WEBVTT 00:00:00.060 --> 00:00:02.430 The following content is provided under a Creative 00:00:02.430 --> 00:00:03.800 Commons license. 00:00:03.800 --> 00:00:06.050 Your support will help MIT OpenCourseWare 00:00:06.050 --> 00:00:10.140 continue to offer high quality educational resources for free. 00:00:10.140 --> 00:00:12.690 To make a donation or to view additional materials 00:00:12.690 --> 00:00:16.590 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.590 --> 00:00:17.260 at ocw.mit.edu. 00:00:21.380 --> 00:00:24.680 PROFESSOR: So Professor Gossard gave the lecture last week. 00:00:24.680 --> 00:00:26.990 I'm going to pick up where he left off. 00:00:26.990 --> 00:00:31.260 But let's talk about the concept questions from the homework 00:00:31.260 --> 00:00:33.200 you've been working on. 00:00:33.200 --> 00:00:35.850 So the first one is our cart. 00:00:35.850 --> 00:00:38.520 You'd expect to be able eliminate the terms involving 00:00:38.520 --> 00:00:41.070 gravity in the equations of motion 00:00:41.070 --> 00:00:43.090 by choosing coordinates with respect 00:00:43.090 --> 00:00:45.180 to the static equilibrium position. 00:00:45.180 --> 00:00:47.060 So we've talked about that. 00:00:47.060 --> 00:00:53.000 And with this one does the restoring force 00:00:53.000 --> 00:00:56.200 on the pendulum, what makes it come back 00:00:56.200 --> 00:01:01.510 to zero after it's a damped out and hanging straight down. 00:01:01.510 --> 00:01:02.260 AUDIENCE: Gravity. 00:01:02.260 --> 00:01:02.926 PROFESSOR: What? 00:01:02.926 --> 00:01:04.349 Gravity does. 00:01:04.349 --> 00:01:09.760 And that gravity term varies with the torque 00:01:09.760 --> 00:01:17.780 that the gravity puts around the pivot is MgL over 2 sine theta. 00:01:17.780 --> 00:01:21.930 So the gravity term is involved with the motion variable theta. 00:01:21.930 --> 00:01:24.080 So in this case, gravity is going 00:01:24.080 --> 00:01:27.270 to be involved in the natural frequency 00:01:27.270 --> 00:01:30.420 and in the equations of motion, no matter what, 00:01:30.420 --> 00:01:36.090 So you will not be able to eliminate the gravity terms. 00:01:36.090 --> 00:01:41.220 Next, this one, this is a vibration isolation question. 00:01:41.220 --> 00:01:44.430 Will the addition of damping increase or reduce 00:01:44.430 --> 00:01:46.570 the vibration of the table in response 00:01:46.570 --> 00:01:49.520 to the floor motion at 30 Hertz? 00:01:56.490 --> 00:02:00.480 I guess this depends on what the natural frequency of the system 00:02:00.480 --> 00:02:00.980 is. 00:02:00.980 --> 00:02:03.300 But we're trying to do vibration isolation. 00:02:03.300 --> 00:02:05.310 And presuming, if you read the problem, 00:02:05.310 --> 00:02:09.880 you're supposed to find a a stiffness such 00:02:09.880 --> 00:02:15.060 that you can reduce the response of the table by 12 dB, 00:02:15.060 --> 00:02:19.460 I think it said, from the motion of the floor. 00:02:19.460 --> 00:02:23.010 So that's something substantially less than 1. 00:02:23.010 --> 00:02:31.240 And you will be-- this one takes a, best described 00:02:31.240 --> 00:02:32.040 with a picture. 00:02:35.650 --> 00:02:42.760 The transfer function for response of the table 00:02:42.760 --> 00:02:44.985 over the motion of the floor, the magnitude 00:02:44.985 --> 00:02:46.360 of that transfer function, that's 00:02:46.360 --> 00:02:49.350 just the ratio of x to y. 00:02:49.350 --> 00:02:55.560 That looks like this, if the damping is zero. 00:02:55.560 --> 00:03:01.150 And as you add damping, all points cross right here. 00:03:01.150 --> 00:03:03.450 Some damping does that. 00:03:03.450 --> 00:03:05.800 More damping because of this. 00:03:05.800 --> 00:03:13.260 And in order to accomplish what's been described-- 00:03:13.260 --> 00:03:16.380 this is 1.0 here. 00:03:16.380 --> 00:03:19.580 If you're trying to make this table respond less 00:03:19.580 --> 00:03:22.710 than the floor, you must be somewhere out here 00:03:22.710 --> 00:03:24.490 where you're below 1. 00:03:24.490 --> 00:03:27.970 So this is omega over omega n. 00:03:27.970 --> 00:03:30.720 And right here at resonance, you're at 1.0. 00:03:30.720 --> 00:03:34.200 So this is, you know, two or three or four for this value 00:03:34.200 --> 00:03:35.210 out here. 00:03:35.210 --> 00:03:38.470 Let's say here's where you find the answer to B. 00:03:38.470 --> 00:03:40.730 And without damping, you're there. 00:03:40.730 --> 00:03:44.570 And that's 12 dB down. 00:03:44.570 --> 00:03:47.620 If you add damping, it pushes you up these curves. 00:03:47.620 --> 00:03:52.830 Does make the response larger, the undesirable response, 00:03:52.830 --> 00:03:55.080 the motion of the table larger or smaller 00:03:55.080 --> 00:03:59.885 as you add damping at that operating point? 00:03:59.885 --> 00:04:02.070 It increases it, right? 00:04:02.070 --> 00:04:05.150 OK, so in this case, will the addition of damping 00:04:05.150 --> 00:04:06.540 increase or reduce the vibration? 00:04:06.540 --> 00:04:07.630 It'll increase it. 00:04:07.630 --> 00:04:10.080 But damping's a necessary evil. 00:04:10.080 --> 00:04:12.009 You need some damping in the system. 00:04:12.009 --> 00:04:14.550 So if you bump it, it doesn't sit there and oscillate all day 00:04:14.550 --> 00:04:15.049 long. 00:04:15.049 --> 00:04:16.760 Next. 00:04:16.760 --> 00:04:17.960 OK, this is a platform. 00:04:17.960 --> 00:04:19.459 Do you think I could actually do it? 00:04:19.459 --> 00:04:21.250 Did you read this? 00:04:21.250 --> 00:04:24.440 So this is a Coast Guard light station off of Cuttyhunk 00:04:24.440 --> 00:04:27.190 down off of Woods Hole-- basically, 00:04:27.190 --> 00:04:30.840 I was doing this-- in time with the motion of the platform. 00:04:30.840 --> 00:04:32.750 Resonance is a wonderful thing. 00:04:32.750 --> 00:04:35.230 If you can make the force be right 00:04:35.230 --> 00:04:37.070 at the natural frequency of the structure, 00:04:37.070 --> 00:04:39.530 it actually doesn't take a lot of force 00:04:39.530 --> 00:04:42.950 to drive the amplitude to pretty large amplitudes, 00:04:42.950 --> 00:04:44.690 if the damping is small. 00:04:44.690 --> 00:04:48.080 So I think the damping in this cases is about 1%. 00:04:48.080 --> 00:04:50.250 And that means the amplification, 00:04:50.250 --> 00:04:53.020 the dynamic amplification 1 over 2 zeta is about 50. 00:04:53.020 --> 00:04:54.970 So I actually could do this. 00:04:54.970 --> 00:04:56.190 This is a true story. 00:04:56.190 --> 00:04:58.566 OK, next. 00:04:58.566 --> 00:05:00.190 For small motions about the horizontal, 00:05:00.190 --> 00:05:04.670 you expect the natural frequency to be a function of gravity. 00:05:04.670 --> 00:05:09.330 So this is, oh, some of you, about equal, yes no. 00:05:09.330 --> 00:05:13.680 But it's just horizontal, the torque that gravity provides 00:05:13.680 --> 00:05:18.170 is some Mg pulling down on the center of mass 00:05:18.170 --> 00:05:20.330 somewhere in that body. 00:05:20.330 --> 00:05:24.950 Not at the pivot, but let's say some distance A away. 00:05:24.950 --> 00:05:27.630 So the torque, the gravity, the restoring torque 00:05:27.630 --> 00:05:34.940 is some Mg cross r cross Mg, Mgr if r is the distance. 00:05:34.940 --> 00:05:36.330 And that's the torque. 00:05:36.330 --> 00:05:39.400 And that length of that moment arm might vary. 00:05:39.400 --> 00:05:42.230 It's going to vary like cosine theta. 00:05:42.230 --> 00:05:46.790 Around horizontal line, if theta is what's, for small angles, 00:05:46.790 --> 00:05:48.612 what's cosine? 00:05:48.612 --> 00:05:49.990 It goes to 1. 00:05:49.990 --> 00:05:52.550 So you find out that this just looks like Mgr. 00:05:52.550 --> 00:05:54.190 It's for small angle of vibration. 00:05:54.190 --> 00:05:56.070 And you can in fact get it out of-- it 00:05:56.070 --> 00:06:01.672 doesn't enter into the equation for the natural frequency. 00:06:01.672 --> 00:06:03.130 So the natural frequency of a thing 00:06:03.130 --> 00:06:08.960 won't be a function of gravity because of this small angle 00:06:08.960 --> 00:06:12.540 vibration around a horizontal point. 00:06:12.540 --> 00:06:13.754 OK. 00:06:13.754 --> 00:06:15.420 One, when the acceleration of the system 00:06:15.420 --> 00:06:19.220 is one half that required to make the mass slide, what's 00:06:19.220 --> 00:06:21.555 the magnitude of the friction force? 00:06:24.180 --> 00:06:28.020 So friction is one of those things that is only 00:06:28.020 --> 00:06:30.670 as big as you need it to be. 00:06:30.670 --> 00:06:34.470 So even the largest friction that this thing can sustain 00:06:34.470 --> 00:06:41.640 is in fact mu mg-- answer A. But f equals ma 00:06:41.640 --> 00:06:43.910 if the acceleration is half of what 00:06:43.910 --> 00:06:47.150 is required to have that thing be just slip. 00:06:47.150 --> 00:06:56.980 It will just slip when you are at a force which is mu mg. 00:06:56.980 --> 00:07:02.430 And so that force is equal to mass times acceleration. 00:07:02.430 --> 00:07:05.190 The acceleration then you can figure out 00:07:05.190 --> 00:07:07.100 what that will be just when it slips. 00:07:07.100 --> 00:07:09.211 But now if you reduce acceleration to half that, 00:07:09.211 --> 00:07:11.710 the friction force required to keep it in place is only half 00:07:11.710 --> 00:07:13.854 as big. 00:07:13.854 --> 00:07:15.270 And it will be that friction force 00:07:15.270 --> 00:07:17.830 and it will be half of mu mg. 00:07:17.830 --> 00:07:19.370 So it's actually B. 00:07:19.370 --> 00:07:21.110 And next-- is that it? 00:07:21.110 --> 00:07:21.910 No. 00:07:21.910 --> 00:07:24.350 OK. 00:07:24.350 --> 00:07:30.890 This is a simple but actually sometimes 00:07:30.890 --> 00:07:33.310 hard to see through question. 00:07:33.310 --> 00:07:35.710 What initial conditions will be required? 00:07:35.710 --> 00:07:38.060 This problem can be solved by initial conditions. 00:07:38.060 --> 00:07:42.100 This mortar launches its shell. 00:07:42.100 --> 00:07:45.980 And the trick to this question, the key to this question 00:07:45.980 --> 00:07:48.750 is for your mathematical model of the system-- 00:07:48.750 --> 00:07:52.070 your equation in motion-- is write the equation of motion 00:07:52.070 --> 00:07:52.975 without the shell. 00:07:55.500 --> 00:07:58.760 Because once it shoots this thing, the shell's gone. 00:07:58.760 --> 00:07:59.620 And it's vibrating. 00:07:59.620 --> 00:08:02.840 It's now a system without that 25 kilogram mortar 00:08:02.840 --> 00:08:04.210 shell part of the system. 00:08:04.210 --> 00:08:05.810 It's gone. 00:08:05.810 --> 00:08:07.580 Now it's just the mass of the system 00:08:07.580 --> 00:08:09.790 without the mortar shell. 00:08:09.790 --> 00:08:11.470 And there are two initial conditions 00:08:11.470 --> 00:08:15.672 that then you can say when you shot the mortar shell, that was 00:08:15.672 --> 00:08:16.880 a certain amount of momentum. 00:08:16.880 --> 00:08:19.150 And from conservation of momentum you can figure out 00:08:19.150 --> 00:08:24.400 what the momentum of the main mass has to be-- equal 00:08:24.400 --> 00:08:26.850 and opposite to the shell you shot. 00:08:26.850 --> 00:08:29.240 So that gives you an initial velocity. 00:08:29.240 --> 00:08:32.770 But there's also an initial displacement in this problem. 00:08:32.770 --> 00:08:34.510 So that's the key to figuring this out. 00:08:34.510 --> 00:08:37.860 So what initial conditions would be required? 00:08:37.860 --> 00:08:40.260 And it's C-- both an initial velocity 00:08:40.260 --> 00:08:42.010 and an initial displacement. 00:08:42.010 --> 00:08:44.940 But the key is to make your mathematical model 00:08:44.940 --> 00:08:48.630 about the system without the shell. 00:08:48.630 --> 00:08:49.440 OK? 00:08:49.440 --> 00:08:50.408 Good. 00:08:50.408 --> 00:08:51.344 Is that it? 00:08:51.344 --> 00:08:53.690 All right. 00:08:53.690 --> 00:08:56.580 So today we're going to pick up where 00:08:56.580 --> 00:08:59.200 Professor Gossard left off. 00:08:59.200 --> 00:09:03.290 But I'm also going to do a bit of a summary 00:09:03.290 --> 00:09:07.390 right now about vibration and modeling 00:09:07.390 --> 00:09:10.130 the different kinds of systems that we talk about when 00:09:10.130 --> 00:09:11.330 we talk about vibration. 00:09:11.330 --> 00:09:17.510 They vary from simple, single degree of freedom oscillators, 00:09:17.510 --> 00:09:19.560 like a simple pendulum-- one degree 00:09:19.560 --> 00:09:24.910 of freedom-- to continuous systems-- beams and vibrate. 00:09:24.910 --> 00:09:26.410 So I'm going to try to give you just 00:09:26.410 --> 00:09:29.340 sort of an overview of vibration just 00:09:29.340 --> 00:09:32.465 to sort of give you a little map of information. 00:09:38.528 --> 00:09:44.200 Kind of to let you know what the body of vibration analysis is 00:09:44.200 --> 00:09:48.860 and what part of it we're covering in this course. 00:09:48.860 --> 00:09:55.723 So I think I will use a little more board. 00:10:18.700 --> 00:10:22.330 So we classify dynamics problems into, 00:10:22.330 --> 00:10:33.450 for convenience, rigid bodies-- rigid body dynamics 00:10:33.450 --> 00:10:35.282 and flexible bodies. 00:10:35.282 --> 00:10:36.240 One way to think of it. 00:10:41.360 --> 00:10:44.310 And this course is basically about rigid body dynamics. 00:10:44.310 --> 00:10:48.640 And under this we then have two categories 00:10:48.640 --> 00:10:53.720 that are convenient-- single degree of freedom systems 00:10:53.720 --> 00:10:58.045 and multiple degree of freedom systems. 00:11:01.420 --> 00:11:03.670 For the purposes of vendors-- talking about vibration. 00:11:06.880 --> 00:11:12.030 Single degree of freedom systems have one equation of motion. 00:11:12.030 --> 00:11:18.300 And if they vibrate, they have-- and if I'll put over here-- 00:11:18.300 --> 00:11:28.195 if vibration occurs then you have one natural frequency. 00:11:33.800 --> 00:11:37.877 And it's sort of silly to talk about a mode 00:11:37.877 --> 00:11:39.710 shape for a single degree of freedom system, 00:11:39.710 --> 00:11:42.090 because it's only relative to itself. 00:11:42.090 --> 00:11:45.920 So one natural frequency and one sort of degenerate mode shape. 00:11:45.920 --> 00:11:47.840 Multiple degree of freedom systems 00:11:47.840 --> 00:11:52.140 have n equations of motion for the number of degrees 00:11:52.140 --> 00:11:53.370 of freedom. 00:11:53.370 --> 00:12:02.510 And if they vibrate they have n omega i's, or m values 00:12:02.510 --> 00:12:08.650 of omega i for i equals 1 to n. 00:12:08.650 --> 00:12:12.640 You get n natural frequencies of the system. 00:12:12.640 --> 00:12:18.155 And you will get with it n mode shapes. 00:12:20.920 --> 00:12:23.240 So a n degree of freedom, this is 00:12:23.240 --> 00:12:27.320 equal to the number of degrees of freedom. 00:12:27.320 --> 00:12:28.830 An n degree of freedom system will 00:12:28.830 --> 00:12:30.570 have n natural frequencies and n mode 00:12:30.570 --> 00:12:33.220 shapes that go along with it. 00:12:33.220 --> 00:12:35.820 Now, what about what about flexible bodies? 00:12:35.820 --> 00:12:42.500 So a taut string like a guitar string. 00:12:47.490 --> 00:12:50.080 And actually I should say over here these rigid body 00:12:50.080 --> 00:12:52.920 things-- we have found what kind of equations of motion? 00:12:52.920 --> 00:13:06.815 These are ordinary differential equations. 00:13:11.230 --> 00:13:13.960 And there's a finite number of them and so forth. 00:13:13.960 --> 00:13:17.840 The flexible bodies like taut strings 00:13:17.840 --> 00:13:24.440 are described by partial differential equations. 00:13:34.050 --> 00:13:36.460 The number of degrees of freedom n 00:13:36.460 --> 00:13:40.960 here is the number of degrees of freedom 00:13:40.960 --> 00:13:42.635 actually goes to infinity. 00:13:45.300 --> 00:13:53.070 And you get an infinite number of omega i's, 00:13:53.070 --> 00:14:00.210 the natural frequencies, and an infinite number 00:14:00.210 --> 00:14:01.675 of corresponding mode shapes. 00:14:10.020 --> 00:14:13.740 So just about everything in the world can be made to vibrate. 00:14:19.860 --> 00:14:25.520 So how do you tell if a-- you've got a mechanical system, 00:14:25.520 --> 00:14:28.007 rigid bodies, you've got three degrees of freedom. 00:14:28.007 --> 00:14:30.215 How do you know whether or not it's going to vibrate? 00:14:32.577 --> 00:14:33.660 It will exhibit vibration? 00:14:40.500 --> 00:14:43.540 Well, one thing you could do is figure out 00:14:43.540 --> 00:14:45.420 all the equations of motion and solve them 00:14:45.420 --> 00:14:48.840 and see if cosine omega t is a solution. 00:14:48.840 --> 00:14:50.460 Right? 00:14:50.460 --> 00:14:51.960 That's the hard way. 00:14:51.960 --> 00:14:55.010 The other way is to go up to it and give it a smack 00:14:55.010 --> 00:14:57.450 and see if it vibrates. 00:14:57.450 --> 00:14:58.499 That's the simple way. 00:14:58.499 --> 00:15:00.790 If you have the mechanical system just give it a whack. 00:15:00.790 --> 00:15:05.850 And if it oscillates around some stable equilibrium position, 00:15:05.850 --> 00:15:07.070 it exhibits vibration. 00:15:07.070 --> 00:15:08.430 So this is a flexible system. 00:15:08.430 --> 00:15:10.620 You can actually probably see this from there. 00:15:10.620 --> 00:15:14.150 Just by giving this frame a smack, it will sit and vibrate. 00:15:14.150 --> 00:15:16.860 And it does it at some natural frequency. 00:15:16.860 --> 00:15:18.360 But that's a continuous system. 00:15:22.060 --> 00:15:25.340 This continuously improving little demo-- 00:15:25.340 --> 00:15:28.060 so Professor Gossard for his lecture last weekend 00:15:28.060 --> 00:15:31.280 had done this really neat embellishment, 00:15:31.280 --> 00:15:33.770 which allows you to figure out and excite the two 00:15:33.770 --> 00:15:35.344 different natural modes. 00:15:35.344 --> 00:15:37.510 But this system you have equations of motion for it. 00:15:37.510 --> 00:15:38.301 You could write it. 00:15:38.301 --> 00:15:41.190 And if you come up and give it a whack, it oscillates. 00:15:41.190 --> 00:15:44.280 And you could also find out that sure enough cosine omega t sine 00:15:44.280 --> 00:15:48.090 omega t are solutions to the equations of motion. 00:15:48.090 --> 00:15:52.320 So systems that vibrate are systems 00:15:52.320 --> 00:15:57.070 that oscillate about static equilibrium positions. 00:15:57.070 --> 00:15:59.020 And another way you can say that is 00:15:59.020 --> 00:16:01.050 when mechanical vibration occurs, 00:16:01.050 --> 00:16:03.960 there's always an exchange of energy 00:16:03.960 --> 00:16:07.860 between kinetic and potential, kinetic and potential. 00:16:07.860 --> 00:16:12.650 So our pendulum it goes-- when it 00:16:12.650 --> 00:16:16.260 reaches zero velocity up here, it's all potential energy. 00:16:16.260 --> 00:16:19.170 It reaches maximum velocity down here, it's all kinetic. 00:16:19.170 --> 00:16:20.905 And it goes back and forth. 00:16:20.905 --> 00:16:23.540 As it sloshes back and forth the energy system 00:16:23.540 --> 00:16:25.550 from kinetic to potential and back again. 00:16:25.550 --> 00:16:29.850 All vibration has that property. 00:16:29.850 --> 00:16:34.310 So that sets some basic properties of vibration. 00:16:34.310 --> 00:16:51.660 And now there's a whole body of knowledge about vibration. 00:16:51.660 --> 00:16:56.160 And we choose, or for the purposes of this class, 00:16:56.160 --> 00:16:58.510 we choose to break it down into two kinds of vibration. 00:17:19.003 --> 00:17:21.694 And one is what we call free vibration. 00:17:30.620 --> 00:17:34.490 And that we've learned already is response, only a response. 00:17:34.490 --> 00:17:46.970 It's a response to initial conditions 00:17:46.970 --> 00:17:49.135 and what we call forced vibrations. 00:17:56.700 --> 00:17:58.820 Now, forces can come of all kinds. 00:17:58.820 --> 00:18:01.890 And for the purposes of this course 00:18:01.890 --> 00:18:05.730 we look at a particular kind of force. 00:18:05.730 --> 00:18:11.310 So we focus on harmonic excitation. 00:18:18.860 --> 00:18:32.740 So excitation that is of the form cosine omega t, 00:18:32.740 --> 00:18:35.260 or e to the i omega t. 00:18:35.260 --> 00:18:41.483 These are external excitations. 00:18:46.270 --> 00:18:48.700 So we choose to break down the analysis 00:18:48.700 --> 00:18:53.550 of the vibration of systems into response 00:18:53.550 --> 00:18:55.430 to initial conditions called free vibration, 00:18:55.430 --> 00:18:58.400 no external forces, and force vibration. 00:18:58.400 --> 00:19:00.940 But we focus on a particular kind-- harmonic. 00:19:00.940 --> 00:19:04.120 And we go even one step further and say we're only 00:19:04.120 --> 00:19:05.450 going to study steady state. 00:19:11.009 --> 00:19:13.050 And steady state means you've waited a long time. 00:19:13.050 --> 00:19:14.960 Turned it on, let it shake for quite awhile. 00:19:14.960 --> 00:19:18.190 All the initial startup transients 00:19:18.190 --> 00:19:19.460 have been damped out. 00:19:19.460 --> 00:19:21.930 And you're left with a steady state vibration. 00:19:21.930 --> 00:19:24.900 And that leads to things like the transfer functions 00:19:24.900 --> 00:19:26.400 for single degree of freedom systems 00:19:26.400 --> 00:19:27.500 that we've talked about. 00:19:30.670 --> 00:19:36.500 Now, there's one other breakdown or subdivision 00:19:36.500 --> 00:19:38.450 that we need to talk about. 00:19:38.450 --> 00:19:41.680 And that is whether systems are linear or non-linear. 00:19:45.680 --> 00:19:49.350 And this is all set up so you can see it. 00:19:49.350 --> 00:19:51.670 This is a double pendulum. 00:19:51.670 --> 00:19:55.130 How many degrees of freedom? 00:19:55.130 --> 00:19:56.414 Two. 00:19:56.414 --> 00:19:58.580 And in general, do you think the equations of motion 00:19:58.580 --> 00:20:01.770 of this thing are going to be non-linear? 00:20:01.770 --> 00:20:03.610 Right. 00:20:03.610 --> 00:20:06.540 Just a simple pendulum is the restoring torque 00:20:06.540 --> 00:20:09.320 is Mgl sine theta. 00:20:09.320 --> 00:20:11.470 So you know it's got sine theta and that. 00:20:11.470 --> 00:20:13.362 And this one gets quite messy. 00:20:13.362 --> 00:20:15.320 And especially if you give it large amplitudes. 00:20:23.020 --> 00:20:26.060 And that really isn't vibration. 00:20:26.060 --> 00:20:26.600 It's not. 00:20:26.600 --> 00:20:29.410 It's looping all over itself and then doing other things. 00:20:29.410 --> 00:20:32.640 So cosine omega t is not a solution. 00:20:32.640 --> 00:20:34.090 It's not a solution to this. 00:20:34.090 --> 00:20:36.600 It's got to be more complicated than that. 00:20:36.600 --> 00:20:39.550 So when this thing is exhibiting large motions, 00:20:39.550 --> 00:20:41.792 the equations of motion are completely non-linear. 00:20:41.792 --> 00:20:43.250 And you're going to need a computer 00:20:43.250 --> 00:20:46.620 to crank out the full solution to integrate 00:20:46.620 --> 00:20:48.770 these non-linear equations of motion. 00:20:48.770 --> 00:20:58.400 But as the amplitude settles down to something pretty small, 00:20:58.400 --> 00:21:02.155 now it's vibrating about an equilibrium position. 00:21:02.155 --> 00:21:05.060 The equilibrium position is straight down. 00:21:05.060 --> 00:21:06.990 And the damping of it has made it such 00:21:06.990 --> 00:21:09.270 that the only motion left is what's called 00:21:09.270 --> 00:21:12.650 its first mode of vibration. 00:21:12.650 --> 00:21:16.970 And so if we linearize the equations of motion, 00:21:16.970 --> 00:21:21.830 assuming small amplitudes around static equilibrium positions, 00:21:21.830 --> 00:21:25.870 then we can find a vibration solution 00:21:25.870 --> 00:21:29.220 and work it out by hand probably. 00:21:29.220 --> 00:21:30.940 That's first mode for this system. 00:21:30.940 --> 00:21:36.110 And if I'm careful-- there's second mode. 00:21:39.390 --> 00:21:42.100 And for small oscillations it has 00:21:42.100 --> 00:21:45.170 a very clear single frequency that it vibrates at. 00:21:45.170 --> 00:21:49.425 The amplitude decays over time because of damping. 00:21:49.425 --> 00:21:51.600 And for every natural frequency there 00:21:51.600 --> 00:21:54.100 is a particular mode shape that goes along 00:21:54.100 --> 00:21:55.800 with that natural frequency. 00:21:55.800 --> 00:21:59.040 The first one for this system-- I 00:21:59.040 --> 00:22:02.800 have to wait for this thing to damp out. 00:22:02.800 --> 00:22:04.410 It's got a little mix of the two, 00:22:04.410 --> 00:22:09.250 but as it-- the second natural frequency motion 00:22:09.250 --> 00:22:11.010 dies out faster than the first because it 00:22:11.010 --> 00:22:13.769 has more cycles per unit time. 00:22:13.769 --> 00:22:14.560 So it settles down. 00:22:14.560 --> 00:22:16.310 This is now mostly first mode vibration. 00:22:16.310 --> 00:22:20.480 And you can see that both move in the same direction, 00:22:20.480 --> 00:22:23.330 the bottom one a little more than the top. 00:22:23.330 --> 00:22:24.500 And that's the first mode. 00:22:24.500 --> 00:22:26.200 It has a unique natural frequency. 00:22:26.200 --> 00:22:30.424 And a mode shape that is specific that goes along 00:22:30.424 --> 00:22:31.590 with that natural frequency. 00:22:48.280 --> 00:23:04.560 So this further break down here, I'll call it, 00:23:04.560 --> 00:23:14.215 is basically into non-linear and linearized. 00:23:20.900 --> 00:23:24.820 So in our discussions of vibration in this course 00:23:24.820 --> 00:23:28.400 we basically only talk about this. 00:23:28.400 --> 00:23:49.810 So we're only doing-- so that's quite a breakdown. 00:23:49.810 --> 00:23:53.220 You start at all possible vibration systems, 00:23:53.220 --> 00:23:54.900 rigid bodies, single degree of freedom, 00:23:54.900 --> 00:23:57.920 multidegree of freedom, finite number of degrees of freedom 00:23:57.920 --> 00:24:00.490 or continuous. 00:24:00.490 --> 00:24:03.330 They can have linear or non-linear equations of motion. 00:24:03.330 --> 00:24:06.590 But if we require them to be linear, 00:24:06.590 --> 00:24:10.550 and that's what we're going to look at then 00:24:10.550 --> 00:24:13.490 you sort of narrow this down what we're looking at to this. 00:24:13.490 --> 00:24:15.720 So there's lots of other things possible to look 00:24:15.720 --> 00:24:18.650 at it, like that really non-linear motion of that two 00:24:18.650 --> 00:24:19.800 dimensional thing. 00:24:19.800 --> 00:24:22.796 But our study of vibration is here. 00:24:22.796 --> 00:24:25.320 So this is what we're doing in 2003. 00:24:25.320 --> 00:24:27.930 But there's a lot of important problems 00:24:27.930 --> 00:24:29.100 that are covered by that. 00:24:31.760 --> 00:24:33.710 Lots of real things in nature that 00:24:33.710 --> 00:24:36.675 are problematic for engineers and problematic for design 00:24:36.675 --> 00:24:41.410 can be analyzed with linear equations of motion. 00:24:41.410 --> 00:24:44.560 And even if they're not linear, if you do the linear solution 00:24:44.560 --> 00:24:46.280 first, it gives you a starting point 00:24:46.280 --> 00:24:48.080 to think about what's the behavior 00:24:48.080 --> 00:24:50.360 of the non-linear systems. 00:24:50.360 --> 00:24:52.680 But this is our study of vibration. 00:24:52.680 --> 00:24:56.890 And we're going to do that for-- and what we had 00:24:56.890 --> 00:25:03.750 started doing-- in two ways. 00:25:03.750 --> 00:25:06.220 We look at the response to initial conditions called 00:25:06.220 --> 00:25:07.450 free vibration. 00:25:07.450 --> 00:25:09.840 And we look at response, steady state 00:25:09.840 --> 00:25:15.720 response of now these linear systems to force vibration. 00:25:15.720 --> 00:25:20.630 And last week you were looking for the first time 00:25:20.630 --> 00:25:24.850 at Professor Gossard's lecture about the free vibration 00:25:24.850 --> 00:25:29.840 response of basically a two degree freedom system. 00:25:29.840 --> 00:25:31.550 So why do two degree of freedom systems? 00:25:31.550 --> 00:25:35.830 Well, it's the simplest next step up from a single degree. 00:25:35.830 --> 00:25:38.050 And they're sort of mathematically tractable. 00:25:38.050 --> 00:25:39.210 You can do them on paper. 00:25:39.210 --> 00:25:42.084 We emphasize looking at two degree freedom systems 00:25:42.084 --> 00:25:43.750 because we can do the math on the board, 00:25:43.750 --> 00:25:44.991 do the math on the paper. 00:25:44.991 --> 00:25:46.740 But as you get to more degrees of freedom, 00:25:46.740 --> 00:25:49.470 you basically are going to have to do-- it's easier 00:25:49.470 --> 00:25:51.460 to do it using the computer. 00:25:51.460 --> 00:25:54.960 And in order to do that you need to know some linear algebra. 00:25:54.960 --> 00:25:57.860 So I'm kind curious. 00:25:57.860 --> 00:26:01.130 In terms of linear algebra, like multiplying two matrices 00:26:01.130 --> 00:26:04.560 together, or finding the determinant of a matrix, 00:26:04.560 --> 00:26:07.780 or inverting a matrix, how many of you 00:26:07.780 --> 00:26:11.980 actually have been taught that? 00:26:11.980 --> 00:26:14.690 How may perhaps have it? 00:26:14.690 --> 00:26:16.770 Do you do that in 1803 now? 00:26:16.770 --> 00:26:17.970 Is that where you do it? 00:26:17.970 --> 00:26:18.470 OK. 00:26:18.470 --> 00:26:18.970 Good. 00:26:18.970 --> 00:26:19.830 So that's helpful. 00:26:19.830 --> 00:26:21.038 I wasn't sure whether or not. 00:26:21.038 --> 00:26:23.670 I can assume that you at least know what 00:26:23.670 --> 00:26:27.040 the determinant of a matrix is. 00:26:27.040 --> 00:26:27.590 That's great. 00:26:27.590 --> 00:26:29.291 That's really helpful. 00:26:29.291 --> 00:26:29.790 OK. 00:26:32.840 --> 00:26:35.630 Let's talk. 00:26:35.630 --> 00:26:38.250 So we want linear equations of motion. 00:26:38.250 --> 00:26:44.200 And I've done a little bit about linearization but not much. 00:26:44.200 --> 00:26:49.600 So let's talk a little bit about that for a second. 00:26:52.530 --> 00:26:59.001 For a pendulum we know the equation of motion for it. 00:26:59.001 --> 00:27:01.500 And actually we could make this a more complicated pendulum. 00:27:01.500 --> 00:27:05.920 It could be a stick or any rigid body swinging about this point 00:27:05.920 --> 00:27:09.140 A. 00:27:09.140 --> 00:27:12.530 We know that we can write the equation of motion Izz 00:27:12.530 --> 00:27:19.270 with respect to A, theta double dot, plus Mgl-- omega L. 00:27:19.270 --> 00:27:32.020 The distance here to wherever g is, Mgl sine theta. 00:27:32.020 --> 00:27:34.780 And for free vibration that's all there is to it. 00:27:34.780 --> 00:27:38.790 And to linearize this equation we just say, 00:27:38.790 --> 00:27:44.660 well, we know that sine theta is equal to theta minus theta 00:27:44.660 --> 00:27:48.990 cubed over 3 factorial plus theta to the fifth over 5 00:27:48.990 --> 00:27:51.080 factorial. 00:27:51.080 --> 00:27:53.890 And the cosine theta-- just to have it available here-- 00:27:53.890 --> 00:28:00.930 is 1 minus theta squared over 2 factorial and so forth. 00:28:00.930 --> 00:28:02.470 And when we say linearize, we really 00:28:02.470 --> 00:28:06.890 mean we want our equations to involve the motion variables 00:28:06.890 --> 00:28:09.580 at most to first order. 00:28:09.580 --> 00:28:12.280 So the first order term for sine is theta. 00:28:12.280 --> 00:28:14.370 There's no first order term for cosine. 00:28:17.610 --> 00:28:18.940 Theta squared is non-linear. 00:28:18.940 --> 00:28:23.190 So the small angle approximation to cosine 00:28:23.190 --> 00:28:24.930 is it's approximately 1. 00:28:24.930 --> 00:28:27.480 And to sine is it's approximately 00:28:27.480 --> 00:28:30.340 theta for small motions. 00:28:30.340 --> 00:28:31.090 Theta-- small. 00:28:33.650 --> 00:28:35.540 So when we linearize this equation, 00:28:35.540 --> 00:28:39.925 we just substitute in for sine theta its linear approximation. 00:28:39.925 --> 00:28:42.200 And we get Mgl theta. 00:28:42.200 --> 00:28:43.920 So we've seen that one many times. 00:28:50.970 --> 00:28:53.330 And that's your linearized equation of motion. 00:28:53.330 --> 00:28:57.122 But on this week's homework you've got a harder problem. 00:28:57.122 --> 00:28:57.955 And that's our cart. 00:29:08.480 --> 00:29:13.620 And here you have a theta and an x are your two equations. 00:29:13.620 --> 00:29:15.710 And you've worked this problem before. 00:29:15.710 --> 00:29:20.520 And you know with the previous homeworks you've 00:29:20.520 --> 00:29:21.840 gotten the equation of motion. 00:29:21.840 --> 00:29:24.670 I'll write one of them down here. 00:29:24.670 --> 00:29:26.490 So one of the equations of motion 00:29:26.490 --> 00:29:34.480 is-- this is m1, m1, k, b. 00:29:39.960 --> 00:29:46.130 And this is a stick, it's l long, g in the middle. 00:29:46.130 --> 00:29:54.050 So the equation of motion for this looks like m1 plus m2, x 00:29:54.050 --> 00:30:08.130 double dot plus m2 l over 2, theta double dot, 00:30:08.130 --> 00:30:18.510 minus m2 l over 2 theta dot squared theta, 00:30:18.510 --> 00:30:27.370 and then plus bx dot, plus kx, and equals, 00:30:27.370 --> 00:30:32.580 and in fact, this one has a force on it. 00:30:32.580 --> 00:30:33.790 It's equal to F of t. 00:30:36.680 --> 00:30:42.480 Now, is that a non-linear equation? 00:30:54.760 --> 00:30:56.690 So this is the force equation mass 00:30:56.690 --> 00:30:58.050 times acceleration or forces. 00:30:58.050 --> 00:30:59.450 You know you've got another equation of motion 00:30:59.450 --> 00:31:01.150 in here, which is the torque one. 00:31:01.150 --> 00:31:02.210 This is just one of them. 00:31:02.210 --> 00:31:03.525 So is it linear or non-linear? 00:31:07.890 --> 00:31:09.620 How many think it's non-linear? 00:31:09.620 --> 00:31:11.060 OK. 00:31:11.060 --> 00:31:19.820 If I have number the terms here-- one, two, three, four, 00:31:19.820 --> 00:31:20.340 five. 00:31:20.340 --> 00:31:22.000 And that's not a motion. 00:31:22.000 --> 00:31:23.260 This doesn't involve motion. 00:31:23.260 --> 00:31:26.424 So if one through five, which ones have a non-linear term? 00:31:26.424 --> 00:31:27.090 AUDIENCE: Three. 00:31:27.090 --> 00:31:27.670 J. KIM VANDIVER: Three. 00:31:27.670 --> 00:31:28.510 OK. 00:31:28.510 --> 00:31:31.080 So how do you linearize that thing? 00:31:36.738 --> 00:31:40.240 AUDIENCE: You make it zero, because it's second order. 00:31:40.240 --> 00:31:42.480 J. KIM VANDIVER: In fact, it's third order. 00:31:42.480 --> 00:31:46.190 So the reason I wanted to mention this today-- 00:31:46.190 --> 00:31:49.210 if you haven't thought about being confronted 00:31:49.210 --> 00:31:52.100 with linearization problems before-- we're 00:31:52.100 --> 00:31:55.740 trying to linearize the system so that we 00:31:55.740 --> 00:32:00.900 can by making it linear we can make cosine omega t a solution. 00:32:00.900 --> 00:32:01.800 Right? 00:32:01.800 --> 00:32:06.120 We want cosine omega t to be a solution to this thing. 00:32:06.120 --> 00:32:15.240 So if you let-- so you've got a term that 00:32:15.240 --> 00:32:24.860 looks like ml over m2l over 2, theta dot squared theta. 00:32:24.860 --> 00:32:28.180 Well, theta in this problem of some function of time 00:32:28.180 --> 00:32:30.460 we're hoping-- we want to find a solution that 00:32:30.460 --> 00:32:37.260 has some amplitude times, say, a cosine omega t. 00:32:37.260 --> 00:32:46.310 And theta dot is minus omega theta naught sine omega t. 00:32:46.310 --> 00:32:48.790 And so that expression up there, the magnitude 00:32:48.790 --> 00:32:52.540 of that expression or the magnitude of theta dot 00:32:52.540 --> 00:32:58.100 squared theta is proportional to-- you get a theta 00:32:58.100 --> 00:33:02.740 naught here and you get omega squared 00:33:02.740 --> 00:33:08.590 theta naught squared here. 00:33:08.590 --> 00:33:12.700 So this term is proportional to theta naught cubed. 00:33:12.700 --> 00:33:18.140 And if the angle theta is small, then a small angle cubed 00:33:18.140 --> 00:33:20.090 is really small. 00:33:20.090 --> 00:33:23.870 And so the way you linearize this equation 00:33:23.870 --> 00:33:27.200 is to throw this out. 00:33:27.200 --> 00:33:30.635 So when you've done all your tricks you can, 00:33:30.635 --> 00:33:35.090 like replacing sine theta with theta and cosine theta 00:33:35.090 --> 00:33:39.630 with one, and you still end up with terms have a higher 00:33:39.630 --> 00:33:44.170 order than one in the motion variable, theta or x, 00:33:44.170 --> 00:33:44.970 you throw it out. 00:33:50.510 --> 00:33:52.700 So if you throw that term out then you 00:33:52.700 --> 00:33:55.580 end up with a nice linear equation in motion. 00:34:09.060 --> 00:34:09.730 OK. 00:34:09.730 --> 00:34:12.100 So now for the rest of today we're 00:34:12.100 --> 00:34:26.380 going to talk about free vibration solution. 00:34:26.380 --> 00:34:29.170 So we're not going to worry for the moment about the force 00:34:29.170 --> 00:34:32.230 vibration steady state transfer function stuff. 00:34:32.230 --> 00:34:34.790 We're talking just about free vibration. 00:34:34.790 --> 00:34:42.130 And this is of linear equations of motion. 00:35:00.490 --> 00:35:03.220 So vibration is a pretty big body of knowledge. 00:35:03.220 --> 00:35:06.990 And we're doing an introduction to vibration in about half 00:35:06.990 --> 00:35:08.930 a dozen lectures here. 00:35:08.930 --> 00:35:13.180 So there's lots of things that I'm not 00:35:13.180 --> 00:35:15.730 going to have time to teach you, but there 00:35:15.730 --> 00:35:19.320 are a few things I really want you to go away with 00:35:19.320 --> 00:35:20.800 and understanding. 00:35:20.800 --> 00:35:29.160 And one of these key concepts is that the vibration 00:35:29.160 --> 00:35:32.950 of a multiple degree of freedom system-- 00:35:32.950 --> 00:35:35.950 say this is a two degree of freedom system. 00:35:35.950 --> 00:35:38.510 That the vibration of this system, the free vibration, 00:35:38.510 --> 00:35:45.670 can be made up of the sum of two parts 00:35:45.670 --> 00:35:49.040 at any vibration of this system at all. 00:35:49.040 --> 00:35:51.240 So at any arbitrary set of initial conditions 00:35:51.240 --> 00:35:53.480 I give it-- I let it go. 00:35:53.480 --> 00:35:56.640 The key concept is the response will 00:35:56.640 --> 00:36:01.410 be made up of two pieces-- vibration 00:36:01.410 --> 00:36:03.940 in each of the two modes. 00:36:03.940 --> 00:36:05.845 And if you can solve the vibration that's 00:36:05.845 --> 00:36:07.970 in first mode-- first mode is the one where they're 00:36:07.970 --> 00:36:09.850 going kind of together, second mode 00:36:09.850 --> 00:36:11.640 they're opposite one another. 00:36:11.640 --> 00:36:13.750 That the total solution can be made up 00:36:13.750 --> 00:36:16.880 of a contribution for mode one and a contribution 00:36:16.880 --> 00:36:18.960 from mode two. 00:36:18.960 --> 00:36:29.200 So this is this concept called mode superposition. 00:36:29.200 --> 00:36:30.650 It's really quite powerful. 00:36:30.650 --> 00:36:34.630 So you can figure out the response of the first mode 00:36:34.630 --> 00:36:37.670 in the system, figure out the response of the second mode's 00:36:37.670 --> 00:36:42.230 contribution, add them together, and that's the total solution. 00:36:42.230 --> 00:36:51.480 And this concept works-- there's all sorts of caveats 00:36:51.480 --> 00:36:57.040 that one gets into-- but basically this is true 00:36:57.040 --> 00:37:11.860 for all lightly damped systems. 00:37:11.860 --> 00:37:15.570 You get into heavy damping and strange damping, 00:37:15.570 --> 00:37:17.230 you have to make some adjustments. 00:37:17.230 --> 00:37:19.770 But for lightly damped systems you'll 00:37:19.770 --> 00:37:21.890 find that this concept of mode superposition 00:37:21.890 --> 00:37:25.970 works out just fine. 00:37:25.970 --> 00:37:31.070 So an illustration of this, a really simple illustration-- 00:37:31.070 --> 00:37:33.500 in some ways easier than this one. 00:37:33.500 --> 00:37:35.840 I don't know if I can get this where you can see it 00:37:35.840 --> 00:37:37.090 in the picture or not. 00:37:37.090 --> 00:37:38.210 Maybe not really. 00:37:38.210 --> 00:37:41.190 This is just two little lead weights. 00:37:41.190 --> 00:37:42.470 This is a double pendulum. 00:37:42.470 --> 00:37:46.080 It has two natural frequencies. 00:37:46.080 --> 00:37:48.190 One is that one. 00:37:48.190 --> 00:37:51.730 You can see the two weights going the same direction. 00:37:51.730 --> 00:37:54.740 The bottom weight at a little bit more larger angle 00:37:54.740 --> 00:37:57.540 than the top weight. 00:37:57.540 --> 00:37:59.960 And it's at a particular frequency. 00:37:59.960 --> 00:38:03.940 And that's the mode shape that goes through this frequency. 00:38:03.940 --> 00:38:07.790 So another key concept is that for free vibration 00:38:07.790 --> 00:38:12.490 the total solution is made up of the free vibration 00:38:12.490 --> 00:38:13.420 of each mode. 00:38:13.420 --> 00:38:15.775 And each mode has a particular frequency 00:38:15.775 --> 00:38:18.740 and a particular shape to it. 00:38:18.740 --> 00:38:23.020 So that's the first mode frequency and the first mode 00:38:23.020 --> 00:38:23.520 shape. 00:38:23.520 --> 00:38:26.440 The second mode-- I have a little harder time getting it 00:38:26.440 --> 00:38:29.560 started-- it looks like that. 00:38:29.560 --> 00:38:31.930 Masses move in opposite directions. 00:38:31.930 --> 00:38:35.225 It's kind of rotating around where you can't see it. 00:38:35.225 --> 00:38:38.920 I have to do it in the plane. 00:38:38.920 --> 00:38:39.870 It's hard to do here. 00:38:39.870 --> 00:38:42.455 It doesn't want to behave like it's confined to a plane. 00:38:47.420 --> 00:38:49.570 They're going opposite directions. 00:38:49.570 --> 00:38:51.520 And the frequency is higher. 00:38:51.520 --> 00:38:53.600 But this motion, that mode shape, 00:38:53.600 --> 00:38:58.530 is a fixed feature of this mode of vibration 00:38:58.530 --> 00:39:00.470 along with this natural frequency. 00:39:00.470 --> 00:39:04.540 So this idea of mode superposition-- 00:39:04.540 --> 00:39:27.310 and a second concept here is that for free vibration 00:39:27.310 --> 00:40:00.740 of each mode it oscillates at a unique frequency for this two 00:40:00.740 --> 00:40:01.970 degree of freedom system. 00:40:01.970 --> 00:40:04.060 You have two natural frequencies-- 00:40:04.060 --> 00:40:07.450 omega 1 and omega 2. 00:40:07.450 --> 00:40:23.220 And at each omega n there is a corresponding mode shape. 00:40:25.790 --> 00:40:31.960 So any vibration of a linear system, free vibration of it, 00:40:31.960 --> 00:40:36.870 any vibration at all is composed of a superposition of the two 00:40:36.870 --> 00:40:37.870 modes. 00:40:37.870 --> 00:40:39.930 Part of this motion is in the first mode 00:40:39.930 --> 00:40:42.370 at its natural frequency and in its shape. 00:40:42.370 --> 00:40:47.440 And part of the motion has a second contribution, 00:40:47.440 --> 00:40:50.030 which is at the natural frequency of the second mode 00:40:50.030 --> 00:40:51.105 and in its shape. 00:40:53.670 --> 00:40:57.360 So I'm going to give you a quick demo 00:40:57.360 --> 00:41:02.450 and ask you-- let's if you can use what I just 00:41:02.450 --> 00:41:08.676 said to analyze a motion. 00:41:16.970 --> 00:41:21.330 So this is just a block on some strings. 00:41:21.330 --> 00:41:23.910 And I'm going to show you a motion. 00:41:23.910 --> 00:41:27.430 And I want you to tell me whether or not 00:41:27.430 --> 00:41:36.580 it could possibly be a natural frequency motion in one mode, 00:41:36.580 --> 00:41:40.552 or the other answer is it's a sum of multiple modes. 00:41:40.552 --> 00:41:42.010 But I'm going to show you a motion, 00:41:42.010 --> 00:41:44.070 and I want you to tell me and argue 00:41:44.070 --> 00:41:47.160 on the basis of what I've just told you whether or not 00:41:47.160 --> 00:41:49.260 you are seeing a single mode of vibration. 00:41:52.810 --> 00:41:54.779 And maybe I'll use the clamp here 00:41:54.779 --> 00:41:56.570 so I don't have to stand there and hold it. 00:42:25.802 --> 00:42:27.010 I'm going to just place this. 00:42:27.010 --> 00:42:28.720 So the way you do free vibration is 00:42:28.720 --> 00:42:30.360 you give it an initial displacement, 00:42:30.360 --> 00:42:32.680 some initial conditions, and let go. 00:42:32.680 --> 00:42:39.975 So I'm going to pull this over and back and let go. 00:42:43.935 --> 00:42:46.480 And just watch closely what you see it do. 00:43:14.794 --> 00:43:16.710 All right, now it's doing more of what I want. 00:43:16.710 --> 00:43:19.224 It's like it's going in a circle right now. 00:43:19.224 --> 00:43:21.390 And now it looks like it's just going back and forth 00:43:21.390 --> 00:43:23.140 on a diagonal. 00:43:23.140 --> 00:43:25.940 And then it's going to start circling the other way. 00:43:25.940 --> 00:43:27.650 It's going in a circle. 00:43:27.650 --> 00:43:31.320 And now it goes to on the diagonal-- left and right. 00:43:31.320 --> 00:43:36.210 And then it starts back into a circle again. 00:43:36.210 --> 00:43:40.950 Are you observing a natural mode of vibration? 00:43:40.950 --> 00:43:42.780 It looks like it's single frequency, right? 00:43:42.780 --> 00:43:45.340 This looks like it's all happening at one frequency. 00:43:45.340 --> 00:43:50.340 But is it a natural mode, a unique natural mode? 00:43:57.150 --> 00:44:00.070 Who wants to make a case for whether it is or isn't? 00:44:00.070 --> 00:44:01.600 How many believe that you're seeing 00:44:01.600 --> 00:44:03.240 a natural mode of vibration? 00:44:03.240 --> 00:44:03.740 None. 00:44:03.740 --> 00:44:05.810 How many think you're not seeing a natural mode of vibration? 00:44:05.810 --> 00:44:06.770 Let's see if you're awake. 00:44:06.770 --> 00:44:07.270 OK. 00:44:07.270 --> 00:44:09.150 So you don't believe that it's natural mode. 00:44:09.150 --> 00:44:10.830 Make the case. 00:44:10.830 --> 00:44:12.460 Why? 00:44:12.460 --> 00:44:16.090 How do you use sort of this definition of a natural mode 00:44:16.090 --> 00:44:19.676 to tell me why this can't be? 00:44:19.676 --> 00:44:21.652 AUDIENCE: It looks like a superposition 00:44:21.652 --> 00:44:25.110 of at least two different kinds of vibration. 00:44:25.110 --> 00:44:26.592 J. KIM VANDIVER: OK. 00:44:26.592 --> 00:44:29.145 The evidence that you see is because what does it do? 00:44:29.145 --> 00:44:30.436 AUDIENCE: It circled sometimes. 00:44:30.436 --> 00:44:33.407 And sometimes it goes straight back and forth on a diagonal. 00:44:33.407 --> 00:44:34.240 J. KIM VANDIVER: OK. 00:44:34.240 --> 00:44:35.490 So it circled around part of the time 00:44:35.490 --> 00:44:38.060 and then goes straight back and forth part of the time. 00:44:38.060 --> 00:44:44.580 Is it a constant mode shape of vibration? 00:44:44.580 --> 00:44:45.220 No. 00:44:45.220 --> 00:44:47.680 And that's all you need to observe. 00:44:47.680 --> 00:44:52.500 If the thing doesn't keep a constant single shape 00:44:52.500 --> 00:44:55.400 at a single frequency, it's not a natural mode. 00:44:55.400 --> 00:44:57.230 So let's do a different case. 00:44:57.230 --> 00:45:00.690 I'll deflect it just this way. 00:45:00.690 --> 00:45:02.510 And ignore that little bit of torsion. 00:45:02.510 --> 00:45:05.670 So it's just going back and forth in line. 00:45:05.670 --> 00:45:09.900 Other than slowly damping out, that has just one motion to it, 00:45:09.900 --> 00:45:12.190 and it's at one natural frequency. 00:45:12.190 --> 00:45:14.740 So do you think that's a mode? 00:45:14.740 --> 00:45:16.100 That probably is two. 00:45:16.100 --> 00:45:17.160 And so is this one. 00:45:20.279 --> 00:45:21.570 And ignore that high frequency. 00:45:21.570 --> 00:45:22.778 Now it's just back and forth. 00:45:22.778 --> 00:45:24.160 It's just a pendulum. 00:45:24.160 --> 00:45:27.480 And it just stays in just pendular motion, no circling 00:45:27.480 --> 00:45:28.790 around or any of that. 00:45:28.790 --> 00:45:31.210 So that's also natural, and it occurs 00:45:31.210 --> 00:45:33.530 at a particular frequency. 00:45:33.530 --> 00:45:37.610 So these are two individual pendular emotions-- 00:45:37.610 --> 00:45:39.330 one this way and one that way. 00:45:39.330 --> 00:45:41.310 And what I was doing at the beginning 00:45:41.310 --> 00:45:44.580 is I pulled it to the side, which 00:45:44.580 --> 00:45:46.470 would start one of those modes. 00:45:46.470 --> 00:45:48.340 And I pulled it back, which will put 00:45:48.340 --> 00:45:51.670 some energy into the other mode, and let it go. 00:45:51.670 --> 00:45:55.540 And now what you have is the sum of these two different motions 00:45:55.540 --> 00:45:57.230 adding up. 00:45:57.230 --> 00:46:00.020 It goes in circles and then in straight lines. 00:46:00.020 --> 00:46:05.700 And the fact that they-- this is a phenomenon called beading. 00:46:05.700 --> 00:46:09.240 And it is because these two pendulums, 00:46:09.240 --> 00:46:13.410 even though they have strings of the same length, 00:46:13.410 --> 00:46:16.654 they actually have slightly different natural frequencies. 00:46:16.654 --> 00:46:18.570 They're each single degree of freedom systems. 00:46:18.570 --> 00:46:21.040 They're two independent single degree of freedom systems, 00:46:21.040 --> 00:46:22.990 each with their own natural frequency. 00:46:22.990 --> 00:46:27.789 But if you mix them then they're going to exhibit this motion. 00:46:27.789 --> 00:46:29.830 So that's something really important to remember. 00:46:29.830 --> 00:46:34.900 A quiz question that I like to ask is-- it's easy to grade 00:46:34.900 --> 00:46:38.850 and it's no math required-- is to literally-- I've often 00:46:38.850 --> 00:46:40.870 done this in exams-- walk in with something 00:46:40.870 --> 00:46:45.330 like that block of wood and say, is this a natural mode? 00:46:53.880 --> 00:47:00.910 Time to do one-- let me see here. 00:47:29.460 --> 00:47:31.930 So now let's pick up where Professor Gossard left off. 00:47:31.930 --> 00:47:34.380 Let's go talking about natural frequencies and mode 00:47:34.380 --> 00:47:39.100 shapes of linearized two degree of freedom systems. 00:47:39.100 --> 00:47:43.880 But I want to generalize a little bit on what he did. 00:47:43.880 --> 00:47:49.400 So he, in his lecture, analyzed this system like this. 00:47:52.710 --> 00:47:54.470 I'll just kind of put the highlights here. 00:48:09.860 --> 00:48:22.100 And this is now solving for natural frequencies and mode 00:48:22.100 --> 00:48:22.600 shapes. 00:48:26.460 --> 00:48:28.845 He came up with a set of equations of motion for this. 00:48:28.845 --> 00:48:33.605 This was, I guess, M1, M2. 00:48:33.605 --> 00:48:40.580 And the equations of motion for this are m1 in matrix form. 00:48:49.436 --> 00:48:53.000 Now I'm going to do this to emphasize something. 00:48:57.480 --> 00:48:59.040 In general there could be damping 00:48:59.040 --> 00:49:01.590 in our linearized system. 00:49:01.590 --> 00:49:08.520 And we have a stiffness matrix-- K1 plus K2 minus K2. 00:49:19.380 --> 00:49:23.570 And in general there could be forces, 00:49:23.570 --> 00:49:28.130 which are functions of time on that system. 00:49:28.130 --> 00:49:30.850 Now, if we want to find natural frequencies in mode shapes, 00:49:30.850 --> 00:49:33.980 we go looking for what we call with the undamped natural 00:49:33.980 --> 00:49:35.700 frequencies in mode shapes. 00:49:35.700 --> 00:49:37.710 So this problem doesn't even have dampers in it. 00:49:37.710 --> 00:49:40.290 But if it did for the purpose of finding 00:49:40.290 --> 00:49:45.280 natural frequencies in mode shapes, you just set to 0. 00:49:45.280 --> 00:49:47.130 And with the forces you do the same thing. 00:49:52.030 --> 00:49:57.850 And now you have undamped, unforced equations of motion. 00:50:01.610 --> 00:50:05.630 And this is then of the form of mass matrix times 00:50:05.630 --> 00:50:11.720 an acceleration vector, X1, X2, plus a stiffness matrix, 00:50:11.720 --> 00:50:15.815 times a displacement vector equals 0. 00:50:15.815 --> 00:50:18.250 So in matrix notation it looks like that. 00:50:20.950 --> 00:50:26.670 This is the way you would do any rigid body vibration problem. 00:50:26.670 --> 00:50:28.490 This is two degrees of freedom. 00:50:28.490 --> 00:50:29.940 But this is the general expression 00:50:29.940 --> 00:50:33.681 for an n degree of freedom system. 00:50:33.681 --> 00:50:36.770 If we had three masses here, then these would be 3 by 3 00:50:36.770 --> 00:50:40.390 matrices instead of 2 by 2's. 00:50:40.390 --> 00:50:44.160 So that's the basic formulation. 00:50:44.160 --> 00:50:47.660 And you went through last time with 2 by 2. 00:50:47.660 --> 00:50:52.890 You can actually go through and find the fourth order equation 00:50:52.890 --> 00:50:59.180 in omega and solve for two roots of omega squared. 00:50:59.180 --> 00:51:02.370 And you've got the two natural frequencies, plug them back in. 00:51:02.370 --> 00:51:05.050 You've got the two mode shapes that go along with them. 00:51:05.050 --> 00:51:08.020 So you did it that way by hand so you 00:51:08.020 --> 00:51:10.650 can see how you can work out the natural frequencies. 00:51:10.650 --> 00:51:14.850 How can you do-- I'm going to show an approach that you'd 00:51:14.850 --> 00:51:16.790 more likely use on a computer. 00:51:16.790 --> 00:51:20.850 And if you get the larger order n degree of freedom systems, 00:51:20.850 --> 00:51:23.630 you're going to want to do this-- instead of by hand-- 00:51:23.630 --> 00:51:25.750 have a computer do the work for you. 00:51:25.750 --> 00:51:30.495 So this is the generic form. 00:51:30.495 --> 00:51:34.580 And let's just assume for a minute 00:51:34.580 --> 00:51:37.050 it's an n degree of freedom system. 00:51:37.050 --> 00:51:41.470 So these are n by n matrices. 00:51:45.750 --> 00:51:48.100 How would we find the natural frequencies and mode 00:51:48.100 --> 00:51:52.490 shapes of this general system? 00:51:52.490 --> 00:52:05.920 So you assume solutions of the form 00:52:05.920 --> 00:52:09.760 of what I've been describing-- a natural mode. 00:52:09.760 --> 00:52:15.110 Any natural mode of the system has a particular shape 00:52:15.110 --> 00:52:18.519 to it and a particular frequency. 00:52:18.519 --> 00:52:19.310 And that's the key. 00:52:19.310 --> 00:52:21.150 That's the key assumption here. 00:52:21.150 --> 00:52:26.647 You assume solutions of the form that this vector x-- 00:52:26.647 --> 00:52:28.480 instead of writing it in brackets like this, 00:52:28.480 --> 00:52:30.532 I'm going to make this. 00:52:30.532 --> 00:52:37.460 So X here is just with a line underneath it. 00:52:37.460 --> 00:52:48.080 So x is of the form X1 of t down to Xn 00:52:48.080 --> 00:52:50.950 if you have n degrees of freedom. 00:52:50.950 --> 00:52:53.890 You're looking for a solution for that thing. 00:52:53.890 --> 00:52:57.900 And it's going to have an amplitude to it-- 00:52:57.900 --> 00:53:02.770 A1 down to An. 00:53:02.770 --> 00:53:06.590 And this is any one mode. 00:53:06.590 --> 00:53:10.710 So any one mode will look like a set of amplitudes 00:53:10.710 --> 00:53:14.610 that govern its mode shape. 00:53:14.610 --> 00:53:17.110 And it will oscillate. 00:53:17.110 --> 00:53:21.110 We can write the oscillation as cosine omega. 00:53:21.110 --> 00:53:24.070 And I'll put an i here, it's the i-th natural frequency 00:53:24.070 --> 00:53:26.970 minus some phase angle. 00:53:26.970 --> 00:53:29.380 So in general, each mode-- assume solutions-- 00:53:29.380 --> 00:53:34.560 I'll say here for each mode. 00:53:37.520 --> 00:53:41.210 So each mode, any mode, mode I will look like this. 00:53:41.210 --> 00:53:43.759 It will have a shape to it governed by this. 00:53:43.759 --> 00:53:45.300 And these are basically on constants. 00:53:45.300 --> 00:53:47.630 Once determined, this is just a constant. 00:53:47.630 --> 00:53:49.510 And here's your time dependence. 00:53:49.510 --> 00:53:53.660 And it's going to-- I left out my t-- oscillate 00:53:53.660 --> 00:53:54.950 at some natural frequency. 00:53:54.950 --> 00:53:57.770 So we know this is what the solution has to look like. 00:53:57.770 --> 00:54:01.485 And we can take this and plug it in to this equation. 00:54:11.710 --> 00:54:18.230 This vector of responses is some vector of amplitude 00:54:18.230 --> 00:54:24.690 times the cosine omega t minus phi. 00:54:24.690 --> 00:54:28.180 And just plug that into this set of matrix equations. 00:54:28.180 --> 00:54:37.590 Note that x double dot is just a-- you get minus omega squared 00:54:37.590 --> 00:54:41.910 a cosine omega t. 00:54:45.720 --> 00:54:49.455 And we now substitute these into here. 00:54:56.040 --> 00:55:01.570 You get minus omega squared for the first term. 00:55:01.570 --> 00:55:04.190 Minus omega squared times the mass 00:55:04.190 --> 00:55:12.280 matrix a cosine omega t minus phi, 00:55:12.280 --> 00:55:19.410 plus this stiffness matrix a-- better consistent notation 00:55:19.410 --> 00:55:26.157 here, excuse me-- a cosine omega t minus phi. 00:55:26.157 --> 00:55:27.365 And all that's equal to zero. 00:55:30.060 --> 00:55:32.120 So these go away. 00:55:32.120 --> 00:55:34.770 You can cancel them out. 00:55:34.770 --> 00:55:38.950 And we can factor out this a quantity. 00:55:38.950 --> 00:55:56.740 And we have minus omega squared m, plus k times a equals 0. 00:55:56.740 --> 00:55:59.067 So you can do this with any linearized n 00:55:59.067 --> 00:56:00.650 degree of freedom system that you know 00:56:00.650 --> 00:56:02.740 has a vibration solution to it. 00:56:09.110 --> 00:56:12.100 These are the unknown mode shapes. 00:56:12.100 --> 00:56:15.010 And so in order to satisfy this equation, 00:56:15.010 --> 00:56:19.290 either this a has to be 0, which is a trivial solution. 00:56:19.290 --> 00:56:21.770 There's no motion, no mode shape-- 00:56:21.770 --> 00:56:35.700 or which this is trivial, not too useful. 00:56:35.700 --> 00:56:48.520 Either a has to be 0, or the determinant of this quantity 00:56:48.520 --> 00:56:49.450 has to be 0. 00:56:58.970 --> 00:57:00.660 But the way you do that on a computer-- 00:57:00.660 --> 00:57:02.220 so that would be beginning the way 00:57:02.220 --> 00:57:05.120 you would analyze this by hand. 00:57:05.120 --> 00:57:07.160 You find the determinant of that matrix. 00:57:07.160 --> 00:57:12.550 And if it is a two degree of freedom system, 00:57:12.550 --> 00:57:16.810 you'll get an equation n omega to the fourth, which has 00:57:16.810 --> 00:57:18.180 two roots for omega squared. 00:57:18.180 --> 00:57:19.888 If it's a three degree of freedom system, 00:57:19.888 --> 00:57:22.930 you'll get an equation of omega to the sixth when you write out 00:57:22.930 --> 00:57:24.180 that determinant. 00:57:24.180 --> 00:57:27.020 And it has three roots for omega squared. 00:57:27.020 --> 00:57:28.740 An n degree of freedom system has 00:57:28.740 --> 00:57:32.850 an equation that's of order 2n omega to the 2nth power. 00:57:32.850 --> 00:57:36.110 And it'll have n solutions or roots 00:57:36.110 --> 00:57:39.684 for the natural frequency for omega squared. 00:57:39.684 --> 00:57:41.100 But that would be if you're trying 00:57:41.100 --> 00:57:42.290 to grind this out by hand. 00:57:46.450 --> 00:57:52.552 The way you do this on a computer-- maybe I 00:57:52.552 --> 00:57:54.111 can get a little bit more on here. 00:57:56.700 --> 00:57:58.080 Come back here. 00:58:15.370 --> 00:58:21.730 So I'll go back to the earlier form I had here plus ka 00:58:21.730 --> 00:58:23.850 equals 0. 00:58:23.850 --> 00:58:35.110 And I'm going to multiply by m inverse. 00:58:35.110 --> 00:58:39.200 So if I invert the mass matrix, if I 00:58:39.200 --> 00:58:43.230 multiply a matrix by its inverse, what do you get? 00:58:43.230 --> 00:58:45.720 So if I multiply m times m inverse? 00:58:45.720 --> 00:58:46.860 AUDIENCE: A unit matrix. 00:58:46.860 --> 00:58:48.776 J. KIM VANDIVER: You get a unit matrix, right? 00:58:48.776 --> 00:58:50.390 It has ones on the diagonal. 00:58:50.390 --> 00:58:53.360 So I'm going to multiply through here by this. 00:58:53.360 --> 00:58:56.240 And so this gives me a minus omega squared. 00:58:56.240 --> 00:59:00.180 And m times m inverse gives me the unit matrix-- 00:59:00.180 --> 00:59:02.460 ones on the diagonal. 00:59:02.460 --> 00:59:15.910 Times a plus M inverse times ka equals 0. 00:59:15.910 --> 00:59:19.210 And this product is just a matrix product-- m inverse 00:59:19.210 --> 00:59:20.110 times k. 00:59:20.110 --> 00:59:21.693 And I'm going to call it the a matrix. 00:59:25.060 --> 00:59:27.790 And I'm going to move this to the other side. 00:59:27.790 --> 00:59:32.440 So I have a linear algebraic expression 00:59:32.440 --> 00:59:42.500 of the form a times the vector equals omega squared 00:59:42.500 --> 00:59:46.500 times the unit matrix times a. 00:59:49.040 --> 00:59:55.300 And I could go ahead and multiply this out. 00:59:55.300 --> 00:59:58.770 For example, this times that and I'll get a vector. 00:59:58.770 --> 01:00:02.100 So this just looks like omega squared 01:00:02.100 --> 01:00:04.250 a if you multiply it out. 01:00:06.940 --> 01:00:11.580 The vector times the matrix is a vector on the left side. 01:00:11.580 --> 01:00:15.550 A vector times a matrix gives me back a vector. 01:00:15.550 --> 01:00:16.720 It's just the unit matrix. 01:00:16.720 --> 01:00:19.630 So it gives me back the vector times omega squared. 01:00:19.630 --> 01:00:24.140 This is in what is known as standard eigenvalue 01:00:24.140 --> 01:00:26.150 formulation. 01:00:26.150 --> 01:00:28.760 It's a standard eigenvalue problem now. 01:00:28.760 --> 01:00:36.020 It's a problem of the form a times a vector equals 01:00:36.020 --> 01:00:40.310 something lambda times a. 01:00:40.310 --> 01:00:42.060 A parameter which we know happens 01:00:42.060 --> 01:00:44.130 to be the frequency squared. 01:00:44.130 --> 01:00:49.580 But this is standard eigenvalue formulation. 01:00:49.580 --> 01:00:50.771 Yeah? 01:00:50.771 --> 01:00:52.645 AUDIENCE: I was just asking if you wrote down 01:00:52.645 --> 01:00:55.340 omega squared a because it's equal to the length up there. 01:00:55.340 --> 01:00:58.660 J. KIM VANDIVER: Ah, good. 01:00:58.660 --> 01:01:00.320 Omega squared a. 01:01:00.320 --> 01:01:04.190 So a times the unit vector you get a back as a vector. 01:01:04.190 --> 01:01:06.060 And I got the omega squared in front of it. 01:01:06.060 --> 01:01:09.640 And oftentimes in a manual for Matlab or something 01:01:09.640 --> 01:01:12.410 they'll describe this as some parameter. 01:01:12.410 --> 01:01:15.460 It's a constant times a. 01:01:15.460 --> 01:01:20.160 And this is standard what they call eigenvalue formulation. 01:01:20.160 --> 01:01:38.720 And in Matlab if you say, for example, E equals EIG of A. 01:01:38.720 --> 01:01:49.455 This returns a vector, which is the natural frequency squared. 01:01:52.392 --> 01:01:54.080 It will return these lambdas. 01:01:54.080 --> 01:01:56.835 And the first one is omega 1 squared down 01:01:56.835 --> 01:01:58.060 to omega n squared. 01:02:00.780 --> 01:02:02.820 And if you go with this function, 01:02:02.820 --> 01:02:12.170 if you go a little further, if you say V comma D is EIG A, 01:02:12.170 --> 01:02:16.570 then this gives you two matrices back. 01:02:16.570 --> 01:02:21.910 It gives you V. And V is a matrix, which 01:02:21.910 --> 01:02:25.310 its columns are the mode shape. 01:02:25.310 --> 01:02:36.420 So A1 to An, this is mode 1 over to A1 to An for mode n. 01:02:36.420 --> 01:02:38.530 It gives you two matrices. 01:02:38.530 --> 01:02:40.070 One that's that. 01:02:40.070 --> 01:02:49.330 And another one a D matrix, which has the lambdas-- lambda 01:02:49.330 --> 01:02:55.650 1 lambda n on the diagonals. 01:02:55.650 --> 01:02:58.210 And it's a diagonal matrix. 01:02:58.210 --> 01:03:00.120 So it gives you two matrices back. 01:03:00.120 --> 01:03:04.270 One that has the eigenvectors, the mode shapes. 01:03:04.270 --> 01:03:06.970 And another matrix whose diagonal elements are 01:03:06.970 --> 01:03:10.190 the natural frequency squared. 01:03:10.190 --> 01:03:14.620 And that's all there is to it if you do this numerically. 01:03:14.620 --> 01:03:17.450 And there's lots of different programs. 01:03:17.450 --> 01:03:20.340 There's multiple ways of doing this in Matlab. 01:03:20.340 --> 01:03:23.720 When you do it this way it doesn't come out sometimes 01:03:23.720 --> 01:03:26.670 nicely ordered and what I call normalized. 01:03:26.670 --> 01:03:31.067 But it does produce the eigenvalues. 01:03:31.067 --> 01:03:32.900 They're called eigenvalues and eigenvectors. 01:03:32.900 --> 01:03:36.210 The eigenvalues are the lambdas, the natural frequencies 01:03:36.210 --> 01:03:37.180 squared. 01:03:37.180 --> 01:03:40.580 And the eigenvectors are these mode shapes that 01:03:40.580 --> 01:03:45.510 go with each natural frequency. 01:03:45.510 --> 01:03:47.760 Once you know the natural frequencies and mode shapes, 01:03:47.760 --> 01:03:50.310 now we want to get back to talking about solutions. 01:03:50.310 --> 01:03:51.950 This idea of mode superposition. 01:03:51.950 --> 01:03:55.070 And if you give it a set of initial conditions, what 01:03:55.070 --> 01:03:55.950 is the response? 01:03:55.950 --> 01:03:59.380 How do you add these two modes together? 01:03:59.380 --> 01:04:01.220 So let's go back now. 01:04:01.220 --> 01:04:08.670 We'll return to two degree of freedom systems 01:04:08.670 --> 01:04:11.415 like this one to do an example. 01:04:14.140 --> 01:04:27.840 And we assume that the solution was sum A1, A2, cosine omega 01:04:27.840 --> 01:04:31.020 t minus a phase angle. 01:04:31.020 --> 01:04:34.750 That each mode would have this character to it. 01:04:34.750 --> 01:04:39.645 And I'm going to normalize my mode shapes. 01:04:51.790 --> 01:04:54.940 So for each mode shape of the system-- so 01:04:54.940 --> 01:04:58.290 this could be for mode one. 01:04:58.290 --> 01:04:59.990 This is the mode shape for mode one. 01:04:59.990 --> 01:05:03.720 This is natural frequency one and phase angle one. 01:05:03.720 --> 01:05:06.610 Each mode shape-- I could write this then 01:05:06.610 --> 01:05:10.460 as-- I could factor out the A1. 01:05:10.460 --> 01:05:13.090 Just pull out A1, divide each member by A1. 01:05:13.090 --> 01:05:22.430 So this can be written as A1 times 1 and A2 over A1. 01:05:22.430 --> 01:05:23.720 So I've just factored out. 01:05:29.670 --> 01:05:33.920 So for mode one it's normalized mode shape-- 01:05:33.920 --> 01:05:41.370 by normalize you just pick some way that you repeatedly use, 01:05:41.370 --> 01:05:42.550 consistent in its use. 01:05:42.550 --> 01:05:47.100 I often say let's make the top element of the vector 1. 01:05:47.100 --> 01:05:49.990 And to make the top one 1 you factor out 01:05:49.990 --> 01:05:52.365 whatever its value is that you get back from the computer 01:05:52.365 --> 01:05:54.080 or from your calculation. 01:05:54.080 --> 01:05:56.930 You factor that out of every member. 01:05:56.930 --> 01:05:59.430 Now you have a normalized mode shape whose top element is 1. 01:05:59.430 --> 01:06:01.263 There's lots of other normalization schemes. 01:06:01.263 --> 01:06:03.380 That's just one way to do it. 01:06:03.380 --> 01:06:12.340 And that's one of the mode shapes. 01:06:12.340 --> 01:06:26.790 The total solution is X1, X2-- and this 01:06:26.790 --> 01:06:29.690 is where the mode superposition part comes in-- 01:06:29.690 --> 01:06:38.080 is some undetermined constant A1 times the mode shape 01:06:38.080 --> 01:06:47.836 A2 over A1 for mode 1 cosine omega 1 t minus phi 1. 01:06:47.836 --> 01:06:49.210 And I'm going to run out of room. 01:07:14.455 --> 01:07:17.780 And now the responses to initial conditions-- this 01:07:17.780 --> 01:07:20.040 has got another term. 01:07:20.040 --> 01:07:21.460 I'm just going to rewrite it here. 01:07:21.460 --> 01:07:24.090 So we're looking at-- our total motion response 01:07:24.090 --> 01:07:39.515 now by mode superposition will be A1, 1 A2 over A1, mode 1. 01:08:01.160 --> 01:08:05.840 So the free vibration response of any two degree 01:08:05.840 --> 01:08:09.920 of freedom system, linearized equation, any two 01:08:09.920 --> 01:08:12.940 degree of freedom linear system can be made up 01:08:12.940 --> 01:08:16.050 of the sum of two terms. 01:08:16.050 --> 01:08:20.810 The motion at its first natural frequency in its first mode 01:08:20.810 --> 01:08:22.990 shape. 01:08:22.990 --> 01:08:26.250 And another term is the motion at a second natural frequency 01:08:26.250 --> 01:08:28.220 and its second mode shape. 01:08:28.220 --> 01:08:31.740 But now you have two undetermined constants 01:08:31.740 --> 01:08:33.350 out here-- A1 and A2. 01:08:33.350 --> 01:08:34.775 Where do they come from? 01:08:38.340 --> 01:08:43.140 You have to use your initial conditions to get those. 01:08:43.140 --> 01:08:44.460 I'll write down-- let's see. 01:08:44.460 --> 01:08:48.875 I have just maybe enough time to write this down. 01:08:51.560 --> 01:08:52.810 These are functions of time. 01:08:55.340 --> 01:09:07.740 So A1 and A2 come from the ICs, the initial conditions. 01:09:07.740 --> 01:09:16.310 So at t equals 0, for example, plug in t equals 0 into here. 01:09:16.310 --> 01:09:20.380 You get cosine phi. 01:09:20.380 --> 01:09:23.890 And over here another phi 1 and another way over here. 01:09:23.890 --> 01:09:26.680 Cosine of minus phi 1 is cosine phi 1. 01:09:26.680 --> 01:09:31.740 So if you put in t equals 0, you find out the X1 01:09:31.740 --> 01:09:39.420 at 0, which I'll write X1 0, and X2 of 0 01:09:39.420 --> 01:09:45.069 is equal to-- I'll actually write them out. 01:09:45.069 --> 01:09:57.750 This is going to be A1 times 1 cosine phi 1, plus A2 times 01:09:57.750 --> 01:10:03.050 1 cosine phi 2. 01:10:03.050 --> 01:10:08.920 And the second equation that this gives you is A1-- 01:10:08.920 --> 01:10:12.030 and now to keep from writing these many, many times, 01:10:12.030 --> 01:10:21.275 I'm going to let this first A2 over A1 for mode 1 01:10:21.275 --> 01:10:28.130 be R1, and the second one A2 over A1 for mode 2. 01:10:28.130 --> 01:10:30.360 I'll just call it R2. 01:10:30.360 --> 01:10:31.710 And then I can write this out. 01:10:31.710 --> 01:10:40.430 So the second equation, this is R1 cosine phi 1, 01:10:40.430 --> 01:10:48.300 plus A2 R2 cosine phi 2. 01:10:48.300 --> 01:10:51.350 And now I have initial conditions 01:10:51.350 --> 01:10:53.790 that are normally given. 01:10:53.790 --> 01:10:56.220 This is an initial displacement on 1 01:10:56.220 --> 01:11:00.180 and a initial displacement on 2. 01:11:00.180 --> 01:11:01.379 The phis I don't know. 01:11:01.379 --> 01:11:02.420 And the A's I don't know. 01:11:02.420 --> 01:11:05.370 I have four unknowns and two equations. 01:11:05.370 --> 01:11:07.215 How do I get two more equations? 01:11:07.215 --> 01:11:11.520 I take the derivative of this expression to get velocity. 01:11:11.520 --> 01:11:16.620 And I get an A omega here and an A omega here. 01:11:16.620 --> 01:11:21.440 And I plug in t equals 0. 01:11:21.440 --> 01:11:24.310 And I get two more equations. 01:11:24.310 --> 01:11:41.050 So X1 dot and X2 dot equal at t equals 0. 01:11:41.050 --> 01:11:44.700 This gives me two more equations. 01:11:44.700 --> 01:11:53.560 And if I have a place to write them-- for example, 01:11:53.560 --> 01:12:00.750 X1 dot-- or X1 0 dot, the initial condition on velocity. 01:12:00.750 --> 01:12:03.875 X2 dot-- this is two equations. 01:12:06.420 --> 01:12:08.790 I've only got time to write down one of them. 01:12:08.790 --> 01:12:11.680 And you could do the other for exercise. 01:12:11.680 --> 01:12:23.970 You find this is A1 omega 1 sine phi 1, plus A2 omega 01:12:23.970 --> 01:12:27.940 2 sine phi 2. 01:12:27.940 --> 01:12:35.430 And the second equation is you get A1 R1 omega 1 sine phi 1, 01:12:35.430 --> 01:12:43.390 plus A2 R2 omega 2 sine phi 2. 01:12:43.390 --> 01:12:47.390 So now you have one, two. 01:12:47.390 --> 01:12:49.865 And this is the initial conditions on velocity. 01:12:49.865 --> 01:12:52.850 These are initial values of velocity. 01:12:52.850 --> 01:12:55.760 Now you have one, two, three, four equations and one, 01:12:55.760 --> 01:13:02.750 two, three, four unknowns-- the A1, A2, phi 1, phi 2. 01:13:02.750 --> 01:13:13.070 And I guess I will give you the answer, so you have it once. 01:13:30.810 --> 01:13:34.980 So a little tedious but this is sort of in the spirit of we 01:13:34.980 --> 01:13:36.940 do two degree of freedom systems so 01:13:36.940 --> 01:13:39.680 that we can see how it works. 01:13:39.680 --> 01:13:43.360 And then for larger degrees of freedom systems 01:13:43.360 --> 01:13:45.580 you would do this with a computer. 01:13:45.580 --> 01:13:52.555 But the solution for A1 is 1 over R2 minus R1. 01:13:55.485 --> 01:13:57.990 It's all in terms now of things you know. 01:13:57.990 --> 01:14:00.910 These R2's and R1's are part of the mode shape, 01:14:00.910 --> 01:14:06.270 and the rest is initial conditions. 01:14:06.270 --> 01:14:23.020 R2 x1 0, minus X2 0 squared, plus R2 V1 0, 01:14:23.020 --> 01:14:31.550 minus V2 0 quantity squared, over omega 1 01:14:31.550 --> 01:14:36.080 squared, the whole thing square root. 01:14:36.080 --> 01:14:37.630 But now this is all stuff you know. 01:14:37.630 --> 01:14:39.790 The given initial conditions on velocity, 01:14:39.790 --> 01:14:43.530 given initial conditions on displacement. 01:14:43.530 --> 01:14:44.930 You know the natural frequency. 01:14:44.930 --> 01:14:47.020 You know the pieces of the mode shapes. 01:14:47.020 --> 01:14:50.940 Just plug it in, you're going to get a number. 01:14:50.940 --> 01:14:54.350 The magnitude of A1 for the given initial conditions. 01:14:54.350 --> 01:15:10.600 A2-- a very similar expression-- minus R1 X1 0 plus x2 0 01:15:10.600 --> 01:15:35.770 squared And you'd solve for A2 and phi 01:15:35.770 --> 01:15:41.240 1-- it's a little simpler-- minus tangent inverse, 01:15:41.240 --> 01:16:00.540 phi 2 0 minus R2 V1 0, all over omega 1 R2 x1 0, minus x2 0. 01:16:00.540 --> 01:16:02.210 That's one of the phase angles. 01:16:02.210 --> 01:16:06.570 And the other phase angle, a very similar expression. 01:16:06.570 --> 01:16:18.650 Minus tangent inverse, minus R1 V1 0, plus V2 0 omega 01:16:18.650 --> 01:16:25.650 2 R1 x1 0, minus x2 0. 01:16:25.650 --> 01:16:30.100 So just expressions in terms of the initial conditions and you 01:16:30.100 --> 01:16:31.869 can get all four quantities. 01:16:31.869 --> 01:16:33.410 You can also do this on the computer. 01:16:33.410 --> 01:16:37.040 But in the few short lectures that we have 01:16:37.040 --> 01:16:38.967 we're not going to get into that. 01:16:38.967 --> 01:16:40.550 But this just shows you where it goes. 01:16:40.550 --> 01:16:41.590 You could do this now. 01:16:41.590 --> 01:16:44.040 There are straightforward ways of doing it 01:16:44.040 --> 01:16:47.420 with matrix algebra on the computer. 01:16:47.420 --> 01:16:52.470 Next time I'll do maybe just a quick example-- I didn't quite 01:16:52.470 --> 01:16:57.830 get to it today-- of a response to initial conditions problem. 01:16:57.830 --> 01:16:59.410 Plug it in there. 01:16:59.410 --> 01:17:00.715 See what happens. 01:17:00.715 --> 01:17:03.260 But we're out of time. 01:17:03.260 --> 01:17:06.680 See you on Thursday.