1 00:00:25,000 --> 00:00:30,864 I will start today with a very interesting phenomenon, 2 00:00:30,864 --> 00:00:34,737 which is known as a beat phenomenon. 3 00:00:34,737 --> 00:00:41,045 Suppose you have two simple harmonic motions with the same 4 00:00:41,045 --> 00:00:45,471 amplitude but with different frequencies. 5 00:00:45,471 --> 00:00:50,340 So, we have X1 is A times cosine omega one T, 6 00:00:50,340 --> 00:00:54,987 and you have X2, which is A times omega 2T. 7 00:00:54,987 --> 00:00:59,635 And, the two are different, same amplitude. 8 00:00:59,635 --> 00:01:04,504 If you sum them, X1 plus X2, you get 2A times 9 00:01:04,504 --> 00:01:11,254 the cosine of half the sum omega one plus omega two divided by 10 00:01:11,254 --> 00:01:19,000 two times T times the cosine of half the difference. 11 00:01:19,000 --> 00:01:25,240 And, when you look at this, you have here a frequency which 12 00:01:25,240 --> 00:01:28,898 is high compared to this frequency. 13 00:01:28,898 --> 00:01:36,000 You can best see that if omega one is very close to omega two. 14 00:01:36,000 --> 00:01:39,739 And, you could simply replace that by omega. 15 00:01:39,739 --> 00:01:43,391 Then, this is simply saying cosine omega T. 16 00:01:43,391 --> 00:01:48,608 and so, if you change from the notation of omega in your head 17 00:01:48,608 --> 00:01:53,130 to frequency in hertz, which is a lot easier to think 18 00:01:53,130 --> 00:01:56,608 of, omega is 2 pi times the frequency, F. 19 00:01:56,608 --> 00:02:00,347 This is in hertz, and this is in radians per 20 00:02:00,347 --> 00:02:04,111 second. Suppose you have an F1, 21 00:02:04,111 --> 00:02:07,416 which is 256 Hz. And, you have an F2, 22 00:02:07,416 --> 00:02:11,547 which is 254 Hz. If you take these numbers now 23 00:02:11,547 --> 00:02:15,586 as a working example, then the period of this 24 00:02:15,586 --> 00:02:20,911 oscillation is 1/256ths of a second, but the period of this 25 00:02:20,911 --> 00:02:26,144 oscillation is only one second because it's the difference 26 00:02:26,144 --> 00:02:34,331 between the two divided by two. And so, what happens is that if 27 00:02:34,331 --> 00:02:40,200 you make a plot of X as a function of time, 28 00:02:40,200 --> 00:02:45,650 I will first sketch this very slow term. 29 00:02:45,650 --> 00:02:54,314 So, this is very slow compared to this one, which is very fast. 30 00:02:54,314 --> 00:03:02,000 I will first make a sketch of the very slow one. 31 00:03:02,000 --> 00:03:07,766 So, this is the slow one. And, I will make also a dotted 32 00:03:07,766 --> 00:03:13,846 line here to guide my hand when I'm going to make the plot. 33 00:03:13,846 --> 00:03:20,137 So, what you're going to see now is that the faster go [MAKES 34 00:03:20,137 --> 00:03:24,225 NOISE], and the slow one goes like this. 35 00:03:24,225 --> 00:03:28,000 And, the net result is this. 36 00:03:28,000 --> 00:03:33,000 37 00:03:33,000 --> 00:03:37,698 This is the fast one. And so, you see right here is 38 00:03:37,698 --> 00:03:42,585 when the slow one is zero. So, the slow one kills the 39 00:03:42,585 --> 00:03:45,780 amplitude. You can think of this as 40 00:03:45,780 --> 00:03:51,325 amplitude modulation that the A is multiplied by that cosine 41 00:03:51,325 --> 00:03:54,426 function. That is just a matter of 42 00:03:54,426 --> 00:03:57,903 definition. We define this as the beat 43 00:03:57,903 --> 00:04:00,910 period when the two are in phase. 44 00:04:00,910 --> 00:04:03,918 And here they are again in phase. 45 00:04:03,918 --> 00:04:07,113 The two are here 180° out of phase. 46 00:04:07,113 --> 00:04:12,000 We define this as the beat period. 47 00:04:12,000 --> 00:04:18,279 For the working example that we have on the blackboard there, 48 00:04:18,279 --> 00:04:24,139 the 256 Hz and the 254 Hz, from here to here would be one 49 00:04:24,139 --> 00:04:27,802 second. That's the time for the slow 50 00:04:27,802 --> 00:04:32,148 one to go 360°. And so the beat period would, 51 00:04:32,148 --> 00:04:34,370 in that case, be half a second. 52 00:04:34,370 --> 00:04:38,296 So, if you want to see in your head what is happening, 53 00:04:38,296 --> 00:04:42,592 then if one oscillation is like this, that when they are in 54 00:04:42,592 --> 00:04:45,185 phase, each one has an amplitude, A. 55 00:04:45,185 --> 00:04:47,703 When they are in phase, you get 2A. 56 00:04:47,703 --> 00:04:52,148 But when they are out of phase, you have to add this 180° out 57 00:04:52,148 --> 00:04:54,962 of phase. Then, the net result is zero. 58 00:04:54,962 --> 00:04:58,814 And, that's the case here. And that's the case there, 59 00:04:58,814 --> 00:05:04,747 and that's the case there. And, this is known in the 60 00:05:04,747 --> 00:05:11,670 literature as a beat phenomenon. And, the frequency of the beat, 61 00:05:11,670 --> 00:05:17,164 so, F beat is then defined as delta F, in our case, 62 00:05:17,164 --> 00:05:21,450 two hertz. Indeed, you see that the beat 63 00:05:21,450 --> 00:05:27,384 period is half a second. A very nice way to demonstrate 64 00:05:27,384 --> 00:05:32,000 this is to make you listen to it. 65 00:05:32,000 --> 00:05:35,173 I have here a tuning fork of 256 Hz. 66 00:05:35,173 --> 00:05:39,162 And, I have another one which is also 256 Hz. 67 00:05:39,162 --> 00:05:44,149 But, I can change this one by putting a little weight on 68 00:05:44,149 --> 00:05:47,504 there. But I can make the frequency of 69 00:05:47,504 --> 00:05:50,767 little lower. I'll put a weight here, 70 00:05:50,767 --> 00:05:55,210 and I will simultaneously make you listen to them. 71 00:05:55,210 --> 00:06:00,288 And if somehow I managed to give them the same amplitude, 72 00:06:00,288 --> 00:06:06,000 your eardrum will reach the signal, which is 256. 73 00:06:06,000 --> 00:06:10,189 And then, there comes a time that you hear nothing. 74 00:06:10,189 --> 00:06:13,625 That means sound plus sound makes silence, 75 00:06:13,625 --> 00:06:17,145 an interesting concept called interference. 76 00:06:17,145 --> 00:06:21,000 You can hear that. So, listen carefully. 77 00:06:21,000 --> 00:06:28,000 78 00:06:28,000 --> 00:06:32,333 That's the beat. If I make the frequency 79 00:06:32,333 --> 00:06:36,666 different larger, the beats come faster. 80 00:06:36,666 --> 00:06:41,333 Can all of you hear this, also in the back? 81 00:06:41,333 --> 00:06:46,000 So now I will make to offset larger. 82 00:06:46,000 --> 00:06:53,000 83 00:06:53,000 --> 00:06:56,481 It's unpleasant, makes you feel like throwing up 84 00:06:56,481 --> 00:06:59,000 or so, very high beat frequency. 85 00:06:59,000 --> 00:07:06,000 86 00:07:06,000 --> 00:07:08,768 If the amplitude of the two are not the same, 87 00:07:08,768 --> 00:07:12,355 then you don't have to be a rocket scientist to anticipate 88 00:07:12,355 --> 00:07:15,123 what will happen. You just never get the zero 89 00:07:15,123 --> 00:07:16,948 here. But you always have some 90 00:07:16,948 --> 00:07:19,402 residual. And so, you will get something 91 00:07:19,402 --> 00:07:22,548 that looks like this. You still have an oscillation 92 00:07:22,548 --> 00:07:24,624 here. And, I can also make you see 93 00:07:24,624 --> 00:07:26,701 that. It's harder to make you hear 94 00:07:26,701 --> 00:07:30,161 that because your ears cannot really tell the difference 95 00:07:30,161 --> 00:07:34,000 between a little bit of sound and no sound. 96 00:07:34,000 --> 00:07:37,672 You hear [SINGS] and that's all you hear. 97 00:07:37,672 --> 00:07:42,904 But I can make you see that. And I'm going to make you see 98 00:07:42,904 --> 00:07:48,688 that right there when we pick up the signals of these two tuning 99 00:07:48,688 --> 00:07:54,104 forks, there's a microphone. And, we will just play it there 100 00:07:54,104 --> 00:07:58,878 with an oscilloscope. So, you will hear it and you'll 101 00:07:58,878 --> 00:08:03,689 see it at the same time. And, it is unlikely when I 102 00:08:03,689 --> 00:08:07,588 strike them that the amplitude has received by the microphone 103 00:08:07,588 --> 00:08:11,032 will be exactly the same. So, it's not likely you will 104 00:08:11,032 --> 00:08:13,891 see it go through zero. But I can, of course, 105 00:08:13,891 --> 00:08:17,920 change the distance through the microphone and try to get it as 106 00:08:17,920 --> 00:08:20,000 close as I can to zero. 107 00:08:20,000 --> 00:08:24,000 108 00:08:24,000 --> 00:08:28,592 Not bad. Now it's close to zero. 109 00:08:28,592 --> 00:08:32,000 I'll do it once more. 110 00:08:32,000 --> 00:08:40,000 111 00:08:40,000 --> 00:08:43,333 Now, I purposely make amplitudes different. 112 00:08:43,333 --> 00:08:46,507 I bring one very close to the microphone. 113 00:08:46,507 --> 00:08:50,000 This is what you saw there on the board. 114 00:08:50,000 --> 00:08:55,000 115 00:08:55,000 --> 00:09:00,000 If I make amplitudes the same, then you get -- 116 00:09:00,000 --> 00:09:06,000 117 00:09:06,000 --> 00:09:08,000 You get the zero back. 118 00:09:08,000 --> 00:09:18,000 119 00:09:18,000 --> 00:09:21,082 Beat phenomenon are actually quite common. 120 00:09:21,082 --> 00:09:25,744 When you see an airplane on the runway and the airplane has two 121 00:09:25,744 --> 00:09:29,954 engines, perhaps you have noticed something irritating at 122 00:09:29,954 --> 00:09:34,766 times that is some crazy humming sound that goes up in volume and 123 00:09:34,766 --> 00:09:38,000 then it goes down and then goes up. 124 00:09:38,000 --> 00:09:42,500 That is almost certainly the two engines beating against each 125 00:09:42,500 --> 00:09:45,274 other. They're not rotating at exactly 126 00:09:45,274 --> 00:09:47,975 the same frequency, but slightly off. 127 00:09:47,975 --> 00:09:52,100 That's my explanation why you have that strangely rising 128 00:09:52,100 --> 00:09:55,774 volume and then decaying volume. It's very common. 129 00:09:55,774 --> 00:09:59,225 When I was a student, I remember there were two 130 00:09:59,225 --> 00:10:03,776 [faucets?] in my bedroom. And, no matter how much I 131 00:10:03,776 --> 00:10:06,619 tightened the faucets, they were leaking. 132 00:10:06,619 --> 00:10:10,172 They were dripping. And, they were dripping in such 133 00:10:10,172 --> 00:10:14,223 a way that the frequency with which the drops came out was 134 00:10:14,223 --> 00:10:16,923 almost the same. So, it went like this: 135 00:10:16,923 --> 00:10:18,274 clunk, clunk, clunk. 136 00:10:18,274 --> 00:10:21,472 But, the frequency was never exactly the same. 137 00:10:21,472 --> 00:10:24,385 So, after awhile it would go clunk, clunk, 138 00:10:24,385 --> 00:10:26,233 clunk, clunk, clunk, clunk, 139 00:10:26,233 --> 00:10:30,000 and then after a while clunk, clunk. 140 00:10:30,000 --> 00:10:32,563 It drove me crazy. But, of course, 141 00:10:32,563 --> 00:10:36,446 I take pride in knowing that it's a big phenomenon, 142 00:10:36,446 --> 00:10:39,941 except that there, sound plus sound never made 143 00:10:39,941 --> 00:10:44,679 silence because when they were 180° out of phase you hear both 144 00:10:44,679 --> 00:10:46,310 drops. You're kidding, 145 00:10:46,310 --> 00:10:50,194 but that's the way it is. Another example of [bit?] 146 00:10:50,194 --> 00:10:54,077 phenomenon whereby two oscillators very closely the 147 00:10:54,077 --> 00:10:58,504 same period, give it enough time, they go out of phase and 148 00:10:58,504 --> 00:11:04,183 then they go back into phase. We have made an effort here 149 00:11:04,183 --> 00:11:07,579 with these two pendulums, really an effort, 150 00:11:07,579 --> 00:11:12,107 about 60 cm length to really make these lengths the same. 151 00:11:12,107 --> 00:11:16,070 And so, when I give them both a certain amplitude, 152 00:11:16,070 --> 00:11:19,385 they will go in unison. They are in phase, 153 00:11:19,385 --> 00:11:24,318 but no matter how hard you try, they will never be exactly the 154 00:11:24,318 --> 00:11:27,148 same. So, as I continue the lecture, 155 00:11:27,148 --> 00:11:32,000 you'll see that they gradually go out of phase. 156 00:11:32,000 --> 00:11:36,516 And, it may take five minutes for them to be out of phase. 157 00:11:36,516 --> 00:11:39,765 That would be, then, the time from here to 158 00:11:39,765 --> 00:11:42,300 here. That would be half a [bit?] 159 00:11:42,300 --> 00:11:44,677 period. So, I'll show you that, 160 00:11:44,677 --> 00:11:49,193 and I'm not going to wait for them to be 180° out of phase 161 00:11:49,193 --> 00:11:54,106 because I can take a long time since the periods are so closely 162 00:11:54,106 --> 00:11:56,879 the same. If you looked at them now, 163 00:11:56,879 --> 00:12:01,000 you'd say, oh boy, are they in unison! 164 00:12:01,000 --> 00:12:04,654 Keep an eye on it. And if you see at minutes from 165 00:12:04,654 --> 00:12:08,005 now, they will no longer be exactly in phase. 166 00:12:08,005 --> 00:12:11,507 When you think of a conductor for an orchestra, 167 00:12:11,507 --> 00:12:15,847 with the conductor is really doing: he is making sure that 168 00:12:15,847 --> 00:12:20,111 all the instruments in phase. He's really a phase locker; 169 00:12:20,111 --> 00:12:23,233 he's a phase keeper. If he didn't do that, 170 00:12:23,233 --> 00:12:26,507 then each person would play at his own pace. 171 00:12:26,507 --> 00:12:31,000 Then, of course, you would end up with chaos. 172 00:12:31,000 --> 00:12:35,663 So, now perhaps you know what the expression comes from: 173 00:12:35,663 --> 00:12:39,479 keep the phase. Now, I go to the heart of this 174 00:12:39,479 --> 00:12:42,701 lecture. And, the heart of this lecture 175 00:12:42,701 --> 00:12:45,923 is damping. You see it's already out of 176 00:12:45,923 --> 00:12:49,570 phase a little? Not completely out of phase, 177 00:12:49,570 --> 00:12:53,301 but they no longer, just keep an eye on them. 178 00:12:53,301 --> 00:12:58,134 You'll see that they grow. At the heart of this lecture is 179 00:12:58,134 --> 00:13:02,580 damping. When we have a string or we 180 00:13:02,580 --> 00:13:08,316 have a pendulum and we let it oscillate, we all note that the 181 00:13:08,316 --> 00:13:14,338 oscillation will come to a halt. That means there is friction or 182 00:13:14,338 --> 00:13:18,926 there is air drag, and it will ultimately die out 183 00:13:18,926 --> 00:13:22,845 or it will stop just like these will stop. 184 00:13:22,845 --> 00:13:27,051 The frictional force always opposes velocity. 185 00:13:27,051 --> 00:13:33,073 And the frictional force in its most general form as a vector is 186 00:13:33,073 --> 00:13:39,000 minus C1 times V, whereby C is a positive number. 187 00:13:39,000 --> 00:13:44,980 So you see that it opposes velocity, minus C2 times V 188 00:13:44,980 --> 00:13:48,659 squared. And, I put here the unit 189 00:13:48,659 --> 00:13:55,100 vector, V, to remind you that it's opposing the velocity. 190 00:13:55,100 --> 00:14:00,504 This term has a name. This is called the viscous 191 00:14:00,504 --> 00:14:04,929 term. And the other term has a name 192 00:14:04,929 --> 00:14:08,262 which is called the pressure term. 193 00:14:08,262 --> 00:14:14,121 One is proportional with the velocity, and the other one is 194 00:14:14,121 --> 00:14:17,757 proportional with the square of that. 195 00:14:17,757 --> 00:14:23,212 In my lecture number 12, which is on OCW which is of my 196 00:14:23,212 --> 00:14:28,363 1999 Newtonian mechanics lectures, I spent an entire 197 00:14:28,363 --> 00:14:33,643 lecture on this equation. And I discussed when the 198 00:14:33,643 --> 00:14:37,136 viscous term is dominant and when the square term is 199 00:14:37,136 --> 00:14:39,602 dominant. I do demonstrations in both 200 00:14:39,602 --> 00:14:41,726 domains. If the velocity is low, 201 00:14:41,726 --> 00:14:44,739 this term dominates. If the velocity is high, 202 00:14:44,739 --> 00:14:47,547 if the speed is high, that term dominates. 203 00:14:47,547 --> 00:14:50,698 When you drive your car 10 to 20 miles an hour, 204 00:14:50,698 --> 00:14:54,465 there is the V squared term that dominates the air drag. 205 00:14:54,465 --> 00:14:57,205 It slows you down. A raindrop that falls, 206 00:14:57,205 --> 00:15:02,000 it's the V squared term that dominates, not this one. 207 00:15:02,000 --> 00:15:06,274 But, if you take ball bearings and you drop them in syrup, 208 00:15:06,274 --> 00:15:08,975 then the velocity is always very low. 209 00:15:08,975 --> 00:15:12,799 The speed is very low, and that this term dominates. 210 00:15:12,799 --> 00:15:16,475 And that's one of the experiments I do during this 211 00:15:16,475 --> 00:15:20,450 lecture to show you that, indeed, the predictions that 212 00:15:20,450 --> 00:15:25,100 follow from this part can really be verified quite beautifully. 213 00:15:25,100 --> 00:15:29,524 So, if you have the patience, I would strongly advise you to 214 00:15:29,524 --> 00:15:33,644 watch that lecture. To manage the mass, 215 00:15:33,644 --> 00:15:38,752 it becomes almost impossible, except where you want to do it 216 00:15:38,752 --> 00:15:42,561 numerically, to take both terms into account. 217 00:15:42,561 --> 00:15:47,235 To be nice to you and 8.03, we always ignore this term, 218 00:15:47,235 --> 00:15:50,525 and we only take that one into account. 219 00:15:50,525 --> 00:15:53,988 And, of course, that is only valid if the 220 00:15:53,988 --> 00:15:57,710 velocities are low, which is often the case, 221 00:15:57,710 --> 00:16:04,274 but of course not always. I will start with an extremely 222 00:16:04,274 --> 00:16:09,076 pedestrian example. I have here a spring on a 223 00:16:09,076 --> 00:16:13,769 horizontal surface. And as it moves there is 224 00:16:13,769 --> 00:16:18,135 friction. And we assume that the friction 225 00:16:18,135 --> 00:16:22,392 has this form. The spring constant is K. 226 00:16:22,392 --> 00:16:27,303 The mass is M. And if I move it the distance X 227 00:16:27,303 --> 00:16:32,761 away from equilibrium, then there is a spring force 228 00:16:32,761 --> 00:16:38,000 that drives it back to equilibrium. 229 00:16:38,000 --> 00:16:40,543 In addition, there is this frictional force, 230 00:16:40,543 --> 00:16:43,913 and I have no idea in which direction it is because if the 231 00:16:43,913 --> 00:16:47,521 object moves in this direction, but personal force is like so. 232 00:16:47,521 --> 00:16:50,596 If the object at this moment moves in this direction, 233 00:16:50,596 --> 00:16:52,430 then the frictional force is so. 234 00:16:52,430 --> 00:16:55,919 So, I cannot put it in there. I have to know the velocity in 235 00:16:55,919 --> 00:17:00,000 order to put in the frictional force, so I won't do that. 236 00:17:00,000 --> 00:17:03,422 But what I can do, I can write down Newton's 237 00:17:03,422 --> 00:17:07,641 second Law, which says that MX double dot is minus KX. 238 00:17:07,641 --> 00:17:10,587 That is this force when X is positive. 239 00:17:10,587 --> 00:17:15,044 The force is in the negative direction, so the minus sign 240 00:17:15,044 --> 00:17:19,104 takes into account that we are dealing with vectors. 241 00:17:19,104 --> 00:17:21,651 And then we have minus this term. 242 00:17:21,651 --> 00:17:26,348 Now, when we deal with it in 8.03, we will call this C1 just 243 00:17:26,348 --> 00:17:28,019 B. That's a tradition. 244 00:17:28,019 --> 00:17:32,000 We won't call it C1. We call it B. 245 00:17:32,000 --> 00:17:36,092 [NOISE OBSCURES] times X dot. X dot is a factor, 246 00:17:36,092 --> 00:17:39,227 is a velocity. And if the velocity is 247 00:17:39,227 --> 00:17:43,320 positive, then the force is in this destruction. 248 00:17:43,320 --> 00:17:48,457 If the velocity is negative, then the frictional force is in 249 00:17:48,457 --> 00:17:51,418 that direction. So, that now is the 250 00:17:51,418 --> 00:17:55,162 differential equation that we have to solve. 251 00:17:55,162 --> 00:18:00,648 I will introduce some shorthand notations that K over M is omega 252 00:18:00,648 --> 00:18:05,920 zero squared. Remember, that is the angular 253 00:18:05,920 --> 00:18:11,158 frequency of the spring oscillating in the absence of 254 00:18:11,158 --> 00:18:15,086 any friction. We've seen that last time. 255 00:18:15,086 --> 00:18:21,028 And, I will also introduce the symbol B over M equals gamma. 256 00:18:21,028 --> 00:18:26,064 That's also a tradition. Gamma is the same units as 257 00:18:26,064 --> 00:18:29,791 omega, seconds to the power minus one. 258 00:18:29,791 --> 00:18:34,525 And, so if I rewrite this differential equation, 259 00:18:34,525 --> 00:18:40,165 then I get X double dot plus gamma times X dot plus omega 260 00:18:40,165 --> 00:18:45,000 zero squared times X equals zero. 261 00:18:45,000 --> 00:18:50,138 And you see, now our task ahead of us, 262 00:18:50,138 --> 00:18:57,083 to solve this differential equation, what is X as a 263 00:18:57,083 --> 00:19:02,245 function of time? Without starting the math, 264 00:19:02,245 --> 00:19:06,994 you can sort of in your head see what's going to happen. 265 00:19:06,994 --> 00:19:10,880 An oscillation that goes on forever, and ever, 266 00:19:10,880 --> 00:19:13,902 and ever is going to come to a halt. 267 00:19:13,902 --> 00:19:17,615 So, instead of a nice cosine that continues, 268 00:19:17,615 --> 00:19:20,724 the cosine will gradually decay away. 269 00:19:20,724 --> 00:19:24,955 And then, this thing will stop. I take the spring, 270 00:19:24,955 --> 00:19:27,718 I offset it, starts to oscillate, 271 00:19:27,718 --> 00:19:33,645 and it comes to a halt. So, the amplitude must die out, 272 00:19:33,645 --> 00:19:36,291 must decay. That's a prediction. 273 00:19:36,291 --> 00:19:41,411 Now, let me try your intuition. Let us assume that the system 274 00:19:41,411 --> 00:19:45,166 will oscillate with a unique frequency omega. 275 00:19:45,166 --> 00:19:50,201 Do you think that this omega will be larger than omega zero, 276 00:19:50,201 --> 00:19:54,212 smaller than omega zero, or equal to omega zero? 277 00:19:54,212 --> 00:19:58,905 Or to put it in the other way, will a whole oscillation, 278 00:19:58,905 --> 00:20:03,514 under the influence of friction, take longer or shorter 279 00:20:03,514 --> 00:20:10,000 with the same amount of time than in the absence of friction? 280 00:20:10,000 --> 00:20:15,271 So, who thinks that if I had damping that the whole period of 281 00:20:15,271 --> 00:20:18,435 a whole oscillation will take longer? 282 00:20:18,435 --> 00:20:21,158 Who thinks it will take shorter? 283 00:20:21,158 --> 00:20:26,606 Who thinks it will be the same? OK, we'll see what comes out of 284 00:20:26,606 --> 00:20:29,418 this. No one thinks that it takes 285 00:20:29,418 --> 00:20:33,337 shorter? Your intuition is very good 286 00:20:33,337 --> 00:20:36,453 there. Now I'm going to give you some 287 00:20:36,453 --> 00:20:41,301 advice, some fatherly advice. During the next 18 minutes, 288 00:20:41,301 --> 00:20:46,495 take no notes because during the next 18 minutes I'm going to 289 00:20:46,495 --> 00:20:51,602 solve this differential equation for you, and I have decided 290 00:20:51,602 --> 00:20:56,277 which I do not always do, to do it exactly the way that 291 00:20:56,277 --> 00:21:00,000 [French?] does it: very elegant. 292 00:21:00,000 --> 00:21:03,597 So, if you take notes, you're wasting your time 293 00:21:03,597 --> 00:21:08,290 because you have in your book. And I would much rather prefer 294 00:21:08,290 --> 00:21:13,061 that you follow each step that I make so you have more time to 295 00:21:13,061 --> 00:21:17,519 concentrate on the methods. So, don't write anything down. 296 00:21:17,519 --> 00:21:21,899 That is my recommendation. 18 minutes: you can time me if 297 00:21:21,899 --> 00:21:25,966 you don't believe me. We're going to jump through the 298 00:21:25,966 --> 00:21:31,176 complex plane. This is a beautiful example 299 00:21:31,176 --> 00:21:36,823 where the complex solutions really come in handy. 300 00:21:36,823 --> 00:21:42,235 And so, I'm going to change this equation to Z, 301 00:21:42,235 --> 00:21:48,000 which is now a complex notation, Z double dot plus 302 00:21:48,000 --> 00:21:54,588 gamma, Z dot plus omega zero squared times Z equals zero. 303 00:21:54,588 --> 00:22:01,647 And then Z, my trial function, is some amplitude A times E to 304 00:22:01,647 --> 00:22:06,000 the power JPT, plus alpha. 305 00:22:06,000 --> 00:22:09,867 Now, clearly this has the smell of an amplitude. 306 00:22:09,867 --> 00:22:13,405 And, this here has the smell of a frequency. 307 00:22:13,405 --> 00:22:18,177 Anything E to the power JPT has the smell of the frequency. 308 00:22:18,177 --> 00:22:21,962 But you'll see there you are in for a surprise. 309 00:22:21,962 --> 00:22:26,240 What happens is this P. Once we have found the proper 310 00:22:26,240 --> 00:22:29,202 solution to Z, we take the real part, 311 00:22:29,202 --> 00:22:34,643 and we're in business. Then we come back to X. 312 00:22:34,643 --> 00:22:41,080 I'm going to take the second derivative of this equation. 313 00:22:41,080 --> 00:22:48,091 So, JP comes out twice so that this J squared times P squared, 314 00:22:48,091 --> 00:22:53,034 and so I'm going to see now minus P squared, 315 00:22:53,034 --> 00:22:58,321 that is the second derivative of this function. 316 00:22:58,321 --> 00:23:04,275 Then the Z comes later. Then I have plus gamma. 317 00:23:04,275 --> 00:23:09,816 And then, what pops out is JP because it's only the first 318 00:23:09,816 --> 00:23:13,971 derivative: JP. You have to be very careful 319 00:23:13,971 --> 00:23:18,918 that you can distinguish your J's from your gammas. 320 00:23:18,918 --> 00:23:24,756 And then you get plus omega zero squared and the whole thing 321 00:23:24,756 --> 00:23:28,416 times Z. And that now must be equal to 322 00:23:28,416 --> 00:23:32,448 zero. Well, if you have an equation 323 00:23:32,448 --> 00:23:35,994 like this, you may think it's one equation, 324 00:23:35,994 --> 00:23:40,976 but it's really two equations because if something like this 325 00:23:40,976 --> 00:23:44,775 has to be zero, and of course Z is zero is not 326 00:23:44,775 --> 00:23:49,250 an acceptable solution because that's the oscillation, 327 00:23:49,250 --> 00:23:54,147 if you have an equation like this which has a real part and 328 00:23:54,147 --> 00:23:58,791 it has an imaginary part, and both must independently be 329 00:23:58,791 --> 00:24:04,276 zero for obvious reasons. These are apples and these are 330 00:24:04,276 --> 00:24:08,526 oranges, and five apples minus five oranges is never zero 331 00:24:08,526 --> 00:24:13,156 oranges, is never zero apples. You cannot subtract apples from 332 00:24:13,156 --> 00:24:15,888 coconuts. So, this has to be zero and 333 00:24:15,888 --> 00:24:19,986 this has to be to be zero. Now, if you made gamma zero, 334 00:24:19,986 --> 00:24:23,098 you'd have no problem. There's no damping. 335 00:24:23,098 --> 00:24:26,209 So, that's ridiculous. If you make P zero, 336 00:24:26,209 --> 00:24:30,611 and you no longer have any oscillation going on because you 337 00:24:30,611 --> 00:24:33,343 smell that PT, that's going to be the 338 00:24:33,343 --> 00:24:39,449 oscillatory thing. So, that's also unacceptable. 339 00:24:39,449 --> 00:24:45,516 And so, this won't work. Therefore, P itself must be 340 00:24:45,516 --> 00:24:49,799 complex. And so, we're going one step 341 00:24:49,799 --> 00:24:56,936 further now and we're going to say the only way that we could 342 00:24:56,936 --> 00:25:04,193 make this work is if we say P equals N plus JS whereby N and S 343 00:25:04,193 --> 00:25:08,000 then, I hope, are real. 344 00:25:08,000 --> 00:25:14,171 So, let's first calculate what P squared is so we go slowly. 345 00:25:14,171 --> 00:25:18,983 So, we don't get into trouble with the algebra. 346 00:25:18,983 --> 00:25:24,004 So, P squared is N squared plus this one squared. 347 00:25:24,004 --> 00:25:29,652 The J squared is minus one. So, that is minus S squared 348 00:25:29,652 --> 00:25:37,214 plus 2N times J times S. So, now I can go back to this 349 00:25:37,214 --> 00:25:45,498 forum and have minus P squared. So, I get minus N squared. 350 00:25:45,498 --> 00:25:54,072 That is this minus P squared plus S squared minus 2N J times 351 00:25:54,072 --> 00:25:56,979 S. So, I'm done here. 352 00:25:56,979 --> 00:26:02,501 Now, I get gamma JP plus gamma times J. 353 00:26:02,501 --> 00:26:10,930 And, now I have to multiply that by P which is N plus gamma 354 00:26:10,930 --> 00:26:19,339 times J times JS. And then, I have plus omega 355 00:26:19,339 --> 00:26:25,125 squared, and this now has to be zero. 356 00:26:25,125 --> 00:26:33,000 I have here J squared. J squared is minus one. 357 00:26:33,000 --> 00:26:38,090 So, I can erase this and replace it by gamma S. 358 00:26:38,090 --> 00:26:43,622 You cannot do that. That's the price you pay if you 359 00:26:43,622 --> 00:26:48,381 didn't take my advice not to take any notes. 360 00:26:48,381 --> 00:26:54,467 So, this is minus gamma S. And now, this has to be zero. 361 00:26:54,467 --> 00:27:00,000 That means the apples must add up to zero. 362 00:27:00,000 --> 00:27:04,862 These are the apples, and the oranges have to be 363 00:27:04,862 --> 00:27:07,241 zero. This is an orange; 364 00:27:07,241 --> 00:27:10,965 this is an orange; this is an orange; 365 00:27:10,965 --> 00:27:15,310 and this is an orange. For this to be zero, 366 00:27:15,310 --> 00:27:18,931 there is a J here. There's a J here. 367 00:27:18,931 --> 00:27:22,655 There's an N here. There's an N here. 368 00:27:22,655 --> 00:27:26,896 2S must be gamma. So, gamma over two is S. 369 00:27:26,896 --> 00:27:30,000 How? We found S. 370 00:27:30,000 --> 00:27:36,040 For this to be zero, we find that N squared minus N 371 00:27:36,040 --> 00:27:43,409 squared plus gamma squared over four because we know what S is 372 00:27:43,409 --> 00:27:51,020 now, so, plus gamma squared over four minus gamma over two times 373 00:27:51,020 --> 00:27:56,093 minus gamma S, which is minus gamma squared 374 00:27:56,093 --> 00:28:03,342 over two minus gamma squared over two plus omega zero squared 375 00:28:03,342 --> 00:28:09,692 equals zero. And, that means that N squared 376 00:28:09,692 --> 00:28:15,969 is omega zero squared minus gamma squared over four. 377 00:28:15,969 --> 00:28:21,876 So, we also have N. So, we now have N and we have 378 00:28:21,876 --> 00:28:25,938 S. And so, now I can go back to my 379 00:28:25,938 --> 00:28:32,461 trial function and I can substitute the values that we 380 00:28:32,461 --> 00:28:38,000 have found in that complex function. 381 00:28:38,000 --> 00:28:46,232 So, we get Z equals A times E to the power J. 382 00:28:46,232 --> 00:28:53,716 And, now we have to put in BT plus alpha. 383 00:28:53,716 --> 00:29:02,509 But, B is N plus JS. So, NT plus JST plus alpha, 384 00:29:02,509 --> 00:29:10,470 yeah, I go slowly. I don't skip too many steps. 385 00:29:10,470 --> 00:29:14,931 But, I know that S is gamma over two. 386 00:29:14,931 --> 00:29:19,764 But, I know that J squared is minus one. 387 00:29:19,764 --> 00:29:24,970 So, therefore, this part here and this part 388 00:29:24,970 --> 00:29:31,290 here can be brought out. So, we get A times E to the 389 00:29:31,290 --> 00:29:38,066 minus gamma over two times T. Here, we have the T. 390 00:29:38,066 --> 00:29:43,749 The J squared makes it a minus, and S is gamma over two. 391 00:29:43,749 --> 00:29:49,121 And then, I get E to the power J times NT plus alpha. 392 00:29:49,121 --> 00:29:54,391 The first thing that you see here, which is amazing, 393 00:29:54,391 --> 00:29:59,143 this term here is an exponential decay in time. 394 00:29:59,143 --> 00:30:04,000 There is nothing oscillatory about it. 395 00:30:04,000 --> 00:30:08,000 It means that this A, which is the original 396 00:30:08,000 --> 00:30:12,857 amplitude, is gradually dying out with a decay time, 397 00:30:12,857 --> 00:30:16,285 [ta?], which is two divided by gamma. 398 00:30:16,285 --> 00:30:20,095 Now remember, gamma has unit second minus 399 00:30:20,095 --> 00:30:23,333 one. So this really has units time. 400 00:30:23,333 --> 00:30:28,857 So, this one is going to be responsible for that decay that 401 00:30:28,857 --> 00:30:33,937 we discussed earlier. What is this now? 402 00:30:33,937 --> 00:30:39,180 E to the power JNT. N is obviously a frequency. 403 00:30:39,180 --> 00:30:45,790 This is a striking example in a complex plane of a rotating 404 00:30:45,790 --> 00:30:49,665 vector, and angular frequency is N. 405 00:30:49,665 --> 00:30:55,022 But, we know what N is. And so, this N really is 406 00:30:55,022 --> 00:31:00,834 nothing but omega squared. N squared is really omega 407 00:31:00,834 --> 00:31:06,294 squared. And so, I can replace this N 408 00:31:06,294 --> 00:31:13,352 now by whatever you have here. And so, if I write this now AE 409 00:31:13,352 --> 00:31:19,352 minus gamma over two, I can write down now J omega T 410 00:31:19,352 --> 00:31:23,823 plus alpha. And I know exactly now what 411 00:31:23,823 --> 00:31:29,117 that omega is. That is the square root of this 412 00:31:29,117 --> 00:31:32,368 number. And so, therefore, 413 00:31:32,368 --> 00:31:36,702 omega squared is omega zero minus gamma squared over four. 414 00:31:36,702 --> 00:31:40,732 So, those of you in the audience who intuitively sense 415 00:31:40,732 --> 00:31:44,990 that the frequency would go down because of the friction, 416 00:31:44,990 --> 00:31:48,336 they were right. You see, omega is lower than 417 00:31:48,336 --> 00:31:50,313 omega zero. Now, of course, 418 00:31:50,313 --> 00:31:53,582 if gamma is very low, the two could be close 419 00:31:53,582 --> 00:31:55,788 together. It depends on gamma. 420 00:31:55,788 --> 00:31:59,209 That's obvious. If you have a lot of friction, 421 00:31:59,209 --> 00:32:04,000 then omega will be a lot less than omega zero. 422 00:32:04,000 --> 00:32:07,979 If you have very little friction, well, 423 00:32:07,979 --> 00:32:12,064 then they will be very closely the same. 424 00:32:12,064 --> 00:32:18,347 And so, we can now write down in all its glory a real part of 425 00:32:18,347 --> 00:32:24,212 this function which then becomes X equals some amplitude, 426 00:32:24,212 --> 00:32:30,496 A, times E to the minus gamma over two times T times E to the 427 00:32:30,496 --> 00:32:35,000 minus J times omega T plus alpha. 428 00:32:35,000 --> 00:32:39,732 Uh-oh, look what I did. I have an equation in the 429 00:32:39,732 --> 00:32:43,084 complex plane. That's perfectly OK. 430 00:32:43,084 --> 00:32:48,309 And then I went back to the real world going from Z to 431 00:32:48,309 --> 00:32:51,760 complex plane to the real part of Z. 432 00:32:51,760 --> 00:32:55,704 And, what did I write down? This is fine. 433 00:32:55,704 --> 00:33:01,225 But then I wrote down E to the power minus J omega T plus 434 00:33:01,225 --> 00:33:04,018 alpha. First of all, 435 00:33:04,018 --> 00:33:07,937 there should not be a minus sign in front of the J. 436 00:33:07,937 --> 00:33:12,248 That's not a major problem. But, I do not take it out of 437 00:33:12,248 --> 00:33:15,618 the complex plane. I leave it in the complex 438 00:33:15,618 --> 00:33:18,440 plane. So, clearly what I should have 439 00:33:18,440 --> 00:33:23,378 written, which is what I will do now, X equals EA times E to the 440 00:33:23,378 --> 00:33:28,081 minus gamma over two times T times the cosine of omega T plus 441 00:33:28,081 --> 00:33:32,000 alpha. Now, we are in the real world. 442 00:33:32,000 --> 00:33:35,647 You're going to see this equation at least for the next 443 00:33:35,647 --> 00:33:39,429 five or 10 minutes of the tape. It is a thorn in my side, 444 00:33:39,429 --> 00:33:41,928 but there's nothing I can do about it. 445 00:33:41,928 --> 00:33:46,047 You just have to live with it. And, think of it as being this. 446 00:33:46,047 --> 00:33:50,100 And, there are two adjustable constants which depend entirely 447 00:33:50,100 --> 00:33:53,274 on the initial condition. At time T equals zero, 448 00:33:53,274 --> 00:33:56,313 there are two things you can do to the object. 449 00:33:56,313 --> 00:34:00,366 You can bring it to a certain point away from equilibrium and 450 00:34:00,366 --> 00:34:05,027 you can give it a kick. We call that a philosophy. 451 00:34:05,027 --> 00:34:09,160 And you're free to choose any way you want to do that. 452 00:34:09,160 --> 00:34:13,761 And so, it's clear that with those two choices that you have 453 00:34:13,761 --> 00:34:18,206 that in your final solution you end up with two adjustable 454 00:34:18,206 --> 00:34:21,871 constants which depend on the initial condition. 455 00:34:21,871 --> 00:34:26,082 And so, you see that this amplitude is going to die out 456 00:34:26,082 --> 00:34:30,137 with the one over EDK time of two over gamma seconds, 457 00:34:30,137 --> 00:34:34,504 and that the frequency is lower than the frequency of the 458 00:34:34,504 --> 00:34:40,013 un-damped system. I try to make a plot. 459 00:34:40,013 --> 00:34:48,194 So here is T and here is X. And to guide my hand I'm going 460 00:34:48,194 --> 00:34:52,787 to first put in this exponential. 461 00:34:52,787 --> 00:35:00,537 And now, I'm going to put in that oscillatory term here 462 00:35:00,537 --> 00:35:08,000 whereby the frequency is uniquely determined. 463 00:35:08,000 --> 00:35:14,500 Omega is uniquely determined, and the period, 464 00:35:14,500 --> 00:35:19,227 T, is that omega divided by 2 pi. 465 00:35:19,227 --> 00:35:21,738 Oops. How could I? 466 00:35:21,738 --> 00:35:28,829 Look what I wrote. I wrote T equals omega divided 467 00:35:28,829 --> 00:35:34,000 by 2 pi. It was not my day. 468 00:35:34,000 --> 00:35:38,822 Clearly it should have been T equals 2 pi divided by omega. 469 00:35:38,822 --> 00:35:42,979 Now, unfortunately for me, very shortly afterwards, 470 00:35:42,979 --> 00:35:45,806 i.e. erase this from the blackboard 471 00:35:45,806 --> 00:35:48,549 so you won't see it for very long. 472 00:35:48,549 --> 00:35:51,709 In any case, this is what it should be. 473 00:35:51,709 --> 00:35:55,284 And therefore, I'm going to put in here zero 474 00:35:55,284 --> 00:36:00,023 crossings to guide myself so that I don't make the mistake 475 00:36:00,023 --> 00:36:05,177 that you often see that people think that this period is slowly 476 00:36:05,177 --> 00:36:10,000 getting shorter in time. That is not true. 477 00:36:10,000 --> 00:36:13,475 It is amplitude that decays away. 478 00:36:13,475 --> 00:36:18,580 And so, if we now try to put in the oscillation, 479 00:36:18,580 --> 00:36:22,382 then this is what's going to happen. 480 00:36:22,382 --> 00:36:28,247 And so, you see that the oscillation dies away in time, 481 00:36:28,247 --> 00:36:32,700 but the period, T, is uniquely determined. 482 00:36:32,700 --> 00:36:39,000 And, that depends on how much friction there is. 483 00:36:39,000 --> 00:36:44,769 It is not uncommon to introduce a quality factor, 484 00:36:44,769 --> 00:36:49,817 Q, that is high if there is little damping, 485 00:36:49,817 --> 00:36:55,105 and that is low if there is a lot of damping. 486 00:36:55,105 --> 00:36:59,312 And, that Q, which is dimensionless, 487 00:36:59,312 --> 00:37:06,500 is omega zero divided by gamma. You see, immediately if gamma 488 00:37:06,500 --> 00:37:11,068 is high, then Q is low. It's a low-quality oscillator. 489 00:37:11,068 --> 00:37:15,465 If you introduce that, you can go back to your omega 490 00:37:15,465 --> 00:37:19,517 squared equation. And that omega squared becomes 491 00:37:19,517 --> 00:37:24,689 omega zero squared times one minus one divided by 4Q squared. 492 00:37:24,689 --> 00:37:27,362 Is this any different from that? 493 00:37:27,362 --> 00:37:32,448 No, so it's a different way writing it because you introduce 494 00:37:32,448 --> 00:37:36,718 this. What it tells you is that if Q 495 00:37:36,718 --> 00:37:40,615 is about ten, and I bet you the Q's are much 496 00:37:40,615 --> 00:37:45,781 higher for these pendulums, then you have here one divided 497 00:37:45,781 --> 00:37:49,315 by 400 that is one quarter of a percent. 498 00:37:49,315 --> 00:37:53,212 But, since omega is the square root of that, 499 00:37:53,212 --> 00:37:57,199 it's only 1/8 of a percent. So, for Q of ten, 500 00:37:57,199 --> 00:38:03,000 omega is only 1/8 of a percent lower than omega zero. 501 00:38:03,000 --> 00:38:11,885 Even if you make the Q as low as two, the frequencies are only 502 00:38:11,885 --> 00:38:18,440 off by about 3.2%. So I want you to appreciate 503 00:38:18,440 --> 00:38:25,724 that most of the time, but not always is omega very 504 00:38:25,724 --> 00:38:32,862 close to omega zero. We can look at the decay in a 505 00:38:32,862 --> 00:38:40,000 different way but we cannot do that here. 506 00:38:40,000 --> 00:38:45,785 I will do that here on the centerboard because we don't 507 00:38:45,785 --> 00:38:50,607 need that anymore. Instead of saying I have to 508 00:38:50,607 --> 00:38:57,035 wait two over gamma seconds for the amplitude to go down by a 509 00:38:57,035 --> 00:39:03,571 factor of E, I can ask myself, how many oscillations do I have 510 00:39:03,571 --> 00:39:08,714 to wait for the amplitude to go down by a factor, 511 00:39:08,714 --> 00:39:12,095 E? How many oscillations? 512 00:39:12,095 --> 00:39:18,095 Well, N oscillations would take this long, T being the period of 513 00:39:18,095 --> 00:39:22,571 one oscillation. But the reasonable values of Q, 514 00:39:22,571 --> 00:39:28,476 T and T zero are the same like omega and omega zero are closely 515 00:39:28,476 --> 00:39:32,873 the same. So, I can write for this that 516 00:39:32,873 --> 00:39:36,440 this is approximately N times T zero. 517 00:39:36,440 --> 00:39:41,889 So, that's approximately N times two pi divided by omega 518 00:39:41,889 --> 00:39:45,357 zero. I can now substitute this time 519 00:39:45,357 --> 00:39:50,608 in this T, and only concentrate on that decay portion, 520 00:39:50,608 --> 00:39:53,778 that early part. So, what I find, 521 00:39:53,778 --> 00:39:58,930 then, that A after N oscillations is A times E to the 522 00:39:58,930 --> 00:40:03,785 minus gamma over 2T, E to the minus gamma over two 523 00:40:03,785 --> 00:40:07,352 times this time after N oscillations, 524 00:40:07,352 --> 00:40:13,000 N divided by 2 pi divided by omega zero. 525 00:40:13,000 --> 00:40:18,111 And, you lose a two. But, omega zero divided by 526 00:40:18,111 --> 00:40:22,444 gamma is Q. And so, now you have it in a 527 00:40:22,444 --> 00:40:26,666 forum which is minus N times pi over Q. 528 00:40:26,666 --> 00:40:33,000 Now, is this any different from what we had before? 529 00:40:33,000 --> 00:40:35,914 No. Here we say I have to wait two 530 00:40:35,914 --> 00:40:41,037 over gamma seconds for the amplitude to go down by a factor 531 00:40:41,037 --> 00:40:44,041 of E. Here we say if N is Q divided 532 00:40:44,041 --> 00:40:48,369 by pi, then the amplitude goes down by one over E. 533 00:40:48,369 --> 00:40:53,403 So, in one case I ask myself how many seconds do I have to 534 00:40:53,403 --> 00:40:56,583 wait? In the other case I ask myself, 535 00:40:56,583 --> 00:41:01,000 how many oscillations to have to wait? 536 00:41:01,000 --> 00:41:06,124 And so, if Q is ten, it tells you that you have to 537 00:41:06,124 --> 00:41:12,294 wait about three oscillations, roughly, for the amplitude to 538 00:41:12,294 --> 00:41:16,686 go down by a factor of E. And, if Q is 100, 539 00:41:16,686 --> 00:41:22,856 you have to wait more like 32 oscillations for the amplitude 540 00:41:22,856 --> 00:41:27,666 to go down by a factor of E. 18 minutes are up. 541 00:41:27,666 --> 00:41:33,000 So, now you can start taking notes again. 542 00:41:33,000 --> 00:41:40,411 I have here two pendulums. And the pendulums have about 543 00:41:40,411 --> 00:41:46,725 the same length. And the objects have about the 544 00:41:46,725 --> 00:41:50,705 same radius. That means the B, 545 00:41:50,705 --> 00:41:58,666 which is this coefficient in front of the velocity is about 546 00:41:58,666 --> 00:42:02,718 the same. But gamma is not the same 547 00:42:02,718 --> 00:42:06,703 because if they have the same B, but if the mass of the two 548 00:42:06,703 --> 00:42:10,414 objects is very different, it's Styrofoam and this is a 549 00:42:10,414 --> 00:42:13,369 billiard ball, there is a huge difference in 550 00:42:13,369 --> 00:42:15,706 gamma. And, since the period of the 551 00:42:15,706 --> 00:42:19,623 two pendulums is very closely the same, they have the same 552 00:42:19,623 --> 00:42:22,097 length. You see that the Q of the two 553 00:42:22,097 --> 00:42:25,876 systems must be very different because if B is the same, 554 00:42:25,876 --> 00:42:30,000 N is very much higher than the billiard ball. 555 00:42:30,000 --> 00:42:33,626 Then the gamma is much lower. And therefore, 556 00:42:33,626 --> 00:42:38,012 the Q of this system is way higher than the Q of this 557 00:42:38,012 --> 00:42:40,963 system. In fact, if I wanted to wait 558 00:42:40,963 --> 00:42:45,939 how many oscillations it would take for that amplitude to go 559 00:42:45,939 --> 00:42:50,240 down by a factor of E, I may have to wait five or 10 560 00:42:50,240 --> 00:42:53,277 minutes. So, I will not attempt that. 561 00:42:53,277 --> 00:42:58,000 But, I will attempt that to measure Q with this pendulum, 562 00:42:58,000 --> 00:43:02,385 if I bring this here, the separation from equilibrium 563 00:43:02,385 --> 00:43:07,311 is about 27 cm. And, by the time that it has 564 00:43:07,311 --> 00:43:09,539 decayed to this, it's 10 cm. 565 00:43:09,539 --> 00:43:14,657 So, that's about a factor of E. And, I want the students in the 566 00:43:14,657 --> 00:43:19,526 audience who are sitting here who can really see it head-on, 567 00:43:19,526 --> 00:43:23,653 I wanted to say stop. Scream the word stop when the 568 00:43:23,653 --> 00:43:27,120 27 has decayed to 10. And, in the meantime, 569 00:43:27,120 --> 00:43:31,000 we count the number of oscillations. 570 00:43:31,000 --> 00:43:33,944 So, we have then counted the number of oscillations. 571 00:43:33,944 --> 00:43:36,368 And therefore, we know what Q is because we 572 00:43:36,368 --> 00:43:38,851 multiplied the number of oscillations by pi. 573 00:43:38,851 --> 00:43:41,679 So, we have measured Q. We could also have done it 574 00:43:41,679 --> 00:43:44,797 through a time measurement, but I would prefer to do it 575 00:43:44,797 --> 00:43:46,990 this way. Now, you guys over there keep 576 00:43:46,990 --> 00:43:50,050 your mouth shut because there's no way you can see it. 577 00:43:50,050 --> 00:43:51,262 Right? You just can't. 578 00:43:51,262 --> 00:43:53,744 You have a projection effect which is awful. 579 00:43:53,744 --> 00:43:55,592 So, I only want to hear from you. 580 00:43:55,592 --> 00:43:58,016 Are you ready for that? Count and say stop. 581 00:43:58,016 --> 00:44:02,000 Not all of you may say stop at the same moment. 582 00:44:02,000 --> 00:44:08,562 Some of you may say stop after ten oscillations. 583 00:44:08,562 --> 00:44:14,147 Other may say stop after 11 oscillations. 584 00:44:14,147 --> 00:44:20,150 That's fine. That is part of the uncertainty 585 00:44:20,150 --> 00:44:23,641 in the measurement. Ready? 586 00:44:23,641 --> 00:44:25,735 27. There we go. 587 00:44:25,735 --> 00:44:30,622 One, two, you see the decay already. 588 00:44:30,622 --> 00:44:34,671 Three, four, five, six, seven, 589 00:44:34,671 --> 00:44:39,000 eight, nine, ten. 590 00:44:39,000 --> 00:44:41,712 My goodness, are you guys crazy? 591 00:44:41,712 --> 00:44:46,962 You're paying a lot of tuition and you can even pay attention 592 00:44:46,962 --> 00:44:51,074 to demonstration. At nine I think it was already 593 00:44:51,074 --> 00:44:53,087 at ten. When it is here, 594 00:44:53,087 --> 00:44:57,025 then it is ten. I'll give you a second chance. 595 00:44:57,025 --> 00:45:00,000 You have to scream stop. 596 00:45:00,000 --> 00:45:07,000 597 00:45:07,000 --> 00:45:15,145 One, two, three, four, five, six, 598 00:45:15,145 --> 00:45:22,018 seven, eight, nine, ten, 11, 599 00:45:22,018 --> 00:45:30,927 12, OK. Yeah, it is somewhere around 600 00:45:30,927 --> 00:45:37,935 ten or 11. And I don't know why you didn't 601 00:45:37,935 --> 00:45:41,344 scream stop, but that's your problem. 602 00:45:41,344 --> 00:45:46,837 So, N is about 11 maybe plus or minus one, 10% uncertainty. 603 00:45:46,837 --> 00:45:50,719 And so, Q then is about five times higher. 604 00:45:50,719 --> 00:45:56,022 It's a crude measurement, but very roughly you would get, 605 00:45:56,022 --> 00:46:00,000 then, times 11. You get about 35. 606 00:46:00,000 --> 00:46:05,541 That means the frequency damped is almost identical to the 607 00:46:05,541 --> 00:46:10,597 frequency un-damped because, remember, this equation, 608 00:46:10,597 --> 00:46:15,361 the one over four, I erase that perhaps but it is, 609 00:46:15,361 --> 00:46:20,027 you have it up there. If Q is 35, you can see how 610 00:46:20,027 --> 00:46:24,694 close omega is to omega zero. On problem set two, 611 00:46:24,694 --> 00:46:30,138 your very first task is to do a take-home experiment with 612 00:46:30,138 --> 00:46:35,000 something similar to what I did today. 613 00:46:35,000 --> 00:46:37,292 Make sure you pick up a kit today. 614 00:46:37,292 --> 00:46:41,253 You can do that between 11:00 and 1:00 and 3:00 to 5:00 in 615 00:46:41,253 --> 00:46:45,282 room 4335, or you can do it tomorrow between 2:00 and 5:00. 616 00:46:45,282 --> 00:46:48,200 You share one kit, so you choose a partner. 617 00:46:48,200 --> 00:46:51,952 And you can do all these experiments together with your 618 00:46:51,952 --> 00:46:54,662 partner. I want to see in your solutions 619 00:46:54,662 --> 00:46:57,580 uncertainties. Any timing measurements that 620 00:46:57,580 --> 00:47:03,000 you have to do has to come with an estimate of your uncertainty. 621 00:47:03,000 --> 00:47:07,042 And all the conclusions they draw must carry on these 622 00:47:07,042 --> 00:47:11,707 uncertainties just as we did last time when we were exploring 623 00:47:11,707 --> 00:47:15,828 the possibility whether the equation of the spring was 624 00:47:15,828 --> 00:47:19,249 indeed accurate. We were only able to come to 625 00:47:19,249 --> 00:47:23,680 the conclusion that it was not accurate because we had our 626 00:47:23,680 --> 00:47:28,423 proper uncertainties in there. So, I want to see uncertainties 627 00:47:28,423 --> 00:47:32,000 in there. Now comes the mini quiz. 628 00:47:32,000 --> 00:47:34,930 Five minute break: it's a bit early. 629 00:47:34,930 --> 00:47:38,028 But that's the best point today to do. 630 00:47:38,028 --> 00:47:43,052 So, if some of you can help me hand it out, and so after five 631 00:47:43,052 --> 00:47:47,992 minutes bring it back to me. It shouldn't take you more than 632 00:47:47,992 --> 00:47:51,509 one minute. That leaves you still with four 633 00:47:51,509 --> 00:47:56,449 minutes to stretch your legs. [SOUND OFF/THEN ON] I will now 634 00:47:56,449 --> 00:47:59,966 take you back to the good old days of 8.02. 635 00:47:59,966 --> 00:48:04,990 And, I will take you back to an RLC circuit [where?] I have a 636 00:48:04,990 --> 00:48:09,633 battery. This is the plus side and this 637 00:48:09,633 --> 00:48:13,232 is the minus side. Here I have a resistor, 638 00:48:13,232 --> 00:48:15,427 R. I have here a pure self 639 00:48:15,427 --> 00:48:18,938 inductor, L. And here I have a capacitor, 640 00:48:18,938 --> 00:48:21,748 C. And then, I have a switch here 641 00:48:21,748 --> 00:48:26,664 and I can throw the switch. And then the battery is going 642 00:48:26,664 --> 00:48:32,273 to charge up the capacitor. And then you'll get 643 00:48:32,273 --> 00:48:37,420 oscillations. This is a wonderful example of 644 00:48:37,420 --> 00:48:42,925 damped oscillations. The way that I solve these 645 00:48:42,925 --> 00:48:50,106 problems is a strict discipline. I assume that when I started 646 00:48:50,106 --> 00:48:55,731 that there is a current going in this direction, 647 00:48:55,731 --> 00:48:59,202 I. That current will make this 648 00:48:59,202 --> 00:49:05,186 side of the capacitor more positive than that side. 649 00:49:05,186 --> 00:49:11,170 So, I equals DQ DT, Q being the charge on this side 650 00:49:11,170 --> 00:49:18,509 of the capacitor. And now I have to do the closed 651 00:49:18,509 --> 00:49:24,849 loop integral of E dot DL. [UNINTELLIGIBLE] does not hold 652 00:49:24,849 --> 00:49:31,867 the closed loop integral is not zero because we have a magnetic 653 00:49:31,867 --> 00:49:39,000 flux going through a surface attached to this closed loop. 654 00:49:39,000 --> 00:49:44,342 So, the closed loop integral of E dot DL is minus D phi DT, 655 00:49:44,342 --> 00:49:49,407 phi being that magnetic flux going through that surface. 656 00:49:49,407 --> 00:49:54,842 And so, I want to know what the E vectors are in each one of 657 00:49:54,842 --> 00:50:00,000 these components is the curl is in this direction. 658 00:50:00,000 --> 00:50:04,545 Then the electric field in the resistor is in this direction. 659 00:50:04,545 --> 00:50:08,257 The self inductor is made of superconducting wire. 660 00:50:08,257 --> 00:50:12,272 So, there is never any E field inside a self inductor. 661 00:50:12,272 --> 00:50:16,136 Unlike what many EE people think, there is never any 662 00:50:16,136 --> 00:50:18,939 electric field inside a self inductor. 663 00:50:18,939 --> 00:50:22,500 So, the EE is zero. This plus and this is minus. 664 00:50:22,500 --> 00:50:25,757 So, the potential, electric field is in this 665 00:50:25,757 --> 00:50:30,000 direction. This is plus and this is minus. 666 00:50:30,000 --> 00:50:34,865 And so, the electric field is opposing me if I go around 667 00:50:34,865 --> 00:50:37,697 clockwise. So, if now I go around 668 00:50:37,697 --> 00:50:42,651 clockwise and I call the integral E dot DL from this side 669 00:50:42,651 --> 00:50:47,252 to this side of the capacitor, if I call that V of C, 670 00:50:47,252 --> 00:50:51,498 which is Q divided by C. That's the definition of 671 00:50:51,498 --> 00:50:54,595 capacitor. If you're ready for this, 672 00:50:54,595 --> 00:50:58,399 and I start here, and I go clockwise around, 673 00:50:58,399 --> 00:51:03,000 then from here to here, I get plus IR. 674 00:51:03,000 --> 00:51:07,142 Going through the self inductor, the integral E dot DL 675 00:51:07,142 --> 00:51:11,285 from here to here is zero because there is no electric 676 00:51:11,285 --> 00:51:15,975 field inside the self inductor. Going from here to here I get 677 00:51:15,975 --> 00:51:18,477 plus VC. That is what we call the 678 00:51:18,477 --> 00:51:21,525 potential difference over the capacitor. 679 00:51:21,525 --> 00:51:24,574 Here, the electric field is opposing me. 680 00:51:24,574 --> 00:51:28,873 I walk into the electric field, so E dot DL is negative. 681 00:51:28,873 --> 00:51:34,144 And, this is minus V zero. And now, I go to Mr. 682 00:51:34,144 --> 00:51:40,901 Maxwell [UNINTELLIGIBLE] and he says that this now is minus L DI 683 00:51:40,901 --> 00:51:43,797 DT. [The hell with Kishov?]. 684 00:51:43,797 --> 00:51:48,086 Minus L DI DT, and now I have my equation 685 00:51:48,086 --> 00:51:52,162 correct. So, now I'm going to replace I 686 00:51:52,162 --> 00:51:56,237 by DQ DT. And now I'm bringing L to the 687 00:51:56,237 --> 00:51:59,991 left. So, I get L times Q double dot 688 00:51:59,991 --> 00:52:04,817 plus R times Q dot because I is Q dot plus VC, 689 00:52:04,817 --> 00:52:12,705 which is Q divided by C. And that now equals V zero. 690 00:52:12,705 --> 00:52:20,941 And, I'm going to divide by L. And so I get Q double dot plus 691 00:52:20,941 --> 00:52:30,000 R over L times Q dot plus Q over LC equals V zero divided by L. 692 00:52:30,000 --> 00:52:34,320 I'm going to replace R over L by gamma. 693 00:52:34,320 --> 00:52:38,868 You will see very shortly why we do that. 694 00:52:38,868 --> 00:52:43,189 And, one over LC is omega zero squared. 695 00:52:43,189 --> 00:52:48,647 And, when we do that, we get an equation which is 696 00:52:48,647 --> 00:52:55,355 almost identical to the one we had on the blackboard for the 697 00:52:55,355 --> 00:52:59,562 spring. We get Q double dot plus gamma 698 00:52:59,562 --> 00:53:05,930 times Q dot plus omega zero squared times Q equals V zero 699 00:53:05,930 --> 00:53:12,705 divided by L. And, now we have to solve this 700 00:53:12,705 --> 00:53:19,252 differential equation. You recognize that gamma is the 701 00:53:19,252 --> 00:53:23,823 damping. R over L is the same function 702 00:53:23,823 --> 00:53:30,000 as the damping had in the case of the spring. 703 00:53:30,000 --> 00:53:33,334 In the case of the spring, it was B over M. 704 00:53:33,334 --> 00:53:36,272 Here, it is R over L. The larger R is, 705 00:53:36,272 --> 00:53:38,972 the more heat dissipation there is. 706 00:53:38,972 --> 00:53:42,545 Heat dissipation goes in terms of I squared R, 707 00:53:42,545 --> 00:53:46,992 and clearly that means you take energy out of the system. 708 00:53:46,992 --> 00:53:49,533 So, that means there is [tempi?]. 709 00:53:49,533 --> 00:53:52,868 And this is, then, the natural frequency of 710 00:53:52,868 --> 00:53:55,409 the RLC circuit if there is no R. 711 00:53:55,409 --> 00:53:59,776 If there is only an L and a C, this is the square of the 712 00:53:59,776 --> 00:54:04,379 frequency. If only this were zero, 713 00:54:04,379 --> 00:54:09,965 then I know the solution because we had it on the board 714 00:54:09,965 --> 00:54:13,172 there. We still have the complex 715 00:54:13,172 --> 00:54:17,310 notation there. Then, Q would be some Q1, 716 00:54:17,310 --> 00:54:23,517 for which we have an A there, times E to the minus gamma over 717 00:54:23,517 --> 00:54:29,517 two times T, gamma is this value, times the cosine of omega 718 00:54:29,517 --> 00:54:35,469 T plus alpha. [And/N?] omega squared is omega 719 00:54:35,469 --> 00:54:40,169 zero squared minus gamma squared over four. 720 00:54:40,169 --> 00:54:45,093 That would be the solution if this were zero. 721 00:54:45,093 --> 00:54:50,689 No, it's not a zero. So, you can take 18.03 and try 722 00:54:50,689 --> 00:54:56,509 to solve if it's not zero. Or, you could think like a 723 00:54:56,509 --> 00:55:00,649 physicist and say, I don't need 18.03. 724 00:55:00,649 --> 00:55:05,082 [LAUGHTER] What is the difference between 725 00:55:05,082 --> 00:55:09,038 this differential equation in the one before is the spring 726 00:55:09,038 --> 00:55:12,370 that we had a zero here. In the case of the zero, 727 00:55:12,370 --> 00:55:16,395 it means that the oscillation would end up at X equals zero 728 00:55:16,395 --> 00:55:20,491 whereas here the oscillation in the charge will end up if we 729 00:55:20,491 --> 00:55:24,586 wait long enough with a fully charged capacitor because when 730 00:55:24,586 --> 00:55:28,542 the oscillation has died out, the capacitor has been fully 731 00:55:28,542 --> 00:55:31,655 charged. And so, clearly, 732 00:55:31,655 --> 00:55:36,413 if you add here, this charge on that capacitor, 733 00:55:36,413 --> 00:55:41,275 then you must be OK. That must take into account 734 00:55:41,275 --> 00:55:46,448 that it is not zero. And, this Q max is simply zero 735 00:55:46,448 --> 00:55:51,724 times C, which means the capacitor is fully charged. 736 00:55:51,724 --> 00:55:57,310 So, do it your 18.03 way, or do it with brains and then 737 00:55:57,310 --> 00:56:02,379 you will immediately agree to this solution to the 738 00:56:02,379 --> 00:56:08,693 differential equation. So, when you see here the 739 00:56:08,693 --> 00:56:14,836 decay, you see the oscillation. And then where you add up, 740 00:56:14,836 --> 00:56:19,793 you must add up with a fully charged capacitor. 741 00:56:19,793 --> 00:56:26,366 You do have to solve depending upon the initial conditions for 742 00:56:26,366 --> 00:56:31,000 alpha. And you have to solve for Q1. 743 00:56:31,000 --> 00:56:34,643 So, that means you have to know your initial conditions. 744 00:56:34,643 --> 00:56:37,492 And, the initial conditions of this problem, 745 00:56:37,492 --> 00:56:40,341 for instance, would be that at T equals zero 746 00:56:40,341 --> 00:56:43,786 when I throw the switch that there's no charge on the 747 00:56:43,786 --> 00:56:46,503 capacitor. There is one way I could do it. 748 00:56:46,503 --> 00:56:48,556 And there is no current flowing. 749 00:56:48,556 --> 00:56:51,935 That could be my initial condition, not the only one 750 00:56:51,935 --> 00:56:55,380 possibly, but that is one possible initial condition. 751 00:56:55,380 --> 00:56:58,759 And now, I will leave you with a little bit of work. 752 00:56:58,759 --> 00:57:01,078 First of all, you substitute in this 753 00:57:01,078 --> 00:57:04,889 equation. Q equal zero when T is zero. 754 00:57:04,889 --> 00:57:08,332 So, you still have Q1, and you still have the cosine 755 00:57:08,332 --> 00:57:10,492 of all. Then you have to take the 756 00:57:10,492 --> 00:57:14,204 derivative of this equation which gives you the current. 757 00:57:14,204 --> 00:57:16,567 And then you have to make that zero. 758 00:57:16,567 --> 00:57:20,144 Now, you've got two terms where you do the derivative. 759 00:57:20,144 --> 00:57:23,182 So, be careful. You get the derivative of this 760 00:57:23,182 --> 00:57:26,219 one times this, and the derivative of this one 761 00:57:26,219 --> 00:57:30,000 times this. So, you have to do it slowly. 762 00:57:30,000 --> 00:57:35,170 When you do that, you'll find that Q1 equals 763 00:57:35,170 --> 00:57:40,221 minus Q max divided by the cosine of alpha. 764 00:57:40,221 --> 00:57:46,835 And you will find that the tangent of alpha equals minus 765 00:57:46,835 --> 00:57:52,246 gamma over two omega. I don't have a very good 766 00:57:52,246 --> 00:57:58,379 feeling for these numbers, but what I do have a good 767 00:57:58,379 --> 00:58:05,594 feeling for it is that I know when Q, say, is larger or equal 768 00:58:05,594 --> 00:58:11,859 to five. Then I know that omega is very 769 00:58:11,859 --> 00:58:16,491 close to omega zero, extremely close. 770 00:58:16,491 --> 00:58:24,339 So, that means I can write for this minus gamma divided by two 771 00:58:24,339 --> 00:58:30,000 omega zero, which is minus one over 2Q. 772 00:58:30,000 --> 00:58:34,074 And, if Q is five or larger, my goodness, 773 00:58:34,074 --> 00:58:38,759 this is one tenth. If we forget the minus sign, 774 00:58:38,759 --> 00:58:44,972 now, so that means that phi is only something like 5.7° with a 775 00:58:44,972 --> 00:58:49,148 minus sign. And, the cosine of 5.7° is one 776 00:58:49,148 --> 00:58:51,694 for a physicist, at least. 777 00:58:51,694 --> 00:58:55,972 So, therefore, for all systems whereby Q is 778 00:58:55,972 --> 00:59:00,351 not absurdly low, you can say that Q1 equals 779 00:59:00,351 --> 00:59:05,799 minus Q max. And I'll put here wiggle, 780 00:59:05,799 --> 00:59:10,277 but it's an extremely good approximation. 781 00:59:10,277 --> 00:59:15,316 And so, therefore, my solution here can now be 782 00:59:15,316 --> 00:59:20,467 changed by simply replacing the Q1 minus Q max. 783 00:59:20,467 --> 00:59:24,498 And truly it should be approximately, 784 00:59:24,498 --> 00:59:28,976 but the approximation is extremely close. 785 00:59:28,976 --> 00:59:35,663 It depends on Q. If now you want to plot this, 786 00:59:35,663 --> 00:59:41,978 you get, of course, something completely similar to 787 00:59:41,978 --> 00:59:49,052 what you have here except that you do not end up at zero. 788 00:59:49,052 --> 00:59:55,115 The Q is not zero, but it's offset by this Q max. 789 00:59:55,115 --> 00:59:59,284 So, if this is T, if this is zero, 790 00:59:59,284 --> 1:00:05,094 and if this is Q max, which is where ultimately 791 1:00:05,094 --> 1:00:12,673 charge will be if you wait long enough, that to plot now this 792 1:00:12,673 --> 1:00:20,000 equation will then give me a curve like this. 793 1:00:20,000 --> 1:00:23,081 Be very careful. The period is uniquely 794 1:00:23,081 --> 1:00:26,162 determined. The period does not change. 795 1:00:26,162 --> 1:00:28,675 That's determined by that omega. 796 1:00:28,675 --> 1:00:33,378 2 pi divided by this omega is the periods, and so you start 797 1:00:33,378 --> 1:00:35,000 with Q zero. 798 1:00:35,000 --> 1:00:41,000 799 1:00:41,000 --> 1:00:44,261 And you see, if you waited long enough, 800 1:00:44,261 --> 1:00:47,436 you end up with Q max. It is this one, 801 1:00:47,436 --> 1:00:52,414 whereas in the case of the spring, if you wait long enough, 802 1:00:52,414 --> 1:00:55,675 you end up with nothing, X equals zero. 803 1:00:55,675 --> 1:01:00,481 And, this is something that I can demonstrate to you in a 804 1:01:00,481 --> 1:01:06,344 very, very nice way I must say. I find this one of the most 805 1:01:06,344 --> 1:01:11,032 intriguing demonstrations. Instead of having a battery, 806 1:01:11,032 --> 1:01:15,459 and throw the battery, in which case you would only, 807 1:01:15,459 --> 1:01:20,581 for an extremely short amount of time, for an amount of time 808 1:01:20,581 --> 1:01:24,315 which is roughly [UNINTELLIGIBLE] divided by 809 1:01:24,315 --> 1:01:28,221 gamma seconds, which may only be milliseconds, 810 1:01:28,221 --> 1:01:34,809 you'd see this thing. But, we want to give you more 811 1:01:34,809 --> 1:01:39,800 than a few milliseconds. And therefore, 812 1:01:39,800 --> 1:01:46,500 what we do: we replace that battery by 7 ms constant 813 1:01:46,500 --> 1:01:52,279 voltage, and we repeat that. And, every time, 814 1:01:52,279 --> 1:01:59,504 we show you this portion. And so, you will see it on the 815 1:01:59,504 --> 1:02:04,650 Tektronix. We trigger it every time that 816 1:02:04,650 --> 1:02:08,488 it comes up here and that it comes up here. 817 1:02:08,488 --> 1:02:13,331 And you will see them. This will be staring you in the 818 1:02:13,331 --> 1:02:16,073 face. I'm going to give you the 819 1:02:16,073 --> 1:02:21,281 values that I will be using so that you can digest at home 820 1:02:21,281 --> 1:02:26,673 exactly what you're going to see, and what all the constants 821 1:02:26,673 --> 1:02:31,151 are in this problem. The R is going to be 50 ohms. 822 1:02:31,151 --> 1:02:37,000 The L is going to be 50 MH, ten to the minus three. 823 1:02:37,000 --> 1:02:45,238 So, that tells me that gamma is 1,000, is R over L. 824 1:02:45,238 --> 1:02:55,289 That tells me that two divided by gamma, which is the one over 825 1:02:55,289 --> 1:03:00,892 EDK time, is then two milliseconds. 826 1:03:00,892 --> 1:03:08,392 C is going to be 0.06 µF. That would give you omega zero 827 1:03:08,392 --> 1:03:11,102 if you're interested in omega zero. 828 1:03:11,102 --> 1:03:14,053 I never had a good feeling for omegas. 829 1:03:14,053 --> 1:03:17,083 I like frequencies, hertz, much better. 830 1:03:17,083 --> 1:03:21,548 But if you're interested in omega zero, which is one over 831 1:03:21,548 --> 1:03:25,694 the square root of LC, that would be 18.3 times 10 to 832 1:03:25,694 --> 1:03:28,405 the third. And, that is radians per 833 1:03:28,405 --> 1:03:34,546 second. The frequency in hertz which is 834 1:03:34,546 --> 1:03:43,199 2 pi lower is that in 29.06 Hz. And so, now you have also Q, 835 1:03:43,199 --> 1:03:48,480 which is omega zero divided by gamma. 836 1:03:48,480 --> 1:03:56,106 And, that Q is now 18.3. And, that means Q divided by 837 1:03:56,106 --> 1:04:03,000 pi, which is, is there a problem Marcos? 838 1:04:03,000 --> 1:04:07,919 Q divided by pie which is also the number of oscillations that 839 1:04:07,919 --> 1:04:12,435 you have to wait for this function to go down by a factor 840 1:04:12,435 --> 1:04:15,177 of E. Remember, we do that with the 841 1:04:15,177 --> 1:04:17,677 pendulum. That is now about 5.8. 842 1:04:17,677 --> 1:04:21,790 So, we're going to take a look at a curve like that, 843 1:04:21,790 --> 1:04:26,306 and then we're going to indeed check that after roughly 6 844 1:04:26,306 --> 1:04:29,854 oscillations, the amplitude is indeed down by 845 1:04:29,854 --> 1:04:33,241 a factor of E. However, while we are at it, 846 1:04:33,241 --> 1:04:38,000 I'm going to kill two birds with one stone. 847 1:04:38,000 --> 1:04:43,869 I'm that going to make us 100 millihenry a little later, 848 1:04:43,869 --> 1:04:48,778 which changes the gamma. The gamma becomes 500. 849 1:04:48,778 --> 1:04:53,581 The decay time than becomes four milliseconds. 850 1:04:53,581 --> 1:04:59,557 I'm not going to change C. So, the frequency goes down by 851 1:04:59,557 --> 1:05:05,000 the square root of two. That becomes 2,055. 852 1:05:05,000 --> 1:05:10,760 And now, the Q also changes. The Q goes up by a factor 853 1:05:10,760 --> 1:05:14,239 square root of two, becomes 25.8, 854 1:05:14,239 --> 1:05:20,869 and now you have to wait about a point to oscillations for the 855 1:05:20,869 --> 1:05:24,891 amplitude to go down by a factor of E. 856 1:05:24,891 --> 1:05:30,000 And that's what I want to show you now. 857 1:05:30,000 --> 1:05:37,450 And, I had to choose my channel unless Marco has already done 858 1:05:37,450 --> 1:05:42,294 that for me. Ah, the surprise is already 859 1:05:42,294 --> 1:05:45,026 there. So, you see now, 860 1:05:45,026 --> 1:05:50,241 here, the oscillations with about 2,900 Hz. 861 1:05:50,241 --> 1:05:57,320 This is the location where the capacitor is fully charged, 862 1:05:57,320 --> 1:06:04,771 that is, that Q max and Q zero is right here at the left side 863 1:06:04,771 --> 1:06:10,302 at the bottom. That is where Q is zero. 864 1:06:10,302 --> 1:06:14,236 And, let's now count the 5.8 oscillations, 865 1:06:14,236 --> 1:06:16,922 roughly six, one, two, three, 866 1:06:16,922 --> 1:06:20,088 four, five, six. And, if you look, 867 1:06:20,088 --> 1:06:24,501 now, at this amplitude, after six oscillations, 868 1:06:24,501 --> 1:06:30,354 and you compared it with this, you can very easily see that it 869 1:06:30,354 --> 1:06:36,272 is roughly a factor of 2.7. So, that works exactly as 870 1:06:36,272 --> 1:06:39,636 predicted. And, you see this beautiful 871 1:06:39,636 --> 1:06:43,000 exponential decay. If now I change the 872 1:06:43,000 --> 1:06:46,545 self-inductance to hundred millihenries, 873 1:06:46,545 --> 1:06:51,363 then the frequency goes down from here 2,906 to 2,055. 874 1:06:51,363 --> 1:06:55,363 The Q goes up, and now you have to wait eight 875 1:06:55,363 --> 1:06:59,090 oscillations. So, Q is zero is on the left 876 1:06:59,090 --> 1:07:03,346 side of the bottom. This is, again, 877 1:07:03,346 --> 1:07:06,756 fully charged Q max. And now we go one, 878 1:07:06,756 --> 1:07:09,807 two, three, four, five, six, seven, 879 1:07:09,807 --> 1:07:12,589 eight. And, the amplitude now is 880 1:07:12,589 --> 1:07:16,897 roughly the same as it was before when we had six 881 1:07:16,897 --> 1:07:20,487 oscillations. So, you see that this works 882 1:07:20,487 --> 1:07:24,974 like a charm if you do it with an L or a C circuit: 883 1:07:24,974 --> 1:07:29,641 really quite remarkable. So dying oscillation and the 884 1:07:29,641 --> 1:07:34,846 damping is the result of the resistor, which is dissipating 885 1:07:34,846 --> 1:07:39,814 energy. If you want sleepless nights, 886 1:07:39,814 --> 1:07:44,472 which is healthy, you may want to try to find for 887 1:07:44,472 --> 1:07:50,683 yourself a convincing reason why the L also comes in the damping. 888 1:07:50,683 --> 1:07:54,177 The L can never dissipate any energy. 889 1:07:54,177 --> 1:07:58,447 Only the R can. Why is it that the gamma is R 890 1:07:58,447 --> 1:08:01,538 over L? I'll be very honest with you. 891 1:08:01,538 --> 1:08:04,351 When I searched for one, it took me 10 minutes to come 892 1:08:04,351 --> 1:08:06,579 up with the right answer. At first you say, 893 1:08:06,579 --> 1:08:08,808 this is weird. But, if you give it a little 894 1:08:08,808 --> 1:08:11,248 bit more thought, and if you got a B+ or higher 895 1:08:11,248 --> 1:08:13,000 for 8.02, you can do it. 896 1:08:13,000 --> 1:08:19,000 897 1:08:19,000 --> 1:08:20,000 All right. 898 1:08:20,000 --> 1:08:25,000 899 1:08:25,000 --> 1:08:30,985 All this is very nice and dandy, but what now happens if 900 1:08:30,985 --> 1:08:37,188 gamma squared over four is larger than omega zero squared? 901 1:08:37,188 --> 1:08:43,391 What now happens if gamma squared over four is larger than 902 1:08:43,391 --> 1:08:47,308 omega zero squared? So, our solution, 903 1:08:47,308 --> 1:08:52,205 omega squared, equals omega zero squared minus 904 1:08:52,205 --> 1:08:58,735 gamma squared over four is utter nonsense because omega would 905 1:08:58,735 --> 1:09:04,865 then be imaginary. So now, we have to go back to 906 1:09:04,865 --> 1:09:09,607 our original differential equation and find a new 907 1:09:09,607 --> 1:09:13,163 solution. We are closer to a solution 908 1:09:13,163 --> 1:09:17,807 than you may think. Remember, we did have at one 909 1:09:17,807 --> 1:09:23,734 point in time N squared equals omega zero squared minus gamma 910 1:09:23,734 --> 1:09:28,477 squared over four. And then, we recognized that N 911 1:09:28,477 --> 1:09:32,725 really is omega. Well, N is therefore also J 912 1:09:32,725 --> 1:09:38,554 times gamma squared over four minus omega squared because if 913 1:09:38,554 --> 1:09:43,000 you square this, you get that. 914 1:09:43,000 --> 1:09:49,077 I did something weird again. I told you, it's not my day. 915 1:09:49,077 --> 1:09:54,395 We had this earlier in the lecture, which is fine. 916 1:09:54,395 --> 1:09:57,000 Then I said, well -- 917 1:09:57,000 --> 1:11:52,911 [SOUND OFF/THEN ON] -- conditions. 918 1:11:52,911 --> 1:11:58,644 [NOISE OBSCURES] Now we get gamma over two, 919 1:11:58,644 --> 1:12:06,425 [NOISE OBSCURES] over four minus omega zero squared to the 920 1:12:06,425 --> 1:12:14,479 power one half times [NOISE OBSCURES] plus A2 times E to the 921 1:12:14,479 --> 1:12:21,987 minus gamma, not minus gamma, minus gamma over two minus 922 1:12:21,987 --> 1:12:30,177 gamma squared over four minus omega zero squared to the power 923 1:12:30,177 --> 1:12:36,651 one half times T. You have to take my word for it 924 1:12:36,651 --> 1:12:40,927 that this is the solution. And when you look at the 925 1:12:40,927 --> 1:12:44,348 solution, nothing is oscillating anymore. 926 1:12:44,348 --> 1:12:49,052 This is a decaying amplitude, K1, and this is a decaying 927 1:12:49,052 --> 1:12:52,388 amplitude, A2. They have different decay 928 1:12:52,388 --> 1:12:55,552 times. This determines one decay time, 929 1:12:55,552 --> 1:13:00,000 and this determines the other decay time. 930 1:13:00,000 --> 1:13:05,069 And, depending upon the initial conditions, will you be able to 931 1:13:05,069 --> 1:13:07,685 find A1 and A2? At T equals zero, 932 1:13:07,685 --> 1:13:12,264 you can specify what X is, and you can specify what X dot 933 1:13:12,264 --> 1:13:13,000 is. 934 1:13:13,000 --> 1:13:18,000 935 1:13:18,000 --> 1:13:25,974 No oscillations: that means I can make some 936 1:13:25,974 --> 1:13:33,000 guess of what the response will be. 937 1:13:33,000 --> 1:13:37,753 Let this be a spring system, and this is X equals zero. 938 1:13:37,753 --> 1:13:41,625 And, I offset the spring to position, X zero, 939 1:13:41,625 --> 1:13:46,731 and I let it go at zero speed. That's my initial condition, 940 1:13:46,731 --> 1:13:50,163 zero speed. So, that means when I let it 941 1:13:50,163 --> 1:13:53,596 go, it must go in this direction, right, 942 1:13:53,596 --> 1:13:57,381 in that diagram? Because DX DT must be zero. 943 1:13:57,381 --> 1:14:01,958 But, there is no oscillation. So will happen is this, 944 1:14:01,958 --> 1:14:07,063 and will asymptotically go to zero with two different decay 945 1:14:07,063 --> 1:14:11,760 times. Now, if you released it here 946 1:14:11,760 --> 1:14:17,084 with a high-speed in this direction, then you might see 947 1:14:17,084 --> 1:14:21,225 this, but not in oscillation. An overshoot: 948 1:14:21,225 --> 1:14:26,845 yes, but not an oscillation. And, you will get a chance in 949 1:14:26,845 --> 1:14:31,971 one of your problems, I think it is 1-7 for this week 950 1:14:31,971 --> 1:14:37,000 where you can play with this a little bit. 951 1:14:37,000 --> 1:14:43,046 There is one more case, which is a very special case. 952 1:14:43,046 --> 1:14:49,209 And that is when omega zero is exactly gamma over two. 953 1:14:49,209 --> 1:14:53,279 And, that's called critical damping. 954 1:14:53,279 --> 1:14:58,511 So, we have three [new means?]: under-damping. 955 1:14:58,511 --> 1:15:04,296 We did that in detail. Then you have over damping 956 1:15:04,296 --> 1:15:09,242 which gives you this solution, and then you have critical 957 1:15:09,242 --> 1:15:12,245 damping. And with critical damping, 958 1:15:12,245 --> 1:15:16,131 the solution, and I give you this without any 959 1:15:16,131 --> 1:15:21,606 derivation, is A plus B times T times E to the minus gamma over 960 1:15:21,606 --> 1:15:24,433 two times T. Only one decay time, 961 1:15:24,433 --> 1:15:29,290 one exponential decay time, but there is something funny 962 1:15:29,290 --> 1:15:33,000 here. There is also a T here. 963 1:15:33,000 --> 1:15:38,308 So, this BT is linearly growing at the beginning, 964 1:15:38,308 --> 1:15:41,737 but of course this one kills it. 965 1:15:41,737 --> 1:15:48,594 And, again, your two adjustable parameters, A and B follow from 966 1:15:48,594 --> 1:15:54,124 the initial conditions. So, now I will show you the 967 1:15:54,124 --> 1:15:59,427 effect of damping. In a case of a mechanical 968 1:15:59,427 --> 1:16:03,506 system, we have a torsional pendulum there. 969 1:16:03,506 --> 1:16:08,167 And, the torsional pendulum has extremely high Q, 970 1:16:08,167 --> 1:16:12,635 very little damping. I'll show it to you there. 971 1:16:12,635 --> 1:16:17,782 But before we show it there, just concentrate on this. 972 1:16:17,782 --> 1:16:22,734 Without any damping, if it oscillates and amplitude, 973 1:16:22,734 --> 1:16:28,561 which is a torsional pendulum is one that has a spring like a 974 1:16:28,561 --> 1:16:34,000 clock, this will die out very, very slowly. 975 1:16:34,000 --> 1:16:36,311 Very high Q, very little damping. 976 1:16:36,311 --> 1:16:38,838 What we have here in electromagnets, 977 1:16:38,838 --> 1:16:42,811 and we can change the current through the electromagnet, 978 1:16:42,811 --> 1:16:47,144 and the copper plate that move through the electromagnet will 979 1:16:47,144 --> 1:16:51,116 now develop any current because there is a change in the 980 1:16:51,116 --> 1:16:54,944 magnetic flux through that metal causes anti-currents, 981 1:16:54,944 --> 1:17:00,000 and the anti-currents will oppose the motion of the metal. 982 1:17:00,000 --> 1:17:04,754 You've seen that with 8.02 where you throw a coin through a 983 1:17:04,754 --> 1:17:09,508 very strong magnetic field. The coin slows down enormously. 984 1:17:09,508 --> 1:17:13,934 Instead of going clunk, a coin can go down this slowly. 985 1:17:13,934 --> 1:17:17,049 It's a very nice demonstration in 8.02. 986 1:17:17,049 --> 1:17:20,245 So, we can also, using the anti-current, 987 1:17:20,245 --> 1:17:23,114 we can introduce artificial damping. 988 1:17:23,114 --> 1:17:25,983 And that's what you're going to see. 989 1:17:25,983 --> 1:17:30,000 One, two, three, this should be it. 990 1:17:30,000 --> 1:17:35,723 Yeah, oh, we have to change the position, but that's not a big 991 1:17:35,723 --> 1:17:38,819 deal. We may have moved it during, 992 1:17:38,819 --> 1:17:41,916 oh, we also need some light on it. 993 1:17:41,916 --> 1:17:45,012 I turned the light off, thank you. 994 1:17:45,012 --> 1:17:49,891 Let's first make sure that we, yeah, there we see it. 995 1:17:49,891 --> 1:17:53,269 So, I will give it a large amplitude. 996 1:17:53,269 --> 1:17:56,834 The amplitude now is in terms of angle. 997 1:17:56,834 --> 1:18:01,338 No damping at all, and if you look closely at the 998 1:18:01,338 --> 1:18:03,777 turning point, you can say, 999 1:18:03,777 --> 1:18:10,786 yeah, yeah, it is decaying. Give it a little bit of time. 1000 1:18:10,786 --> 1:18:14,628 But very slowly, you can see it decaying. 1001 1:18:14,628 --> 1:18:20,489 You have to wait a long time to see the amplitude go down by a 1002 1:18:20,489 --> 1:18:25,676 factor of E because you have such a very high Q system. 1003 1:18:25,676 --> 1:18:30,000 But now, I'm going to give it damping. 1004 1:18:30,000 --> 1:18:36,059 It's still under-damped. First give it a high amplitude. 1005 1:18:36,059 --> 1:18:39,364 There we go. Damping is in now. 1006 1:18:39,364 --> 1:18:43,000 Look how fast it's going now. 1007 1:18:43,000 --> 1:18:49,000 1008 1:18:49,000 --> 1:18:53,634 With this system, I cannot show you an example of 1009 1:18:53,634 --> 1:18:56,048 over damping. We did that, 1010 1:18:56,048 --> 1:19:01,841 and then the current has to be so high that we begin to smell 1011 1:19:01,841 --> 1:19:05,507 smoke. So, we decided not to do that. 1012 1:19:05,507 --> 1:19:10,270 But if I go to the point where you just don't smell smoke, 1013 1:19:10,270 --> 1:19:12,861 then you are still under-damped, 1014 1:19:12,861 --> 1:19:15,870 but you get a huge amount of damping. 1015 1:19:15,870 --> 1:19:20,634 And the reason why I can say with confidence that it still 1016 1:19:20,634 --> 1:19:23,976 under-damped, I release it at zero speed. 1017 1:19:23,976 --> 1:19:28,991 And, if I release it at zero speed, and if it is over-damped, 1018 1:19:28,991 --> 1:19:35,720 it can never overshoot. Yet, you will see that there is 1019 1:19:35,720 --> 1:19:41,479 a minute overshoot. So, it is still under-damped. 1020 1:19:41,479 --> 1:19:47,720 That's getting very close to being critically damped. 1021 1:19:47,720 --> 1:19:53,960 And so, I will jack these up to the maximum possible. 1022 1:19:53,960 --> 1:19:59,960 I can go way beyond 15. Is that the smoking domain? 1023 1:19:59,960 --> 1:20:03,878 It's OK. We don't go to the smoking 1024 1:20:03,878 --> 1:20:06,839 domain, right? So, I will first put in the 1025 1:20:06,839 --> 1:20:10,162 damping, and then I'll offset it and let it go. 1026 1:20:10,162 --> 1:20:12,546 Damping is in now. I'll offset it, 1027 1:20:12,546 --> 1:20:14,713 and let it go. Three, two, one, 1028 1:20:14,713 --> 1:20:16,663 zero. You see the overshoot? 1029 1:20:16,663 --> 1:20:21,142 It's just a teeny-weeny little. That tells you that it is still 1030 1:20:21,142 --> 1:20:25,331 under-damped because it's not supposed to overshoot since I 1031 1:20:25,331 --> 1:20:29,376 released it at zero speed. Of course, if I give it a push 1032 1:20:29,376 --> 1:20:34,000 in that direction, of course it will overshoot. 1033 1:20:34,000 --> 1:20:37,319 That's obvious. But if I do it with zero speed, 1034 1:20:37,319 --> 1:20:41,648 that solution that we have on the blackboard dictates that it 1035 1:20:41,648 --> 1:20:44,534 cannot overshoot. I'll show it once more. 1036 1:20:44,534 --> 1:20:46,771 And, you will see the overshoot. 1037 1:20:46,771 --> 1:20:49,441 That's it. So, that is the whole story 1038 1:20:49,441 --> 1:20:51,966 about damping we have under-damping, 1039 1:20:51,966 --> 1:20:54,997 we have critical damping, and over damping. 1040 1:20:54,997 --> 1:20:57,378 And, let's not put a damper on it. 1041 1:20:57,378 --> 1:21:01,347 I will see you again on Thursday, and make sure you pick 1042 1:21:01,347 up your 8.03 kits.