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 --> 00:00:19,710 at ocw.mit.edu. 8 00:00:23,600 --> 00:00:26,540 YEN-JIE LEE: So welcome back, everybody, to 8.03. 9 00:00:26,540 --> 00:00:28,730 Happy to see you again. 10 00:00:28,730 --> 00:00:32,650 So here is the current status of the 8.03. 11 00:00:32,650 --> 00:00:35,550 So right now, we have finished the discussion 12 00:00:35,550 --> 00:00:37,520 of coupled oscillator. 13 00:00:37,520 --> 00:00:40,830 And then we go to infinite number of coupled oscillators. 14 00:00:40,830 --> 00:00:46,040 And we found that there's a wave equation coming out of it. 15 00:00:46,040 --> 00:00:52,990 And that means, in short, waves are really a group effort. 16 00:00:52,990 --> 00:00:56,810 So many, many objects are working really together, 17 00:00:56,810 --> 00:01:00,380 so that they create a wave phenomena. 18 00:01:00,380 --> 00:01:03,290 And you can also see, there's a close connection 19 00:01:03,290 --> 00:01:06,830 between vibration of a single object 20 00:01:06,830 --> 00:01:10,310 and the formation of the wave structure. 21 00:01:10,310 --> 00:01:12,350 So what we are going to do today is 22 00:01:12,350 --> 00:01:16,250 to give you a short review of what we have done last time. 23 00:01:16,250 --> 00:01:20,030 Then we actually will continue to our understanding 24 00:01:20,030 --> 00:01:24,210 of wave equation today. 25 00:01:24,210 --> 00:01:26,150 So what we have learned last time, 26 00:01:26,150 --> 00:01:30,950 we have learned how to solve infinite system with space 27 00:01:30,950 --> 00:01:32,940 translation symmetry. 28 00:01:32,940 --> 00:01:35,750 And also, we learned how to use it 29 00:01:35,750 --> 00:01:40,790 to solve finite systems by imposing-- or adding 30 00:01:40,790 --> 00:01:42,460 boundary conditions. 31 00:01:42,460 --> 00:01:46,310 That would limit the infinite number of normal modes 32 00:01:46,310 --> 00:01:49,367 to finite number of normal modes based on-- 33 00:01:49,367 --> 00:01:53,720 I mean, it's actually closely related 34 00:01:53,720 --> 00:01:58,030 to how many objects you have in the system. 35 00:01:58,030 --> 00:02:02,580 And also, we went ahead and go to a continuum limit. 36 00:02:02,580 --> 00:02:05,240 And we found out, there is a surprising result coming out 37 00:02:05,240 --> 00:02:06,020 of this. 38 00:02:06,020 --> 00:02:09,500 And this is actually the wave equation. 39 00:02:09,500 --> 00:02:14,310 So what do we mean by going to continuous limit? 40 00:02:14,310 --> 00:02:16,780 So the limit we are talking about 41 00:02:16,780 --> 00:02:20,570 is that, before when we discussed this closed 42 00:02:20,570 --> 00:02:24,160 system of infinite number of objects holding together 43 00:02:24,160 --> 00:02:28,170 by strings, there is a length scale, which 44 00:02:28,170 --> 00:02:30,970 is the separation between objects, which 45 00:02:30,970 --> 00:02:34,390 is called a in my notation. 46 00:02:34,390 --> 00:02:38,440 And to make it continuous, we are taking a limit such 47 00:02:38,440 --> 00:02:42,910 that the a, which is the separation between objects, 48 00:02:42,910 --> 00:02:47,380 is so much smaller than the wavelength. 49 00:02:47,380 --> 00:02:49,870 Basically, the wavelength is actually 50 00:02:49,870 --> 00:02:54,280 the sinusoidal shape you see when I perturb a system. 51 00:02:54,280 --> 00:03:00,220 And what I assuming is that the distance between objects 52 00:03:00,220 --> 00:03:04,090 are so much smaller than the wavelengths. 53 00:03:04,090 --> 00:03:07,650 So that's actually what I called continuous limit. 54 00:03:07,650 --> 00:03:11,050 And that is actually true for most of the example, which we 55 00:03:11,050 --> 00:03:14,080 see in the previous lectures. 56 00:03:14,080 --> 00:03:18,400 For example, I was holding a giant spring. 57 00:03:18,400 --> 00:03:19,930 And I oscillate that. 58 00:03:19,930 --> 00:03:22,430 And all the little components, or, say, 59 00:03:22,430 --> 00:03:28,240 all the little mass on that spring, the space between all 60 00:03:28,240 --> 00:03:32,860 those little mass on the spring are so much smaller 61 00:03:32,860 --> 00:03:34,730 than the length scale we are talking about, 62 00:03:34,730 --> 00:03:38,530 which is at the order of 1 meter. 63 00:03:38,530 --> 00:03:43,000 So that actually is a sensible limit, 64 00:03:43,000 --> 00:03:46,490 which describes the physics we are interested. 65 00:03:46,490 --> 00:03:51,830 When we go to a continuous limit, 66 00:03:51,830 --> 00:03:55,790 we find that something really interesting happens. 67 00:03:55,790 --> 00:04:00,470 So M minus 1 K matrix originally is infinite times 68 00:04:00,470 --> 00:04:02,370 infinite dimension matrix. 69 00:04:02,370 --> 00:04:05,390 It becomes the operator, which is actually 70 00:04:05,390 --> 00:04:10,400 minus T over rho L partial square partial x squared. 71 00:04:10,400 --> 00:04:12,950 And also-- 72 00:04:12,950 --> 00:04:15,030 OK, I changed the notation here. 73 00:04:15,030 --> 00:04:18,769 It was aj, and I changed it to psi, 74 00:04:18,769 --> 00:04:23,700 because what we are going to use later on, when we describe 75 00:04:23,700 --> 00:04:27,800 wave functions, et cetera, especially in 8.04, 76 00:04:27,800 --> 00:04:29,390 we usually use psi. 77 00:04:29,390 --> 00:04:33,940 And the psi j, which were discrete and evaluated 78 00:04:33,940 --> 00:04:38,660 in the individual discrete position in the x direction, 79 00:04:38,660 --> 00:04:40,880 it's becoming a continuous function, 80 00:04:40,880 --> 00:04:46,980 which is psi x and is also a function t. 81 00:04:46,980 --> 00:04:50,750 Therefore, from this exercise, we 82 00:04:50,750 --> 00:04:56,210 found out we see wave equation, actually after we 83 00:04:56,210 --> 00:04:58,040 go to a continuous limit. 84 00:04:58,040 --> 00:04:59,750 And for more information, you can also 85 00:04:59,750 --> 00:05:04,640 take a look at the textbook in the relevant page. 86 00:05:04,640 --> 00:05:07,050 So what are we going to do today? 87 00:05:07,050 --> 00:05:09,350 So today, what we are going to do 88 00:05:09,350 --> 00:05:13,180 is to understand the wave equation, the structure, 89 00:05:13,180 --> 00:05:15,380 and what does that mean, and also 90 00:05:15,380 --> 00:05:20,210 what are the normal modes coming out of this wave equation. 91 00:05:20,210 --> 00:05:22,730 And the next time, in later lectures, 92 00:05:22,730 --> 00:05:26,570 we will also discuss another special kind of motion, which 93 00:05:26,570 --> 00:05:29,960 is progressing wave solutions. 94 00:05:29,960 --> 00:05:32,780 So let's immediately gets started 95 00:05:32,780 --> 00:05:37,800 by looking at a concrete example and also 96 00:05:37,800 --> 00:05:43,970 to derive the normal modes from this wave equation. 97 00:05:43,970 --> 00:05:48,380 Before we do that, let's take a look at this wave equation. 98 00:05:48,380 --> 00:05:54,220 This wave equation is actually equivalent to infinite 99 00:05:54,220 --> 00:05:58,620 number of equations of motion, if you think about it. 100 00:05:58,620 --> 00:05:59,670 Why is that? 101 00:05:59,670 --> 00:06:02,930 That is because each x-- 102 00:06:02,930 --> 00:06:07,640 each partition x you put in will produce a equation of motion. 103 00:06:07,640 --> 00:06:13,100 So basically, originally, when we were doing a discrete case, 104 00:06:13,100 --> 00:06:15,230 those are labeled by c. 105 00:06:15,230 --> 00:06:20,750 c is actually telling you which mass I'm talking about. 106 00:06:20,750 --> 00:06:23,020 Now, it's actually replaced by x. 107 00:06:23,020 --> 00:06:24,860 And what we are actually doing is 108 00:06:24,860 --> 00:06:29,690 to solve infinite number equation of motion in one go. 109 00:06:29,690 --> 00:06:34,230 And that is actually the wave equation. 110 00:06:34,230 --> 00:06:40,730 So the first question we ask is, what is actually normal modes 111 00:06:40,730 --> 00:06:46,790 based on this infinitely long continuous system described 112 00:06:46,790 --> 00:06:48,900 by wave equations? 113 00:06:48,900 --> 00:06:52,690 So let's get started immediately. 114 00:06:52,690 --> 00:06:55,750 So basically, we can first assume 115 00:06:55,750 --> 00:06:59,270 what is actually the functional form for normal modes. 116 00:06:59,270 --> 00:07:01,130 So what we can actually do is we can 117 00:07:01,130 --> 00:07:06,680 assume that psi x, t is actually equal to A, 118 00:07:06,680 --> 00:07:10,610 is actually a function of x, times B, 119 00:07:10,610 --> 00:07:14,550 is actually a function of time. 120 00:07:14,550 --> 00:07:20,070 So what I am doing is actually have a meaning actually. 121 00:07:20,070 --> 00:07:24,450 So A of x actually give you a description 122 00:07:24,450 --> 00:07:27,250 of the functional form-- 123 00:07:27,250 --> 00:07:33,960 the shape of the normal mode as a function of x. 124 00:07:33,960 --> 00:07:40,640 So that's actually giving you the shape 125 00:07:40,640 --> 00:07:42,000 as a function of x-axis. 126 00:07:44,770 --> 00:07:49,990 And B, which is a function of t, is actually 127 00:07:49,990 --> 00:07:55,390 giving you information how individual component 128 00:07:55,390 --> 00:08:00,370 goes up and down or move as a function of time. 129 00:08:00,370 --> 00:08:04,060 So that actually control the time evolution. 130 00:08:09,430 --> 00:08:13,250 And we were using this wave equation 131 00:08:13,250 --> 00:08:18,445 to describe a continuous system, like, for example, 132 00:08:18,445 --> 00:08:20,590 a string with tension t. 133 00:08:23,740 --> 00:08:25,670 So what we could do is the following. 134 00:08:25,670 --> 00:08:29,500 So we are interested in the solution of the wave 135 00:08:29,500 --> 00:08:31,600 equation, which is shown there. 136 00:08:31,600 --> 00:08:35,620 So what we could do is that, OK, let's first assume 137 00:08:35,620 --> 00:08:39,280 this functional form, assuming that every component is 138 00:08:39,280 --> 00:08:44,000 actually following the same time-depend evolution. 139 00:08:44,000 --> 00:08:47,020 And then we can actually plug in this functional form 140 00:08:47,020 --> 00:08:50,780 to the equation of motion and see what we will get. 141 00:08:50,780 --> 00:09:02,170 So if we plug this in into the wave equation, 142 00:09:02,170 --> 00:09:04,750 so what we are going to get is, if you look at the left hand 143 00:09:04,750 --> 00:09:07,770 side, it's actually a partial derivation with respect 144 00:09:07,770 --> 00:09:08,970 to time. 145 00:09:08,970 --> 00:09:10,650 Therefore, what we are going to get 146 00:09:10,650 --> 00:09:18,490 is A of x times partial square of B, which is a function of t, 147 00:09:18,490 --> 00:09:21,050 partial t squared. 148 00:09:21,050 --> 00:09:23,050 The right hand side of the equation, 149 00:09:23,050 --> 00:09:28,260 which is actually equal to vp squared, 150 00:09:28,260 --> 00:09:34,330 is a partial derivative with respect to x. 151 00:09:34,330 --> 00:09:39,870 Therefore, you have B is a function of t 152 00:09:39,870 --> 00:09:47,100 only times partial square A of x partial t squared. 153 00:09:49,810 --> 00:09:55,940 So actually, we can just for convenience-- 154 00:09:55,940 --> 00:09:57,880 oh, sorry, that's supposed to be partial x. 155 00:09:57,880 --> 00:09:59,680 Thank you very much-- 156 00:09:59,680 --> 00:10:01,780 partial x squared. 157 00:10:01,780 --> 00:10:04,600 So just for convenience, we can actually 158 00:10:04,600 --> 00:10:11,790 divide the whole equation by A times B times the vp squared. 159 00:10:11,790 --> 00:10:18,190 We can issue divide the whole equation by A times B, 160 00:10:18,190 --> 00:10:20,650 for example, and the vp squared. 161 00:10:20,650 --> 00:10:23,010 If we do this, then basically I'm 162 00:10:23,010 --> 00:10:25,750 moving this part to the left hand side. 163 00:10:25,750 --> 00:10:35,285 So I get 1 over vp square B of t partial square Bt partial t 164 00:10:35,285 --> 00:10:35,785 square. 165 00:10:38,510 --> 00:10:43,880 And the right hand side, because I also divide AB vp square, 166 00:10:43,880 --> 00:10:53,720 therefore, I get 1 over A of x partial square A of x partial 167 00:10:53,720 --> 00:10:54,720 x square. 168 00:11:00,350 --> 00:11:02,030 So far so good. 169 00:11:02,030 --> 00:11:06,290 And basically, what I'm doing is just 170 00:11:06,290 --> 00:11:10,160 plug in the functional form, which I assume here 171 00:11:10,160 --> 00:11:16,130 and then divide everything by AB times vp square. 172 00:11:16,130 --> 00:11:19,080 And what I immediately find is that left hand 173 00:11:19,080 --> 00:11:26,070 side is a function which only depends on t. 174 00:11:26,070 --> 00:11:29,180 Left hand side only depends on t. 175 00:11:29,180 --> 00:11:32,360 And right hand side is a function 176 00:11:32,360 --> 00:11:36,430 which on the depends on x. 177 00:11:36,430 --> 00:11:41,250 So in short, I have in this situation 178 00:11:41,250 --> 00:11:46,140 f of t, which is left hand side, is equal to g of x. 179 00:11:49,640 --> 00:11:54,560 You will see this over and over again in later lectures related 180 00:11:54,560 --> 00:11:56,830 to physics. 181 00:11:56,830 --> 00:12:01,450 This is actually the so-called separation of variables. 182 00:12:01,450 --> 00:12:04,395 So basically, you are facing a situation 183 00:12:04,395 --> 00:12:08,440 f of t equal to g of x. 184 00:12:08,440 --> 00:12:10,720 If you think about this situation, 185 00:12:10,720 --> 00:12:12,650 that's actually really, really helpful, 186 00:12:12,650 --> 00:12:21,420 because, OK, now what I can do is I can stay at a specific x. 187 00:12:21,420 --> 00:12:24,210 For example, I choose this point. 188 00:12:24,210 --> 00:12:28,890 Then I let the time go forward. 189 00:12:28,890 --> 00:12:33,450 Of course, I cannot stop time, but if it goes forward, 190 00:12:33,450 --> 00:12:37,590 then left hand side equivalent, if it's changing, 191 00:12:37,590 --> 00:12:42,420 you will change, because I change t. 192 00:12:42,420 --> 00:12:45,210 If the left hand side equation is changing, 193 00:12:45,210 --> 00:12:48,620 then that's actually not going to-- 194 00:12:48,620 --> 00:12:50,490 this equation is not going to work, 195 00:12:50,490 --> 00:12:52,260 because I am not changing x. 196 00:12:52,260 --> 00:12:56,940 I am fixing myself at a specific location, and a lot of time 197 00:12:56,940 --> 00:12:57,840 go on. 198 00:12:57,840 --> 00:13:01,590 Then if left hand side is changing, 199 00:13:01,590 --> 00:13:05,470 then this equation cannot work. 200 00:13:05,470 --> 00:13:06,640 You see? 201 00:13:06,640 --> 00:13:10,610 Therefore, what is the consequence of this equation? 202 00:13:10,610 --> 00:13:15,760 That means, left hand side, f of t, must be a constant. 203 00:13:18,570 --> 00:13:22,810 Therefore, no matter what I do, if I change time, 204 00:13:22,810 --> 00:13:25,220 it's not going to change anything. 205 00:13:25,220 --> 00:13:28,430 I can put in whatever time, like 1 billion years 206 00:13:28,430 --> 00:13:33,330 after this lecture or now, it doesn't matter. 207 00:13:33,330 --> 00:13:34,680 It's a constant. 208 00:13:34,680 --> 00:13:35,970 So this must be a constant. 209 00:13:39,780 --> 00:13:42,290 I can do the same trick. 210 00:13:42,290 --> 00:13:45,710 I froze the time. 211 00:13:45,710 --> 00:13:50,300 I fix the time, and then I compare this point 212 00:13:50,300 --> 00:13:54,650 to that point, or, say, something billion billions 213 00:13:54,650 --> 00:13:59,250 of light years away from this class room. 214 00:13:59,250 --> 00:14:03,980 I am changing the x, but I'm not changing the t. 215 00:14:03,980 --> 00:14:06,590 The same argument also holds-- 216 00:14:06,590 --> 00:14:09,950 if this function is changing as a function of x, 217 00:14:09,950 --> 00:14:12,038 then I am screwed, because-- 218 00:14:12,038 --> 00:14:13,280 [LAUGHTER] 219 00:14:13,280 --> 00:14:14,630 --it doesn't work, right? 220 00:14:14,630 --> 00:14:17,060 I mean, this-- 221 00:14:17,060 --> 00:14:20,225 Therefore, it has to be a constant also as well. 222 00:14:23,580 --> 00:14:26,230 Constant also equal to a constant, 223 00:14:26,230 --> 00:14:29,204 that's really lovely, right? 224 00:14:29,204 --> 00:14:33,860 [LAUGHS] That means, I can say this 225 00:14:33,860 --> 00:14:36,850 is equal to that is equal to a constant. 226 00:14:39,780 --> 00:14:44,060 As usual, I call this constant really, really strange 227 00:14:44,060 --> 00:14:49,430 fancy name-- minus Km square, which you will not like it. 228 00:14:49,430 --> 00:14:51,590 But later, you would like it. 229 00:14:51,590 --> 00:14:54,650 [LAUGHTER] 230 00:14:54,650 --> 00:14:55,170 Very good. 231 00:14:55,170 --> 00:14:58,390 So we make tremendous amount of progress. 232 00:14:58,390 --> 00:15:03,360 Originally, we saw that we are in trouble. 233 00:15:03,360 --> 00:15:05,320 It's A times Bt-- 234 00:15:05,320 --> 00:15:09,030 Ax times Bt, sounds really horrible. 235 00:15:09,030 --> 00:15:12,350 Now, actually, you see that this equation 236 00:15:12,350 --> 00:15:16,550 is really simple to solve. 237 00:15:16,550 --> 00:15:20,690 So let's actually take a look at the solution 238 00:15:20,690 --> 00:15:24,380 to the f function and the g function. 239 00:15:24,380 --> 00:15:30,200 So the first thing, if I take the left hand side, 240 00:15:30,200 --> 00:15:35,590 which is a time-dependence part, I can copy there-- 241 00:15:35,590 --> 00:15:36,630 this is actually 1-- 242 00:15:36,630 --> 00:15:39,350 copy the equivalent here-- 243 00:15:39,350 --> 00:15:49,300 is 1 over vp square B of t partial square Bt partial t 244 00:15:49,300 --> 00:15:50,090 square. 245 00:15:50,090 --> 00:15:52,370 And this is equal to a fancy name 246 00:15:52,370 --> 00:15:55,765 of this constant-- minus Km square. 247 00:15:59,420 --> 00:16:04,260 And of course, I can multiply everything by vp squared B. 248 00:16:04,260 --> 00:16:08,740 And I get partial square B partial t square. 249 00:16:08,740 --> 00:16:14,690 And this is equal too minus vp square Km square 250 00:16:14,690 --> 00:16:17,760 B. Wait a second. 251 00:16:17,760 --> 00:16:21,540 We have solved this equation infinite number 252 00:16:21,540 --> 00:16:24,090 of times in this lecture. 253 00:16:26,850 --> 00:16:28,080 You remember the solution? 254 00:16:28,080 --> 00:16:30,550 What is the solution? 255 00:16:30,550 --> 00:16:31,480 Anybody can help me? 256 00:16:34,342 --> 00:16:35,296 AUDIENCE: [INAUDIBLE]. 257 00:16:35,296 --> 00:16:36,727 YEN-JIE LEE: Anybody? 258 00:16:36,727 --> 00:16:39,640 It's sine or cosine function, right? 259 00:16:39,640 --> 00:16:41,750 This is harmonic oscillation. 260 00:16:41,750 --> 00:16:46,160 It's almost like equation of motion of a spring-mass system, 261 00:16:46,160 --> 00:16:47,390 right? 262 00:16:47,390 --> 00:16:49,730 I hope that you are already bored. 263 00:16:49,730 --> 00:16:52,010 And that means I very successful. 264 00:16:52,010 --> 00:16:54,710 [LAUGHTER] 265 00:16:54,710 --> 00:16:58,760 B of t will be equal to-- 266 00:16:58,760 --> 00:17:10,760 [LAUGHS]---- B of m sine omega m t plus beta m, 267 00:17:10,760 --> 00:17:18,210 where omega m is actually equal to vp times Km. 268 00:17:18,210 --> 00:17:23,060 I define omega m equal to vp time Km. 269 00:17:29,030 --> 00:17:36,820 So surprisingly, the solution of B is really simple. 270 00:17:36,820 --> 00:17:42,630 It's actually Bm sine omega m t plus beta m. 271 00:17:42,630 --> 00:17:43,970 So what does that mean? 272 00:17:43,970 --> 00:17:49,940 That means the overall motion or overall time-dependent 273 00:17:49,940 --> 00:17:56,030 evolution of the system is like harmonic motion. 274 00:17:56,030 --> 00:18:03,080 If you do get individual component in this system. 275 00:18:03,080 --> 00:18:06,230 So that's really nice. 276 00:18:06,230 --> 00:18:08,590 So let's take a look at the right hand side. 277 00:18:08,590 --> 00:18:12,370 The right hand side what we have is 278 00:18:12,370 --> 00:18:22,800 1 over A A of x partial square A x partial x squared. 279 00:18:22,800 --> 00:18:26,520 And now, this equal to minus Km squared. 280 00:18:30,630 --> 00:18:32,760 I don't want to go over this again. 281 00:18:32,760 --> 00:18:36,660 This is actually the same thing as number one. 282 00:18:36,660 --> 00:18:38,040 The only thing which is different 283 00:18:38,040 --> 00:18:44,550 is that now the partial derivative is actually the x. 284 00:18:44,550 --> 00:18:47,190 It's partial square A partial x square. 285 00:18:47,190 --> 00:18:50,750 Therefore, I can immediately write down the solution. 286 00:18:50,750 --> 00:18:52,600 A of t-- oh, sorry-- 287 00:18:52,600 --> 00:19:04,200 A of x will be equal to Cm sine Km x plus alpha m. 288 00:19:08,520 --> 00:19:09,430 Any questions so far? 289 00:19:14,180 --> 00:19:23,610 So until now, you accept the fact that f of t and the g of x 290 00:19:23,610 --> 00:19:27,620 has to be the same constant. 291 00:19:27,620 --> 00:19:32,160 And I call it minus Km square. 292 00:19:32,160 --> 00:19:38,540 And I didn't actually tell you what Km I'm choosing. 293 00:19:38,540 --> 00:19:45,700 In reality, according to this result, Km can the anything, 294 00:19:45,700 --> 00:19:50,530 can be any number, as long as it's a constant. 295 00:19:50,530 --> 00:19:56,410 Therefore, I would like to write the corresponding psi, which 296 00:19:56,410 --> 00:20:01,970 is the wave function, is actually labeled by m. 297 00:20:01,970 --> 00:20:07,860 m is it just label which K I was using, nothing fancy. 298 00:20:07,860 --> 00:20:09,920 It's just a label. 299 00:20:09,920 --> 00:20:16,160 Psi of m is a function of x and t. 300 00:20:16,160 --> 00:20:27,520 And that will be equal to Bm times Cm sine omega mt 301 00:20:27,520 --> 00:20:37,830 plus beta m sine Km x plus alpha m. 302 00:20:37,830 --> 00:20:41,520 Bm and Cm are just arbitrary constant. 303 00:20:41,520 --> 00:20:44,110 Therefore, I can merge them. 304 00:20:44,110 --> 00:20:46,650 And I will call it just Am. 305 00:20:53,059 --> 00:20:57,710 And, of course, don't forget we have this condition. 306 00:20:57,710 --> 00:21:03,590 Omega m is actually equal to vp times Km. 307 00:21:03,590 --> 00:21:05,130 So this is actually defined here. 308 00:21:10,880 --> 00:21:14,840 So here, since this is actually a second order differential 309 00:21:14,840 --> 00:21:20,200 equation, you have unknown factor, which is beta m. 310 00:21:20,200 --> 00:21:24,360 You have also Bm is a unknown. 311 00:21:24,360 --> 00:21:26,570 And the right hand side, you also 312 00:21:26,570 --> 00:21:33,834 have a Cm, which is unknown, and alpha m, which is unknown. 313 00:21:36,500 --> 00:21:43,040 When we combine them, I replace Bm times Cm by Am. 314 00:21:43,040 --> 00:21:46,820 Therefore, what we have is that Am is actually 315 00:21:46,820 --> 00:21:50,690 some kind of amplitude, which can be determined 316 00:21:50,690 --> 00:21:53,840 by initial conditions, which I will talk about that later. 317 00:21:56,680 --> 00:22:00,930 Beta m is the unknown coming from the left hand side 318 00:22:00,930 --> 00:22:02,710 derivation. 319 00:22:02,710 --> 00:22:05,650 Alpha m is also unknown, which is actually 320 00:22:05,650 --> 00:22:09,430 coming from the right side derivation related 321 00:22:09,430 --> 00:22:14,080 to the shape of the normal mode of the system. 322 00:22:14,080 --> 00:22:18,580 And finally, there's one additional unknown coefficient, 323 00:22:18,580 --> 00:22:19,940 which is Km-- 324 00:22:19,940 --> 00:22:21,660 it's kind of arbitrary now-- 325 00:22:21,660 --> 00:22:25,660 control actually the wave number, 326 00:22:25,660 --> 00:22:32,490 or say the wavelength of the shape of the normal mode. 327 00:22:32,490 --> 00:22:37,170 So when you see this, doesn't this surprise you? 328 00:22:37,170 --> 00:22:40,110 May not surprise you any more, because we 329 00:22:40,110 --> 00:22:44,390 have solved infinite number of coupled oscillator. 330 00:22:44,390 --> 00:22:48,720 And you have learned that, OK, the normal modes 331 00:22:48,720 --> 00:22:52,720 have a shape of sine function. 332 00:22:52,720 --> 00:22:55,250 It's like a sine function-- 333 00:22:55,250 --> 00:22:58,920 before it was like a sine as a Ka. 334 00:22:58,920 --> 00:23:01,140 And the Ka is performing x. 335 00:23:01,140 --> 00:23:03,030 So what we are actually getting here 336 00:23:03,030 --> 00:23:09,540 is that doesn't surprise you since this system also satisfy 337 00:23:09,540 --> 00:23:12,480 space translational symmetry. 338 00:23:12,480 --> 00:23:19,530 Therefore, the functional form of the shape of the normal mode 339 00:23:19,530 --> 00:23:21,950 is also a sine function. 340 00:23:21,950 --> 00:23:25,050 So that's actually pretty satisfactory 341 00:23:25,050 --> 00:23:30,690 and also come out as what we would expect based on what 342 00:23:30,690 --> 00:23:32,050 we actually have learned. 343 00:23:35,030 --> 00:23:39,330 So let's actually take a look at the structure of this function. 344 00:23:39,330 --> 00:23:42,740 So basically, as I mentioned before, 345 00:23:42,740 --> 00:23:48,760 everything is oscillating at the frequency omega m 346 00:23:48,760 --> 00:23:51,200 with a phase beta m. 347 00:23:51,200 --> 00:23:55,040 So that satisfy the condition of normal mode, 348 00:23:55,040 --> 00:23:56,690 what is the condition. 349 00:23:56,690 --> 00:23:58,580 All the components in the system are 350 00:23:58,580 --> 00:24:03,790 oscillating at the same frequency and the same phase. 351 00:24:03,790 --> 00:24:08,140 Indeed, yes, that's correct. 352 00:24:08,140 --> 00:24:11,870 Also as a function of time, it's actually 353 00:24:11,870 --> 00:24:16,040 going up and down harmonically as we already discussed. 354 00:24:16,040 --> 00:24:19,670 And the relative amplitude, as I said, is a sine function. 355 00:24:19,670 --> 00:24:25,030 And of course, I already demonstrated this before, 356 00:24:25,030 --> 00:24:28,130 that you can see that in this is system, 357 00:24:28,130 --> 00:24:32,610 you can see a sine function when I start to drive it. 358 00:24:32,610 --> 00:24:36,650 So what I am doing is actually to convert the kinetic energy 359 00:24:36,650 --> 00:24:43,780 from my hand to energy stored as potential or kinetic energy 360 00:24:43,780 --> 00:24:50,480 in this string-rod system and in this bell wave machine. 361 00:24:50,480 --> 00:24:54,500 So you can see that beautifully those are sine function. 362 00:24:54,500 --> 00:24:57,800 And, of course, if I do a higher frequency one, 363 00:24:57,800 --> 00:25:01,115 you can see the hand oscillation frequency changed. 364 00:25:03,920 --> 00:25:09,330 And that is actually controlled by this equation, 365 00:25:09,330 --> 00:25:12,800 this dispersion relation. 366 00:25:12,800 --> 00:25:15,470 And this dispersion relation is actually 367 00:25:15,470 --> 00:25:18,770 relating two physical quantity. 368 00:25:18,770 --> 00:25:22,180 One is actually the wave number. 369 00:25:22,180 --> 00:25:26,065 Of course, if you are more familiar with wavelength, 370 00:25:26,065 --> 00:25:29,330 it's actually 2 pi over Km. 371 00:25:29,330 --> 00:25:31,130 Lambda m is actually the wavelength 372 00:25:31,130 --> 00:25:36,080 of the shape of the normal mode. 373 00:25:36,080 --> 00:25:38,000 And the oscillation frequency is actually 374 00:25:38,000 --> 00:25:45,440 controlled by this dispersion relation, this function. 375 00:25:45,440 --> 00:25:47,600 And you can say that, Professor Lee, 376 00:25:47,600 --> 00:25:51,950 I have been so tired of this demo. 377 00:25:51,950 --> 00:25:54,320 I've seen this 1,000 times, right? 378 00:25:54,320 --> 00:25:56,730 And basically, you are showing me 379 00:25:56,730 --> 00:25:59,645 that, OK, you can actually oscillate this system 380 00:25:59,645 --> 00:26:02,480 and excite this system so that it's 381 00:26:02,480 --> 00:26:05,870 oscillating at some natural frequency the system 382 00:26:05,870 --> 00:26:06,630 like, right? 383 00:26:06,630 --> 00:26:09,430 It's actually some kind of resonance behavior. 384 00:26:09,430 --> 00:26:11,090 So I can actually excite-- 385 00:26:11,090 --> 00:26:12,140 I can-- no. 386 00:26:12,140 --> 00:26:15,150 I can randomly shake this system. 387 00:26:15,150 --> 00:26:18,290 And then it's going to be a linear combination of all 388 00:26:18,290 --> 00:26:20,390 the excited normal modes, right? 389 00:26:20,390 --> 00:26:22,880 We have seen this many, many times. 390 00:26:22,880 --> 00:26:28,490 What I am going to show you is that there's another machine 391 00:26:28,490 --> 00:26:31,895 here, which is actually demonstrating 392 00:26:31,895 --> 00:26:36,680 the resonance of some wave. 393 00:26:36,680 --> 00:26:47,360 So here is actually so-called the Rijke tube. 394 00:26:47,360 --> 00:26:49,010 So the structure is like this. 395 00:26:49,010 --> 00:26:54,710 So basically, you have a metal tube, which is red thing there. 396 00:26:54,710 --> 00:26:56,300 And inside the tube-- 397 00:26:56,300 --> 00:26:59,030 you cannot see it now-- but inside the tube, 398 00:26:59,030 --> 00:27:06,710 there's a wire mesh, which the air can flow freely up and down 399 00:27:06,710 --> 00:27:09,080 in this mesh. 400 00:27:09,080 --> 00:27:14,450 And what I'm going to do now is to heat up the mesh 401 00:27:14,450 --> 00:27:16,760 and see what is going to happen. 402 00:27:16,760 --> 00:27:20,480 I will just heat it up by like six second 403 00:27:20,480 --> 00:27:23,400 and see what is going to happen. 404 00:27:23,400 --> 00:27:25,840 So now, I'm going to do this very carefully. 405 00:27:35,640 --> 00:27:38,580 [RESONATING SOUND] 406 00:27:38,580 --> 00:27:40,410 Can you hear that? 407 00:27:47,284 --> 00:27:56,760 [LAUGHS] OK, very good. 408 00:27:56,760 --> 00:28:00,290 So listen, what is happening? 409 00:28:00,290 --> 00:28:07,740 So you hear a mono frequency sound generated from what? 410 00:28:07,740 --> 00:28:13,860 Generated from the heat I gave to the wire mesh. 411 00:28:13,860 --> 00:28:15,540 So what is actually happening? 412 00:28:15,540 --> 00:28:22,380 So when I heat up the mesh, what is going to happen 413 00:28:22,380 --> 00:28:28,000 is that the air around this screen 414 00:28:28,000 --> 00:28:30,410 is going to be heated up. 415 00:28:30,410 --> 00:28:33,350 Therefore, because the air is heated up, 416 00:28:33,350 --> 00:28:35,520 it goes in the upward direction. 417 00:28:35,520 --> 00:28:38,630 And also, the volume of the air is 418 00:28:38,630 --> 00:28:43,510 expanding, because of the increased temperature. 419 00:28:43,510 --> 00:28:47,020 And that actually goes through this system. 420 00:28:47,020 --> 00:28:52,860 And why it does is really like what I'm doing to the bell lab 421 00:28:52,860 --> 00:28:59,180 wave machine, is actually tyring to oscillate-- or excite 422 00:28:59,180 --> 00:29:06,070 any possible normal modes, which this system actually like. 423 00:29:06,070 --> 00:29:10,870 So you can see that, after a while, once the pressure 424 00:29:10,870 --> 00:29:18,520 and also the air inside the tube get then self-organized, 425 00:29:18,520 --> 00:29:25,090 then you hear a very loud sound. 426 00:29:25,090 --> 00:29:28,045 So that means there are energy flowing from the tube 427 00:29:28,045 --> 00:29:29,380 to your ear. 428 00:29:29,380 --> 00:29:31,490 And that is actually coming from what? 429 00:29:31,490 --> 00:29:35,810 Coming from the heat I put into the system. 430 00:29:35,810 --> 00:29:41,410 So it's actually a heat sound wave conversion. 431 00:29:41,410 --> 00:29:43,860 So I hope you enjoyed this demo. 432 00:29:43,860 --> 00:29:46,750 And we will take a five-minute break 433 00:29:46,750 --> 00:29:50,380 to take questions before we move on. 434 00:29:50,380 --> 00:29:52,540 And of course, you are welcome to come here 435 00:29:52,540 --> 00:29:57,370 and to play with the demo if you want. 436 00:29:57,370 --> 00:30:03,440 [LAUGHS] 437 00:30:03,440 --> 00:30:07,010 So welcome back from the break. 438 00:30:07,010 --> 00:30:10,460 So what we are going to do next is 439 00:30:10,460 --> 00:30:14,720 to understand how to determine all those unknown coefficients. 440 00:30:14,720 --> 00:30:21,080 So you get to see here, there are Am, which is the amplitude. 441 00:30:21,080 --> 00:30:25,846 There are beta m, which is basically the phase. 442 00:30:25,846 --> 00:30:30,170 There are Km, which is actually the wave number, 443 00:30:30,170 --> 00:30:34,830 and alpha m, which is the phase for the shape. 444 00:30:34,830 --> 00:30:42,710 So what I'm going to show you is that Am and the beta m, 445 00:30:42,710 --> 00:30:45,770 these two quantity will be determined 446 00:30:45,770 --> 00:30:48,497 by initial conditions. 447 00:30:55,960 --> 00:31:03,460 Well, Km and alpha m, as you may guess, 448 00:31:03,460 --> 00:31:13,170 those can be determined by boundary condition 449 00:31:13,170 --> 00:31:17,560 for the Km and alpha m. 450 00:31:17,560 --> 00:31:20,620 So why don't we just immediately get 451 00:31:20,620 --> 00:31:26,290 started with a concrete example. 452 00:31:26,290 --> 00:31:30,760 So let's take a look at this situation. 453 00:31:30,760 --> 00:31:34,420 So this equation and those all the possible Km 454 00:31:34,420 --> 00:31:40,120 are allowed when we talk about infinitely long system. 455 00:31:40,120 --> 00:31:44,620 So far, we have not imposed any boundary conditions. 456 00:31:44,620 --> 00:31:50,140 And what I'm going to do now is to show you a example boundary 457 00:31:50,140 --> 00:31:55,550 condition and see how we can actually fix Km and alpha m. 458 00:31:55,550 --> 00:32:00,510 So suppose we are interested in this system. 459 00:32:00,510 --> 00:32:05,000 So I have a wall in the left hand side. 460 00:32:05,000 --> 00:32:10,880 And I have a string with length L. 461 00:32:10,880 --> 00:32:13,550 And it's actually connected to massless ring. 462 00:32:23,470 --> 00:32:30,762 And this ring can actually move up and down a long rod 463 00:32:30,762 --> 00:32:31,720 in the right hand side. 464 00:32:35,070 --> 00:32:44,550 And I also assume that this string have a constant tension 465 00:32:44,550 --> 00:32:51,930 T. And also the density is rho L. 466 00:32:51,930 --> 00:32:55,800 So basically, it's a mass per unit length. 467 00:32:55,800 --> 00:32:58,290 So that's the system which I am interested. 468 00:32:58,290 --> 00:33:03,590 And, of course, I need to define my coordinate system as usual. 469 00:33:03,590 --> 00:33:08,040 I define horizontal direction to be x direction. 470 00:33:08,040 --> 00:33:15,240 And I define the vertical direction to be y direction. 471 00:33:15,240 --> 00:33:20,070 And I define y equal to 0 is the equilibrium 472 00:33:20,070 --> 00:33:23,340 portion of the string. 473 00:33:23,340 --> 00:33:28,650 When psi is equal to 0, that means this string is actually 474 00:33:28,650 --> 00:33:30,420 at rest. 475 00:33:30,420 --> 00:33:36,240 And not moving-- is actually in the equilibrium position, 476 00:33:36,240 --> 00:33:41,310 it's not displaced at all with respect to y equal to 0. 477 00:33:41,310 --> 00:33:45,450 And I can also define that x equal to 0 478 00:33:45,450 --> 00:33:50,460 is the position of the left hand side wall. 479 00:33:50,460 --> 00:33:53,170 So this is actually the physical situation. 480 00:33:53,170 --> 00:33:56,940 And I would like to actually find out what 481 00:33:56,940 --> 00:33:59,320 are the boundary conditions. 482 00:33:59,320 --> 00:34:01,510 So what are the boundary condition? 483 00:34:01,510 --> 00:34:04,550 So from what we actually discussed last time, 484 00:34:04,550 --> 00:34:08,670 left hand side, since this string is actually fixed 485 00:34:08,670 --> 00:34:13,280 on the wall, I nailed it there, it cannot move. 486 00:34:13,280 --> 00:34:17,927 Therefore, what is actually the first boundary condition? 487 00:34:22,600 --> 00:34:25,870 Why is actually the first boundary condition? 488 00:34:25,870 --> 00:34:28,350 Anybody can tell me? 489 00:34:28,350 --> 00:34:31,259 Which describes the situation, the physical situation 490 00:34:31,259 --> 00:34:33,231 on your left hand side? 491 00:34:33,231 --> 00:34:35,219 AUDIENCE: y0 is 0. 492 00:34:35,219 --> 00:34:37,290 YEN-JIE LEE: y0 is 0. 493 00:34:37,290 --> 00:34:38,250 Very good. 494 00:34:38,250 --> 00:34:44,699 So when x is equal to 0, y0 is 0. 495 00:34:44,699 --> 00:34:48,060 So on my note, I was using a different notation. 496 00:34:48,060 --> 00:34:50,550 So I would just use psi. 497 00:34:50,550 --> 00:34:55,800 So psi 0 is equal to 0. 498 00:34:55,800 --> 00:35:00,900 Apparently, there's another boundary of this system. 499 00:35:00,900 --> 00:35:03,390 The other boundary condition is actually 500 00:35:03,390 --> 00:35:08,600 happening at x equal to L. What is actually the boundary 501 00:35:08,600 --> 00:35:09,210 condition? 502 00:35:09,210 --> 00:35:10,250 Can somebody help me? 503 00:35:10,250 --> 00:35:10,750 Yes. 504 00:35:10,750 --> 00:35:13,654 AUDIENCE: Is it the derivative of psi is 0? 505 00:35:13,654 --> 00:35:17,380 YEN-JIE LEE: The derivative of the psi is equal to 0. 506 00:35:17,380 --> 00:35:22,270 So we will explain to everybody why is that the case. 507 00:35:22,270 --> 00:35:31,560 The answer proposed is that partial psi partial x L 508 00:35:31,560 --> 00:35:33,120 t is equal to 0. 509 00:35:33,120 --> 00:35:35,580 And this is 0, t, because that has 510 00:35:35,580 --> 00:35:39,880 to be true no matter when I actually invented this boundary 511 00:35:39,880 --> 00:35:42,240 condition. 512 00:35:42,240 --> 00:35:48,360 So what is actually giving us this strange boundary 513 00:35:48,360 --> 00:35:50,820 condition? 514 00:35:50,820 --> 00:35:58,540 So suppose I focus on the force diagram on this ring. 515 00:35:58,540 --> 00:36:00,700 So this ring is actually connected 516 00:36:00,700 --> 00:36:07,490 to a string with string tension T. Also, 517 00:36:07,490 --> 00:36:09,340 there's another force which is actually 518 00:36:09,340 --> 00:36:13,300 trying to balance the string tension, which 519 00:36:13,300 --> 00:36:15,460 is a normal force-- 520 00:36:15,460 --> 00:36:19,440 normal force coming from the rod, which is actually 521 00:36:19,440 --> 00:36:22,210 trying to stop the ring from moving 522 00:36:22,210 --> 00:36:23,990 in the horizontal direction. 523 00:36:23,990 --> 00:36:26,156 So there's normal force then. 524 00:36:28,710 --> 00:36:34,110 And we also know that this ring is actually massless. 525 00:36:34,110 --> 00:36:38,340 So m is equal to 0. 526 00:36:38,340 --> 00:36:46,325 Suppose that this partial psi partial x, the slope is not 0. 527 00:36:46,325 --> 00:36:52,800 If slope is not 0, that means the string may be pulling 528 00:36:52,800 --> 00:36:57,650 this ring to some direction. 529 00:36:57,650 --> 00:37:00,050 What is going to happen? 530 00:37:00,050 --> 00:37:05,960 So if this happens, it is actually clear that the normal 531 00:37:05,960 --> 00:37:09,340 force cannot balance the string for us. 532 00:37:12,270 --> 00:37:14,500 Everybody get it? 533 00:37:14,500 --> 00:37:16,480 What will happen? 534 00:37:16,480 --> 00:37:20,920 If this happened, then this massless ring 535 00:37:20,920 --> 00:37:26,860 will suffer from infinitely large acceleration. 536 00:37:26,860 --> 00:37:30,550 Because F is equal to ma. 537 00:37:30,550 --> 00:37:34,270 And m is 0, so a goes to infinity. 538 00:37:34,270 --> 00:37:37,900 So that means this ring will- peeew- disappear, go 539 00:37:37,900 --> 00:37:40,200 to the edge of the universe. 540 00:37:40,200 --> 00:37:41,950 Did that happen? 541 00:37:41,950 --> 00:37:44,735 No, it didn't happen. 542 00:37:44,735 --> 00:37:49,800 Therefore, this condition must be satisfied. 543 00:37:49,800 --> 00:37:50,300 You see? 544 00:37:50,300 --> 00:37:54,140 So the slope of the string cannot be nonzero. 545 00:37:54,140 --> 00:37:56,900 Otherwise, some crisis will happen. 546 00:37:59,810 --> 00:38:00,450 Very good. 547 00:38:00,450 --> 00:38:03,850 So we have the two conditions. 548 00:38:03,850 --> 00:38:09,410 And the second thing, which we are going to demonstrate you, 549 00:38:09,410 --> 00:38:13,220 is that, OK, I promise you that boundary condition can 550 00:38:13,220 --> 00:38:17,120 fix these two constants. 551 00:38:17,120 --> 00:38:19,510 So therefore, we are going to demonstrate that. 552 00:38:23,740 --> 00:38:29,250 So let's use the first condition we have in the right hand side 553 00:38:29,250 --> 00:38:30,460 board. 554 00:38:30,460 --> 00:38:37,070 And basically, from 1, you can actually get psi m 0, t. 555 00:38:37,070 --> 00:38:40,780 I am plugging in this condition-- 556 00:38:40,780 --> 00:38:47,440 plugging in this solution to boundary condition number 1. 557 00:38:47,440 --> 00:38:49,060 And basically, what I am going to get 558 00:38:49,060 --> 00:38:58,840 is this is equal to Am sine alpha m sine omega m t 559 00:38:58,840 --> 00:39:02,550 plus beta m. 560 00:39:02,550 --> 00:39:05,910 And this is actually equal to 0. 561 00:39:05,910 --> 00:39:10,770 So you only have a alpha m here because I 562 00:39:10,770 --> 00:39:15,000 am setting x to be equal to 0. 563 00:39:15,000 --> 00:39:17,370 I'm setting x to be equal to 0. 564 00:39:17,370 --> 00:39:21,690 Therefore, you already have that functional form. 565 00:39:21,690 --> 00:39:24,000 So now, we are facing a choice. 566 00:39:24,000 --> 00:39:27,750 So you can set Am to be equal to 0 is arbitrary number. 567 00:39:27,750 --> 00:39:31,440 But if you set m equal to 0, everything is 0. 568 00:39:31,440 --> 00:39:33,930 And it's not fun, it's not moving. 569 00:39:33,930 --> 00:39:39,360 Therefore, I don't want to set Am to be equal to 0. 570 00:39:39,360 --> 00:39:43,080 And you can say, huh, maybe this is equal to 0-- 571 00:39:43,080 --> 00:39:46,170 sine omega m t plus beta m is equal to 0. 572 00:39:46,170 --> 00:39:48,780 But this is really a sine function. 573 00:39:48,780 --> 00:39:55,350 And this condition has to be satisfied no matter 574 00:39:55,350 --> 00:39:59,610 at which time you are revisiting this boundary condition. 575 00:39:59,610 --> 00:40:03,450 At all times, this boundary condition has to be satisfied. 576 00:40:03,450 --> 00:40:08,520 Therefore, this cannot be equal to 0. 577 00:40:08,520 --> 00:40:11,520 Therefore, I conclude that this is the 0. 578 00:40:14,372 --> 00:40:15,330 So what does that mean? 579 00:40:15,330 --> 00:40:19,350 That means I can choose alpha m is equal to 0. 580 00:40:21,930 --> 00:40:26,031 So that's actually given by the first boundary condition. 581 00:40:28,860 --> 00:40:32,470 So let's actually take a look at the second boundary condition-- 582 00:40:32,470 --> 00:40:39,660 partial psi partial x evaluated at x equal to L and any time t 583 00:40:39,660 --> 00:40:41,610 is equal to 0. 584 00:40:41,610 --> 00:40:44,310 So now, I can plug in, again, the solution 585 00:40:44,310 --> 00:40:53,100 in the middle board partial psi m L, t, partial x. 586 00:40:53,100 --> 00:41:05,300 And that will be equal to Am Km sine omega m t plus beta m 587 00:41:05,300 --> 00:41:10,990 cosine Km x. 588 00:41:10,990 --> 00:41:14,590 And this is equal to 0. 589 00:41:14,590 --> 00:41:17,950 So I am taking a partial derivative partial psi 590 00:41:17,950 --> 00:41:19,090 partial x. 591 00:41:19,090 --> 00:41:21,190 Therefore, the sine become cosine. 592 00:41:21,190 --> 00:41:27,660 The sine Km x plus alpha m becoming cosine. 593 00:41:27,660 --> 00:41:30,460 And also, I know already from the first boundary condition, 594 00:41:30,460 --> 00:41:33,180 alpha m is equal to 0. 595 00:41:33,180 --> 00:41:36,720 Therefore, I get cosine Km x here. 596 00:41:41,070 --> 00:41:45,790 And this is evaluated at x equal to L. 597 00:41:45,790 --> 00:41:50,410 So that means this thing must be equal to 0 598 00:41:50,410 --> 00:41:53,330 based on the second boundary condition. 599 00:41:53,330 --> 00:41:56,020 Of course, we can have a losing argument-- 600 00:41:56,020 --> 00:41:57,730 Am should not be equal to 0. 601 00:41:57,730 --> 00:42:00,730 Otherwise, you will be equal to 0 all the time, 602 00:42:00,730 --> 00:42:04,500 the whole wave function is 0. 603 00:42:04,500 --> 00:42:06,840 And this is actually changing as a function 604 00:42:06,840 --> 00:42:09,960 of time, the same argument, because this boundary condition 605 00:42:09,960 --> 00:42:13,040 has to be satisfied at all times. 606 00:42:13,040 --> 00:42:14,730 From the beginning of the Universe 607 00:42:14,730 --> 00:42:20,100 to the end of the Universe, this condition has to be satisfied. 608 00:42:20,100 --> 00:42:22,410 Therefore, these cannot be equal to 0. 609 00:42:22,410 --> 00:42:29,150 And what is actually left over is cosine Km x evaluated at L 610 00:42:29,150 --> 00:42:30,250 equal to 0. 611 00:42:30,250 --> 00:42:37,230 So cosine Km L is equal to 0. 612 00:42:37,230 --> 00:42:44,040 So that means you cannot arbitrarily choose Km anymore. 613 00:42:44,040 --> 00:42:47,900 Before we introduced boundary conditions, we were saying, ah, 614 00:42:47,900 --> 00:42:51,480 Km is actually some arbitrary constant. 615 00:42:51,480 --> 00:42:54,720 And now, it's not arbitrary any more. 616 00:42:54,720 --> 00:42:58,380 It has to satisfy this condition. 617 00:42:58,380 --> 00:42:59,890 What does that mean? 618 00:42:59,890 --> 00:43:05,210 This means that Km has to be equal to 2m 619 00:43:05,210 --> 00:43:09,150 minus 1 divided by 2L times pi. 620 00:43:11,860 --> 00:43:13,730 You can actually check this. 621 00:43:13,730 --> 00:43:21,724 And this small m is equal to 1, 2, 3, et cetera, et cetera. 622 00:43:26,180 --> 00:43:29,370 And then you can see that there are 623 00:43:29,370 --> 00:43:32,200 many, many different solutions. 624 00:43:35,560 --> 00:43:40,270 So you can see that, as I mentioned before, 625 00:43:40,270 --> 00:43:45,820 the boundary conditions determine Km and alpha m. 626 00:43:45,820 --> 00:43:49,510 So you can see that the first condition at x equal to 0 627 00:43:49,510 --> 00:43:51,460 determine alpha m. 628 00:43:51,460 --> 00:43:53,620 The second boundary condition also 629 00:43:53,620 --> 00:43:58,600 help us to determine what are the possible Km value. 630 00:43:58,600 --> 00:44:04,210 And that is actually listed here. 631 00:44:04,210 --> 00:44:07,430 Any questions? 632 00:44:07,430 --> 00:44:07,930 No? 633 00:44:12,490 --> 00:44:16,130 So in order to help you to visualize 634 00:44:16,130 --> 00:44:18,370 what we have learned from here, I 635 00:44:18,370 --> 00:44:22,450 can now choose m is equal to 1. 636 00:44:22,450 --> 00:44:26,650 So you can see that I carefully choose my notation 637 00:44:26,650 --> 00:44:27,910 from the beginning. 638 00:44:27,910 --> 00:44:34,480 So therefore, m is now the index of the normal mode 639 00:44:34,480 --> 00:44:36,400 I am referring to. 640 00:44:36,400 --> 00:44:40,930 So now, if I choose m equal to 1, 641 00:44:40,930 --> 00:44:43,300 then I can actually evaluate what 642 00:44:43,300 --> 00:44:49,280 would be the resulting K. So K1, according to last formula 2 643 00:44:49,280 --> 00:44:51,770 minus 1 is giving you 1. 644 00:44:51,770 --> 00:44:53,750 Therefore, you'll get pi over 2L. 645 00:44:56,530 --> 00:44:58,810 And, of course, you can also calculate 646 00:44:58,810 --> 00:45:02,460 based on the wave number what will be the wavelength. 647 00:45:02,460 --> 00:45:08,010 So the wavelength lambda 1 will be equal to 2 pi over K1. 648 00:45:08,010 --> 00:45:11,170 Wine And that will give you 4L. 649 00:45:14,430 --> 00:45:22,730 Don't forget, once you actually already decide K, 650 00:45:22,730 --> 00:45:27,980 the omega is also determined, because omega, 651 00:45:27,980 --> 00:45:31,920 which is the angular frequency of this normal mode, 652 00:45:31,920 --> 00:45:34,980 is determined by that dispersion relation omega 653 00:45:34,980 --> 00:45:38,520 m equal to vp times Km. 654 00:45:38,520 --> 00:45:41,640 So therefore, I can now calculate omega 1. 655 00:45:41,640 --> 00:45:46,100 That will be equal to vp times K1. 656 00:45:46,100 --> 00:45:56,310 And that will you square root of T over rho L pi over 2L. 657 00:45:56,310 --> 00:45:59,730 So this is actually coming from the last lecture, 658 00:45:59,730 --> 00:46:00,750 the formula of vp. 659 00:46:03,840 --> 00:46:10,230 So that means, if you fix the shape of you are normal mode, 660 00:46:10,230 --> 00:46:13,200 then the angular frequency is also fixed, 661 00:46:13,200 --> 00:46:17,880 according to this dispersion relation. 662 00:46:17,880 --> 00:46:21,690 So of course, I can now visualize this situation. 663 00:46:21,690 --> 00:46:28,490 And basically, I can plot this system as a function of time-- 664 00:46:28,490 --> 00:46:31,200 as a function of x, not as a function of time. 665 00:46:31,200 --> 00:46:36,380 So when this system reach the maxima amplitude, 666 00:46:36,380 --> 00:46:38,640 it would look like this. 667 00:46:38,640 --> 00:46:43,940 And this is actually amplitude A1. 668 00:46:43,940 --> 00:46:48,470 Because I am talking about the first normal mode labeled by m 669 00:46:48,470 --> 00:46:52,730 equal to 1, and there is an unknown amplitude A1. 670 00:46:52,730 --> 00:46:54,400 And that is actually showing here. 671 00:46:57,350 --> 00:47:02,186 Of course, I can go ahead and calculate 672 00:47:02,186 --> 00:47:06,380 if m is equal to 2, what is going to happen? 673 00:47:06,380 --> 00:47:09,430 If I increase the m, what is going to happen 674 00:47:09,430 --> 00:47:14,120 is that K is also increased. 675 00:47:14,120 --> 00:47:15,420 So K is increased. 676 00:47:15,420 --> 00:47:19,010 Then that means the wavelength is decreased. 677 00:47:19,010 --> 00:47:21,550 I have calculated the K for you. 678 00:47:21,550 --> 00:47:26,630 And that is equal to 3 pi over 2L. 679 00:47:26,630 --> 00:47:33,300 And those are the lambda 2 will be equal to 4L over 3. 680 00:47:33,300 --> 00:47:38,560 You can actually double check this at home. 681 00:47:38,560 --> 00:47:41,900 And of course, I can now demonstrate 682 00:47:41,900 --> 00:47:47,090 what would be the resulting shape of the massless mode. 683 00:47:47,090 --> 00:47:50,550 It would look like this. 684 00:47:50,550 --> 00:47:57,930 And this is essentially telling you the amplitude A2. 685 00:47:57,930 --> 00:48:01,320 You can also do m equal to 3. 686 00:48:01,320 --> 00:48:03,855 If you doing that, basically what you get 687 00:48:03,855 --> 00:48:10,020 is something like this, et cetera, et cetera. 688 00:48:12,640 --> 00:48:14,790 Any questions? 689 00:48:14,790 --> 00:48:19,880 And the motion of this system that is a function of time 690 00:48:19,880 --> 00:48:22,890 is that this whole shape, this shape, 691 00:48:22,890 --> 00:48:29,350 is multiplied by sine omega m t plus beta m. 692 00:48:29,350 --> 00:48:36,180 So the whole shape is going to scale up and down harmonically. 693 00:48:36,180 --> 00:48:40,830 And so if you focus on one of the point here, 694 00:48:40,830 --> 00:48:45,330 it's going to be going up and down harmonically. 695 00:48:45,330 --> 00:48:49,140 Very important, there's no back and forth movement. 696 00:48:49,140 --> 00:48:51,540 Everything is only moving up and down. 697 00:48:51,540 --> 00:48:56,660 If you focus on only one of the particle in this string, 698 00:48:56,660 --> 00:48:58,900 everything is moving up and down. 699 00:48:58,900 --> 00:48:59,640 Like here, right? 700 00:48:59,640 --> 00:49:02,530 So when I create a curve-- 701 00:49:02,530 --> 00:49:08,140 when I create some kind of wave, all the components are always 702 00:49:08,140 --> 00:49:11,730 moving only up and down, instead of back and forth, 703 00:49:11,730 --> 00:49:12,810 because they can't. 704 00:49:12,810 --> 00:49:14,230 They can't move back and forth. 705 00:49:14,230 --> 00:49:18,530 But you maybe cheated by the shape-- the evolution 706 00:49:18,530 --> 00:49:20,081 as a boundary of time, it seems to me 707 00:49:20,081 --> 00:49:22,330 that, ah, something is actually moving back and forth. 708 00:49:22,330 --> 00:49:26,930 But never-- all the particles are moving up and down-- 709 00:49:26,930 --> 00:49:29,340 very important. 710 00:49:29,340 --> 00:49:36,640 Finally, we have shown you the first three normal modes. 711 00:49:36,640 --> 00:49:41,035 And what is actually the most general solution? 712 00:49:41,035 --> 00:49:42,160 What is a general solution? 713 00:49:48,700 --> 00:49:53,020 Of course, as we had before, general solution 714 00:49:53,020 --> 00:49:58,690 would be a linear combination of all the possible normal modes. 715 00:49:58,690 --> 00:50:02,320 So now, I would like to write psi x, t, 716 00:50:02,320 --> 00:50:07,050 as the general solution is going to be the sum of all 717 00:50:07,050 --> 00:50:10,450 the allowed normal mode. 718 00:50:10,450 --> 00:50:13,960 In this specific case, it's going 719 00:50:13,960 --> 00:50:17,140 to be a summation from 1-- 720 00:50:17,140 --> 00:50:34,630 m equal to 1 to infinity Am sine omega m t plus beta m sine Km x 721 00:50:34,630 --> 00:50:36,420 plus alpha m. 722 00:50:36,420 --> 00:50:40,260 And in this specific case, it become cosine-- 723 00:50:40,260 --> 00:50:42,600 the Km is there-- 724 00:50:42,600 --> 00:50:46,020 cosine 2m minus 1-- 725 00:50:46,020 --> 00:50:48,220 sorry, it should be sine. 726 00:50:48,220 --> 00:50:54,660 It should be sine 2m minus 1 over 2L pi. 727 00:50:54,660 --> 00:50:59,220 And the alpha m in this case is equal to 0. 728 00:50:59,220 --> 00:51:02,820 So the upper formula is the most general case. 729 00:51:02,820 --> 00:51:06,380 You're summing over the possible m's. 730 00:51:06,380 --> 00:51:08,910 And the Km and alpha m can be determined 731 00:51:08,910 --> 00:51:11,250 by boundary conditions. 732 00:51:11,250 --> 00:51:16,080 And in this specific case, the right hand side expression 733 00:51:16,080 --> 00:51:21,861 is reading like sine 2m minus 1 over 2L times pi. 734 00:51:25,370 --> 00:51:27,650 So that's very nice. 735 00:51:27,650 --> 00:51:31,490 And then you can see another sets of example 736 00:51:31,490 --> 00:51:33,500 here in the slide. 737 00:51:33,500 --> 00:51:37,520 So this is another set of normal modes 738 00:51:37,520 --> 00:51:41,900 from m equal to 1 to m equal to 6. 739 00:51:41,900 --> 00:51:45,560 And you can see that in this example 740 00:51:45,560 --> 00:51:51,710 both ends are fixed, instead of one end is actually 741 00:51:51,710 --> 00:51:54,190 attached to a massless ring. 742 00:51:54,190 --> 00:51:57,710 If both ends are fixed, then you get these normal modes. 743 00:51:57,710 --> 00:52:00,290 And of course, in the later-- 744 00:52:00,290 --> 00:52:02,840 in your p set, you will be exercising 745 00:52:02,840 --> 00:52:06,170 this kind of normal modes and solve 746 00:52:06,170 --> 00:52:08,720 the corresponding Km and alpha m. 747 00:52:08,720 --> 00:52:14,180 And you can see that, if you focus on the upper left corner, 748 00:52:14,180 --> 00:52:20,600 you will see that the oscillation frequency is low. 749 00:52:20,600 --> 00:52:21,650 Why is that? 750 00:52:21,650 --> 00:52:27,290 That is because the wave number Km is small, therefore, 751 00:52:27,290 --> 00:52:28,970 wavelength is long. 752 00:52:28,970 --> 00:52:31,760 According to that formula, omega m 753 00:52:31,760 --> 00:52:35,040 is proportional to wave number. 754 00:52:35,040 --> 00:52:38,270 Therefore, you can see that the oscillation frequency 755 00:52:38,270 --> 00:52:43,120 is actually two times slower compared 756 00:52:43,120 --> 00:52:47,510 to m equal to 2 case, which is the upper right corner 757 00:52:47,510 --> 00:52:52,310 result And you can see that, if you increase m more and more, 758 00:52:52,310 --> 00:52:55,520 you get larger and larger K. And therefore, you 759 00:52:55,520 --> 00:52:59,581 see that the oscillation frequency is getting larger 760 00:52:59,581 --> 00:53:00,080 and larger. 761 00:53:06,320 --> 00:53:11,240 So now, we are actually facing an issue here. 762 00:53:11,240 --> 00:53:17,960 Wait a second, so now we have solved the functional 763 00:53:17,960 --> 00:53:19,490 form of the normal mode. 764 00:53:19,490 --> 00:53:23,750 We have learned how to determine Km and alpha m. 765 00:53:23,750 --> 00:53:25,520 But we are facing a difficulty here, 766 00:53:25,520 --> 00:53:28,610 because Am is very difficult to solve, 767 00:53:28,610 --> 00:53:31,910 because you have infinite number of terms here. 768 00:53:31,910 --> 00:53:35,480 And beta m, how do we solve this? 769 00:53:35,480 --> 00:53:39,870 So it's getting really, really difficult. 770 00:53:39,870 --> 00:53:41,990 So what I am going to tell you is 771 00:53:41,990 --> 00:53:45,560 that we can actually, again, use the help from the math 772 00:53:45,560 --> 00:53:46,610 department. 773 00:53:46,610 --> 00:53:49,920 They have actually proposed the solution. 774 00:53:49,920 --> 00:53:52,770 They actually say that, huh, this is actually 775 00:53:52,770 --> 00:53:55,910 identical problem that we solved in the math department, 776 00:53:55,910 --> 00:54:01,430 is just for the decomposition and for the series. 777 00:54:01,430 --> 00:54:03,200 So what is actually for the series? 778 00:54:03,200 --> 00:54:07,220 So you can see, from here, there's a triangular 779 00:54:07,220 --> 00:54:10,160 shape between 0 and 1. 780 00:54:10,160 --> 00:54:13,720 It's a function-- probably is a function of x. 781 00:54:13,720 --> 00:54:16,670 And between 0 and 1, it looks like a triangle. 782 00:54:16,670 --> 00:54:19,850 And if you do for the decomposition, 783 00:54:19,850 --> 00:54:26,390 it can be decomposed as small k sine 784 00:54:26,390 --> 00:54:31,940 function plus the second normal mode 785 00:54:31,940 --> 00:54:33,590 and plus a second massless mode. 786 00:54:33,590 --> 00:54:39,780 And you can see that, if you increase the number of terms 787 00:54:39,780 --> 00:54:43,790 included in this Fourier series, then 788 00:54:43,790 --> 00:54:45,800 you will see that the shape is actually 789 00:54:45,800 --> 00:54:51,290 getting closer and closer to the triangular shape. 790 00:54:51,290 --> 00:54:55,410 In order to help you with the visualization, 791 00:54:55,410 --> 00:54:58,790 here is actually what I prepared. 792 00:54:58,790 --> 00:55:04,820 So this actually extracted from essentially a real example, 793 00:55:04,820 --> 00:55:07,940 which I really used a computer to calculate. 794 00:55:07,940 --> 00:55:12,140 And I tracked the contribution from m equal to 1. 795 00:55:12,140 --> 00:55:16,690 This means that the first term in this summation-- 796 00:55:16,690 --> 00:55:18,830 infinite number of term summation-- 797 00:55:18,830 --> 00:55:21,000 the first term looks like this. 798 00:55:21,000 --> 00:55:25,340 And if you include the first and second and third term, 799 00:55:25,340 --> 00:55:28,310 it becomes something like a plateau. 800 00:55:28,310 --> 00:55:31,260 And then if you increase 1 to 5, it's 801 00:55:31,260 --> 00:55:36,880 evolving as a function of m, becoming more and more-- hm, 802 00:55:36,880 --> 00:55:38,840 strange shape. 803 00:55:38,840 --> 00:55:42,920 And that is actually including the summation from first term 804 00:55:42,920 --> 00:55:50,090 to 11 terms and, finally, 11 to 19. 805 00:55:50,090 --> 00:55:51,990 Huh, what this is-- 806 00:55:51,990 --> 00:55:53,700 what is actually the function I put in? 807 00:55:53,700 --> 00:55:55,860 It's actually a MIT function! 808 00:55:55,860 --> 00:55:58,100 [LAUGHTER] 809 00:55:58,100 --> 00:56:02,040 I put in a MIT function into this again. 810 00:56:02,040 --> 00:56:04,460 And you can see that-- 811 00:56:04,460 --> 00:56:07,940 wow, 1 to 59 term. 812 00:56:07,940 --> 00:56:11,050 I need to use 59 terms to describe 813 00:56:11,050 --> 00:56:14,510 this really wonderful shape. 814 00:56:14,510 --> 00:56:22,490 [LAUGHS] So in order to help you with the visualization, 815 00:56:22,490 --> 00:56:26,810 listen you can see I prepared a little program, which actually 816 00:56:26,810 --> 00:56:34,588 can show you the evolutions as I increase more and more m terms. 817 00:56:34,588 --> 00:56:36,020 Let's take a look. 818 00:56:36,020 --> 00:56:38,840 You can see that, originally it looks-- 819 00:56:38,840 --> 00:56:42,195 doesn't look-- oh, you cannot see anything. 820 00:56:42,195 --> 00:56:43,990 Wait a second. 821 00:56:43,990 --> 00:56:46,422 What is going on? 822 00:56:46,422 --> 00:56:48,776 Let me see if I can-- 823 00:56:48,776 --> 00:56:50,650 I hope I don't screw this up. 824 00:56:56,840 --> 00:56:57,430 Sorry. 825 00:56:57,430 --> 00:56:58,320 I need to restart. 826 00:57:06,550 --> 00:57:08,600 So let's get started. 827 00:57:08,600 --> 00:57:10,980 So you can see that from the first few terms, 828 00:57:10,980 --> 00:57:12,760 it doesn't look anything. 829 00:57:12,760 --> 00:57:17,760 But very soon, when you have 20 terms added to each other, 830 00:57:17,760 --> 00:57:22,630 it looks really pretty much like a MIT dome. 831 00:57:22,630 --> 00:57:26,530 And you can see that this program is really 832 00:57:26,530 --> 00:57:31,230 trying really hard to describe the sharp edge 833 00:57:31,230 --> 00:57:33,620 in the left-hand side and the right-hand side. 834 00:57:33,620 --> 00:57:39,160 You can see that those kind of really infinitely sharp edge 835 00:57:39,160 --> 00:57:42,260 will need infinite number of terms, 836 00:57:42,260 --> 00:57:47,230 so that if your m is really huge, 837 00:57:47,230 --> 00:57:50,630 then the K, the wave number, is going to infinity. 838 00:57:50,630 --> 00:57:55,100 Then you can actually produce infinitely sharp edge 839 00:57:55,100 --> 00:57:56,170 in this function. 840 00:57:56,170 --> 00:57:59,920 And that is actually, you can see from this demonstration 841 00:57:59,920 --> 00:58:04,180 the program is really struggling with this really 842 00:58:04,180 --> 00:58:06,500 super sharp edge. 843 00:58:06,500 --> 00:58:09,100 So look at the left hand side and right hand side corner, 844 00:58:09,100 --> 00:58:15,520 originally the slope is clearly not high enough. 845 00:58:15,520 --> 00:58:21,010 And thus, we include higher, higher m value terms. 846 00:58:21,010 --> 00:58:24,930 And you can see the description becoming much, much better 847 00:58:24,930 --> 00:58:26,400 at the edge but, of course, still 848 00:58:26,400 --> 00:58:29,650 are not perfect, because you need infinite number of terms 849 00:58:29,650 --> 00:58:33,940 to describe the shape of MIT dome. 850 00:58:33,940 --> 00:58:39,040 Of course, we can also take a look at other example, 851 00:58:39,040 --> 00:58:44,950 just testing my eyesight I'm not sure if I-- 852 00:58:44,950 --> 00:58:49,420 OK, so I can increase the speed to save on time. 853 00:58:49,420 --> 00:58:52,540 So this is actually a square pulse, 854 00:58:52,540 --> 00:58:57,400 which you can see from the scope pretty often when 855 00:58:57,400 --> 00:59:00,010 you do experiment. 856 00:59:00,010 --> 00:59:03,430 And you can see that a square pulse is really difficult 857 00:59:03,430 --> 00:59:10,070 to reproduce, as I said before, due to these sharply rising h. 858 00:59:10,070 --> 00:59:12,160 And of course, I can also demonstrate you 859 00:59:12,160 --> 00:59:16,090 another example, which is a triangular shape. 860 00:59:16,090 --> 00:59:18,523 And you can see that-- 861 00:59:18,523 --> 00:59:23,360 ah, still, you can see it works pretty nicely. 862 00:59:23,360 --> 00:59:27,560 And the function doesn't like at all the right hand side 863 00:59:27,560 --> 00:59:34,870 edge, because of exactly the same reason. 864 00:59:34,870 --> 00:59:35,630 OK, very good. 865 00:59:35,630 --> 00:59:38,500 So let's come back to the presentation. 866 00:59:38,500 --> 00:59:40,910 Can you see-- OK, very good. 867 00:59:40,910 --> 00:59:47,720 So the question is, how do we actually extract Am and beta m? 868 00:59:47,720 --> 00:59:51,010 OK, I have done that with a computer program. 869 00:59:51,010 --> 00:59:55,080 And what I'm going to do now is to show you a concrete example. 870 00:59:55,080 --> 00:59:58,220 And we are going to go through it together 871 00:59:58,220 --> 01:00:01,250 to see how we actually can extract Am and beta m. 872 01:00:01,250 --> 01:00:06,630 So suppose I give you an initial condition. 873 01:00:06,630 --> 01:00:10,910 It's exactly the same system I am talking about. 874 01:00:10,910 --> 01:00:13,475 But now, I prepare this system at t 875 01:00:13,475 --> 01:00:17,320 equal to 0 some specific kind of shape. 876 01:00:17,320 --> 01:00:22,460 This L/2 is actually the first half of the system, 877 01:00:22,460 --> 01:00:26,210 is actually untouched. 878 01:00:26,210 --> 01:00:28,920 The first part of the string is actually 879 01:00:28,920 --> 01:00:32,100 at the equilibrium position. 880 01:00:32,100 --> 01:00:35,970 And this is actually x equal to L/2. 881 01:00:35,970 --> 01:00:41,020 Suddenly, I actually move the string sharply up. 882 01:00:41,020 --> 01:00:44,970 And the rest half of the string is actually 883 01:00:44,970 --> 01:00:49,540 at the height of h in this case. 884 01:00:49,540 --> 01:00:56,990 And of course, the right hand side edge is x equal to L. 885 01:00:56,990 --> 01:01:01,880 And this is actually a snapshot, which I actually took 886 01:01:01,880 --> 01:01:06,680 with my camera at t equal to 0. 887 01:01:06,680 --> 01:01:12,350 And also, I assume that at t equal to 0, the string-- 888 01:01:12,350 --> 01:01:17,780 all the components of the string is at rest. 889 01:01:23,740 --> 01:01:27,460 So based on this information, which I give you, 890 01:01:27,460 --> 01:01:31,900 I can now translate this information into mathematics. 891 01:01:31,900 --> 01:01:35,410 So that corresponds to two initial conditions. 892 01:01:42,260 --> 01:01:47,810 The first one is that, since the string 893 01:01:47,810 --> 01:02:00,710 is at rest, that means psi dot x evaluated at t equal to 0 is 0, 894 01:02:00,710 --> 01:02:03,950 because the string is at rest. 895 01:02:03,950 --> 01:02:15,230 The second initial condition is that psi x 0 is known and is 896 01:02:15,230 --> 01:02:17,060 actually shown in this graph. 897 01:02:21,340 --> 01:02:24,850 So from this, we would like to see 898 01:02:24,850 --> 01:02:31,800 if we can actually extract information about capital 899 01:02:31,800 --> 01:02:36,690 Am and the beta m. 900 01:02:36,690 --> 01:02:40,750 So let's immediately get started to see how we can 901 01:02:40,750 --> 01:02:44,610 use those initial conditions. 902 01:02:44,610 --> 01:02:47,790 So from the first initial condition, a, 903 01:02:47,790 --> 01:02:52,530 related to the initial velocity of the string, basically, 904 01:02:52,530 --> 01:02:59,520 we can get psi dot x, t. 905 01:02:59,520 --> 01:03:05,040 And this will be equal to, let's see, the sum over m equal to 1 906 01:03:05,040 --> 01:03:06,090 to infinity. 907 01:03:06,090 --> 01:03:11,250 So basically, I'm taking this equation 908 01:03:11,250 --> 01:03:14,550 when I plug in that equation into 909 01:03:14,550 --> 01:03:17,380 the first initial condition. 910 01:03:17,380 --> 01:03:27,070 So basically, what I have is Am omega m sine omega m 911 01:03:27,070 --> 01:03:31,944 t plus beta m. 912 01:03:31,944 --> 01:03:33,640 Oh, this will become cosine-- 913 01:03:33,640 --> 01:03:38,600 sorry-- because I'm doing a derivative, psi dot. 914 01:03:38,600 --> 01:03:39,880 So this will become cosine. 915 01:03:42,850 --> 01:03:49,518 And sine Km x plus alpha m. 916 01:03:52,930 --> 01:03:59,250 And this is actually equal to 0 when 917 01:03:59,250 --> 01:04:08,480 psi dot x, t, is actually evaluated at t equal to 0. 918 01:04:08,480 --> 01:04:11,330 And this is equal to 0. 919 01:04:11,330 --> 01:04:18,920 So if I plug in t equal to 0 to this equation, this becomes 0. 920 01:04:18,920 --> 01:04:21,630 And then we know that the shape of the normal mode 921 01:04:21,630 --> 01:04:26,520 is some kind of sine function from the previous discussion. 922 01:04:26,520 --> 01:04:33,170 And I am now requiring this thing to be equal to 0. 923 01:04:33,170 --> 01:04:37,790 Of course, I cannot make Am omega m equal to 0. 924 01:04:37,790 --> 01:04:39,830 That's what we discussed before. 925 01:04:39,830 --> 01:04:43,250 And this sine function can be evaluated 926 01:04:43,250 --> 01:04:46,340 at any place, any x value. 927 01:04:46,340 --> 01:04:50,180 Therefore, this cannot be equal to 0. 928 01:04:50,180 --> 01:04:53,000 Therefore, what is actually the result? 929 01:04:53,000 --> 01:04:57,610 The resulting condition is that cosine beta m 930 01:04:57,610 --> 01:05:01,220 will be equal to 0. 931 01:05:01,220 --> 01:05:05,030 Therefore-- huh, from this initial condition, 932 01:05:05,030 --> 01:05:09,590 actually I can conclude that beta m is actually 933 01:05:09,590 --> 01:05:11,390 equal to pi/2, for example. 934 01:05:20,420 --> 01:05:23,540 So therefore, you can see that very clearly 935 01:05:23,540 --> 01:05:26,600 from the first initial condition, the string 936 01:05:26,600 --> 01:05:28,610 is not moving at the beginning, I 937 01:05:28,610 --> 01:05:34,490 can conclude that beta m is equal to pi over 2. 938 01:05:34,490 --> 01:05:38,620 And just as reminder, alpha m is actually equal to 0 939 01:05:38,620 --> 01:05:40,750 from the previous discussion. 940 01:05:40,750 --> 01:05:47,950 And Km is actually equal to 2m minus 1 pi divided by 2L, 941 01:05:47,950 --> 01:05:50,110 because I just want to copy here, 942 01:05:50,110 --> 01:05:55,990 because somehow the board is covered by another board. 943 01:05:55,990 --> 01:06:00,280 So now, I have done with the first initial condition. 944 01:06:00,280 --> 01:06:03,880 And the other initial condition I have 945 01:06:03,880 --> 01:06:07,960 is that, OK, I provided you the picture 946 01:06:07,960 --> 01:06:10,720 I took at the beginning of the experiment. 947 01:06:10,720 --> 01:06:13,560 Therefore, psi x, 0-- 948 01:06:13,560 --> 01:06:15,010 at t equal to 0-- 949 01:06:15,010 --> 01:06:17,910 is known. 950 01:06:17,910 --> 01:06:20,190 So very good. 951 01:06:20,190 --> 01:06:22,500 So I have this condition. 952 01:06:22,500 --> 01:06:25,290 But now, I am facing a difficulty, 953 01:06:25,290 --> 01:06:28,670 because all those terms-- 954 01:06:28,670 --> 01:06:32,610 all the terms, m equal to 1 to infinity-- 955 01:06:32,610 --> 01:06:37,200 contribute, as we demonstrated before, to this shape. 956 01:06:37,200 --> 01:06:43,530 It's very difficult to actually evaluate Am. 957 01:06:43,530 --> 01:06:52,140 So the trick is to make friends from the math department. 958 01:06:52,140 --> 01:06:56,020 So what we could do is that we can 959 01:06:56,020 --> 01:07:00,490 use the orthogonality of the sine function 960 01:07:00,490 --> 01:07:04,290 to overcome this difficulty. 961 01:07:04,290 --> 01:07:06,760 So let me immediately write down what do I 962 01:07:06,760 --> 01:07:10,550 mean by orthogonality of the sine function. 963 01:07:10,550 --> 01:07:27,090 So if I do a integration from 0 to L on dx sine Km x sine Kn x, 964 01:07:27,090 --> 01:07:32,490 if I do this integral, integrating from 0 to L, 965 01:07:32,490 --> 01:07:38,370 so what I am going to get is that basically you either 966 01:07:38,370 --> 01:07:45,480 get L equal to 2 if m is equal to n, 967 01:07:45,480 --> 01:07:51,840 or you get 0, if m is not equal to n. 968 01:07:51,840 --> 01:07:55,390 So basically, I have two sine functions 969 01:07:55,390 --> 01:07:57,660 multiplied to each other. 970 01:07:57,660 --> 01:08:04,620 And I do integration from 0 to L multiplied by delta x. 971 01:08:04,620 --> 01:08:05,880 And this is Km. 972 01:08:05,880 --> 01:08:07,664 This is Kn. 973 01:08:07,664 --> 01:08:13,190 If Km and Kn are different, you can actually 974 01:08:13,190 --> 01:08:15,720 go ahead and do this exercise. 975 01:08:15,720 --> 01:08:19,620 And you will find that, indeed, if they are the same, 976 01:08:19,620 --> 01:08:22,200 then you will get L/2. 977 01:08:22,200 --> 01:08:25,470 On the other hand, if they are not the same-- 978 01:08:25,470 --> 01:08:29,100 the K value are not the same for the first and second sine 979 01:08:29,100 --> 01:08:30,200 function-- 980 01:08:30,200 --> 01:08:33,510 you are going to get 0. 981 01:08:33,510 --> 01:08:40,450 So that's very good news, because if I do this-- 982 01:08:40,450 --> 01:08:48,359 if I do this calculation, I do 2/L integration from 0 983 01:08:48,359 --> 01:09:02,100 to L psi x, 0, sine Km x dx, what 984 01:09:02,100 --> 01:09:06,220 is going to happen if I do this integration? 985 01:09:06,220 --> 01:09:11,529 Remember, psi m is a linear combination 986 01:09:11,529 --> 01:09:21,660 of infinite number of massless modes with different sine Km x. 987 01:09:21,660 --> 01:09:27,840 If I multiplied that by sine Km x, 988 01:09:27,840 --> 01:09:29,850 this is a very crazy thing to do, 989 01:09:29,850 --> 01:09:34,170 because all the other terms will become 0. 990 01:09:34,170 --> 01:09:38,220 If the K value of one of the terms 991 01:09:38,220 --> 01:09:44,279 is not equal to the dictator's value Km, it's 0. 992 01:09:46,790 --> 01:09:51,470 Otherwise, it's L/2, and it is designed here 993 01:09:51,470 --> 01:09:54,380 to cancel that factor. 994 01:09:54,380 --> 01:09:59,340 So you can see that this is like a mode picker. 995 01:09:59,340 --> 01:10:02,270 I'm picking up a mode with this tool. 996 01:10:02,270 --> 01:10:04,780 This is like, this tool, yeah, I'm 997 01:10:04,780 --> 01:10:08,415 picking this mode, which is actually matching my Km. 998 01:10:11,570 --> 01:10:17,120 It's a miracle that this become Am. 999 01:10:17,120 --> 01:10:21,200 I hope you get this idea, even if probably you 1000 01:10:21,200 --> 01:10:25,760 haven't heard about with your decomposition before. 1001 01:10:25,760 --> 01:10:28,150 But essentially, what we are doing 1002 01:10:28,150 --> 01:10:33,830 is that I'm going to evaluate infinite number of integrals. 1003 01:10:33,830 --> 01:10:39,960 And you are going to do that in the exam, hopefully easy. 1004 01:10:39,960 --> 01:10:41,660 [LAUGHS] 1005 01:10:41,660 --> 01:10:44,600 What is going to happen is that, if you do this integral, 1006 01:10:44,600 --> 01:10:47,420 you are going to pick only one mode out of it. 1007 01:10:47,420 --> 01:10:50,870 And you are going to be able to know 1008 01:10:50,870 --> 01:10:53,240 the amplitude of that mode. 1009 01:10:53,240 --> 01:10:56,215 So let's do this immediately in this example. 1010 01:11:02,395 --> 01:11:13,285 So Am is actually equal to 2/L. Since the amplitude 1011 01:11:13,285 --> 01:11:17,190 is actually 0 between 0 and L/2. 1012 01:11:17,190 --> 01:11:23,130 So I can safely just integrate from L/2 to L. 1013 01:11:23,130 --> 01:11:27,130 So I do a integration from L/2 to L, 1014 01:11:27,130 --> 01:11:35,920 because between 0 and L/2, the initial position is 0. 1015 01:11:35,920 --> 01:11:44,180 So what I'm going to get is h sine Km x dx. 1016 01:11:44,180 --> 01:11:47,570 And of course, everybody know how to do this integral. 1017 01:11:47,570 --> 01:11:49,780 It doesn't look that horrible. 1018 01:11:49,780 --> 01:11:58,630 And this would become 2/L minus h over Km. 1019 01:11:58,630 --> 01:12:05,050 And basically, this will become cosine Km evaluated 1020 01:12:05,050 --> 01:12:14,480 at L minus cosine Km evaluated at L/2, 1021 01:12:14,480 --> 01:12:18,400 and where this Km, as just a reminder, 1022 01:12:18,400 --> 01:12:25,450 is basically equal to 2m minus 1 pi divided by 2/L. 1023 01:12:25,450 --> 01:12:33,340 So I hope this actually help you to understand the procedure 1024 01:12:33,340 --> 01:12:37,420 to determining all those unknown coefficients, 1025 01:12:37,420 --> 01:12:40,620 starting from this equation. 1026 01:12:40,620 --> 01:12:43,510 Am and the beta m can be determined 1027 01:12:43,510 --> 01:12:45,640 by initial conditions. 1028 01:12:45,640 --> 01:12:48,250 As we actually show here, you can 1029 01:12:48,250 --> 01:12:50,710 use the initial condition of velocity 1030 01:12:50,710 --> 01:12:53,450 and the initial condition of the shape 1031 01:12:53,450 --> 01:12:56,850 and the help of a mode picker to pick up 1032 01:12:56,850 --> 01:13:01,240 the amplitude from that tool function. 1033 01:13:01,240 --> 01:13:06,040 And also, you can see that Km and alpha m related 1034 01:13:06,040 --> 01:13:11,200 to the shape of the normal mode can be determined 1035 01:13:11,200 --> 01:13:13,540 by boundary conditions-- 1036 01:13:13,540 --> 01:13:18,370 boundary conditions, how this system is actually connected 1037 01:13:18,370 --> 01:13:20,620 to the nearby systems. 1038 01:13:20,620 --> 01:13:25,780 The nearby systems are the rod and the wall. 1039 01:13:25,780 --> 01:13:27,930 So that is actually the two boundary conditions, 1040 01:13:27,930 --> 01:13:32,020 which determine the shape of the normal mode. 1041 01:13:32,020 --> 01:13:35,470 And finally, very important, as usual, 1042 01:13:35,470 --> 01:13:39,980 the most general solution is, of course, 1043 01:13:39,980 --> 01:13:44,790 a linear combination of all of those possible massless modes 1044 01:13:44,790 --> 01:13:47,860 from m equal to 1 to infinity. 1045 01:13:47,860 --> 01:13:53,050 And omega m, don't forget, is determined by the dispersion 1046 01:13:53,050 --> 01:13:56,560 relation, vp times Kl. 1047 01:13:56,560 --> 01:13:59,680 Before the end, I would like to mention to you something 1048 01:13:59,680 --> 01:14:03,310 which you might actually not notice 1049 01:14:03,310 --> 01:14:05,110 when we were discussing this. 1050 01:14:05,110 --> 01:14:11,710 So you can see that omega is now proportional to Km. 1051 01:14:11,710 --> 01:14:16,090 So if you plot omega as a function of k, 1052 01:14:16,090 --> 01:14:19,780 actually you can see that it's becoming a straight line 1053 01:14:19,780 --> 01:14:22,870 in this graph, which is very straightforward. 1054 01:14:22,870 --> 01:14:26,080 And on the other hand, if you remember 1055 01:14:26,080 --> 01:14:30,560 what we got last time with discrete system, 1056 01:14:30,560 --> 01:14:38,620 with length scale between little mass is actually a, 1057 01:14:38,620 --> 01:14:43,000 you get omega square is equal to 4T 1058 01:14:43,000 --> 01:14:47,770 over m sine squared ka over 2. 1059 01:14:47,770 --> 01:14:51,250 So if you plot this omega as a function of k, 1060 01:14:51,250 --> 01:14:54,110 you will get the black curve. 1061 01:14:54,110 --> 01:14:57,050 What does this actually tell you? 1062 01:14:57,050 --> 01:15:01,710 That is actually telling you that, if you prepare 1063 01:15:01,710 --> 01:15:07,750 a system at a specific normal mode based on the oscillation 1064 01:15:07,750 --> 01:15:12,670 frequency, you can actually know the internal length 1065 01:15:12,670 --> 01:15:17,820 scale of individual mass, just in case you 1066 01:15:17,820 --> 01:15:21,100 didn't notice this interesting fact. 1067 01:15:21,100 --> 01:15:22,820 So thank you very much. 1068 01:15:22,820 --> 01:15:24,820 I hope you enjoyed the lecture. 1069 01:15:24,820 --> 01:15:28,600 And I will see you next Thursday-- 1070 01:15:28,600 --> 01:15:32,620 not here in the Walker room-- 1071 01:15:32,620 --> 01:15:34,770 Walker Memorial. 1072 01:15:34,770 --> 01:15:36,340 So good luck, everybody. 1073 01:15:36,340 --> 01:15:40,960 Maybe see some of you in the office hour tomorrow.