1 00:00:00,060 --> 00:00:01,770 The following content is provided 2 00:00:01,770 --> 00:00:04,010 under a Creative Commons license. 3 00:00:04,010 --> 00:00:06,860 Your support will help MIT OpenCourseWare continue 4 00:00:06,860 --> 00:00:10,720 to offer high-quality educational resources for free. 5 00:00:10,720 --> 00:00:13,330 To make a donation, or view additional materials 6 00:00:13,330 --> 00:00:17,209 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,209 --> 00:00:17,834 at ocw.mit.edu. 8 00:00:20,950 --> 00:00:22,470 PROFESSOR: Welcome back. 9 00:00:22,470 --> 00:00:25,500 Today, we will consider solutions 10 00:00:25,500 --> 00:00:29,355 to systems of an infinite number of degrees of freedom. 11 00:00:32,390 --> 00:00:38,960 What we'll consider is when you have identical oscillators, 12 00:00:38,960 --> 00:00:42,600 which are coupled only to the neighbors, 13 00:00:42,600 --> 00:00:47,200 and there is an infinite number of them. 14 00:00:47,200 --> 00:00:53,240 So each one is an infinitely small oscillator, as I say, 15 00:00:53,240 --> 00:00:55,420 coupled to each neighbor. 16 00:00:55,420 --> 00:01:01,410 The simplest example is that of a taut string. 17 00:01:01,410 --> 00:01:06,220 Each piece of that string is a harmonic oscillator. 18 00:01:06,220 --> 00:01:08,130 It oscillates transversely. 19 00:01:08,130 --> 00:01:11,990 And each piece is identical to every other piece. 20 00:01:11,990 --> 00:01:14,500 And it's coupled to its neighbors. 21 00:01:14,500 --> 00:01:18,562 So let's consider the following problem. 22 00:01:18,562 --> 00:01:20,770 The problem we are going to talk about-- the solution 23 00:01:20,770 --> 00:01:22,410 of the following problem. 24 00:01:22,410 --> 00:01:25,830 Suppose you have a string. 25 00:01:25,830 --> 00:01:29,710 For practical purposes, an infinite string. 26 00:01:29,710 --> 00:01:37,970 It has a mass per unit length of new mu, a constant tension T, 27 00:01:37,970 --> 00:01:45,530 and at one end it's connected to a massless ring. 28 00:01:45,530 --> 00:01:48,380 This is an idealized situation. 29 00:01:48,380 --> 00:01:51,705 You could imagine, suppose that ring was on your finger. 30 00:01:51,705 --> 00:01:57,370 You held the string taut and you moved your finger 31 00:01:57,370 --> 00:02:02,770 in some ways, which caused a distortion, which 32 00:02:02,770 --> 00:02:05,680 propagated down this string. 33 00:02:05,680 --> 00:02:12,320 You got a progressive pulse going down this string. 34 00:02:12,320 --> 00:02:15,980 So this is an idealized diagram of that situation. 35 00:02:15,980 --> 00:02:20,300 I've indicated the finger holding the string by, 36 00:02:20,300 --> 00:02:22,770 you could imagine, a massless ring 37 00:02:22,770 --> 00:02:26,050 sliding on a frictionless rod, for example. 38 00:02:26,050 --> 00:02:29,680 And somehow or other, a force is applied to this ring. 39 00:02:29,680 --> 00:02:32,279 But you can think of the actual problem 40 00:02:32,279 --> 00:02:34,195 where you're holding the ring with your finger 41 00:02:34,195 --> 00:02:36,360 and moving this up and down. 42 00:02:36,360 --> 00:02:42,800 Now, somehow or other, you're moving this ring such 43 00:02:42,800 --> 00:02:50,230 that as a result of your motion, there is a progressive pulse. 44 00:02:50,230 --> 00:02:53,220 And the idealized pulse we'll consider 45 00:02:53,220 --> 00:02:56,170 is a triangular shape like this, where 46 00:02:56,170 --> 00:02:58,880 this height is H and the length from here 47 00:02:58,880 --> 00:03:01,643 and here is L. It is idealized. 48 00:03:01,643 --> 00:03:07,610 In reality, you can't have sharp corners like that. 49 00:03:07,610 --> 00:03:11,705 You can imagine there is a very tiny curvature 50 00:03:11,705 --> 00:03:14,200 at those locations. 51 00:03:14,200 --> 00:03:20,480 Now, what we are told is that at some instant t given 52 00:03:20,480 --> 00:03:27,040 by 4L over-- and this is the phase velocity of propagation 53 00:03:27,040 --> 00:03:30,470 down the pulse over square root of t over mu. 54 00:03:30,470 --> 00:03:35,440 At this instant of time, you look at the string 55 00:03:35,440 --> 00:03:40,200 and you see that this is the shape of that pulse. 56 00:03:40,200 --> 00:03:45,880 And as a function of time, that propagates to the right. 57 00:03:45,880 --> 00:03:48,360 OK? 58 00:03:48,360 --> 00:03:52,290 The question is, what is the motion 59 00:03:52,290 --> 00:03:56,560 of this ring as a function of time? 60 00:03:56,560 --> 00:03:58,170 That's one. 61 00:03:58,170 --> 00:04:08,490 Two, what is the power that the force which drives this string 62 00:04:08,490 --> 00:04:11,400 delivers to the string? 63 00:04:11,400 --> 00:04:15,630 So what's the power as a function of time? 64 00:04:15,630 --> 00:04:19,019 All right, next. 65 00:04:19,019 --> 00:04:23,610 What is at any instant of time the potential energy 66 00:04:23,610 --> 00:04:25,055 stored in this pulse? 67 00:04:29,280 --> 00:04:34,470 Four, what is the kinetic energy stored in this pulse 68 00:04:34,470 --> 00:04:37,230 as it's propagating? 69 00:04:37,230 --> 00:04:41,760 And finally the question, are your results 70 00:04:41,760 --> 00:04:50,050 to the power that's delivered, the potential energy 71 00:04:50,050 --> 00:04:52,960 stored in the distortion, the kinetic energy 72 00:04:52,960 --> 00:04:57,180 in the distortion consistent with each other? 73 00:04:59,920 --> 00:05:02,965 These are the things we are told to assume. 74 00:05:02,965 --> 00:05:04,600 And I repeat some of them. 75 00:05:04,600 --> 00:05:07,990 We assume the ring is massless. 76 00:05:07,990 --> 00:05:12,290 The distortion is always sufficiently small 77 00:05:12,290 --> 00:05:16,900 so that this height is much smaller than this. 78 00:05:16,900 --> 00:05:21,560 So that this angle here is sufficiently small 79 00:05:21,560 --> 00:05:27,190 that we can approximate the sine angle the angle 80 00:05:27,190 --> 00:05:30,800 or the tangent of the angle, which of course 81 00:05:30,800 --> 00:05:34,020 is this height divided by that. 82 00:05:34,020 --> 00:05:36,450 So it's h over L over T. 83 00:05:36,450 --> 00:05:39,760 We are told that the mass per unit length is mu 84 00:05:39,760 --> 00:05:42,010 and that the tension is uniform. 85 00:05:42,010 --> 00:05:46,980 In other words, because this distortion is small, 86 00:05:46,980 --> 00:05:51,220 we are making the assumption that the tension in this string 87 00:05:51,220 --> 00:05:58,340 is constant and the same everywhere. 88 00:05:58,340 --> 00:06:01,260 There are no losses, no frictional losses. 89 00:06:01,260 --> 00:06:06,010 And finally, all the motion is-- because this 90 00:06:06,010 --> 00:06:08,255 is so small, is in the transverse direction. 91 00:06:08,255 --> 00:06:14,630 So any piece of the string moves in the transverse direction. 92 00:06:14,630 --> 00:06:18,620 This is a reasonable approximation to reality. 93 00:06:18,620 --> 00:06:21,730 Or put it the other way around, the reality 94 00:06:21,730 --> 00:06:23,920 would be a reasonable approximation 95 00:06:23,920 --> 00:06:27,420 to this idealized situation, which 96 00:06:27,420 --> 00:06:31,890 we will try to understand and solve. 97 00:06:31,890 --> 00:06:37,390 OK, so how do we go about it? 98 00:06:37,390 --> 00:06:38,710 The usual way. 99 00:06:38,710 --> 00:06:44,730 We first have to represent the problem 100 00:06:44,730 --> 00:06:47,840 in terms of mathematics. 101 00:06:47,840 --> 00:06:50,180 So what do we know? 102 00:06:50,180 --> 00:07:07,150 The system is a continuous infinite row of oscillators. 103 00:07:07,150 --> 00:07:11,250 Each one connected only to its neighbor. 104 00:07:11,250 --> 00:07:18,170 And we know that such a system can 105 00:07:18,170 --> 00:07:22,760 be represented by a wave equation. 106 00:07:22,760 --> 00:07:29,110 The equation of motion for this system is a wave equation. 107 00:07:29,110 --> 00:07:34,940 In essence, if you remember when we had coupled oscillators, 108 00:07:34,940 --> 00:07:37,940 three oscillators, four, et cetera. 109 00:07:37,940 --> 00:07:41,550 Every time you add one more oscillator, 110 00:07:41,550 --> 00:07:45,470 you have one more equation of motion. 111 00:07:45,470 --> 00:07:50,670 So if you have an infinite number of oscillators, 112 00:07:50,670 --> 00:07:53,740 you expect to have an infinite number 113 00:07:53,740 --> 00:07:57,710 of coupled equations of motions. 114 00:07:57,710 --> 00:08:00,910 That's what a wave equation is. 115 00:08:00,910 --> 00:08:08,890 For every position x, you have an equation of motion. 116 00:08:08,890 --> 00:08:12,300 Why the distortion is a function of x? 117 00:08:12,300 --> 00:08:17,190 So essentially for every value of x, you have one equation. 118 00:08:17,190 --> 00:08:20,810 Since x is a continuum, this, in essence, 119 00:08:20,810 --> 00:08:24,650 is an infinite number of coupled differential equations. 120 00:08:24,650 --> 00:08:28,070 And we call this the wave equation. 121 00:08:28,070 --> 00:08:33,990 So the equation of motion of the string is that. 122 00:08:33,990 --> 00:08:36,049 What else do we know? 123 00:08:36,049 --> 00:08:39,860 How can we use the laws of nature 124 00:08:39,860 --> 00:08:41,950 to describe this situation? 125 00:08:41,950 --> 00:08:43,200 Well, let's consider the ring. 126 00:08:46,570 --> 00:08:51,470 On the ring, there will be forces. 127 00:08:51,470 --> 00:08:54,675 One is the string is attached, so there 128 00:08:54,675 --> 00:09:00,300 will be the tension in the string pulling on the ring. 129 00:09:00,300 --> 00:09:05,150 We said that the ring cannot-- it moves only up and down, 130 00:09:05,150 --> 00:09:10,520 so there must be some constraining force on it. 131 00:09:10,520 --> 00:09:14,520 If you're imagining it to be on this frictionless rod, 132 00:09:14,520 --> 00:09:18,240 then that constraining force is the reaction 133 00:09:18,240 --> 00:09:21,990 of the rod on the ring. 134 00:09:21,990 --> 00:09:24,520 If you're thinking of it in terms of a finger holding 135 00:09:24,520 --> 00:09:28,870 onto that ring, then your finger is 136 00:09:28,870 --> 00:09:32,170 preventing the ring moving backwards or forwards. 137 00:09:32,170 --> 00:09:34,760 So that's this reaction force. 138 00:09:34,760 --> 00:09:39,600 Now, on top of that, there is the vertical force, 139 00:09:39,600 --> 00:09:43,080 the one we are interested in, which 140 00:09:43,080 --> 00:09:47,140 must be the cause of the distortion. 141 00:09:47,140 --> 00:09:51,830 So those are the three forces. 142 00:09:51,830 --> 00:09:54,550 What else do we know? 143 00:09:54,550 --> 00:09:58,390 We said that we are going to idealize this situation. 144 00:09:58,390 --> 00:10:02,140 We're going to assume that this ring has no mass. 145 00:10:02,140 --> 00:10:04,180 It's only there so you can essentially 146 00:10:04,180 --> 00:10:06,730 hold the tip of the string. 147 00:10:06,730 --> 00:10:11,070 Now if mass is 0, then in Newton's laws of motion 148 00:10:11,070 --> 00:10:14,590 there can be no net force on that mass. 149 00:10:14,590 --> 00:10:18,710 Because if you applied any force to a 0 mass, 150 00:10:18,710 --> 00:10:20,440 it will have an infinite acceleration. 151 00:10:20,440 --> 00:10:22,720 Your ring would disappear. 152 00:10:22,720 --> 00:10:28,900 So the fact that the ring is massless-- in other words, 153 00:10:28,900 --> 00:10:32,760 you're just holding the very tip of the string. 154 00:10:32,760 --> 00:10:35,180 That's what it, in essence, means. 155 00:10:35,180 --> 00:10:39,110 It means there is no net force on the ring. 156 00:10:39,110 --> 00:10:42,000 The ring does not move backwards and forwards. 157 00:10:42,000 --> 00:10:44,315 There's no net force in that direction. 158 00:10:49,760 --> 00:11:04,340 So the reaction of the rod on the string 159 00:11:04,340 --> 00:11:08,470 must be exactly equal to the horizontal component 160 00:11:08,470 --> 00:11:11,980 of the tension. 161 00:11:11,980 --> 00:11:17,610 Actually, we don't need this for the solving of the problem. 162 00:11:17,610 --> 00:11:20,030 On the other hand, in the vertical direction 163 00:11:20,030 --> 00:11:24,800 we also know that there is no net force vertically. 164 00:11:24,800 --> 00:11:30,796 And so the magnitude of the force at any instant of time 165 00:11:30,796 --> 00:11:34,240 will, of course, be equal and opposite 166 00:11:34,240 --> 00:11:39,660 to the magnitude of the force due to the tension 167 00:11:39,660 --> 00:11:44,620 in this string, the component of that in the vertical direction. 168 00:11:44,620 --> 00:11:49,930 So the F of t at any instant of time 169 00:11:49,930 --> 00:11:54,970 will be equal to minus the vertical component 170 00:11:54,970 --> 00:12:00,680 of the tension on the string. 171 00:12:00,680 --> 00:12:04,330 OK, so that's everything we know about the dynamics. 172 00:12:08,070 --> 00:12:15,720 This plus the knowledge that at the time 173 00:12:15,720 --> 00:12:19,180 we were given this was the shape of the distortion, 174 00:12:19,180 --> 00:12:21,370 and it was progressing to the right, 175 00:12:21,370 --> 00:12:26,590 must be enough to be able to predict 176 00:12:26,590 --> 00:12:30,250 the motion of the ring, the power delivered. 177 00:12:32,930 --> 00:12:35,980 Now, why do I say that? 178 00:12:35,980 --> 00:12:37,890 Well, let's go back for a second. 179 00:12:37,890 --> 00:12:40,050 This will tell us what the ring is doing 180 00:12:40,050 --> 00:12:42,690 and what the forces are acting on it. 181 00:12:42,690 --> 00:12:47,070 Now, if you have a force acting on something, 182 00:12:47,070 --> 00:12:51,050 and that something is moving, the power 183 00:12:51,050 --> 00:12:56,030 that you will deliver that to that object 184 00:12:56,030 --> 00:13:00,700 is the dot product between the force 185 00:13:00,700 --> 00:13:04,330 and the velocity of the object. 186 00:13:04,330 --> 00:13:10,380 These we know because power is work per unit time. 187 00:13:10,380 --> 00:13:18,690 Work done by a force is force times the distance moved. 188 00:13:18,690 --> 00:13:23,500 So therefore, work times distance moved per second 189 00:13:23,500 --> 00:13:30,250 is the force times the velocity. 190 00:13:30,250 --> 00:13:37,410 So this is the power generated by a force 191 00:13:37,410 --> 00:13:43,600 if it moves an object with a velocity v of t. 192 00:13:43,600 --> 00:13:47,550 From what I told you earlier, I'm 193 00:13:47,550 --> 00:13:50,600 just thinking through whether we can solve this problem. 194 00:13:50,600 --> 00:13:52,860 From the part I discussed earlier, 195 00:13:52,860 --> 00:13:55,460 we will be able to figure out the force. 196 00:13:55,460 --> 00:13:58,620 We will be able to figure out the motion. 197 00:13:58,620 --> 00:14:02,440 Therefore, we will be able to calculate this and answer 198 00:14:02,440 --> 00:14:04,900 the second part of the problem. 199 00:14:04,900 --> 00:14:08,030 Let's continue. 200 00:14:08,030 --> 00:14:11,210 We want to find the energy stored 201 00:14:11,210 --> 00:14:14,280 in the distortion of this ring. 202 00:14:14,280 --> 00:14:16,220 How would we do that to make sure 203 00:14:16,220 --> 00:14:18,030 that we have everything in place? 204 00:14:18,030 --> 00:14:20,960 Well, if you have some system which 205 00:14:20,960 --> 00:14:26,930 you distort from equilibrium, the potential energy 206 00:14:26,930 --> 00:14:29,450 stored by conservation of energy is 207 00:14:29,450 --> 00:14:38,430 equal to the work you do to distort that system. 208 00:14:38,430 --> 00:14:46,500 In our case, the pulse is a distorted string 209 00:14:46,500 --> 00:14:48,170 from equilibrium. 210 00:14:48,170 --> 00:14:51,780 So if I calculate the work that I 211 00:14:51,780 --> 00:14:56,740 will have to do to produce that shape of the string, 212 00:14:56,740 --> 00:15:00,710 that will be equal to the potential energy stored 213 00:15:00,710 --> 00:15:03,010 in the string. 214 00:15:03,010 --> 00:15:04,890 So potential energy, in other words, 215 00:15:04,890 --> 00:15:07,090 is work to distort the string. 216 00:15:07,090 --> 00:15:13,870 And that is simply the integral of the force times the distance 217 00:15:13,870 --> 00:15:17,285 that will move that force to distort the string. 218 00:15:21,290 --> 00:15:26,000 So that we will be able to do since we-- 219 00:15:26,000 --> 00:15:29,070 as I'll show in a second. 220 00:15:29,070 --> 00:15:33,100 The next part was, what is the kinetic energy in the pulse? 221 00:15:33,100 --> 00:15:36,620 Well kinetic energy is, of course, 222 00:15:36,620 --> 00:15:42,385 1/2 mass times velocity squared of a system. 223 00:15:42,385 --> 00:15:47,350 The string is-- it's a continuous distribution 224 00:15:47,350 --> 00:15:48,980 of masses. 225 00:15:48,980 --> 00:15:53,910 If I take every piece of that string 226 00:15:53,910 --> 00:15:58,050 and calculate its transverse velocity-- 227 00:15:58,050 --> 00:16:00,080 that's the only velocity that it has. 228 00:16:00,080 --> 00:16:03,440 We said that this string has only transverse motion. 229 00:16:03,440 --> 00:16:05,840 So if I calculate the transverse velocity 230 00:16:05,840 --> 00:16:13,630 of every piece of the string and add up all the pieces, 231 00:16:13,630 --> 00:16:16,540 I'll have the total kinetic energy stored. 232 00:16:16,540 --> 00:16:19,440 So at any instant of time, I have 233 00:16:19,440 --> 00:16:23,090 to just find out what every piece of the string is doing, 234 00:16:23,090 --> 00:16:26,640 calculate its velocity squared multiplied 235 00:16:26,640 --> 00:16:33,300 by 1/2 by the mass of that piece, add them all up, 236 00:16:33,300 --> 00:16:34,785 and that's equal to-- therefore, we 237 00:16:34,785 --> 00:16:39,060 will have to integrate across the whole string 1/2 times 238 00:16:39,060 --> 00:16:43,470 the transverse velocity squared times 239 00:16:43,470 --> 00:16:48,010 the density of the-- the string has 240 00:16:48,010 --> 00:16:51,580 certain mass per unit length, mu. 241 00:16:51,580 --> 00:16:56,670 So I have to take this and integrate a mu dx times that. 242 00:16:56,670 --> 00:16:59,330 That'll give us the total kinetic energy. 243 00:16:59,330 --> 00:17:01,160 OK. 244 00:17:01,160 --> 00:17:07,810 So now, here we have the problem in terms of mathematics. 245 00:17:07,810 --> 00:17:11,097 And we'll switch over and try to solve it. 246 00:17:17,250 --> 00:17:17,750 OK. 247 00:17:27,160 --> 00:17:32,540 This wave equation has, as I told you, 248 00:17:32,540 --> 00:17:37,240 is equivalent to an infinite coupled differential equations. 249 00:17:37,240 --> 00:17:42,340 It has infinite number of solutions. 250 00:17:42,340 --> 00:17:45,650 Trying like we did for simpler system, 251 00:17:45,650 --> 00:17:48,150 like system with one degree of freedom 252 00:17:48,150 --> 00:17:51,510 or two degree of freedom is no longer practical. 253 00:17:51,510 --> 00:17:53,990 I can't go through all the possibilities 254 00:17:53,990 --> 00:17:56,670 and then guess which one applies. 255 00:17:56,670 --> 00:18:00,870 So I have to use some more knowledge or experience. 256 00:18:03,930 --> 00:18:08,600 In general, as I say, there's an infinite number of solutions. 257 00:18:08,600 --> 00:18:16,420 But we know that we are looking for a very specific kind 258 00:18:16,420 --> 00:18:18,120 of solution. 259 00:18:18,120 --> 00:18:25,370 We know that this pulse we're told 260 00:18:25,370 --> 00:18:30,140 was generated by this mass-- this ring. 261 00:18:30,140 --> 00:18:31,230 Sorry, not the mass. 262 00:18:31,230 --> 00:18:35,480 By this ring being moved by a finger. 263 00:18:35,480 --> 00:18:39,850 Which gave rise to a progressive wave. 264 00:18:39,850 --> 00:18:46,460 Meaning a shape like this which is moving to the right. 265 00:18:46,460 --> 00:18:50,240 That immediately gives us a clue. 266 00:18:50,240 --> 00:18:55,160 We know that there are some general classes of solutions 267 00:18:55,160 --> 00:18:57,430 of this wave equation. 268 00:18:57,430 --> 00:19:00,540 There are the normal modes, which 269 00:19:00,540 --> 00:19:08,650 are every piece of this string moving with the same frequency 270 00:19:08,650 --> 00:19:11,120 and phase, et cetera. 271 00:19:11,120 --> 00:19:13,800 There is another one which isn't obvious, 272 00:19:13,800 --> 00:19:18,830 but which Professor Walter Lewin discussed in class, 273 00:19:18,830 --> 00:19:22,200 that is actually quite-- I consider it the miraculous one. 274 00:19:22,200 --> 00:19:24,450 It's so not obvious. 275 00:19:24,450 --> 00:19:29,140 There is a solution of this wave equation, 276 00:19:29,140 --> 00:19:32,640 which consists of any distortion. 277 00:19:32,640 --> 00:19:35,780 You can take any shape. 278 00:19:35,780 --> 00:19:42,150 And if somehow or other you manage to make that distortion 279 00:19:42,150 --> 00:19:49,280 move with a uniform velocity corresponding 280 00:19:49,280 --> 00:19:58,020 to this velocity in the wave equation-- 281 00:19:58,020 --> 00:20:04,400 so for our string, the square root of this number-- that 282 00:20:04,400 --> 00:20:08,640 will progress forever. 283 00:20:08,640 --> 00:20:12,400 It's a progressive wave solution. 284 00:20:12,400 --> 00:20:15,960 And it's for any shape. 285 00:20:15,960 --> 00:20:21,870 So here, clearly we are talking of that kind 286 00:20:21,870 --> 00:20:25,940 of a solution of this wave equation. 287 00:20:25,940 --> 00:20:29,460 And as I say, from what you've learned from Professor Walter 288 00:20:29,460 --> 00:20:33,680 Lewin, we know that the wave equation 289 00:20:33,680 --> 00:20:35,940 does have that class of solutions. 290 00:20:40,540 --> 00:20:45,690 So the solution to this problem must 291 00:20:45,690 --> 00:20:48,830 be that of a progressive wave. 292 00:20:48,830 --> 00:20:54,490 Specifically, a shape like this which 293 00:20:54,490 --> 00:20:59,180 moves to the right with a velocity square root of T 294 00:20:59,180 --> 00:21:00,580 over mu. 295 00:21:00,580 --> 00:21:08,390 And I cannot resist emphasizing that no part of the string is 296 00:21:08,390 --> 00:21:12,240 moving here to the right. 297 00:21:12,240 --> 00:21:16,870 The string is only moving in the transverse direction. 298 00:21:16,870 --> 00:21:21,710 But it is the distortion which is moving with that velocity v 299 00:21:21,710 --> 00:21:22,210 . 300 00:21:22,210 --> 00:21:23,501 It's called the phase velocity. 301 00:21:26,220 --> 00:21:27,690 All right. 302 00:21:27,690 --> 00:21:32,100 So our knowledge of the solutions 303 00:21:32,100 --> 00:21:39,080 of wave equations and these descriptions make sense. 304 00:21:39,080 --> 00:21:43,220 And so we know the class of solutions that describe this. 305 00:21:43,220 --> 00:21:48,950 In fact, it is the progressive wave solution of this problem. 306 00:21:48,950 --> 00:21:50,140 Fantastic. 307 00:21:50,140 --> 00:21:59,550 So we know what this string is doing at all times, a earlier 308 00:21:59,550 --> 00:22:03,250 times and later times. 309 00:22:03,250 --> 00:22:08,160 From now on, this triangle will just move forever like that. 310 00:22:08,160 --> 00:22:10,940 That will be the distortion. 311 00:22:10,940 --> 00:22:14,080 I can turn the clock around. 312 00:22:14,080 --> 00:22:18,950 I know that earlier it was here, here, et cetera. 313 00:22:18,950 --> 00:22:25,490 So I know the shape of this string at all times. 314 00:22:25,490 --> 00:22:32,140 Knowing that, I can-- it's a funny way of saying it, 315 00:22:32,140 --> 00:22:34,090 predict because it happened earlier. 316 00:22:34,090 --> 00:22:40,720 But I can tell you what the distortion 317 00:22:40,720 --> 00:22:42,730 must have been at earlier times. 318 00:22:47,880 --> 00:22:54,880 Going back with velocity v until-- up to the times 319 00:22:54,880 --> 00:22:58,530 when this distortion was here. 320 00:22:58,530 --> 00:23:03,580 Of course, the string was doing nothing. 321 00:23:03,580 --> 00:23:09,380 But at some instant of-- so the total picture is the following. 322 00:23:09,380 --> 00:23:12,600 We start off with the taut string. 323 00:23:12,600 --> 00:23:16,600 At some instant of time, the person 324 00:23:16,600 --> 00:23:21,290 must have moved the string with uniform velocity 325 00:23:21,290 --> 00:23:26,610 up, which I call the transverse velocity the transverse. 326 00:23:26,610 --> 00:23:33,260 That'll start this wave front. 327 00:23:33,260 --> 00:23:40,210 You will have come to a stop and start moving downwards again, 328 00:23:40,210 --> 00:23:46,350 formed this triangle, and then I do no more. 329 00:23:46,350 --> 00:23:50,640 This triangle will then progress forward. 330 00:23:50,640 --> 00:23:56,190 It will progress with a phase velocity v 331 00:23:56,190 --> 00:24:01,050 with this velocity, which we know. 332 00:24:01,050 --> 00:24:04,100 It's not something I did in my head. 333 00:24:04,100 --> 00:24:12,650 I can derive this by studying how 334 00:24:12,650 --> 00:24:21,520 a string can be described by the wave equation. 335 00:24:21,520 --> 00:24:23,460 But for this instant, you can say 336 00:24:23,460 --> 00:24:26,200 I took it from a book or somewhere, 337 00:24:26,200 --> 00:24:28,740 or from the lectures of Professor Walter Lewin. 338 00:24:28,740 --> 00:24:32,860 The phase velocity of a pulse on this string 339 00:24:32,860 --> 00:24:37,270 is-- of an ideal string is the square root of the tension 340 00:24:37,270 --> 00:24:40,940 divided in mass per unit length. 341 00:24:40,940 --> 00:24:43,320 OK. 342 00:24:43,320 --> 00:24:50,450 And I was telling you a second ago, when I go back in time 343 00:24:50,450 --> 00:24:56,460 at any instant of time when they ring was moving, 344 00:24:56,460 --> 00:25:02,400 this string has that shape. 345 00:25:02,400 --> 00:25:12,760 And as I move the ring up and generate this wave front, 346 00:25:12,760 --> 00:25:16,610 every piece of the string is moving up 347 00:25:16,610 --> 00:25:21,720 with a transverse velocity v tr, which I don't know. 348 00:25:21,720 --> 00:25:28,660 But the result is that this part of the pulse 349 00:25:28,660 --> 00:25:33,330 moves to the right with velocity [INAUDIBLE], 350 00:25:33,330 --> 00:25:37,330 phase velocity of the pulse. 351 00:25:37,330 --> 00:25:39,880 Don't get confused. 352 00:25:39,880 --> 00:25:42,430 This piece of the string-- and I'm 353 00:25:42,430 --> 00:25:45,640 repeating myself-- is not moving to the right. 354 00:25:45,640 --> 00:25:50,280 This piece is moving up with this transverse velocity. 355 00:25:50,280 --> 00:25:53,900 This piece of the string is moving up. 356 00:25:53,900 --> 00:25:57,050 This piece isn't moving yet. 357 00:25:57,050 --> 00:26:01,080 But shortly afterwards, it will. 358 00:26:01,080 --> 00:26:04,820 And so the result of the movement 359 00:26:04,820 --> 00:26:09,510 of every piece of this string up is 360 00:26:09,510 --> 00:26:15,940 to produce a pulse which is moving to the right. 361 00:26:15,940 --> 00:26:17,770 OK. 362 00:26:17,770 --> 00:26:23,590 So now we want to calculate what is this transverse velocity. 363 00:26:23,590 --> 00:26:29,720 Well, from this picture it will be-- this velocity 364 00:26:29,720 --> 00:26:32,700 is related to that one by that angle. 365 00:26:32,700 --> 00:26:38,550 And so the motion of the ring is the transverse velocity 366 00:26:38,550 --> 00:26:42,680 at time t of the ring-- this part of it. 367 00:26:42,680 --> 00:26:44,730 OK, we're talking about this-- is 368 00:26:44,730 --> 00:26:50,370 equal to v times H over L over 2. 369 00:26:50,370 --> 00:26:52,050 That's the tangent of this angle. 370 00:26:52,050 --> 00:26:54,460 That's the relation between this velocity 371 00:26:54,460 --> 00:26:59,110 and this transverse velocity. 372 00:26:59,110 --> 00:27:02,400 Which is equal to 2v H over L. 373 00:27:02,400 --> 00:27:17,330 And this is the motion of this string from the time when this 374 00:27:17,330 --> 00:27:25,560 is back here until the time when this is moved for the ring 375 00:27:25,560 --> 00:27:32,700 to be at the maximum height H. If we remember that we were 376 00:27:32,700 --> 00:27:36,240 told that the shape of the pulse-- and I 377 00:27:36,240 --> 00:27:40,470 have to refer you now back to do original picture. 378 00:27:40,470 --> 00:27:43,950 If you look at that original picture, 379 00:27:43,950 --> 00:27:49,020 we were told that at time 4L over the square root of T 380 00:27:49,020 --> 00:27:54,480 over mu, that pulse was-- well, the back end of it 381 00:27:54,480 --> 00:28:01,270 was 2L away from the ring and the forward part was 3L. 382 00:28:01,270 --> 00:28:06,610 That pulse is moving with velocity v. 383 00:28:06,610 --> 00:28:10,290 And therefore, we can calculate the time 384 00:28:10,290 --> 00:28:13,550 when the different parts of that pulse 385 00:28:13,550 --> 00:28:17,090 were generated by the ring. 386 00:28:17,090 --> 00:28:27,230 From the time L over v until 3L over 2v, this part of the pulse 387 00:28:27,230 --> 00:28:29,520 was generated. 388 00:28:29,520 --> 00:28:35,230 And so the transverse velocity during that period of the ring 389 00:28:35,230 --> 00:28:40,450 is 2v H over L. 390 00:28:40,450 --> 00:28:43,570 Later on, now the ring is at the top 391 00:28:43,570 --> 00:28:47,910 and just the front edge of the pulse has been generated. 392 00:28:47,910 --> 00:28:54,040 We now suddenly-- this is an idealized situation. 393 00:28:54,040 --> 00:28:58,660 So this ring was going up with uniform velocity. 394 00:28:58,660 --> 00:29:01,240 And it suddenly reverses direction. 395 00:29:01,240 --> 00:29:03,320 Obviously, that's not the physical situation. 396 00:29:03,320 --> 00:29:05,420 That's infinite acceleration. 397 00:29:05,420 --> 00:29:10,250 But we can imagine it happens very fast that it's 398 00:29:10,250 --> 00:29:13,810 almost infinite acceleration. 399 00:29:13,810 --> 00:29:17,800 And so suddenly, we change from the ring, 400 00:29:17,800 --> 00:29:23,870 pulling the string upwards to the ring pulling the string 401 00:29:23,870 --> 00:29:25,120 downwards. 402 00:29:25,120 --> 00:29:30,490 We are now creating the back edge of that pulse. 403 00:29:30,490 --> 00:29:35,820 The situation is symmetric to the previous case. 404 00:29:35,820 --> 00:29:41,030 The only difference is that it's in the opposite direction. 405 00:29:41,030 --> 00:29:43,670 And we know that the magnitude of the two 406 00:29:43,670 --> 00:29:47,060 are the same because the pulse is symmetric. 407 00:29:47,060 --> 00:29:51,450 In one case, we are producing a wave front 408 00:29:51,450 --> 00:29:53,360 which is at some angle like that. 409 00:29:53,360 --> 00:29:56,350 And then the other like this. 410 00:29:56,350 --> 00:29:58,440 But they're symmetric. 411 00:29:58,440 --> 00:30:05,140 The angles are the same, but in the opposite direction. 412 00:30:05,140 --> 00:30:08,160 So the transverse velocity here now 413 00:30:08,160 --> 00:30:12,200 is therefore minus 2v H over L. And this 414 00:30:12,200 --> 00:30:14,680 happens this time interval. 415 00:30:18,370 --> 00:30:26,270 By the time we've reached time 2L over V, 416 00:30:26,270 --> 00:30:35,080 the complete pulse has left the ring. 417 00:30:35,080 --> 00:30:39,960 From then on, the ring must be stationary. 418 00:30:39,960 --> 00:30:42,735 Because if it moved, it would produce new distortions 419 00:30:42,735 --> 00:30:45,170 in the string. 420 00:30:45,170 --> 00:30:57,030 OK, so we have completely described 421 00:30:57,030 --> 00:30:58,280 the motion of the ring. 422 00:30:58,280 --> 00:31:01,270 That was the first part of the problem. 423 00:31:01,270 --> 00:31:06,320 And I repeat quickly, you have a string which is straight. 424 00:31:06,320 --> 00:31:08,870 I'm holding the ring. 425 00:31:08,870 --> 00:31:09,990 I'm waiting, waiting. 426 00:31:09,990 --> 00:31:12,240 At some time when I feel like it, 427 00:31:12,240 --> 00:31:18,716 I start moving it with uniform velocity up. 428 00:31:18,716 --> 00:31:23,610 I then reverse instantaneously and move with uniform velocity 429 00:31:23,610 --> 00:31:25,250 down. 430 00:31:25,250 --> 00:31:30,870 When I'm moving up, I generate the front part of that pulse. 431 00:31:30,870 --> 00:31:35,560 When I'm moving down, I generate the back part. 432 00:31:35,560 --> 00:31:36,650 Then, I stop. 433 00:31:36,650 --> 00:31:38,240 I'm doing nothing. 434 00:31:38,240 --> 00:31:39,870 The ring is doing nothing more. 435 00:31:39,870 --> 00:31:43,350 The string is attached to it horizontally. 436 00:31:43,350 --> 00:31:48,930 But as a result of that motion of the ring, 437 00:31:48,930 --> 00:31:54,710 I've introduced a triangular distortion on the string. 438 00:31:54,710 --> 00:31:59,780 That distortion will now propagate forever 439 00:31:59,780 --> 00:32:03,506 with the phase velocity square root T over mu. 440 00:32:07,600 --> 00:32:15,960 I don't have to discuss how this pulse is progressing forward. 441 00:32:15,960 --> 00:32:18,920 That we've done in general, or Professor Walter Lewin 442 00:32:18,920 --> 00:32:20,180 has done in general. 443 00:32:20,180 --> 00:32:24,620 He has shown that there is this progressive wave 444 00:32:24,620 --> 00:32:31,600 solution to the wave equation, which describes this string. 445 00:32:31,600 --> 00:32:36,160 OK, so we've finally understood the motion 446 00:32:36,160 --> 00:32:44,230 of the ring and the subsequent motion of the pulse progressing 447 00:32:44,230 --> 00:32:45,850 down the string. 448 00:32:45,850 --> 00:32:47,520 End of that. 449 00:32:47,520 --> 00:32:52,140 The next question was the power delivered by the string. 450 00:32:52,140 --> 00:32:56,950 Well, clearly, during the time when the ring is stationary, 451 00:32:56,950 --> 00:33:00,160 isn't doing no work on anything. 452 00:33:00,160 --> 00:33:05,170 My finger doesn't have to do any work on the ring or the string. 453 00:33:05,170 --> 00:33:07,610 There's no power generating. 454 00:33:07,610 --> 00:33:16,300 But during the time when I was moving the ring up, 455 00:33:16,300 --> 00:33:20,240 I had to exert a force. 456 00:33:20,240 --> 00:33:24,155 So I was pulling with a force. 457 00:33:27,190 --> 00:33:29,870 Therefore, I was doing work. 458 00:33:29,870 --> 00:33:36,110 And the rate of doing work is the power that I'm generating. 459 00:33:36,110 --> 00:33:39,220 Similarly when I'm going down, I will 460 00:33:39,220 --> 00:33:43,810 have to do work against the force, the tension 461 00:33:43,810 --> 00:33:48,910 in the string, and I will be exerting power 462 00:33:48,910 --> 00:33:50,290 during that time. 463 00:33:50,290 --> 00:33:52,700 How do we calculate that? 464 00:33:52,700 --> 00:33:57,410 Well, I repeat what I said earlier, 465 00:33:57,410 --> 00:33:59,580 so I don't have to look at the board. 466 00:33:59,580 --> 00:34:07,030 The instantaneous power generated or exerted by a force 467 00:34:07,030 --> 00:34:11,920 is force dot the velocity. 468 00:34:11,920 --> 00:34:13,620 OK. 469 00:34:13,620 --> 00:34:17,320 Now, we did calculate the force. 470 00:34:23,417 --> 00:34:27,420 Sorry, we can calculate the force 471 00:34:27,420 --> 00:34:33,300 that I have to generate to move this ring. 472 00:34:33,300 --> 00:34:36,370 I'm just redrawing the sketch for you. 473 00:34:36,370 --> 00:34:38,139 Here is the ring. 474 00:34:38,139 --> 00:34:39,250 What are the forces? 475 00:34:39,250 --> 00:34:42,150 I am repeating, there is the reaction force here. 476 00:34:42,150 --> 00:34:45,980 That's this force F which I am exerting. 477 00:34:45,980 --> 00:34:49,679 And I've chosen this picture at the time 478 00:34:49,679 --> 00:34:52,770 when the string is like this. 479 00:34:52,770 --> 00:34:57,440 At other times, the string will be in that direction. 480 00:34:57,440 --> 00:34:59,420 But I've sketched this at a time when 481 00:34:59,420 --> 00:35:01,560 the string is in this direction. 482 00:35:01,560 --> 00:35:02,955 That is the tension in it. 483 00:35:02,955 --> 00:35:05,770 It exerts a force. 484 00:35:05,770 --> 00:35:07,940 So those are the three forces. 485 00:35:07,940 --> 00:35:11,970 And as I've discussed earlier, this is a massless ring. 486 00:35:11,970 --> 00:35:14,980 Therefore, there's no net force. 487 00:35:14,980 --> 00:35:19,990 At this instant, this force must be equal 488 00:35:19,990 --> 00:35:25,020 and in the opposite direction than the vertical component 489 00:35:25,020 --> 00:35:31,230 of this tension T. So it is equal to T 490 00:35:31,230 --> 00:35:33,590 times the tangent of this angle. 491 00:35:33,590 --> 00:35:39,260 So at that instant of time, the magnitude of this force-- 492 00:35:39,260 --> 00:35:45,470 and it's in the up direction-- is plus T H over L over 2, 493 00:35:45,470 --> 00:35:49,050 which is 2H T over L. 494 00:35:49,050 --> 00:35:52,320 This corresponds to the period when 495 00:35:52,320 --> 00:35:55,020 the string is like this at the ring. 496 00:35:55,020 --> 00:35:58,530 And earlier on, I told you that corresponds 497 00:35:58,530 --> 00:36:02,190 to this time interval. 498 00:36:02,190 --> 00:36:03,560 OK. 499 00:36:03,560 --> 00:36:05,650 Later, I've reached the top. 500 00:36:05,650 --> 00:36:08,520 I'm starting to move down. 501 00:36:08,520 --> 00:36:13,590 The string now is upwards, but the angle is the same. 502 00:36:13,590 --> 00:36:16,160 And so the force that I have to exert 503 00:36:16,160 --> 00:36:19,670 is the same magnitude as this but in the opposite direction. 504 00:36:19,670 --> 00:36:21,880 Hence, the minus sign. 505 00:36:21,880 --> 00:36:25,040 So it's minus T over that, which is minus 506 00:36:25,040 --> 00:36:29,550 2H T over L. The two are, of course, equal. 507 00:36:29,550 --> 00:36:33,630 So I first have to-- it actually makes 508 00:36:33,630 --> 00:36:35,550 sense if you imagine a string. 509 00:36:35,550 --> 00:36:41,680 If I want to distort it up, I have to pull on that. 510 00:36:41,680 --> 00:36:45,460 Later on, I want to pull it in the opposite. 511 00:36:45,460 --> 00:36:47,390 I pull it down. 512 00:36:47,390 --> 00:36:51,570 Those are the two types. 513 00:36:51,570 --> 00:36:55,070 So we know at every instant of time 514 00:36:55,070 --> 00:36:58,960 what is the force that I exert and the direction of it. 515 00:37:02,430 --> 00:37:09,810 Now, the power that I generate is the force dot velocity. 516 00:37:09,810 --> 00:37:16,580 So I have to take the dot product between this force 517 00:37:16,580 --> 00:37:20,660 and the corresponding velocity. 518 00:37:20,660 --> 00:37:23,290 All the motion is up and down, so I 519 00:37:23,290 --> 00:37:27,200 don't have to-- I'm doing just the components. 520 00:37:27,200 --> 00:37:30,170 And so the dot product is just the multiplication 521 00:37:30,170 --> 00:37:32,370 of the force times the velocity. 522 00:37:32,370 --> 00:37:34,430 And the only thing I have to watch out for 523 00:37:34,430 --> 00:37:36,710 is whether they're in the same direction, 524 00:37:36,710 --> 00:37:38,490 in the opposite direction. 525 00:37:38,490 --> 00:37:45,520 Notice that when the transverse velocity is 526 00:37:45,520 --> 00:37:49,530 positive-- in other words, upwards-- that's 527 00:37:49,530 --> 00:37:52,920 the time when I'm pulling upwards. 528 00:37:52,920 --> 00:37:59,590 And so the force and the velocity have the same sign. 529 00:37:59,590 --> 00:38:04,660 So the power I exert is this 2H T 530 00:38:04,660 --> 00:38:11,130 over L times the transverse velocity 2v H over L. OK? 531 00:38:14,342 --> 00:38:28,650 Now, afterwards when we change to the other-- the back edge, 532 00:38:28,650 --> 00:38:31,820 the force reverses. 533 00:38:31,820 --> 00:38:34,180 It's minus. 534 00:38:34,180 --> 00:38:36,900 So that's the transverse velocity. 535 00:38:36,900 --> 00:38:40,810 And so that product, once again, is positive. 536 00:38:40,810 --> 00:38:48,690 So both on the leading edge of that pulse and the back edge, 537 00:38:48,690 --> 00:38:53,270 I do positive amount of work. 538 00:38:53,270 --> 00:38:57,770 I do work on the ring and not the ring on me. 539 00:38:57,770 --> 00:39:02,390 And of course, immediately that power 540 00:39:02,390 --> 00:39:04,590 is transmitted to the string. 541 00:39:04,590 --> 00:39:08,950 So for both the front edge and the back edge of the piles, 542 00:39:08,950 --> 00:39:12,140 I have to do work on the string. 543 00:39:12,140 --> 00:39:15,080 It's positive. 544 00:39:15,080 --> 00:39:20,260 And at every instant of time during my motion, 545 00:39:20,260 --> 00:39:24,420 this is the power. 546 00:39:24,420 --> 00:39:29,690 It's the product of the force times the transverse velocity. 547 00:39:29,690 --> 00:39:33,950 So this now is true for the complete time 548 00:39:33,950 --> 00:39:40,370 period during which I am generating the pulse. 549 00:39:40,370 --> 00:39:41,760 OK. 550 00:39:41,760 --> 00:39:48,690 I can now plot what is the power that I 551 00:39:48,690 --> 00:39:51,350 have to exert as a function of time. 552 00:39:51,350 --> 00:39:56,110 So here is the power and in this direction is time. 553 00:39:56,110 --> 00:39:58,990 And what we've shown from a time of L 554 00:39:58,990 --> 00:40:04,440 over v to a time 2L over v. And by the way, 555 00:40:04,440 --> 00:40:10,770 just I'm reminding you, halfway through-- this 556 00:40:10,770 --> 00:40:12,840 is when I'm moving it in one direction. 557 00:40:12,840 --> 00:40:13,750 This is in the other. 558 00:40:13,750 --> 00:40:16,690 But the power is positive in both cases. 559 00:40:16,690 --> 00:40:21,820 So this is the power as a function of time. 560 00:40:21,820 --> 00:40:24,890 And so what is the total energy that I 561 00:40:24,890 --> 00:40:32,260 have to give to the string to produce that pulse? 562 00:40:32,260 --> 00:40:37,360 Well, total energy given is the integral 563 00:40:37,360 --> 00:40:42,140 of the power over the time during which I 564 00:40:42,140 --> 00:40:43,720 exert that power. 565 00:40:43,720 --> 00:40:50,360 So it's the integral of P of t dt over that interval. 566 00:40:50,360 --> 00:40:53,690 Well, that's just the area of that pulse. 567 00:40:53,690 --> 00:40:56,480 So it's this height times this time. 568 00:40:56,480 --> 00:41:02,520 And if I do that, I get-- if I take this and multiply it 569 00:41:02,520 --> 00:41:08,540 by that length, I get 4H squared T over L. 570 00:41:08,540 --> 00:41:12,600 All right, so we've now understood what the ring does. 571 00:41:12,600 --> 00:41:18,740 We understand how much energy I had 572 00:41:18,740 --> 00:41:24,510 to apply to it in order to generate that pulse. 573 00:41:24,510 --> 00:41:31,150 OK, finally we were asked about something about the pulse. 574 00:41:31,150 --> 00:41:34,820 I've put some energy into it, where did that energy go? 575 00:41:34,820 --> 00:41:40,690 That energy went into the potential energy stored because 576 00:41:40,690 --> 00:41:44,990 of the shape of the pulse and the kinetic energy in it. 577 00:41:44,990 --> 00:41:49,180 So the essence of the next parts of the question 578 00:41:49,180 --> 00:41:54,320 is, let's now, from first principles-- knowing 579 00:41:54,320 --> 00:41:58,930 what pulse we have, let's try to calculate how much energy we've 580 00:41:58,930 --> 00:42:03,150 stored in the pulse in the form of potential energy. 581 00:42:03,150 --> 00:42:06,230 How much energy we've stored in the pulse 582 00:42:06,230 --> 00:42:08,670 in the form of kinetic energy. 583 00:42:08,670 --> 00:42:11,330 And just to check that we didn't make a mistake, 584 00:42:11,330 --> 00:42:13,140 let's see that energy is conserved. 585 00:42:13,140 --> 00:42:15,930 We know the total energy we've put into the system. 586 00:42:15,930 --> 00:42:19,080 Let's see whether that's equal to the potential energy 587 00:42:19,080 --> 00:42:20,580 plus the kinetic energy. 588 00:42:20,580 --> 00:42:22,550 Of course, it's going to work out right, 589 00:42:22,550 --> 00:42:25,150 so it's more a test I haven't made a mistake. 590 00:42:25,150 --> 00:42:27,930 And we are not going to discover non-conservation of energy 591 00:42:27,930 --> 00:42:29,140 here. 592 00:42:29,140 --> 00:42:33,350 OK, so now let's try to calculate 593 00:42:33,350 --> 00:42:36,930 how much potential energy is stored in the pulse. 594 00:42:44,740 --> 00:42:48,070 OK, you can do it many ways. 595 00:42:48,070 --> 00:42:51,810 But normally, I like going to first principles. 596 00:42:51,810 --> 00:42:55,840 This string normally in equilibrium is straight. 597 00:42:58,360 --> 00:43:01,780 In the moving pulse, it is distorted. 598 00:43:01,780 --> 00:43:04,560 It's this shape. 599 00:43:04,560 --> 00:43:10,000 In order to distort it to this shape-- and ti doesn't matter, 600 00:43:10,000 --> 00:43:12,600 I'm talking at some instant of time. 601 00:43:12,600 --> 00:43:15,560 I don't care whether anything is moving here or not. 602 00:43:15,560 --> 00:43:20,540 At some instant of time, this has a certain amount 603 00:43:20,540 --> 00:43:22,130 of potential energy. 604 00:43:22,130 --> 00:43:25,590 It is the energy stored in the system 605 00:43:25,590 --> 00:43:32,100 because it is distorted from its equilibrium straight path. 606 00:43:32,100 --> 00:43:37,250 I don't care whether any piece here is moving or not. 607 00:43:37,250 --> 00:43:38,880 That's nothing to the potential energy. 608 00:43:38,880 --> 00:43:42,370 How much energy is there if I have a piece of string shaped 609 00:43:42,370 --> 00:43:44,990 like this whether it's part of the pulse or not? 610 00:43:44,990 --> 00:43:46,480 Irrelevant. 611 00:43:46,480 --> 00:43:49,430 OK, how do we do it? 612 00:43:49,430 --> 00:43:54,000 I will do a thought experiment, a Gedanken experiment. 613 00:43:54,000 --> 00:43:59,730 I will imagine I took a string, nailed these points. 614 00:43:59,730 --> 00:44:04,840 They don't move separated by a distance L. In the middle, 615 00:44:04,840 --> 00:44:10,000 I get hold of this and from it being straight, 616 00:44:10,000 --> 00:44:13,040 I start pulling up. 617 00:44:13,040 --> 00:44:14,060 All right? 618 00:44:14,060 --> 00:44:17,510 Until I reach a height H. 619 00:44:17,510 --> 00:44:21,660 At that stage, I've generated this shape. 620 00:44:21,660 --> 00:44:24,950 The potential energy in that shape 621 00:44:24,950 --> 00:44:31,070 must come from the work I did in distorting it. 622 00:44:31,070 --> 00:44:33,070 How much work do I do? 623 00:44:33,070 --> 00:44:38,180 Well, I have to exert a force pulling it up. 624 00:44:38,180 --> 00:44:43,890 And the integral of that force times the total distance, that 625 00:44:43,890 --> 00:44:46,450 will be the total work I have done. 626 00:44:46,450 --> 00:44:48,250 So that's what I have to do. 627 00:44:48,250 --> 00:44:48,750 All right? 628 00:44:48,750 --> 00:44:51,430 So what I'm doing? 629 00:44:51,430 --> 00:45:00,110 I'm going to integrate from this height being 0 to the height 630 00:45:00,110 --> 00:45:03,310 being H. I mean, integral is just 631 00:45:03,310 --> 00:45:09,960 the addition of bits of work as I pull this up. 632 00:45:09,960 --> 00:45:20,270 So integral of the force, which I call little f of y dy. 633 00:45:20,270 --> 00:45:22,325 Now, what is the force? 634 00:45:26,012 --> 00:45:27,720 I'm not [INAUDIBLE] where I'm doing this. 635 00:45:27,720 --> 00:45:32,640 I'm doing it, pulling it up very slowly at uniform velocity. 636 00:45:32,640 --> 00:45:35,690 So I'm not accelerating anything. 637 00:45:35,690 --> 00:45:43,520 So the net force on this must be equal and opposite to the force 638 00:45:43,520 --> 00:45:48,220 exacted on my fingers by this string. 639 00:45:48,220 --> 00:45:51,110 Well, there is a tension in this string 640 00:45:51,110 --> 00:45:53,090 pulling in these directions. 641 00:45:53,090 --> 00:45:55,440 The horizontal components of the tensions 642 00:45:55,440 --> 00:46:00,160 cancel, but the vertical ones add up. 643 00:46:00,160 --> 00:46:03,810 The two vertical components of the string on this side 644 00:46:03,810 --> 00:46:08,422 and that are equal and opposite to my force f. 645 00:46:11,670 --> 00:46:14,720 Now, what we said, the assumption 646 00:46:14,720 --> 00:46:18,650 here is that this distortion is so small 647 00:46:18,650 --> 00:46:21,480 that in the process of me pulling this and changing it, 648 00:46:21,480 --> 00:46:24,020 this tension does not change. 649 00:46:24,020 --> 00:46:27,490 In reality, it would change slightly. 650 00:46:27,490 --> 00:46:30,910 But if this is small enough, it's negligible. 651 00:46:30,910 --> 00:46:32,460 So I'm going to make the assumption 652 00:46:32,460 --> 00:46:37,010 that this tension does not change as I'm moving this up. 653 00:46:37,010 --> 00:46:42,120 So the force I'll be applying will be constant, actually. 654 00:46:42,120 --> 00:46:44,220 And so what that is? 655 00:46:44,220 --> 00:46:49,630 Well, it's twice the tension times the component of it, 656 00:46:49,630 --> 00:46:55,530 this tangent of the angle in the vertical direction from 0 to H 657 00:46:55,530 --> 00:46:56,740 times dy. 658 00:46:56,740 --> 00:47:02,105 But as I say, T will be constant throughout this period. 659 00:47:06,860 --> 00:47:13,310 But y-- there is a y here, which is not a constant. 660 00:47:13,310 --> 00:47:16,710 This goes from 0 to H. And so if I integrate 661 00:47:16,710 --> 00:47:23,830 that with respect to y, I'll get y squared over 2 from 0 to H. 662 00:47:23,830 --> 00:47:30,590 And so this integral is equal 4T H squared over L over 2. 663 00:47:30,590 --> 00:47:33,770 This 2 is from the integral of y dy. 664 00:47:33,770 --> 00:47:38,650 It's y squared over 2 taken from those limits gives me that. 665 00:47:38,650 --> 00:47:43,550 OK, which is 2H squared T over L. 666 00:47:43,550 --> 00:47:48,930 This is the work I did distorting the string. 667 00:47:48,930 --> 00:47:52,880 It must be equal to the energy stored 668 00:47:52,880 --> 00:47:55,070 in the string in that process. 669 00:47:55,070 --> 00:47:58,170 OK, so we've answered that part. 670 00:47:58,170 --> 00:48:00,640 How about the kinetic energy? 671 00:48:04,790 --> 00:48:09,000 There is a pulse which is moving to the right. 672 00:48:09,000 --> 00:48:13,860 Meaning that every piece of this string from here to here 673 00:48:13,860 --> 00:48:16,240 is actually in motion. 674 00:48:16,240 --> 00:48:17,790 Over here, it's stationary. 675 00:48:17,790 --> 00:48:20,200 Over here, it's stationary. 676 00:48:20,200 --> 00:48:25,890 But to the left of center, every piece in here 677 00:48:25,890 --> 00:48:29,240 it's actually moving down. 678 00:48:29,240 --> 00:48:32,580 We discussed that earlier. 679 00:48:32,580 --> 00:48:35,145 Every piece along here is moving up. 680 00:48:38,050 --> 00:48:40,820 The result is you get the impression 681 00:48:40,820 --> 00:48:44,900 of the pulse moving to the right. 682 00:48:44,900 --> 00:48:53,680 But the actual motion of the mass is down here and up here. 683 00:48:53,680 --> 00:48:59,670 So at any instant of time when you have this pulse, 684 00:48:59,670 --> 00:49:03,930 there is kinetic energy of the string in this piece 685 00:49:03,930 --> 00:49:07,020 and in that piece. 686 00:49:07,020 --> 00:49:09,610 And we have to calculate that. 687 00:49:09,610 --> 00:49:11,920 They are both positive kinetic energy. 688 00:49:11,920 --> 00:49:14,690 There is no such thing as negative kinetic energy. 689 00:49:14,690 --> 00:49:15,700 So those are moving. 690 00:49:15,700 --> 00:49:19,550 And the transverse velocity is down and here up. 691 00:49:19,550 --> 00:49:22,470 Kinetic energy is 1/2 mass times velocity squared. 692 00:49:22,470 --> 00:49:24,690 They're both positive, OK? 693 00:49:24,690 --> 00:49:27,960 So let's now just calculate. 694 00:49:27,960 --> 00:49:31,315 Again, it will be the integral, the sum 695 00:49:31,315 --> 00:49:34,230 of the kinetic energy of every piece along here. 696 00:49:34,230 --> 00:49:39,930 So if I take a piece of mass dm multiply it 697 00:49:39,930 --> 00:49:45,400 by 1/2 times its transverse-- the motion of that mass 698 00:49:45,400 --> 00:49:47,250 squared. 699 00:49:47,250 --> 00:49:50,000 That's the total velocity of that mass. 700 00:49:50,000 --> 00:49:51,460 It's not moving horizontally. 701 00:49:51,460 --> 00:49:55,890 We've made that assumption in this idealized case. 702 00:49:55,890 --> 00:50:00,055 So this is the kinetic energy of a piece of the mass. 703 00:50:02,830 --> 00:50:06,940 And I have to integrate that over all the pieces of m 704 00:50:06,940 --> 00:50:08,460 from here to here. 705 00:50:08,460 --> 00:50:11,610 There's no kinetic energy there and there. 706 00:50:11,610 --> 00:50:13,900 OK, this is a constant, actually. 707 00:50:20,480 --> 00:50:23,010 It's only true because this is a straight line 708 00:50:23,010 --> 00:50:25,830 that this pulse consists of two straight lines. 709 00:50:25,830 --> 00:50:29,280 If it wasn't, it would not be constant. 710 00:50:29,280 --> 00:50:34,090 So this integral is that 1/2 times 711 00:50:34,090 --> 00:50:36,540 this transverse velocity, which we 712 00:50:36,540 --> 00:50:41,860 calculated earlier the transverse. 713 00:50:41,860 --> 00:50:46,342 We've calculated over here the transverse velocity. 714 00:50:46,342 --> 00:50:51,740 And you can see it's independent of position of the string. 715 00:50:51,740 --> 00:50:52,790 So back to here. 716 00:50:52,790 --> 00:50:55,660 So we know that transverse velocity. 717 00:50:55,660 --> 00:50:58,180 And what is the piece of mass? 718 00:50:58,180 --> 00:51:01,955 Well, that will be the density of the string, mass 719 00:51:01,955 --> 00:51:06,120 per unit length mu times the little piece of string 720 00:51:06,120 --> 00:51:07,330 of length l. 721 00:51:10,900 --> 00:51:14,360 This is a trivial integral, so they're all constants here. 722 00:51:20,310 --> 00:51:22,320 By the way, this integral-- maybe 723 00:51:22,320 --> 00:51:28,686 I should've said it-- goes over a length from 0 to L. 724 00:51:28,686 --> 00:51:32,370 That's the length of the string we're considering. 725 00:51:32,370 --> 00:51:36,270 So we get up with that equal to that, which 726 00:51:36,270 --> 00:51:45,190 is equal to-- because I could replace v. We 727 00:51:45,190 --> 00:51:50,250 know v. v is the square root of T over mu. 728 00:51:50,250 --> 00:51:55,120 And so if I replace v by v squared by T over mu, 729 00:51:55,120 --> 00:51:58,020 it cancels the mu, but we get a T here. 730 00:51:58,020 --> 00:52:03,760 And so the total energy, kinetic energy, is this. 731 00:52:03,760 --> 00:52:05,900 And we are essentially whole. 732 00:52:05,900 --> 00:52:12,090 Because now we've calculated the total potential energy 733 00:52:12,090 --> 00:52:13,520 stored in the pulse. 734 00:52:13,520 --> 00:52:17,430 We've calculated the total kinetic energy. 735 00:52:17,430 --> 00:52:21,360 So the total energy stored in that policy 736 00:52:21,360 --> 00:52:25,990 is the sum of those two, which is of course, 4H squared 737 00:52:25,990 --> 00:52:33,780 T over L. So in other words, as this pulse moves along, it 738 00:52:33,780 --> 00:52:38,800 has a total energy of this plus that 4H squared T over L. 739 00:52:38,800 --> 00:52:40,610 And it will continue forever. 740 00:52:40,610 --> 00:52:45,190 Energy is conserved, it will go forever. 741 00:52:45,190 --> 00:52:47,150 Where did it come from? 742 00:52:47,150 --> 00:52:55,250 We said it came from the motion of the ring. 743 00:52:55,250 --> 00:53:00,430 And I calculated how much energy I did initially. 744 00:53:00,430 --> 00:53:07,900 It was 4H squared T over L. And lo and behold, big surprise, 745 00:53:07,900 --> 00:53:11,500 it's exactly what's in the pulse. 746 00:53:11,500 --> 00:53:17,300 So the work I do is right at the beginning. 747 00:53:17,300 --> 00:53:21,850 The work I do is at the time when I generate the pulse. 748 00:53:21,850 --> 00:53:26,610 From then on, the energy just progresses forever. 749 00:53:26,610 --> 00:53:32,000 That's the end to that problem. 750 00:53:32,000 --> 00:53:34,710 We'll now deal with the next problem. 751 00:53:34,710 --> 00:53:37,580 So we now come to the second problem 752 00:53:37,580 --> 00:53:40,030 to do with progressive waves. 753 00:53:40,030 --> 00:53:43,670 And the problem I've taken is the following. 754 00:53:43,670 --> 00:53:47,560 You have two coaxial cables. 755 00:53:47,560 --> 00:53:50,530 They're connected at some place. 756 00:53:50,530 --> 00:53:55,820 And we are told that on the left cable-- this one here, 757 00:53:55,820 --> 00:54:02,490 left cable-- there is a voltage at every location given 758 00:54:02,490 --> 00:54:10,240 by v1 cosine omega 1t minus 2 pi over lambda 1t. 759 00:54:10,240 --> 00:54:12,550 And there is some kind of reflection to it. 760 00:54:12,550 --> 00:54:15,170 They don't tell us much about it. 761 00:54:15,170 --> 00:54:17,340 On the right-hand side, they tell 762 00:54:17,340 --> 00:54:22,790 us the voltage wave v2 cosine omega 763 00:54:22,790 --> 00:54:27,060 2t minus 2 pi lambda 2 over t. 764 00:54:27,060 --> 00:54:31,080 And furthermore, they tell us that we 765 00:54:31,080 --> 00:54:36,680 have to assume that these cables are ideal and lossless. 766 00:54:36,680 --> 00:54:37,910 That's one. 767 00:54:37,910 --> 00:54:41,080 That the phase velocity on the left one 768 00:54:41,080 --> 00:54:47,620 is v1 and that whenever there is a voltage on it 769 00:54:47,620 --> 00:54:50,600 at any location, there is a current, 770 00:54:50,600 --> 00:54:55,360 which is v1 divided by some constant z1. 771 00:54:55,360 --> 00:54:59,770 So on the left one, the phase velocity v1 and I1 772 00:54:59,770 --> 00:55:01,360 is v1 over z1. 773 00:55:01,360 --> 00:55:07,340 And on the right cable, the phase velocity is v2 774 00:55:07,340 --> 00:55:12,230 and the current I2 is v2 over z2. 775 00:55:12,230 --> 00:55:18,700 The question is, what is the ratio of omega 2 to omega 1? 776 00:55:18,700 --> 00:55:22,290 What's the ratio of lambda 2 to lambda 1? 777 00:55:22,290 --> 00:55:25,550 And what's the ratio of v2 to v1? 778 00:55:25,550 --> 00:55:30,450 Now, you may have never seen a coaxial cable. 779 00:55:30,450 --> 00:55:32,010 You may have seen a coaxial cable, 780 00:55:32,010 --> 00:55:34,960 but you've never discussed it or learned about it. 781 00:55:34,960 --> 00:55:39,630 Certainly, Professor Walter Lewin did not cover this. 782 00:55:39,630 --> 00:55:41,390 Why did I do this problem? 783 00:55:41,390 --> 00:55:47,665 Precisely to show that by learning how to solve problems 784 00:55:47,665 --> 00:55:50,810 or understanding progressive waves 785 00:55:50,810 --> 00:55:54,020 on one system like this string, we 786 00:55:54,020 --> 00:55:57,130 can do transfer the whole knowledge 787 00:55:57,130 --> 00:56:03,910 we have to understanding how to analyze, in fact, a much more 788 00:56:03,910 --> 00:56:05,200 interesting situation. 789 00:56:05,200 --> 00:56:08,620 A situation you'll come across very often 790 00:56:08,620 --> 00:56:14,090 in life, progressive waves down coaxial cables. 791 00:56:14,090 --> 00:56:19,000 Now, from the way the problem is worded, 792 00:56:19,000 --> 00:56:21,390 we can in fact conclude that we should 793 00:56:21,390 --> 00:56:25,320 be able to solve this problem from our knowledge of what 794 00:56:25,320 --> 00:56:30,560 happens on taut string. 795 00:56:30,560 --> 00:56:33,610 We are told that on this one, there 796 00:56:33,610 --> 00:56:36,590 is a voltage, which has this form. 797 00:56:36,590 --> 00:56:38,640 If you look at this form, this is 798 00:56:38,640 --> 00:56:53,210 a function of some constants times time minus-- I'm sorry, 799 00:56:53,210 --> 00:56:54,930 this should be an x and an x. 800 00:56:54,930 --> 00:57:03,600 Let me immediately correct this, x here and x here. 801 00:57:03,600 --> 00:57:09,710 So it's a function of some constant times 802 00:57:09,710 --> 00:57:13,010 t minus a constant times x. 803 00:57:13,010 --> 00:57:19,470 And we know that such a function represents a progressive wave. 804 00:57:19,470 --> 00:57:21,960 If you plot this at different times 805 00:57:21,960 --> 00:57:24,410 and different locations, what you will see 806 00:57:24,410 --> 00:57:30,960 is a sinusoidal function which is 807 00:57:30,960 --> 00:57:35,570 moving to the right with some, what we call, phase velocity. 808 00:57:35,570 --> 00:57:38,390 Which in the problem, they tell us what it is. 809 00:57:38,390 --> 00:57:40,110 It's v1. 810 00:57:40,110 --> 00:57:42,230 So from this form, we immediately 811 00:57:42,230 --> 00:57:46,460 realize that the voltage on this cable 812 00:57:46,460 --> 00:57:50,450 is a progressive wave, which is going to the right. 813 00:57:50,450 --> 00:57:54,780 Therefore, this system, the cable, 814 00:57:54,780 --> 00:57:59,740 must have a wave equation for it to have 815 00:57:59,740 --> 00:58:05,760 a wave as a solution-- a progressive wave as a solution. 816 00:58:05,760 --> 00:58:06,840 OK, next. 817 00:58:11,790 --> 00:58:15,400 If we know omega 1, what is omega 2? 818 00:58:15,400 --> 00:58:18,400 If we know omega 1, what is lambda 1? 819 00:58:18,400 --> 00:58:19,690 What is lambda 2? 820 00:58:19,690 --> 00:58:24,380 And if we know v1, what is v2? 821 00:58:24,380 --> 00:58:30,410 In other words, we immediately realize 822 00:58:30,410 --> 00:58:35,740 that these two-- the angular frequencies here 823 00:58:35,740 --> 00:58:41,350 or the wavelengths of the progressive waves-- 824 00:58:41,350 --> 00:58:46,420 do not necessarily have the same phase velocity. 825 00:58:46,420 --> 00:58:50,270 And in fact, they tell us that it is different. 826 00:58:50,270 --> 00:58:55,140 So the propagation down these two cables will be different. 827 00:58:55,140 --> 00:58:59,380 And so these cables must be made structurally 828 00:58:59,380 --> 00:59:00,460 in a different way. 829 00:59:00,460 --> 00:59:03,210 They are different cables, but they're just 830 00:59:03,210 --> 00:59:05,590 connected together. 831 00:59:05,590 --> 00:59:06,740 OK. 832 00:59:06,740 --> 00:59:08,300 And of course, electrically-- that's 833 00:59:08,300 --> 00:59:13,190 the other thing I should-- the outside has to be connected. 834 00:59:13,190 --> 00:59:17,120 All right, how do we proceed? 835 00:59:17,120 --> 00:59:21,920 So as I started saying, from the wording of the problem 836 00:59:21,920 --> 00:59:27,780 we immediately conclude that a coaxial cable, in fact, 837 00:59:27,780 --> 00:59:32,730 is the continuum limit of some row of oscillators. 838 00:59:32,730 --> 00:59:34,420 Is that surprising? 839 00:59:34,420 --> 00:59:41,060 Well, if you stop and think and magnified the coaxial cable, 840 00:59:41,060 --> 00:59:45,280 in essence what you see is two parallel wires. 841 00:59:45,280 --> 00:59:47,970 They have a capacitance with each other. 842 00:59:47,970 --> 00:59:52,040 The two conductors have the central wire 843 00:59:52,040 --> 00:59:54,780 and the sheathing around that. 844 00:59:54,780 --> 00:59:58,430 Each have self-inductance. 845 00:59:58,430 --> 01:00:01,950 And so schematically, it looks something like this. 846 01:00:01,950 --> 01:00:07,980 You have a row of inductances, a row of capacitances 847 01:00:07,980 --> 01:00:09,040 connected like this. 848 01:00:09,040 --> 01:00:14,830 So this is a schematic diagram of a coaxial cable. 849 01:00:14,830 --> 01:00:15,650 What is this? 850 01:00:15,650 --> 01:00:19,820 Nothing other than each piece here 851 01:00:19,820 --> 01:00:24,980 is like a simple harmonic oscillator 852 01:00:24,980 --> 01:00:27,100 we've solved many times. 853 01:00:27,100 --> 01:00:30,760 It's simply an inductance in the capacitance. 854 01:00:30,760 --> 01:00:37,110 And they are furthermore, identical harmonic oscillators 855 01:00:37,110 --> 01:00:41,140 only connected to their neighbors. 856 01:00:41,140 --> 01:00:44,950 And so this, from a point of view 857 01:00:44,950 --> 01:00:48,120 of the response of this system, is 858 01:00:48,120 --> 01:00:54,460 identical to a taut string, where at each point in the s 859 01:00:54,460 --> 01:00:58,560 we had a harmonic oscillator which was connected only 860 01:00:58,560 --> 01:01:00,730 to its neighbors. 861 01:01:00,730 --> 01:01:05,060 So the equation of motion for this system-- 862 01:01:05,060 --> 01:01:09,670 well, the variable was the voltage, 863 01:01:09,670 --> 01:01:12,830 the potential difference between the central conductors 864 01:01:12,830 --> 01:01:17,050 and the outside conductor, must satisfy a wave equation 865 01:01:17,050 --> 01:01:23,300 like this where v is the phase velocity 866 01:01:23,300 --> 01:01:26,100 of the propagation of v down the cable. 867 01:01:26,100 --> 01:01:28,360 And we are told on the left-hand side 868 01:01:28,360 --> 01:01:32,290 is v1, on the right-hand side is v2. 869 01:01:32,290 --> 01:01:35,460 Furthermore, in the problem they told us 870 01:01:35,460 --> 01:01:43,470 that if there is a voltage propagating down this cable, 871 01:01:43,470 --> 01:01:46,660 there will also be a current propagating. 872 01:01:46,660 --> 01:01:51,780 And they tell us that the ratio of the voltage to the current 873 01:01:51,780 --> 01:01:53,800 is a constant. 874 01:01:53,800 --> 01:01:57,090 And it's different for the two cables. 875 01:01:57,090 --> 01:02:00,290 In fact, let me sort of digress and tell you 876 01:02:00,290 --> 01:02:02,830 it is the two quantities, the phase 877 01:02:02,830 --> 01:02:05,990 velocity and this constant z, which 878 01:02:05,990 --> 01:02:10,040 is called the characteristic impedance, which 879 01:02:10,040 --> 01:02:12,975 characterizes any coaxial cable. 880 01:02:12,975 --> 01:02:18,100 Or in fact, any system consisting of two parallel 881 01:02:18,100 --> 01:02:19,610 conductors. 882 01:02:19,610 --> 01:02:24,100 So a wire, typical two-conductor wire, 883 01:02:24,100 --> 01:02:27,510 would be a coaxial cable in that sense. 884 01:02:27,510 --> 01:02:32,230 Since I is proportional to the voltage, if this satisfies 885 01:02:32,230 --> 01:02:35,180 the wave equation so must the current. 886 01:02:35,180 --> 01:02:36,340 OK. 887 01:02:36,340 --> 01:02:38,980 Furthermore, what do they tell us? 888 01:02:38,980 --> 01:02:42,620 They tell us that if you look at the first cable, the one 889 01:02:42,620 --> 01:02:46,780 on the left, we see the progressive wave going 890 01:02:46,780 --> 01:02:48,030 to the right. 891 01:02:48,030 --> 01:02:52,280 But also, there may be a reflected wave they tell us. 892 01:02:52,280 --> 01:02:53,950 On the right-hand side, they tell 893 01:02:53,950 --> 01:02:58,210 us there is only a propagating wave to the right. 894 01:02:58,210 --> 01:03:01,000 From that, I can immediately conclude something 895 01:03:01,000 --> 01:03:03,240 about the boundary conditions of this system. 896 01:03:06,100 --> 01:03:08,520 Take the right-hand side. 897 01:03:08,520 --> 01:03:12,550 There's a wave going to the right, but nothing to the left. 898 01:03:12,550 --> 01:03:16,480 Therefore, there must be no reflection 899 01:03:16,480 --> 01:03:23,570 of the progressive wave at the far end of the right cable. 900 01:03:23,570 --> 01:03:25,220 How about the left cable? 901 01:03:25,220 --> 01:03:28,430 Well, the pulse is coming in. 902 01:03:28,430 --> 01:03:34,160 There is a junction between two cables which are not the same. 903 01:03:34,160 --> 01:03:38,420 And whenever you have such a situation where 904 01:03:38,420 --> 01:03:44,710 you have a continuous system of oscillators coming 905 01:03:44,710 --> 01:03:47,180 to another continuous system, like two 906 01:03:47,180 --> 01:03:49,620 wires of different density for example, 907 01:03:49,620 --> 01:03:52,940 or different tensions, or something of that kind, 908 01:03:52,940 --> 01:03:56,730 at the junction you are going to get a reflection. 909 01:03:56,730 --> 01:04:01,230 So where you're going to get a reflection at that junction, 910 01:04:01,230 --> 01:04:04,240 that reflected wave will go all the way to the left, 911 01:04:04,240 --> 01:04:07,220 come to the far end of that cable. 912 01:04:07,220 --> 01:04:09,480 And normally it would be reflected there 913 01:04:09,480 --> 01:04:12,960 unless one works hard to prevent that from happening. 914 01:04:12,960 --> 01:04:16,324 And here, they're telling us nothing 915 01:04:16,324 --> 01:04:20,420 is-- there's no reflection. 916 01:04:20,420 --> 01:04:24,730 There isn't any further reflection from that side. 917 01:04:24,730 --> 01:04:30,700 There is just the original pulse coming from left to the right. 918 01:04:30,700 --> 01:04:39,440 OK, so now, how do we answer the questions they do? 919 01:04:39,440 --> 01:04:43,360 And so now, I will go from what we've 920 01:04:43,360 --> 01:04:46,670 learned about the physical system 921 01:04:46,670 --> 01:04:50,610 to a mathematical description of it. 922 01:04:50,610 --> 01:04:53,910 And at this stage, you probably won't 923 01:04:53,910 --> 01:04:55,630 be able to almost tell the difference 924 01:04:55,630 --> 01:05:04,070 with I'm talking about-- two strings connected together 925 01:05:04,070 --> 01:05:08,110 or I'm talking a coaxial cable. 926 01:05:08,110 --> 01:05:11,212 The behavior, mathematical behavior, will be the same. 927 01:05:11,212 --> 01:05:13,270 The mathematical description will be the same. 928 01:05:15,810 --> 01:05:17,760 I should digress for a second. 929 01:05:17,760 --> 01:05:19,400 If you're curious what happens here, 930 01:05:19,400 --> 01:05:22,750 I was talking about two pulses going down, 931 01:05:22,750 --> 01:05:25,900 the voltage pulse and the current. 932 01:05:25,900 --> 01:05:27,500 What happens on the string? 933 01:05:27,500 --> 01:05:31,760 Well, in a string there are also many pulses. 934 01:05:31,760 --> 01:05:34,740 We normally don't talk about all of them. 935 01:05:34,740 --> 01:05:40,050 For example, if you have a propagating pulse 936 01:05:40,050 --> 01:05:45,440 of displacement on a string, then there 937 01:05:45,440 --> 01:05:49,860 will be also a pulse going down corresponding 938 01:05:49,860 --> 01:05:55,630 to the transverse velocity of the string at every point. 939 01:05:55,630 --> 01:05:57,510 So there's already two. 940 01:05:57,510 --> 01:05:59,220 There's another one. 941 01:05:59,220 --> 01:06:03,350 As the pulse on a string goes along, 942 01:06:03,350 --> 01:06:06,890 the left part of the string acts a force 943 01:06:06,890 --> 01:06:12,760 on the right part of the string to generate that pulse. 944 01:06:12,760 --> 01:06:14,360 And that also propagates. 945 01:06:14,360 --> 01:06:17,620 So even on a simple thing like a string, 946 01:06:17,620 --> 01:06:23,480 there will be simultaneously three progressive waves 947 01:06:23,480 --> 01:06:27,300 going down the string-- the displacement, 948 01:06:27,300 --> 01:06:30,410 the transverse velocity, and the force. 949 01:06:30,410 --> 01:06:35,380 Here, we are talking about two-- the current and the voltage. 950 01:06:35,380 --> 01:06:43,800 All right, so I've translated what I've discussed before. 951 01:06:43,800 --> 01:06:48,620 I'll call x equals 0 the junction between the two 952 01:06:48,620 --> 01:06:50,330 cables. 953 01:06:50,330 --> 01:06:53,350 On the left and on the vertical axis 954 01:06:53,350 --> 01:06:59,830 here, I'm going to either be plotting the voltage 955 01:06:59,830 --> 01:07:03,850 across the cable or the current flowing down the cable. 956 01:07:03,850 --> 01:07:08,400 I'm just too lazy to draw two separate plots. 957 01:07:08,400 --> 01:07:12,130 So what we are told, there is the voltage, 958 01:07:12,130 --> 01:07:15,460 which is given by this expression. 959 01:07:15,460 --> 01:07:19,630 That's a progressive wave, a sinusoidal progressive wave 960 01:07:19,630 --> 01:07:22,720 moving to the right. 961 01:07:22,720 --> 01:07:32,090 There is a reflected one, which is given by some amplitude, 962 01:07:32,090 --> 01:07:34,470 v reflected times cosine. 963 01:07:34,470 --> 01:07:38,410 And this will now have a plus sign 964 01:07:38,410 --> 01:07:41,500 because it's now propagating to the left. 965 01:07:41,500 --> 01:07:45,000 Here, the minus sign tells you it's going to the right. 966 01:07:45,000 --> 01:07:48,700 The plus sign going to the left. 967 01:07:48,700 --> 01:07:54,720 Now, we know from-- for example, our studies 968 01:07:54,720 --> 01:08:01,940 of coupled oscillators, that in a steady state situation, 969 01:08:01,940 --> 01:08:08,365 every oscillator is moving with the same frequency and phase. 970 01:08:08,365 --> 01:08:13,620 So if I extend that to the infinite case, 971 01:08:13,620 --> 01:08:20,609 I know that the frequency of this wave, 972 01:08:20,609 --> 01:08:22,390 which was only one frequency. 973 01:08:22,390 --> 01:08:25,640 So the progressive wave going this way 974 01:08:25,640 --> 01:08:30,420 has only one frequency, omega 1. 975 01:08:30,420 --> 01:08:37,910 If it drives the oscillators as it goes along, 976 01:08:37,910 --> 01:08:41,790 it will drive them to oscillate at the same frequency 977 01:08:41,790 --> 01:08:43,740 as its frequency. 978 01:08:43,740 --> 01:08:46,090 Another way to say it, in this problem 979 01:08:46,090 --> 01:08:48,020 there's only one frequency. 980 01:08:48,020 --> 01:08:51,390 And so the reflected wave will have the same frequency 981 01:08:51,390 --> 01:08:53,460 as the incoming one. 982 01:08:53,460 --> 01:08:59,279 Similarly, the transmitted wave will have the same frequency. 983 01:08:59,279 --> 01:09:02,740 That's why I wrote this to be equal to that. 984 01:09:02,740 --> 01:09:06,365 We can actually formally show that that 985 01:09:06,365 --> 01:09:10,330 has to be the case by considering 986 01:09:10,330 --> 01:09:13,069 what happens at the boundary. 987 01:09:13,069 --> 01:09:17,310 You could never have the two separate halves 988 01:09:17,310 --> 01:09:20,130 moving with different frequencies and the string 989 01:09:20,130 --> 01:09:21,689 not be broken. 990 01:09:21,689 --> 01:09:28,939 Or in this case, the voltage being not the same 991 01:09:28,939 --> 01:09:32,140 on both sides of the string, on both sides 992 01:09:32,140 --> 01:09:34,910 of the boundary of the string. 993 01:09:34,910 --> 01:09:37,620 OK, so everything we learned over there 994 01:09:37,620 --> 01:09:40,960 we can write as this sinusoidal progressive wave 995 01:09:40,960 --> 01:09:44,630 to the right with amplitude v1, the reflected wave 996 01:09:44,630 --> 01:09:48,290 with amplitude v reflected. 997 01:09:48,290 --> 01:09:54,760 And there will be a transmitted wave of some magnitude v2. 998 01:09:54,760 --> 01:10:02,120 And a cosine omega 2t minus 2 pi lambda to the x. 999 01:10:04,970 --> 01:10:07,950 I went ahead of myself to explain 1000 01:10:07,950 --> 01:10:12,360 that this omega 1 has to be the same as that. 1001 01:10:12,360 --> 01:10:15,220 In a second using the same argument, 1002 01:10:15,220 --> 01:10:19,580 I'll show that this has to be equal to that. 1003 01:10:19,580 --> 01:10:22,840 What else do we know about the situation? 1004 01:10:22,840 --> 01:10:26,460 Well, we know that at the boundary, 1005 01:10:26,460 --> 01:10:29,020 the two y's are connected. 1006 01:10:29,020 --> 01:10:33,570 Therefore, there will be the same potential difference 1007 01:10:33,570 --> 01:10:38,210 between the central conductor and the outside conductor 1008 01:10:38,210 --> 01:10:40,870 just to the left of the boundary and to the right. 1009 01:10:40,870 --> 01:10:47,730 So at x equals 0 for all times, the voltage VL 1010 01:10:47,730 --> 01:10:53,090 will be equal to the potential difference 1011 01:10:53,090 --> 01:10:54,520 on the right-hand side. 1012 01:10:54,520 --> 01:11:00,300 So that's one boundary condition. 1013 01:11:00,300 --> 01:11:05,970 The other one, we know that charges are conserved. 1014 01:11:05,970 --> 01:11:10,380 And therefore, if you have junction of two wires, 1015 01:11:10,380 --> 01:11:13,850 the total current going into that junction 1016 01:11:13,850 --> 01:11:16,345 or coming away from it must be 0 or you 1017 01:11:16,345 --> 01:11:18,300 would be creating charges. 1018 01:11:18,300 --> 01:11:23,420 So the sum of all the currents coming in or out 1019 01:11:23,420 --> 01:11:29,670 from the junction for all times and x equals 0 must be 0. 1020 01:11:29,670 --> 01:11:31,910 So this is now all in mathematics. 1021 01:11:31,910 --> 01:11:34,890 You don't need to know this, but this 1022 01:11:34,890 --> 01:11:39,470 is anything to do with voltages, currents, et cetera. 1023 01:11:39,470 --> 01:11:43,000 This is the mathematical description of the situation. 1024 01:11:43,000 --> 01:11:48,830 And let's try to now answer the questions that were posed. 1025 01:11:48,830 --> 01:11:52,190 OK, the first question was, what is 1026 01:11:52,190 --> 01:11:55,070 the ratio of omega 2 to omega 1? 1027 01:11:55,070 --> 01:11:57,480 And I've answered it already. 1028 01:11:57,480 --> 01:12:01,620 The only frequency in this problem is omega 1. 1029 01:12:01,620 --> 01:12:03,310 That's what comes in. 1030 01:12:03,310 --> 01:12:07,930 It'll drive any oscillator anywhere at that frequency 1031 01:12:07,930 --> 01:12:11,070 under the steady-state conditions, which we have here. 1032 01:12:11,070 --> 01:12:17,097 So omega 2 must equal to omega 1 or that ratio is equal to 1. 1033 01:12:17,097 --> 01:12:17,680 First problem. 1034 01:12:17,680 --> 01:12:20,170 The next question we were asked, what 1035 01:12:20,170 --> 01:12:24,310 is the ratio of the wavelengths? 1036 01:12:24,310 --> 01:12:35,820 Now, we know that for any harmonic wave, 1037 01:12:35,820 --> 01:12:39,620 the product of the wavelengths times the frequency 1038 01:12:39,620 --> 01:12:41,560 is the phase velocity. 1039 01:12:41,560 --> 01:12:43,380 You can trivially prove it for yourself. 1040 01:12:43,380 --> 01:12:49,620 Just take that formula for the progressive wave, 1041 01:12:49,620 --> 01:12:51,825 harmonic progressive wave, and calculate 1042 01:12:51,825 --> 01:12:53,410 what the frequency is. 1043 01:12:53,410 --> 01:12:57,000 And if you have difficulty, draw the wave 1044 01:12:57,000 --> 01:13:00,780 as a function of position at 2 times 1045 01:13:00,780 --> 01:13:03,710 and see how far that picture has moved. 1046 01:13:03,710 --> 01:13:09,720 And you'll find that lambda times f is equal to v. OK. 1047 01:13:09,720 --> 01:13:12,570 This is true to the left of the boundary 1048 01:13:12,570 --> 01:13:16,380 and on the right-hand side. 1049 01:13:16,380 --> 01:13:20,890 So I've rewritten this lambda times omega over 2 pi equals 1050 01:13:20,890 --> 01:13:26,940 v. Therefore, if you have two strings of different phase 1051 01:13:26,940 --> 01:13:32,210 velocity, for the first one this equation looks like that. 1052 01:13:32,210 --> 01:13:36,740 For the second one, it's the same but 1's replaced by 2. 1053 01:13:36,740 --> 01:13:40,360 From these, I immediately divide one by the other, 1054 01:13:40,360 --> 01:13:44,120 that lambda 2 over lambda 1 is equal to v2 over v1. 1055 01:13:44,120 --> 01:13:45,930 They are no longer the same. 1056 01:13:45,930 --> 01:13:48,450 The frequencies are the same, but the wavelengths 1057 01:13:48,450 --> 01:13:49,440 are different. 1058 01:13:49,440 --> 01:13:55,380 It depends on the properties of the two cables. 1059 01:13:55,380 --> 01:14:00,880 So we get that the wavelength on the left-hand side, as I say, 1060 01:14:00,880 --> 01:14:03,780 will be a different wavelength to the one on the right. 1061 01:14:03,780 --> 01:14:06,550 And the ratio between them will be 1062 01:14:06,550 --> 01:14:12,100 that of the phase velocity on both sides. 1063 01:14:12,100 --> 01:14:14,650 All right, next. 1064 01:14:14,650 --> 01:14:16,660 The third part of the thing was, what's 1065 01:14:16,660 --> 01:14:19,300 the ratio of the voltages? 1066 01:14:19,300 --> 01:14:23,760 In other words, what is the ratio of v2 to v1? 1067 01:14:23,760 --> 01:14:28,250 v1, I'm reminding you, is the amplitude 1068 01:14:28,250 --> 01:14:33,270 of the voltage wave that comes from the left cable 1069 01:14:33,270 --> 01:14:37,090 to the junction, and then continues out 1070 01:14:37,090 --> 01:14:38,730 with a different amplitude. 1071 01:14:38,730 --> 01:14:41,640 What is that amplitude? 1072 01:14:41,640 --> 01:14:44,930 To answer that, we make use of the boundary conditions, 1073 01:14:44,930 --> 01:14:47,960 which we discussed earlier. 1074 01:14:47,960 --> 01:14:52,220 These boundary conditions immediately tell us that how? 1075 01:14:52,220 --> 01:14:54,320 Well, the first boundary condition 1076 01:14:54,320 --> 01:15:00,160 tells us that the voltage just to the left of x equals 0 1077 01:15:00,160 --> 01:15:02,850 must equal to just to the right to it. 1078 01:15:02,850 --> 01:15:09,070 Now on the left-hand side, you have two waves-- one coming in, 1079 01:15:09,070 --> 01:15:14,500 one coming out, the reflected one. 1080 01:15:14,500 --> 01:15:23,750 If you look at the equation for VL at x equals 0, 1081 01:15:23,750 --> 01:15:30,130 for all times you'll find it's v1 times cosine omega 1082 01:15:30,130 --> 01:15:34,510 1t plus v reflected cosine omega 1t. 1083 01:15:34,510 --> 01:15:40,410 On the right-hand side, you see that it is v2 times-- again, 1084 01:15:40,410 --> 01:15:43,170 with x equals 0 cosine omega 2t. 1085 01:15:43,170 --> 01:15:45,970 But omega 2t is equal to omega 1. 1086 01:15:45,970 --> 01:15:50,050 And so at all times, the cosines cancel 1087 01:15:50,050 --> 01:15:55,020 and you end up that v1 plus v reflected has to be v2. 1088 01:15:55,020 --> 01:15:59,380 At all times that will be true at x equals 0. 1089 01:15:59,380 --> 01:16:01,080 So you have one equation. 1090 01:16:01,080 --> 01:16:04,980 The other boundary condition gives us the other one. 1091 01:16:04,980 --> 01:16:08,150 We know that if there is a wave coming 1092 01:16:08,150 --> 01:16:12,060 in with amplitude v1, the current that comes in, 1093 01:16:12,060 --> 01:16:18,090 it told us in the formulation of the problem, is that divided by 1094 01:16:18,090 --> 01:16:21,580 z1, this characteristic impedance. 1095 01:16:21,580 --> 01:16:23,860 Similarly, for the reflected wave. 1096 01:16:23,860 --> 01:16:24,930 But we have to watch it. 1097 01:16:24,930 --> 01:16:26,055 There is a sign difference. 1098 01:16:26,055 --> 01:16:26,930 Because why? 1099 01:16:26,930 --> 01:16:33,730 If the wave is coming from the left to the right, 1100 01:16:33,730 --> 01:16:37,330 if we call that a positive current. 1101 01:16:37,330 --> 01:16:41,020 After reflection if it goes the other way, 1102 01:16:41,020 --> 01:16:42,310 it will be a negative. 1103 01:16:42,310 --> 01:16:44,910 We'll be subtracting from that junction. 1104 01:16:44,910 --> 01:16:47,630 So that's why there is this minus. 1105 01:16:47,630 --> 01:16:53,880 And then, this is the one-- this is the total current coming 1106 01:16:53,880 --> 01:16:55,290 into the junction. 1107 01:16:55,290 --> 01:16:58,770 And that's got to be equal to the total current coming out 1108 01:16:58,770 --> 01:17:03,210 on the other side, v2 over z2. 1109 01:17:03,210 --> 01:17:07,910 From this, I just multiply by z1, so I get this equation. 1110 01:17:07,910 --> 01:17:10,470 And these are two trivial algebraic equations. 1111 01:17:10,470 --> 01:17:12,920 If I add them, I get 2 v1, et cetera. 1112 01:17:12,920 --> 01:17:15,920 From which, I get that this is the case. 1113 01:17:15,920 --> 01:17:21,070 And again, I notice I forget v1. 1114 01:17:21,070 --> 01:17:22,010 OK. 1115 01:17:22,010 --> 01:17:27,480 And so we get that the v2 over v1 1116 01:17:27,480 --> 01:17:30,330 is that quantity which was given. 1117 01:17:30,330 --> 01:17:33,470 And we've solved the three parts of the problem. 1118 01:17:33,470 --> 01:17:35,020 That's the end. 1119 01:17:35,020 --> 01:17:37,030 And the thing that I would just like 1120 01:17:37,030 --> 01:17:39,300 you to keep in mind, that if you come 1121 01:17:39,300 --> 01:17:41,980 across some kind of a problem to do 1122 01:17:41,980 --> 01:17:43,990 with progressive waves, et cetera. 1123 01:17:43,990 --> 01:17:46,170 If you've never seen the system, it 1124 01:17:46,170 --> 01:17:48,520 doesn't mean you don't know how to solve it. 1125 01:17:48,520 --> 01:17:53,280 Stop, think, see by analogy what it corresponds 1126 01:17:53,280 --> 01:17:56,660 to a system which you have understood. 1127 01:17:56,660 --> 01:18:00,030 The corollary is, suppose you are having trouble 1128 01:18:00,030 --> 01:18:03,520 with progressive waves on strings. 1129 01:18:03,520 --> 01:18:05,680 But you're an electrical engineer 1130 01:18:05,680 --> 01:18:10,620 and you feel very comfortable about waves on cables. 1131 01:18:10,620 --> 01:18:15,380 You can use the study of waves on cables, reflection 1132 01:18:15,380 --> 01:18:19,570 of pulses, reflection coefficients, transmission 1133 01:18:19,570 --> 01:18:22,700 coefficients, et cetera. 1134 01:18:22,700 --> 01:18:25,120 That which you've understood in the electrical system, 1135 01:18:25,120 --> 01:18:27,440 you can translate it to what happens 1136 01:18:27,440 --> 01:18:31,600 if you have mechanical systems connected together. 1137 01:18:31,600 --> 01:18:33,150 Thank you.