1 00:00:24,000 --> 00:00:27,046 To avoid possible misunderstandings, 2 00:00:27,046 --> 00:00:31,138 my lectures start at 9:30 Eastern Standard Time, 3 00:00:31,138 --> 00:00:36,274 which is different from lobby seven time, which you may have 4 00:00:36,274 --> 00:00:39,321 noticed. The clock in lobby seven is 5 00:00:39,321 --> 00:00:43,325 seven minutes slow. Today, we're going to cover 6 00:00:43,325 --> 00:00:47,504 coupled oscillators, which is a big part in 8.03, 7 00:00:47,504 --> 00:00:52,031 though we leave out damping in order to avoid a major 8 00:00:52,031 --> 00:00:55,774 complication. Imagine that I have a pendulum 9 00:00:55,774 --> 00:00:59,604 length L mass M, and I have another pendulum, 10 00:00:59,604 --> 00:01:05,673 also mass M length L. And I connect a spring between 11 00:01:05,673 --> 00:01:09,326 them, as you see there, spring constant K. 12 00:01:09,326 --> 00:01:14,405 Imagine now at time T equals zero that I give this object, 13 00:01:14,405 --> 00:01:19,841 which I call object number one, and this is object number two, 14 00:01:19,841 --> 00:01:24,831 give it a certain position. I give it a certain velocity, 15 00:01:24,831 --> 00:01:30,000 so I have four choices. And, I let the system go. 16 00:01:30,000 --> 00:01:34,175 What you're going to see is something extremely chaotic, 17 00:01:34,175 --> 00:01:38,123 and our task today is to predict what the position of 18 00:01:38,123 --> 00:01:42,527 this one is at any moment in time, and what the position of 19 00:01:42,527 --> 00:01:45,032 that one is at any moment in time. 20 00:01:45,032 --> 00:01:48,297 And, to show you how chaotic that motion is, 21 00:01:48,297 --> 00:01:52,396 I will just show you this. So, I take this one and just 22 00:01:52,396 --> 00:01:56,420 displace it from equilibrium. I displace this one from 23 00:01:56,420 --> 00:02:00,520 equilibrium, and at time T equals zero, I will give the 24 00:02:00,520 --> 00:02:05,000 one, my left hand, just a certain velocity. 25 00:02:05,000 --> 00:02:08,914 And, if you now look at the position of the individual 26 00:02:08,914 --> 00:02:12,903 object, it would seem nearly impossible to come with an 27 00:02:12,903 --> 00:02:16,153 analytic solution, which tells you what these 28 00:02:16,153 --> 00:02:18,960 motions are. You'll see that amplitudes 29 00:02:18,960 --> 00:02:20,806 build up. Of certain ones, 30 00:02:20,806 --> 00:02:24,352 amplitude goes down. This one is hardly moving at 31 00:02:24,352 --> 00:02:26,863 all now. Now it's picking up again. 32 00:02:26,863 --> 00:02:31,000 And so, our task today is to work on that. 33 00:02:31,000 --> 00:02:35,097 What is by no means obvious that I will show that to you, 34 00:02:35,097 --> 00:02:39,487 that any motion no matter how you start it off is going to be 35 00:02:39,487 --> 00:02:43,073 the superposition of two normal [mode?] solutions. 36 00:02:43,073 --> 00:02:47,536 Anyway you start it can always be written as the superposition 37 00:02:47,536 --> 00:02:51,268 of two normal mode solutions. What is a normal mode? 38 00:02:51,268 --> 00:02:53,463 A normal mode is, in this case, 39 00:02:53,463 --> 00:02:57,560 that both objects have exactly the same frequency that is 40 00:02:57,560 --> 00:03:01,878 fundamental to normal mode and that they are either in phase 41 00:03:01,878 --> 00:03:05,317 with each other or out of phase with each other, 42 00:03:05,317 --> 00:03:10,000 nothing in between because there's no damping. 43 00:03:10,000 --> 00:03:13,378 So, it's either in phase or it's out of phase. 44 00:03:13,378 --> 00:03:16,307 That is the normal mode. In other words, 45 00:03:16,307 --> 00:03:19,761 if I call one of those frequencies omega minus, 46 00:03:19,761 --> 00:03:22,614 minus means it is the lowest frequency. 47 00:03:22,614 --> 00:03:27,119 There are two frequencies in the system because there are two 48 00:03:27,119 --> 00:03:28,621 objects. Omega minus, 49 00:03:28,621 --> 00:03:32,000 I call that the lowest frequency. 50 00:03:32,000 --> 00:03:35,500 Then it would mean, if they are in phase with each 51 00:03:35,500 --> 00:03:39,642 other, that they come to a halt at the same moment in time. 52 00:03:39,642 --> 00:03:42,857 That means in phase and in the same direction. 53 00:03:42,857 --> 00:03:47,214 So, they come to a halt at the same moment in time in the same 54 00:03:47,214 --> 00:03:49,785 direction, and at the same frequency. 55 00:03:49,785 --> 00:03:53,714 If, then, I have another frequency which is higher there 56 00:03:53,714 --> 00:03:57,142 are two normal modes because we have two objects. 57 00:03:57,142 --> 00:04:02,000 If we have three objects there are three normal modes. 58 00:04:02,000 --> 00:04:06,599 If we go to the higher normal mode, the higher frequency, 59 00:04:06,599 --> 00:04:11,033 they have the same frequency that there are 180° out of 60 00:04:11,033 --> 00:04:13,826 phase. So, when one comes to a halt 61 00:04:13,826 --> 00:04:17,357 here at the other one comes to a halt there. 62 00:04:17,357 --> 00:04:20,478 That's what it means 180° out of phase. 63 00:04:20,478 --> 00:04:23,024 I can excite, and I will excite, 64 00:04:23,024 --> 00:04:26,144 this system into its normal modes only. 65 00:04:26,144 --> 00:04:31,154 I can excite this normal mode alone and this normal mode alone 66 00:04:31,154 --> 00:04:36,000 if I choose the correct initial conditions. 67 00:04:36,000 --> 00:04:39,613 So, for any randomly chosen initial condition, 68 00:04:39,613 --> 00:04:43,709 the motion of each object can be written as a linear 69 00:04:43,709 --> 00:04:48,448 combination of these two modes. If you take my word for that 70 00:04:48,448 --> 00:04:53,106 for now, but of course I will demonstrate to you and I will 71 00:04:53,106 --> 00:04:55,676 prove that to you, it would mean, 72 00:04:55,676 --> 00:05:00,575 then, that [SOUND OFF/THEN ON] function of time can be written 73 00:05:00,575 --> 00:05:04,028 as having some kind of an amplitude, X zero, 74 00:05:04,028 --> 00:05:10,111 I give it a minus sign. Of course, it's related to that 75 00:05:10,111 --> 00:05:16,034 normal mode frequency times the cosine of omega minus T plus 76 00:05:16,034 --> 00:05:20,451 some [phi?] minus, plus some other amplitude, 77 00:05:20,451 --> 00:05:26,073 which I call X zero plus, which is going to be related to 78 00:05:26,073 --> 00:05:33,000 this frequency times the cosine of omega plus T plus phi plus. 79 00:05:33,000 --> 00:05:37,695 Now, let's look at this. Let's try to see through this, 80 00:05:37,695 --> 00:05:41,347 what this means. It means that if I know my 81 00:05:41,347 --> 00:05:45,173 initial conditions, that I can determine this 82 00:05:45,173 --> 00:05:48,391 amplitude. I can determine this phase. 83 00:05:48,391 --> 00:05:53,782 I can determine this amplitude, and I can determine this phase. 84 00:05:53,782 --> 00:05:58,217 There have to be four adjustable constants because I 85 00:05:58,217 --> 00:06:02,304 have the choice between the positions at T zero, 86 00:06:02,304 --> 00:06:06,875 and the velocity. So, there must be four. 87 00:06:06,875 --> 00:06:10,708 What you see is that omega minus and omega plus, 88 00:06:10,708 --> 00:06:15,437 which are these normal mode frequencies, are independent of 89 00:06:15,437 --> 00:06:19,759 the initial condition. So, now I'm going to write down 90 00:06:19,759 --> 00:06:23,184 the position in time for object number two. 91 00:06:23,184 --> 00:06:27,261 And I know that in this mode, it must have the same 92 00:06:27,261 --> 00:06:30,523 frequency. So there must be here a cosine 93 00:06:30,523 --> 00:06:34,763 minus T plus the same, exactly the same phase because 94 00:06:34,763 --> 00:06:39,085 I told you, normal mode means same frequency in phase, 95 00:06:39,085 --> 00:06:45,478 or same frequency out of phase. And, it is the next one which 96 00:06:45,478 --> 00:06:50,521 is going to be out of phase. But, you will see that shortly. 97 00:06:50,521 --> 00:06:53,769 So, this term must become omega plus T. 98 00:06:53,769 --> 00:06:58,726 For now, I will say plus phi, plus what you're going to see 99 00:06:58,726 --> 00:07:03,000 how the out of phase comes in very shortly. 100 00:07:03,000 --> 00:07:06,241 Notice this omega minus must be that omega minus. 101 00:07:06,241 --> 00:07:08,807 Otherwise it wouldn't be a normal mode. 102 00:07:08,807 --> 00:07:11,441 This omega plus must be this omega plus. 103 00:07:11,441 --> 00:07:14,007 Otherwise it wouldn't be a normal mode. 104 00:07:14,007 --> 00:07:17,587 These phi's are the same to allow them to be in phase. 105 00:07:17,587 --> 00:07:21,098 These phi's are the same, even though they are out of 106 00:07:21,098 --> 00:07:23,462 phase. But you will see very shortly 107 00:07:23,462 --> 00:07:27,311 that I will get a minus sign here, which will take care of 108 00:07:27,311 --> 00:07:29,000 the 180°. 109 00:07:29,000 --> 00:07:34,000 110 00:07:34,000 --> 00:07:37,516 I can, with this system, excite the lowest mode, 111 00:07:37,516 --> 00:07:40,958 so that it only oscillates in that lowest mode. 112 00:07:40,958 --> 00:07:44,848 So, this term is not there, and that this term is not 113 00:07:44,848 --> 00:07:46,344 there. I can do that. 114 00:07:46,344 --> 00:07:50,310 I can always oscillates something in one of the normal 115 00:07:50,310 --> 00:07:53,078 modes. But, if I set it off at random, 116 00:07:53,078 --> 00:07:55,921 then it is going to be a superposition. 117 00:07:55,921 --> 00:08:00,260 Even though it is early for you, can you show me using your 118 00:08:00,260 --> 00:08:03,328 hands and your legs, whatever you want to, 119 00:08:03,328 --> 00:08:07,518 when I set this off at just the right initial conditions, 120 00:08:07,518 --> 00:08:10,735 what will that lowest normal mode look like? 121 00:08:10,735 --> 00:08:15,000 How will these pendulums oscillate? 122 00:08:15,000 --> 00:08:19,430 Very good, very good. So you have a good instinct. 123 00:08:19,430 --> 00:08:21,601 Very good. Let's do that. 124 00:08:21,601 --> 00:08:27,026 So, let's set them off in the same direction and let them go. 125 00:08:27,026 --> 00:08:32,000 Those are ideal initial conditions for this mode. 126 00:08:32,000 --> 00:08:35,500 And you will see that it is a normal mode, namely, 127 00:08:35,500 --> 00:08:39,785 they have the same frequency, and they are in phase with each 128 00:08:39,785 --> 00:08:42,428 other. They come to a halt at the same 129 00:08:42,428 --> 00:08:45,428 moment in time. So, what you see already is 130 00:08:45,428 --> 00:08:48,857 that this value here must, also, be X zero minus. 131 00:08:48,857 --> 00:08:51,428 That's a must. You have just seen it. 132 00:08:51,428 --> 00:08:54,571 In this case, what it means is that you might 133 00:08:54,571 --> 00:08:58,071 as well remove the spring. The spring is not doing 134 00:08:58,071 --> 00:09:00,500 anything. The spring never gets any 135 00:09:00,500 --> 00:09:06,741 longer, never gets any shorter. What do you think the pendulums 136 00:09:06,741 --> 00:09:11,605 will do when I excite the other mode, which has a higher 137 00:09:11,605 --> 00:09:15,231 frequency? Very good, can I see some other 138 00:09:15,231 --> 00:09:17,707 hands or heads or legs? Good. 139 00:09:17,707 --> 00:09:19,564 That's this, symmetry. 140 00:09:19,564 --> 00:09:22,482 There is a symmetry in the system. 141 00:09:22,482 --> 00:09:27,877 And that's why you people are so smart and see immediately the 142 00:09:27,877 --> 00:09:33,008 connection. So, let me make here a drawing 143 00:09:33,008 --> 00:09:38,022 of the omega minus, and then the amplitudes are the 144 00:09:38,022 --> 00:09:41,431 same. And here, I make a drawing of 145 00:09:41,431 --> 00:09:44,640 the omega plus. Higher frequency, 146 00:09:44,640 --> 00:09:50,456 and the amplitudes are now the same, but 180° out of phase. 147 00:09:50,456 --> 00:09:54,568 So, it's easy to excite that. There we go. 148 00:09:54,568 --> 00:09:59,080 Notice they oscillate with the same frequency, 149 00:09:59,080 --> 00:10:06,000 and they're coming to a halt at the same moment in time. 150 00:10:06,000 --> 00:10:10,071 But, if one is here, the other is there. 151 00:10:10,071 --> 00:10:14,978 And so, therefore, this one here must be minus X 152 00:10:14,978 --> 00:10:17,692 zero plus. That minus sign, 153 00:10:17,692 --> 00:10:23,016 now, gives me the 180°. So I can still keep that phi 154 00:10:23,016 --> 00:10:25,000 plus there. 155 00:10:25,000 --> 00:10:31,000 156 00:10:31,000 --> 00:10:34,194 It is utterly trivial, without any work, 157 00:10:34,194 --> 00:10:37,553 to calculate, or tell you what omega minus 158 00:10:37,553 --> 00:10:41,485 is, that just the frequency of a single pendulum, 159 00:10:41,485 --> 00:10:45,090 so that's easy. So, omega minus is simply the 160 00:10:45,090 --> 00:10:48,531 square root of G over L. There's no spring, 161 00:10:48,531 --> 00:10:51,070 right? Each one is doing its own 162 00:10:51,070 --> 00:10:55,084 thing, and the spring is never pushing or pulling. 163 00:10:55,084 --> 00:11:00,000 But we should be able to calculate omega plus. 164 00:11:00,000 --> 00:11:04,862 So let's do that. And I will only make a drawing 165 00:11:04,862 --> 00:11:09,931 of one of those pendulums. I don't need them both. 166 00:11:09,931 --> 00:11:13,344 So, here's one of those pendulums. 167 00:11:13,344 --> 00:11:18,931 And let this separation distance be X from equilibrium. 168 00:11:18,931 --> 00:11:24,827 And, I call this angle theta. And so, here is that spring. 169 00:11:24,827 --> 00:11:30,000 And, the other one is on the other side. 170 00:11:30,000 --> 00:11:37,482 Let's pull in all the forces that we can think of. 171 00:11:37,482 --> 00:11:40,995 So, there's gravity, MG. 172 00:11:40,995 --> 00:11:43,896 There is tension, T. 173 00:11:43,896 --> 00:11:49,394 But there's more. What else is there? 174 00:11:49,394 --> 00:11:54,891 Spring. How much longer is the spring 175 00:11:54,891 --> 00:12:01,000 than wants to be? 2X, exactly. 176 00:12:01,000 --> 00:12:04,649 In other words, there is a force here. 177 00:12:04,649 --> 00:12:10,369 I call that the spring force, FS, which is bringing it also 178 00:12:10,369 --> 00:12:14,413 back to equilibrium. And, that one is 2KX. 179 00:12:14,413 --> 00:12:17,964 Now, the tangent is very close to MG. 180 00:12:17,964 --> 00:12:21,515 We have discussed that several times. 181 00:12:21,515 --> 00:12:25,756 I'm going to introduce a shorthand notation. 182 00:12:25,756 --> 00:12:31,378 Omega zero squared is G divided by L, and omega S squared, 183 00:12:31,378 --> 00:12:37,000 making reference to the spring is K over M. 184 00:12:37,000 --> 00:12:42,647 And now I have to set up the differential equation. 185 00:12:42,647 --> 00:12:46,601 I have to apply Newton's Second Law. 186 00:12:46,601 --> 00:12:51,684 And to remind you, the magnitude of the spring 187 00:12:51,684 --> 00:12:55,863 force is 2KX. So, Newton's Second Law, 188 00:12:55,863 --> 00:13:01,624 MX double dot equals, now, there are two forces that 189 00:13:01,624 --> 00:13:07,498 I have to take into account: first, the spring force, 190 00:13:07,498 --> 00:13:13,360 which is minus 2KX. And then I need the horizontal 191 00:13:13,360 --> 00:13:16,114 component of the [tension/tangent?]. 192 00:13:16,114 --> 00:13:19,891 Remember, that's only one that is important here, 193 00:13:19,891 --> 00:13:22,645 which is T sine theta. But, T is MG, 194 00:13:22,645 --> 00:13:27,131 and sine theta is X divided by L if L is the length of the 195 00:13:27,131 --> 00:13:29,334 pendulum. So, we get minus T, 196 00:13:29,334 --> 00:13:35,000 which is MG times the sine of theta, which is X divided by L. 197 00:13:35,000 --> 00:13:39,495 And so, this is the differential equation. 198 00:13:39,495 --> 00:13:44,758 I'm going to divide M out, and I'm going to bring 199 00:13:44,758 --> 00:13:50,460 everything to one side. So, I'm going to get X double 200 00:13:50,460 --> 00:13:56,929 dot plus two omega S squared times X plus omega zero squared 201 00:13:56,929 --> 00:14:02,824 times X equals zero. And our task now is to solve 202 00:14:02,824 --> 00:14:07,564 this differential equation. And that, of course, 203 00:14:07,564 --> 00:14:12,910 you can do in three seconds because you recognize this 204 00:14:12,910 --> 00:14:17,449 differential equation. It is X double dot plus 205 00:14:17,449 --> 00:14:21,887 something times X. And so, the new frequency, 206 00:14:21,887 --> 00:14:27,737 which I will call omega plus, that new frequency omega plus 207 00:14:27,737 --> 00:14:35,000 is the square root of two omega S squared plus omega squared. 208 00:14:35,000 --> 00:14:40,444 And, what you see is something that I already anticipated, 209 00:14:40,444 --> 00:14:44,265 which was consistent with your intuition. 210 00:14:44,265 --> 00:14:49,710 It is larger than omega zero because this is the effect of 211 00:14:49,710 --> 00:14:53,626 the spring. And so, omega zero is the one. 212 00:14:53,626 --> 00:14:58,976 Here, an omega plus now is the square root of two omega S 213 00:14:58,976 --> 00:15:04,994 squared plus omega zero squared. And, omega S squared is defined 214 00:15:04,994 --> 00:15:07,000 this way. 215 00:15:07,000 --> 00:15:13,000 216 00:15:13,000 --> 00:15:18,982 So, if I now turn to my general solution, if for now you accept 217 00:15:18,982 --> 00:15:23,035 the fact that that is the general solution, 218 00:15:23,035 --> 00:15:28,728 the superposition of the normal modes, then the key point is 219 00:15:28,728 --> 00:15:35,000 that independent of your initial conditions is omega minus. 220 00:15:35,000 --> 00:15:37,875 That was the square root of G over L. 221 00:15:37,875 --> 00:15:40,750 I never put in any initial condition. 222 00:15:40,750 --> 00:15:44,744 Independent of the initial condition is omega plus. 223 00:15:44,744 --> 00:15:48,099 That's this one. I never put in any initial 224 00:15:48,099 --> 00:15:51,054 conditions. Independent of the initial 225 00:15:51,054 --> 00:15:54,488 conditions is this ratio, which is plus one. 226 00:15:54,488 --> 00:16:00,000 And, independent of the initial conditions is this ratio. 227 00:16:00,000 --> 00:16:02,718 That is minus one. In other words, 228 00:16:02,718 --> 00:16:06,507 if I didn't tell you at the initial conditions, 229 00:16:06,507 --> 00:16:10,214 which I haven't, I can predict that this ratio 230 00:16:10,214 --> 00:16:13,427 is plus one. And I can predict that this 231 00:16:13,427 --> 00:16:16,722 ratio is minus one. If I make this three, 232 00:16:16,722 --> 00:16:20,429 then this is minus three. If I make this five, 233 00:16:20,429 --> 00:16:24,300 then this is minus five. The ratio is minus one. 234 00:16:24,300 --> 00:16:28,501 So, the ratios are independent of initial condition, 235 00:16:28,501 --> 00:16:32,208 and the frequencies are independent of initial 236 00:16:32,208 --> 00:16:35,091 condition. If I tell you the initial 237 00:16:35,091 --> 00:16:39,375 conditions, then of course you can also calculate the 238 00:16:39,375 --> 00:16:44,566 individual amplitudes. Suppose, now, 239 00:16:44,566 --> 00:16:51,437 I start this system off in some way at time T equals zero, 240 00:16:51,437 --> 00:16:56,861 which I can choose. And it used the following. 241 00:16:56,861 --> 00:17:00,357 At T equals zero, I make X1 C, 242 00:17:00,357 --> 00:17:05,540 just some number C that we can choose, 3 cm, 243 00:17:05,540 --> 00:17:11,862 whatever you want to choose. But, I make V1 zero. 244 00:17:11,862 --> 00:17:14,562 So, as I release it: no speed. 245 00:17:14,562 --> 00:17:18,846 And, suppose I make X2 zero and I make V2 zero. 246 00:17:18,846 --> 00:17:24,433 So, if you don't know what that means, of course I'm going to 247 00:17:24,433 --> 00:17:28,437 demonstrate it. It means that this one here, 248 00:17:28,437 --> 00:17:33,000 the other one I offset, that's all I do. 249 00:17:33,000 --> 00:17:36,483 And I let this one go, pull my hands off, 250 00:17:36,483 --> 00:17:40,488 and then I want to know what's going to happen, 251 00:17:40,488 --> 00:17:43,710 right? Because I release this one with 252 00:17:43,710 --> 00:17:47,193 zero speed. That's what you see in there. 253 00:17:47,193 --> 00:17:51,983 There is one of an infinite number of initial conditions 254 00:17:51,983 --> 00:17:55,640 that you may choose. So, this now has to be 255 00:17:55,640 --> 00:17:58,775 substituted into my general solution. 256 00:17:58,775 --> 00:18:04,000 And so, you have to take the derivative of X1. 257 00:18:04,000 --> 00:18:07,989 So, you have to do X1 dot and put that equal to zero. 258 00:18:07,989 --> 00:18:10,751 I leave that in your competent hands. 259 00:18:10,751 --> 00:18:14,818 Then you have to take X2, and you have to take X2 dot, 260 00:18:14,818 --> 00:18:17,810 and you have to make that equal to zero. 261 00:18:17,810 --> 00:18:21,033 And I leave that into your competent hands. 262 00:18:21,033 --> 00:18:23,565 That's easy. This and you'll find, 263 00:18:23,565 --> 00:18:27,631 then, that in this particular case for this particular 264 00:18:27,631 --> 00:18:30,930 example, phi minus equals zero, and phi two, 265 00:18:30,930 --> 00:18:36,035 and phi plus equal zero. It will not take you more than 266 00:18:36,035 --> 00:18:39,834 a few minutes to find that. I didn't want waste your time 267 00:18:39,834 --> 00:18:43,090 on taking a derivative of such a simple function. 268 00:18:43,090 --> 00:18:47,025 So, if we take this right now, then I will substitute these 269 00:18:47,025 --> 00:18:50,417 results in the equation. So, you did the hard work. 270 00:18:50,417 --> 00:18:53,605 You did the velocities. I will do the positions. 271 00:18:53,605 --> 00:18:57,133 So, I will substitute in that equation T equals zero. 272 00:18:57,133 --> 00:19:01,000 I know already that the phi's are not there. 273 00:19:01,000 --> 00:19:08,303 And so, then I get that C at time T equals zero is going to 274 00:19:08,303 --> 00:19:13,214 be X zero minus T zero. So, this is one. 275 00:19:13,214 --> 00:19:18,251 T zero, so this is one, plus X zero plus. 276 00:19:18,251 --> 00:19:25,555 That's this initial condition. Then I go through the second 277 00:19:25,555 --> 00:19:31,474 initial condition that is zero, is X zero minus, 278 00:19:31,474 --> 00:19:37,508 minus X zero plus. There is a minus sign here. 279 00:19:37,508 --> 00:19:41,225 And so, what do I find I have solved now? 280 00:19:41,225 --> 00:19:45,871 The general solution, I will find that X zero minus 281 00:19:45,871 --> 00:19:49,681 is one half C, and F zero plus is also one 282 00:19:49,681 --> 00:19:56,000 half C, which of course should not come as a surprise to you. 283 00:19:56,000 --> 00:20:02,000 284 00:20:02,000 --> 00:20:06,398 So, let me write down, now, the general solution that 285 00:20:06,398 --> 00:20:10,036 we have for this specific initial condition. 286 00:20:10,036 --> 00:20:13,589 So, we know everything. We know X zero one. 287 00:20:13,589 --> 00:20:16,803 We know X zero one minus, X zero minus, 288 00:20:16,803 --> 00:20:20,271 not one X zero minus. We know X zero plus. 289 00:20:20,271 --> 00:20:23,063 We know phi's. We know everything. 290 00:20:23,063 --> 00:20:26,024 So, I'm going to write it down here. 291 00:20:26,024 --> 00:20:30,000 So, X1 is going to be one half C. 292 00:20:30,000 --> 00:20:38,480 Remember, we found it at one half C times the cosine of omega 293 00:20:38,480 --> 00:20:46,113 T equals phi zero plus one half C times, this is minus, 294 00:20:46,113 --> 00:20:53,462 by the way, cosine omega minus T cosine omega plus T. 295 00:20:53,462 --> 00:21:02,226 That is X1, and X2 is one half C times the cosine omega minus T 296 00:21:02,226 --> 00:21:10,000 minus one half C times the cosine omega plus T. 297 00:21:10,000 --> 00:21:13,503 Take a deep breath, substitute in there T equals 298 00:21:13,503 --> 00:21:17,753 zero, and you see immediately that X is C and substitute T 299 00:21:17,753 --> 00:21:20,884 equals zero into seconds, and then you see, 300 00:21:20,884 --> 00:21:24,835 indeed, that X2 is zero: no surprise because that's my 301 00:21:24,835 --> 00:21:28,264 initial condition. Now, I remember from my high 302 00:21:28,264 --> 00:21:32,663 school days that the cosine alpha plus the cosine of beta is 303 00:21:32,663 --> 00:21:37,136 twice the cosine of half the sum times the cosine of half the 304 00:21:37,136 --> 00:21:41,993 difference. So, I can write down X1 as 305 00:21:41,993 --> 00:21:48,901 twice the cosine of half the sum times the cosine of half the 306 00:21:48,901 --> 00:21:53,736 difference. So, that two that I get eats up 307 00:21:53,736 --> 00:21:58,917 these one halves. So, I get C times the cosine 308 00:21:58,917 --> 00:22:05,710 omega minus plus omega plus divided by two times T times the 309 00:22:05,710 --> 00:22:11,812 cosine of omega minus, minus omega plus divided by two 310 00:22:11,812 --> 00:22:16,478 times T. I've just rewritten it in a 311 00:22:16,478 --> 00:22:20,608 different form. And, X2 as a function of time, 312 00:22:20,608 --> 00:22:25,657 I now have the cosine of alpha minus the cosine of beta. 313 00:22:25,657 --> 00:22:30,521 That is twice the sign, half the sum times the sign of 314 00:22:30,521 --> 00:22:36,709 half the difference. So, now I get here the sign of 315 00:22:36,709 --> 00:22:44,003 omega minus plus omega plus divided by two times T times the 316 00:22:44,003 --> 00:22:50,309 sign of omega minus, minus omega plus divided by two 317 00:22:50,309 --> 00:22:54,018 times T. Notice when T is zero, 318 00:22:54,018 --> 00:23:01,189 that the sign of this one is zero, consistent with what our 319 00:23:01,189 --> 00:23:07,000 initial conditions, X2 zero, remember? 320 00:23:07,000 --> 00:23:11,205 It's all there. Initial conditions are all 321 00:23:11,205 --> 00:23:14,487 there. Initial conditions are all 322 00:23:14,487 --> 00:23:18,487 there, just written in a different form. 323 00:23:18,487 --> 00:23:24,538 Now, imagine now in your mind that these two frequencies are 324 00:23:24,538 --> 00:23:29,461 not too far apart. Then, these equations smell of 325 00:23:29,461 --> 00:23:31,741 what? Beats. 326 00:23:31,741 --> 00:23:35,574 See, this here is the fast term. 327 00:23:35,574 --> 00:23:41,880 This one and that one, and this one is the slow one. 328 00:23:41,880 --> 00:23:48,556 And so, if these two are close, then what you'll see is 329 00:23:48,556 --> 00:23:53,996 something quite remarkable. At T equals zero, 330 00:23:53,996 --> 00:24:00,301 this one stands still. And, the cosine term is going 331 00:24:00,301 --> 00:24:06,973 to be one because T is one. This one is going to oscillate 332 00:24:06,973 --> 00:24:10,723 happily with this frequency. But, this cosine term is very 333 00:24:10,723 --> 00:24:13,881 gradually going to zero. And, as that cosine term 334 00:24:13,881 --> 00:24:17,434 gradually goes to zero, this one will stop oscillating. 335 00:24:17,434 --> 00:24:19,802 But, this sine term becomes plus one. 336 00:24:19,802 --> 00:24:22,763 And so, the other one will start to oscillate. 337 00:24:22,763 --> 00:24:26,644 And then, a little later in time the cosine term will become 338 00:24:26,644 --> 00:24:30,000 minus one. So, it starts to oscillate. 339 00:24:30,000 --> 00:24:33,753 But when that happens, the sine term is zero again. 340 00:24:33,753 --> 00:24:36,831 So, it stops. So, you see a beautiful beat 341 00:24:36,831 --> 00:24:39,758 phenomenon. The first one will gradually 342 00:24:39,758 --> 00:24:43,211 come to a halt, and the other one will pick up. 343 00:24:43,211 --> 00:24:46,439 And then, the other one will come to a halt, 344 00:24:46,439 --> 00:24:49,217 and then transfers in [a way?] energy. 345 00:24:49,217 --> 00:24:51,919 It is, of course, consistent with the 346 00:24:51,919 --> 00:24:54,396 conservation of mechanical energy. 347 00:24:54,396 --> 00:24:58,000 And, I want to demonstrate that. 348 00:24:58,000 --> 00:25:03,000 349 00:25:03,000 --> 00:25:07,464 So, we have this year. All I have to do is offset X1 350 00:25:07,464 --> 00:25:10,791 over a distance, C, that we can choose, 351 00:25:10,791 --> 00:25:14,468 and then we release this one at zero speed. 352 00:25:14,468 --> 00:25:19,545 And then we'll just watch it. And then, you should see that 353 00:25:19,545 --> 00:25:23,309 strange beat phenomenon. You ready for this? 354 00:25:23,309 --> 00:25:26,023 This will have to hold in place. 355 00:25:26,023 --> 00:25:29,000 Three, two, one, zero. 356 00:25:29,000 --> 00:25:34,000 357 00:25:34,000 --> 00:25:37,343 Look, this one is standing still. 358 00:25:37,343 --> 00:25:41,000 Look, this one is standing still. 359 00:25:41,000 --> 00:25:46,000 360 00:25:46,000 --> 00:25:49,534 That's beating. And you see that the energy is 361 00:25:49,534 --> 00:25:52,125 transferred from one to the other. 362 00:25:52,125 --> 00:25:56,680 And that is a beat phenomenon that follows immediately from 363 00:25:56,680 --> 00:25:59,272 this. What would happen if I moved 364 00:25:59,272 --> 00:26:05,490 the spring up? Suppose I moved the spring here 365 00:26:05,490 --> 00:26:13,023 higher, say, halfway. Anyone without looking too much 366 00:26:13,023 --> 00:26:19,542 at the blackboard, sort of use your intuition. 367 00:26:19,542 --> 00:26:25,771 What would happen? Would the same phenomenon 368 00:26:25,771 --> 00:26:27,509 happen? Yeah? 369 00:26:27,509 --> 00:26:32,000 Yeah? The spring what? 370 00:26:32,000 --> 00:26:34,823 Yes, so would the same phenomenon happen? 371 00:26:34,823 --> 00:26:38,000 But what would happen with the beat period? 372 00:26:38,000 --> 00:26:44,000 373 00:26:44,000 --> 00:26:46,206 Now you may look at the blackboard. 374 00:26:46,206 --> 00:26:49,061 Do you think these have a certain separation? 375 00:26:49,061 --> 00:26:51,591 Right, you make that separation smaller. 376 00:26:51,591 --> 00:26:54,251 You make the effect of the spring smaller. 377 00:26:54,251 --> 00:26:57,950 So, the omega minus becomes even closer to the omega plus. 378 00:26:57,950 --> 00:27:01,000 And so, the big period will increase. 379 00:27:01,000 --> 00:27:05,827 It will take longer for one to stop. 380 00:27:05,827 --> 00:27:12,034 Let's try that. So, are we going to move these 381 00:27:12,034 --> 00:27:17,000 up? I'll put it roughly halfway. 382 00:27:17,000 --> 00:27:22,000 383 00:27:22,000 --> 00:27:26,153 OK, are we ready for this? And so we're doing it again. 384 00:27:26,153 --> 00:27:30,692 And then you'll see the same phenomenon, except it will take 385 00:27:30,692 --> 00:27:33,923 longer for the first one to come to a halt. 386 00:27:33,923 --> 00:27:37,000 So, I have decreased the coupling. 387 00:27:37,000 --> 00:27:48,000 388 00:27:48,000 --> 00:27:54,117 It's still swinging happily. Now it's beginning to change 389 00:27:54,117 --> 00:27:58,159 its mind. You see, and now it standing 390 00:27:58,159 --> 00:28:01,000 still. It took longer. 391 00:28:01,000 --> 00:28:04,874 Do we agree? Now, there is something 392 00:28:04,874 --> 00:28:09,745 mind-boggling, something that you really want 393 00:28:09,745 --> 00:28:13,619 to see. Suppose I bring this all the 394 00:28:13,619 --> 00:28:18,712 way to the top. That's interesting because then 395 00:28:18,712 --> 00:28:24,579 omega minus is exactly the same as omega plus because, 396 00:28:24,579 --> 00:28:31,000 look, this term goes away. What now will happen? 397 00:28:31,000 --> 00:28:33,390 It's almost cutting out the spring. 398 00:28:33,390 --> 00:28:37,468 There is no spring anymore. What now will happen if I start 399 00:28:37,468 --> 00:28:39,226 one here? Three, two, one, 400 00:28:39,226 --> 00:28:40,000 zero. 401 00:28:40,000 --> 00:28:45,000 402 00:28:45,000 --> 00:28:47,005 What do you think? Yeah? 403 00:28:47,005 --> 00:28:49,098 Ah, brilliant. Brilliant. 404 00:28:49,098 --> 00:28:52,151 One will swing, and the other won't. 405 00:28:52,151 --> 00:28:55,901 Isn't that shocking? You have two pendulums, 406 00:28:55,901 --> 00:29:00,000 and they're not even connected anymore. 407 00:29:00,000 --> 00:29:05,510 And you start this one swinging, and the other one will 408 00:29:05,510 --> 00:29:09,591 never start to swing. Isn't that amazing? 409 00:29:09,591 --> 00:29:14,591 For \$30,000 tuition, you learn something fantastic 410 00:29:14,591 --> 00:29:19,183 except that the big period is infinitely long. 411 00:29:19,183 --> 00:29:24,591 So, with a bit of patience, so let's demonstrate that. 412 00:29:24,591 --> 00:29:30,000 So let's demonstrate this. So, there we go. 413 00:29:30,000 --> 00:29:32,691 Unbelievable. Physics works. 414 00:29:32,691 --> 00:29:38,373 And look at this equation. When omega minus is omega plus, 415 00:29:38,373 --> 00:29:43,657 this term is always zero. So, you see X2 remains zero. 416 00:29:43,657 --> 00:29:46,747 Ha, ha, ha, that's what you see. 417 00:29:46,747 --> 00:29:52,928 And, when omega minus is omega plus, this cosine term is always 418 00:29:52,928 --> 00:29:56,517 plus one. Ha, ha, ha, that's what you 419 00:29:56,517 --> 00:29:59,806 see. Isn't it amazing the power of 420 00:29:59,806 --> 00:30:02,000 physics? 421 00:30:02,000 --> 00:30:06,000 422 00:30:06,000 --> 00:30:11,314 So, it is truly remarkable that we can describe for any initial 423 00:30:11,314 --> 00:30:16,714 condition the motion in terms of the linear superposition of the 424 00:30:16,714 --> 00:30:20,742 two normal modes. And so, what originally looked 425 00:30:20,742 --> 00:30:25,885 like an impossibility when I started the first 30 sections of 426 00:30:25,885 --> 00:30:28,714 my lecture with the demonstration, 427 00:30:28,714 --> 00:30:33,857 when I started them off in a random way, it looked so chaotic 428 00:30:33,857 --> 00:30:39,085 that it was almost unimaginable that we would be able to short 429 00:30:39,085 --> 00:30:43,628 that out and be able to predict the motion of each one 430 00:30:43,628 --> 00:30:48,000 individually as a function of time. 431 00:30:48,000 --> 00:30:52,505 I want you to appreciate that we have two coupled oscillators 432 00:30:52,505 --> 00:30:55,209 here. That gives you to normal modes. 433 00:30:55,209 --> 00:30:59,490 If you have three coupled oscillators, we will get back to 434 00:30:59,490 --> 00:31:02,418 that, then there are three normal modes. 435 00:31:02,418 --> 00:31:05,798 And if you have for, then you have four normal 436 00:31:05,798 --> 00:31:09,243 modes. Now, clearly what we need is a 437 00:31:09,243 --> 00:31:11,337 general recipe. In this case, 438 00:31:11,337 --> 00:31:15,750 the problem was a beautifully symmetric we could do all mega 439 00:31:15,750 --> 00:31:19,565 minus just a split second. Omega plus: you could see 440 00:31:19,565 --> 00:31:23,229 exactly the motion. So, in no time [you get?] that 441 00:31:23,229 --> 00:31:27,493 the ratio of the amplitude was plus one and was minus one. 442 00:31:27,493 --> 00:31:30,783 But let me tell you, if I break the symmetry, 443 00:31:30,783 --> 00:31:34,997 oh man, all hell breaks loose. For instance, 444 00:31:34,997 --> 00:31:39,752 if you change the ratio of the masses, you make one longer than 445 00:31:39,752 --> 00:31:43,893 the other, you have three beats connected with springs. 446 00:31:43,893 --> 00:31:47,958 If you break the symmetry, it is very, very hard work. 447 00:31:47,958 --> 00:31:50,949 And, for that, we need a general recipe. 448 00:31:50,949 --> 00:31:54,401 And, I'm going to give you the general recipe. 449 00:31:54,401 --> 00:31:58,619 And then, I'm going to apply that general recipe to this 450 00:31:58,619 --> 00:32:03,636 case. So, everything that you have 451 00:32:03,636 --> 00:32:09,610 seen will then come out after a lot of algebra, 452 00:32:09,610 --> 00:32:15,844 but at least you know that the recipe is working. 453 00:32:15,844 --> 00:32:22,207 So, the first thing that you do that's number one, 454 00:32:22,207 --> 00:32:30,000 you give each object a displacement from equilibrium. 455 00:32:30,000 --> 00:32:35,126 Even though you are free to choose the direction, 456 00:32:35,126 --> 00:32:41,427 I always, matter of discipline, set them all off in the same 457 00:32:41,427 --> 00:32:44,417 direction. That's not a must. 458 00:32:44,417 --> 00:32:48,689 But, that reduces the chance of mistakes. 459 00:32:48,689 --> 00:32:52,213 But, you give them a displacement, 460 00:32:52,213 --> 00:32:56,699 and I always do them in the same direction. 461 00:32:56,699 --> 00:33:01,184 That's number one. Once you have done that, 462 00:33:01,184 --> 00:33:05,776 you write down Newton's Second Law for each, 463 00:33:05,776 --> 00:33:11,116 which means if you have two objects, you have three 464 00:33:11,116 --> 00:33:16,000 unknowns. Well, that is three unknowns. 465 00:33:16,000 --> 00:33:23,000 466 00:33:23,000 --> 00:33:28,288 So, you want to find one of the normal mode frequencies. 467 00:33:28,288 --> 00:33:33,576 You want to find this one, and independently you want to 468 00:33:33,576 --> 00:33:37,711 find that one, which has to come out of that 469 00:33:37,711 --> 00:33:42,326 recipe, what are in principle the three unknowns. 470 00:33:42,326 --> 00:33:48,192 This ratio plus one holds here, but that doesn't hold in other 471 00:33:48,192 --> 00:33:51,653 cases. And so, in principle you have, 472 00:33:51,653 --> 00:33:55,788 as an uncertainty, this amplitude and omega. 473 00:33:55,788 --> 00:34:03,000 So, you always end up with two equations with three unknowns. 474 00:34:03,000 --> 00:34:08,760 So if I see two objects, then you have three unknowns. 475 00:34:08,760 --> 00:34:14,956 And you have two equations. So, that looks like a problem. 476 00:34:14,956 --> 00:34:19,195 Number three: you're going to put in the 477 00:34:19,195 --> 00:34:25,500 condition for normal modes, which means that X1 is C1 times 478 00:34:25,500 --> 00:34:29,847 cosine omega T. But, X2 can be some other 479 00:34:29,847 --> 00:34:35,826 amplitude, which we had to calculate also times the same 480 00:34:35,826 --> 00:34:40,038 omega T. I don't have to put the phase 481 00:34:40,038 --> 00:34:43,763 in there because it's either in phase or out of phase, 482 00:34:43,763 --> 00:34:47,067 and out of phase means that you get minus signs. 483 00:34:47,067 --> 00:34:51,073 And so, you see immediately that if you substitute this in 484 00:34:51,073 --> 00:34:54,658 Newton's Second Law, that you get for two equations, 485 00:34:54,658 --> 00:34:56,696 three unknowns. C1 is unknown, 486 00:34:56,696 --> 00:35:00,000 C2 is unknown, and omega is unknown. 487 00:35:00,000 --> 00:35:04,447 So, you substitute this in here, and then comes the 488 00:35:04,447 --> 00:35:07,828 problem. How are you going to deal with 489 00:35:07,828 --> 00:35:10,763 two equations with three unknowns? 490 00:35:10,763 --> 00:35:14,411 Well, think about what I told you earlier. 491 00:35:14,411 --> 00:35:19,837 If you don't know the initial conditions, you can never find X 492 00:35:19,837 --> 00:35:23,039 zero minus, so you can never find C1. 493 00:35:23,039 --> 00:35:29,000 But the ratio is independent of the initial conditions. 494 00:35:29,000 --> 00:35:34,663 So that means the ratio C1 over C2 cannot be dependent on the 495 00:35:34,663 --> 00:35:37,873 initial conditions. In other words, 496 00:35:37,873 --> 00:35:43,348 you can always solve these equations in terms of C1 divided 497 00:35:43,348 --> 00:35:47,690 by C2 and omega. And, I will show you that that 498 00:35:47,690 --> 00:35:50,805 works. So, you follow this recipe, 499 00:35:50,805 --> 00:35:54,581 and we will do that religiously together. 500 00:35:54,581 --> 00:35:57,884 We will find, then, in this case two 501 00:35:57,884 --> 00:36:03,633 variables for omega We will find an omega minus, 502 00:36:03,633 --> 00:36:07,583 which has an associated value of C over C2. 503 00:36:07,583 --> 00:36:12,191 I'll put a minus there. And, we will find an omega 504 00:36:12,191 --> 00:36:16,705 plus, which has its own ratio associated with it. 505 00:36:16,705 --> 00:36:20,655 In the case of our simple symmetric system, 506 00:36:20,655 --> 00:36:25,357 this would be plus one and this would be minus one. 507 00:36:25,357 --> 00:36:31,000 But that is not the case when you break symmetry. 508 00:36:31,000 --> 00:36:34,382 So, my task is, now, to apply this recipe in 509 00:36:34,382 --> 00:36:37,134 its most general form to the system. 510 00:36:37,134 --> 00:36:39,730 It will be 16 minutes of grinding. 511 00:36:39,730 --> 00:36:44,449 We will go through each step, and out comes something that we 512 00:36:44,449 --> 00:36:47,674 already know. But at least you'll see that 513 00:36:47,674 --> 00:36:52,078 I've made no assumption which I did there about symmetry. 514 00:36:52,078 --> 00:36:56,325 Therefore, this is an ideal moment for you to have your 515 00:36:56,325 --> 00:37:00,966 five-minute break so you can take a deep breath to get ready 516 00:37:00,966 --> 00:37:06,000 for this 16 minute marathon. [SOUND OFF/THEN ON] 517 00:37:06,000 --> 00:37:09,831 All right: general recipe. Someone asked me during the 518 00:37:09,831 --> 00:37:13,807 intermission whether this omega is the same as that one. 519 00:37:13,807 --> 00:37:16,843 Yes, of course, otherwise it wouldn't be in 520 00:37:16,843 --> 00:37:19,807 normal mode. We're going to search for the 521 00:37:19,807 --> 00:37:22,481 normal modes. And, in the normal mode, 522 00:37:22,481 --> 00:37:25,879 the frequencies must be the same of all objects, 523 00:37:25,879 --> 00:37:30,000 whether you have five or six or two or three. 524 00:37:30,000 --> 00:37:33,894 But the ratios are different. That's a different issue. 525 00:37:33,894 --> 00:37:36,923 However, they are in phase or out of phase. 526 00:37:36,923 --> 00:37:41,322 So the ratios can be negative, positive, but you never had any 527 00:37:41,322 --> 00:37:45,504 phase angles other than the 180° or zero because we have no 528 00:37:45,504 --> 00:37:48,245 damping. We have taken the damping out. 529 00:37:48,245 --> 00:37:52,572 So, it is essential that when you substitute that in Newton's 530 00:37:52,572 --> 00:37:57,115 Second Law that these omegas are the same because they will then 531 00:37:57,115 --> 00:38:02,000 give you the normal modes. OK, you ready? 532 00:38:02,000 --> 00:38:07,000 533 00:38:07,000 --> 00:38:12,861 I'm going to offset the first one over a distance, 534 00:38:12,861 --> 00:38:15,732 X1. So, they have mass M. 535 00:38:15,732 --> 00:38:19,559 They have length L to remind you. 536 00:38:19,559 --> 00:38:26,736 And, we are going to have omega zero squared equals G over L, 537 00:38:26,736 --> 00:38:32,000 and omega S squared equals K over M. 538 00:38:32,000 --> 00:38:36,245 And, this one I offset over a distance, X2. 539 00:38:36,245 --> 00:38:39,581 Notice, you don't have to do that. 540 00:38:39,581 --> 00:38:43,826 I always offset them in the same direction. 541 00:38:43,826 --> 00:38:48,779 And then, there is this spring that connects them. 542 00:38:48,779 --> 00:38:53,025 I will just make it very thin, this spring. 543 00:38:53,025 --> 00:39:00,000 Otherwise the picture becomes a little bit too complicated. 544 00:39:00,000 --> 00:39:05,597 So let this angle be theta one. And let this angle be theta 545 00:39:05,597 --> 00:39:08,300 two. And so, we have here MG. 546 00:39:08,300 --> 00:39:11,388 We have here MG. We have here the 547 00:39:11,388 --> 00:39:16,117 [tangent/tension?] is roughly MG for small angles, 548 00:39:16,117 --> 00:39:20,557 and we have here a tangent which is roughly MG. 549 00:39:20,557 --> 00:39:24,707 It's a little bit too big the way I drew it, 550 00:39:24,707 --> 00:39:27,699 but that's not so important now. 551 00:39:27,699 --> 00:39:32,139 And now, there is another force that X on both, 552 00:39:32,139 --> 00:39:38,520 and which force is that? That's the spring force. 553 00:39:38,520 --> 00:39:44,191 Now, do we agree what the magnitude of the spring force 554 00:39:44,191 --> 00:39:47,552 is? And then, we will argue about 555 00:39:47,552 --> 00:39:51,123 the direction. [Nickel?] magnitude, 556 00:39:51,123 --> 00:39:55,009 look very closely. I offset one by X2, 557 00:39:55,009 --> 00:39:59,000 and I offset the other by X1. 558 00:39:59,000 --> 00:40:04,000 559 00:40:04,000 --> 00:40:10,566 I was asking you a question. You can't answer with a 560 00:40:10,566 --> 00:40:13,012 question. Very good. 561 00:40:13,012 --> 00:40:19,450 The magnitude of that force is K times X2 minus X1, 562 00:40:19,450 --> 00:40:22,025 nonnegotiable, right? 563 00:40:22,025 --> 00:40:29,236 Now, if X2 is larger than X1, do we agree that the spring 564 00:40:29,236 --> 00:40:35,290 force is in this direction? And do we agree, 565 00:40:35,290 --> 00:40:40,048 then, that this spring force must be in this direction if X2 566 00:40:40,048 --> 00:40:43,758 is larger than X1? Well, that means I can leave 567 00:40:43,758 --> 00:40:48,516 everything the way it is now. As long as I give this a minus 568 00:40:48,516 --> 00:40:53,032 sign that is a plus sign, I'm OK because if X2 is smaller 569 00:40:53,032 --> 00:40:57,709 than X1, it will automatically flip over, and my algebra is 570 00:40:57,709 --> 00:41:00,320 fine. So, therefore, 571 00:41:00,320 --> 00:41:05,603 in my head I can just think of X to being larger than X1, 572 00:41:05,603 --> 00:41:10,698 set up the differential equations, and I no longer have 573 00:41:10,698 --> 00:41:16,169 to worry about the fact that maybe X2 at certain moments in 574 00:41:16,169 --> 00:41:20,603 time is not larger than X1. So, now I set up the 575 00:41:20,603 --> 00:41:24,094 differential equation, MX1 double dot. 576 00:41:24,094 --> 00:41:28,622 So, that's this one. Let's first do the pendulum. 577 00:41:28,622 --> 00:41:34,000 That is minus T times the sign of theta one. 578 00:41:34,000 --> 00:41:38,765 T is MG, so it is minus MG times X1 divided by L. 579 00:41:38,765 --> 00:41:43,829 Do we agree there is that horizontal component here? 580 00:41:43,829 --> 00:41:48,794 The horizontal component of T is driving it back to 581 00:41:48,794 --> 00:41:53,163 equilibrium minus sign. FS, the spring force, 582 00:41:53,163 --> 00:41:56,340 is driving away from equilibrium. 583 00:41:56,340 --> 00:42:02,000 So, it's going to get plus K times X2 minus X1. 584 00:42:02,000 --> 00:42:07,182 Look at this. Signs are very important now. 585 00:42:07,182 --> 00:42:11,131 Very important. And the next one, 586 00:42:11,131 --> 00:42:14,587 X2 double dot, it has, again, 587 00:42:14,587 --> 00:42:21,868 the restoring force due to the pendulum which is minus MG X2 588 00:42:21,868 --> 00:42:29,395 divided by L because it's the sign of theta two now that comes 589 00:42:29,395 --> 00:42:33,203 in. And now it has the spring 590 00:42:33,203 --> 00:42:36,289 force, which now is a negative sign. 591 00:42:36,289 --> 00:42:40,168 And so, now we get minus K times X2 minus X1. 592 00:42:40,168 --> 00:42:44,223 And, when you have this down on your next exam, 593 00:42:44,223 --> 00:42:48,983 you take a deep breath, and you go over each individual 594 00:42:48,983 --> 00:42:53,303 step to make absolutely sure that this is correct. 595 00:42:53,303 --> 00:42:58,151 If there's anything wrong with this, you are dead in the 596 00:42:58,151 --> 00:43:01,148 waters. The whole problem will fall 597 00:43:01,148 --> 00:43:05,021 apart. It may not even be a simple 598 00:43:05,021 --> 00:43:08,690 harmonic oscillator. You get something ridiculous. 599 00:43:08,690 --> 00:43:11,461 Signs are crucial here. So, therefore, 600 00:43:11,461 --> 00:43:13,857 look at this again. Take a pause. 601 00:43:13,857 --> 00:43:18,425 This one, if there's a positive offset is indeed in a negative 602 00:43:18,425 --> 00:43:21,270 direction, a component of T sine theta. 603 00:43:21,270 --> 00:43:25,838 This one, if X2 is larger than X1 is indeed in that direction. 604 00:43:25,838 --> 00:43:29,207 And you notice, if X2 becomes smaller than X1, 605 00:43:29,207 --> 00:43:34,000 well then, this becomes automatically a negative. 606 00:43:34,000 --> 00:43:36,947 So, are you OK? So this is fine. 607 00:43:36,947 --> 00:43:41,702 This one has a restoring force due to the pendulum, 608 00:43:41,702 --> 00:43:47,027 which is T sine theta two negative sine and the restoring 609 00:43:47,027 --> 00:43:51,211 force of the spring always opposite this one: 610 00:43:51,211 --> 00:43:53,208 I'm happy. I am happy. 611 00:43:53,208 --> 00:43:58,438 So, now, we are going to rearrange this a little bit and 612 00:43:58,438 --> 00:44:05,000 we're going to introduce the omega zero squared notation. 613 00:44:05,000 --> 00:44:13,947 So, I'm going to divide M out, and so I'm going to get X2, 614 00:44:13,947 --> 00:44:20,540 X1 double dot. And then, I'm going to bring 615 00:44:20,540 --> 00:44:28,389 the X1's to the left. And so, I get plus omega zero 616 00:44:28,389 --> 00:44:34,392 squared times X1. But you have a minus KX1. 617 00:44:34,392 --> 00:44:38,221 You divide by M. So, you get plus omega S 618 00:44:38,221 --> 00:44:41,762 squared. And, that's [both?] times X1. 619 00:44:41,762 --> 00:44:45,686 You notice that? You both have an X1 term. 620 00:44:45,686 --> 00:44:50,950 And now, the X2 comes out. The X2 has a plus sign on the 621 00:44:50,950 --> 00:44:54,300 right side, so it gets a minus sign. 622 00:44:54,300 --> 00:44:58,128 So, I get minus omega S squared times X2. 623 00:44:58,128 --> 00:45:01,000 And that is zero. 624 00:45:01,000 --> 00:45:06,000 625 00:45:06,000 --> 00:45:13,461 And then we do the next one. We get X2 double dot. 626 00:45:13,461 --> 00:45:21,532 Ah, I get the same terms. I get an omega zero squared. 627 00:45:21,532 --> 00:45:27,167 And I get an omega S squared times X2. 628 00:45:27,167 --> 00:45:36,000 And then I get this plus; minus times minus is plus. 629 00:45:36,000 --> 00:45:40,055 So I get minus omega S squared times X1 equals zero. 630 00:45:40,055 --> 00:45:44,428 When you have reached this point, you take a deep breath 631 00:45:44,428 --> 00:45:49,278 and you go over each little term to make sure that you haven't 632 00:45:49,278 --> 00:45:54,207 accidentally slipped on a minus sign because you slip up on one 633 00:45:54,207 --> 00:45:56,990 minus sign you're dead in the water. 634 00:45:56,990 --> 00:46:02,000 It may not even become a simple harmonic oscillation. 635 00:46:02,000 --> 00:46:04,776 So let me do that. I agree with that. 636 00:46:04,776 --> 00:46:07,553 I agree with this. I agree with that. 637 00:46:07,553 --> 00:46:11,256 I agree with that. Differential equation is fine. 638 00:46:11,256 --> 00:46:15,961 Notice that I already did point number one, and that already I 639 00:46:15,961 --> 00:46:19,663 did even number two. I've set up the differential 640 00:46:19,663 --> 00:46:23,983 equation, Newton's Second Law. So now comes number three. 641 00:46:23,983 --> 00:46:28,225 And number three means and going to substitute in there, 642 00:46:28,225 --> 00:46:34,395 X1 is C1. Cosine omega T and X2 is C2 643 00:46:34,395 --> 00:46:37,000 cosine omega T. 644 00:46:37,000 --> 00:46:43,000 645 00:46:43,000 --> 00:46:46,822 We go slowly; I want you to follow each step 646 00:46:46,822 --> 00:46:51,888 and I'm going to work on here so that you can see it high. 647 00:46:51,888 --> 00:46:56,866 So, I'm going to substitute this in here so the X1 double 648 00:46:56,866 --> 00:47:00,866 dot will give me a minus omega squared, right, 649 00:47:00,866 --> 00:47:07,000 because you get twice omega out minus omega squared times C1. 650 00:47:07,000 --> 00:47:09,520 And, to hell with cosine omega T. 651 00:47:09,520 --> 00:47:11,962 Why to hell with cosine omega T? 652 00:47:11,962 --> 00:47:15,349 Because every term will have cosine omega T. 653 00:47:15,349 --> 00:47:19,208 So you might as well divide out right away, right? 654 00:47:19,208 --> 00:47:24,092 Every term that you're going to have in here will have a cosine 655 00:47:24,092 --> 00:47:27,636 omega T in it. So, I'll leave the cosine omega 656 00:47:27,636 --> 00:47:34,610 T already out now. So, I get minus omega squared 657 00:47:34,610 --> 00:47:39,220 C1. Then I get plus omega zero 658 00:47:39,220 --> 00:47:45,102 squared plus omega S squared times C1. 659 00:47:45,102 --> 00:47:52,733 That is this term, minus omega S squared times C2 660 00:47:52,733 --> 00:48:01,000 equals zero because, remember, X2 has the C2. 661 00:48:01,000 --> 00:48:04,099 Am I going to fast? Beautiful. 662 00:48:04,099 --> 00:48:09,872 So, I go to the second one. The second one is X2 double 663 00:48:09,872 --> 00:48:13,293 dot. So, I also get a minus omega 664 00:48:13,293 --> 00:48:17,889 squared times C2. And then, here I get a C2. 665 00:48:17,889 --> 00:48:23,555 So, I get plus omega zero squared plus omega S squared 666 00:48:23,555 --> 00:48:27,617 times C2. And then, I get minus omega S 667 00:48:27,617 --> 00:48:34,400 squared times C1 equals zero. And, when you have reached this 668 00:48:34,400 --> 00:48:37,653 point in your exam, you take a deep breath. 669 00:48:37,653 --> 00:48:41,526 And, you make sure that all your signs are correct. 670 00:48:41,526 --> 00:48:45,940 If not, you're dead in the water and the problem will fall 671 00:48:45,940 --> 00:48:47,799 apart. So, let's do that. 672 00:48:47,799 --> 00:48:51,207 Minus omega squared C1, I can live with that, 673 00:48:51,207 --> 00:48:53,298 plus that term, which is C1, 674 00:48:53,298 --> 00:48:57,945 yes, minus omega S squared C2. Even if you make a little slip 675 00:48:57,945 --> 00:49:02,747 of the pen, and you change this into a one and this into a two, 676 00:49:02,747 --> 00:49:06,000 it's all over, of course. 677 00:49:06,000 --> 00:49:09,745 And then, the next equation, minus omega squared C2. 678 00:49:09,745 --> 00:49:13,711 Then I have this times C2, and then at minus that times 679 00:49:13,711 --> 00:49:15,327 C1. We're almost there, 680 00:49:15,327 --> 00:49:18,632 even though it doesn't look that way, does it? 681 00:49:18,632 --> 00:49:22,378 Remember what I said. You cannot solve two equations 682 00:49:22,378 --> 00:49:26,858 with three unknowns there is no way that you can solve for C1, 683 00:49:26,858 --> 00:49:30,531 C2, and for this omega, which is what you're really 684 00:49:30,531 --> 00:49:34,252 after. This is the omega you are 685 00:49:34,252 --> 00:49:37,405 after. But you cancel for C1 divided 686 00:49:37,405 --> 00:49:42,810 by C2 and omega knowing that, already, I'm going to eliminate 687 00:49:42,810 --> 00:49:46,504 C1 over C2. And, you'll see how I do that. 688 00:49:46,504 --> 00:49:50,468 C1 divided by C2: I go to the first equation. 689 00:49:50,468 --> 00:49:53,531 This is not so hard what I'm doing. 690 00:49:53,531 --> 00:49:57,045 I put in my mind this on the right side. 691 00:49:57,045 --> 00:50:02,000 So, I get a plus omega S squared times C2. 692 00:50:02,000 --> 00:50:06,906 And then, I divide C1 by C2. And so, when I get then is the 693 00:50:06,906 --> 00:50:10,374 following. I get omega S squared up stairs 694 00:50:10,374 --> 00:50:14,435 which is this one. And all this comes downstairs, 695 00:50:14,435 --> 00:50:19,087 minus omega squared plus omega zero squared plus omega S 696 00:50:19,087 --> 00:50:20,779 squared. Do we agree? 697 00:50:20,779 --> 00:50:25,347 That's the first equation. I've simply written it as C1 698 00:50:25,347 --> 00:50:28,308 divided by C2. I can always do that, 699 00:50:28,308 --> 00:50:32,531 right? Bring the C1's to one side, 700 00:50:32,531 --> 00:50:36,000 the C2's to one side, and divide them. 701 00:50:36,000 --> 00:50:39,843 That's this. And, I'm going to do the same 702 00:50:39,843 --> 00:50:44,625 with the next equation. I bring this up to the right 703 00:50:44,625 --> 00:50:48,562 side, that I have C1's there and C2's here. 704 00:50:48,562 --> 00:50:54,187 And now, this becomes upstairs. So, I get minus omega squared 705 00:50:54,187 --> 00:50:58,312 plus omega zero squared plus omega S squared, 706 00:50:58,312 --> 00:51:03,000 and I get downstairs omega S squared. 707 00:51:03,000 --> 00:51:07,614 And now, I have eliminated C1 and C2 because if I solve this 708 00:51:07,614 --> 00:51:10,586 equation, I get my solutions for omega. 709 00:51:10,586 --> 00:51:13,871 This is one equation. There is one unknown. 710 00:51:13,871 --> 00:51:16,452 That's omega. You've got to admit, 711 00:51:16,452 --> 00:51:19,033 and there better be two solutions. 712 00:51:19,033 --> 00:51:23,648 There better be an omega minus, and there better be an omega 713 00:51:23,648 --> 00:51:27,871 plus coming out of this untouched by young human hands. 714 00:51:27,871 --> 00:51:34,369 I must find two solutions. Let's first make sure that this 715 00:51:34,369 --> 00:51:37,956 is correct, and the answer is yes. 716 00:51:37,956 --> 00:51:43,173 So, what do I do now? Well, I'm going to simplify 717 00:51:43,173 --> 00:51:48,065 this one step further, which is now very easy. 718 00:51:48,065 --> 00:51:52,739 I multiply this by this, and this with this. 719 00:51:52,739 --> 00:51:59,043 So, we get omega S through the power of four is minus omega 720 00:51:59,043 --> 00:52:06,000 squared plus omega zero squared plus omega S squared. 721 00:52:06,000 --> 00:52:13,033 One equation with one unknown; omega is the only unknown. 722 00:52:13,033 --> 00:52:17,429 Take the square root left and right. 723 00:52:17,429 --> 00:52:24,589 So, I get minus omega squared plus omega zero squared plus 724 00:52:24,589 --> 00:52:32,000 omega S squared equals plus or minus omega S squared. 725 00:52:32,000 --> 00:52:38,279 Do not forget the plus or minus because the square root of omega 726 00:52:38,279 --> 00:52:43,064 S to the fourth is plus or minus omega S squared. 727 00:52:43,064 --> 00:52:47,948 And now, you're going to see the light of the day. 728 00:52:47,948 --> 00:52:53,032 You won't believe this. I bring omega squared to the 729 00:52:53,032 --> 00:52:59,012 other side, and I get omega zero squared plus omega S squared 730 00:52:59,012 --> 00:53:05,106 minus or plus omega S squared. And, the simplicity is 731 00:53:05,106 --> 00:53:09,127 overpowering. I feel it over my whole body. 732 00:53:09,127 --> 00:53:13,914 It is unbelievable. When I have a minus sign here I 733 00:53:13,914 --> 00:53:18,893 find that omega is omega zero. That's my omega minus. 734 00:53:18,893 --> 00:53:22,436 That's the one. That's my omega minus. 735 00:53:22,436 --> 00:53:28,276 When there is a plus sign here, I find exactly what we predict 736 00:53:28,276 --> 00:53:32,916 before. I find that omega plus is the 737 00:53:32,916 --> 00:53:38,648 square root of omega zero squared plus two omega S squared 738 00:53:38,648 --> 00:53:42,167 because the plus sign make this two. 739 00:53:42,167 --> 00:53:45,988 Isn't that elegant? Isn't it beautiful? 740 00:53:45,988 --> 00:53:51,117 So, they just pop out. And I predict that there will 741 00:53:51,117 --> 00:53:54,737 be two solutions. You have them both. 742 00:53:54,737 --> 00:54:00,368 So, by substituting this in there, automatically come out 743 00:54:00,368 --> 00:54:06,000 two solutions. But, what now is C1 over C2? 744 00:54:06,000 --> 00:54:12,620 We haven't solved for that yet. But I promise you can always 745 00:54:12,620 --> 00:54:18,567 solve it in terms of omega and in terms of C1 over C2. 746 00:54:18,567 --> 00:54:21,933 Any volunteers? Any volunteers? 747 00:54:21,933 --> 00:54:27,881 Look at the blackboard. It's somewhere hidden in those 748 00:54:27,881 --> 00:54:31,471 equations. I said we are going to 749 00:54:31,471 --> 00:54:37,082 eliminate C1 over C2. You think C1 over C2 likes to 750 00:54:37,082 --> 00:54:41,884 be eliminated? Let's call it back. 751 00:54:41,884 --> 00:54:47,242 And look at this equation. Substitute in here for omega 752 00:54:47,242 --> 00:54:50,615 squared. Substitute in omega minus. 753 00:54:50,615 --> 00:54:54,285 Then you get an answer for C1 over C2. 754 00:54:54,285 --> 00:54:57,460 What do you think that answer is? 755 00:54:57,460 --> 00:55:00,634 What do you think that answer is? 756 00:55:00,634 --> 00:55:02,817 Plus one. But even one. 757 00:55:02,817 --> 00:55:07,300 Plus one. So, you get C1 over C2 is plus 758 00:55:07,300 --> 00:55:09,984 one. If you take this omega zero and 759 00:55:09,984 --> 00:55:13,819 you put it in here, you get omega zero squared with 760 00:55:13,819 --> 00:55:16,119 a minus sign, with a plus sign. 761 00:55:16,119 --> 00:55:20,030 You have omega S squared divided by omega S squared, 762 00:55:20,030 --> 00:55:23,558 and it's plus one. And now, you take the second 763 00:55:23,558 --> 00:55:26,012 solution, and you put it in here. 764 00:55:26,012 --> 00:55:30,000 What do you think you're going to find? 765 00:55:30,000 --> 00:55:33,518 Minus one. C1 over C2 is now minus one. 766 00:55:33,518 --> 00:55:39,259 And so, what you have seen now is that the general recipe comes 767 00:55:39,259 --> 00:55:43,981 up with frequencies, comes up with the ratios of the 768 00:55:43,981 --> 00:55:48,888 amplitudes, not with the individual amplitudes because 769 00:55:48,888 --> 00:55:52,222 you don't know the initial condition. 770 00:55:52,222 --> 00:55:56,296 It comes up with the ratio of the amplitudes, 771 00:55:56,296 --> 00:56:01,574 and now you can write any solution, provided that you know 772 00:56:01,574 --> 00:56:06,842 the initial conditions. If you know the initial 773 00:56:06,842 --> 00:56:09,807 conditions, then you can also find C1. 774 00:56:09,807 --> 00:56:12,932 And therefore, since you know the ratio, 775 00:56:12,932 --> 00:56:17,099 you automatically have C2. And, you can find this C1. 776 00:56:17,099 --> 00:56:21,666 So, I'll give this one a minus sign, and this a plus sign. 777 00:56:21,666 --> 00:56:24,951 And then, you know the ratio in that mode. 778 00:56:24,951 --> 00:56:30,000 So, now you may think erroneously that life is easy. 779 00:56:30,000 --> 00:56:35,935 That's far from the truth. Suppose you have something as 780 00:56:35,935 --> 00:56:38,741 simple as this: L, M, L, M. 781 00:56:38,741 --> 00:56:44,244 It's called a double pendulum. There is no symmetry. 782 00:56:44,244 --> 00:56:50,179 I want to test your intuition. There are going to be two 783 00:56:50,179 --> 00:56:55,791 modes, two normal modes: a lower one and a higher one 784 00:56:55,791 --> 00:57:02,440 because they are two objects. In the lowest frequency, 785 00:57:02,440 --> 00:57:08,135 use your hands in your legs. What will this pendulum look 786 00:57:08,135 --> 00:57:11,593 like? You are all doing sort of the 787 00:57:11,593 --> 00:57:15,254 right thing. Would it look like this, 788 00:57:15,254 --> 00:57:19,525 or would it look like this? In other words, 789 00:57:19,525 --> 00:57:25,627 let me make this one a little bit, let me make the difference 790 00:57:25,627 --> 00:57:29,694 a little larger. Would it look like this, 791 00:57:29,694 --> 00:57:36,000 or would it look like this? Slightly exaggerated. 792 00:57:36,000 --> 00:57:41,636 Who is for this one? OK, that means that the ratio, 793 00:57:41,636 --> 00:57:46,370 C2 over C1, is going to be plus two, right? 794 00:57:46,370 --> 00:57:50,992 Who is for this one? That's the way it is, 795 00:57:50,992 --> 00:57:56,290 believe it or not. C2 over C1 is going to be one 796 00:57:56,290 --> 00:58:02,152 plus the square root of two. And, it will take you 30 797 00:58:02,152 --> 00:58:07,000 minutes of grinding to find that. 798 00:58:07,000 --> 00:58:12,235 And I'm not exaggerating when I say 30 minutes. 799 00:58:12,235 --> 00:58:18,951 You have to go to the whole procedure and then you will find 800 00:58:18,951 --> 00:58:24,528 that this ratio is 2.4. And I will demonstrate it. 801 00:58:24,528 --> 00:58:29,536 The highest mode must be something like this, 802 00:58:29,536 --> 00:58:33,292 right? Because they must be out of 803 00:58:33,292 --> 00:58:37,947 pace. Any idea of the ratio, 804 00:58:37,947 --> 00:58:41,832 C1 over C2? I'll call this C1, 805 00:58:41,832 --> 00:58:45,583 call this C2? Any one of you, 806 00:58:45,583 --> 00:58:49,066 any intuition? Minus, yeah? 807 00:58:49,066 --> 00:58:52,952 Boy, you're good. You're good. 808 00:58:52,952 --> 00:58:58,980 It is minus 2.4. This one will be much further 809 00:58:58,980 --> 00:59:05,148 away than this one. Do any one of you want to make 810 00:59:05,148 --> 00:59:09,086 a guess what omega minus is? Anyone want to make a guess 811 00:59:09,086 --> 00:59:12,237 what omega plus is? No way on Earth have you, 812 00:59:12,237 --> 00:59:15,459 or I, or anyone else can look at this and say, 813 00:59:15,459 --> 00:59:18,180 oh yes, of course, omega minus is this. 814 00:59:18,180 --> 00:59:21,259 It's a long road. 30 minutes of calculating, 815 00:59:21,259 --> 00:59:24,481 and then out of that popped these frequencies. 816 00:59:24,481 --> 00:59:27,202 And out of that, then, pops the ratios. 817 00:59:27,202 --> 00:59:32,000 And now I'm going to demonstrate you just this case. 818 00:59:32,000 --> 00:59:35,780 And the way I'm going to do that, even though we have not 819 00:59:35,780 --> 00:59:39,493 discussed driven coupled oscillators, in order to set it 820 00:59:39,493 --> 00:59:42,869 off in these normal modes, a normal modes is also a 821 00:59:42,869 --> 00:59:45,097 resonance frequency. We call that, 822 00:59:45,097 --> 00:59:48,742 often, natural frequency. It's what the system likes to 823 00:59:48,742 --> 00:59:51,983 do if you leave it alone. And, it is easy for me, 824 00:59:51,983 --> 00:59:55,628 when I just tease it a little bit, to excite it in that 825 00:59:55,628 --> 00:59:57,586 resonance mode. So, therefore, 826 00:59:57,586 --> 1:00:00,624 I am going to drive it, but only very briefly. 827 1:00:00,624 --> 1:00:04,000 And then, I will leave it alone. 828 1:00:04,000 --> 1:00:08,302 And I will get it into the state that you will see this 829 1:00:08,302 --> 1:00:11,170 that is [really?] in the normal mode. 830 1:00:11,170 --> 1:00:13,799 There's one at only one frequency. 831 1:00:13,799 --> 1:00:17,464 They come to a halt at the same moment in time. 832 1:00:17,464 --> 1:00:21,208 They are in phase, but you will clearly see that 833 1:00:21,208 --> 1:00:24,554 this distance is more than double this one. 834 1:00:24,554 --> 1:00:27,024 And then, I will go to this one. 835 1:00:27,024 --> 1:00:30,450 I will try that. All right, double pendulum. 836 1:00:30,450 --> 1:00:41,000 Oh, my goodness. This is a double pendulum. Yeah. 837 1:00:41,000 --> 1:00:48,000 838 1:00:48,000 --> 1:00:51,076 So, I have to drive it a little. 839 1:00:51,076 --> 1:00:55,541 But it's really only very little at resonance. 840 1:00:55,541 --> 1:00:58,717 And then, I will stop writing it. 841 1:00:58,717 --> 1:01:03,412 This is the mode. Do you see that the bottom one 842 1:01:03,412 --> 1:01:06,469 is further away than twice the top one? 843 1:01:06,469 --> 1:01:10,410 This is a normal mode. OK, you see that it's not a 844 1:01:10,410 --> 1:01:13,868 straight line? Now, it's hard to say that it 845 1:01:13,868 --> 1:01:17,487 is one plus the square root of two, of course, 846 1:01:17,487 --> 1:01:20,544 from your seats. But that's what it is. 847 1:01:20,544 --> 1:01:24,163 And now, I will excite the second normal mode, 848 1:01:24,163 --> 1:01:26,656 which is a resonance. That's it. 849 1:01:26,656 --> 1:01:31,000 And now, I can almost stop swinging it. 850 1:01:31,000 --> 1:01:35,570 You notice that your upper one has a much larger amplitude than 851 1:01:35,570 --> 1:01:38,739 the lower one, and the ratio is one plus the 852 1:01:38,739 --> 1:01:42,351 square root of two. Did you notice they are out of 853 1:01:42,351 --> 1:01:44,267 phase? It's the minus sign. 854 1:01:44,267 --> 1:01:47,511 They are out of phase. And in the other mode, 855 1:01:47,511 --> 1:01:51,196 they were in phase. And so, all of that will follow 856 1:01:51,196 --> 1:01:54,439 from the recipe: if I gave this problem on an 857 1:01:54,439 --> 1:01:58,862 exam, you would have reasons to kill me because it would take 858 1:01:58,862 --> 1:02:02,536 you too long. I would certainly wear a 859 1:02:02,536 --> 1:02:06,119 bulletproof vest on campus. That is easy because you have 860 1:02:06,119 --> 1:02:08,103 symmetry. This doesn't have that 861 1:02:08,103 --> 1:02:09,000 symmetry. 862 1:02:09,000 --> 1:02:14,000 863 1:02:14,000 --> 1:02:16,345 Yeah, C2 over C1. This is C2. 864 1:02:16,345 --> 1:02:19,947 I haven't specified what I call one and two. 865 1:02:19,947 --> 1:02:24,219 Is that the problem? I call this one and I call this 866 1:02:24,219 --> 1:02:26,816 two. So, C2 over C1 is plus 2.4. 867 1:02:26,816 --> 1:02:32,011 It's good that you asked that. And here, I changed the 868 1:02:32,011 --> 1:02:36,997 sequence because I knew that this one was larger than this 869 1:02:36,997 --> 1:02:40,058 one. So here, you have C1 over C2 is 870 1:02:40,058 --> 1:02:43,294 minus 2.4. The minus sign means out of 871 1:02:43,294 --> 1:02:46,530 phase. Yeah: good that you asked that. 872 1:02:46,530 --> 1:02:48,892 Very good. Well, I'm slowly, 873 1:02:48,892 --> 1:02:54,227 slowly going to turn you into experts, and now we are going to 874 1:02:54,227 --> 1:03:00,000 try something else to see how good your intuition is. 875 1:03:00,000 --> 1:03:03,458 I want you to understand, though, that your intuition is 876 1:03:03,458 --> 1:03:06,413 no better, no worse than my own. In other words, 877 1:03:06,413 --> 1:03:09,997 there is no way that I could have done any better than you 878 1:03:09,997 --> 1:03:11,884 did. When I was in high school, 879 1:03:11,884 --> 1:03:14,902 or whatever it was, maybe in college when I first 880 1:03:14,902 --> 1:03:17,920 saw this, my first impression was, what the hell? 881 1:03:17,920 --> 1:03:20,624 A straight line? But if you give it a little 882 1:03:20,624 --> 1:03:23,957 thought, you will come to the conclusion it cannot be. 883 1:03:23,957 --> 1:03:26,661 Then, it ultimately comes out this solution. 884 1:03:26,661 --> 1:03:31,000 So, don't feel bad if your intuition lets you down. 885 1:03:31,000 --> 1:03:35,562 I'm as bad as you are when it comes to that. 886 1:03:35,562 --> 1:03:42,247 So, now we're going to evaluate the system, which would take you 887 1:03:42,247 --> 1:03:47,765 an hour and a half to work out with a general recipe. 888 1:03:47,765 --> 1:03:51,903 And that is four springs and three cars. 889 1:03:51,903 --> 1:03:56,678 One car, two cars, three cars, and all springs 890 1:03:56,678 --> 1:04:01,028 have spring constant, K, nicely symmetric, 891 1:04:01,028 --> 1:04:07,436 you would think. And all objects have mass M. 892 1:04:07,436 --> 1:04:13,449 And, I want to know, I want to see whether we have 893 1:04:13,449 --> 1:04:20,445 any intuition without being too quantitative for the three 894 1:04:20,445 --> 1:04:26,581 normal mode solutions. There must be an omega minus 895 1:04:26,581 --> 1:04:33,074 where all three are in phase. Then, there must be an omega 896 1:04:33,074 --> 1:04:37,370 plus, which is more complicated. And then, there must be an 897 1:04:37,370 --> 1:04:40,629 omega plus, plus, which is the highest of all 898 1:04:40,629 --> 1:04:42,777 three. In the highest possible 899 1:04:42,777 --> 1:04:46,851 frequency, every object, two adjacent objects are always 900 1:04:46,851 --> 1:04:49,962 out of phase. So, if you have 20 objects in 901 1:04:49,962 --> 1:04:53,518 the highest mode, number one is out of phase with 902 1:04:53,518 --> 1:04:56,629 number two. Number two is out of phase with 903 1:04:56,629 --> 1:04:59,666 number three. Number three is out of phase 904 1:04:59,666 --> 1:05:03,000 with number four, and so on. 905 1:05:03,000 --> 1:05:05,569 Let's first look at omega minus. 906 1:05:05,569 --> 1:05:10,292 I will even dare asking you whether you have any idea what 907 1:05:10,292 --> 1:05:13,193 omega minus is. You don't have that; 908 1:05:13,193 --> 1:05:17,171 I don't have that. Where would your object be if, 909 1:05:17,171 --> 1:05:20,734 for instance, I displaced the first one over 910 1:05:20,734 --> 1:05:24,795 a distance which I just normalized to be plus one? 911 1:05:24,795 --> 1:05:28,856 I call that plus one. Any idea where this one will 912 1:05:28,856 --> 1:05:31,425 be? Any idea where that one will 913 1:05:31,425 --> 1:05:35,453 be? What to think the second one 914 1:05:35,453 --> 1:05:36,942 will be? Plus one. 915 1:05:36,942 --> 1:05:39,658 Will be the third one? Plus one. 916 1:05:39,658 --> 1:05:42,725 Wrong. Why does it have to be wrong? 917 1:05:42,725 --> 1:05:48,157 Suppose this one is a plus one. And, suppose this one is also a 918 1:05:48,157 --> 1:05:51,574 plus one. Then, this spring is no longer 919 1:05:51,574 --> 1:05:55,779 than it wants to be, and this spring is no longer 920 1:05:55,779 --> 1:06:00,072 than it wants to be. So, there is no force on this 921 1:06:00,072 --> 1:06:03,866 object. So, that's not possible. 922 1:06:03,866 --> 1:06:08,533 The object cannot go anywhere. Any suggestions which of those 923 1:06:08,533 --> 1:06:10,477 is wrong? Is this correct, 924 1:06:10,477 --> 1:06:13,822 and it's plus one? How about the middle one? 925 1:06:13,822 --> 1:06:17,477 Where do you want it, a little further away or a 926 1:06:17,477 --> 1:06:20,277 little less? All have to be in phase. 927 1:06:20,277 --> 1:06:24,944 Don't use the word minus sign. If you use the word minus sign 928 1:06:24,944 --> 1:06:30,000 then, how can you do that? They have to be in phase. 929 1:06:30,000 --> 1:06:37,690 They must be on this side. Do we make the distance larger 930 1:06:37,690 --> 1:06:40,849 or smaller? It's larger. 931 1:06:40,849 --> 1:06:48,815 Should I tell you what it is? Should I tell you what it is? 932 1:06:48,815 --> 1:06:52,248 You want to know? Tell me. 933 1:06:52,248 --> 1:06:56,369 You really want to know, right? 934 1:06:56,369 --> 1:06:59,939 I knew it. Is that obvious? 935 1:06:59,939 --> 1:07:04,276 No. Now, the next one. 936 1:07:04,276 --> 1:07:09,588 The next one is interesting. I will, again, 937 1:07:09,588 --> 1:07:16,039 put this one in plus one because that is my starting 938 1:07:16,039 --> 1:07:20,592 point. I always put that at plus one. 939 1:07:20,592 --> 1:07:24,640 Now what? This one you can guess. 940 1:07:24,640 --> 1:07:30,458 You really can. Think now about the symmetry of 941 1:07:30,458 --> 1:07:34,000 the system. Yeah? 942 1:07:34,000 --> 1:07:36,006 Very good. Very good. 943 1:07:36,006 --> 1:07:41,422 The middle one stays still, and this one comes to minus 944 1:07:41,422 --> 1:07:43,629 one. Now the third one. 945 1:07:43,629 --> 1:07:49,045 Remember, the highest mode, neighbors are always out of 946 1:07:49,045 --> 1:07:53,960 phase with each other no matter how many you have. 947 1:07:53,960 --> 1:07:59,778 So, it's already certain that this one must be on the side, 948 1:07:59,778 --> 1:08:07,000 and it's already certain that this one must be on that side. 949 1:08:07,000 --> 1:08:11,847 Any volunteers? Would you like to try that 950 1:08:11,847 --> 1:08:16,221 again? Would you do plus one and minus 951 1:08:16,221 --> 1:08:20,004 one? Help me put this one at plus 952 1:08:20,004 --> 1:08:25,561 one, sorry, minus one, and this one at plus one, 953 1:08:25,561 --> 1:08:31,000 or shall we change the minus one here? 954 1:08:31,000 --> 1:08:35,380 Don't feel bad. I wouldn't have been able to do 955 1:08:35,380 --> 1:08:39,000 either. Shall I tell you what it is? 956 1:08:39,000 --> 1:08:45,000 957 1:08:45,000 --> 1:08:49,299 Now, I can demonstrate this. And the way I'm going to 958 1:08:49,299 --> 1:08:53,846 demonstrate this is with the air track that I have here. 959 1:08:53,846 --> 1:08:58,807 See, the beauty with this air track is that I can place these 960 1:08:58,807 --> 1:09:02,940 cars wherever I want, and they're not going to move 961 1:09:02,940 --> 1:09:08,071 because I don't have any air. Then, I turn on the air and 962 1:09:08,071 --> 1:09:10,757 they are and immediately free to go. 963 1:09:10,757 --> 1:09:13,826 In other words, if I offset this one in a 964 1:09:13,826 --> 1:09:16,818 positive direction over a distance, one, 965 1:09:16,818 --> 1:09:19,196 we've just called this positive. 966 1:09:19,196 --> 1:09:23,109 I apologize for that because I called this positive. 967 1:09:23,109 --> 1:09:27,099 And the other one has to be offset by one in the same 968 1:09:27,099 --> 1:09:31,473 direction, which I just did. But this one has to be offset 969 1:09:31,473 --> 1:09:36,000 by the square root of two, which I just did. 970 1:09:36,000 --> 1:09:40,095 And now, turn on the air, and I just have exactly the 971 1:09:40,095 --> 1:09:43,089 right initial conditions for this mode. 972 1:09:43,089 --> 1:09:47,578 So, it is going to oscillate in the superposition of three 973 1:09:47,578 --> 1:09:50,335 normal modes. But there is only one, 974 1:09:50,335 --> 1:09:54,825 and that's the one you're going to see because I chose the 975 1:09:54,825 --> 1:09:59,000 initial conditions. Are you ready for this? 976 1:09:59,000 --> 1:10:01,974 Now, admit it. This is fantastic. 977 1:10:01,974 --> 1:10:06,901 They are all in phase. That was a criterion for normal 978 1:10:06,901 --> 1:10:09,039 mode. They are in phase. 979 1:10:09,039 --> 1:10:13,315 They come to a halt at the same moment in time. 980 1:10:13,315 --> 1:10:18,335 They all have the same frequency, and amplitude here is 981 1:10:18,335 --> 1:10:23,354 the square of two times larger. That is not so obvious, 982 1:10:23,354 --> 1:10:28,002 but you see this is the criterion for normal modes. 983 1:10:28,002 --> 1:10:32,000 Same frequency, and in phase. 984 1:10:32,000 --> 1:10:35,297 This one, OK, we are going to put this one at 985 1:10:35,297 --> 1:10:37,770 zero. I'm going to put this one at 986 1:10:37,770 --> 1:10:40,617 plus one. And I'm going to put this one 987 1:10:40,617 --> 1:10:43,839 at minus one. Now, this one is offset by one 988 1:10:43,839 --> 1:10:47,286 in this direction. This one is offset by one in 989 1:10:47,286 --> 1:10:50,434 this direction, and this one is not offset. 990 1:10:50,434 --> 1:10:54,031 I turn on the air; what do you think will happen? 991 1:10:54,031 --> 1:10:57,103 That's right, it's going to oscillate in a 992 1:10:57,103 --> 1:11:01,000 superposition of three normal modes. 993 1:11:01,000 --> 1:11:06,000 But there is only one. That's the omega plus. 994 1:11:06,000 --> 1:11:11,000 995 1:11:11,000 --> 1:11:16,588 Impressive, isn't it? Next one, this one. 996 1:11:16,588 --> 1:11:22,455 So, we are first going to put this at zero, 997 1:11:22,455 --> 1:11:28,463 this at zero. And we're going to put this at 998 1:11:28,463 --> 1:11:32,182 zero. And now, what we have to do, 999 1:11:32,182 --> 1:11:35,379 this has to be put over plus one plus one. 1000 1:11:35,379 --> 1:11:39,744 The other one has to be put minus the square root of two. 1001 1:11:39,744 --> 1:11:44,422 And we have measured that here. We've come so close that this 1002 1:11:44,422 --> 1:11:46,916 spring is not going to like that. 1003 1:11:46,916 --> 1:11:51,594 And now, this one is going to be, again, plus one in the same 1004 1:11:51,594 --> 1:11:55,569 direction as this one. That's the highest frequency. 1005 1:11:55,569 --> 1:11:59,000 And there we go. Are you ready? 1006 1:11:59,000 --> 1:12:06,000 1007 1:12:06,000 --> 1:12:07,919 Three, two, one, zero. 1008 1:12:07,919 --> 1:12:13,220 One, and only one frequency. And, every adjacent one is out 1009 1:12:13,220 --> 1:12:17,516 of phase with the other. These are out of phase. 1010 1:12:17,516 --> 1:12:23,000 But these are in phase because these are out of phase. 1011 1:12:23,000 --> 1:12:34,000 1012 1:12:34,000 --> 1:12:37,623 Now, I had one student calculate this for me. 1013 1:12:37,623 --> 1:12:41,823 And I'll be frank with you, I even paid him for that 1014 1:12:41,823 --> 1:12:44,952 because I was too lazy to do it myself. 1015 1:12:44,952 --> 1:12:49,729 I put the job on the market, and he came within a few days. 1016 1:12:49,729 --> 1:12:54,670 There were a few people that said, oh, and an A plus for 8.03 1017 1:12:54,670 --> 1:12:58,376 and he was going to do that in no time at all. 1018 1:12:58,376 --> 1:13:02,000 20 hours later, he had it right. 1019 1:13:02,000 --> 1:13:07,665 It cost me a fortune. Now, I'm going to test you once 1020 1:13:07,665 --> 1:13:12,459 more to show you that these things are not so 1021 1:13:12,459 --> 1:13:17,143 straightforward, not to make you feel bad at 1022 1:13:17,143 --> 1:13:19,322 all. On the contrary, 1023 1:13:19,322 --> 1:13:24,988 my intuition is no better. I'm going to make a triple 1024 1:13:24,988 --> 1:13:30,000 pendulum. This is a triple pendulum. 1025 1:13:30,000 --> 1:13:37,000 1026 1:13:37,000 --> 1:13:41,203 Omega minus, what do you think it would look 1027 1:13:41,203 --> 1:13:44,428 like? My favorite student is here. 1028 1:13:44,428 --> 1:13:48,240 Do you think it will be a straight line? 1029 1:13:48,240 --> 1:13:53,323 I don't think so either. I think what you're going to 1030 1:13:53,323 --> 1:13:58,308 see without making any predictions about amplitudes, 1031 1:13:58,308 --> 1:14:03,000 exaggerated, you're going to see this. 1032 1:14:03,000 --> 1:14:13,064 That's what I think. No idea what it is. 1033 1:14:13,064 --> 1:14:19,000 How about omega plus? 1034 1:14:19,000 --> 1:14:26,000 1035 1:14:26,000 --> 1:14:31,816 This is not going to work. It's a whole different system. 1036 1:14:31,816 --> 1:14:36,489 Two of them are probably going to be in phase, 1037 1:14:36,489 --> 1:14:41,474 and one out of phase. Which two will be in phase? 1038 1:14:41,474 --> 1:14:46,252 Will you see this? Now, these two are in phase, 1039 1:14:46,252 --> 1:14:51,341 and this is out of phase. Or really, you see this. 1040 1:14:51,341 --> 1:14:57,261 Now, these two are in phase, and this one is out of phase. 1041 1:14:57,261 --> 1:15:03,588 Who is in favor of this? Who is in favor of that? 1042 1:15:03,588 --> 1:15:09,303 I want to see the hands long. Who is in favor of this? 1043 1:15:09,303 --> 1:15:14,264 Who is in favor of that? Yeah, well, we'll see. 1044 1:15:14,264 --> 1:15:18,254 [LAUGHTER] How about the highest mode? 1045 1:15:18,254 --> 1:15:24,401 That's the easiest one because now they must go like this. 1046 1:15:24,401 --> 1:15:28,392 Now, of course, the ratio, C1 over C2, 1047 1:15:28,392 --> 1:15:34,000 and C1 over C3, that's a different story. 1048 1:15:34,000 --> 1:15:38,812 That is your 20 hours of grinding if you get paid. 1049 1:15:38,812 --> 1:15:43,821 If you don't get paid, you can probably do it in two 1050 1:15:43,821 --> 1:15:47,160 hours. So, I'm going to demonstrate 1051 1:15:47,160 --> 1:15:51,187 this to you. I'm going to find these three 1052 1:15:51,187 --> 1:15:54,526 modes. Again, I do that by a little 1053 1:15:54,526 --> 1:15:58,651 bit of driving because of their resonances. 1054 1:15:58,651 --> 1:16:02,973 And then, I want you to eyeball these ratios, 1055 1:16:02,973 --> 1:16:08,431 C1 over C2, and C1 over C3. Oh, no problem. 1056 1:16:08,431 --> 1:16:10,625 No problem. There we go. 1057 1:16:10,625 --> 1:16:15,301 All right, so now, this is not where the length is 1058 1:16:15,301 --> 1:16:18,832 the same. So, to get this one into the 1059 1:16:18,832 --> 1:16:23,412 lowest mode is easy for me. Just give it a swing, 1060 1:16:23,412 --> 1:16:26,656 and let it go. And just do nothing. 1061 1:16:26,656 --> 1:16:29,805 Then, I will stop even driving it. 1062 1:16:29,805 --> 1:16:32,000 That's it. 1063 1:16:32,000 --> 1:16:38,000 1064 1:16:38,000 --> 1:16:42,461 Now, it's nicely in that mode. You see, the bottom one [has 1065 1:16:42,461 --> 1:16:45,230 it?]. Notice that the lines are never 1066 1:16:45,230 --> 1:16:48,846 in the same direction. You see the offset of the 1067 1:16:48,846 --> 1:16:51,000 angles? Can you all see that? 1068 1:16:51,000 --> 1:16:55,384 So, our first picture on the blackboard is quite accurate. 1069 1:16:55,384 --> 1:17:00,000 Now, the second one, that's harder for me to find. 1070 1:17:00,000 --> 1:17:04,070 I mean, we first make sure that they are not moving. 1071 1:17:04,070 --> 1:17:07,262 OK, so I'm going to try to search for it. 1072 1:17:07,262 --> 1:17:08,859 I got it. This is it. 1073 1:17:08,859 --> 1:17:12,131 So, the first solution is the correct one. 1074 1:17:12,131 --> 1:17:15,802 The upper two are in phase, and the bottom one, 1075 1:17:15,802 --> 1:17:19,394 look at the large amplitude of the bottom one. 1076 1:17:19,394 --> 1:17:23,704 That's not so intuitive. The bottom one is out of phase 1077 1:17:23,704 --> 1:17:26,417 with the upper two. Can you see it? 1078 1:17:26,417 --> 1:17:29,291 Normal mode, why is it a normal mode? 1079 1:17:29,291 --> 1:17:34,000 Because they all have the same frequency. 1080 1:17:34,000 --> 1:17:38,904 And, they are in phase or they are out of phase. 1081 1:17:38,904 --> 1:17:44,643 Now, the highest frequency, let me see whether I can get 1082 1:17:44,643 --> 1:17:46,000 that one. 1083 1:17:46,000 --> 1:17:52,000 1084 1:17:52,000 --> 1:18:05,063 Got it. That's it. 1085 1:18:05,063 --> 1:18:46,557 Now, look how small the amplitude of the bottom one is. 1086 1:18:46,557 --> 1:19:10,378 And, that can all be calculated. 1087 1:19:10,378 --> 1:19:41,883 They are beginning to rotate a little bit. 1088 1:19:41,883 --> 1:20:08,009 There we go. That's the third mode. 1089 1:20:08,009 --> 1:20:48,735 So, now you know everything there is to be known about 1090 1:20:48,735 --> 1:21:26,387 coupled oscillators. The most important thing that 1091 1:21:26,387 --> 1:22:10,955 you probably have learned is that if you really want to get 1092 1:22:10,955 --> 1:22:49,376 the general solution for a system, it is hard work. 1093 1:22:49,376 --> 1:23:05,513 See you next Tuesday. Have a good weekend.