:52:58,920 with many, many more degrees of freedom. 757 00:52:58,920 --> 00:53:00,770 Thank you.