WEBVTT 00:00:00.060 --> 00:00:01.770 The following content is provided 00:00:01.770 --> 00:00:04.010 under a Creative Commons license. 00:00:04.010 --> 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. 00:00:20.906 --> 00:00:23.590 PROFESSOR: Welcome back, and today I'm 00:00:23.590 --> 00:00:28.110 going to do for you three problems all related 00:00:28.110 --> 00:00:31.310 to driven harmonic oscillators. 00:00:31.310 --> 00:00:32.930 Now one of the things you'll notice 00:00:32.930 --> 00:00:36.080 as I'm doing these problems, that it seems 00:00:36.080 --> 00:00:39.010 I'm doing the same thing over and over again. 00:00:39.010 --> 00:00:41.850 You may have also noticed that in the other problems 00:00:41.850 --> 00:00:45.280 I did earlier for you in harmonic oscillators, 00:00:45.280 --> 00:00:47.440 et cetera. 00:00:47.440 --> 00:00:50.350 So at this stage, I have to remind you about something. 00:00:50.350 --> 00:00:54.090 This is no accident that it is like that. 00:00:54.090 --> 00:00:56.510 It's to do with the scientific method, 00:00:56.510 --> 00:00:59.690 and I would want to briefly remind you 00:00:59.690 --> 00:01:03.590 about how we are solving problems. 00:01:03.590 --> 00:01:07.270 Here is a diagram of a generic problem. 00:01:07.270 --> 00:01:08.440 What does it consists of? 00:01:08.440 --> 00:01:14.940 A typical problem is we are told of a physical situation 00:01:14.940 --> 00:01:17.710 in words, ordinary language. 00:01:17.710 --> 00:01:23.090 And in ordinary language, we are asked questions about given 00:01:23.090 --> 00:01:25.330 that physical situation, what will 00:01:25.330 --> 00:01:27.470 happen as a function of time? 00:01:27.470 --> 00:01:33.870 What kind of emotion it has, questions of that kind. 00:01:33.870 --> 00:01:38.680 What problem solving is, doing the scientific method, 00:01:38.680 --> 00:01:39.970 is the following. 00:01:39.970 --> 00:01:50.300 The first thing we have to do is, by studying this situation 00:01:50.300 --> 00:01:57.270 and using the laws of nature, describing 00:01:57.270 --> 00:02:01.421 the physical situation in terms of mathematical equations. 00:02:01.421 --> 00:02:02.420 This is always possible. 00:02:05.610 --> 00:02:09.820 So starting with as I say this description, 00:02:09.820 --> 00:02:13.150 using laws of nature, we end up with a description 00:02:13.150 --> 00:02:16.260 in terms of mathematical equations. 00:02:16.260 --> 00:02:19.310 The next thing we do is, we now forget 00:02:19.310 --> 00:02:21.650 that this has anything to do with physics. 00:02:21.650 --> 00:02:24.260 We have a set of mathematical equations, 00:02:24.260 --> 00:02:25.750 which we have to solve. 00:02:25.750 --> 00:02:27.180 This is mathematics. 00:02:27.180 --> 00:02:29.560 We use mathematics to solve this problem. 00:02:29.560 --> 00:02:30.700 All right? 00:02:30.700 --> 00:02:34.810 Once we have found the solution to this mathematical equation, 00:02:34.810 --> 00:02:41.380 there's only one thing left, we identify the answer 00:02:41.380 --> 00:02:46.360 with the questions posed in the problem. 00:02:46.360 --> 00:02:50.930 So understanding of physics comes 00:02:50.930 --> 00:02:57.720 in this step, which I call step 1, and in the last one. 00:02:57.720 --> 00:03:01.580 The middle one, step 2, is pure mathematics. 00:03:01.580 --> 00:03:14.050 Now, it so happens that if you take all driven harmonic 00:03:14.050 --> 00:03:18.800 oscillators, the mathematics that you end up with 00:03:18.800 --> 00:03:19.710 is identical. 00:03:19.710 --> 00:03:24.820 In other words, if I look at the description of the problem 00:03:24.820 --> 00:03:30.820 in mathematical terms, I cannot tell you which problem that 00:03:30.820 --> 00:03:32.230 description belongs. 00:03:32.230 --> 00:03:38.330 That's a generic description of many, many, many situations. 00:03:38.330 --> 00:03:42.570 Today I'll take three of them, but all three situations, 00:03:42.570 --> 00:03:45.130 all three problems-- although to you, they 00:03:45.130 --> 00:03:49.600 may look completely different-- after I've translated them 00:03:49.600 --> 00:03:53.020 in terms of mathematical equations, are identical. 00:03:53.020 --> 00:03:56.460 And so it's not surprising that from here on when we then 00:03:56.460 --> 00:03:59.580 try to solve them, et cetera, we are doing the same problem, 00:03:59.580 --> 00:04:03.450 or at least it appears so. 00:04:03.450 --> 00:04:05.600 Having said that, let's immediately 00:04:05.600 --> 00:04:10.090 dive in and take one of the problems. 00:04:10.090 --> 00:04:12.810 So the first problem I'm going to consider 00:04:12.810 --> 00:04:16.089 is an idealized-driven RLC circuit. 00:04:18.750 --> 00:04:21.000 Why do I see idealized? 00:04:21.000 --> 00:04:22.690 Because as I've mentioned before, 00:04:22.690 --> 00:04:27.390 in real life situations, things extremely complicated, 00:04:27.390 --> 00:04:29.570 and if you take everything into account, 00:04:29.570 --> 00:04:32.770 it will take you essentially forever 00:04:32.770 --> 00:04:35.010 to get an exact solution. 00:04:35.010 --> 00:04:37.680 But we're often not most of the time not interested 00:04:37.680 --> 00:04:39.290 in the very exact one. 00:04:39.290 --> 00:04:42.740 So we simplify a physical situation 00:04:42.740 --> 00:04:48.480 to the bear essential parts and solve that problem. 00:04:48.480 --> 00:04:54.690 So this is a problem where you have a circuit consisting 00:04:54.690 --> 00:04:58.050 all a capacitance, inductance, and a resistance, 00:04:58.050 --> 00:05:03.360 in series, which is driven by an alternating voltage source V0 00:05:03.360 --> 00:05:04.670 cosine omega t. 00:05:07.330 --> 00:05:12.220 Before time t equals 0, I'm assuming the circuit is open. 00:05:12.220 --> 00:05:15.330 In other words, there is no charges anywhere, 00:05:15.330 --> 00:05:16.750 no current flowing. 00:05:16.750 --> 00:05:20.700 At t equals 0, we close the circuit 00:05:20.700 --> 00:05:25.690 so the charge on the capacitor t equals 0, is 0. 00:05:25.690 --> 00:05:27.790 The current flowing in the circuit is 0. 00:05:27.790 --> 00:05:28.750 All right? 00:05:28.750 --> 00:05:31.750 And the question we want to answer 00:05:31.750 --> 00:05:36.640 is, what will the current be in the circuit at some time t 00:05:36.640 --> 00:05:42.570 a specific one, in 17 seconds or whatever, some specific time. 00:05:42.570 --> 00:05:45.550 Since this is completely specified, 00:05:45.550 --> 00:05:48.480 I can predict what will happen. 00:05:48.480 --> 00:05:51.730 That's what it means in this case solving this problem. 00:05:51.730 --> 00:05:52.540 So let's do it. 00:06:01.270 --> 00:06:05.670 So we are now entering step 1. 00:06:05.670 --> 00:06:09.330 We're going to take the physical situation 00:06:09.330 --> 00:06:16.050 and convert it to a description in terms of mathematics. 00:06:16.050 --> 00:06:22.980 This is the conceptually hardest step of any problem. 00:06:22.980 --> 00:06:28.240 It is the one where you need to understand the physics. 00:06:28.240 --> 00:06:31.660 It may not necessarily be the hardest or the longest. 00:06:31.660 --> 00:06:35.870 Often the solution of the equations 00:06:35.870 --> 00:06:40.900 are incredibly tedious and long and take all your time. 00:06:40.900 --> 00:06:43.860 But they are routine mathematics. 00:06:43.860 --> 00:06:49.370 Here you've got to conceptualize what's going on 00:06:49.370 --> 00:06:54.430 and translate it as I say to the language of mathematics. 00:06:54.430 --> 00:07:00.230 So in the case of I've had have to define 00:07:00.230 --> 00:07:02.240 coordinates for this problem. 00:07:02.240 --> 00:07:08.530 So what I have is, I'll assume that at some instant of time 00:07:08.530 --> 00:07:10.590 that in this circuit, it's closed 00:07:10.590 --> 00:07:12.770 because it's past equal to 0, there 00:07:12.770 --> 00:07:18.230 is a current I flowing that on this capacitor on the left 00:07:18.230 --> 00:07:20.830 there is plus charge Q and minus charge 00:07:20.830 --> 00:07:23.240 Q on the right side that is this inductance. 00:07:26.010 --> 00:07:31.300 And over here, we have an alternating potential applied. 00:07:31.300 --> 00:07:34.000 Now I have to be very specific at any instant of time which 00:07:34.000 --> 00:07:35.540 side is positive and negative. 00:07:35.540 --> 00:07:37.700 I'll assume that t equals 0. 00:07:37.700 --> 00:07:40.710 It's 0 over here, and at this point 00:07:40.710 --> 00:07:45.340 here, it's plus V0 cosine omega t. 00:07:45.340 --> 00:07:50.930 So these are the coordinates of this problem. 00:07:50.930 --> 00:07:54.350 And now what drives the current? 00:07:54.350 --> 00:07:58.320 What decides what happens here? 00:07:58.320 --> 00:08:00.560 What are the law of natures applicable? 00:08:00.560 --> 00:08:04.750 In this case, are the Maxwell's equations or to be specific, 00:08:04.750 --> 00:08:06.810 the Faraday equation. 00:08:06.810 --> 00:08:08.340 All right? 00:08:08.340 --> 00:08:15.490 One more thing I should add that I is a current here flowing 00:08:15.490 --> 00:08:18.430 charge by conservation of charge. 00:08:18.430 --> 00:08:20.630 It's law of physics. 00:08:20.630 --> 00:08:23.930 The current at the instant of time, 00:08:23.930 --> 00:08:28.700 will be equal to the rate of change of the charge. 00:08:28.700 --> 00:08:32.450 So at the instant time, I is, in my language, 00:08:32.450 --> 00:08:35.240 Q with a dot on top of it. 00:08:35.240 --> 00:08:38.559 This is plus because of the way in which I 00:08:38.559 --> 00:08:40.270 define my coordinates. 00:08:40.270 --> 00:08:43.640 Beware if you by mistake say suppose 00:08:43.640 --> 00:08:47.300 I define my coordinates this being minus and this plus, 00:08:47.300 --> 00:08:49.890 then this equation would be a minus here, 00:08:49.890 --> 00:08:51.920 conservation of charge. 00:08:51.920 --> 00:08:54.350 But with the way I've defined these coordinates, 00:08:54.350 --> 00:08:56.720 this is plus. 00:08:56.720 --> 00:08:57.850 So that's one. 00:08:57.850 --> 00:09:00.670 Next I said what the law of nature 00:09:00.670 --> 00:09:03.120 which governs the motion of the charges 00:09:03.120 --> 00:09:07.390 is the Faraday's law, which states 00:09:07.390 --> 00:09:12.350 that if I take the line integral of the electric field 00:09:12.350 --> 00:09:17.320 around a closed loop, that's this, 00:09:17.320 --> 00:09:23.480 it equals to minus the rate of change 00:09:23.480 --> 00:09:27.160 of total magnetic flux linking this circuit, 00:09:27.160 --> 00:09:29.980 going through this circuit. 00:09:29.980 --> 00:09:34.580 Now, you know that the magnetic flux through a circuit 00:09:34.580 --> 00:09:40.520 is equal to the current times the self-inductance 00:09:40.520 --> 00:09:45.130 of that circuit or the total inductance in that circuit. 00:09:45.130 --> 00:09:49.770 So taking this law of nature, this law of nature, 00:09:49.770 --> 00:09:53.240 I can now translate this into mathematics, 00:09:53.240 --> 00:09:54.970 and that's what I'm going to do. 00:09:54.970 --> 00:09:58.450 So first I'm going to take the line integral of E dl 00:09:58.450 --> 00:10:01.700 around this circuit. 00:10:01.700 --> 00:10:06.490 If I go around from here to here, 00:10:06.490 --> 00:10:10.300 this is positive relative to this point, 00:10:10.300 --> 00:10:13.105 so the electric field from here to here 00:10:13.105 --> 00:10:19.300 will be in this direction, and so the line integral of E dl 00:10:19.300 --> 00:10:24.250 along here will be minus, and by definition of a voltage, 00:10:24.250 --> 00:10:27.100 it will be just minus the voltage drop 00:10:27.100 --> 00:10:30.360 across here, which is V0 cosine omega t. 00:10:30.360 --> 00:10:33.725 So this is minus V0 cosine omega t. 00:10:33.725 --> 00:10:35.910 I continue going around. 00:10:35.910 --> 00:10:39.530 From here to here, the electric field is in this direction. 00:10:39.530 --> 00:10:42.117 And so that will be the integral V dl 00:10:42.117 --> 00:10:46.210 is Q over C, that's from the definition of a capacitance, 00:10:46.210 --> 00:10:47.840 of course. 00:10:47.840 --> 00:10:55.730 Then as we go around in these idealized diagrams, 00:10:55.730 --> 00:11:00.250 all wires have 0 resistance and therefore there cannot be field 00:11:00.250 --> 00:11:00.750 in them. 00:11:00.750 --> 00:11:03.400 That's why there's no contribution to this 00:11:03.400 --> 00:11:06.060 from the field inside the wires. 00:11:06.060 --> 00:11:07.800 I continue. 00:11:07.800 --> 00:11:11.370 This inductance is just coiled wire, 00:11:11.370 --> 00:11:15.320 so there's still no contribution to this in the L. I continue. 00:11:15.320 --> 00:11:19.690 Here I know there is a current flowing like this means 00:11:19.690 --> 00:11:22.590 the reason the electric field in this direction 00:11:22.590 --> 00:11:27.170 and once again knowing the definition of our Ohms law, 00:11:27.170 --> 00:11:30.400 I get that the integral of the electric fields from here 00:11:30.400 --> 00:11:33.300 to here is I times R. 00:11:33.300 --> 00:11:37.160 So this is this quantity, the integral 00:11:37.160 --> 00:11:39.800 of the electric field around this closed loop. 00:11:39.800 --> 00:11:45.930 That must equal by Faraday's law to minus the rate of change 00:11:45.930 --> 00:11:49.260 of magnetic flux, which is minus L dI dt. 00:11:53.080 --> 00:11:55.170 We're almost finished. 00:11:55.170 --> 00:11:56.690 Now we do algebra. 00:11:56.690 --> 00:11:57.660 We're playing around. 00:11:57.660 --> 00:12:02.270 I can rewrite this like this knowing that the current is 00:12:02.270 --> 00:12:07.490 the rate of change of Q, and the rate of change of current 00:12:07.490 --> 00:12:09.200 is the second derivative. 00:12:09.200 --> 00:12:12.100 So this is just straightforward algebra. 00:12:12.100 --> 00:12:15.510 Now, I'm going to rewrite just moving terms around 00:12:15.510 --> 00:12:18.830 to make it look in the way I prefer. 00:12:18.830 --> 00:12:24.190 And so I've rewritten this equation now in this form. 00:12:24.190 --> 00:12:29.980 And now, I will just redefine some constants 00:12:29.980 --> 00:12:32.720 so that I recognize the equation better. 00:12:32.720 --> 00:12:38.350 So I will define R over L by a constant gamma. 00:12:38.350 --> 00:12:43.430 And I will define 1 over LC by the constant omega 0 squared. 00:12:43.430 --> 00:12:47.390 And I define V0 L by the constant f. 00:12:47.390 --> 00:12:53.270 And so this equation rewritten looks like that. 00:12:53.270 --> 00:12:54.740 Eureka. 00:12:54.740 --> 00:12:58.340 This is an equation I've seen millions of times, 00:12:58.340 --> 00:12:59.840 and I know how to solve it. 00:13:03.950 --> 00:13:08.540 So what we've done is, we've taken this physical situation, 00:13:08.540 --> 00:13:13.440 translated it into a mathematical equation. 00:13:13.440 --> 00:13:16.530 This is the equation of motion, this just 00:13:16.530 --> 00:13:20.490 defines the constants, and this is the boundary conditions. 00:13:20.490 --> 00:13:23.870 It tells us that in our particular problem, 00:13:23.870 --> 00:13:27.630 we switch that circuit at t equals 0. 00:13:27.630 --> 00:13:29.680 At that time, what was the condition 00:13:29.680 --> 00:13:31.570 of the current and charge? 00:13:31.570 --> 00:13:34.650 So Q at 0 we said was 0. 00:13:34.650 --> 00:13:37.090 Q dot at 0 was 0. 00:13:37.090 --> 00:13:41.530 This completely, mathematically defines that problem. 00:13:41.530 --> 00:13:48.270 This is the mathematical description of this problem. 00:13:48.270 --> 00:13:51.740 From now on, I don't have to even remember 00:13:51.740 --> 00:13:55.380 that this has anything to do with physics. 00:13:55.380 --> 00:13:58.790 This is now a problem in mathematics. 00:13:58.790 --> 00:14:02.610 What is the solution of this equation 00:14:02.610 --> 00:14:05.250 satisfying these boundary conditions? 00:14:05.250 --> 00:14:07.380 We'll do that later. 00:14:07.380 --> 00:14:11.150 I'm now going to immediately go to the next problem. 00:14:11.150 --> 00:14:12.990 See in each case, I've done the physics. 00:14:12.990 --> 00:14:15.780 This is the end of the physics in essence. 00:14:15.780 --> 00:14:18.220 From now on, I just have to do mathematics. 00:14:18.220 --> 00:14:22.010 So let me take the next problem, a different one. 00:14:22.010 --> 00:14:23.450 It seems completely different. 00:14:23.450 --> 00:14:27.240 Here we were dealing with charges, currents, circuits, et 00:14:27.240 --> 00:14:31.290 cetera, and now I'm going to look at the physical one. 00:14:31.290 --> 00:14:34.610 I could even bring you a model of that problem. 00:14:34.610 --> 00:14:39.670 What I'm going to consider here is an ideal pendulum, 00:14:39.670 --> 00:14:44.560 simple pendulum, ideal, which is a heavy mass, considered 00:14:44.560 --> 00:14:49.400 to be a point mass, a string, which is considered 00:14:49.400 --> 00:14:54.860 to be always taught but yet massless. 00:14:54.860 --> 00:14:57.970 So it's an ideal pendulum. 00:14:57.970 --> 00:15:01.360 What I have here is an approximation to an ideal one, 00:15:01.360 --> 00:15:05.970 and if I understand what the ideal one does, 00:15:05.970 --> 00:15:11.070 this in reality, will you do something pretty close to it. 00:15:11.070 --> 00:15:14.010 And so what I'm going to consider, 00:15:14.010 --> 00:15:19.290 I am going to be oscillating my hand backwards and forwards 00:15:19.290 --> 00:15:24.790 sinusoidally starting this at some definite time, 00:15:24.790 --> 00:15:28.480 and I'll try to predict the motion of this. 00:15:28.480 --> 00:15:31.770 And notice I can make all sorts of motion out of this. 00:15:31.770 --> 00:15:35.580 If I do it slowly, look, my hand and the ball 00:15:35.580 --> 00:15:37.820 are going in the same direction. 00:15:37.820 --> 00:15:42.720 If I go fast, they're going in opposite direction. 00:15:42.720 --> 00:15:48.470 And the way this is oscillating is different, 00:15:48.470 --> 00:15:52.870 depends on how I am moving my hand backward and forward. 00:15:52.870 --> 00:15:56.450 So let me tell you very precisely what is 00:15:56.450 --> 00:16:02.270 this idealized situation which I am trying to solve and predict 00:16:02.270 --> 00:16:03.310 what will happen. 00:16:03.310 --> 00:16:07.120 So I have this ideal pendulum. 00:16:07.120 --> 00:16:11.270 It has a length, L, mass, m. 00:16:11.270 --> 00:16:12.590 It's in the vertical plane. 00:16:12.590 --> 00:16:15.500 Gravity is down there. 00:16:15.500 --> 00:16:21.750 At one end of the string, I will be moving my hand 00:16:21.750 --> 00:16:25.740 in perfect sinusoidal fashion. 00:16:25.740 --> 00:16:30.340 It will have an amplitude of x0 and angular frequency omega, 00:16:30.340 --> 00:16:32.896 so it's x0 sine omega t. 00:16:32.896 --> 00:16:36.160 I am making the following assumptions, 00:16:36.160 --> 00:16:37.610 that's why it's idealized. 00:16:37.610 --> 00:16:40.730 I'm assuming it's a massless string. 00:16:40.730 --> 00:16:42.960 I'm assuming it's always taught. 00:16:42.960 --> 00:16:48.420 I'm going to assume that it's all making small oscillations 00:16:48.420 --> 00:16:53.400 so that if in any calculation, I can take a sine 00:16:53.400 --> 00:16:55.360 to be an angle to be equal to the angle. 00:16:59.940 --> 00:17:04.210 Furthermore I'm going to assume that there is a bit of drag, 00:17:04.210 --> 00:17:09.010 there is friction, and I will assume 00:17:09.010 --> 00:17:14.210 that the frictional force on this that mass 00:17:14.210 --> 00:17:18.160 is proportional to the velocity, and the constant 00:17:18.160 --> 00:17:24.349 of that velocity is B. Now you ask me, why choose that? 00:17:24.349 --> 00:17:28.300 Well, the simple answer is, I want 00:17:28.300 --> 00:17:31.800 to find this situation when I know I can solve it. 00:17:31.800 --> 00:17:34.710 And I know how to solve this problem 00:17:34.710 --> 00:17:37.160 if it's proportional to velocity. 00:17:37.160 --> 00:17:40.090 If it's something more complicated, 00:17:40.090 --> 00:17:43.160 then the equation may be correct at the end, 00:17:43.160 --> 00:17:47.300 but I will not be able to just like that on the board, 00:17:47.300 --> 00:17:48.290 solve it for you. 00:17:48.290 --> 00:17:52.260 I could do it numerically, use a computer, et cetera, 00:17:52.260 --> 00:17:56.620 but at the same time, I know that if I have something which 00:17:56.620 --> 00:18:00.680 is very close to being what the actual situation is, 00:18:00.680 --> 00:18:05.790 the prediction of the real world compared to this idealized one 00:18:05.790 --> 00:18:07.490 will not be very different. 00:18:07.490 --> 00:18:10.410 Qualitatively, I will get a good understanding 00:18:10.410 --> 00:18:11.220 of what's going on. 00:18:15.680 --> 00:18:19.870 So this is the assumption, and what is the question? 00:18:19.870 --> 00:18:21.690 What do I want to predict? 00:18:21.690 --> 00:18:24.590 I want to now take a very specific situation. 00:18:24.590 --> 00:18:28.980 I want to take a situation that at t equals 0 initially, 00:18:28.980 --> 00:18:36.750 the mass is stationary, and it's exactly below my hand, 00:18:36.750 --> 00:18:39.880 but my hand is not stationary, it's moving. 00:18:39.880 --> 00:18:45.790 So I am moving this backwards and forwards sinusoidally, 00:18:45.790 --> 00:18:46.790 all right? 00:18:46.790 --> 00:18:52.450 And in the instant, when my hand is over the ball, 00:18:52.450 --> 00:18:55.980 I let go of the ball and then I continue moving it. 00:18:55.980 --> 00:19:03.010 So I have specified this exactly what this ball is doing at t 00:19:03.010 --> 00:19:05.760 equals 0, what my hand is doing, and I 00:19:05.760 --> 00:19:12.480 want to predict what will happen at some arbitrary later time. 00:19:12.480 --> 00:19:15.820 So my question is what will happen, 00:19:15.820 --> 00:19:18.860 and this time to make it a little more interesting also, 00:19:18.860 --> 00:19:26.870 I want to know in qualitative terms what kind of motion 00:19:26.870 --> 00:19:31.450 will it have if I move this slowly and quickly 00:19:31.450 --> 00:19:34.850 and check against reality at the end. 00:19:34.850 --> 00:19:36.760 And [? originally ?] I could tell you 00:19:36.760 --> 00:19:39.640 that things are different when I'm doing it slowly. 00:19:39.640 --> 00:19:41.620 As I told you before, if I do this slowly 00:19:41.620 --> 00:19:45.560 like this, my hand and the ball going in the same direction. 00:19:45.560 --> 00:19:49.310 If I go fast, I can make them go in opposite directions. 00:19:49.310 --> 00:19:53.830 All that should come out of this. 00:19:53.830 --> 00:19:57.670 So the part of the question is to qualitatively understand 00:19:57.670 --> 00:20:00.060 this phenomenon. 00:20:00.060 --> 00:20:02.780 So that's my problem number 2. 00:20:02.780 --> 00:20:07.100 The third problem that I'm going to do. 00:20:07.100 --> 00:20:08.930 Sorry. 00:20:08.930 --> 00:20:09.860 I haven't done it. 00:20:09.860 --> 00:20:13.420 That's what I want to do, and I have to do it. 00:20:13.420 --> 00:20:18.660 I wanted to get the exact things over quicker, but no such luck. 00:20:21.229 --> 00:20:22.770 Now, I'm going to solve this problem. 00:20:22.770 --> 00:20:24.360 So back again. 00:20:24.360 --> 00:20:27.370 Look at that diagram and what's the first thing to do. 00:20:27.370 --> 00:20:38.150 Step 1 use the laws of nature to restate this problem 00:20:38.150 --> 00:20:41.270 in terms of math and language of mathematics. 00:20:41.270 --> 00:20:42.740 That's what we have to do. 00:20:42.740 --> 00:20:43.240 All right. 00:20:43.240 --> 00:20:45.430 Let's go a little fast. 00:20:45.430 --> 00:20:48.660 So I have to define some coordinate system 00:20:48.660 --> 00:20:52.460 so I redraw my picture and now I'm 00:20:52.460 --> 00:20:56.640 going to tell you that I'll take some vertical stationary axis 00:20:56.640 --> 00:21:01.630 here, and I'll say it instant t, this mass 00:21:01.630 --> 00:21:03.700 is at some position, y. 00:21:07.090 --> 00:21:11.600 My hand holding here is at position 00:21:11.600 --> 00:21:15.700 is at x0 sine omega t, time t. 00:21:15.700 --> 00:21:16.370 Fine. 00:21:16.370 --> 00:21:18.420 That this string is taught. 00:21:18.420 --> 00:21:21.760 I said the assumption is it's massless 00:21:21.760 --> 00:21:24.900 so therefore it has to be taught. 00:21:24.900 --> 00:21:30.235 That's one of the reasons why we have to make a massless. 00:21:30.235 --> 00:21:32.860 If it wasn't massless, it could start getting kinks and I'll be 00:21:32.860 --> 00:21:34.957 doing this, but the massless one cannot, 00:21:34.957 --> 00:21:37.540 and that's a good question for you to think why that statement 00:21:37.540 --> 00:21:38.340 is correct. 00:21:38.340 --> 00:21:39.515 It's nontrivial. 00:21:39.515 --> 00:21:40.890 All right. 00:21:40.890 --> 00:21:46.230 Then at some instant of time, this angle I'll called theta. 00:21:46.230 --> 00:21:51.030 And now this is a problem in dynamics. 00:21:51.030 --> 00:21:55.490 Basically, I can draw my free body diagram, a force diagram. 00:21:55.490 --> 00:21:57.990 I have this force, mass m. 00:21:57.990 --> 00:21:59.470 What forces are acting on it? 00:21:59.470 --> 00:22:03.510 Well, there is gravity pulling it down, the force mg. 00:22:03.510 --> 00:22:06.170 There is a frictional force which 00:22:06.170 --> 00:22:12.360 we assume will be proportional to the velocity, 00:22:12.360 --> 00:22:13.920 so this is the frictional force which 00:22:13.920 --> 00:22:18.200 will be minus b times the velocity. 00:22:18.200 --> 00:22:21.230 There will be a tension in this string pulling it 00:22:21.230 --> 00:22:22.670 in that direction. 00:22:22.670 --> 00:22:26.760 And the sum of forces, by Newton's laws of motion, 00:22:26.760 --> 00:22:29.440 will give rise to the acceleration. 00:22:29.440 --> 00:22:32.450 The acceleration will be net force 00:22:32.450 --> 00:22:38.280 divided by C inertia, the system which is the mass. 00:22:38.280 --> 00:22:39.270 We have now experience. 00:22:39.270 --> 00:22:41.390 I'm not going to go slowly. 00:22:41.390 --> 00:22:45.380 I'll immediately take the components of forces. 00:22:45.380 --> 00:22:47.710 First let me consider vertical motion. 00:22:47.710 --> 00:22:50.610 I told you the angle is small. 00:22:50.610 --> 00:22:52.800 If I make the angle sufficiently small 00:22:52.800 --> 00:22:57.420 and that ball is oscillating, it is not moving up and down. 00:22:57.420 --> 00:23:00.770 The distance is tiny insignificant in this problem. 00:23:00.770 --> 00:23:03.790 So I can ignore the vertical motion. 00:23:03.790 --> 00:23:07.400 So the vertical acceleration is 0. 00:23:07.400 --> 00:23:11.880 The net vertical force here is the component 00:23:11.880 --> 00:23:14.680 of t in the up direction which is t cosine 00:23:14.680 --> 00:23:18.300 theta minus the gravity down, and that's 00:23:18.300 --> 00:23:20.250 equal to 0, no acceleration. 00:23:20.250 --> 00:23:22.120 Therefore t is mg. 00:23:22.120 --> 00:23:24.080 T is the tension in this string. 00:23:24.080 --> 00:23:25.650 How about horizontal? 00:23:25.650 --> 00:23:29.100 For horizontal motion, I have the horizontal component 00:23:29.100 --> 00:23:32.820 of this tension here which is t sine theta. 00:23:32.820 --> 00:23:36.430 This is theta, so t sine theta is the horizontal component. 00:23:36.430 --> 00:23:41.300 Now, this is moving here, therefore, 00:23:41.300 --> 00:23:44.580 there will be a frictional force proportional to that. 00:23:44.580 --> 00:23:49.520 So that's minus, this is the velocity, times b, 00:23:49.520 --> 00:23:50.860 so this is the frictional force. 00:23:50.860 --> 00:23:53.330 It's always minus, it's always in opposition 00:23:53.330 --> 00:23:55.280 to the actual velocity. 00:23:55.280 --> 00:23:58.800 And that by Newton's law is equal to mass 00:23:58.800 --> 00:24:01.900 times the horizontal acceleration. 00:24:01.900 --> 00:24:06.480 Beware of my symbols. 00:24:06.480 --> 00:24:12.060 This whole problem for me is really in one dimension. 00:24:12.060 --> 00:24:18.320 So I called x the horizontal motion of my hand, 00:24:18.320 --> 00:24:23.620 and I'm calling y the horizontal motion of the mass. 00:24:23.620 --> 00:24:26.400 This has nothing to do with x, y, z, one being 00:24:26.400 --> 00:24:27.930 up and the other horizontal. 00:24:27.930 --> 00:24:29.650 It's all horizontal. 00:24:29.650 --> 00:24:30.440 I had a choice. 00:24:30.440 --> 00:24:33.700 I could have called this x1 and this x2. 00:24:33.700 --> 00:24:35.940 I chose to call them x and y. 00:24:35.940 --> 00:24:38.822 You're smart enough to figure the difference, 00:24:38.822 --> 00:24:40.280 to follow that kind of terminology. 00:24:43.590 --> 00:24:47.660 So now I can rewrite this. 00:24:47.660 --> 00:24:53.570 I know what sine theta is if I look at this diagram. 00:24:53.570 --> 00:25:03.650 Sine theta is simply equal to the distance. 00:25:03.650 --> 00:25:05.740 It's x minus y. 00:25:05.740 --> 00:25:06.875 It's sine theta. 00:25:09.570 --> 00:25:13.290 This angle equals, I should draw this here. 00:25:13.290 --> 00:25:16.180 This angle equals that one, so sine theta 00:25:16.180 --> 00:25:19.750 is this distance divided by that length, 00:25:19.750 --> 00:25:23.740 and this distance is x minus y over l. 00:25:23.740 --> 00:25:27.530 So this is sine theta minus by equals this. 00:25:27.530 --> 00:25:30.140 Quickly, I can play with the algebra, 00:25:30.140 --> 00:25:32.040 and I end up with this equation. 00:25:32.040 --> 00:25:35.610 And lo and behold, Eureka, once again I 00:25:35.610 --> 00:25:40.230 get almost the same equation. 00:25:40.230 --> 00:25:43.860 I purposely chose it so it didn't come out exactly right. 00:25:43.860 --> 00:25:46.480 Before I ended up with a cosine, now I 00:25:46.480 --> 00:25:48.690 have ended up with a sine. 00:25:48.690 --> 00:25:49.470 Intention. 00:25:49.470 --> 00:25:50.190 All right. 00:25:50.190 --> 00:25:59.650 So to show that something which seems different often is not. 00:25:59.650 --> 00:26:04.020 So rewriting this in terms of constant, which I like, 00:26:04.020 --> 00:26:07.030 I do the same as before. 00:26:07.030 --> 00:26:12.310 I define this b over m as gamma, this g over l as omega 00:26:12.310 --> 00:26:17.720 squared, this thing here as f, and I end up 00:26:17.720 --> 00:26:23.150 with this equation where the constants are here 00:26:23.150 --> 00:26:27.940 and the initial conditions are that the position 00:26:27.940 --> 00:26:32.934 of this mass at time 0 is 0, and that the velocity was 0. 00:26:32.934 --> 00:26:35.100 You remember I considered the mass being stationary. 00:26:38.170 --> 00:26:43.010 This is now a set of mathematical equations. 00:26:43.010 --> 00:26:50.210 I have described this problem in terms of mathematics. 00:26:50.210 --> 00:26:53.720 From now on, I can forget that this 00:26:53.720 --> 00:26:57.210 has anything to do with that problem. 00:26:57.210 --> 00:26:59.650 I have to solve now a mathematical problem. 00:27:03.530 --> 00:27:05.360 Finally. 00:27:05.360 --> 00:27:07.070 One more. 00:27:07.070 --> 00:27:10.840 Let me consider one more problem, 00:27:10.840 --> 00:27:18.010 and once we've done all three, at least part 00:27:18.010 --> 00:27:25.820 of step 1 of all three we'll go and solve them all in one go. 00:27:25.820 --> 00:27:33.380 And I repeat what I've done here I've done all the physics. 00:27:33.380 --> 00:27:38.620 This is the part which causes most difficulty, 00:27:38.620 --> 00:27:42.740 needs most understanding of the physical situation, et cetera. 00:27:42.740 --> 00:27:46.720 It may or may not take a very long time. 00:27:46.720 --> 00:27:50.930 The next step is routine mathematics, 00:27:50.930 --> 00:27:54.044 which can be the part which takes you a whole night solving 00:27:54.044 --> 00:27:55.460 because you make algebraic errors, 00:27:55.460 --> 00:27:58.480 and you sweat and you go back, et cetera. 00:27:58.480 --> 00:28:02.080 That's the part you may be cursing when you're doing it, 00:28:02.080 --> 00:28:05.530 but it is routine from then on. 00:28:05.530 --> 00:28:09.890 It doesn't need an understanding of the physics 00:28:09.890 --> 00:28:13.590 of the situation, and what I'm trying to help you understand 00:28:13.590 --> 00:28:17.050 is physics, not mathematics at this stage. 00:28:17.050 --> 00:28:18.690 So let's take another problem, which 00:28:18.690 --> 00:28:22.100 again, seems completely different. 00:28:22.100 --> 00:28:23.270 All right. 00:28:23.270 --> 00:28:26.810 Here it's a seismograph, in other words, 00:28:26.810 --> 00:28:32.470 a device for detecting tremors of the Earth. 00:28:32.470 --> 00:28:34.640 And the following would work. 00:28:34.640 --> 00:28:39.910 Take attached to the floor a spring. 00:28:39.910 --> 00:28:46.470 Put the mass, m, on it, and just look at the mass 00:28:46.470 --> 00:28:48.380 how it's behaving. 00:28:48.380 --> 00:28:53.550 If the Earth starts oscillating, an earthquake, 00:28:53.550 --> 00:28:55.540 the mass will oscillate. 00:28:55.540 --> 00:28:58.130 How much it oscillates will give you 00:28:58.130 --> 00:29:01.630 a measure of how badly the Earth is oscillating. 00:29:01.630 --> 00:29:08.800 So I'm going to here do an idealized seismograph. 00:29:08.800 --> 00:29:17.280 I will model it by a mass, m, sitting on top of a massless, 00:29:17.280 --> 00:29:25.830 ideal spring, which is attached to the floor. 00:29:25.830 --> 00:29:29.190 I will idealize the earthquake and saying 00:29:29.190 --> 00:29:31.510 it's oscillating with a single frequency 00:29:31.510 --> 00:29:32.710 with a single amplitude. 00:29:32.710 --> 00:29:39.092 The floor is oscillating by an amplitude y0 sine omega t. 00:29:39.092 --> 00:29:39.800 It's oscillating. 00:29:42.550 --> 00:29:46.440 So my assumptions are this spring is ideal massless. 00:29:46.440 --> 00:29:49.480 The base, Hooke's law, that's the ideal part. 00:29:49.480 --> 00:29:53.570 I will again assume that there is a damping, which 00:29:53.570 --> 00:29:57.130 will be proportional to the velocity, for the same reasons 00:29:57.130 --> 00:30:00.560 as I took the other one to be proportional to velocity. 00:30:00.560 --> 00:30:03.050 And the question this time will be 00:30:03.050 --> 00:30:07.510 not to predict where that mass is at a given interval of time, 00:30:07.510 --> 00:30:12.330 but the question is, for what values of the mass 00:30:12.330 --> 00:30:14.960 and the spring constant, k, will I 00:30:14.960 --> 00:30:18.360 get the most sensitive instrument? 00:30:18.360 --> 00:30:20.560 So this seems completely different 00:30:20.560 --> 00:30:23.780 and the question seems different the other ones et cetera, 00:30:23.780 --> 00:30:25.880 but as you will see it all boils down 00:30:25.880 --> 00:30:29.170 to one and the same question. 00:30:29.170 --> 00:30:31.750 So how do we do this? 00:30:31.750 --> 00:30:35.860 So now let me come down to step 1, 00:30:35.860 --> 00:30:38.930 understanding the situation, the problem 00:30:38.930 --> 00:30:41.310 in terms of mathematics. 00:30:41.310 --> 00:30:42.890 All right. 00:30:42.890 --> 00:30:47.520 So I've re-drawn here the situation. 00:30:47.520 --> 00:30:49.940 This is my coordinate system of the problem. 00:30:52.930 --> 00:30:56.580 The surface of the floor, I'll assume it's here. 00:30:56.580 --> 00:30:58.080 It's moving. 00:30:58.080 --> 00:30:59.670 It is not a good reference place. 00:30:59.670 --> 00:31:00.570 It's moving. 00:31:00.570 --> 00:31:01.430 All right. 00:31:01.430 --> 00:31:03.590 So it's here. 00:31:03.590 --> 00:31:06.230 The mass, m, is there attached to the spring 00:31:06.230 --> 00:31:09.450 of natural length, l0, spring constant, k. 00:31:09.450 --> 00:31:12.420 There is the mass sitting on it. 00:31:12.420 --> 00:31:17.420 Now, we've got to take a coordinate system which 00:31:17.420 --> 00:31:18.090 isn't moving. 00:31:18.090 --> 00:31:20.430 Got to have an inertial one. 00:31:20.430 --> 00:31:22.690 The surface of the Earth is no good for this. 00:31:22.690 --> 00:31:24.500 It's vibrating. 00:31:24.500 --> 00:31:28.220 So the best thing we have are the fixed stars. 00:31:28.220 --> 00:31:32.750 So let's take the fixed stars as a reference. 00:31:32.750 --> 00:31:37.640 So this is a height in that room, which 00:31:37.640 --> 00:31:43.370 is fixed in position relative to the fixed stars. 00:31:43.370 --> 00:31:47.350 Now, the surface of the Earth can be moving relative to that, 00:31:47.350 --> 00:31:51.280 and I am making the assumption it's moving sinusoidally 00:31:51.280 --> 00:31:52.660 with amplitude y0. 00:31:52.660 --> 00:31:56.990 So this is the floor, position of the floor at time, t. 00:31:56.990 --> 00:32:05.430 The mass relative to this, which I defined as my y equal 0, 00:32:05.430 --> 00:32:07.025 is at position y of m. 00:32:09.970 --> 00:32:15.440 So this now is again a problem in dynamics, 00:32:15.440 --> 00:32:21.320 a mass with forces acting on it, what is the acceleration, 00:32:21.320 --> 00:32:24.280 and what is the solution of the dynamic equation, 00:32:24.280 --> 00:32:25.820 the equation of motion. 00:32:25.820 --> 00:32:30.040 So the force diagram or free body diagram 00:32:30.040 --> 00:32:36.120 is a point mass, m, with a force fg due to gravity on it, 00:32:36.120 --> 00:32:43.050 a force, f, due to the spring acting on it, 00:32:43.050 --> 00:32:45.740 and I forgot to put this. 00:32:45.740 --> 00:32:54.560 Of course, there will be a force here due to friction 00:32:54.560 --> 00:32:55.795 acting on it. 00:32:55.795 --> 00:32:57.670 This is probably not a good place to draw it. 00:32:57.670 --> 00:33:00.480 Let me draw it so it doesn't confuse you. 00:33:00.480 --> 00:33:06.570 I'll draw it next to this, friction. 00:33:06.570 --> 00:33:10.692 So these are in same direction et cetera. 00:33:10.692 --> 00:33:11.460 All right. 00:33:11.460 --> 00:33:16.110 So now let's use Newton's laws of motion, 00:33:16.110 --> 00:33:21.260 f equals ma, and relate the forces 00:33:21.260 --> 00:33:24.310 to the resulting acceleration. 00:33:24.310 --> 00:33:29.560 So first of all what is the net up-going force? 00:33:29.560 --> 00:33:32.980 I'm taking a coordinate system where this is a one dimensional 00:33:32.980 --> 00:33:35.320 problem, so I don't have to use vectors. 00:33:35.320 --> 00:33:37.380 I'll consider just the magnitude. 00:33:37.380 --> 00:33:42.430 My positive force is upwards. 00:33:42.430 --> 00:33:46.800 So at any instant of time, this mass 00:33:46.800 --> 00:33:50.410 will have a force due to the spring and it will depend. 00:33:50.410 --> 00:33:55.790 If this spring is compressed, the force will be up. 00:33:55.790 --> 00:34:00.530 If it's elongated, stretched, the force will be down. 00:34:00.530 --> 00:34:05.780 And the magnitude would be k times the change of length 00:34:05.780 --> 00:34:09.570 from the natural length, so the upward force 00:34:09.570 --> 00:34:14.190 will be the difference between those two subtracted 00:34:14.190 --> 00:34:16.909 from the natural length. 00:34:16.909 --> 00:34:21.449 So k times the natural length minus this quantity, which 00:34:21.449 --> 00:34:23.170 is the difference between this and this, 00:34:23.170 --> 00:34:28.900 which therefore this is how much the string is compressed, 00:34:28.900 --> 00:34:31.139 how much is compressed and is forced upwards. 00:34:31.139 --> 00:34:34.230 If this is a positive number, it's a negative one, 00:34:34.230 --> 00:34:36.580 it's stretched, it's putting it down. 00:34:36.580 --> 00:34:39.810 So this is the force up due to the spring, 00:34:39.810 --> 00:34:42.330 this is the force down due to gravity, 00:34:42.330 --> 00:34:45.723 and this is the force in opposition to the velocity. 00:34:45.723 --> 00:34:47.264 I don't know whether it's up or down. 00:34:47.264 --> 00:34:52.190 It depends which way it's moving due to the friction. 00:34:52.190 --> 00:34:54.239 This is the net force, and that's 00:34:54.239 --> 00:34:56.429 got to be equal to the inertia of the system, which 00:34:56.429 --> 00:34:59.220 is the mass times the acceleration of it 00:34:59.220 --> 00:35:01.570 to the upward direction. 00:35:01.570 --> 00:35:05.500 And I told you this spring is massless, et cetera, 00:35:05.500 --> 00:35:07.850 so I don't have to take into account 00:35:07.850 --> 00:35:09.020 the motion of the spring. 00:35:09.020 --> 00:35:11.395 And that's why I have to take an idealized situation. 00:35:13.910 --> 00:35:15.590 Back now to algebra. 00:35:15.590 --> 00:35:19.810 From here, I manipulate this, and I end up 00:35:19.810 --> 00:35:20.970 with this equation. 00:35:20.970 --> 00:35:25.920 You can do the same, and I've replaced 00:35:25.920 --> 00:35:30.790 for example, the motion of the floor by y0 sine omega t, 00:35:30.790 --> 00:35:33.690 et cetera, from here. 00:35:33.690 --> 00:35:35.850 And at first I say, oh, hell. 00:35:39.390 --> 00:35:45.220 This looks different than the one equation I've seen before. 00:35:45.220 --> 00:35:50.450 But then I don't have to be an Einstein to realize, hold on. 00:35:50.450 --> 00:36:00.140 If I redefine this as my variable, in other words, 00:36:00.140 --> 00:36:08.460 if I take define y as ym plus mg over k minus l0, this quantity, 00:36:08.460 --> 00:36:13.310 the rate of change of y is the rate of change of ym 00:36:13.310 --> 00:36:17.350 because these are constants, and the second derivative of y 00:36:17.350 --> 00:36:20.710 is the second derivative of this, is a constant. 00:36:20.710 --> 00:36:23.360 And therefore, I can rewrite the equation 00:36:23.360 --> 00:36:28.350 as well y double dot plus gamma y dot plus omega 0 00:36:28.350 --> 00:36:33.750 squared y is equal to a constant and sine omega t 00:36:33.750 --> 00:36:40.700 where these constants are written here from this equation 00:36:40.700 --> 00:36:43.590 and I redefine my variable. 00:36:43.590 --> 00:36:47.720 And now, Eureka, I'm back again to the same problem. 00:36:47.720 --> 00:36:50.100 So once, again, what we've done. 00:36:50.100 --> 00:36:54.470 We've taken a physical situation and a physical question 00:36:54.470 --> 00:36:57.210 from our understanding of the laws of nature 00:36:57.210 --> 00:36:59.840 which apply to that situation. 00:36:59.840 --> 00:37:04.580 We have translated it to a problem in mathematics. 00:37:04.580 --> 00:37:06.200 And here it is. 00:37:06.200 --> 00:37:10.110 That's the problem, and I can now, 00:37:10.110 --> 00:37:12.580 at least for the next few minutes, 00:37:12.580 --> 00:37:16.040 forget that this is physics, treat it 00:37:16.040 --> 00:37:18.060 as a problem in mathematics. 00:37:18.060 --> 00:37:20.190 And that's what I'll do it now, but first I've 00:37:20.190 --> 00:37:22.120 got to make some room for myself. 00:37:22.120 --> 00:37:25.570 I have nowhere to write, so I have to erase some things, 00:37:25.570 --> 00:37:29.440 and we'll continue in a few minutes. 00:37:29.440 --> 00:37:30.270 All right. 00:37:30.270 --> 00:37:34.720 We've erased the boards, and we can therefore continue. 00:37:34.720 --> 00:37:37.450 Let me remind you what we have shown 00:37:37.450 --> 00:37:48.140 is we've taken three completely different systems, described 00:37:48.140 --> 00:37:51.440 them in terms of mathematics. 00:37:51.440 --> 00:37:53.800 The three systems with an electrical circuit, 00:37:53.800 --> 00:37:56.600 an RLC circuit. 00:37:56.600 --> 00:38:03.240 We took a pendulum, driven on a pendulum, and a seismograph. 00:38:03.240 --> 00:38:11.175 Each one of these ended up with almost identical equation. 00:38:11.175 --> 00:38:15.110 In fact, the only difference is one of them 00:38:15.110 --> 00:38:20.380 ended up with a cosine here, the other two with a sine. 00:38:20.380 --> 00:38:24.520 I will comment about that in a second. 00:38:24.520 --> 00:38:28.720 We are now, forgetting about the physics, 00:38:28.720 --> 00:38:31.690 we are now in the world of mathematics. 00:38:31.690 --> 00:38:36.110 These equations of motion need to be 00:38:36.110 --> 00:38:38.460 solved if we are to predict what will 00:38:38.460 --> 00:38:41.270 happen in those situations. 00:38:41.270 --> 00:38:46.300 Now, there are many ways of solving differential equations. 00:38:46.300 --> 00:38:51.810 It is not my job to teach you how 00:38:51.810 --> 00:38:55.360 to solve differential equations. 00:38:55.360 --> 00:38:58.470 For an example, you can go to the lectures of Professor 00:38:58.470 --> 00:39:02.350 Walter Lewin, and he does illustrate 00:39:02.350 --> 00:39:05.790 how one can solve these using complex amplitudes. 00:39:08.740 --> 00:39:14.060 I am satisfied if I find the solutions. 00:39:14.060 --> 00:39:15.530 I don't care how. 00:39:15.530 --> 00:39:17.880 I can take them from what he did, 00:39:17.880 --> 00:39:21.190 I can take them from a book of mathematics, et cetera, 00:39:21.190 --> 00:39:25.070 and what you will find is that if you 00:39:25.070 --> 00:39:29.490 have a generic second ordered differential 00:39:29.490 --> 00:39:35.960 equation of this form, it has solution 00:39:35.960 --> 00:39:40.030 which looks like this, the most general solution that 00:39:40.030 --> 00:39:41.800 looks like that. 00:39:41.800 --> 00:39:46.000 There is no other solution in the universe which 00:39:46.000 --> 00:39:48.690 is not represented by this equation. 00:39:48.690 --> 00:39:51.250 You can manipulate it, make it look different, 00:39:51.250 --> 00:39:53.720 but it will boil down to this equation. 00:39:53.720 --> 00:39:55.500 How do I know that? 00:39:55.500 --> 00:39:57.250 Because of something I mentioned earlier, 00:39:57.250 --> 00:40:00.180 the so-called uniqueness theorem. 00:40:00.180 --> 00:40:05.240 This is a second order differential equation. 00:40:05.240 --> 00:40:09.110 This equation satisfies it. 00:40:09.110 --> 00:40:11.740 Try it, take it, differentiate it twice this, 00:40:11.740 --> 00:40:16.590 et cetera, u [? dissatisfied. ?] Furthermore, 00:40:16.590 --> 00:40:19.030 it has two arbitrary constants. 00:40:22.650 --> 00:40:27.530 This C can be anything and this phi can be anything. 00:40:27.530 --> 00:40:29.210 They are arbitrary. 00:40:29.210 --> 00:40:34.560 Everything else in this equation is completely determined. 00:40:34.560 --> 00:40:41.490 For example, a of omega depends on omega. 00:40:41.490 --> 00:40:48.450 This amplitude, a, is equal to f divided 00:40:48.450 --> 00:40:52.460 by the square root of omega 0 squared minus omega squared 00:40:52.460 --> 00:40:56.760 0 squared plus gamma omega 0 squared. 00:40:56.760 --> 00:41:02.310 All these constants-- omega, gamma, f, omega 0-- 00:41:02.310 --> 00:41:04.285 are known in each of the problems. 00:41:07.270 --> 00:41:10.630 Let me just as an example take the last problem we 00:41:10.630 --> 00:41:15.310 did When we derived the equation of motion, 00:41:15.310 --> 00:41:20.090 we furthermore knew exactly what each one of those constants 00:41:20.090 --> 00:41:20.630 meant. 00:41:20.630 --> 00:41:24.140 They are known, so I don't need to solve for them. 00:41:24.140 --> 00:41:28.500 So a of omega is this quantity. 00:41:28.500 --> 00:41:33.330 The delta, the phase there is, again, a known quantity. 00:41:33.330 --> 00:41:35.800 The time of delta is this. 00:41:35.800 --> 00:41:38.965 It is only under these conditions 00:41:38.965 --> 00:41:43.090 that this satisfies that equation. 00:41:43.090 --> 00:41:46.690 And find the omega prime here also is a known quantity. 00:41:46.690 --> 00:41:48.400 Nothing here is unknown. 00:41:48.400 --> 00:41:51.660 The only unknowns are the C, and the phi, 00:41:51.660 --> 00:41:54.220 and they are arbitrary constants. 00:41:54.220 --> 00:41:57.790 This works for any value of C and any value of phi. 00:42:02.340 --> 00:42:05.760 Now, one more comment. 00:42:05.760 --> 00:42:08.080 These look equations. 00:42:08.080 --> 00:42:11.000 One looks like a cosine theta sine. 00:42:11.000 --> 00:42:13.310 If you stop and think for a second, 00:42:13.310 --> 00:42:16.670 they're actually one and the same equation 00:42:16.670 --> 00:42:20.840 where you've redefined what is t 0. 00:42:20.840 --> 00:42:26.015 So by just changing the phase of this, you'll get to this. 00:42:26.015 --> 00:42:28.040 You've changed it by 90 degrees. 00:42:28.040 --> 00:42:31.610 Just redefine what t0 is, your clock, 00:42:31.610 --> 00:42:34.780 and you'll get from this to that. 00:42:34.780 --> 00:42:39.150 So if this equation I've written here 00:42:39.150 --> 00:42:44.380 is a solution to the cosine part, 00:42:44.380 --> 00:42:46.850 if my driver has this phase, in other words, 00:42:46.850 --> 00:42:51.400 sine omega t, then the answer will be displaced with respect 00:42:51.400 --> 00:42:53.290 to this one by that same thing. 00:42:53.290 --> 00:42:56.740 Instead of having a cosine here, you have a sign. 00:42:56.740 --> 00:43:01.570 So from my perspective, these two are identical equations, 00:43:01.570 --> 00:43:06.140 and the solution for them is identical. 00:43:06.140 --> 00:43:10.420 It's also worth looking at what the mathematics of this looks 00:43:10.420 --> 00:43:11.070 like. 00:43:11.070 --> 00:43:15.840 What this looks like is the solution to these equations 00:43:15.840 --> 00:43:22.460 is some function with a certain amplitude, which is oscillatory 00:43:22.460 --> 00:43:28.350 with the same frequency as the driver. 00:43:28.350 --> 00:43:30.760 So this is just an oscillatory function. 00:43:30.760 --> 00:43:32.870 It's got a different phase to that. 00:43:32.870 --> 00:43:35.330 They lag will lead it, different phase, 00:43:35.330 --> 00:43:39.740 but it's just constant amplitude oscillatory function. 00:43:39.740 --> 00:43:45.270 Added to it is another oscillatory function 00:43:45.270 --> 00:43:49.870 with a slightly different frequency, 00:43:49.870 --> 00:43:55.920 but which with time decays, because of this e to the minus 00:43:55.920 --> 00:43:57.910 gamma over 2t. 00:43:57.910 --> 00:44:03.240 And by the way, one more thing I should just emphasize. 00:44:03.240 --> 00:44:07.400 In writing this solution, I've made the assumption 00:44:07.400 --> 00:44:09.510 that this oscillator does oscil. 00:44:09.510 --> 00:44:12.775 You're driving something which oscillates and is 00:44:12.775 --> 00:44:18.230 an overdamped system in which case 00:44:18.230 --> 00:44:22.560 this term here would be different and not interesting 00:44:22.560 --> 00:44:25.730 when normally you wouldn't drive things and be interested in how 00:44:25.730 --> 00:44:30.540 it responds if it's incredibly damped system. 00:44:30.540 --> 00:44:33.820 So I am making the assumption that this 00:44:33.820 --> 00:44:38.110 is an underdamped system, which means that this quantity here 00:44:38.110 --> 00:44:42.420 is greater than this quantity here. 00:44:42.420 --> 00:44:46.000 And if you're having difficulty with this part, 00:44:46.000 --> 00:44:54.080 I urge you to review what happens with just not driven 00:44:54.080 --> 00:44:57.860 oscillators under conditions of being underdamped 00:44:57.860 --> 00:45:00.875 or overdamped. 00:45:00.875 --> 00:45:01.875 That's the same physics. 00:45:04.860 --> 00:45:08.000 So we found the most general solution. 00:45:08.000 --> 00:45:12.100 This solution must apply to every problem we've done. 00:45:12.100 --> 00:45:15.150 So I'll now take one of them. 00:45:15.150 --> 00:45:19.130 For example, let's consider this problem. 00:45:19.130 --> 00:45:19.950 We did this. 00:45:19.950 --> 00:45:22.170 We have an RLC circuit. 00:45:22.170 --> 00:45:24.890 They are an equation of motion. 00:45:24.890 --> 00:45:28.300 From here, I can immediately write 00:45:28.300 --> 00:45:30.930 what is the solution of this. 00:45:30.930 --> 00:45:37.060 As a function of time, the charge Q 00:45:37.060 --> 00:45:47.650 will be equal to-- just look at that-- a of omega times 00:45:47.650 --> 00:45:58.990 cosine omega t minus delta-- I'm just copying from over there-- 00:45:58.990 --> 00:46:09.640 plus some arbitrary constant e to the minus gamma over 2t. 00:46:09.640 --> 00:46:16.710 And here we have a cosine, so I took the cosine here, and here 00:46:16.710 --> 00:46:26.850 is cosine of omega prime t plus this arbitrary phase phi. 00:46:26.850 --> 00:46:31.460 So this is a general equation, which 00:46:31.460 --> 00:46:38.820 satisfies that one at all times. 00:46:38.820 --> 00:46:41.310 And I'm repeating myself. 00:46:41.310 --> 00:46:51.450 This, you know, it will be given by A of omega 00:46:51.450 --> 00:46:55.030 will be equal to f, and what is f? 00:46:55.030 --> 00:47:02.950 f is v0 over L-- I just read it off there-- divided 00:47:02.950 --> 00:47:10.470 by square root of omega 0 squared, 00:47:10.470 --> 00:47:22.490 which is 1 over LC minus omega squared 00:47:22.490 --> 00:47:28.790 That's just the frequencies with which the voltage is driving 00:47:28.790 --> 00:47:37.260 there all squared plus gamma omega. 00:47:37.260 --> 00:47:39.277 Gamma is R over L omega all squared. 00:47:44.840 --> 00:47:52.530 And time delta is equal to gamma omega. 00:47:52.530 --> 00:47:59.780 Gamma is R over L divided by omega 0 00:47:59.780 --> 00:48:06.070 squared, which is 1 over LC minus omega squared. 00:48:11.560 --> 00:48:17.750 So far we know everything except the C and the phi. 00:48:17.750 --> 00:48:22.710 The question actually was, I've now erased, 00:48:22.710 --> 00:48:27.030 it but the question was what is the current flowing? 00:48:27.030 --> 00:48:33.780 And the current flowing I of t, is 00:48:33.780 --> 00:48:39.560 equal to dq dt with the way we define 00:48:39.560 --> 00:48:48.100 the signs with a part plus here, which is equal to Q dot, which 00:48:48.100 --> 00:48:52.520 I can get from this by differentiating this equation, 00:48:52.520 --> 00:48:57.630 and I get differentiate this minus omega 00:48:57.630 --> 00:49:08.930 A of omega sine of omega t minus delta. 00:49:08.930 --> 00:49:13.720 Go to differentiate this, I get minus gamma 00:49:13.720 --> 00:49:25.220 over 2 Ce to the minus gamma over 2t times cosine omega 00:49:25.220 --> 00:49:29.090 prime t plus phi. 00:49:29.090 --> 00:49:31.350 So I've differentiated this times that. 00:49:31.350 --> 00:49:34.220 Now I've got to take this differentiated 00:49:34.220 --> 00:49:41.570 by that, which minus Ce e to the minus gamma over 2t times 00:49:41.570 --> 00:49:52.291 omega prime times sine omega prime t plus phi. 00:49:52.291 --> 00:49:52.790 Sorry. 00:49:52.790 --> 00:49:55.430 That's so. 00:49:55.430 --> 00:50:00.660 So and again, we know every quantity 00:50:00.660 --> 00:50:09.030 here except for C and phi. 00:50:09.030 --> 00:50:10.065 Those are unknowns. 00:50:14.340 --> 00:50:17.640 So this would be the answer to my question 00:50:17.640 --> 00:50:20.970 if I knew C and phi, then everything is known. 00:50:20.970 --> 00:50:22.960 Then I could predict exactly what 00:50:22.960 --> 00:50:25.320 would happen with any instant of time. 00:50:25.320 --> 00:50:27.390 How do I find those two? 00:50:27.390 --> 00:50:31.190 This is where the boundary conditions come in. 00:50:31.190 --> 00:50:33.820 That's why you need at least two boundary 00:50:33.820 --> 00:50:36.850 conditions or initial conditions, 00:50:36.850 --> 00:50:41.080 two facts about this system to be able to determine 00:50:41.080 --> 00:50:43.300 those constants, and we're always 00:50:43.300 --> 00:50:47.610 in the physical situation of this kind, 00:50:47.610 --> 00:50:49.340 you will have two bits of information. 00:50:49.340 --> 00:50:53.490 For example, in this case, I told you 00:50:53.490 --> 00:50:55.650 that we will assume that t equals 00:50:55.650 --> 00:50:59.340 0 the charge and the capacitance is 0 00:50:59.340 --> 00:51:01.972 and the current in the circuit is 0. 00:51:01.972 --> 00:51:04.180 I could have told you something else at the beginning 00:51:04.180 --> 00:51:06.350 and the answer would be different. 00:51:06.350 --> 00:51:07.170 All right. 00:51:07.170 --> 00:51:09.310 And the thing that would be different 00:51:09.310 --> 00:51:11.050 would be those constants. 00:51:11.050 --> 00:51:13.270 So how do we do it? 00:51:13.270 --> 00:51:22.520 Simply by taking this equation and writing it at the time t 00:51:22.520 --> 00:51:23.380 equals 0. 00:51:23.380 --> 00:51:31.510 So at t equals 0, we know what Q of 0 is, 00:51:31.510 --> 00:51:35.170 and we know what Q dot of 0 is. 00:51:39.290 --> 00:51:45.340 So I take this equation, set t to 0, 00:51:45.340 --> 00:51:49.270 and write down this equal to 0. 00:51:53.000 --> 00:51:53.500 Sorry. 00:51:53.500 --> 00:51:55.965 I take this equation, the Q. This one set 00:51:55.965 --> 00:52:01.740 t to 0 and write 0 equals that with t equals 0. 00:52:01.740 --> 00:52:04.670 I write another equation from here. 00:52:04.670 --> 00:52:11.790 At t equals 0, Q dot is 0, so I take this one, set t to 0, 00:52:11.790 --> 00:52:13.510 and write another equation. 00:52:13.510 --> 00:52:17.680 I will end up with two algebraic equations 00:52:17.680 --> 00:52:21.430 with two unknowns which you can solve. 00:52:21.430 --> 00:52:26.660 Now remember I told you, the hard part conceptually 00:52:26.660 --> 00:52:32.280 is getting the equations of motion right. 00:52:32.280 --> 00:52:38.710 But the slog, the sweat, and a hard labor 00:52:38.710 --> 00:52:44.320 is getting this solution right, and I'm not 00:52:44.320 --> 00:52:48.760 going to go here now and embarrass myself making error 00:52:48.760 --> 00:52:53.110 after algebraic error solving those equations. 00:52:53.110 --> 00:52:55.920 I know you can solve them if you keep your cool 00:52:55.920 --> 00:52:59.020 and follow all the constants, et cetera. 00:52:59.020 --> 00:53:04.070 You take these two algebraic equations, solve them, 00:53:04.070 --> 00:53:10.340 you come out with a value for C and a value for the phase phi, 00:53:10.340 --> 00:53:12.250 and then I have the final answer. 00:53:12.250 --> 00:53:19.350 I then have this equation I of t and Q of t for any T. 00:53:19.350 --> 00:53:24.310 And so I can predict at any time with this will do. 00:53:24.310 --> 00:53:27.200 End of story. 00:53:27.200 --> 00:53:33.204 Now, let me take another case and to look at this one here, 00:53:33.204 --> 00:53:34.370 and I'll go a little faster. 00:53:38.110 --> 00:53:45.380 This is the mathematical description of the pendulum. 00:53:45.380 --> 00:53:49.880 So again, I'll take this, initially I'll 00:53:49.880 --> 00:53:59.110 do the same as I did before, so I will write that y at time t 00:53:59.110 --> 00:54:03.650 is equal to-- and I have to once again just follow 00:54:03.650 --> 00:54:12.320 the thing-- it's f, which is gx over L divided 00:54:12.320 --> 00:54:21.140 by the square root of omega squared, 00:54:21.140 --> 00:54:35.160 which is g over L minus omega squared all squared plus gamma, 00:54:35.160 --> 00:54:40.460 which is b over m this time gamma omega all squared. 00:54:46.100 --> 00:54:49.010 So times. 00:54:49.010 --> 00:54:52.210 Now, here I have a sign so it's going 00:54:52.210 --> 00:55:08.650 to be the sine omega t minus delta plus-- 00:55:08.650 --> 00:55:16.067 and to save time, not rewrite it, the full thing, 00:55:16.067 --> 00:55:17.400 but I'll start at the beginning. 00:55:17.400 --> 00:55:23.190 There's the Ce to the minus gamma over 2t 00:55:23.190 --> 00:55:30.220 here at cosine et cetera. 00:55:30.220 --> 00:55:36.550 So what we see here as before, we 00:55:36.550 --> 00:55:42.920 can get what this mass will do as a function of time, 00:55:42.920 --> 00:55:48.600 and as before, I don't want to just go around and around 00:55:48.600 --> 00:55:50.370 repeating the thing is. 00:55:50.370 --> 00:55:55.590 In order to solve, everything here is known except the C 00:55:55.590 --> 00:56:02.070 and this phase here-- maybe I should at least write that in. 00:56:02.070 --> 00:56:07.780 That is omega prime t plus phi. 00:56:07.780 --> 00:56:11.350 And in principle, this and that you 00:56:11.350 --> 00:56:16.570 can get, again, by using the initial conditions. 00:56:16.570 --> 00:56:17.930 I don't want to focus on that. 00:56:17.930 --> 00:56:21.630 In this case, I want to focus what the problem said. 00:56:21.630 --> 00:56:25.410 What it said it was, discuss what 00:56:25.410 --> 00:56:29.030 you would expect the motion to be for that. 00:56:29.030 --> 00:56:32.030 In other words, when we took this special, 00:56:32.030 --> 00:56:40.530 we took this and took in given initial conditions 00:56:40.530 --> 00:56:42.720 and let's go. 00:56:42.720 --> 00:56:46.280 This will describe what goes on. 00:56:46.280 --> 00:56:51.850 And let's look at this solution. 00:56:51.850 --> 00:56:58.660 What we see is that there are two terms here as before. 00:56:58.660 --> 00:57:05.890 This is an oscillation of the same frequency as a driver. 00:57:05.890 --> 00:57:09.850 So other words, that term is something 00:57:09.850 --> 00:57:17.050 which has this moving with the same frequency as my hand. 00:57:17.050 --> 00:57:21.910 This is a slightly different frequency, 00:57:21.910 --> 00:57:26.200 which is exponentially decaying. 00:57:26.200 --> 00:57:30.090 When you add two oscillating functions, 00:57:30.090 --> 00:57:33.150 you've learned from Professor Walter Lewin's class, 00:57:33.150 --> 00:57:35.130 you get beats. 00:57:35.130 --> 00:57:42.820 So at first when this is large, these two wave motions, 00:57:42.820 --> 00:57:44.720 these oscillatory motions will beat 00:57:44.720 --> 00:57:47.365 minutes one against the other and you 00:57:47.365 --> 00:57:50.450 will get some kind of a crazy motion. 00:57:50.450 --> 00:57:52.850 You can't make sense of it. 00:57:52.850 --> 00:57:57.840 As a function of time, as t becomes bigger and bigger, 00:57:57.840 --> 00:58:03.410 this exponential kills this term, and after a long time, 00:58:03.410 --> 00:58:07.670 you'll end up with only this term, which 00:58:07.670 --> 00:58:09.970 is simple sinusoidal motion. 00:58:09.970 --> 00:58:12.550 Try it for yourself. 00:58:12.550 --> 00:58:15.900 If I start any old way and I then 00:58:15.900 --> 00:58:19.000 move with a given frequency, you'll 00:58:19.000 --> 00:58:25.340 find this has a constant frequency, some amplitude given 00:58:25.340 --> 00:58:29.860 by this, and it continues doing this forever 00:58:29.860 --> 00:58:32.740 consistent with it. 00:58:32.740 --> 00:58:35.810 But there is another interesting feature, 00:58:35.810 --> 00:58:38.350 which I would like you to focus on. 00:58:38.350 --> 00:58:44.630 If the friction is very small, like this, this 00:58:44.630 --> 00:58:49.980 will have very little damping, after a long time, 00:58:49.980 --> 00:58:54.470 this has gone and this term I can neglect. 00:58:54.470 --> 00:59:01.040 And what I see here is that this, 00:59:01.040 --> 00:59:07.050 the amplitude of the oscillations of this not only 00:59:07.050 --> 00:59:12.680 depends on the amplitude-- and again, I notice the time here; 00:59:12.680 --> 00:59:18.150 this is the node quantity here-- not only depends 00:59:18.150 --> 00:59:24.140 on the amplitude of how far my head is moving, 00:59:24.140 --> 00:59:28.900 but it depends critically on the difference between g 00:59:28.900 --> 00:59:31.920 over l and omega squared. 00:59:31.920 --> 00:59:37.850 This, if you remember from the simple pendulum-- that's 00:59:37.850 --> 00:59:44.150 this omega 0 squared-- gives the frequency 00:59:44.150 --> 00:59:49.400 of oscillation of a free pendulum of this length. 00:59:49.400 --> 00:59:56.710 So what happens depends critically how different 00:59:56.710 --> 00:59:59.950 and of which magnitude is between the frequency 00:59:59.950 --> 01:00:02.840 of oscillation of my hand compared 01:00:02.840 --> 01:00:06.820 to the natural frequency of oscillation. 01:00:06.820 --> 01:00:11.110 If-- and here I won't do the algebra. 01:00:11.110 --> 01:00:13.590 I'll let you do this as an exercise-- play 01:00:13.590 --> 01:00:19.280 for yourself changing the value of this making 01:00:19.280 --> 01:00:22.030 it bigger than that or smaller than that. 01:00:24.710 --> 01:00:35.250 And what you will find is that the motion of that and sorry, 01:00:35.250 --> 01:00:37.180 there's one more thing I should have 01:00:37.180 --> 01:00:40.940 said I didn't, and let me just say it now. 01:00:40.940 --> 01:00:44.590 There is also this question of this phase delta, 01:00:44.590 --> 01:00:53.920 tan delta is equal to gamma omega 01:00:53.920 --> 01:01:01.900 over omega 0 squared minus omega squared. 01:01:01.900 --> 01:01:09.350 So the sine of delta depends on whether it's bigger or smaller. 01:01:09.350 --> 01:01:13.670 Play for yourself changing this value. 01:01:13.670 --> 01:01:16.960 See what happens to the amplitude here 01:01:16.960 --> 01:01:21.730 and what happens to this phase, and what you'll see 01:01:21.730 --> 01:01:28.880 is that if I go slowly, the phase is such 01:01:28.880 --> 01:01:31.830 that the two go in phase together 01:01:31.830 --> 01:01:36.570 if my oscillating frequency is lower 01:01:36.570 --> 01:01:40.020 than the natural frequency of oscillation of this, 01:01:40.020 --> 01:01:44.480 while if I go faster, in other words, higher frequency, 01:01:44.480 --> 01:01:49.790 this becomes a negative and the phase gets out 01:01:49.790 --> 01:01:53.610 of phase by 180 degrees, the two. 01:01:53.610 --> 01:01:57.710 Furthermore, if you look at this amplitude, 01:01:57.710 --> 01:02:01.290 you can completely understand it in the following way. 01:02:06.130 --> 01:02:10.490 Imagine that I had this string and I move it slowly. 01:02:15.100 --> 01:02:18.910 I'm moving this slowly, and what you'll see 01:02:18.910 --> 01:02:29.750 is it's going in phase with that, and the angle of this 01:02:29.750 --> 01:02:34.890 will be such that this string appears 01:02:34.890 --> 01:02:40.180 to be the continuation of a single string, which 01:02:40.180 --> 01:02:41.275 has a longer length. 01:02:44.060 --> 01:02:48.760 And the string with the longer length has a lower frequency. 01:02:48.760 --> 01:02:53.410 So if I move here with the lower frequency 01:02:53.410 --> 01:02:56.360 than the natural frequency, this will 01:02:56.360 --> 01:03:02.810 appear to be part of a string which has just the right length 01:03:02.810 --> 01:03:07.970 to give me the frequency of my hand. 01:03:07.970 --> 01:03:12.010 While if I move with the higher frequency, 01:03:12.010 --> 01:03:16.700 what the string will do is do this. 01:03:16.700 --> 01:03:19.580 Here is my hand moving backwards and forwards. 01:03:19.580 --> 01:03:21.290 There is the mass, and the mass is 01:03:21.290 --> 01:03:23.060 moving backwards and forwards. 01:03:23.060 --> 01:03:24.210 All right. 01:03:24.210 --> 01:03:30.200 This string is oscillating as if it 01:03:30.200 --> 01:03:35.120 was a free string of a shorter length. 01:03:35.120 --> 01:03:39.880 I urge you as an exercise, to play with the numbers 01:03:39.880 --> 01:03:45.540 here and consider these two cases. 01:03:45.540 --> 01:03:48.790 It is a beautiful, educational tool 01:03:48.790 --> 01:03:53.720 to understand why if you have a driven system, 01:03:53.720 --> 01:03:57.740 response can be in phase or out of phase 01:03:57.740 --> 01:04:01.430 depending on the relative frequency you're driving 01:04:01.430 --> 01:04:04.380 the system compared to the natural frequency 01:04:04.380 --> 01:04:07.170 of oscillation of the oscillator. 01:04:07.170 --> 01:04:11.810 And this example brings everything out. 01:04:11.810 --> 01:04:15.010 Take this example, consider very little 01:04:15.010 --> 01:04:19.120 damping so that you can do the calculations easily. 01:04:19.120 --> 01:04:21.170 See what's the relative amplitude 01:04:21.170 --> 01:04:25.010 of the driver compared to the response 01:04:25.010 --> 01:04:28.350 and what's the relative phase of the driver compared 01:04:28.350 --> 01:04:32.110 to the response, and all this characteristics you'll 01:04:32.110 --> 01:04:35.720 see which you can play for yourselves with a little model. 01:04:35.720 --> 01:04:38.490 I really urge you to do that. 01:04:38.490 --> 01:04:43.570 Finally today, I will just comment. 01:04:43.570 --> 01:04:46.410 You see we're now in a world of repeating. 01:04:46.410 --> 01:04:49.260 We're doing the same thing over and over again. 01:04:49.260 --> 01:05:03.680 I would just like to comment about the seismograph. 01:05:03.680 --> 01:05:11.580 Once again, this is an equation of this form. 01:05:11.580 --> 01:05:13.620 It's the sine version. 01:05:13.620 --> 01:05:18.160 So this is the solution to it with a sine, 01:05:18.160 --> 01:05:23.510 and the question is, in this case was, 01:05:23.510 --> 01:05:30.290 what the value of k over m makes it most sensitive? 01:05:30.290 --> 01:05:34.130 That's equivalent to asking the question, 01:05:34.130 --> 01:05:39.460 after a long time-- this earthquake 01:05:39.460 --> 01:05:45.930 has lasted a long time-- this so-called transient, 01:05:45.930 --> 01:05:48.770 this term dies away. 01:05:48.770 --> 01:05:50.290 So this we can forget. 01:05:50.290 --> 01:05:56.360 This is the response of the mass to the earthquake which 01:05:56.360 --> 01:06:05.540 we assumed it's a uniform frequency, sinusoidal driver. 01:06:05.540 --> 01:06:10.640 And so the question is, what is the amplitude of the response? 01:06:10.640 --> 01:06:14.080 How big is this term? 01:06:14.080 --> 01:06:19.920 This term we know how big it is, and if you 01:06:19.920 --> 01:06:26.080 look at this function, you see that this amplitude depends 01:06:26.080 --> 01:06:27.570 on omega, on the driver. 01:06:30.660 --> 01:06:38.860 Now let's assume the damping is not very big. 01:06:38.860 --> 01:06:41.790 This will be small compared to that. 01:06:41.790 --> 01:06:45.600 And so if you want a big response, 01:06:45.600 --> 01:06:49.330 you want this quantity to be very small 01:06:49.330 --> 01:06:53.680 so that f divided by this is a big number. 01:06:53.680 --> 01:06:56.610 When will this be small? 01:06:56.610 --> 01:06:59.500 When these two are almost equal. 01:07:02.180 --> 01:07:07.090 So what you want, if you want to have an instrument which 01:07:07.090 --> 01:07:16.930 is very sensitive, you want to study so the dominant frequency 01:07:16.930 --> 01:07:20.630 of oscillations of earthquakes in your area 01:07:20.630 --> 01:07:27.870 and adjust the natural frequency of oscillation close to that. 01:07:27.870 --> 01:07:32.470 And the natural frequency is of course given by k over m. 01:07:32.470 --> 01:07:38.490 So what matters is, it doesn't matter what m is, what k is, 01:07:38.490 --> 01:07:41.870 but what's the ratio of k over m? 01:07:41.870 --> 01:07:46.910 You want that to be close to the natural frequency, 01:07:46.910 --> 01:07:49.920 that the natural frequency of this oscillator, that value 01:07:49.920 --> 01:07:54.220 k over m, to be close to the frequency of the oscillations 01:07:54.220 --> 01:07:56.390 of the earth in the earthquake. 01:08:01.030 --> 01:08:10.500 If this driver really had a definite frequency, 01:08:10.500 --> 01:08:15.060 I could solve this problem to find the maximum. 01:08:15.060 --> 01:08:18.960 If I plotted this as a function of omega, 01:08:18.960 --> 01:08:23.439 I would get a curve something like this. 01:08:23.439 --> 01:08:28.260 This is a function of omega, the amplitude, 01:08:28.260 --> 01:08:31.810 the response of that oscillator. 01:08:31.810 --> 01:08:38.609 if you do that, you'll find that if the natural frequency 01:08:38.609 --> 01:08:55.040 of oscillation is over here, the peak 01:08:55.040 --> 01:08:58.000 is slightly at the lower frequency 01:08:58.000 --> 01:09:01.199 than the natural frequency of oscillation omega 0. 01:09:03.750 --> 01:09:09.819 And so you could do this exactly by plotting it out 01:09:09.819 --> 01:09:16.330 to find out what is the optimum value of the frequency 01:09:16.330 --> 01:09:20.620 that you want to adjust so that you get the best response. 01:09:20.620 --> 01:09:23.910 So much for driven oscillators. 01:09:23.910 --> 01:09:30.130 We will now move to couple oscillators, 01:09:30.130 --> 01:09:32.040 and that we'll do next time, some problems. 01:09:32.040 --> 01:09:33.490 Thank you.