1 00:00:31,000 --> 00:00:34,990 I want to start with the physical pendulum, 2 00:00:34,990 --> 00:00:40,310 which is exactly the same one that I discussed during the 3 00:00:40,310 --> 00:00:44,490 first lecture. This is a hoop with mass M and 4 00:00:44,490 --> 00:00:48,005 radius R. And, we were calculating the 5 00:00:48,005 --> 00:00:51,425 period of this hoop as it oscillates. 6 00:00:51,425 --> 00:00:56,840 And we did that using the famous stork equation from 8.01, 7 00:00:56,840 --> 00:01:01,969 the torque relative to point P is the moment of inertia 8 00:01:01,969 --> 00:01:05,579 relative to point, P, times the angular 9 00:01:05,579 --> 00:01:09,000 acceleration, alpha. 10 00:01:09,000 --> 00:01:13,296 Today, I will do this again, but I will use the conservation 11 00:01:13,296 --> 00:01:17,229 of energy to show you that in case there is no damping, 12 00:01:17,229 --> 00:01:21,308 when mechanical energy is conserved, but you can find the 13 00:01:21,308 --> 00:01:25,313 correct to fragile equations through the conservation of 14 00:01:25,313 --> 00:01:27,644 energy. If the thing is swinging, 15 00:01:27,644 --> 00:01:30,193 in general there are two components. 16 00:01:30,193 --> 00:01:35,000 You have kinetic energy and you have potential energy. 17 00:01:35,000 --> 00:01:39,815 And the kinetic energy, K, is one half times the moment 18 00:01:39,815 --> 00:01:43,738 of inertia about point P times omega squared. 19 00:01:43,738 --> 00:01:48,821 You remember that from 8.01. And, this omega is theta dot. 20 00:01:48,821 --> 00:01:51,764 We call that the angular velocity. 21 00:01:51,764 --> 00:01:55,242 This angular velocity changes with time. 22 00:01:55,242 --> 00:01:59,878 When the object goes through equilibrium, to a new or 23 00:01:59,878 --> 00:02:05,054 velocity is at maximum. But when object comes to a 24 00:02:05,054 --> 00:02:07,740 halt, the angular velocity is zero. 25 00:02:07,740 --> 00:02:10,980 Do not confuse this omega with omega zero. 26 00:02:10,980 --> 00:02:15,009 I will give that a zero, now, to distinguish it from 27 00:02:15,009 --> 00:02:17,933 omega, which is the angular frequency. 28 00:02:17,933 --> 00:02:21,804 The angular frequency is a constant of the motion, 29 00:02:21,804 --> 00:02:26,703 and that is two pi divided by T zero if T zero is the period of 30 00:02:26,703 --> 00:02:30,101 oscillation. So, this is the kinetic energy, 31 00:02:30,101 --> 00:02:35,000 and this is the square of the angular velocity. 32 00:02:35,000 --> 00:02:37,805 And then, we have potential energy. 33 00:02:37,805 --> 00:02:41,188 Let this be point A, and when we are here, 34 00:02:41,188 --> 00:02:43,828 the center of mass is at point B. 35 00:02:43,828 --> 00:02:48,778 And so, the potential energy, U, or actually I should say the 36 00:02:48,778 --> 00:02:53,399 potential energy at point B minus the potential energy at 37 00:02:53,399 --> 00:02:57,937 point, A, it's always the difference in potential energy 38 00:02:57,937 --> 00:03:02,000 that matters. That equals MGH, 39 00:03:02,000 --> 00:03:09,125 H being the difference in height between point B and point 40 00:03:09,125 --> 00:03:12,000 A. So, this here is MGH, 41 00:03:12,000 --> 00:03:19,625 Massachusetts General Hospital. That's the way to remember it. 42 00:03:19,625 --> 00:03:25,625 Now, H is very easy. H is the same as R times one 43 00:03:25,625 --> 00:03:32,596 minus the cosine of theta. We went through that many 44 00:03:32,596 --> 00:03:36,438 times, so you can easily confirm that. 45 00:03:36,438 --> 00:03:41,630 R is the radius of this circle. This is the radius. 46 00:03:41,630 --> 00:03:46,823 Now, for small angles, the cosine of theta is theta 47 00:03:46,823 --> 00:03:51,911 squared divided by two. And so, I can rewrite this 48 00:03:51,911 --> 00:03:57,000 differential equation now, that E total. 49 00:03:57,000 --> 00:04:04,000 50 00:04:04,000 --> 00:04:08,567 I see nothing wrong with that. I'm sorry. 51 00:04:08,567 --> 00:04:12,108 Is there anything wrong with it? 52 00:04:12,108 --> 00:04:16,219 The cosine theta alone, you're right, 53 00:04:16,219 --> 00:04:20,216 is one minus theta squared over two. 54 00:04:20,216 --> 00:04:23,641 Thank you very much. Thank you. 55 00:04:23,641 --> 00:04:30,608 So, I'm going to write it now as a total energy is one half IP 56 00:04:30,608 --> 00:04:36,774 times theta dot squared. And then, I get plus MGR times 57 00:04:36,774 --> 00:04:42,308 theta squared over two. And, what you do now, 58 00:04:42,308 --> 00:04:46,048 since this is a constant of the motion, you always do that if 59 00:04:46,048 --> 00:04:48,541 you work with the conservation of energy. 60 00:04:48,541 --> 00:04:51,345 You take the time derivative of this equation, 61 00:04:51,345 --> 00:04:54,898 which must be zero because mechanical energy is conserved. 62 00:04:54,898 --> 00:04:58,450 And, out pops the differential equation that we also found 63 00:04:58,450 --> 00:05:01,753 during my lecture number one when I used the different 64 00:05:01,753 --> 00:05:06,184 method. So, I take the time derivative. 65 00:05:06,184 --> 00:05:12,143 So, this [T?] each of this half, and so we get I if P times 66 00:05:12,143 --> 00:05:15,431 theta dot times theta double dot. 67 00:05:15,431 --> 00:05:20,671 I had used the chain rule. And then, I get plus MGR, 68 00:05:20,671 --> 00:05:26,732 these two, each of these two, and so I get theta times theta 69 00:05:26,732 --> 00:05:33,000 dot, and now this equals zero because DE DT is zero. 70 00:05:33,000 --> 00:05:38,644 And whenever you do that, you will always see that the 71 00:05:38,644 --> 00:05:44,607 theta dot term disappears. Or if you have the equation in 72 00:05:44,607 --> 00:05:48,122 X, then the X dot term disappears. 73 00:05:48,122 --> 00:05:51,636 And, you see that. This term goes. 74 00:05:51,636 --> 00:05:56,641 You can divide it out. And so, your differential 75 00:05:56,641 --> 00:06:01,327 equation takes on, now, a very familiar form. 76 00:06:01,327 --> 00:06:07,610 Let me write these two here. So, I get theta double dot plus 77 00:06:07,610 --> 00:06:14,000 MGR divided by I of P times theta equals zero. 78 00:06:14,000 --> 00:06:20,781 And this differential equation you should recognize. 79 00:06:20,781 --> 00:06:27,696 The solution to that equation is immediately obvious. 80 00:06:27,696 --> 00:06:34,345 Omega zero squared, I use now the omega zero equals 81 00:06:34,345 --> 00:06:41,925 MG times R divided by I of P. And so, the general solution 82 00:06:41,925 --> 00:06:47,358 for this oscillation then becomes that theta is theta 83 00:06:47,358 --> 00:06:51,119 zero. That is the amplitude times the 84 00:06:51,119 --> 00:06:55,089 cosine or the sine, if you prefer that, 85 00:06:55,089 --> 00:07:00,000 omega zero T plus some phase angle, phi. 86 00:07:00,000 --> 00:07:04,549 That is the general solution. And if you knew the initial 87 00:07:04,549 --> 00:07:08,691 conditions, what the situation was at T equals zero, 88 00:07:08,691 --> 00:07:12,753 if you know the velocity, the angular velocity at T 89 00:07:12,753 --> 00:07:17,383 equals zero, and if you know where it is at T equals zero, 90 00:07:17,383 --> 00:07:22,257 you can solve for this theta zero, and you can solve for this 91 00:07:22,257 --> 00:07:25,019 phi. But, omega zero is independent 92 00:07:25,019 --> 00:07:29,000 of the initial conditions. Very well. 93 00:07:29,000 --> 00:07:32,567 Now, I would like to cover a case whereby I am going to 94 00:07:32,567 --> 00:07:35,739 introduce damping. Whenever we deal with damping, 95 00:07:35,739 --> 00:07:38,976 there are two terms that are important in physics. 96 00:07:38,976 --> 00:07:42,478 That is the speed itself. That is a damping term which 97 00:07:42,478 --> 00:07:45,187 opposes the velocity, which is when you're 98 00:07:45,187 --> 00:07:48,821 proportional with the speed. And then there is a damping 99 00:07:48,821 --> 00:07:52,917 term which opposes the velocity, which is proportional with the 100 00:07:52,917 --> 00:07:56,287 square of the speed. We will always leave the square 101 00:07:56,287 --> 00:07:59,855 out with 8.03 because the differential equations become 102 00:07:59,855 --> 00:08:03,026 impossible to solve. Maybe numerically you can do 103 00:08:03,026 --> 00:08:07,883 it, but not analytically. However, if you're interested 104 00:08:07,883 --> 00:08:10,352 in all the physics, which is wonderful, 105 00:08:10,352 --> 00:08:13,600 with a V squared and V, my lecture number 12 on OCW 106 00:08:13,600 --> 00:08:16,589 Open Courseware from 1999, Newtonian mechanics, 107 00:08:16,589 --> 00:08:19,512 I deal with the V squared and with the V term, 108 00:08:19,512 --> 00:08:23,540 and I do many demonstrations to show you that there are certain 109 00:08:23,540 --> 00:08:26,009 domains where the V term is important . 110 00:08:26,009 --> 00:08:29,842 We call that the viscous term, and there are certain domains 111 00:08:29,842 --> 00:08:34,000 in physics where the V squared term is important. 112 00:08:34,000 --> 00:08:39,189 So, I will now simply restrict myself, then, 113 00:08:39,189 --> 00:08:45,344 to the damping force, which is linearly proportional 114 00:08:45,344 --> 00:08:51,137 with the velocity, and we will write it down as F 115 00:08:51,137 --> 00:08:56,689 equals minus B times V. We will use a shorthand 116 00:08:56,689 --> 00:09:02,000 notation that gamma equals B over M. 117 00:09:02,000 --> 00:09:04,332 That is only a shorthand notation. 118 00:09:04,332 --> 00:09:07,513 I will erase this. We don't need this anymore. 119 00:09:07,513 --> 00:09:11,117 We need so many blackboards today that I'll use this 120 00:09:11,117 --> 00:09:13,450 blackboard from a damping problem. 121 00:09:13,450 --> 00:09:17,337 When you deal with damping, we recognize three different 122 00:09:17,337 --> 00:09:19,882 domains. One domain, whereby gamma is 123 00:09:19,882 --> 00:09:23,628 smaller than the omega zero, we call that underdamped. 124 00:09:23,628 --> 00:09:29,000 Then we have a domain whereby gamma is larger than omega zero. 125 00:09:29,000 --> 00:09:31,512 We call that overdamped. And then, you have a very 126 00:09:31,512 --> 00:09:33,820 special case where gamma equals of omega zero. 127 00:09:33,820 --> 00:09:35,000 [SOUND OFF/THEN ON] 128 00:09:35,000 --> 00:09:43,000 129 00:09:43,000 --> 00:09:49,606 And the behavior of these three different kinds is very 130 00:09:49,606 --> 00:09:54,255 different. I will only discuss with you 131 00:09:54,255 --> 00:10:00,250 today the underdamped case. So, I have a pendulum, 132 00:10:00,250 --> 00:10:06,000 and the pendulum has mass M and length L. 133 00:10:06,000 --> 00:10:11,576 And I will assume that the mass in the string is negligibly 134 00:10:11,576 --> 00:10:13,596 small. I have damping, 135 00:10:13,596 --> 00:10:19,173 and this is the case that gamma is smaller than omega zero. 136 00:10:19,173 --> 00:10:23,788 This is the equilibrium position of the pendulum. 137 00:10:23,788 --> 00:10:29,365 I will therefore call this position X, and I'm going to put 138 00:10:29,365 --> 00:10:33,692 all the forces on this object in this picture, 139 00:10:33,692 --> 00:10:37,395 which is MG. And, that is T. 140 00:10:37,395 --> 00:10:41,395 And, there are no other forces on that mass. 141 00:10:41,395 --> 00:10:46,604 Now, if we deal with small angles, as we have done before 142 00:10:46,604 --> 00:10:49,581 more than once, then the tension, 143 00:10:49,581 --> 00:10:54,139 T, is very close to MG. So, the only force that is 144 00:10:54,139 --> 00:10:59,348 driving it back to equilibrium, the only restoring force, 145 00:10:59,348 --> 00:11:04,000 then, is the horizontal component of T. 146 00:11:04,000 --> 00:11:07,952 And so, there is this horizontal component. 147 00:11:07,952 --> 00:11:12,376 That's the only one that we are concerned about. 148 00:11:12,376 --> 00:11:17,647 So, the differential equation, then, in terms of Newton's 149 00:11:17,647 --> 00:11:20,658 second law becomes MX double dot. 150 00:11:20,658 --> 00:11:26,399 And then, we get minus T times the sine of theta minus B times 151 00:11:26,399 --> 00:11:28,752 X dot. That's the damping, 152 00:11:28,752 --> 00:11:34,023 and this is the restoring force due to the tension in the 153 00:11:34,023 --> 00:11:41,259 string. And, the sine of theta is X 154 00:11:41,259 --> 00:11:49,831 divided by L. So, I get MX double dot plus MG 155 00:11:49,831 --> 00:11:58,987 divided by L times X. That is the sine of theta, 156 00:11:58,987 --> 00:12:08,307 plus B times X dot equals zero. Make sure I have this M double 157 00:12:08,307 --> 00:12:10,615 dot. I have the plus sign here. 158 00:12:10,615 --> 00:12:13,000 I have MG X over L. That's fine. 159 00:12:13,000 --> 00:12:16,923 So now, I divide M out, and I also use the shorthand 160 00:12:16,923 --> 00:12:20,769 notation that omega zero squared is G divided by L. 161 00:12:20,769 --> 00:12:25,461 These are shorthand notations which give you a little bit more 162 00:12:25,461 --> 00:12:29,000 insight when you see the solutions. 163 00:12:29,000 --> 00:12:33,444 And so, that gives me, now, the differential equation 164 00:12:33,444 --> 00:12:38,572 that X double dot plus gamma times X dot plus I divide M out, 165 00:12:38,572 --> 00:12:42,846 right, plus omega zero squared times X equals zero. 166 00:12:42,846 --> 00:12:47,974 And, that is the differential equation that you'll recognize. 167 00:12:47,974 --> 00:12:52,846 And, I would never want you to derive the solution to this 168 00:12:52,846 --> 00:12:56,264 differential equation. That's, of course, 169 00:12:56,264 --> 00:13:00,367 way too time consuming to do that during an exam. 170 00:13:00,367 --> 00:13:04,897 The solutions to this are given on your formula sheet, 171 00:13:04,897 --> 00:13:09,000 which will be part of your exam. 172 00:13:09,000 --> 00:13:15,753 And so, let me write down here what that solution is. 173 00:13:15,753 --> 00:13:23,545 So, X as a function of time is a certain amplitude times E to 174 00:13:23,545 --> 00:13:27,701 the minus gamma over two times T. 175 00:13:27,701 --> 00:13:35,623 And then, we have a cosine or a sine, cosine omega T plus some 176 00:13:35,623 --> 00:13:40,716 phase angle alpha. If you knew the initial 177 00:13:40,716 --> 00:13:44,522 conditions, then you can find what the amplitude is, 178 00:13:44,522 --> 00:13:47,358 and you can find with the angle, alpha, 179 00:13:47,358 --> 00:13:49,746 is. If you don't know the initial 180 00:13:49,746 --> 00:13:52,507 conditions, then you do not know this. 181 00:13:52,507 --> 00:13:56,164 There is another way that you can write this form, 182 00:13:56,164 --> 00:14:00,343 which is sometimes better. And, I cannot tell you when it 183 00:14:00,343 --> 00:14:04,000 is better and when it is not better. 184 00:14:04,000 --> 00:14:06,827 It depends on the initial conditions. 185 00:14:06,827 --> 00:14:11,383 But, I want you to appreciate that you can also write this, 186 00:14:11,383 --> 00:14:14,604 for instance, as E to the minus gamma over 187 00:14:14,604 --> 00:14:19,317 two times T, and then B times cosine omega T plus C times the 188 00:14:19,317 --> 00:14:23,009 sine of omega T. From the physics point of view, 189 00:14:23,009 --> 00:14:26,858 there is no difference. But from the math point of 190 00:14:26,858 --> 00:14:32,209 view, there is a difference. You now have these as the two 191 00:14:32,209 --> 00:14:36,552 adjustable constants depending upon the initial condition. 192 00:14:36,552 --> 00:14:39,066 And sometimes, if you assume this, 193 00:14:39,066 --> 00:14:42,038 it works faster than if you assume that. 194 00:14:42,038 --> 00:14:44,933 And as I said, it really depends on the 195 00:14:44,933 --> 00:14:49,123 initial conditions which goes faster, that they are very 196 00:14:49,123 --> 00:14:52,323 similar, of course. Omega, which is now the 197 00:14:52,323 --> 00:14:56,209 frequency with which this object, angular frequency, 198 00:14:56,209 --> 00:14:59,866 is going to move, that omega is always lower than 199 00:14:59,866 --> 00:15:04,056 omega zero. And, that shouldn't surprise 200 00:15:04,056 --> 00:15:08,244 you because when there is damping, there is something that 201 00:15:08,244 --> 00:15:12,063 is opposing the motion. And so, it shouldn't surprise 202 00:15:12,063 --> 00:15:16,104 you that when you solve for omega, that omega squared is 203 00:15:16,104 --> 00:15:20,071 omega zero squared minus gamma squared divided by four. 204 00:15:20,071 --> 00:15:24,478 That is also something that we would probably give you on the 205 00:15:24,478 --> 00:15:29,033 formula sheet because you will only find that if you substitute 206 00:15:29,033 --> 00:15:33,000 this back into the differential equation. 207 00:15:33,000 --> 00:15:37,389 Often, particularly French, likes to write down Q equals 208 00:15:37,389 --> 00:15:41,618 omega divided by gamma. This is the quality omega zero 209 00:15:41,618 --> 00:15:44,571 divided by gamma. And, if you do that, 210 00:15:44,571 --> 00:15:49,279 then omega squared can also be written as omega zero squared 211 00:15:49,279 --> 00:15:52,391 times one minus one over four Q squared. 212 00:15:52,391 --> 00:15:55,982 And so, you see that if Q is for instance ten, 213 00:15:55,982 --> 00:16:00,531 which is by no means absurdly high, that omega is only one 214 00:16:00,531 --> 00:16:05,000 eighth of a percent lower than omega zero. 215 00:16:05,000 --> 00:16:08,818 So, there that close. And, even when Q is two, 216 00:16:08,818 --> 00:16:13,399 the difference is only 3% between omega and omega zero. 217 00:16:13,399 --> 00:16:18,321 And so, what happens here is that the amplitude is decaying 218 00:16:18,321 --> 00:16:22,224 at a time constant, a one over E time constant, 219 00:16:22,224 --> 00:16:25,363 which is two divided by gamma seconds. 220 00:16:25,363 --> 00:16:28,503 If you put in T, two divided by gamma, 221 00:16:28,503 --> 00:16:33,000 the amplitude goes down by a factor of E. 222 00:16:33,000 --> 00:16:37,605 Since energy is always proportional to amplitude 223 00:16:37,605 --> 00:16:43,680 squared, the decay time of the energy is not two over gamma but 224 00:16:43,680 --> 00:16:48,384 is one divided by gamma. Now, I'm going to make a 225 00:16:48,384 --> 00:16:52,108 change. I'm going to drive this system. 226 00:16:52,108 --> 00:16:56,223 This is no longer the equilibrium position, 227 00:16:56,223 --> 00:17:00,535 but the equilibrium position was really here. 228 00:17:00,535 --> 00:17:06,218 And, I am driving the top of this pendulum with a function, 229 00:17:06,218 --> 00:17:12,000 eta, is eta zero times the cosine of omega T. 230 00:17:12,000 --> 00:17:16,806 So, I am driving it now. So, this eta is in terms of 231 00:17:16,806 --> 00:17:21,518 inches, millimeters. This is the motion of my hand. 232 00:17:21,518 --> 00:17:25,193 And, this omega is no longer negotiable. 233 00:17:25,193 --> 00:17:30,000 This omega has nothing to do with this omega. 234 00:17:30,000 --> 00:17:33,493 This is the frequency with which the system likes to 235 00:17:33,493 --> 00:17:36,027 oscillate. That, now, is the frequency 236 00:17:36,027 --> 00:17:38,835 with which I want the system to oscillate. 237 00:17:38,835 --> 00:17:41,780 They are totally unrelated. This is my will. 238 00:17:41,780 --> 00:17:45,068 This is nonnegotiable. I can make this anything I 239 00:17:45,068 --> 00:17:46,712 want. I can make it zero. 240 00:17:46,712 --> 00:17:50,136 I can make it large. I can make it also that value, 241 00:17:50,136 --> 00:17:52,465 of course. So, now, the equilibrium 242 00:17:52,465 --> 00:17:55,000 position is not here. 243 00:17:55,000 --> 00:18:01,000 244 00:18:01,000 --> 00:18:05,034 And I will call, now, this distance from the 245 00:18:05,034 --> 00:18:07,661 equilibrium position, [eta?]. 246 00:18:07,661 --> 00:18:12,540 You always use a U coordinate system, the equilibrium 247 00:18:12,540 --> 00:18:17,512 position, as your zero. And so, the only thing that is 248 00:18:17,512 --> 00:18:23,048 going to change now is that the sine of theta is no longer X 249 00:18:23,048 --> 00:18:26,800 divided by L. But, the sine of theta is X 250 00:18:26,800 --> 00:18:31,867 minus eta divided by L because, notice, if this is eta, 251 00:18:31,867 --> 00:18:36,933 which is the position of my hand, then the sine of this 252 00:18:36,933 --> 00:18:42,000 angle is now X minus eta divided by L. 253 00:18:42,000 --> 00:18:46,918 In this, now, you had to carry through your 254 00:18:46,918 --> 00:18:52,072 differential equations. And, what you'll see, 255 00:18:52,072 --> 00:18:57,576 then, if you do that, that this now is not zero. 256 00:18:57,576 --> 00:19:04,486 But this now becomes eta zero times omega zero squared times 257 00:19:04,486 --> 00:19:09,287 the cosine of omega T. My omega, so, 258 00:19:09,287 --> 00:19:12,952 instead of sine theta, X divided by L, 259 00:19:12,952 --> 00:19:18,500 all you have to do is this. You know that eta is eta zero 260 00:19:18,500 --> 00:19:22,264 cosine omega T. You carry that through, 261 00:19:22,264 --> 00:19:27,316 and you will see that your differential equation now 262 00:19:27,316 --> 00:19:31,103 changes. Notice that X double dot is an 263 00:19:31,103 --> 00:19:34,047 acceleration. Notice that gamma times X dot 264 00:19:34,047 --> 00:19:37,131 is an acceleration. Gamma is one over second, 265 00:19:37,131 --> 00:19:40,987 and X dot is meters per second. This is an acceleration. 266 00:19:40,987 --> 00:19:43,160 And, X dot is meters per second. 267 00:19:43,160 --> 00:19:46,385 This is an acceleration. Notice that this is an 268 00:19:46,385 --> 00:19:49,820 acceleration that has the dimension omega squared, 269 00:19:49,820 --> 00:19:52,624 which is one second squared times meters. 270 00:19:52,624 --> 00:19:55,849 That is an acceleration. Notice that this is an 271 00:19:55,849 --> 00:19:58,092 acceleration. Eta zero is meters, 272 00:19:58,092 --> 00:20:03,000 and omega squared is one divided by second squared. 273 00:20:03,000 --> 00:20:06,814 So, these are apples. These are apples. 274 00:20:06,814 --> 00:20:10,628 These are apples. And these are apples. 275 00:20:10,628 --> 00:20:14,643 So, the equation looks very kosher to me. 276 00:20:14,643 --> 00:20:19,059 Now, the solutions: that's a different story. 277 00:20:19,059 --> 00:20:24,178 That is something that I wouldn't want you to derive 278 00:20:24,178 --> 00:20:30,000 either, but the solutions now become as follows. 279 00:20:30,000 --> 00:20:35,861 I now get X as a function of T is an amplitude, 280 00:20:35,861 --> 00:20:40,957 which is a very strong function of omega. 281 00:20:40,957 --> 00:20:47,073 But, I will simply write it down as an amplitude, 282 00:20:47,073 --> 00:20:53,571 A, times the cosine of my omega, T, minus some phase 283 00:20:53,571 --> 00:20:58,922 angle, delta. And, this is what we call the 284 00:20:58,922 --> 00:21:05,293 steady state solution. And this A becomes, 285 00:21:05,293 --> 00:21:09,517 then, a rather complicated function. 286 00:21:09,517 --> 00:21:15,189 Upstairs, I get the eta zero omega zero squared. 287 00:21:15,189 --> 00:21:19,051 I get this part. And, downstairs, 288 00:21:19,051 --> 00:21:25,568 I get the square root of omega zero squared minus omega 289 00:21:25,568 --> 00:21:30,637 squared, squared, plus omega gamma squared. 290 00:21:30,637 --> 00:21:38,000 And, that is the amplitude as a function of omega. 291 00:21:38,000 --> 00:21:47,226 And the tangent of delta becomes omega gamma divided by 292 00:21:47,226 --> 00:21:53,718 omega zero squared minus omega squared. 293 00:21:53,718 --> 00:22:02,432 If we plot that function, A, as a function of omega, 294 00:22:02,432 --> 00:22:10,291 so here's omega, and here I'm going to plot the 295 00:22:10,291 --> 00:22:14,932 A. Then, I can recognize that if 296 00:22:14,932 --> 00:22:19,887 omega goes to zero, that means if I move this very 297 00:22:19,887 --> 00:22:25,651 slowly, that for sure the amplitude of this object must be 298 00:22:25,651 --> 00:22:30,000 the same as my hand, must be eta zero. 299 00:22:30,000 --> 00:22:34,325 If it takes me one week to go from here to here, 300 00:22:34,325 --> 00:22:39,754 then of course the pendulum is always hanging below my hand. 301 00:22:39,754 --> 00:22:45,000 So, when omega goes to zero, I can always check the result 302 00:22:45,000 --> 00:22:49,049 here that A must go to eta zero. And, indeed, 303 00:22:49,049 --> 00:22:54,202 if you put in omega equals zero, you will see that is the 304 00:22:54,202 --> 00:22:57,055 case. If omega goes to infinity, 305 00:22:57,055 --> 00:23:02,194 then A goes to zero. And then, there is a very 306 00:23:02,194 --> 00:23:05,880 special case. That is when omega happens to 307 00:23:05,880 --> 00:23:09,655 be omega zero, which is the frequency of the 308 00:23:09,655 --> 00:23:12,463 system in the absence of damping. 309 00:23:12,463 --> 00:23:15,974 This is my omega zero. It's not this one. 310 00:23:15,974 --> 00:23:20,714 But it is the omega zero. Then, you'll get an amplitude 311 00:23:20,714 --> 00:23:25,630 which is Q times eta zero. And, again, I just happened to 312 00:23:25,630 --> 00:23:30,553 remember that. If you substitute this back in 313 00:23:30,553 --> 00:23:33,702 here, this omega equals of omega zero. 314 00:23:33,702 --> 00:23:37,872 You will see that. And, that's always very nice to 315 00:23:37,872 --> 00:23:40,255 remember. At that omega zero, 316 00:23:40,255 --> 00:23:44,000 you always get Q times more than your driver. 317 00:23:44,000 --> 00:23:47,659 If Q is 100, you get 100 times the amplitude 318 00:23:47,659 --> 00:23:49,957 of the driver. And so, here, 319 00:23:49,957 --> 00:23:52,765 you get eta zero if omega is zero. 320 00:23:52,765 --> 00:23:57,021 And then, when you reach that omega zero frequency, 321 00:23:57,021 --> 00:24:02,697 you come out high. And then it goes back to zero. 322 00:24:02,697 --> 00:24:06,204 And this point here is Q times eta zero. 323 00:24:06,204 --> 00:24:11,059 And, we discussed before, I'll never make too much of a 324 00:24:11,059 --> 00:24:15,376 deal out of that, that the actual maximum of this 325 00:24:15,376 --> 00:24:19,692 curve is not at omega zero but is always a lower. 326 00:24:19,692 --> 00:24:24,727 That is really more of an algebraic interest than it is a 327 00:24:24,727 --> 00:24:28,414 physical interest because if gamma is low, 328 00:24:28,414 --> 00:24:33,000 if Q is high, the two almost coincide. 329 00:24:33,000 --> 00:24:37,492 So, the steady state solution has no adjustable constants. 330 00:24:37,492 --> 00:24:41,039 In other words, the system has lost its memory 331 00:24:41,039 --> 00:24:44,428 of what happened when we started driving it. 332 00:24:44,428 --> 00:24:47,738 So, at T equals zero, if we know what X is, 333 00:24:47,738 --> 00:24:52,467 and if we know what X dot is, and if we know what the driving 334 00:24:52,467 --> 00:24:56,724 term is at T equals zero, then the general solution for 335 00:24:56,724 --> 00:25:01,532 all times larger than zero is the sum of the trangent solution 336 00:25:01,532 --> 00:25:06,500 and the steady state solution. And so, here you see the 337 00:25:06,500 --> 00:25:09,687 [trangent?] solution. You can write it in this form, 338 00:25:09,687 --> 00:25:11,625 or you can read it in this form. 339 00:25:11,625 --> 00:25:14,250 And here you see the steady state solution. 340 00:25:14,250 --> 00:25:16,500 This A has nothing to do with that A. 341 00:25:16,500 --> 00:25:20,375 This A follows from the initial conditions just like the alpha, 342 00:25:20,375 --> 00:25:23,500 this A does not follow from the initial conditions. 343 00:25:23,500 --> 00:25:26,937 This A follows from eta zero, what my amplitude is of my 344 00:25:26,937 --> 00:25:30,000 driver, and it follows from omega. 345 00:25:30,000 --> 00:25:33,756 So, the two A's don't confuse the two. 346 00:25:33,756 --> 00:25:39,544 And so, the general solution is, then, the sum of the two. 347 00:25:39,544 --> 00:25:45,433 The trangent one will die out. If you weigh it a few times, 348 00:25:45,433 --> 00:25:49,089 two over gamma, the trangent is gone. 349 00:25:49,089 --> 00:25:54,267 And so, you would end up, then, only with the steady 350 00:25:54,267 --> 00:25:58,532 state solution. And, you had some chance in 351 00:25:58,532 --> 00:26:05,002 your homework to work with that. If we are not driving the 352 00:26:05,002 --> 00:26:09,169 system, and if we have one object on a spring or on a 353 00:26:09,169 --> 00:26:13,816 pendulum, or a floating object in liquid, then there is one 354 00:26:13,816 --> 00:26:17,821 frequency, one normal mode frequency, one resonance 355 00:26:17,821 --> 00:26:21,667 frequency, we call it also the natural frequency. 356 00:26:21,667 --> 00:26:26,234 The moment that you couple oscillators, if you couple two, 357 00:26:26,234 --> 00:26:30,000 you get two normal mode frequencies. 358 00:26:30,000 --> 00:26:32,692 You get to resonance frequencies. 359 00:26:32,692 --> 00:26:37,319 And if you have three objects, you get three normal mode 360 00:26:37,319 --> 00:26:40,600 frequencies. And so, now I would like to 361 00:26:40,600 --> 00:26:44,387 discuss with you the case whereby I couple two 362 00:26:44,387 --> 00:26:47,920 oscillators. If I gave you on an exam three 363 00:26:47,920 --> 00:26:51,622 coupled oscillators, that would be very nasty 364 00:26:51,622 --> 00:26:54,735 because it's extremely time consuming. 365 00:26:54,735 --> 00:26:59,110 If I gave you four coupled oscillators, that would be 366 00:26:59,110 --> 00:27:05,000 criminal because you cannot finish that in 85 minutes. 367 00:27:05,000 --> 00:27:07,870 So, two is certainly within reason. 368 00:27:07,870 --> 00:27:10,656 Three is marginally within reason. 369 00:27:10,656 --> 00:27:14,709 Four is out of the question. When we have coupled 370 00:27:14,709 --> 00:27:18,087 oscillators, we always leave damping out. 371 00:27:18,087 --> 00:27:21,126 And yet, we will learn a lot from it. 372 00:27:21,126 --> 00:27:24,588 Even without damping, we will learn a lot. 373 00:27:24,588 --> 00:27:29,823 And so, what I've chosen to do with you is a spring system with 374 00:27:29,823 --> 00:27:34,298 two objects, two masses, and the springs have no mass, 375 00:27:34,298 --> 00:27:39,817 negligible mass. The spring constants are K, 376 00:27:39,817 --> 00:27:43,469 and the masses of the objects are M. 377 00:27:43,469 --> 00:27:49,417 So, this is the equilibrium position of these two objects, 378 00:27:49,417 --> 00:27:53,173 and the ends are fixed of the string. 379 00:27:53,173 --> 00:27:59,330 I will introduce a shorthand notation, zero squared equals K 380 00:27:59,330 --> 00:28:02,452 over M. Now, what I do, 381 00:28:02,452 --> 00:28:07,974 I offset both these objects from equilibrium just as I did 382 00:28:07,974 --> 00:28:11,558 that here. I always offset them in the 383 00:28:11,558 --> 00:28:15,433 same direction, and I call that positive. 384 00:28:15,433 --> 00:28:17,176 Is that a must? No. 385 00:28:17,176 --> 00:28:19,017 Is that useful? Yes. 386 00:28:19,017 --> 00:28:22,891 So, I offset them. And so, this position, 387 00:28:22,891 --> 00:28:28,703 now this is object number one, and this is object number two. 388 00:28:28,703 --> 00:28:35,000 This is now a distance, X1, away from equilibrium. 389 00:28:35,000 --> 00:28:38,493 And this one is now at a position, X2, 390 00:28:38,493 --> 00:28:43,590 away from its equilibrium. You always want to know what 391 00:28:43,590 --> 00:28:48,405 the displacement is relative to its own equilibrium. 392 00:28:48,405 --> 00:28:54,069 And, its own equilibrium for one is here, and equilibrium for 393 00:28:54,069 --> 00:28:57,562 two is there. I now make the following 394 00:28:57,562 --> 00:29:02,000 assumption that X2 is larger than X1. 395 00:29:02,000 --> 00:29:03,719 Is that a must? No. 396 00:29:03,719 --> 00:29:07,828 Is it useful? Yes, but if you want to assume 397 00:29:07,828 --> 00:29:11,554 that X2 is smaller than X1, be my guest. 398 00:29:11,554 --> 00:29:15,280 I will assume that X2 is larger than X1. 399 00:29:15,280 --> 00:29:20,439 Now, follow me closely now. So, we have an object here, 400 00:29:20,439 --> 00:29:25,598 and we have an object here, and this is now one spring. 401 00:29:25,598 --> 00:29:32,000 This is the other spring, and this is the third spring. 402 00:29:32,000 --> 00:29:35,806 It is immediately obvious that this spring is longer than it 403 00:29:35,806 --> 00:29:38,193 wants to be. So, there is a force that 404 00:29:38,193 --> 00:29:41,677 drives it back to equilibrium. If X2 is larger than X1, 405 00:29:41,677 --> 00:29:44,645 this spring is also longer than it wants to be. 406 00:29:44,645 --> 00:29:48,000 So, it wants to contract, so there is a force in that 407 00:29:48,000 --> 00:29:50,516 direction. If this spring is longer than 408 00:29:50,516 --> 00:29:52,838 it wants to be, it wants to contract. 409 00:29:52,838 --> 00:29:55,677 So, there is a force in X2 in this direction. 410 00:29:55,677 --> 00:30:00,000 This spring is clearly shorter than it wants to be. 411 00:30:00,000 --> 00:30:05,632 So, it's pushing. So there is a force in this 412 00:30:05,632 --> 00:30:10,751 direction. And so, now I'm going to write 413 00:30:10,751 --> 00:30:18,175 down the differential equation first for object number one. 414 00:30:18,175 --> 00:30:25,087 So I get M X1 double dot. So, this one is minus K times 415 00:30:25,087 --> 00:30:29,183 X1. And, this one is plus K times 416 00:30:29,183 --> 00:30:34,278 X2 minus X1. Is it a disaster if it turns 417 00:30:34,278 --> 00:30:36,963 out that X2 is not larger than X1? 418 00:30:36,963 --> 00:30:38,997 Not at all. This equation, 419 00:30:38,997 --> 00:30:41,763 now, is correct for all situations. 420 00:30:41,763 --> 00:30:46,482 The fact that I have assumed that X2 is larger than X1 gave 421 00:30:46,482 --> 00:30:49,981 me a plus sign here, and so, my differential 422 00:30:49,981 --> 00:30:54,212 equation is safe no matter what X1 is relative to X2. 423 00:30:54,212 --> 00:30:57,385 So, you can always make that assumption. 424 00:30:57,385 --> 00:31:03,000 And you don't have to worry later anymore about signs. 425 00:31:03,000 --> 00:31:07,190 I can divide, now, M out, and then I get that 426 00:31:07,190 --> 00:31:11,952 X1 double dot plus two omega zero squared times X1. 427 00:31:11,952 --> 00:31:16,714 But you notice I have one here and I have one here. 428 00:31:16,714 --> 00:31:21,952 They have both a minus sign. And then, I have here plus, 429 00:31:21,952 --> 00:31:26,619 so that becomes minus omega zero squared times X2. 430 00:31:26,619 --> 00:31:29,761 That is zero. So, that is my first 431 00:31:29,761 --> 00:31:35,965 differential equation. So, now I go to the second one. 432 00:31:35,965 --> 00:31:41,034 I get MX two double dot. Now I have two forces both in 433 00:31:41,034 --> 00:31:45,626 the negative direction. First I have the one that 434 00:31:45,626 --> 00:31:49,165 drives this one away from equilibrium. 435 00:31:49,165 --> 00:31:52,417 So I get minus K times X2 minus X1. 436 00:31:52,417 --> 00:31:57,965 And then, I have the force due to the fact that this one is 437 00:31:57,965 --> 00:32:02,173 shorter than it wants to be by an amount, K2. 438 00:32:02,173 --> 00:32:08,524 So, I get minus K times X2. It's shorter by an amount, 439 00:32:08,524 --> 00:32:13,477 X2, than it wants to be. And so, I can divide M out, 440 00:32:13,477 --> 00:32:19,207 and so I get MX2 double dot. And then, I get plus omega zero 441 00:32:19,207 --> 00:32:23,480 squared times X2. And then, I get minus omega 442 00:32:23,480 --> 00:32:26,685 zero squared times X1 equals zero. 443 00:32:26,685 --> 00:32:30,083 Now, compare this one with this one. 444 00:32:30,083 --> 00:32:36,007 Notice the incredible symmetry. I could have found this one by 445 00:32:36,007 --> 00:32:43,000 changing a one to two here, and by changing the two to one. 446 00:32:43,000 --> 00:32:46,034 Of course, the system is so symmetric that, 447 00:32:46,034 --> 00:32:49,069 yes, thank you very much for the two right? 448 00:32:49,069 --> 00:32:52,465 Yeah, thank you. The system is so symmetric that 449 00:32:52,465 --> 00:32:56,945 clearly nature cannot make any distinction between one and two. 450 00:32:56,945 --> 00:33:01,063 So, it is in this particular case because of the beautiful 451 00:33:01,063 --> 00:33:05,398 symmetry, it is obvious that these two differential equations 452 00:33:05,398 --> 00:33:10,230 look very, very similar. Now, I'm going to make an 453 00:33:10,230 --> 00:33:14,334 important step. I'm going to substitute in here 454 00:33:14,334 --> 00:33:18,617 X1 is C times cosine omega T, and X2, this is C1, 455 00:33:18,617 --> 00:33:23,702 is C2 times cosine omega T. I want to know what the normal 456 00:33:23,702 --> 00:33:26,914 mode frequencies are for this system. 457 00:33:26,914 --> 00:33:32,000 I want to solve for omega. I want to find omega. 458 00:33:32,000 --> 00:33:35,870 I'm not driving the system. Since I have no damping, 459 00:33:35,870 --> 00:33:40,120 in which case at normal mode solutions, either of the two 460 00:33:40,120 --> 00:33:44,674 objects are in phase with each other or they are out of phase 461 00:33:44,674 --> 00:33:48,165 with each other. But if there are out of phase, 462 00:33:48,165 --> 00:33:51,883 we can always take care of that with a minus sign. 463 00:33:51,883 --> 00:33:55,223 So, I'm now going to substitute that in here. 464 00:33:55,223 --> 00:33:59,245 And I may want to continue working on the centerboard, 465 00:33:59,245 --> 00:34:03,647 in which case I might as well erase this so that it stays a 466 00:34:03,647 --> 00:34:06,000 little compact. 467 00:34:06,000 --> 00:34:13,000 468 00:34:13,000 --> 00:34:17,330 So, I'm going to substitute this trial function into my 469 00:34:17,330 --> 00:34:21,339 differential equations. And, every term will have a 470 00:34:21,339 --> 00:34:24,948 cosine omega T, so I dump all the cosine omega 471 00:34:24,948 --> 00:34:27,273 T's. So, I go to the equation, 472 00:34:27,273 --> 00:34:32,695 which has a big one there. And so, that one becomes C1. 473 00:34:32,695 --> 00:34:36,930 X1 double dot gives me a minus omega squared, 474 00:34:36,930 --> 00:34:40,395 right? If I take cosine omega T and I 475 00:34:40,395 --> 00:34:45,497 do the second derivative, I get minus omega squared in 476 00:34:45,497 --> 00:34:49,347 front of it. So, I get C1 times two omega 477 00:34:49,347 --> 00:34:53,197 zero squared. That is that two omega zero 478 00:34:53,197 --> 00:34:56,566 squared. And then, I get minus omega 479 00:34:56,566 --> 00:34:59,262 squared. And then I get plus, 480 00:34:59,262 --> 00:35:06,000 not plus, then I get minus omega zero squared times C2. 481 00:35:06,000 --> 00:35:10,392 And that equals zero. And then I go to my second 482 00:35:10,392 --> 00:35:14,037 differential equation which is this one. 483 00:35:14,037 --> 00:35:18,523 But I'm going to rearrange the C1's and the C2's. 484 00:35:18,523 --> 00:35:21,700 So, I'm going to put the C1's here. 485 00:35:21,700 --> 00:35:26,747 So, I'm going to get minus omega zero squared times C1. 486 00:35:26,747 --> 00:35:32,261 And here, I'm going to get plus two omega zero squared minus 487 00:35:32,261 --> 00:35:37,886 omega zero squared times C2. And that equals zero. 488 00:35:37,886 --> 00:35:43,095 So, I can leave this plus out. So, what do I have here now? 489 00:35:43,095 --> 00:35:47,137 I have here two equations with three unknowns. 490 00:35:47,137 --> 00:35:51,449 I want to know what omega is in the normal modes, 491 00:35:51,449 --> 00:35:55,670 but I also would like to know what C1 and C2 is. 492 00:35:55,670 --> 00:36:00,431 Well, that is tough luck. You can't have it both ways. 493 00:36:00,431 --> 00:36:06,000 We do not know at all what C1 and C2 is separately. 494 00:36:06,000 --> 00:36:09,498 But you can always find in the normal modes, 495 00:36:09,498 --> 00:36:13,810 without knowing anything else what the ratios of those 496 00:36:13,810 --> 00:36:17,959 amplitudes, C1 and C2. And, you can always solve for 497 00:36:17,959 --> 00:36:21,050 omega, and you will see how that works. 498 00:36:21,050 --> 00:36:24,386 So, with two equations and three unknowns, 499 00:36:24,386 --> 00:36:28,128 you can only solve for omega and for the ratio, 500 00:36:28,128 --> 00:36:32,828 C1 over C2. So now, you may do that anyway 501 00:36:32,828 --> 00:36:36,505 you want to. This is not so difficult to 502 00:36:36,505 --> 00:36:39,051 solve this. I will, however, 503 00:36:39,051 --> 00:36:42,917 use Kramer's rule, and I will introduce D, 504 00:36:42,917 --> 00:36:48,197 which is the determinant of the following: two omega zero 505 00:36:48,197 --> 00:36:53,288 squared minus omega squared. And then I have here minus 506 00:36:53,288 --> 00:36:57,625 omega zero squared. And then, I have here minus 507 00:36:57,625 --> 00:37:03,000 omega zero squared minus omega zero squared. 508 00:37:03,000 --> 00:37:06,739 And this, now, must be zero if I'm not looking 509 00:37:06,739 --> 00:37:09,150 for solutions, which I am not, 510 00:37:09,150 --> 00:37:13,638 C1 is zero and C2 is zero. It's clear that if C1 and C2 511 00:37:13,638 --> 00:37:17,045 are zero, the two equations are satisfied. 512 00:37:17,045 --> 00:37:20,536 Zero is zero. But that's not an interesting 513 00:37:20,536 --> 00:37:23,445 case. And, the only way that you can 514 00:37:23,445 --> 00:37:27,766 get a solution which is interesting is value's for C1 515 00:37:27,766 --> 00:37:34,000 and C2 which are not zero is by making this determinant zero. 516 00:37:34,000 --> 00:37:39,157 And so, that means that D, which now becomes two omega 517 00:37:39,157 --> 00:37:43,147 zero squared minus omega squared, squared, 518 00:37:43,147 --> 00:37:48,791 minus omega zero to the fourth, you have to make that zero. 519 00:37:48,791 --> 00:37:53,755 And when you do that, you find two values for omega. 520 00:37:53,755 --> 00:37:59,593 You find that omega squared is two omega zero squared plus or 521 00:37:59,593 --> 00:38:05,486 minus omega zero squared. Those are the two solutions. 522 00:38:05,486 --> 00:38:10,726 And, when we evaluate those two solutions, you find a result 523 00:38:10,726 --> 00:38:15,523 which is so embarrassingly simple that you could almost 524 00:38:15,523 --> 00:38:18,986 have said that without any work, almost. 525 00:38:18,986 --> 00:38:23,782 Notice that omega minus, which is the lowest one of the 526 00:38:23,782 --> 00:38:30,000 two, is the same as omega zero because that's the minus sign. 527 00:38:30,000 --> 00:38:33,380 What does that mean if omega minus is omega zero? 528 00:38:33,380 --> 00:38:36,619 Well, it means, of course, that the two objects 529 00:38:36,619 --> 00:38:40,563 are just oscillating like this. The inner spring is never 530 00:38:40,563 --> 00:38:43,802 stretched, is never longer than it wants to be, 531 00:38:43,802 --> 00:38:47,746 is never shorter than it wants to be, so each one is only 532 00:38:47,746 --> 00:38:50,563 driven, so to speak, by the outer spring. 533 00:38:50,563 --> 00:38:54,507 So, that's immediately obvious that that's a normal mode. 534 00:38:54,507 --> 00:38:58,591 And, you can make a prediction that C1 over C2 must be plus 535 00:38:58,591 --> 00:39:02,057 one. They must go in unison like 536 00:39:02,057 --> 00:39:04,495 this. And you can confirm that by 537 00:39:04,495 --> 00:39:08,914 substituting this omega one. If you substitute that in this 538 00:39:08,914 --> 00:39:12,495 equation, you'll find the ratio see one over C2. 539 00:39:12,495 --> 00:39:15,542 If you prefer this equation, be my guest. 540 00:39:15,542 --> 00:39:19,428 You can do that too. And you'll find no matter which 541 00:39:19,428 --> 00:39:22,704 one of the two you take that it is plus one. 542 00:39:22,704 --> 00:39:26,742 Now, omega plus is less obvious, is the square root of 543 00:39:26,742 --> 00:39:32,000 three times omega zero. That's when you take the plus. 544 00:39:32,000 --> 00:39:36,142 Even though it's not obvious that it is the square root of 545 00:39:36,142 --> 00:39:38,759 three, it is clear what is happening. 546 00:39:38,759 --> 00:39:42,611 Here are the two objects, and now they are doing this. 547 00:39:42,611 --> 00:39:46,245 They are exactly 180° out of phase with each other. 548 00:39:46,245 --> 00:39:49,952 So, now you can make comfortably the prediction that 549 00:39:49,952 --> 00:39:53,804 C1 over C2 is minus one. And indeed, if you substitute 550 00:39:53,804 --> 00:39:58,383 the square at a free omega zero in either this or that equation, 551 00:39:58,383 --> 00:40:04,015 this indeed comes out. So, the general solution now 552 00:40:04,015 --> 00:40:09,830 for a given initial condition, so if now I give you the 553 00:40:09,830 --> 00:40:15,000 initial condition, then the general solution will 554 00:40:15,000 --> 00:40:21,030 be for X1 would be X0 minus, the minus makes reference to 555 00:40:21,030 --> 00:40:27,599 that lowest frequency times the cosine omega minus T plus some 556 00:40:27,599 --> 00:40:32,338 phase angle minus, all these minus signs make 557 00:40:32,338 --> 00:40:38,393 reference to this mode. And then, I have plus X0 plus 558 00:40:38,393 --> 00:40:43,262 times the cosine of omega plus T plus some phase angle plus. 559 00:40:43,262 --> 00:40:47,223 Of course, if you prefer instead of cosine-signs, 560 00:40:47,223 --> 00:40:49,038 that's fine, of course. 561 00:40:49,038 --> 00:40:52,669 And, you have four adjustable constants: one, 562 00:40:52,669 --> 00:40:56,383 two, three, four. And, if you know the initial 563 00:40:56,383 --> 00:41:00,839 conditions, if you know what X1 is, and what X1 dot is, 564 00:41:00,839 --> 00:41:04,388 and what X2 is, and what X2 dot is at time T 565 00:41:04,388 --> 00:41:10,000 equals zero, you can solve for all four in principle. 566 00:41:10,000 --> 00:41:14,558 There is no longer any freedom from number two because all four 567 00:41:14,558 --> 00:41:17,720 adjustable constants have now been consumed. 568 00:41:17,720 --> 00:41:20,441 And so, therefore, you can write down, 569 00:41:20,441 --> 00:41:24,558 now, the solution for X2 by simply making this one a two. 570 00:41:24,558 --> 00:41:27,279 And, this whole term here is the same. 571 00:41:27,279 --> 00:41:31,323 The only difference is that this plus sign now becomes a 572 00:41:31,323 --> 00:41:35,000 minus sign. Nothing else is different. 573 00:41:35,000 --> 00:41:39,206 The frequencies are the same. Otherwise they wouldn't be 574 00:41:39,206 --> 00:41:43,184 normal mode frequencies. And, it is the ratio here of 575 00:41:43,184 --> 00:41:47,926 the C1 over C2 that is plus one. And, it is the ratio here that 576 00:41:47,926 --> 00:41:51,215 is the minus one. That's the reason why this 577 00:41:51,215 --> 00:41:54,810 becomes a minus, and why the plus here remains a 578 00:41:54,810 --> 00:41:56,186 plus. So that, now, 579 00:41:56,186 --> 00:42:00,087 is the general solution if you also know the initial 580 00:42:00,087 --> 00:42:06,720 conditions. So now, we are going to drive 581 00:42:06,720 --> 00:42:13,395 the system. So now, we go back to where we 582 00:42:13,395 --> 00:42:19,418 were, and I'm going to drive this end. 583 00:42:19,418 --> 00:42:26,418 And on going to drive this and with my hand. 584 00:42:26,418 --> 00:42:33,093 This is eta, and eta equals eta zero times 585 00:42:33,093 --> 00:42:40,387 cosine omega T. There is only one term in all 586 00:42:40,387 --> 00:42:46,112 of this on the blackboard that is going to change. 587 00:42:46,112 --> 00:42:52,303 And that is this term. This spring here on the left is 588 00:42:52,303 --> 00:42:59,663 no longer shorter by the amount X1, but is shorter by the amount 589 00:42:59,663 --> 00:43:04,722 X1 minus eta. And so, it's only that term. 590 00:43:04,722 --> 00:43:10,166 It's only this minus KX1 that now have to be changed into 591 00:43:10,166 --> 00:43:14,833 minus K times X1 minus eta. Nothing else changes. 592 00:43:14,833 --> 00:43:17,944 But, that has major consequences, 593 00:43:17,944 --> 00:43:20,277 of course. For one thing, 594 00:43:20,277 --> 00:43:25,819 if you're going to substitute, now, these trial functions, 595 00:43:25,819 --> 00:43:30,000 omega is no longer a negotiable. 596 00:43:30,000 --> 00:43:33,458 Omega is my omega now. I set this omega. 597 00:43:33,458 --> 00:43:37,004 You're not going to solve for that omega. 598 00:43:37,004 --> 00:43:41,793 That would be an insult to me. I dictate what omega is. 599 00:43:41,793 --> 00:43:46,758 So, that means I now get two equations with two unknowns, 600 00:43:46,758 --> 00:43:50,039 C1 and C2. Omega is no longer unknown. 601 00:43:50,039 --> 00:43:55,359 So, now I get solutions for C1 and they get solutions for C2. 602 00:43:55,359 --> 00:44:00,413 And of all the equations on the blackboards that you have, 603 00:44:00,413 --> 00:44:06,000 the one that is going to change is the first one. 604 00:44:06,000 --> 00:44:12,422 And when you carry through, that eta, which is eta zero 605 00:44:12,422 --> 00:44:17,894 cosine omega T, this zero here changes into eta 606 00:44:17,894 --> 00:44:24,555 zero times omega zero squared. The cosine omega T is gone 607 00:44:24,555 --> 00:44:31,334 because I've divided cosine omega T out in all those other 608 00:44:31,334 --> 00:44:35,993 terms. So, now I use creamers rule, 609 00:44:35,993 --> 00:44:41,660 and now I can actually come up with a solution for C1, 610 00:44:41,660 --> 00:44:47,754 no longer ratio C1 over C2. No, I can actually come up now 611 00:44:47,754 --> 00:44:52,779 with a solution. So, C1 now using Kramer's rule, 612 00:44:52,779 --> 00:44:58,660 this now as my first column. Eta zero omega zero squared 613 00:44:58,660 --> 00:45:04,968 zero, and this now becomes my second column minus omega zero 614 00:45:04,968 --> 00:45:09,966 squared. And I get here two omega zero 615 00:45:09,966 --> 00:45:14,059 squared minus omega squared divided by D. 616 00:45:14,059 --> 00:45:19,072 If I pick a random value for omega, D is not zero. 617 00:45:19,072 --> 00:45:24,085 D is only zero at those two resonance frequencies. 618 00:45:24,085 --> 00:45:29,303 And then, C2 becomes, the first column is like this. 619 00:45:29,303 --> 00:45:35,544 That is two omega zero squared minus omega squared minus omega 620 00:45:35,544 --> 00:45:41,683 zero squared. And now, the second column 621 00:45:41,683 --> 00:45:49,758 becomes eta zero omega zero squared, and a zero divided by 622 00:45:49,758 --> 00:45:53,583 D. If I work C1 out a little 623 00:45:53,583 --> 00:46:01,516 further, then I get C1 equals eta zero omega zero squared 624 00:46:01,516 --> 00:46:10,016 times two omega zero squared minus omega zero squared divided 625 00:46:10,016 --> 00:46:14,923 by D. And here, I get that C2, 626 00:46:14,923 --> 00:46:19,916 this is zero. So, I get plus omega zero to 627 00:46:19,916 --> 00:46:24,544 the fourth times eta zero divided by D. 628 00:46:24,544 --> 00:46:31,000 And D, now, is this determinant but is not zero. 629 00:46:31,000 --> 00:46:35,076 It's only zero for those two special frequencies. 630 00:46:35,076 --> 00:46:40,087 If you want to see what the amplitudes now are as a function 631 00:46:40,087 --> 00:46:44,164 of omega, it's a very, very interesting behavior, 632 00:46:44,164 --> 00:46:47,986 then what helps is you go omega first to zero. 633 00:46:47,986 --> 00:46:52,912 And then, you see what happens. And, it is not so intuitive 634 00:46:52,912 --> 00:46:56,224 what happens. If you put in omega equals 635 00:46:56,224 --> 00:47:02,000 zero, you will find that C1 is plus two thirds eta zero. 636 00:47:02,000 --> 00:47:06,142 You can confirm that by substituting that into C1. 637 00:47:06,142 --> 00:47:09,270 You will see that that's what happens. 638 00:47:09,270 --> 00:47:13,582 And, you'll find that C2 is plus one third eta zero. 639 00:47:13,582 --> 00:47:16,287 So, that is at omega equals zero. 640 00:47:16,287 --> 00:47:20,091 And, when you go to infinity with frequencies, 641 00:47:20,091 --> 00:47:24,065 then it should not surprise you that C1 is zero, 642 00:47:24,065 --> 00:47:28,123 and that C2 is zero. And then, there is one case, 643 00:47:28,123 --> 00:47:32,012 which is pathetic. And, that is the case that I 644 00:47:32,012 --> 00:47:37,000 make omega squared, two omega zero squared. 645 00:47:37,000 --> 00:47:40,017 At that frequency, C1 becomes zero, 646 00:47:40,017 --> 00:47:44,098 but C2 is not zero. And, I spent almost a whole 647 00:47:44,098 --> 00:47:48,624 lecture with you on that demonstrating that in three 648 00:47:48,624 --> 00:47:52,972 different ways that, indeed, there is this bizarre 649 00:47:52,972 --> 00:47:56,344 solution. So, when omega squared equals 650 00:47:56,344 --> 00:48:00,249 two omega zero squared, then C1 becomes zero, 651 00:48:00,249 --> 00:48:05,835 but C2 is not zero. And so, now you can make a plot 652 00:48:05,835 --> 00:48:10,774 of C's as a function of omega. And during that lecture, 653 00:48:10,774 --> 00:48:13,792 I showed you three of those plots. 654 00:48:13,792 --> 00:48:18,914 And, I used the convention, then, that when the object is 655 00:48:18,914 --> 00:48:24,402 moving in phase with the driver, I put it positive and out of 656 00:48:24,402 --> 00:48:28,426 phase negative. And, I plot here C divided by 657 00:48:28,426 --> 00:48:33,000 eta zero. And, I will use a color code. 658 00:48:33,000 --> 00:48:37,348 I will do C1 in red, and I will do, 659 00:48:37,348 --> 00:48:42,593 C1 is red and C2 I will do in white chalk. 660 00:48:42,593 --> 00:48:47,965 So, here is zero omega. Here is omega zero, 661 00:48:47,965 --> 00:48:54,744 which is my omega minus, and then my omega plus was at 662 00:48:54,744 --> 00:49:01,523 three, the square root of three. So, that's about 1.7. 663 00:49:01,523 --> 00:49:08,757 So, here's my omega plus. And then things go nuts. 664 00:49:08,757 --> 00:49:14,961 That is when D goes to zero. So, C1 is two thirds here. 665 00:49:14,961 --> 00:49:18,982 And then, it goes to infinity there. 666 00:49:18,982 --> 00:49:23,693 And then, here, when it is 1.4 times omega 667 00:49:23,693 --> 00:49:30,357 zero, the square root of two, then it goes to a zero and it 668 00:49:30,357 --> 00:49:35,916 comes up here. And then, I get a curve here, 669 00:49:35,916 --> 00:49:41,346 without being too precise. And for C2, I get one third. 670 00:49:41,346 --> 00:49:44,765 And then, it goes up. And then, C2, 671 00:49:44,765 --> 00:49:49,793 something like this. And then, I get something like 672 00:49:49,793 --> 00:49:51,000 that. 673 00:49:51,000 --> 00:49:59,000 674 00:49:59,000 --> 00:50:02,825 Yeah, I can live with that. Now, we ignore damping, 675 00:50:02,825 --> 00:50:06,957 and because we ignore damping, we get these on physical 676 00:50:06,957 --> 00:50:11,395 infinities when you did the frequency omega minus and omega 677 00:50:11,395 --> 00:50:13,920 plus, these resonance frequencies. 678 00:50:13,920 --> 00:50:17,976 Now, if you include damping, then the solutions become 679 00:50:17,976 --> 00:50:21,802 extremely complicated. But, of course you avoid the 680 00:50:21,802 --> 00:50:25,168 infinity values. But, the resonant amplitudes 681 00:50:25,168 --> 00:50:30,213 can still be very high. And so, these plots are still 682 00:50:30,213 --> 00:50:34,639 very useful provided that you don't interpret infinities as 683 00:50:34,639 --> 00:50:37,998 being real, but something that is very large. 684 00:50:37,998 --> 00:50:41,203 If Q is high, then of course the amplitudes 685 00:50:41,203 --> 00:50:44,332 are enormously high. And, this can lead to 686 00:50:44,332 --> 00:50:47,309 distraction. We've seen the movie of the 687 00:50:47,309 --> 00:50:50,133 Tacoma Bridge, and then we have seen a 688 00:50:50,133 --> 00:50:53,567 dramatic experiment of the breaking wineglass. 689 00:50:53,567 --> 00:50:56,696 You remember that wineglass demonstration. 690 00:50:56,696 --> 00:51:00,970 Because of the catastrophic success, to use Bush's words, 691 00:51:00,970 --> 00:51:04,863 of that demonstration, students have asked me for an 692 00:51:04,863 --> 00:51:09,241 encore. They would like to see it 693 00:51:09,241 --> 00:51:12,289 again. And that's a very reasonable 694 00:51:12,289 --> 00:51:15,965 thing to do. It fits very nicely into this 695 00:51:15,965 --> 00:51:21,344 concept of coupled oscillators. A wineglass would have a huge 696 00:51:21,344 --> 00:51:25,468 number of oscillators coupled, and a wineglass, 697 00:51:25,468 --> 00:51:29,682 I have one here, can be made to oscillate easily 698 00:51:29,682 --> 00:51:35,506 in its lowest normal mode. I have to wash my hands to make 699 00:51:35,506 --> 00:51:39,307 you listen to it. So, because my hands are now a 700 00:51:39,307 --> 00:51:43,755 little greasy from the chalk, and this is the frequency. 701 00:51:43,755 --> 00:51:47,637 That is the lowest mode. It is a circle like this 702 00:51:47,637 --> 00:51:50,952 looking from above, it becomes an ellipse, 703 00:51:50,952 --> 00:51:55,804 and then it becomes an ellipse like this, and an ellipse like 704 00:51:55,804 --> 00:52:00,575 this, and an ellipse like that. And if now you dry that with 705 00:52:00,575 --> 00:52:05,347 sound at exactly that frequency and you put in enough power, 706 00:52:05,347 --> 00:52:09,875 another way of saying it is as if you make eta zero large 707 00:52:09,875 --> 00:52:14,000 enough, then the system can break. 708 00:52:14,000 --> 00:52:16,526 It's making a lot of noise. I'll warn you. 709 00:52:16,526 --> 00:52:19,546 Those were sitting here, I really think you should 710 00:52:19,546 --> 00:52:22,319 protect your ears. The sound can be deafening. 711 00:52:22,319 --> 00:52:25,092 So, be careful, and those who are a little bit 712 00:52:25,092 --> 00:52:27,680 further away, make sure that you close your 713 00:52:27,680 --> 00:52:29,714 ears. I have the luxury that I can 714 00:52:29,714 --> 00:52:32,980 turn my hearing aids off, but without my hearing aids, 715 00:52:32,980 --> 00:52:36,000 believe me, I can still hear a lot. 716 00:52:36,000 --> 00:52:39,513 So, I also will have to protect myself. 717 00:52:39,513 --> 00:52:43,582 So, one hearing aid is off. The other is off, 718 00:52:43,582 --> 00:52:46,818 and I'm going to put this on anyway. 719 00:52:46,818 --> 00:52:52,366 I'm going to stroke this glass with a frequency which is only 720 00:52:52,366 --> 00:52:57,174 slightly different from this frequency of the sounds. 721 00:52:57,174 --> 00:53:03,000 That is about 427 Hz, which you can very easily hear. 722 00:53:03,000 --> 00:53:09,804 And then you'll see the glass in slow motion because of the 723 00:53:09,804 --> 00:53:16,374 fact that the stroboscopic light has a slightly different 724 00:53:16,374 --> 00:53:20,949 frequency. That should be coming up now. 725 00:53:20,949 --> 00:53:24,000 And you say it will. 726 00:53:24,000 --> 00:53:35,000 727 00:53:35,000 --> 00:53:38,977 They do very much. I'm going to make it very dark 728 00:53:38,977 --> 00:53:44,033 now because this is important that you can see this very well. 729 00:53:44,033 --> 00:53:47,762 So, we'll make it completely dark in the room. 730 00:53:47,762 --> 00:53:52,569 I'm going to protect my ears. I hope you can still hear me. 731 00:53:52,569 --> 00:53:56,795 I can hardly hear myself. And I'm going to drive the 732 00:53:56,795 --> 00:54:01,436 system, now, at very low amplitudes, at a frequency close 733 00:54:01,436 --> 00:54:05,000 to its resonance. [SOUND PLAYS] 734 00:54:05,000 --> 00:54:09,529 You can already see that the glass is moving. 735 00:54:09,529 --> 00:54:13,132 I'm going to increase the amplitude. 736 00:54:13,132 --> 00:54:18,073 You see it moving? I can go a little bit over the 737 00:54:18,073 --> 00:54:21,470 resonance and under the resonance. 738 00:54:21,470 --> 00:54:25,382 You will see that it's not moving then. 739 00:54:25,382 --> 00:54:30,529 So, I do that purposely now. Now I'm off resonance. 740 00:54:30,529 --> 00:54:36,324 I'm over it. And now I'm off resonance. 741 00:54:36,324 --> 00:54:40,287 I'm under resonance. I am below. 742 00:54:40,287 --> 00:54:46,168 I'm now at 423 Hz when the resonance is at 427. 743 00:54:46,168 --> 00:54:50,643 So, I'm going to put it back at 427. 744 00:54:50,643 --> 00:54:55,630 And now I'm going to increase the sound. 745 00:54:55,630 --> 00:55:01,000 So I warn you. And there it goes. 746 00:55:01,000 --> 00:55:04,149 And that's it. It broke. 747 00:55:04,149 --> 00:55:11,406 And it broke very fast. [APPLAUSE] So this is an ideal 748 00:55:11,406 --> 00:55:18,116 moment for a break. Since you had such a good time 749 00:55:18,116 --> 00:55:25,510 with the breaking glass, let's settle for four minutes. 750 00:55:25,510 --> 00:55:30,029 We will reconvene in four minutes. 751 00:55:30,029 --> 00:55:37,375 [SOUND OFF/THEN ON] OK, I now want to discuss with 752 00:55:37,375 --> 00:55:42,102 you continuous media. If you have N coupled 753 00:55:42,102 --> 00:55:46,603 oscillators, then you get N normal modes. 754 00:55:46,603 --> 00:55:52,792 When you make N infinitely high, then you get continuous 755 00:55:52,792 --> 00:55:56,956 media. And, if we have one dimensional 756 00:55:56,956 --> 00:56:02,583 continuous media like a string, or a pipe with air, 757 00:56:02,583 --> 00:56:09,447 sound, then we have studied the transverse motion of a string, 758 00:56:09,447 --> 00:56:15,299 and then we derived that now you have to use the wave 759 00:56:15,299 --> 00:56:20,758 equation. We derive the wave equation. 760 00:56:20,758 --> 00:56:26,275 And then you get the D2Y DX squared is one over V squared 761 00:56:26,275 --> 00:56:31,103 times D2Y DT squared, Y being now the displacement 762 00:56:31,103 --> 00:56:36,325 away from equilibrium in this position if X is in this 763 00:56:36,325 --> 00:56:39,792 direction. And, that is fine. 764 00:56:39,792 --> 00:56:44,471 And so, excuse me? It's amazing how you can look 765 00:56:44,471 --> 00:56:48,653 at it and think that's fine, and it is not. 766 00:56:48,653 --> 00:56:54,528 All right, so a V in the case of a string is the square root 767 00:56:54,528 --> 00:56:58,610 of T over mu. That was just a special case 768 00:56:58,610 --> 00:57:03,753 for the string. We also examined longitudinal 769 00:57:03,753 --> 00:57:05,367 motion, again, 1D. 770 00:57:05,367 --> 00:57:10,873 Sound, so longitudinal wave, and if you deal with pressure, 771 00:57:10,873 --> 00:57:14,955 you think of sound as being a pressure wave. 772 00:57:14,955 --> 00:57:20,462 Then you get D2P DX squared equals one over V squared times 773 00:57:20,462 --> 00:57:24,639 D2P DT squared. P, then, being positive would 774 00:57:24,639 --> 00:57:31,000 be overpressure over and above the ambient 1 atmosphere. 775 00:57:31,000 --> 00:57:33,877 And if it's negative, then it is below. 776 00:57:33,877 --> 00:57:38,269 So it's not the total pressure, but it is the overpressure. 777 00:57:38,269 --> 00:57:40,844 And V, then, is the speed of sound, 778 00:57:40,844 --> 00:57:45,008 which in air room temperature is about 340 m per second. 779 00:57:45,008 --> 00:57:49,325 If you prefer not to work and pressure but in terms of the 780 00:57:49,325 --> 00:57:53,489 actual position of the air molecules in analogy with the 781 00:57:53,489 --> 00:57:57,276 position of the string, then you can write down the 782 00:57:57,276 --> 00:58:01,819 same equation in terms of psi or D2 psi DX squared one over V 783 00:58:01,819 --> 00:58:07,154 squared D2 psi DT squared. That, then, gives you the 784 00:58:07,154 --> 00:58:11,289 position, the actual motion of the air molecules. 785 00:58:11,289 --> 00:58:15,942 I often prefer the pressure, and I will follow the also 786 00:58:15,942 --> 00:58:18,613 today. So, there are an infinite 787 00:58:18,613 --> 00:58:22,404 number of normal modes. And the ratios of the 788 00:58:22,404 --> 00:58:27,143 amplitudes of two adjacent oscillators, they are coupled 789 00:58:27,143 --> 00:58:31,193 now, reflects itself, of course, in terms of the 790 00:58:31,193 --> 00:58:35,070 overall shape, which you can best see when you 791 00:58:35,070 --> 00:58:38,000 deal with a string. 792 00:58:38,000 --> 00:58:45,000 793 00:58:45,000 --> 00:58:49,846 As I said, it's easiest to see with transfers, 794 00:58:49,846 --> 00:58:54,153 oscillations, what the displacements look 795 00:58:54,153 --> 00:58:57,707 like. It's harder to see that with 796 00:58:57,707 --> 00:59:01,787 sound. I also discussed with you a 797 00:59:01,787 --> 00:59:06,227 special situation that I connected to media, 798 00:59:06,227 --> 00:59:11,492 medium one and medium two, and I gave these to media 799 00:59:11,492 --> 00:59:14,796 different masses per unit length. 800 00:59:14,796 --> 00:59:19,339 When you set up a traveling wave on a string, 801 00:59:19,339 --> 00:59:23,469 or you set it up, you can set up a pulse, 802 00:59:23,469 --> 00:59:29,044 it reflects at the end. And, how it reflects depends on 803 00:59:29,044 --> 00:59:34,000 the boundary conditions at the end. 804 00:59:34,000 --> 00:59:38,730 And when you connect them to media, then you get not only 805 00:59:38,730 --> 00:59:42,446 your reflection, but you also get some of the 806 00:59:42,446 --> 00:59:45,149 pulse. Some of the wave goes into 807 00:59:45,149 --> 00:59:48,359 medium to. And so, let's assume that we 808 00:59:48,359 --> 00:59:51,822 have an incident wave coming in like this. 809 00:59:51,822 --> 00:59:56,215 And we have here mu one, and we have here V1 given by 810 00:59:56,215 --> 00:59:59,340 this equation. They both have the same 811 00:59:59,340 --> 1:00:05,000 tangent, T, and here you have mu two and you have V2. 812 1:00:05,000 --> 1:00:09,023 We discussed that. We used the wave equation to 813 1:00:09,023 --> 1:00:14,534 solve what happens when we have an incident harmonic wave coming 814 1:00:14,534 --> 1:00:17,333 in. We derived even this speed of 815 1:00:17,333 --> 1:00:20,394 propagation using the wave equation. 816 1:00:20,394 --> 1:00:23,281 None of this came out of the blue. 817 1:00:23,281 --> 1:00:26,955 We always derived that. And then, we found, 818 1:00:26,955 --> 1:00:31,153 by using the boundary conditions at the junction, 819 1:00:31,153 --> 1:00:36,314 namely that the string is not breaking, and that DYDX on the 820 1:00:36,314 --> 1:00:42,000 left side is the same as DY DX on the right side. 821 1:00:42,000 --> 1:00:47,111 [UNINTELLIGIBLE] boundary conditions, we found that the 822 1:00:47,111 --> 1:00:52,884 amplitude of the reflected wave, and the same would hold for a 823 1:00:52,884 --> 1:00:58,374 pulse, divided by the amplitude of the incident wave was V2 824 1:00:58,374 --> 1:01:03,012 minus V1 divided by V1 plus V2. And, I call that R 825 1:01:03,012 --> 1:01:07,494 reflectivity. And, we found that the 826 1:01:07,494 --> 1:01:13,845 amplitude of the transmitted wave or pulse divided by the 827 1:01:13,845 --> 1:01:20,536 amplitude of the incident one was 2V2 divided by V1 plus V2. 828 1:01:20,536 --> 1:01:27,000 And, I called that shorthand notation transmitivity. 829 1:01:27,000 --> 1:01:32,782 And when we had done that, I put in some simple test cases 830 1:01:32,782 --> 1:01:36,028 where our intuition is very good. 831 1:01:36,028 --> 1:01:41,202 The first thing I did was I said, suppose mu two was 832 1:01:41,202 --> 1:01:45,869 infinitely high. So, mu two is infinitely high. 833 1:01:45,869 --> 1:01:50,028 In other words, that medium two is a wall. 834 1:01:50,028 --> 1:01:55,202 That means, number one, the string number one cannot 835 1:01:55,202 --> 1:01:57,942 move. It's fixed at the end. 836 1:01:57,942 --> 1:02:04,296 So, that means V2 is zero. And, we go to this equation and 837 1:02:04,296 --> 1:02:07,481 we find that R equals minus one. V2 is zero. 838 1:02:07,481 --> 1:02:11,555 You get minus V1 over V1. And, we like that because what 839 1:02:11,555 --> 1:02:16,000 it means is that when a mountain rolls in, it comes back as a 840 1:02:16,000 --> 1:02:18,592 valley. And then, a valley rolls in. 841 1:02:18,592 --> 1:02:22,740 It comes back as a mountain, and I demonstrated that with 842 1:02:22,740 --> 1:02:25,555 strings. It was very pleasing that T of 843 1:02:25,555 --> 1:02:28,444 R is then zero. Well, it better be zero, 844 1:02:28,444 --> 1:02:32,795 right? If everything comes back at you 845 1:02:32,795 --> 1:02:38,114 upside down, but nevertheless, everything comes back at you. 846 1:02:38,114 --> 1:02:41,901 You expect that nothing goes into the wall. 847 1:02:41,901 --> 1:02:45,868 And, you see when V2 is zero that TR is zero. 848 1:02:45,868 --> 1:02:50,467 And, we were all very happy. And we could all sleep. 849 1:02:50,467 --> 1:02:54,163 But then, and you guessed it, then I said, 850 1:02:54,163 --> 1:02:57,139 let's suppose mu two becomes zero. 851 1:02:57,139 --> 1:03:03,000 So, I attach that string to nothing, to empty space. 852 1:03:03,000 --> 1:03:04,579 Is that practical? Yes. 853 1:03:04,579 --> 1:03:07,595 I can do it. I can take a nickel wire and I 854 1:03:07,595 --> 1:03:11,976 have here a magnetic field very strong, and I can pull on that 855 1:03:11,976 --> 1:03:15,063 nickel wire. And, the end of the nickel wire 856 1:03:15,063 --> 1:03:17,505 ends up in nothing, in empty space. 857 1:03:17,505 --> 1:03:21,885 But, the I keep the tension on. So, it's completely practical. 858 1:03:21,885 --> 1:03:24,614 It can be done. It's not just nonsense. 859 1:03:24,614 --> 1:03:28,204 So, mu two goes to zero. That means that V2 goes to 860 1:03:28,204 --> 1:03:32,272 infinity. And then, we looked at R, 861 1:03:32,272 --> 1:03:35,909 and we say, well, if V2 goes to infinity, 862 1:03:35,909 --> 1:03:40,545 then R equals plus one. And, we were all very happy. 863 1:03:40,545 --> 1:03:44,727 A mountain comes back as a mountain, and I even 864 1:03:44,727 --> 1:03:49,090 demonstrated that. We were still able to sleep at 865 1:03:49,090 --> 1:03:52,727 that point. But then, then came the awful 866 1:03:52,727 --> 1:03:57,545 thing that if we substitute V2 equals infinity in this 867 1:03:57,545 --> 1:04:00,636 equation, that TR becomes plus two. 868 1:04:00,636 --> 1:04:07,000 And now, we can no longer sleep because this is absurd. 869 1:04:07,000 --> 1:04:10,372 All the energy that rolls in comes back. 870 1:04:10,372 --> 1:04:14,954 But there is something in addition that goes into that 871 1:04:14,954 --> 1:04:17,375 second medium. Now, admit it. 872 1:04:17,375 --> 1:04:22,131 Who could not sleep that night? You should all fail this 873 1:04:22,131 --> 1:04:26,109 course, by the way. [LAUGHTER] but in any case, 874 1:04:26,109 --> 1:04:30,000 I don't have to feel guilty, right? 875 1:04:30,000 --> 1:04:36,129 Who thought about this and said, there is something weird? 876 1:04:36,129 --> 1:04:39,892 I must find an explanation for that. 877 1:04:39,892 --> 1:04:46,129 In my case, I had to find an explanation because I couldn't 878 1:04:46,129 --> 1:04:49,462 sleep. Who found an explanation? 879 1:04:49,462 --> 1:04:54,086 Who could say, oh yes, don't worry about it? 880 1:04:54,086 --> 1:04:57,634 What was your solution? Excuse me? 881 1:04:57,634 --> 1:05:01,947 Very good. That's a very nice way of 882 1:05:01,947 --> 1:05:05,062 looking at it. So that's probably why you 883 1:05:05,062 --> 1:05:08,178 could sleep. Well, actually I'll tell you 884 1:05:08,178 --> 1:05:11,761 why I couldn't sleep. But your solution is even 885 1:05:11,761 --> 1:05:14,565 shorter. But the reason why I want to 886 1:05:14,565 --> 1:05:19,005 show you what I'm going to show you is that I want to also 887 1:05:19,005 --> 1:05:22,977 expose you to the idea which you already alluded to, 888 1:05:22,977 --> 1:05:27,105 namely that there is energy involved when we deal with 889 1:05:27,105 --> 1:05:31,000 pulses and when we deal with waves. 890 1:05:31,000 --> 1:05:36,688 You remember when we have a traveling wave that the total 891 1:05:36,688 --> 1:05:42,479 energy per wavelength lambda, I only did it per wavelength 892 1:05:42,479 --> 1:05:47,355 that the total energy, perhaps you remember that, 893 1:05:47,355 --> 1:05:52,638 equals 2A squared times pi squared times the tension, 894 1:05:52,638 --> 1:05:56,803 T, divided by lambda. A was the amplitude. 895 1:05:56,803 --> 1:06:03,000 Energy is always proportional to amplitude squared. 896 1:06:03,000 --> 1:06:06,500 This was the tension, and this was the wavelength. 897 1:06:06,500 --> 1:06:10,428 And, if V2 goes to infinity, then the wavelength goes to 898 1:06:10,428 --> 1:06:12,142 infinity. That's obvious, 899 1:06:12,142 --> 1:06:14,500 right? If something moves with the 900 1:06:14,500 --> 1:06:18,428 speed of light even faster, infinity is even faster than 901 1:06:18,428 --> 1:06:21,857 the speed of light, then lambda goes to infinity. 902 1:06:21,857 --> 1:06:25,357 So, this goes to zero. And so, we came to the same 903 1:06:25,357 --> 1:06:29,214 conclusion, some decadent solution, TR equals plus two, 904 1:06:29,214 --> 1:06:34,000 has no meaning because there is no energy in there. 905 1:06:34,000 --> 1:06:41,307 And so, I was able to sleep. Let's now turn to normal modes 906 1:06:41,307 --> 1:06:46,850 of continuous media. And I suggest that we go 907 1:06:46,850 --> 1:06:53,149 longitudinal because we did so much transfer stuff. 908 1:06:53,149 --> 1:06:59,700 Let's go longitudinal. I have here a pi has length L, 909 1:06:59,700 --> 1:07:06,000 and it is open here, and it is open there. 910 1:07:06,000 --> 1:07:09,733 That means the overpressure, under pressure here, 911 1:07:09,733 --> 1:07:13,622 can never build up. It's connected to the universe. 912 1:07:13,622 --> 1:07:17,744 So, at these two boundary conditions, the speed that I 913 1:07:17,744 --> 1:07:21,711 have there must be zero. If I write down the general 914 1:07:21,711 --> 1:07:26,222 equation for a standing wave, because normal mode solutions 915 1:07:26,222 --> 1:07:30,033 are standing waves, I can write down P equals some 916 1:07:30,033 --> 1:07:34,000 amplitude times the sine or the cosine. 917 1:07:34,000 --> 1:07:38,454 Let me take a sine, two pi divided by lambda times 918 1:07:38,454 --> 1:07:40,545 X. I'll be very general. 919 1:07:40,545 --> 1:07:44,181 I will introduce some phase angle, alpha, 920 1:07:44,181 --> 1:07:48,272 for which I will find a solution very shortly. 921 1:07:48,272 --> 1:07:53,181 And then, cosine omega T, or sine omega T if you prefer 922 1:07:53,181 --> 1:07:56,454 that. This is a standing wave in very 923 1:07:56,454 --> 1:08:00,363 general terms. Everything here is the space. 924 1:08:00,363 --> 1:08:04,181 X, this is X, and here, all the information 925 1:08:04,181 --> 1:08:08,909 here deals with time, which is typical for a standing 926 1:08:08,909 --> 1:08:13,724 wave. [NOISE OBSCURES] always to 927 1:08:13,724 --> 1:08:17,393 write down for two pi over lambda K. 928 1:08:17,393 --> 1:08:23,052 And I must observe the situation that P must be zero at 929 1:08:23,052 --> 1:08:26,930 X equals zero, and also at X equals L. 930 1:08:26,930 --> 1:08:33,218 If P is zero at X equals zero, immediately you see that alpha 931 1:08:33,218 --> 1:08:37,765 is zero. So I'm going to rewrite it now. 932 1:08:37,765 --> 1:08:42,582 I'm going to write down P, P zero times the sine of KX. 933 1:08:42,582 --> 1:08:45,525 So I'm going to replace this by K. 934 1:08:45,525 --> 1:08:50,253 I know that alpha is zero times the cosine of omega T. 935 1:08:50,253 --> 1:08:55,070 And now, I must meet the boundary condition that when X 936 1:08:55,070 --> 1:08:57,924 equals L, that P is, again, zero. 937 1:08:57,924 --> 1:09:01,938 And that, now, breaks open a whole spectrum of 938 1:09:01,938 --> 1:09:06,399 possibilities in which I introduce this normal mode 939 1:09:06,399 --> 1:09:10,502 number N as in Nancy whereby N can be one, two, 940 1:09:10,502 --> 1:09:15,895 three, etc. And then I get solutions when K 941 1:09:15,895 --> 1:09:20,939 of N equals N pi divided by L. You see that immediately 942 1:09:20,939 --> 1:09:26,170 because if I make X now L, then I get the sine of M pi no 943 1:09:26,170 --> 1:09:30,000 matter what N is. I always get zero. 944 1:09:30,000 --> 1:09:36,342 And so, my lambda of N, which is two pi divided by K is 945 1:09:36,342 --> 1:09:41,862 then 2L divided by N. So, I can rewrite now this 946 1:09:41,862 --> 1:09:48,439 equation as P zero times the sine of N pi X divided by L. 947 1:09:48,439 --> 1:09:55,251 And now, I have cosine omega NT because omega N are now the 948 1:09:55,251 --> 1:10:01,241 frequencies which are associated with the N'th mode, 949 1:10:01,241 --> 1:10:07,437 N being N as in Nancy. What now is the connection 950 1:10:07,437 --> 1:10:13,250 between this omega N and this K? Well, that connection you will 951 1:10:13,250 --> 1:10:19,062 find through the wave equation. You now have to substitute this 952 1:10:19,062 --> 1:10:24,406 result back into the wave equation, which is this one's to 953 1:10:24,406 --> 1:10:28,343 solve or omega. And, I want to do that with 954 1:10:28,343 --> 1:10:32,000 you. It is not that much work. 955 1:10:32,000 --> 1:10:35,428 I go to 2PDX squared. So, here it is: 956 1:10:35,428 --> 1:10:39,809 D2P DX squared. So, all I get is I get this out 957 1:10:39,809 --> 1:10:43,238 twice. So, I get N squared pi squared 958 1:10:43,238 --> 1:10:47,619 over L squared. I get a minus sign because if I 959 1:10:47,619 --> 1:10:52,761 take twice the derivative, I always end up with a minus 960 1:10:52,761 --> 1:10:55,904 sign. But the sine comes back as a 961 1:10:55,904 --> 1:10:58,952 sine. And, not only does the sine 962 1:10:58,952 --> 1:11:04,000 come back, but all the rest comes back. 963 1:11:04,000 --> 1:11:06,495 And so, I will just write P here. 964 1:11:06,495 --> 1:11:11,175 So, this P is exactly that P. So, this is the only thing that 965 1:11:11,175 --> 1:11:15,153 was added by taking the second derivative against X. 966 1:11:15,153 --> 1:11:19,130 Now, what is D2PDT squared? Now I have to go to this 967 1:11:19,130 --> 1:11:22,172 function. [I have?] partial derivatives. 968 1:11:22,172 --> 1:11:25,526 X is constant here, but T is the one that is 969 1:11:25,526 --> 1:11:29,426 changing, whereas here, we had that X was changing. 970 1:11:29,426 --> 1:11:33,964 The T was constant. So now, we are going to get 971 1:11:33,964 --> 1:11:36,657 minus omega N squared, again, times P. 972 1:11:36,657 --> 1:11:40,805 The whole function comes back. And now, we are in business 973 1:11:40,805 --> 1:11:44,880 because now the wave equation will tell us the connection 974 1:11:44,880 --> 1:11:47,936 between the two. It tells us that N squared 975 1:11:47,936 --> 1:11:51,866 times pi squared divided by L squared is now one over V 976 1:11:51,866 --> 1:11:54,631 squared. That is the one over V squared 977 1:11:54,631 --> 1:11:58,124 that I had there. And, there's a minus sign here. 978 1:11:58,124 --> 1:12:03,000 And there's a minus sign here times omega N squared. 979 1:12:03,000 --> 1:12:09,437 So, this is not connected. And so, you see the solution 980 1:12:09,437 --> 1:12:15,158 for omega N just is being presented to you [NOISE 981 1:12:15,158 --> 1:12:20,165 OBSCURES] is now N pi times V divided by L. 982 1:12:20,165 --> 1:12:24,337 That follows from the wave equation. 983 1:12:24,337 --> 1:12:29,582 And so, if you prefer the frequency in hertz, 984 1:12:29,582 --> 1:12:34,112 then you have to divide this by two pi. 985 1:12:34,112 --> 1:12:41,373 So, you get NV divided by 2L. So, if I want to plot now, 986 1:12:41,373 --> 1:12:45,482 so this is X equals zero, and this X equals L, 987 1:12:45,482 --> 1:12:49,865 I can plot now here the pressure in terms of this 988 1:12:49,865 --> 1:12:52,878 overpressure or under pressure, P. 989 1:12:52,878 --> 1:12:57,352 And, you get a curve which looks very similar to a 990 1:12:57,352 --> 1:13:01,118 transverse solution. But, of course, 991 1:13:01,118 --> 1:13:04,550 it is not transverse. It's really longitudinal. 992 1:13:04,550 --> 1:13:07,534 But you get, then, that for N equals one, 993 1:13:07,534 --> 1:13:11,487 you would get this mode. There must be a pressure node 994 1:13:11,487 --> 1:13:15,814 here and there because we can never build a pressure that's 995 1:13:15,814 --> 1:13:19,693 connected with the universe. You can never build that 996 1:13:19,693 --> 1:13:22,752 overpressure. And so, the pressure buildup 997 1:13:22,752 --> 1:13:26,332 here is positive. And then, later in time it will 998 1:13:26,332 --> 1:13:29,987 be negative, and then positive, and then negative. 999 1:13:29,987 --> 1:13:34,314 And, when you go to N equals two, then you get another node 1000 1:13:34,314 --> 1:13:38,903 in pressure here. And so, now you get this, 1001 1:13:38,903 --> 1:13:43,245 always a pressure node here, always a pressure node there. 1002 1:13:43,245 --> 1:13:47,129 But now you end up with another pressure node there. 1003 1:13:47,129 --> 1:13:51,089 And, if you have any difficulties to see what the air 1004 1:13:51,089 --> 1:13:54,821 molecules are doing, I would recommend you go back 1005 1:13:54,821 --> 1:14:00,000 to psi space which is the actual position of the molecules. 1006 1:14:00,000 --> 1:14:03,786 And when you do that, you will always find that where 1007 1:14:03,786 --> 1:14:06,990 the pressure has an anti-node, which is here, 1008 1:14:06,990 --> 1:14:10,703 psi always has a node, and where the pressure has as 1009 1:14:10,703 --> 1:14:13,106 node, psi always has an anti-node. 1010 1:14:13,106 --> 1:14:16,820 Of course, the molecules can freely move in and out. 1011 1:14:16,820 --> 1:14:20,315 There is no problem. So, the molecules can freely 1012 1:14:20,315 --> 1:14:23,956 move in and out here. So, where psi has its largest 1013 1:14:23,956 --> 1:14:27,669 amplitude, its anti-node, that is where the pressure 1014 1:14:27,669 --> 1:14:33,687 cannot build up. But going to psi space actually 1015 1:14:33,687 --> 1:14:41,182 often helps me to see precisely what is going on with the motion 1016 1:14:41,182 --> 1:14:46,535 of the molecules. I have here a linear system, 1017 1:14:46,535 --> 1:14:52,126 which is a sound cavity. It's made of magnesium. 1018 1:14:52,126 --> 1:14:56,646 [There's?] no air, but it is magnesium. 1019 1:14:56,646 --> 1:15:02,000 And, it is open, open on both sides. 1020 1:15:02,000 --> 1:15:09,563 You can't have it any better. I'll make a drawing for you. 1021 1:15:09,563 --> 1:15:14,739 So, here is my magnesium. This is a rod. 1022 1:15:14,739 --> 1:15:21,241 It's one-dimensional, and the length is 122 cm and 1023 1:15:21,241 --> 1:15:30,000 the speed of sound in magnesium is about 5,000 m per second. 1024 1:15:30,000 --> 1:15:36,137 When I hit that magnesium rot on the side, it wants to go into 1025 1:15:36,137 --> 1:15:40,362 standing waves. It prefers the lowest mode. 1026 1:15:40,362 --> 1:15:45,493 It almost always does. But it may also create second 1027 1:15:45,493 --> 1:15:49,920 and third harmonics. And so, the lowest mode, 1028 1:15:49,920 --> 1:15:55,253 F1, is then V divided by 2L. So, the lowest frequency, 1029 1:15:55,253 --> 1:16:00,182 F1, is V divided by 2L, which when I calculate it, 1030 1:16:00,182 --> 1:16:05,591 is about 250 Hz. And the actual value we measure 1031 1:16:05,591 --> 1:16:08,767 is about 2,044. And, there may also be, 1032 1:16:08,767 --> 1:16:13,614 when I hit it with a hammer, there may also be some that is 1033 1:16:13,614 --> 1:16:16,791 twice as high. So that may be 4,100 Hz. 1034 1:16:16,791 --> 1:16:19,632 That would be double the frequency. 1035 1:16:19,632 --> 1:16:24,564 That would be the second mode. And this is quite remarkable. 1036 1:16:24,564 --> 1:16:29,077 So this is not filled with air, but this is filled with 1037 1:16:29,077 --> 1:16:32,947 magnesium. But, by exciting it here, 1038 1:16:32,947 --> 1:16:35,751 it's like blowing on the flute there. 1039 1:16:35,751 --> 1:16:40,346 It goes into these normal mode solutions, and it is this one 1040 1:16:40,346 --> 1:16:42,917 that you will hear loud and clear. 1041 1:16:42,917 --> 1:16:46,811 It's a beautiful tone. And this one you may hear in 1042 1:16:46,811 --> 1:16:50,238 the beginning, but the higher harmonics often 1043 1:16:50,238 --> 1:16:53,276 die out faster than the lower harmonics. 1044 1:16:53,276 --> 1:16:56,158 So, are you ready for this? Open-open. 1045 1:16:56,158 --> 1:17:00,442 It's open on both sides, and I just bang the hell out of 1046 1:17:00,442 --> 1:17:02,898 it. 2,044. 1047 1:17:02,898 --> 1:17:10,085 Isn't that beautiful? It's oscillating like this. 1048 1:17:10,085 --> 1:17:18,770 Exactly oscillating the way that I derived for you for air. 1049 1:17:18,770 --> 1:17:24,459 And here, you see that it [holds for?]. 1050 1:17:24,459 --> 1:17:33,314 Only the fundamental is there. OK, I wish you luck on 1051 1:17:33,314 --> 1:17:34,670 Thursday. I'll see you then.