WEBVTT 00:00:00.070 --> 00:00:01.770 The following content is provided 00:00:01.770 --> 00:00:04.019 under a Creative Commons license. 00:00:04.019 --> 00:00:06.860 Your support will help MIT OpenCourseWare continue 00:00:06.860 --> 00:00:10.720 to offer high quality educational resources for free. 00:00:10.720 --> 00:00:13.330 To make a donation or view additional materials 00:00:13.330 --> 00:00:17.209 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:17.209 --> 00:00:17.834 at ocw.mit.edu. 00:00:21.380 --> 00:00:23.240 PROFESSOR: Welcome back. 00:00:23.240 --> 00:00:27.660 Today, we will consider problems where 00:00:27.660 --> 00:00:30.940 the solutions are standing waves. 00:00:30.940 --> 00:00:36.550 So the first problem I'm going to do is shown over here. 00:00:36.550 --> 00:00:47.298 What we have is a string attached at both ends. 00:00:47.298 --> 00:00:50.990 The length of the string is L. And it 00:00:50.990 --> 00:00:55.800 is distorted in this strange shape. 00:00:55.800 --> 00:00:57.720 The dimensions are given everywhere. 00:00:57.720 --> 00:01:02.210 It's symmetric about the middle, OK? 00:01:02.210 --> 00:01:07.040 Somehow miraculously, it's held fixed. 00:01:07.040 --> 00:01:10.120 You could imagine maybe sort of nails nailed here and here 00:01:10.120 --> 00:01:11.890 and fingers holding there. 00:01:11.890 --> 00:01:17.020 And then miraculously, we let go, all right? 00:01:17.020 --> 00:01:20.950 And the question is after we let go, 00:01:20.950 --> 00:01:25.210 what is the wavelength of the lowest mode that is not 00:01:25.210 --> 00:01:32.860 excited, the amplitude of the lowest mode which is excited? 00:01:32.860 --> 00:01:36.110 And finally, the question is, what 00:01:36.110 --> 00:01:41.640 is the shape of this string at this strange time? 00:01:41.640 --> 00:01:44.712 Now, then what do they tell us? 00:01:44.712 --> 00:01:49.180 That this is an ideal lossless string, 00:01:49.180 --> 00:01:53.800 which has a mass per unit length of mu. 00:01:53.800 --> 00:01:59.460 It has a uniform tension T. For all the equations to work 00:01:59.460 --> 00:02:01.760 and the derivation of the wave equation, 00:02:01.760 --> 00:02:08.090 you have to assume that the angle is sine angle, et cetera, 00:02:08.090 --> 00:02:11.720 is equal like this, which is-- you look at this 00:02:11.720 --> 00:02:13.100 and say, holy smoke. 00:02:13.100 --> 00:02:14.930 How can this be the case? 00:02:14.930 --> 00:02:19.870 Well, of course, this is an idealized situation. 00:02:19.870 --> 00:02:23.310 In reality, this would be very, very small 00:02:23.310 --> 00:02:25.240 compared to that, all right? 00:02:25.240 --> 00:02:31.042 And one would have to have some curvature at these points. 00:02:31.042 --> 00:02:35.450 The differences will be, the real situation 00:02:35.450 --> 00:02:38.420 from this idealised one, will only 00:02:38.420 --> 00:02:41.150 differ in some very high frequencies. 00:02:41.150 --> 00:02:45.760 So the first approximation, one can almost 00:02:45.760 --> 00:02:49.520 imagine that it's possible to set this up. 00:02:49.520 --> 00:02:52.010 And this assumption, as I told you, initially 00:02:52.010 --> 00:02:58.970 is the system, at t equals 0, is stationary. 00:02:58.970 --> 00:02:59.470 OK. 00:03:03.220 --> 00:03:10.430 The above, as I was just saying, is really highly idealized. 00:03:10.430 --> 00:03:14.500 It is almost a mathematical representation 00:03:14.500 --> 00:03:18.620 already of a real situation. 00:03:18.620 --> 00:03:20.790 So you could imagine this almost being 00:03:20.790 --> 00:03:24.700 a mathematical description of this situation. 00:03:24.700 --> 00:03:30.470 And what makes it physics rather than the mathematics 00:03:30.470 --> 00:03:35.040 is that string obeys the wave equation. 00:03:35.040 --> 00:03:38.470 It obeys this wave equation. 00:03:38.470 --> 00:03:46.010 Once I've told you that that picture on that diagram 00:03:46.010 --> 00:03:50.420 satisfies the wave equation where V is a constant given 00:03:50.420 --> 00:03:53.590 by this quantity, we've now converted this problem 00:03:53.590 --> 00:03:54.750 completely to mathematics. 00:03:58.610 --> 00:04:03.930 One thing to notice, this is the phase velocity 00:04:03.930 --> 00:04:08.080 of propagation of a progressive wave on a string. 00:04:08.080 --> 00:04:11.150 So here, that time, they ask us in part 00:04:11.150 --> 00:04:19.290 C-- is the shape of the string at a time L over V from time 00:04:19.290 --> 00:04:21.209 equals 0. 00:04:21.209 --> 00:04:25.350 So we're now solving this problem, mathematics problem. 00:04:29.870 --> 00:04:42.110 We know that if you have a coupled oscillator, 00:04:42.110 --> 00:04:46.700 there are normal mode solutions, solutions 00:04:46.700 --> 00:04:49.010 where every oscillator is oscillating 00:04:49.010 --> 00:04:53.170 with the same frequency and phase. 00:04:53.170 --> 00:04:57.050 For a system, a continuous system 00:04:57.050 --> 00:05:00.700 like this is, an infinite number of identical oscillators 00:05:00.700 --> 00:05:06.910 in a straight line for a string like that, the most 00:05:06.910 --> 00:05:12.540 general solution will also be-- one 00:05:12.540 --> 00:05:16.610 will be able to represent it as a superposition 00:05:16.610 --> 00:05:19.110 of normal modes. 00:05:19.110 --> 00:05:23.330 Of course, if you look at such a system 00:05:23.330 --> 00:05:25.430 and you distort this from equilibrium, 00:05:25.430 --> 00:05:28.950 you can set up an infinite different varieties 00:05:28.950 --> 00:05:30.220 of solutions. 00:05:30.220 --> 00:05:33.330 But what I'm saying is that any one of those solutions 00:05:33.330 --> 00:05:36.180 however complicated, one could always 00:05:36.180 --> 00:05:41.520 reduce it to a superposition of normal modes. 00:05:41.520 --> 00:05:45.180 In particular, if you study the solutions 00:05:45.180 --> 00:05:50.620 of a system like this, one dimensional system 00:05:50.620 --> 00:05:56.740 like that, you will find that the normal modes are, 00:05:56.740 --> 00:06:00.830 in fact, standing waves which are sinusoidal. 00:06:07.610 --> 00:06:16.210 So I've written for you here a completely general solution 00:06:16.210 --> 00:06:17.405 to a string. 00:06:20.510 --> 00:06:26.390 And you can write it, as I said, as a superposition 00:06:26.390 --> 00:06:31.855 of an infinite number of normal modes, pretty complicated. 00:06:34.550 --> 00:06:38.270 It tells you at every position on the string 00:06:38.270 --> 00:06:41.130 and at every time what is the displacement. 00:06:41.130 --> 00:06:43.480 You can calculate the transverse velocity 00:06:43.480 --> 00:06:45.710 by differentiating, et cetera. 00:06:45.710 --> 00:06:48.230 It just tells you everything about the oscillations 00:06:48.230 --> 00:06:50.260 of a string. 00:06:50.260 --> 00:06:52.760 This is not any old string. 00:06:52.760 --> 00:06:55.520 Because we are told some things about it. 00:06:55.520 --> 00:06:58.810 We are told it's attached at both ends 00:06:58.810 --> 00:07:02.500 so that string has no displacement in the Y 00:07:02.500 --> 00:07:08.770 direction at x equals 0 and at x equals L. For all times 00:07:08.770 --> 00:07:18.280 t this equation will satisfy that string-- doesn't matter 00:07:18.280 --> 00:07:21.760 even what distortion it is-- provided 00:07:21.760 --> 00:07:27.060 we put some constraints on these various constants 00:07:27.060 --> 00:07:28.510 such that this is true. 00:07:31.550 --> 00:07:33.850 Well, you can almost do it in your head. 00:07:33.850 --> 00:07:41.370 If you look at this, all these-- if for all times, at x 00:07:41.370 --> 00:07:49.980 equals 0, y is 0, this term, B of N, has to be 0. 00:07:49.980 --> 00:07:51.790 So this, you could forget about. 00:07:51.790 --> 00:07:53.995 And you're down to this. 00:07:56.550 --> 00:07:58.310 Furthermore, you know it has to be 00:07:58.310 --> 00:08:06.980 0 at x equals L. That puts a constraint on k of n. 00:08:06.980 --> 00:08:14.410 So your solution then is of this form. 00:08:14.410 --> 00:08:16.960 This is any arbitrary constant. 00:08:21.210 --> 00:08:24.320 For the wave equation, you can show for a string 00:08:24.320 --> 00:08:29.880 that these two are related through the phase velocity. 00:08:29.880 --> 00:08:32.440 And so once you've established this, 00:08:32.440 --> 00:08:34.820 then we've got that constant. 00:08:34.820 --> 00:08:37.450 So you're still left with quite a number of constants 00:08:37.450 --> 00:08:43.260 and an infinite series where the An can be anything. 00:08:43.260 --> 00:08:44.710 This phase can be anything. 00:08:48.200 --> 00:08:50.470 It's always useful to check yourself. 00:08:50.470 --> 00:08:51.980 There are some things you can check. 00:08:51.980 --> 00:08:58.870 So for example, let's consider the n equals 1 harmonic, 00:08:58.870 --> 00:09:03.660 that normal mode, the first normal mode. 00:09:03.660 --> 00:09:08.390 If you look at this, it's clear that the wavelength 00:09:08.390 --> 00:09:11.790 of that harmonic is 2L. 00:09:11.790 --> 00:09:18.790 If you look at this, the period of that harmonic is 2L over v. 00:09:18.790 --> 00:09:23.900 Period is 1 over frequency, or frequency is 1 over period. 00:09:23.900 --> 00:09:25.210 So the frequency is v/2L. 00:09:29.470 --> 00:09:36.080 We know that for harmonic waves or standing waves 00:09:36.080 --> 00:09:40.090 that lambda times F is the phase velocity. 00:09:40.090 --> 00:09:43.360 Let's check it here, multiply lambda by F, 00:09:43.360 --> 00:09:45.980 and you get that this is satisfied. 00:09:45.980 --> 00:09:47.730 So at least this is a quick check 00:09:47.730 --> 00:09:51.070 that we haven't made some stupid mistake. 00:09:51.070 --> 00:09:59.770 All right, so this satisfies any string 00:09:59.770 --> 00:10:06.380 of length L which is tied at both ends. 00:10:06.380 --> 00:10:09.830 But our string is not any string. 00:10:09.830 --> 00:10:17.810 We've said that initially, at t equals 0 for all positions, 00:10:17.810 --> 00:10:20.350 the string is stationary. 00:10:20.350 --> 00:10:26.020 So our description of it must take into account the equation 00:10:26.020 --> 00:10:29.215 which describes this shape of the string 00:10:29.215 --> 00:10:35.800 and it must have this boundary condition, OK? 00:10:35.800 --> 00:10:40.510 So we take this equation-- so far, 00:10:40.510 --> 00:10:42.880 we've reduced it to that-- differentiate it 00:10:42.880 --> 00:10:45.980 with respect to time, get this. 00:10:45.980 --> 00:10:49.650 And that's an easy-- we just have to differentiate-- 00:10:49.650 --> 00:10:52.380 the cosine gives you minus sine. 00:10:52.380 --> 00:10:57.230 OK, so you end up with this. 00:10:57.230 --> 00:11:00.210 And this, we know is 0. 00:11:00.210 --> 00:11:05.705 It's 0 at-- I'm sorry, this is a t equals 0. 00:11:05.705 --> 00:11:07.180 I'm sorry. 00:11:07.180 --> 00:11:13.490 We know that at t equals 0 at every position 00:11:13.490 --> 00:11:15.510 x-- we said that. 00:11:15.510 --> 00:11:19.060 We're holding this string with nothing moving. 00:11:19.060 --> 00:11:20.930 At t equals 0, it's stationary. 00:11:20.930 --> 00:11:21.980 We then let go. 00:11:21.980 --> 00:11:23.780 So this is 0. 00:11:23.780 --> 00:11:26.310 And if you look at this, this will 00:11:26.310 --> 00:11:32.710 be 0 at all values of x at t equals 0, 00:11:32.710 --> 00:11:36.100 only if all values of phi n is 0. 00:11:36.100 --> 00:11:40.370 So notice, gradually, by making use of all the information we 00:11:40.370 --> 00:11:44.590 have about the situation-- what string it is, where it's 00:11:44.590 --> 00:11:50.540 attached, how long it is, what is its motion at t 00:11:50.540 --> 00:11:55.740 equals 0-- we're gradually getting rid of the constants 00:11:55.740 --> 00:11:59.400 or determining them. 00:11:59.400 --> 00:12:03.370 Now that phi n is 0, I go back to this 00:12:03.370 --> 00:12:05.050 and rewrite the equation. 00:12:05.050 --> 00:12:12.030 And now, we've ended up that the general equation which 00:12:12.030 --> 00:12:18.530 describes a string which has length L tied 00:12:18.530 --> 00:12:25.800 at both ends, stationary at t equals 0 is this. 00:12:25.800 --> 00:12:29.810 An is still to be determined. 00:12:29.810 --> 00:12:34.200 We have not made use, so far, of the information what 00:12:34.200 --> 00:12:36.900 is the shape of the string at t equals 0. 00:12:36.900 --> 00:12:38.560 That's what we'll do next. 00:12:38.560 --> 00:12:42.230 But before I do this, I just want to make a comment. 00:12:42.230 --> 00:12:45.360 In many books, problems, et cetera, 00:12:45.360 --> 00:12:49.190 you'll find people starting with this equation. 00:12:49.190 --> 00:12:53.500 And you'll be wondering, how on Earth-- is this the most 00:12:53.500 --> 00:12:56.910 general equation for string problems? 00:12:56.910 --> 00:12:58.480 No. 00:12:58.480 --> 00:13:02.090 The answer why they started with this 00:13:02.090 --> 00:13:06.510 is, often without telling you, doing all this analysis 00:13:06.510 --> 00:13:08.580 in their head. 00:13:08.580 --> 00:13:14.420 This is only true for a situation 00:13:14.420 --> 00:13:18.550 where this string is tied at both ends 00:13:18.550 --> 00:13:20.720 and initially stationary. 00:13:20.720 --> 00:13:23.260 That forces you to that. 00:13:23.260 --> 00:13:28.410 And An now is now not determined. 00:13:28.410 --> 00:13:31.750 And in order to find out the value of An-- in other words, 00:13:31.750 --> 00:13:35.340 the amplitude of the different normal modes-- 00:13:35.340 --> 00:13:41.340 we have to make use of the shape of the string at t equals 0. 00:13:41.340 --> 00:13:42.710 And that's what we'll do now. 00:13:51.390 --> 00:13:55.080 All right, so now, I'm going to make 00:13:55.080 --> 00:13:57.800 use of the last bit of information I have. 00:13:57.800 --> 00:14:02.000 And since that defines the physical situation completely, 00:14:02.000 --> 00:14:04.890 there'd better be enough to find all the constants. 00:14:04.890 --> 00:14:06.550 There should be nothing left. 00:14:06.550 --> 00:14:09.720 They should be then completely determined. 00:14:09.720 --> 00:14:14.270 So first of all, I said we didn't make use of the shape 00:14:14.270 --> 00:14:15.630 at t equals 0. 00:14:15.630 --> 00:14:19.790 At t equals 0, cosine of 0 is 1. 00:14:19.790 --> 00:14:28.790 So what we know is that the string shape is given by, 00:14:28.790 --> 00:14:32.420 from here, by this equation. 00:14:32.420 --> 00:14:39.820 So this tells us that-- this is a description 00:14:39.820 --> 00:14:43.320 of the shape of the string in terms 00:14:43.320 --> 00:14:46.380 of an infinite series of sinusoidal functions. 00:14:49.290 --> 00:15:00.360 OK, we notice this tells us the shape of the string at every X 00:15:00.360 --> 00:15:03.280 position at t equals 0. 00:15:03.280 --> 00:15:07.540 So I look at that, the original shape. 00:15:07.540 --> 00:15:10.870 And I can describe it mathematically. 00:15:10.870 --> 00:15:17.170 y of x at time equals 0 is 0 at all points from one 00:15:17.170 --> 00:15:20.190 end of the string to where the distortion starts. 00:15:20.190 --> 00:15:26.760 And it starts at 3/8 L if you look at the diagram. 00:15:26.760 --> 00:15:35.530 Then from position x of 3/8 L to x is 5/8 L, 00:15:35.530 --> 00:15:40.470 the string is straight again, and its displaced by a distance 00:15:40.470 --> 00:15:46.530 H. And afterwards, past the 5/8 L, it's back to 0. 00:15:46.530 --> 00:15:51.170 For a second, let's go back and just look at that. 00:15:51.170 --> 00:15:54.690 So we're talking about this shape, right? 00:15:54.690 --> 00:16:03.890 It's 0, 0 in Y, and then jumps to H between 3/8 L and 5/8 L. 00:16:03.890 --> 00:16:11.040 This shape, when you let go, that 00:16:11.040 --> 00:16:17.260 will determine the amplitude of the different normal modes 00:16:17.260 --> 00:16:24.370 that are important in the oscillation of that system. 00:16:24.370 --> 00:16:31.150 OK, so we must determine these infinite number of constants. 00:16:31.150 --> 00:16:35.970 At first sight, it looks like an impossible task 00:16:35.970 --> 00:16:39.550 if it wasn't for a brilliant mathematician 00:16:39.550 --> 00:16:42.850 in the 18th century, Fourier, who 00:16:42.850 --> 00:16:46.860 made a relatively simple but incredibly important 00:16:46.860 --> 00:16:48.730 observation. 00:16:48.730 --> 00:16:53.960 When applied to this problem, it is the following. 00:16:58.410 --> 00:17:03.850 He realized that one can do the following trick. 00:17:08.230 --> 00:17:12.770 If you take this function, sine n pi 00:17:12.770 --> 00:17:17.290 over L x to 1 which is describing our shape 00:17:17.290 --> 00:17:22.220 over there, if you multiply it-- and by the way, 00:17:22.220 --> 00:17:25.780 n-- I'm just reminding you-- goes from 1 to infinity. 00:17:25.780 --> 00:17:29.200 It's 1, 2, 3, 4, 5, 6, et cetera. 00:17:29.200 --> 00:17:33.350 If you take this function and you multiply it 00:17:33.350 --> 00:17:39.330 by a similar function, but with different value of m, 00:17:39.330 --> 00:17:46.120 then this integral from one end of the string to the other, 00:17:46.120 --> 00:17:53.390 is equal to 0 if this, the m you've chosen here, 00:17:53.390 --> 00:17:55.010 is different than n. 00:17:58.300 --> 00:17:58.970 Try it. 00:17:58.970 --> 00:18:02.600 I don't want-- a good exercise for yourself. 00:18:02.600 --> 00:18:05.630 Take this function. 00:18:05.630 --> 00:18:10.970 And do the integration, and you'll find it comes out to 0. 00:18:10.970 --> 00:18:13.320 It's not hard to do. 00:18:13.320 --> 00:18:17.740 So that is 0 if n is not equal to m. 00:18:17.740 --> 00:18:23.740 On the other hand, if m equals n-- 00:18:23.740 --> 00:18:29.200 so this is the integral of sine squared n pi over L-- 00:18:29.200 --> 00:18:30.885 it comes out to be L/2. 00:18:30.885 --> 00:18:32.010 You can do it for yourself. 00:18:35.350 --> 00:18:43.400 This trick allows us to solve for all values of An. 00:18:43.400 --> 00:18:45.880 Now, let me tell you, I did not do 00:18:45.880 --> 00:18:53.320 this in-- the Fourier trick-- in general. 00:18:53.320 --> 00:18:59.330 If you go into books, you'll find the formulation 00:18:59.330 --> 00:19:04.250 of this trick known as Fourier's theorem 00:19:04.250 --> 00:19:06.090 for any periodic function. 00:19:09.800 --> 00:19:13.850 The advantage of learning it, how to do it for any function, 00:19:13.850 --> 00:19:17.710 you don't have to do this integral every time. 00:19:17.710 --> 00:19:19.840 Because suppose our function turned out 00:19:19.840 --> 00:19:23.710 there to be a cosine, we would have to then figure out, 00:19:23.710 --> 00:19:28.225 what do I multiply it by to make this come out so simply? 00:19:35.505 --> 00:19:37.240 As I say, the advantage is then you 00:19:37.240 --> 00:19:41.670 can use formulate in books, et cetera. 00:19:41.670 --> 00:19:43.310 There's nothing wrong with that. 00:19:43.310 --> 00:19:44.590 It is good. 00:19:44.590 --> 00:19:49.940 And the way one changes a physical situation into one 00:19:49.940 --> 00:19:53.840 where you can apply Fourier's theorem, which only applies 00:19:53.840 --> 00:20:00.650 to periodic functions, is to take your original shape, 00:20:00.650 --> 00:20:05.260 and out of it, create a periodic function. 00:20:05.260 --> 00:20:09.450 Once you've done that, you can use Fourier's theorem. 00:20:09.450 --> 00:20:15.140 I've, in this particular case because I find it easier 00:20:15.140 --> 00:20:19.230 to explain it, simply essentially derived 00:20:19.230 --> 00:20:24.380 the Fourier theorem for this particular problem. 00:20:24.380 --> 00:20:29.960 For this particular problem, the function 00:20:29.960 --> 00:20:34.940 by which you multiply this is sine m pi. 00:20:34.940 --> 00:20:39.860 OK, once you've realized that this is true, 00:20:39.860 --> 00:20:50.420 I can go back and use it to determine every An. 00:20:57.890 --> 00:20:59.160 I go to this function. 00:21:03.510 --> 00:21:12.390 I multiply both sides in turn by sine m pi over L 00:21:12.390 --> 00:21:21.670 for every n where I want to find the value of An, for each one. 00:21:21.670 --> 00:21:24.070 So I have to do an infinite number 00:21:24.070 --> 00:21:27.010 of integrals in principle. 00:21:27.010 --> 00:21:31.100 But often you don't need to know every value of n. 00:21:31.100 --> 00:21:34.920 You may want to know each one of An 00:21:34.920 --> 00:21:39.030 is the amplitude of the harmonic. 00:21:39.030 --> 00:21:41.830 You may be interested in what is the amplitude 00:21:41.830 --> 00:21:45.320 of the first harmonic solution or the second or third. 00:21:45.320 --> 00:21:49.600 However many you want, so many integrals you have to do. 00:21:49.600 --> 00:21:51.730 No magic. 00:21:51.730 --> 00:21:54.545 You're basically solving infinite number of equations. 00:21:58.260 --> 00:22:02.890 And if I apply this to our problem, 00:22:02.890 --> 00:22:11.410 we find that An is equal to 2/L. It's from here. 00:22:11.410 --> 00:22:19.400 The integral of 0 of 2L of y sine at dx. 00:22:19.400 --> 00:22:19.910 Right? 00:22:19.910 --> 00:22:25.200 All I'm taking, I'm multiplying both sides of this equation 00:22:25.200 --> 00:22:32.540 by sine of m pi L over x and integrating them 00:22:32.540 --> 00:22:37.680 both from 0 to L. On one side, you 00:22:37.680 --> 00:22:42.870 end up with An and on the other, 2/L times this integral. 00:22:42.870 --> 00:22:47.890 Then once you've done this integral, you know the answer. 00:22:47.890 --> 00:22:55.680 In our case, y of x of 0 is particularly easy. 00:22:55.680 --> 00:22:59.500 It's 0 in this range, 0 in this range, 00:22:59.500 --> 00:23:02.960 and a constant for the rest. 00:23:02.960 --> 00:23:07.760 Well, over the part of the range where it's 0, 00:23:07.760 --> 00:23:10.490 if I multiply something by 0, it's 0. 00:23:10.490 --> 00:23:15.240 And so I only have to do this integral over the range 00:23:15.240 --> 00:23:24.780 where y of x for 0 time is H. So I have to do this integral. 00:23:24.780 --> 00:23:31.350 And now I have reduced the situation that every n 00:23:31.350 --> 00:23:33.840 that we are interested in, we can 00:23:33.840 --> 00:23:36.460 calculate by doing this integral. 00:23:36.460 --> 00:23:42.476 So in principle, we've solved it for all values of n. 00:23:46.650 --> 00:23:50.410 So finally, I can answer the question. 00:23:50.410 --> 00:23:54.990 So the first question was what lowest value of n 00:23:54.990 --> 00:23:59.460 for which An is 0. 00:23:59.460 --> 00:24:00.220 Well, let's start. 00:24:00.220 --> 00:24:03.390 Take number 1, 2, 3. 00:24:03.390 --> 00:24:13.870 And for n equals 1, this integral is not equal to 0, 00:24:13.870 --> 00:24:17.360 because what you have to do is you basically 00:24:17.360 --> 00:24:21.260 are integrating this-- for n equals 00:24:21.260 --> 00:24:26.390 1, that's what the sine pi over L x looks like. 00:24:26.390 --> 00:24:30.930 You're integrating that and multiply it by 0. 00:24:30.930 --> 00:24:33.610 And here, you're multiplying by a constant and 0. 00:24:33.610 --> 00:24:37.210 Clearly, this is not going to give you a 0. 00:24:37.210 --> 00:24:40.840 Let's take the second one, the second harmonic 00:24:40.840 --> 00:24:42.090 in this calculation. 00:24:42.090 --> 00:24:48.260 You're multiplying this y by this function 00:24:48.260 --> 00:24:50.910 and integrating it, OK? 00:24:50.910 --> 00:24:55.830 Notice the sine function is symmetric about 0 00:24:55.830 --> 00:24:57.940 and so is this symmetric. 00:24:57.940 --> 00:25:01.710 Except here, this is an anti-symmetric function. 00:25:01.710 --> 00:25:06.370 So the sine is negative to the right of the middle 00:25:06.370 --> 00:25:08.110 and positive to the left. 00:25:08.110 --> 00:25:12.320 If I multiply this positive by that, I get a positive number. 00:25:12.320 --> 00:25:15.370 If I multiply this by that, I get a negative number. 00:25:15.370 --> 00:25:17.030 The two are equal. 00:25:17.030 --> 00:25:20.900 And so the integral across there is 0. 00:25:20.900 --> 00:25:28.200 So n equals 2 is the lowest harmonic for which An is 0. 00:25:28.200 --> 00:25:37.940 When I let that string go, the second harmonic 00:25:37.940 --> 00:25:40.150 will not be excited. 00:25:40.150 --> 00:25:44.750 Which is the lowest n for which it is excited-- in other words 00:25:44.750 --> 00:25:46.010 that An is not equal 0. 00:25:46.010 --> 00:25:52.370 Well, I told you when n equals 1, clearly An is not 0. 00:25:52.370 --> 00:25:54.300 So that will be excited. 00:25:54.300 --> 00:25:58.920 So the first harmonic will be excited. 00:25:58.920 --> 00:26:00.250 What is the amplitude? 00:26:00.250 --> 00:26:02.630 Well, we have to calculate it. 00:26:02.630 --> 00:26:05.130 I told you, every An-- I told you what it is. 00:26:05.130 --> 00:26:06.880 I have to do some work. 00:26:06.880 --> 00:26:12.210 So you take A of 1 is 2/L. The integral from that to the H 00:26:12.210 --> 00:26:17.560 times this function where n is 1, all right? 00:26:17.560 --> 00:26:21.090 And that, fortunately, is an easy integral to do. 00:26:21.090 --> 00:26:23.140 Integrate the sine, you get the cosine. 00:26:23.140 --> 00:26:26.480 And putting in the limits, you get this answer. 00:26:26.480 --> 00:26:31.490 So that is our answer to that part of the problem. 00:26:31.490 --> 00:26:34.860 The final part of the problem was 00:26:34.860 --> 00:26:39.350 they said, OK, I have this shape. 00:26:39.350 --> 00:26:40.800 I let go. 00:26:40.800 --> 00:26:46.160 What will that shape look like at a time 00:26:46.160 --> 00:26:52.360 was L-- in the problem, the time was L times 00:26:52.360 --> 00:26:57.550 the square root of mu over T. T's the tension in the string. 00:26:57.550 --> 00:27:00.570 But I know that square root of mu over t 00:27:00.570 --> 00:27:03.490 is 1/d, the phase velocity. 00:27:03.490 --> 00:27:08.900 So they ask us to calculate the shape of that string 00:27:08.900 --> 00:27:12.560 at the time which is the length of the string divided 00:27:12.560 --> 00:27:15.470 by the phase velocity of progressive waves 00:27:15.470 --> 00:27:17.410 on the string. 00:27:17.410 --> 00:27:27.720 OK, now we know completely what is the shape of the string. 00:27:27.720 --> 00:27:28.980 It is this formula. 00:27:28.980 --> 00:27:33.330 This is the formula that tells us 00:27:33.330 --> 00:27:37.545 what the string looks like at all places, all times. 00:27:41.130 --> 00:27:46.650 From our knowledge of this integral, 00:27:46.650 --> 00:27:52.520 depending where it's symmetric or anti-symmetric 00:27:52.520 --> 00:27:56.290 sinusoidal function about the center, 00:27:56.290 --> 00:28:00.050 we know that every second one will be 0. 00:28:00.050 --> 00:28:05.330 So this sum now, I've only summed 00:28:05.330 --> 00:28:09.830 for n equals 1, 3, 5, 7, et cetera. 00:28:09.830 --> 00:28:17.880 The terms n equals 2, 4, 6, et cetera, will have 0 amplitude. 00:28:17.880 --> 00:28:25.540 But this now describes our situation in its entirety-- 00:28:25.540 --> 00:28:29.950 because I've now not only know the spatial shape, but also 00:28:29.950 --> 00:28:32.380 as a function of time. 00:28:32.380 --> 00:28:38.240 So this describes what it was at time t. 00:28:38.240 --> 00:28:45.130 What will it be if I increase the time-- let's start from 0, 00:28:45.130 --> 00:28:48.690 and I go to a time L/v? 00:28:48.690 --> 00:28:56.280 All right, well, if this changes from 0 to L/v, 00:28:56.280 --> 00:29:01.150 this changes to cosine n pi. 00:29:01.150 --> 00:29:05.550 But you know that for n equals 1, 3, 5, et cetera, 00:29:05.550 --> 00:29:09.540 cosine of n pi-- in other words, at pi, at 3 pi, 00:29:09.540 --> 00:29:12.710 et cetera-- is minus 1. 00:29:12.710 --> 00:29:21.690 So at this later time, every term in this expansion 00:29:21.690 --> 00:29:23.500 has changed signs. 00:29:23.500 --> 00:29:30.250 But it's exactly the same series, except it's minus 1 00:29:30.250 --> 00:29:31.220 here. 00:29:31.220 --> 00:29:33.190 So what has happened to the string? 00:29:33.190 --> 00:29:38.210 It is exactly the same shape with an opposite sign. 00:29:38.210 --> 00:29:42.660 So with the string, instead of being up here like that 00:29:42.660 --> 00:29:47.800 like it was over here, it's flipped over like that. 00:29:47.800 --> 00:29:50.900 Without doing any more work, I can conclude that simply 00:29:50.900 --> 00:29:54.460 because I have the same series as before with a negative sign. 00:29:54.460 --> 00:29:56.360 So that is the answer. 00:29:56.360 --> 00:29:57.550 So I've solved the problem. 00:29:57.550 --> 00:30:00.960 But at this instance, I just want to digress for a second 00:30:00.960 --> 00:30:04.900 and tell you this actually is a problem 00:30:04.900 --> 00:30:10.000 that we could have almost done all by thinking alone. 00:30:10.000 --> 00:30:15.160 It wasn't necessary to do big fraction of the work we did. 00:30:15.160 --> 00:30:18.680 The only part which there was no choice 00:30:18.680 --> 00:30:21.330 is this amplitude [INAUDIBLE]. 00:30:21.330 --> 00:30:28.140 Because I could have gone back, scratched my memory 00:30:28.140 --> 00:30:30.310 and knowledge, and said, look, hold on. 00:30:30.310 --> 00:30:33.440 When I have been a string, I could always 00:30:33.440 --> 00:30:37.560 analyze it in terms of normal modes. 00:30:37.560 --> 00:30:45.380 But also, we know that we can describe the solutions 00:30:45.380 --> 00:30:48.370 in terms of progressive waves. 00:30:48.370 --> 00:30:50.890 The two are equivalent to each other. 00:30:50.890 --> 00:30:52.550 It's not extra solutions. 00:30:52.550 --> 00:30:56.400 But by adding appropriately normal modes, 00:30:56.400 --> 00:31:02.220 I can get a superposition of progressive waves. 00:31:02.220 --> 00:31:06.080 I can use the uniqueness theorem to see 00:31:06.080 --> 00:31:13.110 if I've got a solution which represents 00:31:13.110 --> 00:31:16.760 the situation at hand. 00:31:16.760 --> 00:31:24.580 If you look at the original problem, 00:31:24.580 --> 00:31:31.920 if you look at this problem, this is stationary, all right? 00:31:31.920 --> 00:31:39.530 So I could imagine this solution being the superposition 00:31:39.530 --> 00:31:45.020 of two progressive waves that look like this but half 00:31:45.020 --> 00:31:49.080 the amplitude, one moving to the left and one 00:31:49.080 --> 00:31:51.090 moving to the right. 00:31:51.090 --> 00:31:57.190 As they overlap-- in other words, 00:31:57.190 --> 00:32:02.750 I have two waves, one like this moving to the right, 00:32:02.750 --> 00:32:06.250 and one like this moving to the left. 00:32:06.250 --> 00:32:11.610 If I add them, I'll get this distortion, 00:32:11.610 --> 00:32:14.610 which is stationary. 00:32:14.610 --> 00:32:18.370 Therefore, if I solved the problem 00:32:18.370 --> 00:32:25.860 for each one of these progressive waves and add them, 00:32:25.860 --> 00:32:28.340 I'll get the answer to this. 00:32:28.340 --> 00:32:31.300 It'll satisfy all the boundary conditions. 00:32:31.300 --> 00:32:32.840 By the uniqueness theorem, it will 00:32:32.840 --> 00:32:38.720 be the correct description of what happened. 00:32:38.720 --> 00:32:44.320 And then with that, I can predict 00:32:44.320 --> 00:32:49.000 what will be the shape at any time. 00:32:49.000 --> 00:32:54.220 So for example, in the time they ask L/v, 00:32:54.220 --> 00:33:00.220 a progressive wave will move a distance L. 00:33:00.220 --> 00:33:05.010 So the wave over here, which was going 00:33:05.010 --> 00:33:10.310 to do the right would hit this boundary-- that pulse-- would 00:33:10.310 --> 00:33:12.180 flip over. 00:33:12.180 --> 00:33:16.940 And you know that this string is fixed there. 00:33:16.940 --> 00:33:21.250 What happens if a pulse comes to a fixed end? 00:33:21.250 --> 00:33:26.440 Imagine you are causing the pulse. 00:33:26.440 --> 00:33:29.340 You are the forcewave moving along 00:33:29.340 --> 00:33:31.300 which causes the distortion. 00:33:31.300 --> 00:33:34.810 When you come to the rigid end, you're pulling. 00:33:34.810 --> 00:33:36.300 It won't give. 00:33:36.300 --> 00:33:39.550 So the only way to happen, the string will pull you down. 00:33:39.550 --> 00:33:41.450 So you'll flip over. 00:33:41.450 --> 00:33:45.040 So this pulse, when it hits this end, 00:33:45.040 --> 00:33:47.340 will flip over upside down, and then 00:33:47.340 --> 00:33:49.600 progress this way upside down. 00:33:49.600 --> 00:33:54.710 This one will go to this end, flip over, and progress back. 00:33:54.710 --> 00:34:02.370 By the time t has changed by L over the velocity, 00:34:02.370 --> 00:34:05.670 those flipped pulses would have come back here and overlapped, 00:34:05.670 --> 00:34:07.130 but upside down. 00:34:07.130 --> 00:34:08.909 And so this would have flipped. 00:34:08.909 --> 00:34:12.730 By this kind of analysis, if you use your wits about it, 00:34:12.730 --> 00:34:20.060 sometimes you can solve more difficult problems 00:34:20.060 --> 00:34:25.159 quickly by using all the knowledge you have about waves, 00:34:25.159 --> 00:34:26.912 the propagation, et cetera. 00:34:26.912 --> 00:34:28.120 Probably a good time to stop. 00:34:28.120 --> 00:34:29.970 Thank you.