1 00:00:00,060 --> 00:00:01,770 The following content is provided 2 00:00:01,770 --> 00:00:04,019 under a Creative Commons license. 3 00:00:04,019 --> 00:00:06,860 Your support will help MIT OpenCourseWare continue 4 00:00:06,860 --> 00:00:10,720 to offer high-quality educational resources for free. 5 00:00:10,720 --> 00:00:13,330 To make a donation or view additional materials 6 00:00:13,330 --> 00:00:17,209 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,209 --> 00:00:17,834 at ocw.mit.edu. 8 00:00:21,280 --> 00:00:22,280 PROFESSOR: Welcome back. 9 00:00:22,280 --> 00:00:25,550 And today we will continue doing some more 10 00:00:25,550 --> 00:00:27,570 problems to do with standing waves. 11 00:00:27,570 --> 00:00:30,640 It's a very important topic, so I thought it worthwhile 12 00:00:30,640 --> 00:00:34,380 considering some different kinds of problems. 13 00:00:34,380 --> 00:00:36,660 So the first problem I want to discuss with you 14 00:00:36,660 --> 00:00:38,270 is the following. 15 00:00:38,270 --> 00:00:43,210 Suppose you have two equal-length strings, 16 00:00:43,210 --> 00:00:47,010 each of length L, but they are very different. 17 00:00:47,010 --> 00:00:49,460 One is very massive. 18 00:00:49,460 --> 00:00:52,920 The other one is very light. 19 00:00:52,920 --> 00:00:57,620 And we tie the massive one at one end. 20 00:00:57,620 --> 00:01:00,070 It's connected to the other one in the middle. 21 00:01:00,070 --> 00:01:02,530 And the other one is tied here. 22 00:01:02,530 --> 00:01:05,190 The whole system is taut. 23 00:01:05,190 --> 00:01:13,120 And the question is, what are two lowest normal modes 24 00:01:13,120 --> 00:01:14,900 of vibrations? 25 00:01:14,900 --> 00:01:17,910 Calculate the frequencies of these 26 00:01:17,910 --> 00:01:20,770 and sketch what they look like. 27 00:01:20,770 --> 00:01:24,930 And I like this problem because it's counter-intuitive. 28 00:01:24,930 --> 00:01:27,950 You could take a few seconds for yourself 29 00:01:27,950 --> 00:01:30,290 and think whether you could predict by guessing. 30 00:01:30,290 --> 00:01:34,920 And I'm almost certain your guess will be wrong. 31 00:01:37,480 --> 00:01:41,210 Certainly, the first time I tried to figure this out, 32 00:01:41,210 --> 00:01:45,810 my intuition gave me the wrong answer. 33 00:01:45,810 --> 00:01:46,700 OK. 34 00:01:46,700 --> 00:01:50,570 Now, again, it's idealized. 35 00:01:50,570 --> 00:01:58,235 We'll assume that these strings are ideal, lossless strings. 36 00:01:58,235 --> 00:02:01,120 All the displacements are very small. 37 00:02:01,120 --> 00:02:04,860 So in the derivations of the equations of motion, et cetera, 38 00:02:04,860 --> 00:02:07,210 the usual sine theta equals theta, 39 00:02:07,210 --> 00:02:11,640 there is a constant tension in this string T. 40 00:02:11,640 --> 00:02:15,730 Because they are connected, of course, like this, 41 00:02:15,730 --> 00:02:19,260 the tension everywhere must be the same. 42 00:02:19,260 --> 00:02:23,490 And all we know is that the density 43 00:02:23,490 --> 00:02:27,760 per unit length of this string, which I call mu sub H, 44 00:02:27,760 --> 00:02:33,470 this is the heavy string, is much bigger 45 00:02:33,470 --> 00:02:39,860 than the density, the mass per unit length, of the light one. 46 00:02:39,860 --> 00:02:41,550 Now, the fact that they don't give me 47 00:02:41,550 --> 00:02:45,730 the values of these, just say this, 48 00:02:45,730 --> 00:02:48,090 basically is telling me they want 49 00:02:48,090 --> 00:02:50,490 almost a qualitative answer. 50 00:02:50,490 --> 00:02:55,865 I cannot calculate it exactly if I don't know these two numbers. 51 00:02:55,865 --> 00:02:56,500 All right? 52 00:02:59,020 --> 00:03:03,480 Now, the other thing we are told is, of course, 53 00:03:03,480 --> 00:03:07,420 a system like that can oscillate in three dimensions. 54 00:03:07,420 --> 00:03:10,860 It could oscillate up and down, out of the board, 55 00:03:10,860 --> 00:03:11,810 in, et cetera. 56 00:03:11,810 --> 00:03:16,400 They're telling us limit ourselves 57 00:03:16,400 --> 00:03:20,955 to oscillations of the string in the plane of the board. 58 00:03:20,955 --> 00:03:21,550 OK? 59 00:03:21,550 --> 00:03:26,200 So it's a 2D problem, not a three-dimensional problem. 60 00:03:26,200 --> 00:03:27,200 OK. 61 00:03:27,200 --> 00:03:29,460 So how do we do this? 62 00:03:33,920 --> 00:03:37,060 We immediately recognize that when 63 00:03:37,060 --> 00:03:43,560 you have a continuous line of harmonic oscillators 64 00:03:43,560 --> 00:03:48,560 in the continuum limit, and that's what the string is, 65 00:03:48,560 --> 00:03:55,150 the equation of motion of the string is this. 66 00:03:55,150 --> 00:03:57,260 This is the standard wave equation. 67 00:03:59,950 --> 00:04:03,980 It's the same equation for the heavy string 68 00:04:03,980 --> 00:04:05,690 and the light string. 69 00:04:05,690 --> 00:04:12,580 And the only difference is the phase velocity v, 70 00:04:12,580 --> 00:04:23,710 which depends on the tension and the mass density of the string. 71 00:04:23,710 --> 00:04:27,570 So the heavy one we will call v sub H, 72 00:04:27,570 --> 00:04:30,170 and it's the square root of T over mu H. 73 00:04:30,170 --> 00:04:37,470 And the light one I call v sub L, square root of T over mu L. 74 00:04:37,470 --> 00:04:38,380 OK. 75 00:04:38,380 --> 00:04:45,750 Since we know that mu H is much bigger than mu L, 76 00:04:45,750 --> 00:04:50,860 we immediately know that vH is much less than vL. 77 00:04:50,860 --> 00:04:53,390 That's the information we have about those strings. 78 00:04:53,390 --> 00:04:56,130 OK. 79 00:04:56,130 --> 00:04:58,610 What else do we know about this system? 80 00:04:58,610 --> 00:05:00,940 I mean, we are now in the usual process 81 00:05:00,940 --> 00:05:03,520 of translating a physical situation 82 00:05:03,520 --> 00:05:06,603 into the mathematical description of the system. 83 00:05:06,603 --> 00:05:07,350 OK? 84 00:05:07,350 --> 00:05:09,920 In this particular idealized case, 85 00:05:09,920 --> 00:05:11,710 this is almost a mathematical description 86 00:05:11,710 --> 00:05:13,500 up there, but still. 87 00:05:13,500 --> 00:05:15,110 So this is the equation of motion. 88 00:05:15,110 --> 00:05:17,520 And the other thing we know about the system 89 00:05:17,520 --> 00:05:19,930 are the boundary conditions. 90 00:05:19,930 --> 00:05:30,500 We know that at both ends, at x equals 0 and at x equals 2L, 91 00:05:30,500 --> 00:05:32,850 the string is tied down. 92 00:05:32,850 --> 00:05:36,660 So that for all time, therefore, the displacement will be 0, 93 00:05:36,660 --> 00:05:39,890 y equals 0. 94 00:05:39,890 --> 00:05:45,520 We know that at the boundary, the strings are connected. 95 00:05:45,520 --> 00:05:50,880 So at all times, y of H must be equal to yL. 96 00:05:50,880 --> 00:05:53,300 Otherwise, it would be broken. 97 00:05:53,300 --> 00:05:54,890 So they're connected. 98 00:05:54,890 --> 00:05:58,680 The displacement of the string in the middle, 99 00:05:58,680 --> 00:06:01,980 both the heavy one and the light one, is the same, 100 00:06:01,980 --> 00:06:05,250 and this is true at all times. 101 00:06:05,250 --> 00:06:09,910 And the other thing is that the slope 102 00:06:09,910 --> 00:06:14,590 of the string on both sides must be the same. 103 00:06:14,590 --> 00:06:15,830 How do I know that? 104 00:06:15,830 --> 00:06:22,430 Well, imagine the junction to be some object. 105 00:06:22,430 --> 00:06:25,440 Then that junction has 0 mass. 106 00:06:28,005 --> 00:06:33,620 If it has 0 mass, you cannot have a net vertical force 107 00:06:33,620 --> 00:06:34,520 on that. 108 00:06:34,520 --> 00:06:37,190 You cannot have a net force on it, period. 109 00:06:37,190 --> 00:06:41,800 Or because of Newton's laws, force on the 0 mass object 110 00:06:41,800 --> 00:06:44,630 will give it an infinite acceleration. 111 00:06:44,630 --> 00:06:46,630 It will just disappear there. 112 00:06:46,630 --> 00:06:52,060 So at the junction, there can be no net force. 113 00:06:52,060 --> 00:06:55,370 The tension on the string on both sides is the same. 114 00:06:55,370 --> 00:07:00,640 And therefore, the slope of the strings at all times, 115 00:07:00,640 --> 00:07:03,740 on the left it must equal the right. 116 00:07:03,740 --> 00:07:07,010 So this is the complete mathematical description. 117 00:07:07,010 --> 00:07:15,560 And we are asked to, with this information, 118 00:07:15,560 --> 00:07:19,570 figure out, what are the two lowest mode 119 00:07:19,570 --> 00:07:23,880 frequencies and what are their shapes? 120 00:07:23,880 --> 00:07:24,780 OK? 121 00:07:24,780 --> 00:07:26,640 So that's what now I will do. 122 00:07:32,160 --> 00:07:34,150 OK. 123 00:07:34,150 --> 00:07:38,510 Now, we know that what we're interested 124 00:07:38,510 --> 00:07:42,260 is in the normal mode. 125 00:07:42,260 --> 00:07:48,580 If you have a normal system oscillating in the normal mode, 126 00:07:48,580 --> 00:07:52,340 you know that the all parts oscillate 127 00:07:52,340 --> 00:07:57,070 with the same frequency and same phase. 128 00:07:57,070 --> 00:08:00,070 That's the meaning of being in the normal mode. 129 00:08:00,070 --> 00:08:01,250 OK? 130 00:08:01,250 --> 00:08:05,800 So we know that when this system is oscillating 131 00:08:05,800 --> 00:08:09,940 in the normal mode, both sides of the string 132 00:08:09,940 --> 00:08:13,685 will be oscillating with the same omega, same frequency 133 00:08:13,685 --> 00:08:14,805 and phase. 134 00:08:14,805 --> 00:08:15,380 All right? 135 00:08:18,140 --> 00:08:25,020 And also, on both sides we have a string, a row, 136 00:08:25,020 --> 00:08:28,270 a continuous row of oscillators, and we 137 00:08:28,270 --> 00:08:32,510 know what the normal mode oscillations of such a system 138 00:08:32,510 --> 00:08:33,059 is. 139 00:08:33,059 --> 00:08:35,270 We've seen it over and over again. 140 00:08:35,270 --> 00:08:38,360 It is sinusoidal in fall. 141 00:08:38,360 --> 00:08:43,770 So the general expression of the displacement of the string, 142 00:08:43,770 --> 00:08:46,760 whether it's on the left side or on the right side, 143 00:08:46,760 --> 00:08:51,140 will be the displacement of it from equilibrium 144 00:08:51,140 --> 00:08:58,250 is a sinusoidal function in position multiplied 145 00:08:58,250 --> 00:09:03,930 by a sinusoidal function in time. 146 00:09:03,930 --> 00:09:10,650 This function will be different for the left and the right, 147 00:09:10,650 --> 00:09:16,901 the spatial distribution. 148 00:09:16,901 --> 00:09:18,710 This part will be the same. 149 00:09:22,040 --> 00:09:25,320 Why would this be different for the two sides? 150 00:09:25,320 --> 00:09:33,290 Well, this equation satisfies the wave equation. 151 00:09:35,980 --> 00:09:37,180 It is the solution. 152 00:09:37,180 --> 00:09:38,760 I am looking for that wave equation. 153 00:09:42,370 --> 00:09:46,490 If this is to satisfy the wave equation, 154 00:09:46,490 --> 00:09:50,520 you'll find that omega/k, which you 155 00:09:50,520 --> 00:09:54,880 can write like lambda times f where lambda is the wavelength 156 00:09:54,880 --> 00:09:57,160 and f is the frequency-- this is the angular 157 00:09:57,160 --> 00:10:00,930 frequency on the wave number-- has 158 00:10:00,930 --> 00:10:03,440 to be equal to v. Otherwise, this 159 00:10:03,440 --> 00:10:06,470 would not satisfy the wave equation. 160 00:10:06,470 --> 00:10:07,240 OK? 161 00:10:07,240 --> 00:10:12,126 And v is different for the two strings. 162 00:10:12,126 --> 00:10:13,050 All right? 163 00:10:13,050 --> 00:10:13,890 We showed that. 164 00:10:13,890 --> 00:10:15,810 We discussed it a second ago. 165 00:10:15,810 --> 00:10:19,110 v, the phase velocity for the heavy string, 166 00:10:19,110 --> 00:10:23,670 is much smaller than for the light one. 167 00:10:23,670 --> 00:10:27,150 And therefore, if omega is the same, 168 00:10:27,150 --> 00:10:30,430 the k would be different in the two cases. 169 00:10:30,430 --> 00:10:32,340 OK. 170 00:10:32,340 --> 00:10:33,240 Fine. 171 00:10:33,240 --> 00:10:42,910 Now, using this, we know that the wavelength is given by v/f. 172 00:10:42,910 --> 00:10:45,420 Just I've copied it from there. 173 00:10:45,420 --> 00:10:51,420 And since the frequency on both sides is the same 174 00:10:51,420 --> 00:10:53,900 but v is different, I can immediately 175 00:10:53,900 --> 00:10:58,270 conclude that the wavelength of the part of the string which 176 00:10:58,270 --> 00:11:03,530 is massive, the heavy one, over the wavelength of the string 177 00:11:03,530 --> 00:11:08,755 on the light side is simply equal to vH over vL. 178 00:11:08,755 --> 00:11:18,660 But we've been told that the density of the heavy one 179 00:11:18,660 --> 00:11:21,250 is much greater than the light one. 180 00:11:21,250 --> 00:11:23,470 Therefore, the velocity, this one, 181 00:11:23,470 --> 00:11:26,100 is much smaller than this one. 182 00:11:26,100 --> 00:11:27,580 And therefore, we immediately can 183 00:11:27,580 --> 00:11:32,490 conclude that the wavelength in the normal mode 184 00:11:32,490 --> 00:11:35,250 oscillation of the heavy one is much 185 00:11:35,250 --> 00:11:39,220 shorter than of the light one. 186 00:11:39,220 --> 00:11:40,660 All right. 187 00:11:40,660 --> 00:11:46,120 So now if we were given more information 188 00:11:46,120 --> 00:11:49,780 exactly what these quantities are, et cetera, 189 00:11:49,780 --> 00:11:56,980 we could write the equations of the string on both sides 190 00:11:56,980 --> 00:11:58,170 and try to solve it. 191 00:11:58,170 --> 00:11:59,640 And in the next problem, I'll try 192 00:11:59,640 --> 00:12:01,580 to do an example which is exact. 193 00:12:01,580 --> 00:12:05,140 Here I'm treating the problem of it more qualitatively. 194 00:12:05,140 --> 00:12:09,730 And let's see what we can figure out just from this fact 195 00:12:09,730 --> 00:12:13,240 that the wavelength of the massive one 196 00:12:13,240 --> 00:12:16,900 is much smaller than of the light one. 197 00:12:16,900 --> 00:12:23,900 We are now looking for a solution to this problem 198 00:12:23,900 --> 00:12:28,680 where everything is moving with the same phase and frequency. 199 00:12:28,680 --> 00:12:30,490 All right? 200 00:12:30,490 --> 00:12:34,980 And we are looking for the lowest normal mode, 201 00:12:34,980 --> 00:12:37,730 in other words, when everything is moving 202 00:12:37,730 --> 00:12:42,662 as slowly as possible subject to the boundary conditions. 203 00:12:45,500 --> 00:12:50,360 Now, moving slowly means big wavelengths. 204 00:12:50,360 --> 00:12:53,150 So what we have to try to figure out-- here 205 00:12:53,150 --> 00:12:58,184 is a sketch of this string. 206 00:12:58,184 --> 00:12:59,100 This is the heavy one. 207 00:12:59,100 --> 00:13:01,440 This is the light one. 208 00:13:01,440 --> 00:13:08,150 And what we want, we want to find, on each side, 209 00:13:08,150 --> 00:13:10,670 this string is just like any old string. 210 00:13:10,670 --> 00:13:14,210 In the normal mode-- in a single mode, 211 00:13:14,210 --> 00:13:17,070 will have a sinusoidal function, all right, 212 00:13:17,070 --> 00:13:20,390 of a single wavelength on both sides. 213 00:13:20,390 --> 00:13:24,890 So this must be a part of a sine curve or a cosine curve. 214 00:13:24,890 --> 00:13:25,925 So must this. 215 00:13:29,180 --> 00:13:33,580 And we want to make it as long wavelength as possible, 216 00:13:33,580 --> 00:13:37,060 and we want to satisfy all the boundary conditions. 217 00:13:37,060 --> 00:13:41,360 So here it's located. 218 00:13:41,360 --> 00:13:43,570 Here this one is located. 219 00:13:43,570 --> 00:13:44,900 OK? 220 00:13:44,900 --> 00:13:49,220 The one on the right, you want to make as long a wavelength as 221 00:13:49,220 --> 00:13:56,610 possible, and so this is almost a straight line. 222 00:13:56,610 --> 00:13:58,710 This is part of a sine curve, but it's 223 00:13:58,710 --> 00:14:01,430 almost a straight line. 224 00:14:01,430 --> 00:14:06,050 This one is part of a sine curve. 225 00:14:06,050 --> 00:14:08,840 And at the boundary, we must have 226 00:14:08,840 --> 00:14:11,910 the two touch, so the string is not broken, 227 00:14:11,910 --> 00:14:13,980 and the slope to be the same. 228 00:14:13,980 --> 00:14:17,350 So the answer is obvious. 229 00:14:17,350 --> 00:14:21,270 The solution will be something like a sine that 230 00:14:21,270 --> 00:14:25,540 just turns around near the top here. 231 00:14:25,540 --> 00:14:29,100 And when the curvature of this sine curve 232 00:14:29,100 --> 00:14:32,690 is such that it points towards this point, 233 00:14:32,690 --> 00:14:36,750 then we have found the lowest frequency solution. 234 00:14:36,750 --> 00:14:40,000 All right? 235 00:14:40,000 --> 00:14:41,040 This is approximate. 236 00:14:41,040 --> 00:14:44,590 This will have very slight curvature but approximate. 237 00:14:44,590 --> 00:14:48,770 And so what we see, when the wavelength 238 00:14:48,770 --> 00:14:51,710 on the left-hand side is basically 239 00:14:51,710 --> 00:14:54,340 4 times this, which is 4L. 240 00:14:54,340 --> 00:14:57,530 So this will be the lowest normal mode 241 00:14:57,530 --> 00:15:00,550 that we figured out without solving any equations, 242 00:15:00,550 --> 00:15:05,140 from just our knowledge of the boundary conditions 243 00:15:05,140 --> 00:15:10,740 and what normal mode solutions are on the string like this. 244 00:15:10,740 --> 00:15:11,400 All right? 245 00:15:11,400 --> 00:15:14,920 Now, I can now, knowing the wavelength, 246 00:15:14,920 --> 00:15:21,350 I can calculate the frequency this corresponds to because we 247 00:15:21,350 --> 00:15:28,080 know that-- we've got it here-- omega is simply 248 00:15:28,080 --> 00:15:30,440 k times v. All right? 249 00:15:30,440 --> 00:15:34,810 k is 2 pi over the wavelength. 250 00:15:34,810 --> 00:15:36,220 So we have omega 1. 251 00:15:36,220 --> 00:15:38,980 This is the lowest normal mode. 252 00:15:38,980 --> 00:15:43,370 Angular frequency, 2 pi over lambda 1 times vH. 253 00:15:43,370 --> 00:15:45,520 We know what lambda 1 is. 254 00:15:45,520 --> 00:15:50,180 We just argued ourselves to show it's approximately 4L. 255 00:15:50,180 --> 00:15:54,030 It's actually a little less than 4L, but approximately 4L. 256 00:15:54,030 --> 00:15:58,420 And so this is the lowest normal mode frequency. 257 00:15:58,420 --> 00:15:59,010 All right. 258 00:15:59,010 --> 00:16:00,470 Now, what's the next normal? 259 00:16:00,470 --> 00:16:03,710 They ask us for the two lowest normal modes. 260 00:16:03,710 --> 00:16:06,010 We repeat the argument. 261 00:16:06,010 --> 00:16:11,930 We are now trying to figure out-- each of these strings 262 00:16:11,930 --> 00:16:19,390 now we would like to oscillate with slightly higher frequency. 263 00:16:19,390 --> 00:16:25,710 This, again, will be a sine curve on both sides. 264 00:16:25,710 --> 00:16:28,770 And again, we've got to find a situation where 265 00:16:28,770 --> 00:16:34,060 the curvature-- the sine has shorter wavelength-- 266 00:16:34,060 --> 00:16:38,610 but the minimum shorter that I can make it 267 00:16:38,610 --> 00:16:43,200 so that I satisfy the boundary conditions as before. 268 00:16:43,200 --> 00:16:46,820 The strings must not break, and the slope must be the same. 269 00:16:46,820 --> 00:16:48,950 Well, if you play around, you'll see 270 00:16:48,950 --> 00:16:53,150 that the next lowest normal mode will look like this. 271 00:16:53,150 --> 00:16:56,150 The string goes like this. 272 00:16:56,150 --> 00:16:57,010 All right? 273 00:16:57,010 --> 00:17:02,480 And here, the massive one is just curving. 274 00:17:02,480 --> 00:17:05,220 And when its curvature is such that it 275 00:17:05,220 --> 00:17:10,119 points towards this origin, where this is almost straight, 276 00:17:10,119 --> 00:17:15,240 that will be the sketch of the normal mode. 277 00:17:15,240 --> 00:17:16,640 So that's what it will look like. 278 00:17:16,640 --> 00:17:18,839 That will be the next lowest normal mode. 279 00:17:18,839 --> 00:17:20,190 Now, what is the wavelength? 280 00:17:20,190 --> 00:17:22,550 You can calculate that from this length. 281 00:17:22,550 --> 00:17:27,800 It's just a little less than 4/3 L. OK? 282 00:17:27,800 --> 00:17:33,870 And again, like before, since we know the vH and the wavelength, 283 00:17:33,870 --> 00:17:36,580 we can calculate the frequency. 284 00:17:36,580 --> 00:17:39,510 So these are the two normal mode frequencies. 285 00:17:39,510 --> 00:17:41,970 And why did I say it's counter-intuitive? 286 00:17:41,970 --> 00:17:44,240 For you, maybe it is intuitive. 287 00:17:44,240 --> 00:17:48,870 But I would have thought that if you connect a heavy string 288 00:17:48,870 --> 00:17:51,370 to a light one, it's the light one that 289 00:17:51,370 --> 00:17:55,090 would be wobbling up and down and not the heavy one. 290 00:17:55,090 --> 00:17:58,880 And in the reality, it's the opposite that happens. 291 00:17:58,880 --> 00:18:00,480 It's the little guy that seems to be 292 00:18:00,480 --> 00:18:02,670 deciding what the big guy is doing. 293 00:18:02,670 --> 00:18:05,540 And this is the one which is-- also 294 00:18:05,540 --> 00:18:10,240 this one-- all the time is nearly straight. 295 00:18:10,240 --> 00:18:13,960 The role of this one is simply to keep 296 00:18:13,960 --> 00:18:18,920 this end of the heavy string tied down, 297 00:18:18,920 --> 00:18:24,680 and it determines the boundary condition here. 298 00:18:24,680 --> 00:18:25,450 OK? 299 00:18:25,450 --> 00:18:29,580 But it's an interesting case to think about. 300 00:18:29,580 --> 00:18:30,140 OK. 301 00:18:30,140 --> 00:18:34,910 So that's the first problem I wanted to do. 302 00:18:34,910 --> 00:18:38,260 Now let's go to another one. 303 00:18:38,260 --> 00:18:42,700 In some ways, it's similar to the one we've done. 304 00:18:42,700 --> 00:18:48,500 But just in order to show you how general these discussions 305 00:18:48,500 --> 00:18:54,930 are of these problems, I'm taking a situation 306 00:18:54,930 --> 00:18:59,230 where the displacement from equilibrium 307 00:18:59,230 --> 00:19:04,970 is not transverse to the location of the oscillator. 308 00:19:04,970 --> 00:19:09,025 I'm going to consider an example of longitudinal waves. 309 00:19:15,710 --> 00:19:17,670 Let me discuss the problem here. 310 00:19:28,230 --> 00:19:31,390 This is the problem we're going to consider. 311 00:19:31,390 --> 00:19:37,520 We're going to consider a solid rod attached to a wall. 312 00:19:37,520 --> 00:19:42,280 So this is just a round rod. 313 00:19:45,600 --> 00:19:48,650 It is connected to another rod. 314 00:19:51,305 --> 00:19:55,850 Let's say take some case, maybe a tungsten, a lead alloy, 315 00:19:55,850 --> 00:19:59,200 a rod of tungsten lead, which has 316 00:19:59,200 --> 00:20:04,010 a certain Young's modulus y. 317 00:20:04,010 --> 00:20:06,860 And Young's modulus, let me remind you, 318 00:20:06,860 --> 00:20:09,730 it is the extension per unit length 319 00:20:09,730 --> 00:20:14,190 if you apply to it a force, certain force per unit area. 320 00:20:14,190 --> 00:20:16,220 So it's extension per unit length 321 00:20:16,220 --> 00:20:18,100 divided by force per unit length. 322 00:20:18,100 --> 00:20:24,020 It tells you how, under the system, 323 00:20:24,020 --> 00:20:28,400 what strain you get into the system. 324 00:20:28,400 --> 00:20:32,060 This is analogous to like, in a spring, 325 00:20:32,060 --> 00:20:35,960 we talk of the constant k for the spring. 326 00:20:35,960 --> 00:20:38,070 Tells you how much the spring [INAUDIBLE] 327 00:20:38,070 --> 00:20:39,510 when you apply a force. 328 00:20:39,510 --> 00:20:42,090 So the y will tell you how much this 329 00:20:42,090 --> 00:20:46,330 extends under a force per unit area. 330 00:20:46,330 --> 00:20:52,940 And I'm going to assume this rod has a density rho, mass density 331 00:20:52,940 --> 00:20:54,390 rho. 332 00:20:54,390 --> 00:20:57,620 It's connected to another, but this time much 333 00:20:57,620 --> 00:21:01,790 lighter rod, a rod which has three 334 00:21:01,790 --> 00:21:06,690 lengths, 3L length, same area. 335 00:21:06,690 --> 00:21:09,320 They are welded together here, connected together, 336 00:21:09,320 --> 00:21:10,640 glued together. 337 00:21:10,640 --> 00:21:13,210 OK? 338 00:21:13,210 --> 00:21:15,050 I've chosen two materials. 339 00:21:15,050 --> 00:21:16,970 And if you look in the books, you 340 00:21:16,970 --> 00:21:23,490 can find different materials having same y. 341 00:21:23,490 --> 00:21:25,950 This is a carbon fiber composite. 342 00:21:25,950 --> 00:21:29,630 This is a tungsten-lead alloy. 343 00:21:29,630 --> 00:21:32,920 And they both have the same Young's modulus. 344 00:21:32,920 --> 00:21:35,930 But this is much, much lighter. 345 00:21:35,930 --> 00:21:40,610 The density of this is 1/9 of the density of this material. 346 00:21:40,610 --> 00:21:44,890 And I chose these-- I wanted to come up with numbers where 347 00:21:44,890 --> 00:21:48,440 one can solve the problem completely. 348 00:21:48,440 --> 00:21:54,790 In the previous problem, I want to do this qualitatively 349 00:21:54,790 --> 00:21:58,330 so I didn't have to pick up numbers which gave you 350 00:21:58,330 --> 00:22:02,755 a final mathematical description which has an analytic solution. 351 00:22:07,360 --> 00:22:08,600 OK. 352 00:22:08,600 --> 00:22:12,490 So we have these two rods connected like this. 353 00:22:12,490 --> 00:22:14,225 This one is attached to a wall. 354 00:22:14,225 --> 00:22:16,830 It cannot move. 355 00:22:16,830 --> 00:22:19,306 And this is free. 356 00:22:19,306 --> 00:22:21,605 I purposely wanted a problem where 357 00:22:21,605 --> 00:22:23,530 there are new things coming in so 358 00:22:23,530 --> 00:22:25,960 that you can see how one tackles it. 359 00:22:25,960 --> 00:22:27,570 OK. 360 00:22:27,570 --> 00:22:29,180 As usual, there's some assumption. 361 00:22:29,180 --> 00:22:32,310 We're going to assume that this is a lossless system, 362 00:22:32,310 --> 00:22:37,480 while the rods are oscillating that they're not losing energy, 363 00:22:37,480 --> 00:22:40,640 that the oscillations are small so 364 00:22:40,640 --> 00:22:44,290 that we can do all the usual approximations. 365 00:22:44,290 --> 00:22:48,400 We'll make the assumption that this radius is small 366 00:22:48,400 --> 00:22:50,740 compared to this length so that we 367 00:22:50,740 --> 00:22:53,590 don't have to worry about transverse oscillations 368 00:22:53,590 --> 00:22:54,590 of the system. 369 00:22:54,590 --> 00:22:58,080 We're only talking about longitudinal oscillations. 370 00:22:58,080 --> 00:23:01,870 So we're talking about small oscillations 371 00:23:01,870 --> 00:23:05,310 in the longitudinal direction. 372 00:23:05,310 --> 00:23:08,860 And we are told that for such a system-- 373 00:23:08,860 --> 00:23:12,830 and you could derive it for yourselves-- 374 00:23:12,830 --> 00:23:17,270 that the phase velocity is the square root of Young's 375 00:23:17,270 --> 00:23:20,070 modulus divided by the density. 376 00:23:20,070 --> 00:23:22,500 So it's different for the two sides. 377 00:23:22,500 --> 00:23:23,240 OK? 378 00:23:23,240 --> 00:23:27,570 And the question is, can we calculate 379 00:23:27,570 --> 00:23:30,270 for this-- knowing all these quantities, 380 00:23:30,270 --> 00:23:32,950 these are now all given quantities-- 381 00:23:32,950 --> 00:23:37,030 can we calculate what is the angular frequency omega 382 00:23:37,030 --> 00:23:43,200 1 of the first normal mode as this oscillates? 383 00:23:43,200 --> 00:23:46,610 The problem, in some ways, is similar to the previous one. 384 00:23:46,610 --> 00:23:51,160 But as I say, normally we more often do things 385 00:23:51,160 --> 00:23:55,560 like strings oscillating because it's easier 386 00:23:55,560 --> 00:24:03,740 to plot on the board the displacement of the string, 387 00:24:03,740 --> 00:24:06,750 which is in the y direction, and the position 388 00:24:06,750 --> 00:24:09,540 in the x direction. 389 00:24:09,540 --> 00:24:13,030 Here we have the displacement from equilibrium 390 00:24:13,030 --> 00:24:17,000 at any point in the same direction, 391 00:24:17,000 --> 00:24:21,160 in the same as the position, and that makes it harder to plot. 392 00:24:21,160 --> 00:24:24,240 That's why one normally doesn't take this example. 393 00:24:27,170 --> 00:24:28,990 We tend to take examples which are 394 00:24:28,990 --> 00:24:33,290 easier to illustrate on the board. 395 00:24:33,290 --> 00:24:34,140 OK. 396 00:24:34,140 --> 00:24:39,560 So let's now, as usual-- this is the physical situation-- 397 00:24:39,560 --> 00:24:43,170 what does it correspond to as a description in terms 398 00:24:43,170 --> 00:24:44,980 of mathematics? 399 00:24:44,980 --> 00:24:50,020 Well, let's define, at any point x and time 400 00:24:50,020 --> 00:24:54,320 t, the displacement of the material 401 00:24:54,320 --> 00:24:58,450 of the rod from the position of equilibrium 402 00:24:58,450 --> 00:25:00,610 by the Greek letter psi. 403 00:25:00,610 --> 00:25:06,120 So it's psi at x of t 404 00:25:06,120 --> 00:25:13,150 Now, this system is, again, an ideal system 405 00:25:13,150 --> 00:25:19,930 of a continuous distribution of harmonic oscillators. 406 00:25:19,930 --> 00:25:23,630 Each piece of mass oscillates backwards and forwards 407 00:25:23,630 --> 00:25:24,620 longitudinally. 408 00:25:24,620 --> 00:25:26,290 That's the oscillator. 409 00:25:26,290 --> 00:25:31,090 It's continuous, that is, an oscillator at every x. 410 00:25:31,090 --> 00:25:31,590 OK? 411 00:25:34,420 --> 00:25:38,060 And each oscillator is coupled to its neighbors. 412 00:25:38,060 --> 00:25:43,360 So this is mathematically exactly the same situation 413 00:25:43,360 --> 00:25:46,600 as a string, where the oscillators are there 414 00:25:46,600 --> 00:25:50,850 oscillating up and down instead of longitudinally. 415 00:25:50,850 --> 00:25:54,470 So the equation of motion of this system 416 00:25:54,470 --> 00:25:56,348 will be, as always, the wave equation. 417 00:25:58,976 --> 00:26:01,970 This is the longitudinal displacement, 418 00:26:01,970 --> 00:26:03,190 how it's oscillating. 419 00:26:03,190 --> 00:26:06,150 v is the phase velocity. 420 00:26:09,040 --> 00:26:14,500 And we know that the phase velocity 421 00:26:14,500 --> 00:26:18,560 is the square root of Young's modulus over density. 422 00:26:18,560 --> 00:26:21,590 Since we know the ratio of the densities, 423 00:26:21,590 --> 00:26:25,520 and the y is the same for both materials, 424 00:26:25,520 --> 00:26:29,560 you know that the ratio of the phase velocity in the two 425 00:26:29,560 --> 00:26:34,710 materials, v2 over v1, will be 3, the square root of 9. 426 00:26:34,710 --> 00:26:36,320 OK? 427 00:26:36,320 --> 00:26:43,420 And as always, we know that omega/k is the phase velocity. 428 00:26:43,420 --> 00:26:48,560 Otherwise, it wouldn't satisfy this equation. 429 00:26:48,560 --> 00:26:53,720 So we know that the k2/k1-- and I'm reminding you, 430 00:26:53,720 --> 00:26:56,380 k is 2 pi over the wavelength, right? 431 00:26:56,380 --> 00:26:57,910 So k2/k1 is 1/3. 432 00:27:00,820 --> 00:27:01,320 OK. 433 00:27:01,320 --> 00:27:05,630 So this is the equations of motion and the constants of it. 434 00:27:05,630 --> 00:27:08,020 What are the boundary conditions? 435 00:27:08,020 --> 00:27:10,410 Well, they're slightly different to the one 436 00:27:10,410 --> 00:27:13,840 we've had of strings attached at both ends. 437 00:27:17,830 --> 00:27:23,140 We know that, in the middle, the two rods 438 00:27:23,140 --> 00:27:25,840 are attached to each other. 439 00:27:25,840 --> 00:27:28,790 So the displacement from equilibrium, 440 00:27:28,790 --> 00:27:31,860 whether you're on the left rod or on the right rod, 441 00:27:31,860 --> 00:27:33,262 will be the same. 442 00:27:33,262 --> 00:27:37,590 So the one on the left, which I call 443 00:27:37,590 --> 00:27:40,880 psi 1, at position L [INAUDIBLE] t, 444 00:27:40,880 --> 00:27:43,120 will be equal to psi 2 at L of t. 445 00:27:43,120 --> 00:27:45,690 So that's one boundary condition. 446 00:27:45,690 --> 00:27:48,460 Another boundary condition is the one 447 00:27:48,460 --> 00:27:51,530 analogous to when you have two strings joined, 448 00:27:51,530 --> 00:27:55,350 that the slope of the strings must be the same on both sides 449 00:27:55,350 --> 00:27:57,360 because the junction has no mass. 450 00:27:57,360 --> 00:28:00,540 Similarly here, the junction between the two rods has no 451 00:28:00,540 --> 00:28:04,920 mass, so there cannot be a net force on that junction. 452 00:28:04,920 --> 00:28:10,720 And since the y, the Young's modulus, is the same, 453 00:28:10,720 --> 00:28:14,070 the tension is the same inside, then 454 00:28:14,070 --> 00:28:21,940 the slope of d psi 1 dx at that junction 455 00:28:21,940 --> 00:28:24,920 must equal to d psi dx at the ther. 456 00:28:24,920 --> 00:28:29,730 As I repeat, this is analogous to a junction between two 457 00:28:29,730 --> 00:28:32,960 strings when you're talking about transverse oscillations 458 00:28:32,960 --> 00:28:33,800 of strings. 459 00:28:33,800 --> 00:28:38,670 The slope at the junction must be the same on both sides, 460 00:28:38,670 --> 00:28:42,990 unless there is a massive object at that junction. 461 00:28:42,990 --> 00:28:44,360 And we don't have such a thing. 462 00:28:44,360 --> 00:28:47,715 If there was a heavy bead or something at that junction, 463 00:28:47,715 --> 00:28:50,250 this would no longer be true. 464 00:28:50,250 --> 00:28:52,600 The other thing is, what are the boundary conditions 465 00:28:52,600 --> 00:28:54,360 of the ends? 466 00:28:54,360 --> 00:28:56,690 On the left end, it's easier. 467 00:28:56,690 --> 00:28:59,890 We said the left end is attached to the wall, 468 00:28:59,890 --> 00:29:02,080 therefore it cannot move. 469 00:29:02,080 --> 00:29:02,780 OK? 470 00:29:02,780 --> 00:29:08,120 Therefore, the displacement from equilibrium of this left end 471 00:29:08,120 --> 00:29:11,610 is equal to 0 at all times. 472 00:29:11,610 --> 00:29:13,930 The right end is different. 473 00:29:13,930 --> 00:29:17,470 It's free, and so it's certainly not 0. 474 00:29:17,470 --> 00:29:19,180 What do we know about that? 475 00:29:19,180 --> 00:29:25,390 Well, at the end, the rod ends, and there 476 00:29:25,390 --> 00:29:27,430 is no mass or anything there. 477 00:29:27,430 --> 00:29:31,710 So the end of the rod is not pushing anything. 478 00:29:31,710 --> 00:29:36,680 There's effective, it's pushing a 0 mass at the end of it. 479 00:29:36,680 --> 00:29:42,100 Therefore, there cannot be a net force on that. 480 00:29:42,100 --> 00:29:47,070 And so the rod cannot be compressed at the end 481 00:29:47,070 --> 00:29:51,590 or elongated because then it would be exerting like a spring 482 00:29:51,590 --> 00:29:55,790 attached to nothing with exerted force on it, 483 00:29:55,790 --> 00:29:57,270 which is not possible. 484 00:29:57,270 --> 00:30:02,410 And so the rate of change of psi 1 with x 485 00:30:02,410 --> 00:30:04,345 of the displacement from equilibrium 486 00:30:04,345 --> 00:30:11,950 from x at the-- sorry, I'm here, I'm here. 487 00:30:11,950 --> 00:30:19,980 The slope of psi 2 with respect to x at the end, which is-- I 488 00:30:19,980 --> 00:30:20,895 am sorry. 489 00:30:20,895 --> 00:30:24,240 This will be 4L. 490 00:30:24,240 --> 00:30:25,190 This is my mistake. 491 00:30:25,190 --> 00:30:26,260 This is 4L here. 492 00:30:30,850 --> 00:30:31,570 OK. 493 00:30:31,570 --> 00:30:36,290 At the end of the rod, it has to be 0. 494 00:30:36,290 --> 00:30:36,940 OK. 495 00:30:36,940 --> 00:30:39,950 So this is now the mathematical description. 496 00:30:39,950 --> 00:30:45,840 And all we have to do is solve the equation of motion subject 497 00:30:45,840 --> 00:30:49,150 to these boundary conditions and find out 498 00:30:49,150 --> 00:30:54,480 what is the lowest frequency that satisfies the boundary 499 00:30:54,480 --> 00:30:58,280 conditions and the wave equation. 500 00:30:58,280 --> 00:30:59,110 OK? 501 00:30:59,110 --> 00:31:05,430 And again, I have written out the solution for you. 502 00:31:05,430 --> 00:31:07,610 And let me go a little faster now. 503 00:31:10,500 --> 00:31:12,760 We now have experience. 504 00:31:12,760 --> 00:31:18,420 We know that the wave equation has standing wave solutions, 505 00:31:18,420 --> 00:31:24,510 which are some amplitude times a sinusoidal function in x 506 00:31:24,510 --> 00:31:28,980 and a sinusoidal function of time. 507 00:31:28,980 --> 00:31:37,290 And now to save time, I am going to inspect immediately 508 00:31:37,290 --> 00:31:40,610 the boundary conditions, and using, 509 00:31:40,610 --> 00:31:44,010 in particular, these boundary conditions, 510 00:31:44,010 --> 00:31:49,640 the ones at x equals 0 and 4L for all time. 511 00:31:49,640 --> 00:31:56,100 And I'm going to figure out sinusoidal functions of x 512 00:31:56,100 --> 00:32:03,250 and t, which satisfy these boundary conditions. 513 00:32:03,250 --> 00:32:06,380 Now, we know that we are talking about standing waves. 514 00:32:06,380 --> 00:32:16,360 So for both rods, the angular frequency, 515 00:32:16,360 --> 00:32:20,120 the [INAUDIBLE] oscillations in time, will be the same. 516 00:32:20,120 --> 00:32:23,550 So this part will be the same. 517 00:32:23,550 --> 00:32:27,965 This part must satisfy the boundary condition 518 00:32:27,965 --> 00:32:30,880 at the extreme left and right, which 519 00:32:30,880 --> 00:32:36,840 forces me to write here a sine k and here cosine 520 00:32:36,840 --> 00:32:41,790 k2 of what happens at the boundaries. 521 00:32:41,790 --> 00:32:42,290 OK? 522 00:32:51,470 --> 00:32:54,660 A good exercise for you is slowly 523 00:32:54,660 --> 00:32:57,410 think through the boundary conditions 524 00:32:57,410 --> 00:33:01,530 and the general solutions, the normal mode solutions, 525 00:33:01,530 --> 00:33:02,890 of the wave equation. 526 00:33:02,890 --> 00:33:04,350 And you'll come to the conclusion 527 00:33:04,350 --> 00:33:14,010 that this describes the standing waves in the two rods. 528 00:33:16,670 --> 00:33:17,640 OK. 529 00:33:17,640 --> 00:33:27,130 We know further that omega is equal to k1 v1 in the left rod, 530 00:33:27,130 --> 00:33:30,230 and equal to k2 v2 in the right rod. 531 00:33:30,230 --> 00:33:33,160 This has to be the same in both, as I said, 532 00:33:33,160 --> 00:33:35,860 because we are in the normal mode. 533 00:33:35,860 --> 00:33:39,610 We're in a normal mode, where the whole system is oscillating 534 00:33:39,610 --> 00:33:42,430 with the same frequency and phase. 535 00:33:42,430 --> 00:33:42,930 OK. 536 00:33:42,930 --> 00:33:48,320 Now, we will use the other bit of information, the boundary 537 00:33:48,320 --> 00:33:54,280 conditions at the junction, which I've written here. 538 00:33:54,280 --> 00:34:07,810 And from that, try to find a solution that satisfies these 539 00:34:07,810 --> 00:34:11,489 and see what is the angular frequency of oscillation 540 00:34:11,489 --> 00:34:13,719 for that. 541 00:34:13,719 --> 00:34:31,179 So we can now write at x equals L-- 542 00:34:31,179 --> 00:34:34,159 these are the two equations. 543 00:34:34,159 --> 00:34:38,010 We'll take the displacement of the rod on the left 544 00:34:38,010 --> 00:34:41,179 and make it equal to the right and the slope on the left 545 00:34:41,179 --> 00:34:43,020 equal to the right. 546 00:34:43,020 --> 00:34:44,920 On the left, this is the equation. 547 00:34:44,920 --> 00:34:46,639 On the right is that. 548 00:34:46,639 --> 00:34:50,139 At that boundary to make the two equal, 549 00:34:50,139 --> 00:34:54,440 I get that A1 sine k1 L at x equals 550 00:34:54,440 --> 00:35:01,230 L is equal to A2 cosine times this, all right, at x. 551 00:35:01,230 --> 00:35:02,800 When we're talking about x2 equals 552 00:35:02,800 --> 00:35:06,750 L, L minus 4L is minus 3, et cetera. 553 00:35:06,750 --> 00:35:07,990 OK? 554 00:35:07,990 --> 00:35:13,580 The cosine terms cancel because the time part 555 00:35:13,580 --> 00:35:16,410 cancels because they're the same on both sides. 556 00:35:16,410 --> 00:35:19,010 So this is the equation we have to satisfy 557 00:35:19,010 --> 00:35:20,680 so that the string is not broken. 558 00:35:20,680 --> 00:35:23,160 And I'm just writing it like this 559 00:35:23,160 --> 00:35:29,110 because we know that-- we did that earlier-- that k2/k1 560 00:35:29,110 --> 00:35:30,340 is 1/3. 561 00:35:30,340 --> 00:35:34,640 And the cosine of a plus and the minus angle is the same, 562 00:35:34,640 --> 00:35:38,890 so I've just replaced that with plus k1 L. 563 00:35:38,890 --> 00:35:46,860 This is now looking at the slopes 564 00:35:46,860 --> 00:35:52,355 of the displacement psi at that boundary. 565 00:35:52,355 --> 00:35:55,350 So I differentiate these with respect 566 00:35:55,350 --> 00:35:58,540 to x, all right, this one and that one, and equate them. 567 00:35:58,540 --> 00:36:00,930 And I get this equation, all right? 568 00:36:00,930 --> 00:36:05,110 And so in order to satisfy the boundary conditions 569 00:36:05,110 --> 00:36:10,330 and the wave equation, I end up with these two equations. 570 00:36:10,330 --> 00:36:13,260 I've chosen things such that these 571 00:36:13,260 --> 00:36:16,330 can be solved analytically. 572 00:36:16,330 --> 00:36:17,240 It's not hard. 573 00:36:17,240 --> 00:36:18,620 You can do it for yourself. 574 00:36:18,620 --> 00:36:21,840 You have two equations, two unknowns. 575 00:36:21,840 --> 00:36:25,240 The unknown is the ratio of A2 to A1. 576 00:36:25,240 --> 00:36:26,440 That's one unknown. 577 00:36:26,440 --> 00:36:28,790 And the other unknown is k1. 578 00:36:28,790 --> 00:36:31,780 You can solve these equations trigonometrically 579 00:36:31,780 --> 00:36:34,720 only because I've chosen numbers of rho, 580 00:36:34,720 --> 00:36:37,830 et cetera, which make the answer come out. 581 00:36:37,830 --> 00:36:42,940 And you end up that you get k1 from it. 582 00:36:42,940 --> 00:36:46,130 Knowing k1, you can calculate omega. 583 00:36:46,130 --> 00:36:48,080 And this is the answer you will get. 584 00:36:48,080 --> 00:36:50,680 I won't waste your time here. 585 00:36:50,680 --> 00:36:54,405 You can do this algebra for yourselves. 586 00:36:58,580 --> 00:37:02,730 So here I've got the final answer 587 00:37:02,730 --> 00:37:04,490 because I had all the input. 588 00:37:04,490 --> 00:37:09,490 If I had tried to do the same with the string one, 589 00:37:09,490 --> 00:37:13,350 if I chose some random ratio of the density of the string 590 00:37:13,350 --> 00:37:17,350 on both sides, everything else would have been similar. 591 00:37:17,350 --> 00:37:21,750 But here I would have ended up with a transcendental equation, 592 00:37:21,750 --> 00:37:25,180 in general, which I cannot solve analytically. 593 00:37:25,180 --> 00:37:29,170 And so if you take any random case for yourself 594 00:37:29,170 --> 00:37:33,320 and try to solve it, you would find that, at this point, 595 00:37:33,320 --> 00:37:34,487 you would be stuck. 596 00:37:34,487 --> 00:37:36,570 You would be stuck with a transcendental equation, 597 00:37:36,570 --> 00:37:39,160 which, of course, you could solve numerically 598 00:37:39,160 --> 00:37:40,990 over the computer, et cetera. 599 00:37:40,990 --> 00:37:41,500 OK. 600 00:37:41,500 --> 00:37:42,280 So that's the end. 601 00:37:42,280 --> 00:37:46,290 And now I'll come to the third, yet another example 602 00:37:46,290 --> 00:37:49,760 of standing waves, this time in three dimensions. 603 00:37:49,760 --> 00:37:54,260 So now we come to the third standing wave problem. 604 00:37:54,260 --> 00:37:57,127 And just for variety, I decided to take one 605 00:37:57,127 --> 00:37:57,960 in three dimensions. 606 00:38:00,474 --> 00:38:02,820 So the problem is the following. 607 00:38:02,820 --> 00:38:06,900 Suppose you have a room, which is 2 meters in size 608 00:38:06,900 --> 00:38:08,290 by 3 meters by 4. 609 00:38:11,710 --> 00:38:17,990 What is the lowest standing wave frequency 610 00:38:17,990 --> 00:38:20,170 that you can get in such a room? 611 00:38:20,170 --> 00:38:21,870 OK? 612 00:38:21,870 --> 00:38:29,360 So it's basically a problem to do with pressure waves, 613 00:38:29,360 --> 00:38:33,200 because that's what sound is, standing pressure 614 00:38:33,200 --> 00:38:36,610 waves in three dimensions. 615 00:38:36,610 --> 00:38:40,250 We are told that, in this particular room, 616 00:38:40,250 --> 00:38:43,440 what the density of the air is. 617 00:38:43,440 --> 00:38:47,410 We're also told what the bulk modulus, in other words, 618 00:38:47,410 --> 00:38:50,250 how compressed the air can get and what's 619 00:38:50,250 --> 00:38:55,630 the resultant pressure-- what kind of compression 620 00:38:55,630 --> 00:38:57,880 gives rise to air. 621 00:38:57,880 --> 00:39:01,340 And we are asked to calculate lowest normal mode 622 00:39:01,340 --> 00:39:03,770 frequency of this system. 623 00:39:03,770 --> 00:39:04,390 OK. 624 00:39:04,390 --> 00:39:05,430 So how do we do this? 625 00:39:08,290 --> 00:39:10,315 Not surprisingly, you'll remember already 626 00:39:10,315 --> 00:39:14,980 when we were talking about one-dimensional system, way 627 00:39:14,980 --> 00:39:20,510 the standing waves of strings or on a rod, et cetera, 628 00:39:20,510 --> 00:39:22,670 it's quite complex. 629 00:39:22,670 --> 00:39:24,870 If you have three dimensions, the situation 630 00:39:24,870 --> 00:39:26,660 is getting more and more complicated. 631 00:39:29,410 --> 00:39:32,260 In this course, we don't actually 632 00:39:32,260 --> 00:39:37,560 derive the three-dimensional wave equation. 633 00:39:37,560 --> 00:39:42,730 But by analogy with the one-dimensional one 634 00:39:42,730 --> 00:39:47,930 and two-dimensional one, we sort of indicate 635 00:39:47,930 --> 00:39:50,290 what form you would expect this to be. 636 00:39:50,290 --> 00:39:52,360 And it actually is the case. 637 00:39:52,360 --> 00:39:56,310 The three-dimensional wave equation 638 00:39:56,310 --> 00:39:59,730 for a scalar quantity like pressure 639 00:39:59,730 --> 00:40:04,150 that has no direction, all right, is written here. 640 00:40:04,150 --> 00:40:09,590 It's the d2p dt squared, like for the one dimension, 641 00:40:09,590 --> 00:40:13,650 is equal to v squared times the gradient squared 642 00:40:13,650 --> 00:40:15,970 of the pressure, which, if I write it 643 00:40:15,970 --> 00:40:21,770 in Cartesian coordinates, it's written here. 644 00:40:21,770 --> 00:40:27,140 Clearly, if I take a three-dimensional system 645 00:40:27,140 --> 00:40:31,080 but reduced two dimensions with some small distances 646 00:40:31,080 --> 00:40:34,450 so that, in essence, it's a one-dimensional problem, 647 00:40:34,450 --> 00:40:36,770 this gives you the wave equation you would expect. 648 00:40:36,770 --> 00:40:39,700 If I remove two of these terms, what I 649 00:40:39,700 --> 00:40:45,160 have is the wave equation in one dimension. 650 00:40:45,160 --> 00:40:46,010 OK. 651 00:40:46,010 --> 00:40:53,165 In this equation, p is the excess pressure 652 00:40:53,165 --> 00:40:58,420 of the air above its equilibrium Imagine the room. 653 00:40:58,420 --> 00:40:59,960 Everything is at equilibrium. 654 00:40:59,960 --> 00:41:01,260 The air is not moving. 655 00:41:01,260 --> 00:41:03,470 The pressure is not changing, et cetera. 656 00:41:03,470 --> 00:41:05,940 There will be some pressure. 657 00:41:05,940 --> 00:41:08,740 What we are interested in is what 658 00:41:08,740 --> 00:41:13,290 happens if we displace that air from equilibrium, 659 00:41:13,290 --> 00:41:18,650 change its pressure, and let go, what will happen? 660 00:41:18,650 --> 00:41:21,390 And so it's that excess pressure we are interested. 661 00:41:21,390 --> 00:41:23,050 That's the p. 662 00:41:23,050 --> 00:41:27,600 And we are interested to see if there is a way 663 00:41:27,600 --> 00:41:31,070 to do that such that the air in the room 664 00:41:31,070 --> 00:41:34,080 everywhere, the pressure, oscillates 665 00:41:34,080 --> 00:41:36,750 with the same frequency and phase. 666 00:41:36,750 --> 00:41:41,260 If it does, then in that room the air 667 00:41:41,260 --> 00:41:45,530 will be in its normal mode, oscillating in its normal mode. 668 00:41:48,130 --> 00:41:51,850 One can show when one derives this-- and I did not derive it, 669 00:41:51,850 --> 00:41:55,890 of course-- is that this v is the phase 670 00:41:55,890 --> 00:42:00,610 velocity of propagation of pressure waves in air, 671 00:42:00,610 --> 00:42:06,790 the sound velocity, which is the square root of the bulk modulus 672 00:42:06,790 --> 00:42:08,892 divided by the density. 673 00:42:08,892 --> 00:42:10,860 OK? 674 00:42:10,860 --> 00:42:11,520 OK. 675 00:42:11,520 --> 00:42:15,540 So the translation of this problem 676 00:42:15,540 --> 00:42:19,565 is this is the equation of motion of the system. 677 00:42:23,540 --> 00:42:26,400 And what are the boundary conditions? 678 00:42:26,400 --> 00:42:32,510 The boundary conditions are that at the edges of the room, 679 00:42:32,510 --> 00:42:37,940 at the walls, the pressure does not 680 00:42:37,940 --> 00:42:43,050 change perpendicular to the wall. 681 00:42:43,050 --> 00:42:50,240 So for example, the two walls, at the end in the x direction, 682 00:42:50,240 --> 00:42:57,640 at the end there, at one side of the wall, the rate of change 683 00:42:57,640 --> 00:43:01,570 of pressure with position and at the other, it will not change. 684 00:43:01,570 --> 00:43:05,380 Top and bottom, perpendicular, it will not change this way. 685 00:43:05,380 --> 00:43:06,455 It will not change. 686 00:43:09,700 --> 00:43:11,865 Why is this the boundary condition? 687 00:43:16,020 --> 00:43:20,580 The reason for that is that if you imagine at the wall, 688 00:43:20,580 --> 00:43:26,885 the molecules cannot move, right, because of the wall. 689 00:43:26,885 --> 00:43:30,490 So therefore, you cannot there, near the end, 690 00:43:30,490 --> 00:43:36,870 increase the number of molecules. 691 00:43:36,870 --> 00:43:41,290 So perpendicular to it, that the number 692 00:43:41,290 --> 00:43:45,530 will be independent of the position, which is dp dx, 693 00:43:45,530 --> 00:43:47,840 is equal to 0. 694 00:43:47,840 --> 00:43:48,880 OK? 695 00:43:48,880 --> 00:43:57,330 And this is analogous to the string situation, which 696 00:43:57,330 --> 00:44:00,420 is attached to a massless ring where 697 00:44:00,420 --> 00:44:08,640 the slope is constant at the boundary. 698 00:44:08,640 --> 00:44:09,300 OK. 699 00:44:09,300 --> 00:44:12,670 So with these boundary conditions, 700 00:44:12,670 --> 00:44:16,186 what are the solutions of this equation? 701 00:44:16,186 --> 00:44:16,685 OK? 702 00:44:24,180 --> 00:44:27,080 To find the most general solution of this equation, 703 00:44:27,080 --> 00:44:29,440 it's obviously very, very complicated. 704 00:44:29,440 --> 00:44:31,345 There are infinite possibilities. 705 00:44:34,080 --> 00:44:40,910 But from my experience of what happens in one dimension, 706 00:44:40,910 --> 00:44:44,130 we can do a pretty good, intelligent guess. 707 00:44:44,130 --> 00:44:46,740 And what I've written here, we can 708 00:44:46,740 --> 00:44:50,810 guess that the solution will look like that. 709 00:44:50,810 --> 00:44:52,840 Why is it like this? 710 00:44:52,840 --> 00:44:56,460 Well, first of all, we want a solution 711 00:44:56,460 --> 00:45:02,910 of this equation, which is a normal mode, meaning everything 712 00:45:02,910 --> 00:45:06,790 oscillating with the same frequency and phase. 713 00:45:06,790 --> 00:45:10,500 And so independent of where you are, 714 00:45:10,500 --> 00:45:14,060 the time dependence must be the same for everything 715 00:45:14,060 --> 00:45:16,990 or else it would not be a normal mode. 716 00:45:16,990 --> 00:45:19,280 So that's why that. 717 00:45:19,280 --> 00:45:23,380 There is some amplitude, so that's p maximum. 718 00:45:23,380 --> 00:45:28,630 You would expect symmetry between the three directions. 719 00:45:28,630 --> 00:45:30,980 There's nothing special between them. 720 00:45:30,980 --> 00:45:35,250 And you want some function here which 721 00:45:35,250 --> 00:45:39,000 satisfies the boundary conditions. 722 00:45:39,000 --> 00:45:46,430 And what you will find is that this, a cosine kx, 723 00:45:46,430 --> 00:45:49,630 the slope of this with respect to x gives you a sine. 724 00:45:49,630 --> 00:45:55,120 And at x equals 0, you will get the boundary condition 725 00:45:55,120 --> 00:45:57,750 satisfied. 726 00:45:57,750 --> 00:46:01,140 And we'll see in a second what constraint 727 00:46:01,140 --> 00:46:06,830 the boundary puts on this at the other end. 728 00:46:06,830 --> 00:46:11,990 For this to be a solution of our wave equation, 729 00:46:11,990 --> 00:46:18,790 we must satisfy this and that. 730 00:46:18,790 --> 00:46:21,920 And so in some ways, you could say, look, I've guessed it. 731 00:46:21,920 --> 00:46:25,100 But then I'll use the uniqueness there, say, OK, I've guessed it 732 00:46:25,100 --> 00:46:26,440 or I knew it. 733 00:46:26,440 --> 00:46:28,510 I guessed this was the solution. 734 00:46:28,510 --> 00:46:33,850 I will now check, does it satisfy everything I know, 735 00:46:33,850 --> 00:46:36,660 the wave equation and the boundary conditions? 736 00:46:36,660 --> 00:46:38,922 And the answer is yes, it does. 737 00:46:38,922 --> 00:46:40,380 And then I use the uniqueness there 738 00:46:40,380 --> 00:46:44,500 that says therefore that is the only solution that satisfies 739 00:46:44,500 --> 00:46:47,170 everything I know about the situation. 740 00:46:47,170 --> 00:46:48,300 OK. 741 00:46:48,300 --> 00:47:00,200 Having done that, to get this I used the boundary condition 742 00:47:00,200 --> 00:47:02,220 as x equals 0. 743 00:47:02,220 --> 00:47:09,110 But I want this slope also to be 0 at the edges of the room. 744 00:47:09,110 --> 00:47:15,830 And that gives me a constraint on what each one of these cases 745 00:47:15,830 --> 00:47:16,800 can be. 746 00:47:16,800 --> 00:47:29,395 And you find that kx has to be some constant pi over Lx, 747 00:47:29,395 --> 00:47:36,060 where Lx is the length of the room in the x direction, 748 00:47:36,060 --> 00:47:43,330 so that the slope of this at the edge of the room is 0. 749 00:47:43,330 --> 00:47:46,800 Similarly for ky and kz. 750 00:47:46,800 --> 00:47:53,550 And so you find that these L's, m's, and n's, have to be 751 00:47:53,550 --> 00:47:56,120 0, 1, 2, or 3. 752 00:47:56,120 --> 00:47:58,770 They cannot all be 0. 753 00:47:58,770 --> 00:48:01,980 If they are all 0, then nothing goes on. 754 00:48:01,980 --> 00:48:08,826 This means that k is 0, which makes omega 0. 755 00:48:08,826 --> 00:48:12,230 It's a trivial, uninteresting solution to our wave equation. 756 00:48:12,230 --> 00:48:14,995 So at least one of these has to be non-zero. 757 00:48:17,780 --> 00:48:23,190 So which one will give us the lowest frequency? 758 00:48:23,190 --> 00:48:24,580 All right? 759 00:48:24,580 --> 00:48:33,670 And so it will be the one where the k, all right, 760 00:48:33,670 --> 00:48:43,630 is as small as possible, which means that the wavelength is 761 00:48:43,630 --> 00:48:46,060 as long as possible. 762 00:48:46,060 --> 00:48:47,510 OK? 763 00:48:47,510 --> 00:48:56,750 So if we make the L's and m, which 764 00:48:56,750 --> 00:49:02,030 correspond to the room, the 2 meter times 3 meters, 765 00:49:02,030 --> 00:49:06,520 those to be 0, the only direction we accept 766 00:49:06,520 --> 00:49:11,840 is in the direction of z, which is the longest 767 00:49:11,840 --> 00:49:16,980 dimension of the room, then that will give us 768 00:49:16,980 --> 00:49:22,686 the longest wavelength and therefore lowest frequency. 769 00:49:22,686 --> 00:49:23,185 OK? 770 00:49:34,280 --> 00:49:44,310 So k, the largest value of the wavelength, 771 00:49:44,310 --> 00:49:50,510 will be when kz is pi over 4m, in other words, the wavelength 772 00:49:50,510 --> 00:49:55,420 in that direction being 8 meters. 773 00:49:55,420 --> 00:50:02,790 And since we know that v is the square root of the bulk modulus 774 00:50:02,790 --> 00:50:05,840 divided by the density, I could take the numbers that 775 00:50:05,840 --> 00:50:09,535 are given, I get that the phase velocity, or the velocity 776 00:50:09,535 --> 00:50:13,940 of sound in this room, is 342 meters. 777 00:50:13,940 --> 00:50:17,160 The frequency, of course, is equal to the phase velocity 778 00:50:17,160 --> 00:50:17,685 over lambda. 779 00:50:17,685 --> 00:50:20,380 Or the way I normally remember, it's frequency times 780 00:50:20,380 --> 00:50:22,180 lambda is the phase velocity. 781 00:50:22,180 --> 00:50:23,660 All right? 782 00:50:23,660 --> 00:50:29,390 And I found the longest wavelength is 8 meters. 783 00:50:29,390 --> 00:50:35,190 So from that, I get that the lowest normal mode 784 00:50:35,190 --> 00:50:40,100 will oscillate at almost 43 hertz. 785 00:50:40,100 --> 00:50:45,560 I'll just sketch for you here what we've essentially-- we've 786 00:50:45,560 --> 00:50:55,370 got this room, and we are looking 787 00:50:55,370 --> 00:51:00,270 for solutions which have the longest wavelength. 788 00:51:00,270 --> 00:51:06,950 See, in principle, one could have 789 00:51:06,950 --> 00:51:12,650 a wave like this in this direction. 790 00:51:12,650 --> 00:51:15,930 This, the slope, has to be perpendicular 791 00:51:15,930 --> 00:51:17,640 at the boundaries. 792 00:51:17,640 --> 00:51:19,550 You could have, in this direction, 793 00:51:19,550 --> 00:51:25,230 for example, a wave like that. 794 00:51:25,230 --> 00:51:30,310 And in this direction, you could have [INAUDIBLE] or a higher 795 00:51:30,310 --> 00:51:31,170 moment. 796 00:51:31,170 --> 00:51:32,800 So it's hard for me to draw it. 797 00:51:32,800 --> 00:51:36,020 But you can imagine this pressure 798 00:51:36,020 --> 00:51:38,890 changing in all directions. 799 00:51:38,890 --> 00:51:42,310 But it has to be represented, as I say, 800 00:51:42,310 --> 00:51:48,020 by a sinusoidal function whose slope at the boundaries 801 00:51:48,020 --> 00:51:52,600 is always perpendicular like this. 802 00:51:52,600 --> 00:51:57,730 And the wave number is, of course, 803 00:51:57,730 --> 00:51:59,940 the sum of the wave number of this squared 804 00:51:59,940 --> 00:52:01,790 plus this squared plus this squared, 805 00:52:01,790 --> 00:52:04,830 taken the square root over. 806 00:52:04,830 --> 00:52:08,620 The lowest frequency will be when, 807 00:52:08,620 --> 00:52:12,060 in this direction and this direction, it's straight. 808 00:52:12,060 --> 00:52:15,680 There's no change of pressure, no change in this direction. 809 00:52:15,680 --> 00:52:19,990 And the only direction is in this direction, 810 00:52:19,990 --> 00:52:22,290 and this is the 4-meter direction. 811 00:52:22,290 --> 00:52:24,740 And that corresponds to a wavelength 812 00:52:24,740 --> 00:52:28,130 of 8 meter, which I calculated there. 813 00:52:28,130 --> 00:52:31,200 And so this will behave in the same way 814 00:52:31,200 --> 00:52:33,450 as a one-dimensional system. 815 00:52:33,450 --> 00:52:35,920 Nothing changing in two dimensions. 816 00:52:35,920 --> 00:52:38,150 And only in one dimension it's changing. 817 00:52:38,150 --> 00:52:42,620 That will give you the lowest normal mode. 818 00:52:42,620 --> 00:52:47,500 And the other normal modes, you can calculate the frequency 819 00:52:47,500 --> 00:52:53,270 as you change the little l, m, and n, play around. 820 00:52:53,270 --> 00:52:57,060 Always remember, at least one of them has to be non-zero. 821 00:52:57,060 --> 00:53:00,300 And you can calculate all the frequencies like this. 822 00:53:00,300 --> 00:53:00,840 OK. 823 00:53:00,840 --> 00:53:04,910 That's enough for today on standing waves. 824 00:53:04,910 --> 00:53:06,760 Thank you.