1 00:00:02,120 --> 00:00:04,460 The following content is provided under a Creative 2 00:00:04,460 --> 00:00:05,880 Commons license. 3 00:00:05,880 --> 00:00:08,090 Your support will help MIT OpenCourseWare 4 00:00:08,090 --> 00:00:12,180 continue to offer high quality educational resources for free. 5 00:00:12,180 --> 00:00:14,720 To make a donation or to view additional materials 6 00:00:14,720 --> 00:00:18,680 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:18,680 --> 00:00:19,560 at ocw.mit.edu. 8 00:00:23,610 --> 00:00:28,170 PROFESSOR: OK, welcome back, everybody, to 8.03. 9 00:00:28,170 --> 00:00:33,540 Today we are going to continue the discussion of wave equation 10 00:00:33,540 --> 00:00:37,360 starting from last lecture. 11 00:00:37,360 --> 00:00:38,930 So what have we learned last time? 12 00:00:38,930 --> 00:00:45,330 As a reminder, we have started to study the behavior of a wave 13 00:00:45,330 --> 00:00:46,680 equation. 14 00:00:46,680 --> 00:00:54,700 We understood basic behavior of the wave equation, 15 00:00:54,700 --> 00:00:58,800 and also are trying to solve the general solution of the wave 16 00:00:58,800 --> 00:00:59,860 equation. 17 00:00:59,860 --> 00:01:01,440 The first thing which we learned, 18 00:01:01,440 --> 00:01:03,900 as usual, is the normal modes. 19 00:01:03,900 --> 00:01:06,210 What are the normal modes in the case 20 00:01:06,210 --> 00:01:11,880 of wave equation over continuous translations in metric system. 21 00:01:11,880 --> 00:01:15,180 What we found last time is that they are standing waves. 22 00:01:15,180 --> 00:01:21,990 And of course, as usual, the full solution, 23 00:01:21,990 --> 00:01:25,370 which is a general description of this system, 24 00:01:25,370 --> 00:01:29,820 is the superposition of infinite number of normal modes. 25 00:01:29,820 --> 00:01:32,850 And that means one can understand 26 00:01:32,850 --> 00:01:34,990 this kind of system systematically 27 00:01:34,990 --> 00:01:38,980 the using Fourier series. 28 00:01:38,980 --> 00:01:40,560 So this is just a reminder. 29 00:01:40,560 --> 00:01:44,010 So what we have done is that basically we 30 00:01:44,010 --> 00:01:48,490 start with infinite number of coupled oscillators. 31 00:01:48,490 --> 00:01:54,120 And we make the space between those massive objects 32 00:01:54,120 --> 00:01:56,910 in the system smaller and smaller 33 00:01:56,910 --> 00:01:59,510 until it become continuous, right? 34 00:01:59,510 --> 00:02:03,440 And a very interesting thing happened 35 00:02:03,440 --> 00:02:08,139 that automatically already give you wave equations, OK? 36 00:02:08,139 --> 00:02:09,539 Which is as you're shown here. 37 00:02:12,440 --> 00:02:15,750 Last time, as I mentioned, we discussed the normal modes, 38 00:02:15,750 --> 00:02:17,010 which are standing waves. 39 00:02:20,210 --> 00:02:22,685 Those are the first few normal modes, 40 00:02:22,685 --> 00:02:27,270 and those are the functional form of the normal modes, which 41 00:02:27,270 --> 00:02:28,450 are standing waves. 42 00:02:28,450 --> 00:02:30,670 So the structure just looks like this. 43 00:02:30,670 --> 00:02:34,890 So you have A m, which is the amplitude, sin k 44 00:02:34,890 --> 00:02:38,610 m x plus alpha m, which can be determined 45 00:02:38,610 --> 00:02:41,490 by boundary conditions. 46 00:02:41,490 --> 00:02:44,410 And the sin omega m t plus beta m, 47 00:02:44,410 --> 00:02:47,600 that means all the points in the system 48 00:02:47,600 --> 00:02:51,000 are oscillating at the same frequency, omega m, 49 00:02:51,000 --> 00:02:54,330 and as the same phase, which is beta m. 50 00:02:54,330 --> 00:02:57,680 And those are based on the wave equation. 51 00:02:57,680 --> 00:03:01,410 If you plug this solution back into the equation, 52 00:03:01,410 --> 00:03:05,680 you will find that omega m is actually not a free parameter. 53 00:03:05,680 --> 00:03:11,370 It's actually proportional to k m, which is the wave 54 00:03:11,370 --> 00:03:14,100 number of m's normal mode. 55 00:03:14,100 --> 00:03:18,930 And this constant of v p, we will find out today, 56 00:03:18,930 --> 00:03:23,010 this is actually the speed of the wave 57 00:03:23,010 --> 00:03:24,955 in the case of traveling wave. 58 00:03:24,955 --> 00:03:28,440 And if you look at the individual normal modes 59 00:03:28,440 --> 00:03:33,180 and make product of those normal mode as a function of time, 60 00:03:33,180 --> 00:03:36,570 you can see from here there are six different normal modes. 61 00:03:36,570 --> 00:03:40,430 And they are like a sinusoidal shape in terms of amplitude 62 00:03:40,430 --> 00:03:45,070 as a functional position on the string. 63 00:03:45,070 --> 00:03:48,590 And there you can see that if you distort 64 00:03:48,590 --> 00:03:52,770 the string more, then you get a higher oscillation frequency, 65 00:03:52,770 --> 00:03:58,610 as you see from m equal to 1 to m equals to 6 case. 66 00:03:58,610 --> 00:04:01,520 So as I mentioned today, we will talk 67 00:04:01,520 --> 00:04:05,860 about another interesting kind of solution, which is 68 00:04:05,860 --> 00:04:08,680 progressing wave solution, OK? 69 00:04:08,680 --> 00:04:12,490 Why is this solution exciting? 70 00:04:12,490 --> 00:04:15,370 Not just because this actually matches 71 00:04:15,370 --> 00:04:18,790 what we actually already learned about waves, right? 72 00:04:18,790 --> 00:04:22,630 And this solution should enable us 73 00:04:22,630 --> 00:04:26,620 to send, for example, energy from one point 74 00:04:26,620 --> 00:04:27,590 to another point. 75 00:04:27,590 --> 00:04:29,540 This is actually what I am doing now, right? 76 00:04:29,540 --> 00:04:33,460 I'm sending energy from my mouth to your ear 77 00:04:33,460 --> 00:04:36,730 so that you can hear what I have been talking about, right? 78 00:04:36,730 --> 00:04:39,370 About 8.03, right? 79 00:04:39,370 --> 00:04:40,360 So that's really cool. 80 00:04:40,360 --> 00:04:45,670 And we will try to understand how actually we 81 00:04:45,670 --> 00:04:50,890 can get this solution out of this strange wave equation, OK? 82 00:04:50,890 --> 00:04:52,780 So what is showing here is the normal mode. 83 00:04:52,780 --> 00:04:57,310 And we are going to talk about a second kind of solution, 84 00:04:57,310 --> 00:05:01,360 which is the progressing wave solution. 85 00:05:01,360 --> 00:05:06,376 And this solution have this form, psi (x, t) 86 00:05:06,376 --> 00:05:11,260 will be equal to some kind of function, f. 87 00:05:11,260 --> 00:05:18,130 And this is actually a function of x minus v p times t. 88 00:05:18,130 --> 00:05:23,990 This is actually a general form of the progressing wave. 89 00:05:23,990 --> 00:05:29,220 And f function is some kind of well-behaved function 90 00:05:29,220 --> 00:05:31,790 of your choice. 91 00:05:31,790 --> 00:05:35,230 OK, so the first thing which I would like to do 92 00:05:35,230 --> 00:05:40,900 is to show that this functional form is actually the solution 93 00:05:40,900 --> 00:05:42,620 of the wave equation, right? 94 00:05:42,620 --> 00:05:44,290 So fairly straightforward. 95 00:05:44,290 --> 00:05:48,580 We can actually go ahead and plug them into this equation. 96 00:05:48,580 --> 00:05:57,321 And before that, I will define tau to be x minus v p times t, 97 00:05:57,321 --> 00:05:57,820 OK? 98 00:05:57,820 --> 00:06:02,385 So in order to prepare for plugging 99 00:06:02,385 --> 00:06:07,090 in this functional form to a wave equation, 100 00:06:07,090 --> 00:06:11,110 I would calculate using chain law. 101 00:06:11,110 --> 00:06:16,500 Partial f partial x will be equal to partial f, 102 00:06:16,500 --> 00:06:17,815 partial tau. 103 00:06:17,815 --> 00:06:21,310 Partial tau, partial x. 104 00:06:21,310 --> 00:06:27,070 And this will give you partial f, partial tau times, 105 00:06:27,070 --> 00:06:33,100 in this case, partial tau, partial x will give you 1, OK? 106 00:06:33,100 --> 00:06:41,800 And that will give you f prime tau, OK? 107 00:06:41,800 --> 00:06:43,900 Therefore we can go ahead and calculate 108 00:06:43,900 --> 00:06:50,170 as well partial square f, partial x squared, OK? 109 00:06:50,170 --> 00:06:56,390 That will give you f double prime tau. 110 00:06:56,390 --> 00:06:58,390 So this is actually the first set of equation 111 00:06:58,390 --> 00:07:03,910 I need in order to describe the right side of the wave 112 00:07:03,910 --> 00:07:05,880 equation. 113 00:07:05,880 --> 00:07:10,260 The other equation which I need in the preparation for plugging 114 00:07:10,260 --> 00:07:13,640 in the whole thing into the wave equation 115 00:07:13,640 --> 00:07:18,460 is to calculate partial f, partial t. 116 00:07:18,460 --> 00:07:21,170 And according to chain law, partial f, 117 00:07:21,170 --> 00:07:25,630 partial t is equal to partial f, partial tau, partial tau, 118 00:07:25,630 --> 00:07:28,780 partial t. 119 00:07:28,780 --> 00:07:33,520 In this case, f is actually a function of x minus v p t, 120 00:07:33,520 --> 00:07:37,390 and tau is defined as x minus v p t. 121 00:07:37,390 --> 00:07:40,030 Therefore you can actually immediately conclude 122 00:07:40,030 --> 00:07:42,670 that these would be equal to minus v 123 00:07:42,670 --> 00:07:47,110 p, partial f, partial tau. 124 00:07:47,110 --> 00:07:51,606 And now it is actually equal to minus v p f prime. 125 00:07:55,360 --> 00:07:58,180 Similarly, you can calculate partial square 126 00:07:58,180 --> 00:08:01,930 of f, partial t squared, and that would give you 127 00:08:01,930 --> 00:08:05,474 a v p squared f prime. 128 00:08:05,474 --> 00:08:10,960 Oh, f double prime because we did the double differential. 129 00:08:10,960 --> 00:08:12,940 OK, all right. 130 00:08:12,940 --> 00:08:18,540 So from the first equation and second equation, 131 00:08:18,540 --> 00:08:21,810 which I have on the board, we can actually plug that 132 00:08:21,810 --> 00:08:24,710 into the wave equation. 133 00:08:24,710 --> 00:08:29,820 And what I'm going to get is partial square f, partial t 134 00:08:29,820 --> 00:08:30,580 square. 135 00:08:30,580 --> 00:08:35,676 That would be equal to v p square, partial square f, 136 00:08:35,676 --> 00:08:38,360 partial x square. 137 00:08:38,360 --> 00:08:47,390 So that is actually exactly the wave equation, which 138 00:08:47,390 --> 00:08:49,220 we showed from the beginning. 139 00:08:49,220 --> 00:08:56,580 So that means this functional form satisfy the wave equation. 140 00:08:56,580 --> 00:09:01,860 And I didn't even specify what is actually f function. 141 00:09:01,860 --> 00:09:05,850 f function is some kind of well-behaved function. 142 00:09:05,850 --> 00:09:11,010 And it can have all kinds of different shape, which 143 00:09:11,010 --> 00:09:12,880 we will discuss later. 144 00:09:12,880 --> 00:09:14,750 But that is actually pretty encouraging. 145 00:09:14,750 --> 00:09:19,210 And that means if you try to distort the string 146 00:09:19,210 --> 00:09:25,485 and if this shape is actually propagating at-- 147 00:09:25,485 --> 00:09:29,634 have the functional form of x minus-- 148 00:09:29,634 --> 00:09:31,800 If you have any function which you have a functional 149 00:09:31,800 --> 00:09:35,490 form of f of x minus Vpt-- 150 00:09:35,490 --> 00:09:40,740 no matter what kind of shape it is, this shape is going to be-- 151 00:09:40,740 --> 00:09:43,960 first of all, this function is a solution to the wave equation. 152 00:09:43,960 --> 00:09:46,140 Secondly, we will show you that this shape 153 00:09:46,140 --> 00:09:50,820 is going to be propagating at the speed of Vp. 154 00:09:50,820 --> 00:09:52,500 And the shape will not change. 155 00:09:52,500 --> 00:09:56,970 It will stay like that forever according to the wave equation. 156 00:10:00,450 --> 00:10:03,630 There's another function of form which 157 00:10:03,630 --> 00:10:09,750 can also be the functional form of the progressing wave. 158 00:10:09,750 --> 00:10:17,820 We usually write it as f KX plus or minus omega t. 159 00:10:17,820 --> 00:10:25,320 And in this case if omega is actually Vp times k, 160 00:10:25,320 --> 00:10:28,860 which is actually already required 161 00:10:28,860 --> 00:10:32,780 from the discussion of normal modes. 162 00:10:32,780 --> 00:10:37,280 Then this equation-- this kind of functional form 163 00:10:37,280 --> 00:10:43,190 is also the solution of wave equation. 164 00:10:43,190 --> 00:10:46,440 And of course you can actually go ahead and prove that 165 00:10:46,440 --> 00:10:50,010 probably after the lecture. 166 00:10:50,010 --> 00:10:53,194 The proof will be very similar to what we have done here. 167 00:10:55,980 --> 00:10:59,910 So now, actually try to understand 168 00:10:59,910 --> 00:11:03,400 what does this function mean? 169 00:11:03,400 --> 00:11:09,370 So I have a function which is f of x minus Vp times t. 170 00:11:09,370 --> 00:11:12,745 Where x is actually the position of the-- 171 00:11:12,745 --> 00:11:13,920 on the string. 172 00:11:13,920 --> 00:11:18,040 And the t is actually the time, which I go ahead and check 173 00:11:18,040 --> 00:11:20,800 this function. 174 00:11:20,800 --> 00:11:25,140 And for example, if I can give you an example-- 175 00:11:25,140 --> 00:11:27,990 a function, for example, I can make it 176 00:11:27,990 --> 00:11:30,850 like a triangular shape like this. 177 00:11:30,850 --> 00:11:36,510 And this is actually plotted as a function of Tau. 178 00:11:36,510 --> 00:11:40,020 So f of Tau is actually giving you 179 00:11:40,020 --> 00:11:44,870 the information of the shape of the progressing wave. 180 00:11:47,960 --> 00:11:54,800 So let's discuss if I write f of x minus Vt, 181 00:11:54,800 --> 00:11:57,820 what does that mean? 182 00:11:57,820 --> 00:12:02,560 OK, first of all, I can take t equal to 0. 183 00:12:02,560 --> 00:12:06,450 What will what happen is that a lot instants of time, 184 00:12:06,450 --> 00:12:10,850 t equal to 0, x is equal to Tau. 185 00:12:10,850 --> 00:12:18,650 That means at t equal to 0, the shape of the string-- 186 00:12:18,650 --> 00:12:20,690 well, it looks like this-- like this function. 187 00:12:23,500 --> 00:12:30,220 Now, I would like to know how this string is going to evolve 188 00:12:30,220 --> 00:12:31,630 as a function of time. 189 00:12:31,630 --> 00:12:35,160 So that's actually all we care, right? 190 00:12:35,160 --> 00:12:41,850 And that means I'm going to increase t to a larger value. 191 00:12:41,850 --> 00:12:47,430 So suppose originally, I have a t equal to 0, 192 00:12:47,430 --> 00:12:50,880 I'm sitting on this point. 193 00:12:50,880 --> 00:12:54,870 I sample the f at this position. 194 00:12:54,870 --> 00:12:59,050 If I increase t-- 195 00:12:59,050 --> 00:13:03,550 if I increase t from 0 to a larger value where 196 00:13:03,550 --> 00:13:09,670 I will start to sample, will I move to a left-hand side, 197 00:13:09,670 --> 00:13:11,170 or a right-hand side? 198 00:13:11,170 --> 00:13:13,060 Anybody can help me. 199 00:13:13,060 --> 00:13:16,600 Because this functional form is actually x minus Vt. 200 00:13:16,600 --> 00:13:19,600 What will happen if I increase t? 201 00:13:19,600 --> 00:13:23,530 Tau will increase or decrease? 202 00:13:23,530 --> 00:13:24,797 AUDIENCE: Decrease. 203 00:13:24,797 --> 00:13:25,630 PROFESSOR: Decrease. 204 00:13:25,630 --> 00:13:25,930 AUDIENCE: Yeah. 205 00:13:25,930 --> 00:13:26,620 Move to the left. 206 00:13:26,620 --> 00:13:27,495 PROFESSOR: Very good. 207 00:13:27,495 --> 00:13:28,510 So decrease, right? 208 00:13:28,510 --> 00:13:33,350 So that means at the fixed x, I am going to sample this point. 209 00:13:36,440 --> 00:13:37,480 What does that mean? 210 00:13:37,480 --> 00:13:43,830 That means originally, if I plot everything in terms of x-- 211 00:13:43,830 --> 00:13:47,600 originally, it have this shape. 212 00:13:47,600 --> 00:13:51,130 Now, if I increase t, what is going to happen 213 00:13:51,130 --> 00:13:55,750 is that originally I was sampling this shape here. 214 00:13:55,750 --> 00:13:58,760 And now I am sampling the shape here. 215 00:13:58,760 --> 00:14:06,560 That means if t equal to t prime, which is larger than t, 216 00:14:06,560 --> 00:14:10,650 this shape first of all is unchanged. 217 00:14:10,650 --> 00:14:13,570 Secondly, it's actually moving. 218 00:14:13,570 --> 00:14:17,500 The shape looks like as if it's moving 219 00:14:17,500 --> 00:14:20,410 in the horizontal direction. 220 00:14:20,410 --> 00:14:24,130 And the direction of this movement 221 00:14:24,130 --> 00:14:27,010 is in the positive x direction. 222 00:14:27,010 --> 00:14:31,690 If I'd write my progressing wave solution in a functional 223 00:14:31,690 --> 00:14:36,700 form of f of x minus Vt-- 224 00:14:36,700 --> 00:14:38,930 this should be Vt here, sorry for that. 225 00:14:42,020 --> 00:14:45,410 Any questions? 226 00:14:45,410 --> 00:14:46,790 So look at what we have done. 227 00:14:46,790 --> 00:14:50,980 First of all, we have proved that f 228 00:14:50,980 --> 00:14:54,730 of x minus Vpt, this function of form 229 00:14:54,730 --> 00:14:58,260 is a solution to the wave equation. 230 00:14:58,260 --> 00:14:59,920 It's a solution. 231 00:14:59,920 --> 00:15:02,590 Secondly, we also discussed the property 232 00:15:02,590 --> 00:15:04,210 of this functional form. 233 00:15:04,210 --> 00:15:06,790 So that's essentially describing a shape. 234 00:15:06,790 --> 00:15:09,920 And the whole shape is going to move 235 00:15:09,920 --> 00:15:13,600 as if it's moving as a function of time 236 00:15:13,600 --> 00:15:18,120 and to the positive x direction. 237 00:15:18,120 --> 00:15:20,810 So let me ask you another question. 238 00:15:20,810 --> 00:15:24,230 So what will happen if I write the solution 239 00:15:24,230 --> 00:15:29,720 in the form of f of x plus Vt? 240 00:15:29,720 --> 00:15:32,800 Anybody can help me with the direction 241 00:15:32,800 --> 00:15:35,481 of the propagation of the wave. 242 00:15:35,481 --> 00:15:36,522 AUDIENCE: Go to the left. 243 00:15:36,522 --> 00:15:37,188 PROFESSOR: Yeah. 244 00:15:37,188 --> 00:15:39,340 Go to the left-hand side of the board. 245 00:15:39,340 --> 00:15:43,850 That means if I have another expression. 246 00:15:43,850 --> 00:15:48,260 I'm using the same f function which is defined here. 247 00:15:48,260 --> 00:15:52,870 In this case, if I write f of x plus Vt, 248 00:15:52,870 --> 00:15:56,480 that means the shape will be moving 249 00:15:56,480 --> 00:16:00,440 in the negative x direction. 250 00:16:00,440 --> 00:16:01,910 It's symmetric. 251 00:16:01,910 --> 00:16:03,910 Of course you can also discuss what 252 00:16:03,910 --> 00:16:06,320 will happen if you take this functional form 253 00:16:06,320 --> 00:16:14,488 and you are going to get exactly the same conclusion. 254 00:16:17,300 --> 00:16:23,230 So right now I have been talking about moving shape. 255 00:16:23,230 --> 00:16:27,660 So what is actually really moving? 256 00:16:27,660 --> 00:16:30,190 Professor Lee, you just told us before 257 00:16:30,190 --> 00:16:34,720 that every point on the string can only move up and down. 258 00:16:34,720 --> 00:16:36,220 Now, you are talking about something 259 00:16:36,220 --> 00:16:39,820 moving in the positive x and negative x direction. 260 00:16:39,820 --> 00:16:43,010 What does that mean? 261 00:16:43,010 --> 00:16:46,360 So that mean-- take y example. 262 00:16:46,360 --> 00:16:51,970 So if I have a Gaussian pulse, and I write this thing 263 00:16:51,970 --> 00:16:56,720 in the form of x minus Vt. 264 00:16:56,720 --> 00:17:02,130 At t equal to 0, this shape looks like this. 265 00:17:02,130 --> 00:17:06,280 In the next moment, if I increase time 266 00:17:06,280 --> 00:17:09,339 to t equal to 1, what is going to happen 267 00:17:09,339 --> 00:17:16,170 is that this shape move toward the positive x direction. 268 00:17:16,170 --> 00:17:20,260 And of course 0's are the equilibrium 269 00:17:20,260 --> 00:17:23,780 position of this waves-- 270 00:17:23,780 --> 00:17:26,700 of the string. 271 00:17:26,700 --> 00:17:33,020 And what is happening is it's like this-- so basically, 272 00:17:33,020 --> 00:17:37,950 all the points on the string are really working together 273 00:17:37,950 --> 00:17:41,970 to produce this shifting-- 274 00:17:41,970 --> 00:17:45,430 this progress Gaussian wave. 275 00:17:45,430 --> 00:17:49,770 So what is happening is that if I focus on this point, 276 00:17:49,770 --> 00:17:51,480 this point will go down. 277 00:17:51,480 --> 00:17:55,170 And this point will go down, go down, go down, 278 00:17:55,170 --> 00:17:58,230 until I touch this point. 279 00:17:58,230 --> 00:18:00,660 So basically, what is happening is 280 00:18:00,660 --> 00:18:03,210 like this-- all the points are only moving 281 00:18:03,210 --> 00:18:06,480 up and down horizontally. 282 00:18:06,480 --> 00:18:13,680 But they all move in a manner such 283 00:18:13,680 --> 00:18:20,800 that if you look at just the shape of the amplitude-- 284 00:18:20,800 --> 00:18:23,350 the amplitude is a function of x-- 285 00:18:23,350 --> 00:18:27,280 it looks as if the amplitude-- the shape 286 00:18:27,280 --> 00:18:32,140 is actually moving toward positive x direction. 287 00:18:32,140 --> 00:18:36,380 Moving toward the right-hand side of the board. 288 00:18:36,380 --> 00:18:38,510 So what is actually moving? 289 00:18:38,510 --> 00:18:43,160 What is moving is actually all the point like mass 290 00:18:43,160 --> 00:18:43,940 on the string-- 291 00:18:43,940 --> 00:18:46,070 they are only moving up and done. 292 00:18:46,070 --> 00:18:50,090 But they are moving together so nicely such 293 00:18:50,090 --> 00:18:53,750 that it looks as if the whole shape is actually 294 00:18:53,750 --> 00:18:58,010 shifting toward the positive x direction. 295 00:18:58,010 --> 00:18:59,590 Any questions so far? 296 00:19:02,140 --> 00:19:04,930 So I hope that's straight forward enough. 297 00:19:04,930 --> 00:19:11,470 And I would like to discuss with you a interesting situation. 298 00:19:14,940 --> 00:19:19,740 So we have learned that, OK, I can have, 299 00:19:19,740 --> 00:19:23,520 for example, triangular pulse. 300 00:19:23,520 --> 00:19:26,520 And I can have this triangular pulse moving 301 00:19:26,520 --> 00:19:30,350 in the positive x direction. 302 00:19:30,350 --> 00:19:34,000 And that we will also find out that the speed 303 00:19:34,000 --> 00:19:39,025 of the propagation is actually Vp because that's actually 304 00:19:39,025 --> 00:19:44,970 if you increase t and that's actually the speed of movement 305 00:19:44,970 --> 00:19:49,080 when you sample the shape and the f of Tau. 306 00:19:49,080 --> 00:19:52,110 So therefore, we can conclude that the speed 307 00:19:52,110 --> 00:19:58,770 of this triangular pulse is going to be Vp. 308 00:19:58,770 --> 00:20:03,180 If I have another triangular pulse starting 309 00:20:03,180 --> 00:20:06,600 from the right-hand side end of the string, 310 00:20:06,600 --> 00:20:09,950 they have exactly the same shape, exactly 311 00:20:09,950 --> 00:20:12,230 the same amplitude. 312 00:20:16,080 --> 00:20:21,790 So of course, according to what we actually wrote there, 313 00:20:21,790 --> 00:20:24,540 they are going to move forever. 314 00:20:24,540 --> 00:20:29,250 And at some point they will actually meet each other. 315 00:20:29,250 --> 00:20:32,460 And what is going to happen is that you 316 00:20:32,460 --> 00:20:34,350 will have some pulse, which is actually 317 00:20:34,350 --> 00:20:38,130 two times of the shape at some point. 318 00:20:38,130 --> 00:20:44,220 Because of the linearity of the wave equation. 319 00:20:44,220 --> 00:20:46,340 So that's actually pretty straight forward. 320 00:20:49,000 --> 00:20:53,130 However, if I consider another case, which is like this. 321 00:20:53,130 --> 00:20:55,580 So I have two progressing waves. 322 00:20:58,850 --> 00:21:02,120 One is actually going in the right-hand side direction. 323 00:21:02,120 --> 00:21:05,220 The other one is going to the left-hand side direction. 324 00:21:05,220 --> 00:21:09,410 They have exactly the same shape, but they have-- 325 00:21:09,410 --> 00:21:13,700 the amplitude is actually taking the minus sign. 326 00:21:13,700 --> 00:21:17,610 So they actually are exactly-- 327 00:21:17,610 --> 00:21:19,410 they have exactly the same amplitude, 328 00:21:19,410 --> 00:21:22,180 but pointing to a different direction. 329 00:21:22,180 --> 00:21:25,127 One is actually pointing upward. 330 00:21:25,127 --> 00:21:26,960 The other one is actually pointing downward. 331 00:21:31,460 --> 00:21:37,440 So at some point these two waves is going to overlap each other. 332 00:21:41,500 --> 00:21:45,070 When they overlap each other, what is going to happen? 333 00:21:45,070 --> 00:21:51,240 It's like this-- they are going to overlap each other. 334 00:21:51,240 --> 00:21:56,340 That means the amplitude will be cancelling each other. 335 00:21:56,340 --> 00:22:01,840 Then from the experiment you will see something like this. 336 00:22:01,840 --> 00:22:06,340 So now, this is the question I would like to ask you. 337 00:22:06,340 --> 00:22:09,860 What will happen next? 338 00:22:09,860 --> 00:22:12,380 The first possibility is that they cancel. 339 00:22:16,580 --> 00:22:18,590 Completely, they disappear. 340 00:22:21,120 --> 00:22:25,320 The second possibility is that, OK, they pass each other. 341 00:22:31,330 --> 00:22:34,270 The third possibility is that, OK, 342 00:22:34,270 --> 00:22:37,570 it depends on the mood of the string. 343 00:22:37,570 --> 00:22:40,680 Maybe something interesting is popping out. 344 00:22:40,680 --> 00:22:47,290 Maybe it decide to produce two circular waves. 345 00:22:47,290 --> 00:22:48,690 Get creative, right? 346 00:22:48,690 --> 00:22:49,190 Creative. 347 00:22:54,370 --> 00:22:55,870 OK, everybody have to vote. 348 00:22:55,870 --> 00:22:57,100 OK? 349 00:22:57,100 --> 00:23:01,930 How many of you think that you will cancel and disappear. 350 00:23:04,960 --> 00:23:06,760 Anybody? 351 00:23:06,760 --> 00:23:07,630 Nobody? 352 00:23:07,630 --> 00:23:08,130 Really? 353 00:23:11,060 --> 00:23:14,570 So you can see that here, nothings there. 354 00:23:14,570 --> 00:23:16,250 Right? 355 00:23:16,250 --> 00:23:18,300 Why didn't you think that will be canceled? 356 00:23:18,300 --> 00:23:20,620 OK, nobody think that will cancel. 357 00:23:20,620 --> 00:23:22,190 Very good. 358 00:23:22,190 --> 00:23:23,552 Maybe we are all wrong, right? 359 00:23:23,552 --> 00:23:25,160 [LAUGHTER] 360 00:23:25,160 --> 00:23:28,880 Second, they'll pass each other. 361 00:23:28,880 --> 00:23:30,300 How many of you think so? 362 00:23:44,750 --> 00:23:46,950 Very good. 363 00:23:46,950 --> 00:23:50,360 Finally, how many of you think that will be, 364 00:23:50,360 --> 00:23:53,330 oh, no, it depends on the mood of the string. 365 00:23:53,330 --> 00:23:55,010 Get creative. 366 00:23:55,010 --> 00:23:58,700 One, two, three-- thank you for the support. 367 00:23:58,700 --> 00:24:00,050 [LAUGHTER] 368 00:24:00,050 --> 00:24:01,320 There are four people. 369 00:24:01,320 --> 00:24:01,820 OK. 370 00:24:05,150 --> 00:24:10,900 So let's discuss these three situations carefully. 371 00:24:10,900 --> 00:24:15,470 So the first situation, if they cancel exactly, 372 00:24:15,470 --> 00:24:21,290 which sounds logical because if you look at this string 373 00:24:21,290 --> 00:24:25,965 how could this string remember what happened before? 374 00:24:29,180 --> 00:24:32,150 How could it remember? 375 00:24:32,150 --> 00:24:40,450 Therefore, shouldn't answer number one be a logical choice? 376 00:24:40,450 --> 00:24:45,760 The catch is, OK, if they cancel then that means energy is not 377 00:24:45,760 --> 00:24:47,410 conserved. 378 00:24:47,410 --> 00:24:50,230 So somehow the energy I put in-- 379 00:24:50,230 --> 00:24:55,750 I work really hard to shake the string, use my energy. 380 00:24:55,750 --> 00:24:57,070 And it disappear. 381 00:24:57,070 --> 00:24:58,180 Oh my god, disappear. 382 00:24:58,180 --> 00:25:01,420 [LAUGHTER] 383 00:25:01,420 --> 00:25:05,290 Then the energy is sad. 384 00:25:05,290 --> 00:25:12,860 The second one is, OK, I believe in energy conservation. 385 00:25:12,860 --> 00:25:14,870 So they will pass each other. 386 00:25:14,870 --> 00:25:20,680 But that means the string have memory because right 387 00:25:20,680 --> 00:25:22,880 now there's nothing there. 388 00:25:22,880 --> 00:25:25,630 What is going on? 389 00:25:25,630 --> 00:25:29,450 OK, since most of you think that is actually what is happening, 390 00:25:29,450 --> 00:25:34,010 can some of you explain to me how this string actually 391 00:25:34,010 --> 00:25:37,270 remember what happened before? 392 00:25:37,270 --> 00:25:40,965 Anybody can help me. 393 00:25:40,965 --> 00:25:42,462 AUDIENCE: Maybe the two light forms 394 00:25:42,462 --> 00:25:44,957 reflect off of each other. 395 00:25:44,957 --> 00:25:46,672 Bounce off of each other or something. 396 00:25:46,672 --> 00:25:48,380 PROFESSOR: Yeah, they balance each other, 397 00:25:48,380 --> 00:25:57,720 but how is this different from a stationary string at rest? 398 00:25:57,720 --> 00:26:00,420 Of course, I mean at some point it looks identical, right? 399 00:26:00,420 --> 00:26:04,890 But there's something which is different between this one 400 00:26:04,890 --> 00:26:07,420 and that one. 401 00:26:07,420 --> 00:26:09,308 AUDIENCE: There's no [INAUDIBLE].. 402 00:26:09,308 --> 00:26:11,668 No [INAUDIBLE] in the string. 403 00:26:11,668 --> 00:26:13,100 [INAUDIBLE] 404 00:26:13,100 --> 00:26:14,900 PROFESSOR: Very good point. 405 00:26:14,900 --> 00:26:21,560 This one, which is actually unperturbed, has zero velocity. 406 00:26:21,560 --> 00:26:24,050 And this one, no. 407 00:26:24,050 --> 00:26:26,600 It actually have a got velocity. 408 00:26:26,600 --> 00:26:30,500 Actually, this string is already or starting to-- 409 00:26:30,500 --> 00:26:34,100 it's already ready to move down. 410 00:26:34,100 --> 00:26:39,980 And this part of the string is already ready to move up. 411 00:26:39,980 --> 00:26:43,700 So that is actually how the string can 412 00:26:43,700 --> 00:26:45,940 remember what happened before. 413 00:26:45,940 --> 00:26:52,310 It remembered it by the velocity. 414 00:26:52,310 --> 00:26:55,710 So what is actually not plotted here is a trick. 415 00:26:55,710 --> 00:26:58,400 It's actually the velocity. 416 00:26:58,400 --> 00:27:04,530 The velocity is already nonzero compared to this situation. 417 00:27:04,530 --> 00:27:08,340 And what is going to happen is that afterward you 418 00:27:08,340 --> 00:27:14,480 will produce two corresponding triangular pulse, 419 00:27:14,480 --> 00:27:18,300 continue and then pass each other. 420 00:27:18,300 --> 00:27:21,980 Finally, the third condition, creative. 421 00:27:21,980 --> 00:27:27,200 That may not happen because all the memory is still there 422 00:27:27,200 --> 00:27:31,710 in the form of kinetic energy. 423 00:27:34,440 --> 00:27:41,820 So we can actually go ahead and do a small demonstration here. 424 00:27:41,820 --> 00:27:46,290 So let's focus on the right-hand side part of this setup. 425 00:27:46,290 --> 00:27:51,810 So this is actually the Bell Lab machine we had before. 426 00:27:51,810 --> 00:27:53,790 So now, what I'm going to do is now 427 00:27:53,790 --> 00:27:57,640 I'm going to create a square pulse-- positive square 428 00:27:57,640 --> 00:27:59,350 pulse from the lab inside. 429 00:27:59,350 --> 00:28:04,200 And the negative pulse in the right-hand side 430 00:28:04,200 --> 00:28:07,150 and see what is going to happen. 431 00:28:07,150 --> 00:28:08,981 No, not like this. 432 00:28:08,981 --> 00:28:09,480 Stop. 433 00:28:09,480 --> 00:28:10,050 Stop. 434 00:28:10,050 --> 00:28:11,130 OK. 435 00:28:11,130 --> 00:28:11,760 All right. 436 00:28:11,760 --> 00:28:14,160 Let's do it. 437 00:28:14,160 --> 00:28:15,140 You see? 438 00:28:15,140 --> 00:28:17,180 They pass each other. 439 00:28:17,180 --> 00:28:19,490 And the shape actually continues. 440 00:28:19,490 --> 00:28:22,500 So let's do that again. 441 00:28:22,500 --> 00:28:26,700 They cancel at some point, but they do pass each other 442 00:28:26,700 --> 00:28:27,370 and continue. 443 00:28:27,370 --> 00:28:29,430 And there are some refractions, et cetera, 444 00:28:29,430 --> 00:28:31,770 which we are going to discuss afterward. 445 00:28:31,770 --> 00:28:34,350 Let's do that again. 446 00:28:34,350 --> 00:28:35,140 You see? 447 00:28:35,140 --> 00:28:37,260 At some point, they cancel. 448 00:28:37,260 --> 00:28:40,410 But the positive pulse continue traveling 449 00:28:40,410 --> 00:28:42,570 to your left-hand side. 450 00:28:42,570 --> 00:28:46,630 And then the negative pulse travel to your right-hand side. 451 00:28:46,630 --> 00:28:47,567 Continue, please. 452 00:28:50,140 --> 00:28:55,420 So based on the experiment most of you actually were correct. 453 00:28:55,420 --> 00:28:59,020 The answer is number two. 454 00:28:59,020 --> 00:29:03,150 And I would like to show you a few more examples 455 00:29:03,150 --> 00:29:07,360 based on my little simulation. 456 00:29:07,360 --> 00:29:12,790 So first of all, I would like to show you a triangular pulse. 457 00:29:12,790 --> 00:29:14,890 They pass each other. 458 00:29:14,890 --> 00:29:18,550 And you can see that they pass each other, 459 00:29:18,550 --> 00:29:22,540 and the shape is actually changing as a function of time. 460 00:29:22,540 --> 00:29:25,930 And actually, afterward, they continue, 461 00:29:25,930 --> 00:29:29,140 and they keep the same shape based 462 00:29:29,140 --> 00:29:33,250 on this computer simulation. 463 00:29:33,250 --> 00:29:35,070 Another interesting thing to notice 464 00:29:35,070 --> 00:29:40,780 is that if you focus on the point at x equal to 0, 465 00:29:40,780 --> 00:29:44,880 you will see that at this point actually never change 466 00:29:44,880 --> 00:29:46,100 amplitude. 467 00:29:46,100 --> 00:29:49,890 Because those two pulse are really symmetric. 468 00:29:49,890 --> 00:29:54,190 One is positive, the other one is negative. 469 00:29:54,190 --> 00:29:57,470 As usual, we can actually change the shape of the pulse. 470 00:29:57,470 --> 00:30:00,250 For example, I can changed it to circular shape 471 00:30:00,250 --> 00:30:02,870 and see what will happen. 472 00:30:02,870 --> 00:30:03,600 Oh. 473 00:30:03,600 --> 00:30:05,940 [LAUGHTER] 474 00:30:05,940 --> 00:30:11,500 And again, the position at x equal to 0 is unchanged. 475 00:30:11,500 --> 00:30:15,070 Let's take a look at that again. 476 00:30:15,070 --> 00:30:17,600 It really does something really funny. 477 00:30:17,600 --> 00:30:18,541 It looks like, voom. 478 00:30:21,190 --> 00:30:22,890 And then you can see that it is actually 479 00:30:22,890 --> 00:30:29,960 the velocity of the individual component of the string. 480 00:30:29,960 --> 00:30:32,880 Which it remember though in the original shape. 481 00:30:32,880 --> 00:30:37,260 So it can see the velocity by eye looks 482 00:30:37,260 --> 00:30:41,650 different from what you see it before in the first example. 483 00:30:41,650 --> 00:30:46,110 And finally, as usual, we have the MIT waves. 484 00:30:46,110 --> 00:30:47,610 [LAUGHTER] 485 00:30:47,610 --> 00:30:51,330 And it does really, really crazy things. 486 00:30:51,330 --> 00:30:54,990 And the amazing thing is that the string 487 00:30:54,990 --> 00:30:58,160 have such a good memory. 488 00:30:58,160 --> 00:31:03,570 It really remember what is going to happen 489 00:31:03,570 --> 00:31:05,490 before they touch each other. 490 00:31:11,700 --> 00:31:14,650 So what is going to happen to these two MIT waves? 491 00:31:14,650 --> 00:31:17,820 They are going to be propagating forever. 492 00:31:17,820 --> 00:31:21,240 Cannot stop until the edge of the universe. 493 00:31:21,240 --> 00:31:25,090 Maybe they dig out of the universe, but not my problem 494 00:31:25,090 --> 00:31:27,970 anymore. 495 00:31:27,970 --> 00:31:31,180 So we talk about the energy stored 496 00:31:31,180 --> 00:31:34,060 in the string and et cetera. 497 00:31:34,060 --> 00:31:38,140 So how about we go ahead and calculate the kinetic energy 498 00:31:38,140 --> 00:31:39,860 and the potential energy. 499 00:31:39,860 --> 00:31:42,570 So the first part is the kinetic energy. 500 00:31:50,670 --> 00:31:55,350 Only one is actually half mv square. 501 00:31:55,350 --> 00:32:02,740 So if I consider a small segment on the string, which 502 00:32:02,740 --> 00:32:05,640 have a width of delta x. 503 00:32:05,640 --> 00:32:08,580 And I can now calculate delta m. 504 00:32:08,580 --> 00:32:13,720 If I assume this string have a mass per unit length Rho 505 00:32:13,720 --> 00:32:17,100 L, and the string tension t. 506 00:32:17,100 --> 00:32:21,810 If that's actually given to you when we set up the experiment, 507 00:32:21,810 --> 00:32:28,320 then we can actually calculate the mass of this small portion 508 00:32:28,320 --> 00:32:29,430 of the string. 509 00:32:29,430 --> 00:32:35,160 Then delta m, the mass, will be equal to Rho L times dx. 510 00:32:35,160 --> 00:32:40,050 Because Rho L is the mass per unit length. 511 00:32:40,050 --> 00:32:42,930 Therefore, what is actually the kinetic energy 512 00:32:42,930 --> 00:32:44,880 is becoming pretty straight forward. 513 00:32:44,880 --> 00:32:49,350 It's the integration over the whole string. 514 00:32:49,350 --> 00:32:53,280 Integration over the whole string is 1/2-- 515 00:32:53,280 --> 00:32:55,656 based on this equation-- 516 00:32:55,656 --> 00:33:01,970 Rho L, dx, and times v. 517 00:33:01,970 --> 00:33:04,330 But what is actually v here? 518 00:33:04,330 --> 00:33:09,600 v is actually the velocity of individual point-like mass 519 00:33:09,600 --> 00:33:10,320 on the string. 520 00:33:13,180 --> 00:33:16,110 And we actually already talked about that. 521 00:33:16,110 --> 00:33:19,470 The velocity of individual mass is actually only 522 00:33:19,470 --> 00:33:21,690 in the y direction. 523 00:33:21,690 --> 00:33:24,690 And the position of individual mass 524 00:33:24,690 --> 00:33:30,930 is described by the function Psi. 525 00:33:30,930 --> 00:33:33,030 Therefore, what is actually velocity? 526 00:33:33,030 --> 00:33:38,820 Velocity is actually partial Psi, partial t. 527 00:33:38,820 --> 00:33:44,010 So that is actually giving you the velocity of individual mass 528 00:33:44,010 --> 00:33:45,240 on the string. 529 00:33:45,240 --> 00:33:48,900 And then if you square that, that is actually giving you 530 00:33:48,900 --> 00:33:51,580 the total kinetic energy-- 531 00:33:51,580 --> 00:33:53,216 is in this functional form. 532 00:33:57,020 --> 00:33:59,910 Let's also discuss what is actually the potential energy. 533 00:34:04,800 --> 00:34:12,989 The potential energy as you remember delta W, the work, 534 00:34:12,989 --> 00:34:18,460 is equal to F times delta S, the displacement. 535 00:34:18,460 --> 00:34:23,730 F is the force, and the delta S is the displacement. 536 00:34:23,730 --> 00:34:28,510 So originally, before we actually 537 00:34:28,510 --> 00:34:31,800 perturb and make some displacement with respect 538 00:34:31,800 --> 00:34:34,020 to equilibrium position-- 539 00:34:34,020 --> 00:34:37,010 this string have originally-- if I 540 00:34:37,010 --> 00:34:41,960 look at this small part of the string I join in this region. 541 00:34:41,960 --> 00:34:43,940 This looks like this. 542 00:34:43,940 --> 00:34:50,750 This is delta x, and it has a constant string tension t. 543 00:34:50,750 --> 00:34:57,710 Now, I can actually introduce some displacement. 544 00:34:57,710 --> 00:35:00,010 And what is going to happen is, look, 545 00:35:00,010 --> 00:35:02,390 it's going to look like this. 546 00:35:02,390 --> 00:35:08,030 This string is actually a little bit stretched. 547 00:35:08,030 --> 00:35:13,430 And this is actually the original delta x. 548 00:35:13,430 --> 00:35:16,880 The width of this little segment. 549 00:35:16,880 --> 00:35:23,450 And this direction is actually a small change 550 00:35:23,450 --> 00:35:33,210 in the y direction, which is actually showing us delta Psi. 551 00:35:33,210 --> 00:35:36,060 And of course, we can calculate the length. 552 00:35:36,060 --> 00:35:39,400 The length of this string and that 553 00:35:39,400 --> 00:35:44,310 will give you square root of delta x square plus delta Psi 554 00:35:44,310 --> 00:35:44,810 square. 555 00:35:49,980 --> 00:35:56,800 We can now go ahead and calculate the delta W. 556 00:35:56,800 --> 00:36:03,160 So delta W will be equal to F, which is the force, times delta 557 00:36:03,160 --> 00:36:08,680 S. We know in the force-- the magnitude of the force is what? 558 00:36:08,680 --> 00:36:11,970 Is the string tension. 559 00:36:11,970 --> 00:36:14,690 So therefore, I put T here. 560 00:36:14,690 --> 00:36:17,320 And delta S, what is delta S? 561 00:36:17,320 --> 00:36:22,700 Is how much I stretch this string. 562 00:36:22,700 --> 00:36:24,320 So this is actually the difference 563 00:36:24,320 --> 00:36:32,120 between the resulting length and the original length, delta x. 564 00:36:32,120 --> 00:36:38,540 So that is actually giving you the delta S. So 565 00:36:38,540 --> 00:36:43,100 that means I can write it in this functional form. 566 00:36:43,100 --> 00:36:48,640 dx square plus d Psi square minus dx. 567 00:36:53,220 --> 00:37:01,080 I can of course take delta x out of this square root thing, 568 00:37:01,080 --> 00:37:05,380 and basically I get delta x, square root of 1 569 00:37:05,380 --> 00:37:11,600 plus d Psi dx square minus dx. 570 00:37:14,130 --> 00:37:20,280 Remember what we have been discussing until now, 571 00:37:20,280 --> 00:37:24,510 we were always discussing small amplitude-- 572 00:37:24,510 --> 00:37:27,570 or small vibration. 573 00:37:27,570 --> 00:37:35,220 Therefore, that means I can use a small angle approximation. 574 00:37:35,220 --> 00:37:38,670 That means delta Psi is going to be very, very 575 00:37:38,670 --> 00:37:42,720 small with respect to delta x. 576 00:37:42,720 --> 00:37:54,490 So that means the first turn will be roughly delta x 1 577 00:37:54,490 --> 00:38:03,220 plus d Psi dx squared 1/2 because you 578 00:38:03,220 --> 00:38:08,110 have a square root of that, plus higher order turn. 579 00:38:15,970 --> 00:38:19,780 And of course we assume that delta Psi is actually 580 00:38:19,780 --> 00:38:21,850 much smaller than delta x. 581 00:38:21,850 --> 00:38:27,580 Therefore, we ignore all those higher order terms. 582 00:38:27,580 --> 00:38:30,810 So if we actually replace this expression back 583 00:38:30,810 --> 00:38:33,220 into the original equation, you will 584 00:38:33,220 --> 00:38:39,790 see that the first turn, 1 cancel with this minus dx turn. 585 00:38:39,790 --> 00:38:42,556 This actually cancel that. 586 00:38:42,556 --> 00:38:44,950 They actually cancel. 587 00:38:44,950 --> 00:38:53,020 Therefore, I can calculate dW will be equal to T times delta 588 00:38:53,020 --> 00:39:01,690 x, times 1/2 d Psi dx square. 589 00:39:01,690 --> 00:39:05,630 Therefore, what will be the total potential energy. 590 00:39:05,630 --> 00:39:08,470 The total potential energy will be 591 00:39:08,470 --> 00:39:17,850 in the equation of the work dW over the whole range from-- 592 00:39:17,850 --> 00:39:20,140 of the system. 593 00:39:20,140 --> 00:39:29,010 And basically, you can actually write it down as 1/2 T Psi. 594 00:39:33,260 --> 00:39:37,610 Partial Psi, partial x square dx. 595 00:39:40,890 --> 00:39:45,810 All right, so we can actually understand and calculate 596 00:39:45,810 --> 00:39:49,600 the kinetic energy and the potential energy. 597 00:39:49,600 --> 00:39:55,170 So before we take a break, let's take a short example 598 00:39:55,170 --> 00:39:59,640 to check if we understand what we have learned so far. 599 00:39:59,640 --> 00:40:03,180 So for example, if I have a function 600 00:40:03,180 --> 00:40:11,070 Psi xt, and that is actually equal to 1 over 1 601 00:40:11,070 --> 00:40:15,820 plus x minus 3t to the fourth. 602 00:40:15,820 --> 00:40:17,430 It's a crazy function. 603 00:40:17,430 --> 00:40:22,460 If I assume that I can do a very precise thing, 604 00:40:22,460 --> 00:40:26,670 manipulate this string so that I produce a wave function 605 00:40:26,670 --> 00:40:28,080 of this functional form. 606 00:40:28,080 --> 00:40:32,580 1 over 1 plus x minus 3t to the fourth. 607 00:40:32,580 --> 00:40:35,100 Can somebody tell me what is actually going 608 00:40:35,100 --> 00:40:40,270 to be the velocity of the wave? 609 00:40:40,270 --> 00:40:41,520 Can anybody tell me? 610 00:40:45,882 --> 00:40:47,340 The first thing which you can do is 611 00:40:47,340 --> 00:40:50,610 to express this crazy function in a functional 612 00:40:50,610 --> 00:40:54,480 form of fx minus Vpt, right? 613 00:40:54,480 --> 00:40:59,220 And the Vp is actually the speed of the wave, right? 614 00:40:59,220 --> 00:41:01,590 So anybody know what is actually-- yes? 615 00:41:01,590 --> 00:41:02,550 AUDIENCE: Three. 616 00:41:02,550 --> 00:41:05,340 PROFESSOR: Yeah, the three because the whole function 617 00:41:05,340 --> 00:41:08,060 can be written as f x minus 3t. 618 00:41:10,830 --> 00:41:16,140 Therefore, the velocity Vp will be 3. 619 00:41:16,140 --> 00:41:18,720 Of course if you are not sure, you 620 00:41:18,720 --> 00:41:22,820 can actually calculate Vp square by the ratio 621 00:41:22,820 --> 00:41:30,190 of partial square Psi, partial t square, and partial square Psi, 622 00:41:30,190 --> 00:41:31,830 partial x square. 623 00:41:31,830 --> 00:41:34,800 And that will give you of course the Vp square, 624 00:41:34,800 --> 00:41:38,640 according to that wave equation. 625 00:41:38,640 --> 00:41:42,600 All right, so we will take a five minute break from now. 626 00:41:42,600 --> 00:41:49,080 And during the break I will try to return the exam to you. 627 00:41:49,080 --> 00:41:53,130 So we will come back at 24-- 628 00:41:53,130 --> 00:41:53,940 12:24. 629 00:42:03,880 --> 00:42:06,050 So welcome back, everybody. 630 00:42:06,050 --> 00:42:09,790 So we will continue the discussion of traveling wave. 631 00:42:12,680 --> 00:42:14,710 So we have the very interesting discussion 632 00:42:14,710 --> 00:42:17,750 of two waves that are canceling each other. 633 00:42:17,750 --> 00:42:20,950 And somehow the string have a way 634 00:42:20,950 --> 00:42:23,890 to remember what happened before, which is actually 635 00:42:23,890 --> 00:42:27,850 the velocity of each individual point on the string 636 00:42:27,850 --> 00:42:31,340 as a function of x-- 637 00:42:31,340 --> 00:42:33,190 that instance of time. 638 00:42:33,190 --> 00:42:36,620 So let's actually take a look at this example. 639 00:42:36,620 --> 00:42:41,640 So make use of what we have learned so far. 640 00:42:41,640 --> 00:42:46,410 As we see here there is a triangular shape, 641 00:42:46,410 --> 00:42:49,990 which I create in the lab. 642 00:42:49,990 --> 00:42:55,420 And this triangular shape is actually there 643 00:42:55,420 --> 00:42:57,350 and it's stationary. 644 00:42:57,350 --> 00:42:59,110 It's not moving. 645 00:42:59,110 --> 00:43:04,870 The strings are at rest, but have a triangular shape, 646 00:43:04,870 --> 00:43:07,180 which I setup there. 647 00:43:07,180 --> 00:43:10,190 So based on what we have learned so far-- 648 00:43:10,190 --> 00:43:13,110 we have learned normal modes, we have 649 00:43:13,110 --> 00:43:17,290 learned about traveling wave. 650 00:43:17,290 --> 00:43:21,670 I believe before we learned this class, the first reaction 651 00:43:21,670 --> 00:43:25,020 to you is to do what? 652 00:43:25,020 --> 00:43:26,784 What kind of decomposition. 653 00:43:26,784 --> 00:43:27,700 AUDIENCE: [INAUDIBLE]. 654 00:43:27,700 --> 00:43:30,706 PROFESSOR: Fourier decomposition. 655 00:43:30,706 --> 00:43:34,210 So what you are going to do is, OK, very good. 656 00:43:34,210 --> 00:43:34,990 I have this shape. 657 00:43:34,990 --> 00:43:37,420 So I do a Fourier decomposition and I 658 00:43:37,420 --> 00:43:39,310 have infinite number of terms. 659 00:43:39,310 --> 00:43:43,120 And I am going to evolve infinite number of term 660 00:43:43,120 --> 00:43:47,930 as a function of time and see what will happen to the system. 661 00:43:47,930 --> 00:43:51,590 So that's actually what you would do before we 662 00:43:51,590 --> 00:43:52,750 learned traveling wave. 663 00:43:55,300 --> 00:43:58,900 What I would like to say today is that if I really 664 00:43:58,900 --> 00:44:06,185 prepare this string at rest, stationary, at t equal to 0-- 665 00:44:09,070 --> 00:44:13,260 in contrast to what I just said before, 666 00:44:13,260 --> 00:44:16,140 brute force measure, which I used computer 667 00:44:16,140 --> 00:44:20,280 to decompose it and evolve all the infinite number of terms. 668 00:44:20,280 --> 00:44:25,800 What we could do is that I can show you 669 00:44:25,800 --> 00:44:30,950 that this situation is a superposition of two traveling 670 00:44:30,950 --> 00:44:31,450 waves. 671 00:44:34,410 --> 00:44:36,450 Then the question becomes super simple. 672 00:44:39,000 --> 00:44:45,420 So instead of doing a brute force calculation 673 00:44:45,420 --> 00:44:46,810 using computer-- 674 00:44:46,810 --> 00:44:50,430 decompose it to infinite number of normal modes-- 675 00:44:50,430 --> 00:44:52,860 what I can actually show you is that, OK, 676 00:44:52,860 --> 00:45:08,340 if I have a g function, which is equal to f x plus Vpt plus f x 677 00:45:08,340 --> 00:45:11,190 minus Vpt. 678 00:45:11,190 --> 00:45:16,110 So these three function is superposition of two 679 00:45:16,110 --> 00:45:19,050 traveling waves. 680 00:45:19,050 --> 00:45:22,920 The shape is described by f function. 681 00:45:22,920 --> 00:45:26,910 And one of them is traveling to the right-hand side. 682 00:45:26,910 --> 00:45:30,360 The other one is actually traveling to a left-hand side. 683 00:45:30,360 --> 00:45:34,020 If I assume that the superposition of these two 684 00:45:34,020 --> 00:45:40,990 traveling wave is g then I can now calculate the velocity, 685 00:45:40,990 --> 00:45:43,590 partial g, partial t. 686 00:45:43,590 --> 00:45:50,190 And that will give you Vp f prime minus-- 687 00:45:50,190 --> 00:45:52,950 right, because here it's actually x minus Vp-- 688 00:45:52,950 --> 00:45:57,760 so I got minus Vp out of it, f prime. 689 00:45:57,760 --> 00:46:01,230 The first term, if I do this partial differentiation, 690 00:46:01,230 --> 00:46:04,755 then basically I get the positive Vp out of it. 691 00:46:04,755 --> 00:46:09,540 And then the second term I get minus Vp out of it. 692 00:46:09,540 --> 00:46:14,950 And these two terms cancel exactly. 693 00:46:14,950 --> 00:46:16,660 What does that mean? 694 00:46:16,660 --> 00:46:21,100 That means all the points-- 695 00:46:21,100 --> 00:46:24,730 this g is actually a function of x and t-- 696 00:46:24,730 --> 00:46:32,320 on the point at t equal to 0 have initial d 697 00:46:32,320 --> 00:46:34,752 velocity equal to 0. 698 00:46:37,760 --> 00:46:43,970 So in other word, if I have any random kind of shape-- 699 00:46:43,970 --> 00:46:48,260 in this case is a triangular shape-- 700 00:46:48,260 --> 00:46:53,890 I can always decompose this stationary shape 701 00:46:53,890 --> 00:46:58,190 into two traveling wave. 702 00:46:58,190 --> 00:47:02,310 One is actually traveling in the positive direction. 703 00:47:02,310 --> 00:47:06,110 The other one is traveling in the negative direction. 704 00:47:06,110 --> 00:47:08,720 So that means what is happening? 705 00:47:08,720 --> 00:47:17,690 This is equal to a superposition of two traveling waves. 706 00:47:17,690 --> 00:47:24,800 If I assume the height of this mountain to be h then 707 00:47:24,800 --> 00:47:30,230 I need to have h over 2 as the height 708 00:47:30,230 --> 00:47:33,020 for the individual traveling waves. 709 00:47:33,020 --> 00:47:37,430 One is actually traveling to the right-hand side. 710 00:47:37,430 --> 00:47:39,470 The other one is actually traveling to he 711 00:47:39,470 --> 00:47:42,140 left-hand side of the board. 712 00:47:42,140 --> 00:47:46,640 So based on this trick, actually we 713 00:47:46,640 --> 00:47:51,750 can see that I don't need to do infinite number of terms 714 00:47:51,750 --> 00:47:53,120 anymore. 715 00:47:53,120 --> 00:47:55,520 I don't need to do a Fourier decomposition 716 00:47:55,520 --> 00:47:59,980 and get really crazy and take forever 717 00:47:59,980 --> 00:48:03,410 to write the code on your computer. 718 00:48:03,410 --> 00:48:06,830 And maybe there are some bug in your code, 719 00:48:06,830 --> 00:48:09,360 which is frustrating. 720 00:48:09,360 --> 00:48:14,390 And what we could do is to simply decompose it 721 00:48:14,390 --> 00:48:16,310 into two traveling wave. 722 00:48:16,310 --> 00:48:19,180 And I can now predict what will happened 723 00:48:19,180 --> 00:48:24,080 at time equal to, for example, 5 what is going to happen. 724 00:48:24,080 --> 00:48:28,670 So what is going to happen is that you have two triangular 725 00:48:28,670 --> 00:48:31,870 shape waves. 726 00:48:31,870 --> 00:48:38,090 Each of them actually traveled a distance of 5 times Vp. 727 00:48:45,030 --> 00:48:50,190 So that is actually a very interesting fact. 728 00:48:50,190 --> 00:48:52,560 And of course we can see from here 729 00:48:52,560 --> 00:48:56,430 if I quickly create a triangular shape, 730 00:48:56,430 --> 00:49:00,610 and you will see that it really did become two triangular shape 731 00:49:00,610 --> 00:49:01,440 wave. 732 00:49:01,440 --> 00:49:02,400 So I can do this. 733 00:49:06,400 --> 00:49:08,240 I can do this. 734 00:49:08,240 --> 00:49:08,910 You see? 735 00:49:08,910 --> 00:49:15,840 So originally, I'm creating some stationary shape. 736 00:49:15,840 --> 00:49:17,670 And I release that. 737 00:49:17,670 --> 00:49:22,020 It does become two traveling waves 738 00:49:22,020 --> 00:49:24,810 with amplitude half of the original height-- 739 00:49:24,810 --> 00:49:26,330 original displacement. 740 00:49:26,330 --> 00:49:32,140 I can also do it in the opposite direction, a positive wave. 741 00:49:32,140 --> 00:49:34,100 You see? 742 00:49:34,100 --> 00:49:35,040 It does work. 743 00:49:35,040 --> 00:49:37,050 And of course, after the class you 744 00:49:37,050 --> 00:49:40,690 can make even more complicated shape 745 00:49:40,690 --> 00:49:43,630 if I have many more than two hands. 746 00:49:43,630 --> 00:49:46,215 Maybe I can do that, but unfortunately I'm human. 747 00:49:49,670 --> 00:49:53,900 You can see that I can create different slope 748 00:49:53,900 --> 00:49:55,970 in the positive and negative h. 749 00:49:55,970 --> 00:50:00,110 And it does create two traveling wave. 750 00:50:00,110 --> 00:50:03,860 And that's amazing because this is actually 751 00:50:03,860 --> 00:50:07,400 looks like just some kind of mathematical trick. 752 00:50:07,400 --> 00:50:13,250 And it really match with what we can do experimentally. 753 00:50:13,250 --> 00:50:19,620 So finally, I would like to talk about the last topic 754 00:50:19,620 --> 00:50:29,370 of the lecture today, which is to connect 755 00:50:29,370 --> 00:50:32,790 two strings together. 756 00:50:32,790 --> 00:50:35,720 So suppose I have two strings-- 757 00:50:35,720 --> 00:50:39,470 the left-hand side string is actually thinner. 758 00:50:39,470 --> 00:50:47,130 It has mass per unit length Rho L, and string tension t. 759 00:50:47,130 --> 00:50:56,060 In the right-hand side you can have a thicker string with mass 760 00:50:56,060 --> 00:51:00,980 per unit length 4 times Rho L, and the string tension 761 00:51:00,980 --> 00:51:02,190 is as you keep constant t. 762 00:51:05,250 --> 00:51:08,310 Based on what we have learned before, 763 00:51:08,310 --> 00:51:17,400 the velocity Vp, is equal to square root of t over Rho L. 764 00:51:17,400 --> 00:51:21,100 So that's actually from the last few lectures. 765 00:51:21,100 --> 00:51:24,600 So left-hand side you will have V1, 766 00:51:24,600 --> 00:51:28,780 which is the velocity of the traveling wave, equal to square 767 00:51:28,780 --> 00:51:31,410 root of t over Rho L. 768 00:51:31,410 --> 00:51:35,450 And the right-hand side you will have square root of t 769 00:51:35,450 --> 00:51:44,160 over 4 Rho L. And that will give you one half of V1. 770 00:51:44,160 --> 00:51:45,580 So what does that mean? 771 00:51:45,580 --> 00:51:47,500 This means that if I have a traveling 772 00:51:47,500 --> 00:51:52,370 wave in the left-hand side, the speed of the traveling wave 773 00:51:52,370 --> 00:51:56,935 will be 2 times the speed of the traveling 774 00:51:56,935 --> 00:52:02,640 wave in the right-hand side based on this calculation. 775 00:52:02,640 --> 00:52:07,110 So what I would like to do is the following-- 776 00:52:07,110 --> 00:52:11,880 so I would like to ask a question about this system. 777 00:52:11,880 --> 00:52:18,950 What will happen if I introduce a displacement and a traveling 778 00:52:18,950 --> 00:52:22,340 wave from the left-hand side. 779 00:52:22,340 --> 00:52:24,770 And the question is, what is going 780 00:52:24,770 --> 00:52:27,780 to happen to this system as a function of time 781 00:52:27,780 --> 00:52:31,530 once I actually give this input traveling wave. 782 00:52:31,530 --> 00:52:35,200 And the answer is that this traveling wave 783 00:52:35,200 --> 00:52:39,410 is going to pass through the boundary of two systems. 784 00:52:39,410 --> 00:52:41,630 And there may be refraction. 785 00:52:41,630 --> 00:52:44,400 There may be transmission, et cetera. 786 00:52:44,400 --> 00:52:47,130 And that we are actually in the good position 787 00:52:47,130 --> 00:52:50,510 to understand this phenomena. 788 00:52:50,510 --> 00:52:54,300 So let's take a look at this situation carefully. 789 00:52:54,300 --> 00:53:01,070 So now I define here the position of the boundary 790 00:53:01,070 --> 00:53:05,330 is at x equal to 0. 791 00:53:05,330 --> 00:53:12,200 And I can now go ahead and write down the conditions, which is-- 792 00:53:12,200 --> 00:53:14,930 need to be satisfied in order to connect these two 793 00:53:14,930 --> 00:53:17,030 systems properly. 794 00:53:17,030 --> 00:53:21,620 Which you actually already see this several times, 795 00:53:21,620 --> 00:53:24,700 the boundary condition. 796 00:53:28,230 --> 00:53:30,720 So what are the boundary conditions 797 00:53:30,720 --> 00:53:34,660 which I need in order to connect the left-hand side 798 00:53:34,660 --> 00:53:37,920 and the right-hand side systems? 799 00:53:37,920 --> 00:53:40,320 So the first boundary condition is 800 00:53:40,320 --> 00:53:43,950 that the string is continuous. 801 00:53:43,950 --> 00:53:47,835 Therefore, if I have some kind of y 802 00:53:47,835 --> 00:53:51,540 is actually-- y of x t is describing 803 00:53:51,540 --> 00:53:56,150 the displacement of all the time mass on the string 804 00:53:56,150 --> 00:53:58,560 in a horizontal direction. 805 00:53:58,560 --> 00:54:03,010 Then that means y the left-hand side evaluated 806 00:54:03,010 --> 00:54:10,290 at 0 minus in the slightly left-hand side of the boundary 807 00:54:10,290 --> 00:54:17,700 will be equal to YR, which is actually evaluated 808 00:54:17,700 --> 00:54:21,260 at the slightly right-hand side of the boundary 809 00:54:21,260 --> 00:54:23,700 at x equal to 0. 810 00:54:23,700 --> 00:54:29,700 And the YL is actually the wave function for the left-hand side 811 00:54:29,700 --> 00:54:31,590 thinner string. 812 00:54:31,590 --> 00:54:37,630 And the YR is actually the wave function, 813 00:54:37,630 --> 00:54:41,940 which describe the right-hand side of the string. 814 00:54:41,940 --> 00:54:46,740 So this means that the boundary condition tells us that 815 00:54:46,740 --> 00:54:53,970 the string cannot break. 816 00:54:53,970 --> 00:54:57,120 It should match carefully so that these two 817 00:54:57,120 --> 00:55:00,730 systems are connected to each other properly. 818 00:55:00,730 --> 00:55:04,370 The second condition is that, OK, 819 00:55:04,370 --> 00:55:13,130 since this boundary actually have no massive particles left, 820 00:55:13,130 --> 00:55:17,520 I can't actually assume that this is massless ring there. 821 00:55:17,520 --> 00:55:22,450 Therefore, the slope of the left-hand side, 822 00:55:22,450 --> 00:55:28,350 partial YL, partial x, s equal to 0, 823 00:55:28,350 --> 00:55:32,670 will have to be equal to the slope at the right-hand side. 824 00:55:38,030 --> 00:55:41,080 If the slope doesn't match between the left-hand side 825 00:55:41,080 --> 00:55:44,900 and right-hand side, that means since they have constant 826 00:55:44,900 --> 00:55:49,040 tension that means the tension-- the string tension cannot 827 00:55:49,040 --> 00:55:50,510 cancel each other. 828 00:55:50,510 --> 00:55:54,080 Then the massless ring will be transferred to, 829 00:55:54,080 --> 00:55:56,510 for example Mars, in a second. 830 00:55:56,510 --> 00:55:59,840 Because it has few infinite amount of acceleration. 831 00:55:59,840 --> 00:56:01,130 And that didn't happen. 832 00:56:01,130 --> 00:56:07,340 When I actually tried to actually displace 833 00:56:07,340 --> 00:56:12,410 the string or the Bell Labs system, 834 00:56:12,410 --> 00:56:14,730 I didn't see crazy things happen. 835 00:56:14,730 --> 00:56:18,950 Therefore, the tension at this, which 836 00:56:18,950 --> 00:56:22,110 acting on this massless ring must cancel each other. 837 00:56:22,110 --> 00:56:25,676 So that's the second boundary condition we have. 838 00:56:28,760 --> 00:56:33,410 So now, I would like to make some assumption. 839 00:56:33,410 --> 00:56:39,350 So first of all I have an input pulse, which is actually 840 00:56:39,350 --> 00:56:43,310 coming into this system. 841 00:56:43,310 --> 00:56:44,180 Looks like this. 842 00:56:44,180 --> 00:56:48,980 And I call it fi, is traveling toward 843 00:56:48,980 --> 00:56:53,150 the positive x direction. 844 00:56:53,150 --> 00:56:56,526 So therefore, I can of course write it down 845 00:56:56,526 --> 00:57:00,670 as minus k1x plus omega t. 846 00:57:05,850 --> 00:57:12,030 So this is actually the incident pulse, I call it fi. 847 00:57:12,030 --> 00:57:15,600 And after it pass the boundary-- 848 00:57:15,600 --> 00:57:18,870 so I can actually expect that there 849 00:57:18,870 --> 00:57:24,185 may be some kind of refraction, which happened at the boundary, 850 00:57:24,185 --> 00:57:26,610 fr, I call it fr. 851 00:57:26,610 --> 00:57:31,250 And this time this fr is going to be traveling 852 00:57:31,250 --> 00:57:33,870 to the negative x direction. 853 00:57:33,870 --> 00:57:37,260 Therefore, I can express this function 854 00:57:37,260 --> 00:57:43,560 as fr is a function of plus k1x plus omega t. 855 00:57:47,440 --> 00:57:52,730 Finally, there can be also transmission wave. 856 00:57:52,730 --> 00:57:54,740 So you get the refraction and there 857 00:57:54,740 --> 00:57:57,080 can be some energy, which somehow 858 00:57:57,080 --> 00:57:58,970 pass through the boundary. 859 00:57:58,970 --> 00:58:03,080 And I call this transmission wave ft, 860 00:58:03,080 --> 00:58:08,990 which is actually in a form of minus k2x plus omega t. 861 00:58:12,310 --> 00:58:19,920 And in this case, I assume that the system is actually 862 00:58:19,920 --> 00:58:22,860 having a k1 in the left-hand side, 863 00:58:22,860 --> 00:58:24,930 and k2 in right-hand side. 864 00:58:24,930 --> 00:58:26,690 Which is actually the wave number 865 00:58:26,690 --> 00:58:31,940 and the k1 is actually equal to omega over V1. 866 00:58:31,940 --> 00:58:35,420 And the k2 is actually equal to omega over V2. 867 00:58:38,810 --> 00:58:40,370 So that is actually the set up. 868 00:58:40,370 --> 00:58:43,330 And then also the three traveling 869 00:58:43,330 --> 00:58:48,030 waves, which we actually demonstrate the situation. 870 00:58:48,030 --> 00:58:52,910 So we can now go ahead and plug those three traveling 871 00:58:52,910 --> 00:58:56,870 wave solution into the boundary conditions. 872 00:58:56,870 --> 00:58:58,790 And then we will be able to solve 873 00:58:58,790 --> 00:59:00,870 their relative amplitudes. 874 00:59:04,600 --> 00:59:07,926 So let's make use of the first boundary condition. 875 00:59:12,980 --> 00:59:21,390 So YL is now a superposition of fi and fr. 876 00:59:21,390 --> 00:59:28,260 YR will be just the transmission wave, ft. 877 00:59:28,260 --> 00:59:31,860 So now, I can plug this expression back 878 00:59:31,860 --> 00:59:34,390 into the equation number one. 879 00:59:34,390 --> 00:59:41,880 Then basically, what I get is fi omega t. 880 00:59:41,880 --> 00:59:45,390 Originally, it's actually minus k1x plus omega t, 881 00:59:45,390 --> 00:59:50,690 but this thing is actually evaluated at x equal to 0. 882 00:59:50,690 --> 00:59:53,610 The wave function has to be continuous 883 00:59:53,610 --> 00:59:57,780 between the negative side of 0 and the positive side of 0. 884 00:59:57,780 --> 01:00:03,600 Therefore, if I plug in x equal to 0, minus k1 turn disappear. 885 01:00:03,600 --> 01:00:07,020 And what is left over is omega t. 886 01:00:07,020 --> 01:00:12,250 And this is the second turn fr, I can write down expressively. 887 01:00:12,250 --> 01:00:16,150 You get fr omega t. 888 01:00:16,150 --> 01:00:20,050 And then right-hand side of the expression is YR, 889 01:00:20,050 --> 01:00:25,263 only have 1 turn, ft And now you are going to get ft omega t. 890 01:00:29,070 --> 01:00:34,700 So now we can also go ahead and plug in this equation 891 01:00:34,700 --> 01:00:36,480 to equation number two. 892 01:00:39,810 --> 01:00:44,940 What is going to happen is that I do a partial differentiation 893 01:00:44,940 --> 01:00:45,960 with respect to x. 894 01:00:45,960 --> 01:00:50,530 And the plug in x equal to 0 to the expression. 895 01:00:50,530 --> 01:00:54,900 And what I'm going to get is minus k1 896 01:00:54,900 --> 01:01:00,200 f prime i, as a function of omega t, 897 01:01:00,200 --> 01:01:07,250 plus k1 fr prime omega t. 898 01:01:07,250 --> 01:01:10,930 And this will be equal to minus k2. 899 01:01:10,930 --> 01:01:15,030 In the right-hand side you only have one turn, which is ft. 900 01:01:15,030 --> 01:01:21,894 So you are going to have minus k2 ft, the function of omega t. 901 01:01:24,540 --> 01:01:28,030 Any questions so far? 902 01:01:28,030 --> 01:01:31,530 AUDIENCE: [INAUDIBLE]. 903 01:01:31,530 --> 01:01:34,200 PROFESSOR: Yes, thank you very much. 904 01:01:34,200 --> 01:01:35,490 OK, very good. 905 01:01:35,490 --> 01:01:37,450 So we are making progress here. 906 01:01:37,450 --> 01:01:42,840 And what I can do now is to do a integration over t 907 01:01:42,840 --> 01:01:45,450 for the equation number two. 908 01:01:45,450 --> 01:01:48,270 So if I do a integration basically, 909 01:01:48,270 --> 01:01:55,120 what I'm going to get is minus k1 over 2 fi omega t. 910 01:01:55,120 --> 01:02:00,020 I do a integration over t, plus-- 911 01:02:00,020 --> 01:02:11,060 over omega, sorry-- and the plus K1 over omega fr omega t. 912 01:02:11,060 --> 01:02:18,887 And this is actually equal to minus K2 over omega ft omega t. 913 01:02:21,560 --> 01:02:25,150 Based on the equation, which we have before-- 914 01:02:25,150 --> 01:02:30,300 K1 over omega is actually 1 over V1. 915 01:02:30,300 --> 01:02:33,600 So basically, what we have is actually-- 916 01:02:33,600 --> 01:02:37,600 this is actually 1 over V1. 917 01:02:37,600 --> 01:02:40,500 This is actually 1 over V1. 918 01:02:40,500 --> 01:02:45,630 And this is actually 1 over V2. 919 01:02:45,630 --> 01:02:48,810 So in short while we are going to get 920 01:02:48,810 --> 01:03:00,600 in the second equation will become minus V2 fi omega t 921 01:03:00,600 --> 01:03:07,600 plus fr omega t. 922 01:03:07,600 --> 01:03:12,440 If I multiply both sides by V1 and V2 then I get the minus V2 923 01:03:12,440 --> 01:03:13,460 here. 924 01:03:13,460 --> 01:03:15,410 And this will be equal to-- 925 01:03:19,170 --> 01:03:21,905 there should be a minus here because I 926 01:03:21,905 --> 01:03:26,810 am taking out minus V2 there. 927 01:03:26,810 --> 01:03:30,680 And this will be equal to the right-hand side 928 01:03:30,680 --> 01:03:33,505 because I multiply both side by V1 and V2. 929 01:03:33,505 --> 01:03:39,944 Actually, I get minus V1 ft omega t. 930 01:03:43,680 --> 01:03:46,060 So what is actually left over is that now I 931 01:03:46,060 --> 01:03:51,280 have equation number one, and I have equation number two. 932 01:03:51,280 --> 01:03:58,080 Those are just functions of fi, fr, and ft. So 933 01:03:58,080 --> 01:04:03,070 that means we can actually easily solve the equation 934 01:04:03,070 --> 01:04:06,530 and write everything in terms of fi. 935 01:04:06,530 --> 01:04:13,450 So we can now solve one and two and write 936 01:04:13,450 --> 01:04:23,400 in terms of fi, which is actually the incident wave. 937 01:04:23,400 --> 01:04:25,110 That's actually what we could do. 938 01:04:25,110 --> 01:04:26,250 So if I do that-- 939 01:04:26,250 --> 01:04:28,980 if I solved the equation one and two, 940 01:04:28,980 --> 01:04:33,060 basically I get fr omega t, will be 941 01:04:33,060 --> 01:04:46,360 equal to V2 minus V1 divided by V2 plus V1 times fi omega t. 942 01:04:46,360 --> 01:04:50,770 If you trust me, if I try to solve one and two, 943 01:04:50,770 --> 01:04:56,950 and express fr, and ft in terms of fi-- 944 01:04:56,950 --> 01:05:01,180 then basically the second thing which I get from this solution 945 01:05:01,180 --> 01:05:13,610 is that ft will be equal to 2V2 divided by V1 plus V2, fi omega 946 01:05:13,610 --> 01:05:16,150 t. 947 01:05:16,150 --> 01:05:19,580 So look at what we have done. 948 01:05:19,580 --> 01:05:21,670 Basically, the first thing which we did 949 01:05:21,670 --> 01:05:25,650 is to identify what are the boundary conditions. 950 01:05:25,650 --> 01:05:27,890 Under the condition one is the string 951 01:05:27,890 --> 01:05:30,900 doesn't break at the boundary. 952 01:05:30,900 --> 01:05:33,670 The slope match between the two boundary 953 01:05:33,670 --> 01:05:37,200 because you have constant tension. 954 01:05:37,200 --> 01:05:41,750 Then I assume the solution have the functional form of three 955 01:05:41,750 --> 01:05:42,800 traveling wave. 956 01:05:42,800 --> 01:05:46,040 The incident traveling wave, fi, traveling 957 01:05:46,040 --> 01:05:47,855 to the positive direction. 958 01:05:47,855 --> 01:05:52,690 The refraction is expressed as fr going 959 01:05:52,690 --> 01:05:54,470 to the negative direction. 960 01:05:54,470 --> 01:05:58,440 And finally, ft is going to-- is the transmission wave going 961 01:05:58,440 --> 01:06:00,870 to the positive direction. 962 01:06:00,870 --> 01:06:06,400 Then I plug those equation in to the boundary condition. 963 01:06:06,400 --> 01:06:12,700 And I solve everything, fr and ft, in terms of fi 964 01:06:12,700 --> 01:06:14,880 and this is actually what I get. 965 01:06:14,880 --> 01:06:18,850 So that's actually in short what I have been doing. 966 01:06:18,850 --> 01:06:21,610 So basically, this expression is actually 967 01:06:21,610 --> 01:06:31,040 equal to R time fi, where R is actually V2 minus V1, divided 968 01:06:31,040 --> 01:06:34,480 by V1 plus V2. 969 01:06:34,480 --> 01:06:40,570 And in this case this is equal to transmission, 970 01:06:40,570 --> 01:06:52,510 which I am writing as Tau, times fi, the initial incident wave. 971 01:06:52,510 --> 01:07:00,070 And this Tau is equal to 2 times V2 divided by V1 plus V2. 972 01:07:00,070 --> 01:07:07,090 So in this example, V2 is equal to 1 over 2 V1. 973 01:07:07,090 --> 01:07:10,455 So I can now plug it in and see what I get. 974 01:07:10,455 --> 01:07:15,550 Basically, V2 will be equal to V1 over 2. 975 01:07:15,550 --> 01:07:21,160 Then I can evaluate, will be the R and Tau. 976 01:07:21,160 --> 01:07:26,140 So the R will be minus 1/3. 977 01:07:26,140 --> 01:07:28,870 It's a negative value. 978 01:07:28,870 --> 01:07:32,080 And the Tau will be equal to 2 over 3. 979 01:07:35,440 --> 01:07:39,250 So what have we learned from here? 980 01:07:39,250 --> 01:07:46,840 So if I create a pulse starting from the one which is actually 981 01:07:46,840 --> 01:07:47,680 have-- 982 01:07:47,680 --> 01:07:52,060 which is lighter or have smaller Rho L-- 983 01:07:52,060 --> 01:07:54,490 smaller mass per unit length-- 984 01:07:54,490 --> 01:07:57,630 when it passed through the boundary 985 01:07:57,630 --> 01:08:03,010 there will be a refracted wave, which the amplitude will 986 01:08:03,010 --> 01:08:06,417 change it's sign. 987 01:08:06,417 --> 01:08:08,000 So what is going to happen is that you 988 01:08:08,000 --> 01:08:13,060 will get a reflective wave and the amplitude changes sign. 989 01:08:13,060 --> 01:08:16,130 And then there will be a transmitted wave, 990 01:08:16,130 --> 01:08:19,939 which is actually going to the positive direction. 991 01:08:19,939 --> 01:08:23,899 So this is actually a demonstration we have here. 992 01:08:23,899 --> 01:08:28,260 So left-hand side is the system, which I was talking about-- 993 01:08:28,260 --> 01:08:31,160 the smaller Rho L system. 994 01:08:31,160 --> 01:08:36,109 And right-hand side is the larger Rho L system. 995 01:08:36,109 --> 01:08:39,090 And now I can do the experiment and see what happen. 996 01:08:39,090 --> 01:08:51,100 And I connect the two system with this ring, 997 01:08:51,100 --> 01:08:52,880 so that they are coupled to each other. 998 01:08:59,720 --> 01:09:00,801 I hope you will work. 999 01:09:07,870 --> 01:09:11,560 All right, so now I can create-- 1000 01:09:11,560 --> 01:09:15,460 oh, I'm in trouble now. 1001 01:09:15,460 --> 01:09:15,960 One second. 1002 01:09:18,779 --> 01:09:32,710 Hopefully will-- this is not easy. 1003 01:09:32,710 --> 01:09:38,144 OK, now I can create a pulse from the left-hand side. 1004 01:09:38,144 --> 01:09:39,142 Oh, no. 1005 01:09:47,140 --> 01:09:48,859 That is the pressure. 1006 01:09:48,859 --> 01:09:51,410 So now I can create a pulse from the left-hand side. 1007 01:09:51,410 --> 01:09:53,870 And you can see that there is a small pulse actually 1008 01:09:53,870 --> 01:09:57,120 that pass through the median-- 1009 01:09:57,120 --> 01:10:01,980 pass through the boundary, but unfortunately this demo 1010 01:10:01,980 --> 01:10:04,156 is not setup already. 1011 01:10:11,551 --> 01:10:13,040 Ah, gosh. 1012 01:10:13,040 --> 01:10:16,300 OK, so we will see what we can get from here. 1013 01:10:16,300 --> 01:10:17,440 Now, it works. 1014 01:10:17,440 --> 01:10:18,850 Very good. 1015 01:10:18,850 --> 01:10:25,180 So now I can actually create a pulse from the left-hand side, 1016 01:10:25,180 --> 01:10:28,930 and you can see that it does pass through this boundary 1017 01:10:28,930 --> 01:10:29,980 if I setup the ring. 1018 01:10:29,980 --> 01:10:34,390 The ring was falling down somehow during the lecture. 1019 01:10:34,390 --> 01:10:37,930 And this ring is actually presenting the boundary 1020 01:10:37,930 --> 01:10:41,680 and connect these two system. 1021 01:10:41,680 --> 01:10:44,920 Based on what we predict from the equation-- 1022 01:10:44,920 --> 01:10:49,450 basically you will see that if I have a positive amplitude 1023 01:10:49,450 --> 01:10:52,270 passing through the boundary there 1024 01:10:52,270 --> 01:10:56,740 will be a negative pulse going backward 1025 01:10:56,740 --> 01:11:02,890 and a positive pulse going through the boundary, which 1026 01:11:02,890 --> 01:11:04,330 is the transmission wave. 1027 01:11:04,330 --> 01:11:06,910 And let's see what is going to happen. 1028 01:11:06,910 --> 01:11:07,750 You see? 1029 01:11:07,750 --> 01:11:11,790 It does have a negative pulse going backward. 1030 01:11:11,790 --> 01:11:15,430 And you do see that there is a pulse, which is actually 1031 01:11:15,430 --> 01:11:17,280 going through this system. 1032 01:11:17,280 --> 01:11:19,480 Let's see this again. 1033 01:11:19,480 --> 01:11:23,080 You see that there's a positive amplitude pulse going 1034 01:11:23,080 --> 01:11:24,815 through the boundary and that there's 1035 01:11:24,815 --> 01:11:28,020 a refraction through this-- 1036 01:11:28,020 --> 01:11:32,530 which is actually going backward in the left-hand side system. 1037 01:11:32,530 --> 01:11:37,850 So on the other hand, if I start a traveling wave 1038 01:11:37,850 --> 01:11:44,750 from your left-hand side, that means 1039 01:11:44,750 --> 01:11:50,290 V2 is going to be larger than V1. 1040 01:11:50,290 --> 01:11:54,820 V2 is actually going to be larger than the V1. 1041 01:11:54,820 --> 01:12:00,340 So what are we going to get is a positive amplitude 1042 01:12:00,340 --> 01:12:06,730 refraction and also a positive value transmission wave. 1043 01:12:06,730 --> 01:12:10,510 And let's see what is going to happen. 1044 01:12:10,510 --> 01:12:11,380 You see? 1045 01:12:11,380 --> 01:12:14,930 The refraction is positive this time. 1046 01:12:14,930 --> 01:12:18,490 And the transmission wave have also positive amplitudes. 1047 01:12:18,490 --> 01:12:22,430 Let's take a look at this thing again. 1048 01:12:22,430 --> 01:12:25,850 A very nice pulse and you can see the refraction 1049 01:12:25,850 --> 01:12:28,410 because of this mathematics. 1050 01:12:28,410 --> 01:12:32,420 Interesting thing is that it match with experimental result. 1051 01:12:32,420 --> 01:12:34,880 And the prediction was that you are 1052 01:12:34,880 --> 01:12:38,780 going to get the positive amplitude reflective wave, 1053 01:12:38,780 --> 01:12:44,120 and it does agree with the experimental data. 1054 01:12:44,120 --> 01:12:46,130 So this is actually what we have learned. 1055 01:12:46,130 --> 01:12:49,800 So we have learned traveling wave solution. 1056 01:12:49,800 --> 01:12:52,970 Energy of a oscillating string, and also 1057 01:12:52,970 --> 01:12:55,460 the potential kinetic energy. 1058 01:12:55,460 --> 01:13:00,320 And also we learn how to actually match two media 1059 01:13:00,320 --> 01:13:03,660 and passings-- how this traveling wave pass 1060 01:13:03,660 --> 01:13:05,960 through the median, et cetera. 1061 01:13:05,960 --> 01:13:10,190 And next time we will talk about more systems described 1062 01:13:10,190 --> 01:13:12,290 by wave equation. 1063 01:13:12,290 --> 01:13:16,430 And also dispersion relation, what does that mean, et cetera. 1064 01:13:16,430 --> 01:13:20,080 Thank you very much, and see you on Thursday.