WEBVTT 00:00:00.080 --> 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 --> 00:00:17.305 at ocw.mit.edu. 00:00:21.650 --> 00:00:26.150 J. KIM VANDIVER: Today's lecture is not mathematically hard, 00:00:26.150 --> 00:00:30.932 but it's really important to establish vocabulary today. 00:00:30.932 --> 00:00:32.390 We're going to talk about vibration 00:00:32.390 --> 00:00:34.690 for the rest of the term. 00:00:34.690 --> 00:00:40.110 And vibration is essentially applied dynamics. 00:00:40.110 --> 00:00:44.740 So up until now, we've been finding equations of motion, 00:00:44.740 --> 00:00:46.330 but not solving them. 00:00:46.330 --> 00:00:48.190 Did you notice that? 00:00:48.190 --> 00:00:52.660 I've almost never asked you to solve the equation of motion 00:00:52.660 --> 00:00:55.890 that you've just discovered using Lagrange or whatever. 00:00:55.890 --> 00:00:57.390 The rest of the term, we're actually 00:00:57.390 --> 00:01:00.980 going to be talking mostly about the resulting motion. 00:01:00.980 --> 00:01:03.170 The equations of motion are pretty easy to find. 00:01:03.170 --> 00:01:07.020 You have all the techniques that you need to know for finding. 00:01:07.020 --> 00:01:11.140 And now, we're going to talk about how things vibrate. 00:01:11.140 --> 00:01:14.130 So why do we choose vibration? 00:01:14.130 --> 00:01:20.690 Vibration, one, is an incredibly common phenomenon. 00:01:20.690 --> 00:01:23.010 We wouldn't have speech without vibration. 00:01:23.010 --> 00:01:26.980 You wouldn't have musical instruments without vibration. 00:01:26.980 --> 00:01:30.640 It's a positive thing when it's making good music. 00:01:30.640 --> 00:01:33.190 It's a negative thing when it's keeping you awake at night 00:01:33.190 --> 00:01:35.620 because the air conditioner in the next room 00:01:35.620 --> 00:01:38.130 is causing something to rattle in the room 00:01:38.130 --> 00:01:40.190 and it's driving you nuts. 00:01:40.190 --> 00:01:44.000 So you can want it, it can be desirable, 00:01:44.000 --> 00:01:45.880 and you cannot want it. 00:01:45.880 --> 00:01:49.060 And you need to know ways of getting rid of it. 00:01:49.060 --> 00:01:54.340 And so we're going to talk about vibration, about making 00:01:54.340 --> 00:01:56.870 vibration, about suppressing vibration, 00:01:56.870 --> 00:02:01.160 about isolating sensitive instruments from the vibration 00:02:01.160 --> 00:02:03.240 of the floor, things like that. 00:02:03.240 --> 00:02:05.570 So that's the topic of the rest of the term. 00:02:05.570 --> 00:02:09.190 And today, we're going to talk about single degree of freedom 00:02:09.190 --> 00:02:11.490 systems. 00:02:11.490 --> 00:02:17.460 And you might think that we're spending an awful lot of time 00:02:17.460 --> 00:02:19.720 on single degree of freedom systems. 00:02:19.720 --> 00:02:24.500 But actually, there's a reason for that. 00:02:24.500 --> 00:02:27.980 Lots of things in real life, like-- this 00:02:27.980 --> 00:02:30.000 is just an aluminum rod. 00:02:30.000 --> 00:02:32.400 This will vibrate. 00:02:32.400 --> 00:02:35.470 And continuous systems, which this is, 00:02:35.470 --> 00:02:40.390 have a theoretically infinite number of degrees of freedom. 00:02:40.390 --> 00:02:44.950 Yet when it comes to talking about its vibration, 00:02:44.950 --> 00:02:50.320 it is conceptually easy to think about the vibration 00:02:50.320 --> 00:02:54.740 of an object like this, one natural frequency, 00:02:54.740 --> 00:02:57.280 one natural mode at a time. 00:02:57.280 --> 00:02:59.950 And in fact, you can model that natural mode 00:02:59.950 --> 00:03:03.600 with its single degree of freedom equivalent. 00:03:03.600 --> 00:03:06.300 And that's the way I approach vibration. 00:03:06.300 --> 00:03:08.600 So if you can isolate one particular mode, 00:03:08.600 --> 00:03:11.560 you can literally model it as a Mass-Spring-Dashpot. 00:03:11.560 --> 00:03:15.950 So you need to understand the Mass-Spring-Dashpot behavior 00:03:15.950 --> 00:03:21.590 inside and out, because it's the vocabulary we use to do 00:03:21.590 --> 00:03:23.470 much more complicated things. 00:03:23.470 --> 00:03:26.680 So a single degree of freedom system, 00:03:26.680 --> 00:03:32.962 like the simple pendulum, has a natural frequency. 00:03:32.962 --> 00:03:34.295 In this case, it has mode shape. 00:03:38.710 --> 00:03:44.170 Here's another one, kind of fun, single degree of freedom. 00:03:44.170 --> 00:03:46.196 This obviously involves rotation. 00:03:58.356 --> 00:03:59.730 And you can figure that out using 00:03:59.730 --> 00:04:02.500 Lagrange or whatever, single degree of freedom systems. 00:04:02.500 --> 00:04:05.365 But now, I'm going to excite one mode of vibration of this. 00:04:08.638 --> 00:04:09.634 [CLANG] 00:04:09.634 --> 00:04:11.630 [HIGH-PITCHED TONE] 00:04:11.630 --> 00:04:13.225 Hear the real high pitch? 00:04:16.730 --> 00:04:19.079 I'll get it down here by the mic so that people at home 00:04:19.079 --> 00:04:24.430 can hear it-- about a kilohertz, way up there. 00:04:24.430 --> 00:04:27.300 And that's one natural mode of this thing 00:04:27.300 --> 00:04:28.620 in longitudinal vibration. 00:04:28.620 --> 00:04:32.511 When I thump it sideways-- 00:04:32.511 --> 00:04:33.010 [CLANG] 00:04:33.010 --> 00:04:33.900 [LOWER TONE] 00:04:33.900 --> 00:04:34.870 you hear a lower tone. 00:04:34.870 --> 00:04:35.750 Hear that? 00:04:35.750 --> 00:04:36.760 [HUMS LOW] 00:04:36.760 --> 00:04:37.647 rather than-- 00:04:37.647 --> 00:04:38.581 [HUMS HIGH] 00:04:38.581 --> 00:04:40.920 [CLANG] 00:04:40.920 --> 00:04:43.250 That's bending vibration of this thing. 00:04:43.250 --> 00:04:45.040 But each mode of vibration I can think 00:04:45.040 --> 00:04:47.930 of in terms of its equivalent single degree of freedom 00:04:47.930 --> 00:04:48.900 oscillate. 00:04:48.900 --> 00:04:51.430 So we'll get to talking about these things 00:04:51.430 --> 00:04:54.140 a little bit-- continuous systems-- in the last couple 00:04:54.140 --> 00:04:55.760 lectures of the term. 00:04:55.760 --> 00:04:57.940 But for today then, we're really going 00:04:57.940 --> 00:05:01.090 to develop this vocabulary around the vibration 00:05:01.090 --> 00:05:04.620 of single degree of freedom systems. 00:05:04.620 --> 00:05:05.486 So let's start. 00:05:24.643 --> 00:05:27.640 All right. 00:05:27.640 --> 00:05:31.310 So to keep it from being totally boring, 00:05:31.310 --> 00:05:35.280 I'm going to start with a little Mass-Spring-Dashpot that 00:05:35.280 --> 00:05:36.305 has two springs. 00:05:45.280 --> 00:05:49.210 And they're of such a length that unstretched, 00:05:49.210 --> 00:05:52.780 they just meet in the middle. 00:05:52.780 --> 00:06:00.050 And then, I'm going to take a mass 00:06:00.050 --> 00:06:04.170 and I'm going to squeeze it in between these two springs-- 00:06:04.170 --> 00:06:08.170 I can't draw a spring very well today-- and this is k1 00:06:08.170 --> 00:06:11.580 and this is k2 and here's m. 00:06:11.580 --> 00:06:14.150 And we'll put it on a roller so it's obviously constrained 00:06:14.150 --> 00:06:17.910 to motion in one direction. 00:06:17.910 --> 00:06:24.830 And I'll pick this point here as the place 00:06:24.830 --> 00:06:28.244 I'm going to put my inertial coordinate. 00:06:28.244 --> 00:06:30.160 So my inertial coordinate's just measured from 00:06:30.160 --> 00:06:32.740 or happens to be where the endpoints of these two springs 00:06:32.740 --> 00:06:34.080 were. 00:06:34.080 --> 00:06:36.520 Now, to squeeze the spring in here, 00:06:36.520 --> 00:06:40.720 I have this clearly pre-compression 00:06:40.720 --> 00:06:41.990 in these springs. 00:06:41.990 --> 00:06:49.050 So we are no longer in a zero-force state, right? 00:06:49.050 --> 00:06:51.660 So and I want to get the equations of motion in this. 00:06:51.660 --> 00:06:53.772 And moreover, I want to predict-- 00:06:53.772 --> 00:06:55.730 I want to find out what's the natural frequency 00:06:55.730 --> 00:06:58.320 of this spring. 00:06:58.320 --> 00:07:00.610 So let's check your intuition. 00:07:00.610 --> 00:07:02.690 So write down on your piece of paper 00:07:02.690 --> 00:07:08.120 whether or not the natural frequency will be different 00:07:08.120 --> 00:07:11.090 because there's pre-compression, or whether or not 00:07:11.090 --> 00:07:12.810 that pre-compression in the springs 00:07:12.810 --> 00:07:16.880 has nothing to do with the natural frequency. 00:07:16.880 --> 00:07:19.640 So write down on your paper "natural frequency 00:07:19.640 --> 00:07:25.442 is different" or "natural frequency is the same." 00:07:25.442 --> 00:07:26.650 Let's have a prediction here. 00:07:30.379 --> 00:07:32.170 And then, we'll set about figuring this out 00:07:32.170 --> 00:07:33.669 and in the course of doing it, we'll 00:07:33.669 --> 00:07:35.680 develop a little vocabulary. 00:07:35.680 --> 00:07:37.610 All through the course so far, when 00:07:37.610 --> 00:07:39.160 we've done equations of motion, we've 00:07:39.160 --> 00:07:42.590 usually picked the zero-spring-force position. 00:07:42.590 --> 00:07:46.310 And we sort of led you down this rosy path that suggests 00:07:46.310 --> 00:07:47.500 that's the way we do it. 00:07:47.500 --> 00:07:50.340 But there are other ways that you're 00:07:50.340 --> 00:07:52.800 going to find that are preferable to that, sometimes. 00:07:52.800 --> 00:07:56.150 So that's one of the reasons I'm doing this example. 00:07:56.150 --> 00:07:57.665 So let's do a free body diagram. 00:08:06.030 --> 00:08:08.520 And if I held this mass, for example, 00:08:08.520 --> 00:08:10.970 right at the center when I put the springs in, 00:08:10.970 --> 00:08:13.590 it's obvious that this spring gets compressed 00:08:13.590 --> 00:08:16.200 by half of the length of the mass 00:08:16.200 --> 00:08:18.010 and this spring gets compressed by half 00:08:18.010 --> 00:08:20.160 of the length of the mass, right? 00:08:20.160 --> 00:08:23.980 So this is going to be L long. 00:08:23.980 --> 00:08:26.340 So if I held it right in the middle, 00:08:26.340 --> 00:08:28.715 it would compress L/2 and L/2. 00:08:28.715 --> 00:08:30.590 But then, when I release it, if these springs 00:08:30.590 --> 00:08:32.580 are a different spring constant, it's 00:08:32.580 --> 00:08:33.840 going to move a little bit. 00:08:33.840 --> 00:08:37.260 So the force on this side pushing back 00:08:37.260 --> 00:08:43.330 is sum k1 times L/2 minus the distance 00:08:43.330 --> 00:08:46.660 that I move in that direction, which would relieve it. 00:08:46.660 --> 00:08:49.690 And the force on this side also pushes back. 00:08:49.690 --> 00:08:56.840 It's k2 times L/2 over 2 plus x, because when 00:08:56.840 --> 00:09:00.550 I go in that direction, I'm compressing it even further. 00:09:00.550 --> 00:09:04.360 And those are the total forces in the x-direction 00:09:04.360 --> 00:09:05.220 on this body. 00:09:05.220 --> 00:09:08.040 There's an N and an mg, which we know 00:09:08.040 --> 00:09:09.930 we don't have to deal with because we're only 00:09:09.930 --> 00:09:12.750 interested in motion left and right. 00:09:12.750 --> 00:09:13.250 All right? 00:09:13.250 --> 00:09:18.980 So we can say sum of the forces in the x-direction, 00:09:18.980 --> 00:09:22.810 mass times the acceleration. 00:09:22.810 --> 00:09:39.130 And those forces are k1 L/2 minus x minus k2 L/2 plus x. 00:09:39.130 --> 00:09:44.840 And that's the complete equation of motion for this problem. 00:09:44.840 --> 00:09:47.060 And rearrange it so that I get the functions 00:09:47.060 --> 00:09:50.006 of x together here. 00:09:50.006 --> 00:10:00.160 So mx double dot plus k1 plus k2 times x 00:10:00.160 --> 00:10:10.690 equals L/2 times k1 minus k2. 00:10:10.690 --> 00:10:13.410 And that's your equation of motion. 00:10:13.410 --> 00:10:15.050 It's non-homogeneous. 00:10:15.050 --> 00:10:19.620 This is all constants on the right-hand side. 00:10:19.620 --> 00:10:25.672 And on the left-hand side are the functions of x, right? 00:10:25.672 --> 00:10:27.930 So what's the natural frequency of the system? 00:10:34.888 --> 00:10:37.380 AUDIENCE: Square root of k1 plus k2 over m. 00:10:37.380 --> 00:10:40.530 J. KIM VANDIVER: I hear a square root of the quantity k1 00:10:40.530 --> 00:10:44.300 plus k2, the stiffness, divided by m, k over m, 00:10:44.300 --> 00:10:46.899 a usual Mass-Spring-Dashpot system. 00:10:46.899 --> 00:10:48.440 Did the pre-compression have anything 00:10:48.440 --> 00:10:50.490 to do with the natural frequency? 00:10:50.490 --> 00:10:52.960 I won't ask you to embarrass yourselves, but a few of you 00:10:52.960 --> 00:10:56.030 probably got that wrong, all right? 00:10:56.030 --> 00:11:00.440 So there's a lesson in this that I want you to go away with. 00:11:08.090 --> 00:11:09.740 and I'll say it once. 00:11:09.740 --> 00:11:17.090 And that is when an external force 00:11:17.090 --> 00:11:23.000 has nothing to do with the motion coordinates 00:11:23.000 --> 00:11:24.739 in the problem. 00:11:24.739 --> 00:11:26.405 It doesn't affect the natural frequency. 00:11:29.900 --> 00:11:31.320 These come from external forces. 00:11:31.320 --> 00:11:34.340 These are these pre-compressions, right? 00:11:34.340 --> 00:11:39.000 And I can separate them out and they are not functions of x. 00:11:39.000 --> 00:11:41.215 The stuff on the right-hand side of the equation, 00:11:41.215 --> 00:11:44.010 that's not a function of the motion variable-- 00:11:44.010 --> 00:11:47.820 cannot affect the natural frequency. 00:11:47.820 --> 00:11:49.550 So I'll give you another one. 00:11:52.070 --> 00:11:59.910 This is our common thing hanging from a stick. 00:12:02.890 --> 00:12:05.940 I've taken my system I built the other day 00:12:05.940 --> 00:12:11.620 for a different purpose, but now, it's just a mass 00:12:11.620 --> 00:12:13.910 hanging from a spring. 00:12:13.910 --> 00:12:16.420 And it's right now at its equilibrium position 00:12:16.420 --> 00:12:21.862 or there's non-zero force in the spring. 00:12:21.862 --> 00:12:23.320 It clearly has a natural frequency. 00:12:26.070 --> 00:12:29.205 And is that natural frequency a function of gravity? 00:12:32.420 --> 00:12:34.870 And so if you go to write the equation to motion 00:12:34.870 --> 00:12:48.940 of this system, you would find mx double dot plus kx equals mg 00:12:48.940 --> 00:12:49.630 g. 00:12:49.630 --> 00:12:51.350 But the mg is not a function of x. 00:12:51.350 --> 00:12:53.600 The natural frequency's again, the square root of k/m. 00:13:01.470 --> 00:13:04.360 Now, we want to talk about solving this differential 00:13:04.360 --> 00:13:05.540 equation. 00:13:05.540 --> 00:13:07.380 And because it's got this constant term 00:13:07.380 --> 00:13:09.350 in the right-hand side, it's non-homogeneous, 00:13:09.350 --> 00:13:11.760 which is kind of a nuisance term in terms of dealing 00:13:11.760 --> 00:13:13.390 with a differential equation. 00:13:13.390 --> 00:13:19.610 It'd be a lot nicer if the right-hand side were 0. 00:13:19.610 --> 00:13:26.010 So I want to make the right-hand side of this one 0. 00:13:26.010 --> 00:13:32.180 And draw a use of a conclusion from that. 00:13:50.540 --> 00:13:51.920 First thing I need to know is I'd 00:13:51.920 --> 00:13:56.060 like to know what is the static equilibrium position of this. 00:14:07.740 --> 00:14:09.674 And when you go to compute static equilibrium, 00:14:09.674 --> 00:14:11.090 you look at the equation of motion 00:14:11.090 --> 00:14:13.630 and you say, make all motion variables things 00:14:13.630 --> 00:14:15.560 that are functions of time 0. 00:14:15.560 --> 00:14:19.160 So no acceleration-- you're left with this. 00:14:19.160 --> 00:14:21.730 So you just solve this for whatever the value of x is 00:14:21.730 --> 00:14:24.670 and I'll call it x of s for x-static. 00:14:24.670 --> 00:14:26.290 And you'll find that, oh, well, it's 00:14:26.290 --> 00:14:34.770 that term divided by k1 plus k2, k1 00:14:34.770 --> 00:14:40.100 minus k2 all over k1 plus k2. 00:14:40.100 --> 00:14:41.465 And that's the static position. 00:14:47.970 --> 00:14:50.910 So now, let's say, ah, well, we started off 00:14:50.910 --> 00:14:54.340 with this motion variable that wasn't arbitrarily defined 00:14:54.340 --> 00:14:56.540 at the middle. 00:14:56.540 --> 00:15:00.380 And let's say that, well, it's made up 00:15:00.380 --> 00:15:05.060 of a static component, which is a constant, just 00:15:05.060 --> 00:15:09.720 a value, plus a dynamic component I'll 00:15:09.720 --> 00:15:11.450 call x of d, which moves. 00:15:11.450 --> 00:15:14.310 This is the function of time. 00:15:14.310 --> 00:15:15.240 This is a constant. 00:15:15.240 --> 00:15:17.660 It's not a function of time. 00:15:17.660 --> 00:15:19.480 And that means if I take its derivative, 00:15:19.480 --> 00:15:21.835 I might need a value for x dot. 00:15:21.835 --> 00:15:23.280 That goes away. 00:15:23.280 --> 00:15:25.220 It's just xd dot. 00:15:25.220 --> 00:15:31.440 And x double dot is xd double dot. 00:15:31.440 --> 00:15:36.230 And let's substitute this into my equation of motion. 00:15:36.230 --> 00:15:51.300 So it becomes m xd double dot plus k1 plus k2 times-- 00:15:51.300 --> 00:16:00.200 and now, this term has got two pieces now-- times xd plus k1 00:16:00.200 --> 00:16:11.420 plus k2 times xs equals L/2 k1 minus k2. 00:16:15.418 --> 00:16:17.760 All right? 00:16:17.760 --> 00:16:22.340 Now if I say, well, let's examine the static case, 00:16:22.340 --> 00:16:24.900 then this goes away. 00:16:24.900 --> 00:16:27.740 For the static equilibrium case, this term is 0. 00:16:27.740 --> 00:16:31.790 This term is 0 because the dynamic motion 00:16:31.790 --> 00:16:35.710 is 0 in the static case. 00:16:35.710 --> 00:16:39.460 That xd is motion about the static equilibrium position. 00:16:39.460 --> 00:16:41.860 So for static case, these two terms go away 00:16:41.860 --> 00:16:44.920 and we know that this equals that. 00:16:44.920 --> 00:16:47.110 But if that's true, we can get rid of these. 00:16:47.110 --> 00:16:49.530 They cancel one another. 00:16:49.530 --> 00:16:55.360 These terms cancel and I'm left with m xd double dot 00:16:55.360 --> 00:17:03.460 plus k equivalent, I'll call it, xd equals 0. 00:17:03.460 --> 00:17:05.589 So the k equivalent's just the total stiffnesses 00:17:05.589 --> 00:17:07.297 in the system, whatever works out, right? 00:17:07.297 --> 00:17:10.960 In this case, it's k1 plus k2 and the natural frequency, 00:17:10.960 --> 00:17:16.440 omega n, is the square root of k equivalent divided by m. 00:17:19.450 --> 00:17:23.709 So most often, if you're interested in vibration, 00:17:23.709 --> 00:17:25.500 you're interested in natural frequencies, 00:17:25.500 --> 00:17:29.360 you're interested in solving the differential equation, 00:17:29.360 --> 00:17:33.570 you will find it advantageous to write your equations of motion 00:17:33.570 --> 00:17:37.065 around the static equilibrium position. 00:17:40.030 --> 00:17:42.840 So I could have started this problem by saying, 00:17:42.840 --> 00:17:44.990 whatever the static equilibrium position is 00:17:44.990 --> 00:17:48.190 of this thing, that's what I'm measuring x from. 00:17:50.850 --> 00:17:54.874 And then, I would have come to this equation eventually. 00:17:54.874 --> 00:17:57.540 You'd have to figure out what is the static equilibrium position 00:17:57.540 --> 00:18:00.560 and know what you're doing, but once you know it, 00:18:00.560 --> 00:18:02.050 then you have the answer. 00:18:02.050 --> 00:18:06.830 Now, the same thing is true of that problem. 00:18:06.830 --> 00:18:09.600 That's a non-homogeneous differential equation 00:18:09.600 --> 00:18:14.610 for the hanging mass. 00:18:14.610 --> 00:18:19.610 And we derive the equations of motion things 00:18:19.610 --> 00:18:22.780 for this many different ways this term, all right? 00:18:22.780 --> 00:18:25.250 But we usually said, zero-spring force. 00:18:25.250 --> 00:18:28.010 But now, if you started from here and said, 00:18:28.010 --> 00:18:29.940 this is the static equilibrium position, 00:18:29.940 --> 00:18:33.110 what's the motion about this position, 00:18:33.110 --> 00:18:40.636 then you'd get the equation with 0 on the right-hand side-- 00:18:40.636 --> 00:18:42.260 lots of advantages there to using that. 00:19:07.450 --> 00:19:10.620 All single degree of freedom oscillators 00:19:10.620 --> 00:19:14.120 will boil down to this equation. 00:19:14.120 --> 00:19:16.540 This is one involving translation, 00:19:16.540 --> 00:19:19.390 but for a simple pendulum. 00:19:19.390 --> 00:19:23.975 This object, for example, is a pendulum, but it's rotational. 00:19:23.975 --> 00:19:26.100 So it's a pendulum, but it's one degree of freedom. 00:19:28.650 --> 00:19:31.210 All pendulum problems, if you do them 00:19:31.210 --> 00:19:33.700 about equilibrium positions, boil 00:19:33.700 --> 00:19:39.130 down to some I with respect to the point 00:19:39.130 --> 00:19:44.580 that they're rocking about, theta double 00:19:44.580 --> 00:19:52.030 dot plus some Kt, torsional spring constant theta, 00:19:52.030 --> 00:19:52.990 equals 0. 00:19:52.990 --> 00:19:55.300 They take the same form. 00:19:55.300 --> 00:19:57.545 So all translational single degree 00:19:57.545 --> 00:20:00.020 of freedom systems, all rotational single degree 00:20:00.020 --> 00:20:03.070 of freedom systems, it's the same differential equation-- 00:20:03.070 --> 00:20:06.000 just this involves mass and linear acceleration. 00:20:06.000 --> 00:20:08.370 This involves mass moment of inertia and rotational 00:20:08.370 --> 00:20:10.270 acceleration. 00:20:10.270 --> 00:20:13.260 So everything that I say about the solution 00:20:13.260 --> 00:20:15.710 to single degree of freedom systems 00:20:15.710 --> 00:20:19.390 applies to both types of problems. 00:20:19.390 --> 00:20:27.900 So let's look into the solution of this equation briefly. 00:20:33.040 --> 00:20:35.545 Mostly, I'm doing this to establish some terminology. 00:20:38.960 --> 00:20:45.320 So a solution I know or I can show that xd of t, 00:20:45.320 --> 00:20:47.590 the solution to this problem-- notice, 00:20:47.590 --> 00:20:50.090 are there any external forces, by the way, excitations, 00:20:50.090 --> 00:20:51.320 f of t's or anything? 00:20:51.320 --> 00:20:52.350 No. 00:20:52.350 --> 00:20:55.170 So this thing has no external excitation 00:20:55.170 --> 00:20:56.620 that's going to make it move. 00:20:56.620 --> 00:21:02.640 So it's only source of vibration or motion is what? 00:21:02.640 --> 00:21:08.040 Comes from-- I hear initial conditions, right? 00:21:08.040 --> 00:21:10.670 You have to do something to perturb it 00:21:10.670 --> 00:21:13.980 and then it will vibrate. 00:21:13.980 --> 00:21:15.050 So here it is. 00:21:15.050 --> 00:21:18.960 It's about its equilibrium position. 00:21:18.960 --> 00:21:23.630 I give it an initial deflection and let go. 00:21:23.630 --> 00:21:26.560 Or it's around its initial condition 00:21:26.560 --> 00:21:29.540 and I give it an initial velocity. 00:21:29.540 --> 00:21:32.130 It also responds to some combination of the two. 00:21:32.130 --> 00:21:34.190 So initial conditions are the only things 00:21:34.190 --> 00:21:37.300 that account for motion of something 00:21:37.300 --> 00:21:39.500 without external excitation. 00:21:39.500 --> 00:21:42.430 And that motion, I can write that solution 00:21:42.430 --> 00:21:49.360 as A cosine omega t. 00:21:49.360 --> 00:21:53.070 You'll find this is a possible solution. 00:21:53.070 --> 00:21:59.000 B sine omega t is another possible solution. 00:21:59.000 --> 00:22:06.690 Sum A cosine omega t minus phase angle's also a solution. 00:22:06.690 --> 00:22:15.210 And sum A e to the i omega t you'll find is also a solution. 00:22:15.210 --> 00:22:17.150 Any of those things you could throw in 00:22:17.150 --> 00:22:20.350 and the precise values of these things, the A's, the B's, the 00:22:20.350 --> 00:22:23.965 phi's, and so forth depend on-- 00:22:23.965 --> 00:22:25.340 AUDIENCE: the initial conditions. 00:22:25.340 --> 00:22:26.740 J. KIM VANDIVER: The initial conditions. 00:22:26.740 --> 00:22:28.040 So let's do this one quickly. 00:22:37.950 --> 00:22:40.370 All right. 00:22:40.370 --> 00:22:54.600 And I'll choose And I'm going to stop writing the x sub d here. 00:22:54.600 --> 00:23:02.356 This is now my position from the equilibrium point. 00:23:02.356 --> 00:23:05.660 So x of t-- I'm going to say, let 00:23:05.660 --> 00:23:16.690 it be an A1 cosine omega t plus a B1 sine omega t 00:23:16.690 --> 00:23:18.460 and plug it in. 00:23:18.460 --> 00:23:28.670 When I plug it into the equation of motion, 00:23:28.670 --> 00:23:31.120 x double dot requires you to take two derivatives of each 00:23:31.120 --> 00:23:31.830 of these terms. 00:23:31.830 --> 00:23:34.940 Two derivatives of cosine gives you minus omega cosine. 00:23:34.940 --> 00:23:39.690 Two derivative sine minus omega squared cosine minus omega 00:23:39.690 --> 00:23:40.850 squared sine. 00:23:40.850 --> 00:23:46.390 So the answer comes out minus m omega 00:23:46.390 --> 00:23:56.300 squared plus k equivalent here times A1 cosine 00:23:56.300 --> 00:24:07.836 plus B1 sine-- omega t's obviously in them-- equals 0. 00:24:07.836 --> 00:24:09.710 So I just plugged in that equation of motion. 00:24:09.710 --> 00:24:11.790 I get this back. 00:24:11.790 --> 00:24:13.250 This is what I started with. 00:24:13.250 --> 00:24:15.670 That's x. 00:24:15.670 --> 00:24:22.990 In general, it is not equal to 0, can take on all 00:24:22.990 --> 00:24:23.730 sorts of values. 00:24:23.730 --> 00:24:27.350 So that's not generally 0 and that means this must be. 00:24:30.470 --> 00:24:32.600 And from this, then, when we solve this, 00:24:32.600 --> 00:24:37.907 we find that omega what we call n squared is k over m. 00:24:37.907 --> 00:24:40.490 And that's, of course, where our natural frequency comes from. 00:24:40.490 --> 00:24:54.349 This is called the undamped natural frequency, 00:24:54.349 --> 00:24:56.390 because there's no dampening in this problem yet. 00:24:56.390 --> 00:24:57.806 We get the square root of k over m 00:24:57.806 --> 00:25:01.650 is the natural frequency of the system. 00:25:01.650 --> 00:25:04.420 Let's find out what are A1 and B1. 00:25:04.420 --> 00:25:13.864 Well, let's let x0 be x at t equals 0 here. 00:25:16.850 --> 00:25:19.950 And if we just plug that in here, put t equals 0 here, 00:25:19.950 --> 00:25:21.270 cosine goes to 1. 00:25:21.270 --> 00:25:23.140 This term goes away. 00:25:23.140 --> 00:25:30.080 So this implies that A1 equals x0. 00:25:30.080 --> 00:25:33.970 So we find out right away that the A1 cosine omega 00:25:33.970 --> 00:25:38.100 t takes care of the response to an initial deflection. 00:25:38.100 --> 00:25:54.700 And we need a x dot here minus A1 omega sine omega t plus B1 00:25:54.700 --> 00:25:58.440 omega cosine omega t. 00:25:58.440 --> 00:26:02.230 That's the derivative of x. 00:26:02.230 --> 00:26:04.550 You know the solution's that, so its first derivative, 00:26:04.550 --> 00:26:06.640 the velocity, must look like this. 00:26:06.640 --> 00:26:13.450 And let's let v0 equals x dot at t equals 0. 00:26:13.450 --> 00:26:17.180 When we plug that in, this term goes away 00:26:17.180 --> 00:26:23.570 and we get B1 omega and cosine is 1. 00:26:23.570 --> 00:26:29.750 So therefore, B1 is v0 over omega. 00:26:29.750 --> 00:26:33.120 But in fact, the omega's omega n, 00:26:33.120 --> 00:26:38.740 because we already found that, that the only frequency that 00:26:38.740 --> 00:26:42.920 satisfies the equation of motion when 00:26:42.920 --> 00:26:46.530 you have only initial conditions in the system, 00:26:46.530 --> 00:26:49.360 the only frequency that is allowed in the answer 00:26:49.360 --> 00:26:52.460 is the natural frequency. 00:26:52.460 --> 00:27:02.630 So we now know B1 is v0 over omega n and A1 is x0. 00:27:02.630 --> 00:27:08.300 So if I give you any combination of initial displacement 00:27:08.300 --> 00:27:12.120 and initial velocity, you can write out for me the exact time 00:27:12.120 --> 00:27:14.510 history of the motion. 00:27:14.510 --> 00:27:19.370 X0 to cosine omega t plus v0 over omega n sine omega t 00:27:19.370 --> 00:27:23.740 is the complete solution for a response to initial conditions. 00:27:55.780 --> 00:28:05.920 So any translational oscillator one degree 00:28:05.920 --> 00:28:13.950 of freedom where you have a translational coordinate 00:28:13.950 --> 00:28:18.730 measured from its equilibrium position 00:28:18.730 --> 00:28:23.290 has the equation of motion-- actually, 00:28:23.290 --> 00:28:24.290 you've done this enough. 00:28:24.290 --> 00:28:27.620 But if we added a force here and we added some damping 00:28:27.620 --> 00:28:29.680 and I wanted the equation of motion of this, 00:28:29.680 --> 00:28:34.830 you know that it's m x double dot plus b x 00:28:34.830 --> 00:28:42.430 dot plus kx equals F of t. 00:28:42.430 --> 00:28:45.034 And so you're going to be confronted with problems-- 00:28:45.034 --> 00:28:46.700 find the equation of motion in a system. 00:28:46.700 --> 00:28:47.950 It comes up looking like that and they say, 00:28:47.950 --> 00:28:49.585 what's the natural frequency? 00:28:49.585 --> 00:28:50.850 And I've been a little sloppy. 00:28:50.850 --> 00:28:55.310 I really mean, what's the undamped natural frequency? 00:28:55.310 --> 00:28:56.890 And so to find the undamped-- when 00:28:56.890 --> 00:28:59.360 one says that, what's the undamped natural frequency, 00:28:59.360 --> 00:29:11.500 you just temporarily let b and F be 0, just temporarily, 00:29:11.500 --> 00:29:19.932 and solve then for omega n equals square root of k/n. 00:29:19.932 --> 00:29:21.180 It's what you do. 00:29:21.180 --> 00:29:26.890 And then, so we know this is a parameter that tells us 00:29:26.890 --> 00:29:29.700 about the behavior of the system, which we always 00:29:29.700 --> 00:29:32.320 want to know for the single degree of freedom systems. 00:29:32.320 --> 00:29:34.390 What is the natural frequency of the system? 00:29:42.712 --> 00:29:49.050 And we know for b equals 0 and F of 0, 00:29:49.050 --> 00:29:52.590 then the response can be only due to initial conditions. 00:29:52.590 --> 00:29:54.636 So we have x of t. 00:29:54.636 --> 00:29:59.770 We know it's going to be some x0 cosine omega n 00:29:59.770 --> 00:30:08.110 t plus v0 over omega n sine omega n t. 00:30:08.110 --> 00:30:11.510 And every simple vibration system in the world 00:30:11.510 --> 00:30:14.310 behaves basically like this from initial conditions. 00:30:14.310 --> 00:30:17.570 It'll be some part responding to the initial displacement, 00:30:17.570 --> 00:30:19.830 some part to the initial velocity. 00:30:19.830 --> 00:30:23.780 And damping is going to make it a little bit more complex, 00:30:23.780 --> 00:30:25.470 but not actually by much. 00:30:25.470 --> 00:30:29.180 The same basic terms appear even when you have damping in it. 00:30:33.940 --> 00:30:45.640 This can be expressed as sum A cosine omega, in this case, 00:30:45.640 --> 00:30:48.110 n t minus the phase angle. 00:30:48.110 --> 00:30:53.820 And it's useful to know this trigonometric identity 00:30:53.820 --> 00:30:55.980 to be able to put things together 00:30:55.980 --> 00:30:57.660 into an expression like that. 00:30:57.660 --> 00:31:07.100 And you'll find out that A is just the square root of the two 00:31:07.100 --> 00:31:07.630 pieces. 00:31:07.630 --> 00:31:09.270 It's a sine and cosine term. 00:31:09.270 --> 00:31:16.910 So you have an x0 squared plus a v0 over omega n squared 00:31:16.910 --> 00:31:18.970 square root. 00:31:18.970 --> 00:31:21.620 Remember, this is any A and B. It's just a square root 00:31:21.620 --> 00:31:23.620 of A squared plus B squared. 00:31:23.620 --> 00:31:25.100 That's what we're doing here. 00:31:25.100 --> 00:31:34.360 And the phase angle, the tangent inverse 00:31:34.360 --> 00:31:38.270 of this-- we've been calling this like an A and this 00:31:38.270 --> 00:31:39.680 is the B quantity. 00:31:39.680 --> 00:31:46.230 So tangent inverse of-- get my signs right-- 00:31:46.230 --> 00:31:49.480 B over A, which in this case then 00:31:49.480 --> 00:32:12.710 is tangent inverse of v0 over x0 omega n. 00:32:12.710 --> 00:32:15.310 That's all there is to it. 00:32:15.310 --> 00:32:20.030 And finally, another trig thing that you 00:32:20.030 --> 00:32:22.910 need to know-- we're going to use it quite a bit-- 00:32:22.910 --> 00:32:29.090 is that if you have an expression A cosine omega 00:32:29.090 --> 00:32:46.278 t minus phi, that's equal to the real part of A e to the i omega 00:32:46.278 --> 00:32:46.778 t. 00:32:50.200 --> 00:32:57.710 And if A is real and-- I don't want to write it that way-- 00:32:57.710 --> 00:33:09.550 when A is real, it's A times e to the i omega t minus phi, 00:33:09.550 --> 00:33:13.530 because Euler's formula says e to the i theta 00:33:13.530 --> 00:33:19.850 equals cosine theta plus i sine of theta. 00:33:19.850 --> 00:33:22.590 So if you have an i omega t minus phi 00:33:22.590 --> 00:33:26.070 here, you get back a cosine omega 00:33:26.070 --> 00:33:30.240 t minus phi and another term, an i sine omega t minus phi. 00:33:30.240 --> 00:33:36.130 So you can always express that as the real part of that. 00:33:36.130 --> 00:33:38.130 So we're going to need that little trig identity 00:33:38.130 --> 00:33:39.230 as we go through the term. 00:33:44.320 --> 00:33:54.925 Now, I've found in many years of teaching vibration 00:33:54.925 --> 00:33:58.800 that something that many students 00:33:58.800 --> 00:34:02.810 find a little confusing is this notion of phase angle. 00:34:02.810 --> 00:34:05.400 What does "phase angle" really mean? 00:34:05.400 --> 00:34:09.080 So I'll try to explain it to you in a couple different ways. 00:34:09.080 --> 00:34:15.159 So let's look at what this vibration 00:34:15.159 --> 00:34:18.070 that we're talking about here, x0 cosine omega 00:34:18.070 --> 00:34:24.179 t plus v0 over omega n sine-- what's it look like? 00:34:24.179 --> 00:34:32.530 So that's-- we've just got our-- and we see what it looks like. 00:34:32.530 --> 00:34:37.520 But if you plot the motion of this thing just versus time, 00:34:37.520 --> 00:34:42.899 what's it look like and where does phase angle come into it? 00:34:42.899 --> 00:34:49.560 So this is now x of t and this is 00:34:49.560 --> 00:34:58.894 t equals 0 and this undamped system 00:34:58.894 --> 00:35:00.860 is essentially going to look like that. 00:35:04.540 --> 00:35:07.730 And this is the value x0, the amplitude, 00:35:07.730 --> 00:35:11.900 the initial condition on x that you began with. 00:35:11.900 --> 00:35:21.420 And right here, the slope-- v0 is the slope, the initial slope 00:35:21.420 --> 00:35:27.140 of this curve, right, because the time derivative is F x dot. 00:35:27.140 --> 00:35:32.320 If we were plotting x dot, the initial velocity is omega x0. 00:35:32.320 --> 00:35:36.200 And so it's just the slope is v0 here. 00:35:36.200 --> 00:35:38.450 So this is your initial velocity. 00:35:38.450 --> 00:35:42.540 This is the-- and I didn't-- yeah, that's right. 00:35:50.850 --> 00:35:54.570 This is the initial displacement. 00:35:54.570 --> 00:35:56.290 The total written out mathematically, 00:35:56.290 --> 00:35:58.100 it looks like that. 00:35:58.100 --> 00:36:04.040 And I'm plotting this function, A cosine omega t minus phi. 00:36:04.040 --> 00:36:05.660 Yeah? 00:36:05.660 --> 00:36:06.510 Did I see a hand up? 00:36:06.510 --> 00:36:12.310 AUDIENCE: Does x0 at t equals 0 or is it a little bit after? 00:36:12.310 --> 00:36:14.050 J. KIM VANDIVER: Well, I was just 00:36:14.050 --> 00:36:16.174 looking at it myself and said, this can't be right. 00:36:18.160 --> 00:36:22.290 This has got to be the initial condition on x and this 00:36:22.290 --> 00:36:24.250 has to be the initial condition on v. 00:36:24.250 --> 00:36:27.380 Now, whatever this turns out to be is 00:36:27.380 --> 00:36:28.589 whatever it turns out to be. 00:36:28.589 --> 00:36:29.880 You have some initial velocity. 00:36:29.880 --> 00:36:31.550 You have some initial displacement. 00:36:31.550 --> 00:36:34.640 The system can actually peak out sometime later 00:36:34.640 --> 00:36:36.550 at a maximum value, right? 00:36:36.550 --> 00:36:41.550 And that maximum value is that. 00:36:41.550 --> 00:36:46.800 So this over here is the square root 00:36:46.800 --> 00:36:57.210 of x0 squared plus v0 over omega n squared square root. 00:36:57.210 --> 00:36:58.570 That's what the peak value is. 00:36:58.570 --> 00:37:01.680 And this system's undamped, so it just goes on forever. 00:37:01.680 --> 00:37:07.120 So the question is, though, what is this gap here 00:37:07.120 --> 00:37:11.400 between when it starts and when it meets its maximum? 00:37:11.400 --> 00:37:19.680 Well, when we use an expression like-- we 00:37:19.680 --> 00:37:24.400 said we can express this as some A cosine omega t minus phi. 00:37:24.400 --> 00:37:27.250 It's just the point at which the cosine then 00:37:27.250 --> 00:37:28.185 reaches its maximum. 00:37:32.290 --> 00:37:40.870 So if this axis here is omega t, if we plot this actually 00:37:40.870 --> 00:37:45.350 versus omega t, then one full cycle 00:37:45.350 --> 00:37:51.150 here is 2 pi or 360 degrees. 00:37:51.150 --> 00:37:55.470 So if you plot it versus omega t, then this gap in here 00:37:55.470 --> 00:37:58.510 is just phi. 00:37:58.510 --> 00:38:02.370 That's the delay in angle, if you will, 00:38:02.370 --> 00:38:04.680 that the system goes through between getting 00:38:04.680 --> 00:38:06.740 from the initial conditions to getting 00:38:06.740 --> 00:38:08.160 to the peak of the cosine. 00:38:14.460 --> 00:38:20.250 And phi must also then be equal to some omega 00:38:20.250 --> 00:38:28.080 n times a delta tau, I'll call it, some time delay. 00:38:28.080 --> 00:38:35.500 So if this is plotted-- if this axis is time-- 00:38:35.500 --> 00:38:50.580 not omega t, but time-- then x the same plot, this delay here, 00:38:50.580 --> 00:38:52.840 this is a time delay. 00:38:52.840 --> 00:38:55.060 And when you plot it against time, 00:38:55.060 --> 00:38:58.640 it's a delay in time to get to the peak. 00:38:58.640 --> 00:39:04.750 And omega n delta tau, this delay, 00:39:04.750 --> 00:39:07.390 must be equal to the phase angle. 00:39:07.390 --> 00:39:13.730 So the delta tau, this time delay, is phi over omega n. 00:39:16.560 --> 00:39:20.220 So you can think about this as a delay in time 00:39:20.220 --> 00:39:23.890 or as a shift in phase angle, depending on whether or not 00:39:23.890 --> 00:39:27.130 you want to plot this thing as a function of omega t 00:39:27.130 --> 00:39:29.340 or as a function of time. 00:39:29.340 --> 00:39:33.820 But you're going to need this concept of phase angle 00:39:33.820 --> 00:39:36.690 the rest of the term. 00:39:36.690 --> 00:39:39.035 Want to ask any questions about phase? 00:39:45.190 --> 00:39:51.170 Because we're doing vibration for the remainder of the term, 00:39:51.170 --> 00:39:55.510 this is an introduction to a topic called linear systems. 00:39:55.510 --> 00:39:58.960 And so this is basically the fundamental stuff in which you 00:39:58.960 --> 00:40:01.230 then, when you go on to 2004, which 00:40:01.230 --> 00:40:03.040 is controls and that sort of thing, 00:40:03.040 --> 00:40:06.900 this is the basic intro to it. 00:40:06.900 --> 00:40:11.350 And we'll talk more about linear system behavior as we go along. 00:40:20.600 --> 00:40:23.210 Now, we're going to do something that you've-- much of this 00:40:23.210 --> 00:40:24.920 stuff I know you've seen before. 00:40:24.920 --> 00:40:29.060 Some of the new parts is just vocabulary and ways 00:40:29.060 --> 00:40:31.020 of thinking about vibration that engineers 00:40:31.020 --> 00:40:33.880 do that mathematicians tend not to. 00:40:33.880 --> 00:40:36.896 So you have seen most of this stuff before where? 00:40:36.896 --> 00:40:37.651 AUDIENCE: 1.803. 00:40:37.651 --> 00:40:38.900 J. KIM VANDIVER: 1.803, right? 00:40:38.900 --> 00:40:40.340 You've done all this. 00:40:40.340 --> 00:40:49.210 And a year ago last May, in May, I taught the 1803 lecture 00:40:49.210 --> 00:40:51.120 with Professor Haynes Miller. 00:40:51.120 --> 00:40:52.600 Now, if you had 1.803 last spring, 00:40:52.600 --> 00:40:54.799 I think you had somebody different. 00:40:54.799 --> 00:40:56.090 But he invited me to come here. 00:40:56.090 --> 00:40:57.620 It was in the same classroom and we 00:40:57.620 --> 00:41:02.210 taught the second-order ordinary differential equation together. 00:41:02.210 --> 00:41:03.720 It was really a lot of fun. 00:41:03.720 --> 00:41:05.529 He said, well, here's what we do. 00:41:05.529 --> 00:41:07.070 And then, I said, oh, well, engineers 00:41:07.070 --> 00:41:08.890 look at it the following way. 00:41:08.890 --> 00:41:10.940 So what I'm going to show you is what 00:41:10.940 --> 00:41:12.270 he and I did in class that day. 00:41:12.270 --> 00:41:13.940 You can go back and watch that on video. 00:41:13.940 --> 00:41:15.150 It's kind of fun. 00:41:15.150 --> 00:41:17.080 But I'll give you my take on it today. 00:41:17.080 --> 00:41:18.980 So this is the engineer's view of what 00:41:18.980 --> 00:41:21.310 you've already seen in 1.803. 00:41:21.310 --> 00:41:25.100 So we have that system and we have that equation of motion. 00:41:25.100 --> 00:41:27.270 And the engineers and mathematicians 00:41:27.270 --> 00:41:31.840 would more or less agree to that m x double dot plus bx. 00:41:31.840 --> 00:41:34.120 But I went and looked at the web page last night. 00:41:34.120 --> 00:41:37.240 Last spring, the person used c instead of b. 00:41:37.240 --> 00:41:40.010 Haynes Miller the year before used b. 00:41:40.010 --> 00:41:45.730 So you can't depend on any absolute consistency. 00:41:45.730 --> 00:41:51.020 So let's start off with our homogeneous equation here. 00:41:51.020 --> 00:41:55.380 And I'm looking now for the response to initial conditions 00:41:55.380 --> 00:41:56.480 with damping. 00:41:56.480 --> 00:41:58.470 You've done this in 1.803. 00:41:58.470 --> 00:42:02.180 You know that you can solve this by assuming a solution 00:42:02.180 --> 00:42:05.310 of a form Ae to the st. 00:42:05.310 --> 00:42:10.060 Plugging it in gives you a quadratic equation 00:42:10.060 --> 00:42:18.170 that looks like s squared plus sb plus k equals 0. 00:42:18.170 --> 00:42:19.050 This has roots. 00:42:23.770 --> 00:42:27.700 I left out my m here, so it starts off looking like that. 00:42:27.700 --> 00:42:30.210 You divide through by the m. s squared 00:42:30.210 --> 00:42:36.875 plus b/m s plus k/m equals 0. 00:42:36.875 --> 00:42:38.500 And that's where Haynes would leave it. 00:42:38.500 --> 00:42:41.910 And he'd give you the entire answer in terms of b/m and k/m 00:42:41.910 --> 00:42:43.650 and that kind of thing. 00:42:43.650 --> 00:42:47.870 Engineers, we like to call that the natural frequency squared. 00:42:47.870 --> 00:42:50.920 And this term, we'd modify to put it 00:42:50.920 --> 00:42:56.060 in a terminology that is more convenient to engineering. 00:42:56.060 --> 00:42:59.140 So I'll show you how that works out. 00:42:59.140 --> 00:43:01.100 When you solve this quadratic just using 00:43:01.100 --> 00:43:04.385 the quadratic equation, you get the following. 00:43:15.167 --> 00:43:17.000 You get that the roots, there's two of them. 00:43:17.000 --> 00:43:20.060 I'll call them S1 and 2. 00:43:20.060 --> 00:43:21.900 The roots to this equation look like 00:43:21.900 --> 00:43:34.290 minus b over 2m plus or minus square root of b squared 00:43:34.290 --> 00:43:39.490 over 4m squared minus k/m. 00:43:42.480 --> 00:43:51.811 And that's what you'd get to do in 1.803. 00:43:51.811 --> 00:43:56.680 And an engineer would say, well, let's change that a little bit. 00:43:56.680 --> 00:44:03.210 So my roots that I would use for S1 and 2, 00:44:03.210 --> 00:44:07.120 I just factor out-- that's omega n squared. 00:44:07.120 --> 00:44:10.290 I can factor that out and it becomes omega n on the outside. 00:44:10.290 --> 00:44:13.000 And I put an omega n in the numerator and denominator here, 00:44:13.000 --> 00:44:14.310 as well. 00:44:14.310 --> 00:44:49.030 So I get roots that look like-- so I've just 00:44:49.030 --> 00:44:50.610 manipulated that a little bit. 00:44:53.834 --> 00:44:57.175 I have a name for this term. 00:44:57.175 --> 00:45:04.240 I use the Greek letter zeta is b over 2 omega n 00:45:04.240 --> 00:45:07.626 m is the way I remember it in my brain. 00:45:07.626 --> 00:45:08.875 It's called the damping ratio. 00:45:19.900 --> 00:45:24.320 And if I say that, then the roots, S1 and 2 for this, 00:45:24.320 --> 00:45:32.060 look like minus zeta omega n plus or minus 00:45:32.060 --> 00:45:38.788 omega n times the square root of zeta squared minus 1. 00:45:45.340 --> 00:45:48.810 And those are the roots that a vibration engineer 00:45:48.810 --> 00:45:55.160 would use to describe this second-order linear 00:45:55.160 --> 00:45:58.135 differential equation solution homogeneous solution. 00:46:00.720 --> 00:46:03.870 Those are the roots of the equation. 00:46:03.870 --> 00:46:07.970 And when you have no damping, then this term goes away 00:46:07.970 --> 00:46:12.010 and you're left with-- and I left an i out of here, I think. 00:46:15.300 --> 00:46:16.800 No, I'm fine. 00:46:16.800 --> 00:46:17.900 The i comes out of here. 00:46:33.730 --> 00:46:37.240 So for one thing to absolutely take away from today 00:46:37.240 --> 00:46:38.810 is to remember this. 00:46:41.800 --> 00:46:44.850 That's our definition of damping called the damping ratio. 00:46:44.850 --> 00:46:47.400 When that's 1, it's a number we call critical damping. 00:46:47.400 --> 00:46:50.060 I'll show you what that means in a second. 00:46:50.060 --> 00:46:55.100 And when it's greater than 1, the system won't vibrate. 00:46:55.100 --> 00:46:57.660 It just has exponential decay. 00:46:57.660 --> 00:46:59.630 If it's less than 1, you get vibration. 00:46:59.630 --> 00:47:02.320 And that's why we like to use it this way as it's meaningful. 00:47:02.320 --> 00:47:04.040 Its value, you instantly know if it's 00:47:04.040 --> 00:47:05.080 greater than or less than 1, it's 00:47:05.080 --> 00:47:06.830 going to change the behavior of the system 00:47:06.830 --> 00:47:10.190 from vibrating to not vibrating. 00:47:10.190 --> 00:47:14.580 So now, there's four possible solutions to this. 00:47:14.580 --> 00:47:20.690 I'm not going to elaborate on all of them, but zeta equals 0, 00:47:20.690 --> 00:47:21.795 we've already done. 00:47:21.795 --> 00:47:23.870 We know the answer to that. 00:47:23.870 --> 00:47:26.210 Response to initial conditions-- simple. 00:47:26.210 --> 00:47:27.840 We know that one. 00:47:27.840 --> 00:47:31.960 We have another solution when zeta's greater than 1. 00:47:31.960 --> 00:47:34.200 When zeta's greater than 1, this quantity here 00:47:34.200 --> 00:47:38.780 is the inside is greater than 1, so it's a real positive number. 00:47:38.780 --> 00:47:42.580 And both the roots of this thing are completely real. 00:47:46.840 --> 00:47:52.200 And you know that the-- remember the response, 00:47:52.200 --> 00:47:54.920 we hypothesize in the beginning that response 00:47:54.920 --> 00:47:58.210 looks like some Ae to the st. So now, we just plug back in. 00:47:58.210 --> 00:47:59.890 This is our st value. 00:47:59.890 --> 00:48:01.690 We can plug them back in and we will get 00:48:01.690 --> 00:48:04.880 the motion of the system back. 00:48:04.880 --> 00:48:11.420 So for zeta greater than 1, st comes out 00:48:11.420 --> 00:48:21.490 looking like minus zeta omega n t plus or minus 00:48:21.490 --> 00:48:29.040 square root of zeta squared minus 1 times t. 00:48:29.040 --> 00:48:32.590 And you just plug this in, and x is just e to the st. 00:48:32.590 --> 00:48:35.750 But these are just pure real values. 00:48:35.750 --> 00:48:40.930 And you'll find out that the system from initial conditions 00:48:40.930 --> 00:48:43.280 on velocity and displacement just-- 00:48:43.280 --> 00:48:44.039 [WINDS DOWN] 00:48:44.039 --> 00:48:44.580 and dies out. 00:48:48.010 --> 00:48:51.180 Zeta equals to 1. 00:48:51.180 --> 00:49:00.940 Then, st is just minus-- you get a double root-- minus omega nt, 00:49:00.940 --> 00:49:01.440 twice. 00:49:04.340 --> 00:49:07.710 And the solution for this, I can write out the whole thing. 00:49:07.710 --> 00:49:16.700 x of t here is just some A1 plus t A2 e 00:49:16.700 --> 00:49:21.200 to the minus zeta omega n t. 00:49:21.200 --> 00:49:24.320 And again, it looks-- it's just some kind 00:49:24.320 --> 00:49:28.740 of damp, not very interesting response, no oscillations. 00:49:28.740 --> 00:49:31.590 And then finally, zeta less than 1. 00:49:31.590 --> 00:49:34.216 And this is the only one-- this one produces oscillation. 00:49:37.670 --> 00:49:52.000 And the solution for st is plus or minus-- minus zeta omega n 00:49:52.000 --> 00:50:04.950 t, a real part, plus or minus i omega n t 00:50:04.950 --> 00:50:07.710 times the square root of 1 minus zeta squared. 00:50:07.710 --> 00:50:13.780 Now, I've turned around this zeta squared minus 1. 00:50:13.780 --> 00:50:15.220 This is now a negative number. 00:50:15.220 --> 00:50:18.270 A square root of a negative number gives me i. 00:50:18.270 --> 00:50:20.220 And now, I turn this around, so this is just 00:50:20.220 --> 00:50:22.240 a real positive number. 00:50:22.240 --> 00:50:24.510 So when you get i into this answer, 00:50:24.510 --> 00:50:29.274 what does it tell you that the solution looks like? 00:50:29.274 --> 00:50:30.440 AUDIENCE: Sines and cosines. 00:50:30.440 --> 00:50:32.460 J. KIM VANDIVER: Sines and cosines, right? 00:50:32.460 --> 00:50:36.490 So now, this gives you sines and cosines with a decay. 00:50:36.490 --> 00:50:39.215 This is an exponential to e to the minus zeta omega n 00:50:39.215 --> 00:50:42.420 t multiplied by a sine and a cosine. 00:50:42.420 --> 00:50:44.410 And so this is the interesting part. 00:50:44.410 --> 00:50:47.740 So most of the work of the rest of this term, 00:50:47.740 --> 00:50:50.585 we're only interested in this final solution. 00:50:53.340 --> 00:51:13.336 And what it looks like for this one-- so for zeta less than 1, 00:51:13.336 --> 00:51:20.820 x of t is some Ae to the minus zeta omega n 00:51:20.820 --> 00:51:32.310 t times a cosine omega d t-- make it d times 00:51:32.310 --> 00:51:36.770 t minus a phase angle-- come out looking like that. 00:51:36.770 --> 00:51:41.720 And if you draw it, depends on initial conditions, so again, 00:51:41.720 --> 00:51:44.290 a positive velocity and a positive displacement. 00:51:44.290 --> 00:51:46.240 It does this, but then it dies out. 00:51:49.120 --> 00:51:50.900 It's very similar to the undamped case, 00:51:50.900 --> 00:51:53.900 except that it has this damping that 00:51:53.900 --> 00:51:55.690 causes it to die out with time. 00:51:55.690 --> 00:51:59.160 But this right here, this is still the initial slope is v0 00:51:59.160 --> 00:52:01.040 and the initial displacement here is x0. 00:52:15.050 --> 00:52:19.750 And I'm now going to give you the exact expressions for this 00:52:19.750 --> 00:52:20.780 and we'll talk about it. 00:52:27.810 --> 00:52:30.410 Another way of writing this then in terms 00:52:30.410 --> 00:52:41.170 of the initial conditions is this looks like x0 cosine omega 00:52:41.170 --> 00:52:49.680 d t plus v0 over omega d. 00:53:46.110 --> 00:53:49.560 So expanding this out, this result clearly 00:53:49.560 --> 00:53:52.360 has to depend on the initial displacement 00:53:52.360 --> 00:53:54.470 and on the initial velocity. 00:53:54.470 --> 00:53:55.230 Now, what's this? 00:53:55.230 --> 00:53:57.470 I keep writing this omega d. 00:53:57.470 --> 00:53:59.530 So notice in here in the solution, 00:53:59.530 --> 00:54:03.370 it's omega n times the square root of 1 minus zeta squared. 00:54:03.370 --> 00:54:06.570 So the frequency that's in here isn't exactly omega n. 00:54:06.570 --> 00:54:10.470 It's omega n altered by a bit. 00:54:10.470 --> 00:54:23.250 Omega sub d is called the damped natural frequency. 00:54:28.050 --> 00:54:32.250 And it's equal to omega n times the square root 00:54:32.250 --> 00:54:33.995 of 1 minus theta squared. 00:54:36.880 --> 00:54:40.790 The system actually oscillates at a slightly different 00:54:40.790 --> 00:54:41.400 frequency. 00:54:41.400 --> 00:54:47.140 And for most systems that vibrate at all, 00:54:47.140 --> 00:54:49.260 this damping term is quite small. 00:54:49.260 --> 00:54:52.090 And when you square it, it gets even smaller. 00:54:52.090 --> 00:54:58.310 So this is usually a number that's 0.99, oftentimes, 00:54:58.310 --> 00:54:59.930 or even bigger than that. 00:54:59.930 --> 00:55:04.960 This is very close to 1 for all small amounts of damping. 00:55:04.960 --> 00:55:08.350 But being really careful about this 00:55:08.350 --> 00:55:11.190 in including it everywhere, that's 00:55:11.190 --> 00:55:13.610 what this result looks like. 00:55:13.610 --> 00:55:18.080 And this little thing, psi, this little phase angle here, 00:55:18.080 --> 00:55:30.860 is tangent inverse of theta over the square root of 1 00:55:30.860 --> 00:55:33.220 minus theta squared. 00:55:33.220 --> 00:55:36.550 And this number-- when damping is small, 00:55:36.550 --> 00:55:39.530 this is a very small number. 00:55:39.530 --> 00:55:43.160 And most of the time of problems that we deal with, 00:55:43.160 --> 00:55:44.510 the damping will be small. 00:55:44.510 --> 00:55:54.230 So let's say, for small damping-- 00:55:54.230 --> 00:55:58.640 and by that, I mean zeta, say, less than 10%, 00:55:58.640 --> 00:56:01.990 what we call 10%, 0.1. 00:56:01.990 --> 00:56:06.560 And if you have a little more-- you don't care too much 00:56:06.560 --> 00:56:09.480 about the precision, it might even be 20%. 00:56:09.480 --> 00:56:16.470 Actually, if it were 0.2, squared is 0.04, right? 00:56:16.470 --> 00:56:25.090 1 minus 0.04-- 0.96 square root, 0.98. 00:56:25.090 --> 00:56:27.890 So even with 20% damping, the difference 00:56:27.890 --> 00:56:30.370 between the undamped natural frequency 00:56:30.370 --> 00:56:31.972 and the damped natural frequency's 2%. 00:56:34.480 --> 00:56:39.740 So for most cases with any kind of small damping at all, 00:56:39.740 --> 00:56:42.870 we can write an approximation which is easier to remember. 00:56:42.870 --> 00:56:45.410 And it's all I carry around in my head. 00:56:45.410 --> 00:56:47.800 I can't remember this, quite frankly. 00:56:47.800 --> 00:56:51.290 Don't try to and I would instead express 00:56:51.290 --> 00:57:01.160 the answer to this as just x0 cosine omega 00:57:01.160 --> 00:57:15.530 d t plus v0 over omega d sine omega damped times time 00:57:15.530 --> 00:57:21.260 times e to the minus zeta omega n t. 00:57:21.260 --> 00:57:24.630 So why do I bother to carry the omega d's along 00:57:24.630 --> 00:57:29.590 if I just said that they're almost exactly the same. 00:57:29.590 --> 00:57:33.582 For light damping, then omega n's approximately omega d. 00:57:33.582 --> 00:57:36.760 Well, you need to keep this one in here 00:57:36.760 --> 00:57:43.840 because even though it's only 2% difference at 20% damping, 00:57:43.840 --> 00:57:50.530 if you say the solution is omega n when it's really omega d, 00:57:50.530 --> 00:57:56.040 this thing will accumulate a phase error over time. 00:57:56.040 --> 00:58:00.380 So it's gets bigger and bigger, this error here, 00:58:00.380 --> 00:58:02.720 because you haven't taken care of that little 2%. 00:58:02.720 --> 00:58:06.550 That 2% can bite you after you go through enough cycles. 00:58:06.550 --> 00:58:10.092 So I keep omega d in the expression here. 00:58:10.092 --> 00:58:14.300 But other than that, it's almost exactly the same expression 00:58:14.300 --> 00:58:17.280 that we just came up to for the simple response 00:58:17.280 --> 00:58:20.100 of an undamped system to initial conditions, 00:58:20.100 --> 00:58:25.740 x0 cosine plus v0 over omega n sine. 00:58:25.740 --> 00:58:30.000 And now, all we've added to it is put the transient decay 00:58:30.000 --> 00:58:33.770 and the fact that it decays into the expression 00:58:33.770 --> 00:58:38.030 and changed the frequency it oscillates at to omega 00:58:38.030 --> 00:58:39.490 d instead of omega n. 00:59:13.770 --> 00:59:17.280 So I'm going to try to impress something on you. 00:59:17.280 --> 00:59:33.750 If I took this pendulum and my stopwatch, 00:59:33.750 --> 00:59:39.690 measured the natural frequency of this thing, 00:59:39.690 --> 00:59:43.080 I could get a very accurate value if I do it carefully. 00:59:43.080 --> 00:59:47.010 Then, I take the same object and I dunk it in water 00:59:47.010 --> 00:59:49.404 and it goes back and forth. 00:59:49.404 --> 00:59:51.070 And it conspicuously goes back and forth 00:59:51.070 --> 00:59:52.944 but dies down now after a while, because it's 00:59:52.944 --> 00:59:54.890 got that water damping it. 00:59:54.890 --> 01:00:00.180 But I measure that frequency and it's 01:00:00.180 --> 01:00:05.120 10% different, 20% different. 01:00:05.120 --> 01:00:09.670 And I have seen people make this mistake dozens of times. 01:00:09.670 --> 01:00:12.150 You say, that's the experiment. 01:00:12.150 --> 01:00:14.396 Explain why. 01:00:14.396 --> 01:00:21.250 What's the reason that that measured frequency has changed? 01:00:21.250 --> 01:00:25.469 Got any ocean engineers in the audience? 01:00:25.469 --> 01:00:27.380 All right. 01:00:27.380 --> 01:00:30.799 So why does-- if you put the pendulum in water-- 01:00:30.799 --> 01:00:32.090 and it's still oscillating now. 01:00:32.090 --> 01:00:34.330 So it isn't so damp that it's-- 01:00:34.330 --> 01:00:35.029 [BLOWS] 01:00:35.029 --> 01:00:36.070 So it's got some damping. 01:00:36.070 --> 01:00:38.050 It's dying out and the natural frequency's 01:00:38.050 --> 01:00:39.820 changed by 15% or 20%. 01:00:39.820 --> 01:00:42.040 What's the explanation? 01:00:42.040 --> 01:00:49.020 And the answer you always get from people is, damping. 01:00:49.020 --> 01:00:49.520 Why? 01:00:49.520 --> 01:00:55.665 Because everybody's been taught this thing, right? 01:00:55.665 --> 01:00:58.040 And they all then assume that the change in the frequency 01:00:58.040 --> 01:00:58.914 is caused by damping. 01:00:58.914 --> 01:01:01.500 But damping couldn't possibly be the reason, 01:01:01.500 --> 01:01:03.720 because with 20% damping, this thing'll die out 01:01:03.720 --> 01:01:06.750 in about two swings and it's done. 01:01:06.750 --> 01:01:08.890 That's a lot of damping, actually, 01:01:08.890 --> 01:01:11.730 but it only accounts for 2% change in natural frequency, 01:01:11.730 --> 01:01:14.160 not 15%. 01:01:14.160 --> 01:01:14.710 Hmmm. 01:01:14.710 --> 01:01:16.571 So what causes the change in the frequency? 01:01:20.892 --> 01:01:22.350 AUDIENCE: Buoyancy of the pendulum? 01:01:22.350 --> 01:01:23.766 J. KIM VANDIVER: No, not buoyancy. 01:01:27.220 --> 01:01:30.000 That could actually have an effect. 01:01:30.000 --> 01:01:33.957 That's actually-- I should say, yes, you're partly right. 01:01:33.957 --> 01:01:34.915 There's another reason. 01:01:39.190 --> 01:01:43.230 When the thing is swinging back and forth there in the water, 01:01:43.230 --> 01:01:46.110 it actually carries some water with it. 01:01:46.110 --> 01:01:48.740 Effectively, the kinetic energy-- you 01:01:48.740 --> 01:01:51.546 now know how to do vibration problems. 01:01:51.546 --> 01:01:53.170 Find the equations of motion accounting 01:01:53.170 --> 01:01:56.420 for the potential energy and the kinetic energy. 01:01:56.420 --> 01:01:59.250 The kinetic energy changes, because some water 01:01:59.250 --> 01:02:03.010 moves with the object and it's called added mass. 01:02:03.010 --> 01:02:05.450 It literally-- there is water moving with the object that 01:02:05.450 --> 01:02:07.670 has kinetic energy associated with the motion 01:02:07.670 --> 01:02:10.270 and it acts like it's more massive. 01:02:10.270 --> 01:02:11.940 It is dynamically more massive. 01:02:11.940 --> 01:02:15.400 There's water moving with it. 01:02:15.400 --> 01:02:17.820 So trying to impress on you that damping 01:02:17.820 --> 01:02:21.920 doesn't cause much of a change in systems 01:02:21.920 --> 01:02:22.940 that actually vibrate. 01:02:22.940 --> 01:02:24.430 Really observe the vibration. 01:02:24.430 --> 01:02:26.200 If you can observe the vibration, 01:02:26.200 --> 01:02:29.290 damping cannot possibly account for a very large shift 01:02:29.290 --> 01:02:30.830 in frequency. 01:02:30.830 --> 01:02:34.252 What's the motion look like? 01:02:34.252 --> 01:02:36.130 Let's move on a little bit here. 01:02:44.110 --> 01:02:46.160 So that's what this solution looks like. 01:02:46.160 --> 01:02:48.870 We know it depends on initial conditions. 01:02:48.870 --> 01:02:57.020 The distance from here to here will make this a time axis. 01:02:57.020 --> 01:02:59.110 This is one period. 01:02:59.110 --> 01:03:00.640 So this is tau d. 01:03:00.640 --> 01:03:03.810 That's the damped period of vibration. 01:03:07.060 --> 01:03:20.820 And we know that x of t is some Ae to the minus theta omega n t 01:03:20.820 --> 01:03:26.740 cosine omega d t minus a phase angle. 01:03:26.740 --> 01:03:29.370 We can write that expression like this. 01:03:31.880 --> 01:03:39.230 And this term, this is just a cosine. 01:03:39.230 --> 01:03:44.780 This term repeats every period, right? 01:03:47.560 --> 01:03:52.210 If it's at maximum value here, exactly one period later, 01:03:52.210 --> 01:03:54.090 it's again at its maximum. 01:03:54.090 --> 01:04:00.020 So the cosine term goes to 1 every 2 pi 01:04:00.020 --> 01:04:03.530 or every period of motion, right? 01:04:03.530 --> 01:04:11.440 So I want to take-- I'm going to define this as the value 01:04:11.440 --> 01:04:16.110 at x at some time t. 01:04:16.110 --> 01:04:18.970 I'll call it t0. 01:04:18.970 --> 01:04:32.220 And out here is x at t0 plus n tau d, n periods later. 01:04:32.220 --> 01:04:34.280 So this is the period, defined as period. 01:04:36.960 --> 01:04:44.900 Remember, omega d is the same thing 01:04:44.900 --> 01:04:51.090 as 2 pi times the frequency in hertz. 01:04:51.090 --> 01:05:01.980 And frequency is 1 over period, 2 pi over the period. 01:05:01.980 --> 01:05:04.170 So remember, there's a relationship 01:05:04.170 --> 01:05:08.180 that you need to remember now that relates radian 01:05:08.180 --> 01:05:12.450 frequency to frequency in cycles per second in hertz 01:05:12.450 --> 01:05:15.815 to frequency expressed in period. 01:05:15.815 --> 01:05:17.030 All right? 01:05:17.030 --> 01:05:21.170 This would be tau d here and this would be an f d. 01:05:21.170 --> 01:05:23.910 For any frequency, you can say that. 01:05:23.910 --> 01:05:27.652 At omega is 2 pi f is 2 pi over tau. 01:05:27.652 --> 01:05:29.110 So you've got to be good with that. 01:05:29.110 --> 01:05:35.190 But now, so here we are, two peaks separated by n periods. 01:05:35.190 --> 01:05:44.050 And I want to take the ratio of x of t to x of t plus n tau d 01:05:44.050 --> 01:05:45.680 here. 01:05:45.680 --> 01:05:49.600 And that's just going to be then my-- 01:05:49.600 --> 01:05:54.260 when I take that ratio, x of t has cosine omega d t 01:05:54.260 --> 01:05:56.150 minus phi in it. 01:05:56.150 --> 01:05:59.810 And n periods later, exactly the same thing appears, right? 01:05:59.810 --> 01:06:02.300 So the cosine term just cancels out. 01:06:02.300 --> 01:06:05.390 This just is e-- and the A's cancel out. 01:06:05.390 --> 01:06:07.135 That's the initial conditions. 01:06:07.135 --> 01:06:11.660 It's e to the minus zeta omega n t-- 01:06:11.660 --> 01:06:20.480 and I guess I called it t0-- over e 01:06:20.480 --> 01:06:31.225 to the minus zeta omega n t0 plus n damped periods. 01:06:33.950 --> 01:06:36.500 And if I bring this into the numerator, 01:06:36.500 --> 01:06:38.540 the exponent becomes positive. 01:06:38.540 --> 01:06:45.630 The t0 terms, minus zeta omega and t0 plus, those cancel. 01:06:45.630 --> 01:06:53.530 And this expression is just e to the plus zeta omega 01:06:53.530 --> 01:06:56.844 n times n td. 01:07:02.640 --> 01:07:06.390 And the last step that I want to do to this, what I'm coming up 01:07:06.390 --> 01:07:10.570 with is a way of estimating-- purposely 01:07:10.570 --> 01:07:20.520 doing this-- is this transient curve we know is controlled 01:07:20.520 --> 01:07:22.850 by a damping, by zeta. 01:07:22.850 --> 01:07:28.340 I want to have an experimental way to determine what is zeta. 01:07:28.340 --> 01:07:31.530 And I do it by computing something called 01:07:31.530 --> 01:07:33.300 the logarithmic decrement. 01:07:33.300 --> 01:07:41.060 So if I take the natural log of x of t over x of t 01:07:41.060 --> 01:07:50.270 plus n periods, it's the natural log of this expression. 01:07:50.270 --> 01:07:53.580 So I just get the exponent back. 01:07:53.580 --> 01:08:07.526 This then is n zeta omega-- I guess 01:08:07.526 --> 01:08:13.530 I better to do it carefully-- omega n n tau d. 01:08:13.530 --> 01:08:18.439 The tau d is 2 pi over omega and I get some nice things 01:08:18.439 --> 01:08:19.640 to cancel out here. 01:08:29.630 --> 01:08:38.689 So this natural log over the ratio-- this is n zeta omega n 01:08:38.689 --> 01:08:44.040 and this is 2 pi over omega d, which 01:08:44.040 --> 01:08:49.649 is omega n times the square root of 1 minus zeta squared. 01:08:49.649 --> 01:08:52.260 Omega n's go away. 01:08:52.260 --> 01:09:09.149 And for zeta small, this term's approximately 1, in which case 01:09:09.149 --> 01:09:19.120 this then becomes n 2 pi zeta. 01:09:19.120 --> 01:09:27.200 And zeta equals 1 over 2 pi n natural log 01:09:27.200 --> 01:09:36.729 of this ratio of x of t over x of t plus nt. 01:09:44.060 --> 01:09:50.310 So experimentally, if you just go in and measure your-- 01:09:50.310 --> 01:09:55.960 if you plot out the response, you measure a peak value, 01:09:55.960 --> 01:09:58.990 you measure the peak value n periods later, 01:09:58.990 --> 01:10:03.950 compute the log of that ratio, divide by 1 over 2 01:10:03.950 --> 01:10:06.570 pi n, the number of periods, you have 01:10:06.570 --> 01:10:08.840 an estimate of the natural frequency-- 01:10:08.840 --> 01:10:10.752 estimate of the damping ratio, excuse me. 01:10:15.380 --> 01:10:20.000 And to give you one quick little rule of thumb here, 01:10:20.000 --> 01:10:23.790 so this is an experimental way that very quickly, you 01:10:23.790 --> 01:10:29.360 can estimate the damping of a pendulum or whatever 01:10:29.360 --> 01:10:31.380 by just doing a quick measurement. 01:11:05.070 --> 01:11:08.550 So if it happens that after n periods, 01:11:08.550 --> 01:11:12.610 this value is half of the initial value, 01:11:12.610 --> 01:11:15.960 then this ratio is 2, right? 01:11:15.960 --> 01:11:18.570 So x of t-- some n periods later, this is only half 01:11:18.570 --> 01:11:19.930 as big. 01:11:19.930 --> 01:11:20.950 This value's 2. 01:11:20.950 --> 01:11:24.910 The natural log of 2 is some number you can calculate. 01:11:24.910 --> 01:11:27.030 So there's a little rule. 01:11:27.030 --> 01:11:29.700 If you just work that out, you find 01:11:29.700 --> 01:11:44.300 that zeta equals 1 over 2 pi n 50% times the natural log of 2. 01:11:44.300 --> 01:11:49.150 And you end up here was 0-- let me do this 01:11:49.150 --> 01:11:54.960 carefully-- 1 over 2 pi, n 50%, natural log of 2. 01:11:54.960 --> 01:12:03.450 And that is 0.11 over n 50%. 01:12:03.450 --> 01:12:07.130 That's a really handy little engineer tool 01:12:07.130 --> 01:12:09.140 to carry around in your head. 01:12:09.140 --> 01:12:15.190 So if I have an oscillator, this little end here, 01:12:15.190 --> 01:12:17.920 I can do an experiment. 01:12:17.920 --> 01:12:22.020 Give it initial deflection and it starts off 01:12:22.020 --> 01:12:24.940 at six inches or three inches amplitude. 01:12:24.940 --> 01:12:27.040 And you let it oscillate until you see it die down 01:12:27.040 --> 01:12:28.800 to half of that value. 01:12:28.800 --> 01:12:33.795 So let's say, one, two, about four cycles 01:12:33.795 --> 01:12:38.710 this thing decays by about 50%. 01:12:38.710 --> 01:12:42.580 Four cycles-- plug 4 into that formula. 01:12:42.580 --> 01:12:47.240 You get about 0.025. 01:12:47.240 --> 01:12:48.762 Agree? 01:12:48.762 --> 01:12:52.490 2 and a 1/2% damping. 01:12:52.490 --> 01:12:55.720 Really very convenient little thing 01:12:55.720 --> 01:12:58.260 to carry around with you-- measure pendulum, 01:12:58.260 --> 01:12:59.510 how much damping does it have? 01:12:59.510 --> 01:13:00.960 And now, this is what I'm saying. 01:13:00.960 --> 01:13:03.850 Most things that have any substantial amount 01:13:03.850 --> 01:13:08.078 of vibration, the damping is going to be way less than 10%. 01:13:11.330 --> 01:13:15.950 If it dies, if it takes one cycle for the amplitude 01:13:15.950 --> 01:13:20.530 to decrease, one cycle for the amplitude to decrease by 50%, 01:13:20.530 --> 01:13:22.222 how much damping does it have? 01:13:22.222 --> 01:13:23.550 AUDIENCE: 11%. 01:13:23.550 --> 01:13:25.930 J. KIM VANDIVER: 11%. 01:13:25.930 --> 01:13:28.320 So 11% damping is a lot of damping. 01:13:28.320 --> 01:13:32.189 The thing starts out here and the next cycle, it's half gone, 01:13:32.189 --> 01:13:34.230 and the next cycle after that, it's half of that. 01:13:34.230 --> 01:13:37.520 And so in about three cycles, it's gone. 01:13:37.520 --> 01:13:40.060 So if you see anything that's vibrating any length of time 01:13:40.060 --> 01:13:43.990 at all, its damping is way less than 10% 01:13:43.990 --> 01:13:50.520 and this notion of small damping is a perfectly good one. 01:13:50.520 --> 01:13:52.730 And I'll close by just saying one other thing. 01:13:52.730 --> 01:13:56.705 If something vibrates a lot, the damping's small. 01:14:01.030 --> 01:14:03.160 You need small damping for things 01:14:03.160 --> 01:14:04.890 to actually vibrate very much. 01:14:04.890 --> 01:14:06.770 This thing, this is vibrating-- 01:14:06.770 --> 01:14:07.630 [HIGH TONE] 01:14:07.630 --> 01:14:12.320 that high-pitched one, that's about a kilohertz. 01:14:12.320 --> 01:14:14.730 How many cycles do you think it's gone through 01:14:14.730 --> 01:14:19.270 to get down to 50% of that initial amplitude 01:14:19.270 --> 01:14:21.062 that you could hear? 01:14:21.062 --> 01:14:23.730 A few thousand? 01:14:23.730 --> 01:14:27.900 How much damping do you think this rod has? 01:14:27.900 --> 01:14:32.080 Really tiny, really tiny. 01:14:32.080 --> 01:14:32.640 All right. 01:14:32.640 --> 01:14:34.290 So even though all we talked about today was 01:14:34.290 --> 01:14:35.831 single degree of freedom oscillators, 01:14:35.831 --> 01:14:39.840 I hope you learned a few things that we'll carry now 01:14:39.840 --> 01:14:41.230 through the rest of the term. 01:14:41.230 --> 01:14:44.750 We'll use all these concepts that we did today 01:14:44.750 --> 01:14:47.150 to talk about more complicated vibration. 01:14:47.150 --> 01:14:51.820 Good luck on your 2.001 quiz. 01:14:51.820 --> 01:14:55.500 See you on Tuesday.