WEBVTT 00:00:00.070 --> 00:00:02.430 The following content is provided under a Creative 00:00:02.430 --> 00:00:03.810 Commons license. 00:00:03.810 --> 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 --> 00:00:26.430 PROFESSOR: Are we ready with the concept questions 00:00:26.430 --> 00:00:29.740 from the homework this week? 00:00:29.740 --> 00:00:32.284 How do we get different-- there we go. 00:00:32.284 --> 00:00:33.700 I looked at one, it was one thing. 00:00:33.700 --> 00:00:35.270 I looked at the other, [INAUDIBLE]. 00:00:35.270 --> 00:00:37.260 Does g enter into the expression for 00:00:37.260 --> 00:00:39.950 the undamped natural frequency? 00:00:39.950 --> 00:00:45.705 And most people said no, but about a third of you said yes. 00:00:45.705 --> 00:00:47.330 If you have worked on that problem now, 00:00:47.330 --> 00:00:49.830 you have already discovered the answer. 00:00:49.830 --> 00:00:56.520 So you'll find that g does not come into the expression. 00:00:56.520 --> 00:01:02.200 When you do a pendulum, g is in the expression. 00:01:02.200 --> 00:01:05.090 And there's a question on the homework 00:01:05.090 --> 00:01:07.420 about what's the difference. 00:01:07.420 --> 00:01:09.830 How can you predict when g is going 00:01:09.830 --> 00:01:12.880 to be involved in a natural frequency expression 00:01:12.880 --> 00:01:14.180 and when it is not? 00:01:16.562 --> 00:01:18.270 I want you to think about that one a bit, 00:01:18.270 --> 00:01:20.170 maybe talk about it at-- if there's still 00:01:20.170 --> 00:01:21.670 questions about it, we talk about it 00:01:21.670 --> 00:01:23.480 in recitation on Thursday, Friday. 00:01:23.480 --> 00:01:25.020 OK. 00:01:25.020 --> 00:01:27.205 So does g enter into the expression here? 00:01:27.205 --> 00:01:31.170 I'm sure you know this simple pendulum, the natural frequency 00:01:31.170 --> 00:01:34.000 square root of g over l. 00:01:34.000 --> 00:01:35.910 For a simple mass-spring dashpot, 00:01:35.910 --> 00:01:38.070 the natural frequency is k over m 00:01:38.070 --> 00:01:41.070 whether or not it's affected by gravity. 00:01:41.070 --> 00:01:44.840 So there's something different about these two. 00:01:44.840 --> 00:01:46.510 OK, let's go onto the third one. 00:01:51.340 --> 00:01:54.180 "In an experiment, this system is given initial velocity 00:01:54.180 --> 00:01:57.240 observed to decay in amplitude of vibration 00:01:57.240 --> 00:01:59.620 by 50% in 10 cycles. 00:01:59.620 --> 00:02:01.230 You can estimate the damping ratio 00:02:01.230 --> 00:02:02.742 to be approximately," what? 00:02:02.742 --> 00:02:05.200 Well, it gives a little rule of thumb I gave you last week. 00:02:05.200 --> 00:02:06.930 I'll go over it again today. 00:02:06.930 --> 00:02:11.360 0.11 divided by the number of cycles did decay 50%. 00:02:11.360 --> 00:02:14.290 So it took 10 cycles, 0.11 divided 00:02:14.290 --> 00:02:19.670 by 10, 0.01, 1.1% damping. 00:02:19.670 --> 00:02:22.460 OK, next. 00:02:22.460 --> 00:02:25.160 "At which of the three excitation frequencies 00:02:25.160 --> 00:02:28.380 will the response magnitude be greatest?" 00:02:28.380 --> 00:02:33.210 You've done oscillators excited by cosine omega t 00:02:33.210 --> 00:02:34.820 kind of things before. 00:02:34.820 --> 00:02:38.142 So at the ratio 1, most people said 00:02:38.142 --> 00:02:39.600 it's where it would be the largest. 00:02:39.600 --> 00:02:40.880 Why at 1? 00:02:40.880 --> 00:02:43.870 Anybody want to give a shout here? 00:02:43.870 --> 00:02:46.750 What happens when you drive the system 00:02:46.750 --> 00:02:49.384 at its natural frequency? 00:02:49.384 --> 00:02:52.050 It's called resonance, and we're going to talk about that today. 00:02:52.050 --> 00:02:54.540 So it's when the frequency ratio is 00:02:54.540 --> 00:02:59.820 one that-- and for the system being lightly damped that you 00:02:59.820 --> 00:03:01.540 get the largest response. 00:03:01.540 --> 00:03:05.749 Finally-- which one are we on here? 00:03:05.749 --> 00:03:06.290 Oh, this one. 00:03:06.290 --> 00:03:10.680 Can all the kinetic energy be accounted for by an expression 00:03:10.680 --> 00:03:12.765 of the form IZ omega squared. 00:03:17.040 --> 00:03:19.310 By the way, I brought this system if you haven't. 00:03:22.190 --> 00:03:23.940 So there's a really simple demo, but it 00:03:23.940 --> 00:03:26.180 has all sorts of-- so in this case, 00:03:26.180 --> 00:03:32.340 we're talking about that motion. 00:03:32.340 --> 00:03:36.740 It certainly has some I omega squared kind of kinetic energy, 00:03:36.740 --> 00:03:40.255 but does the center of gravity translate as it's oscillating? 00:03:43.841 --> 00:03:45.590 What's the potential energy in the system? 00:03:48.970 --> 00:03:52.610 By the way, any time you get an oscillation, 00:03:52.610 --> 00:03:57.060 energy flows from potential to kinetic, potential kinetic. 00:03:57.060 --> 00:03:58.580 That's what oscillation is. 00:03:58.580 --> 00:04:01.790 So there has to be an exchange going on between kinetic energy 00:04:01.790 --> 00:04:02.930 and potential energy. 00:04:02.930 --> 00:04:04.640 And if there's no losses in the system, 00:04:04.640 --> 00:04:08.590 the total energy is constant. 00:04:08.590 --> 00:04:12.150 So the kinetic energy's certainly in the motion, 00:04:12.150 --> 00:04:14.450 but when it reaches maximum amplitude, what's 00:04:14.450 --> 00:04:17.459 its velocity when it's right here? 00:04:17.459 --> 00:04:18.230 Zero. 00:04:18.230 --> 00:04:22.070 So all of its energy must be where? 00:04:22.070 --> 00:04:22.880 In the potential. 00:04:22.880 --> 00:04:25.840 And where's the potential in this system? 00:04:25.840 --> 00:04:26.960 Pardon? 00:04:26.960 --> 00:04:27.590 In the string. 00:04:27.590 --> 00:04:29.427 That's not stored in the string. 00:04:29.427 --> 00:04:31.260 There's only two sources of potential energy 00:04:31.260 --> 00:04:33.360 we talk about in this class. 00:04:33.360 --> 00:04:33.920 Gravity and-- 00:04:33.920 --> 00:04:34.670 AUDIENCE: Strings. 00:04:34.670 --> 00:04:35.461 PROFESSOR: Strings. 00:04:35.461 --> 00:04:38.700 Well, these strings don't stretch, 00:04:38.700 --> 00:04:40.995 so there's no spring kinetic energy. 00:04:40.995 --> 00:04:42.120 We've got potential energy. 00:04:42.120 --> 00:04:47.490 There must be gravitational potential energy. 00:04:47.490 --> 00:04:49.215 How is it coming into this system? 00:04:52.880 --> 00:04:58.180 So he says when it turns, the center of gravity 00:04:58.180 --> 00:05:02.440 has to raise up a little bit, and that's 00:05:02.440 --> 00:05:04.590 the potential energy in this system. 00:05:04.590 --> 00:05:07.010 The center of gravity goes up and down a tiny bit. 00:05:07.010 --> 00:05:10.560 So is there any velocity in the vertical direction? 00:05:10.560 --> 00:05:12.270 Is there any kinetic energy associated 00:05:12.270 --> 00:05:13.710 with up and down motion? 00:05:13.710 --> 00:05:16.480 Yeah, so that doesn't entirely capture it, 00:05:16.480 --> 00:05:17.841 1/2 I omega squared. 00:05:17.841 --> 00:05:19.340 Is it an important amount of energy? 00:05:19.340 --> 00:05:23.530 I don't know, but there is some velocity up and down. 00:05:23.530 --> 00:05:26.170 My guess is that it actually isn't important, 00:05:26.170 --> 00:05:28.270 that the answer is it does move up and down. 00:05:28.270 --> 00:05:30.850 It has to, or you would not have any potential energy 00:05:30.850 --> 00:05:32.810 exchange in the system. 00:05:32.810 --> 00:05:35.030 OK. 00:05:35.030 --> 00:05:37.130 Is that it? 00:05:37.130 --> 00:05:39.312 OK. 00:05:39.312 --> 00:05:40.270 Let's keep moving here. 00:05:42.960 --> 00:05:44.710 Got a lot of fun things to show you today. 00:05:47.400 --> 00:05:50.862 So last time, we talked about response to initial conditions. 00:05:50.862 --> 00:05:52.820 I'm going to finish up with that and then go on 00:05:52.820 --> 00:05:57.360 to talking about excitation of harmonic forces. 00:05:57.360 --> 00:05:59.110 So last time, we were considering a system 00:05:59.110 --> 00:06:02.260 like this-- X is measured from the zero spring 00:06:02.260 --> 00:06:04.510 force in this case. 00:06:04.510 --> 00:06:07.050 Give you an equation of motion of that sort. 00:06:07.050 --> 00:06:12.320 And we've found that you could express x of T 00:06:12.320 --> 00:06:19.150 as a-- I'll give you the exact expression-- x0 square root 00:06:19.150 --> 00:06:20.893 of 1 minus zeta squared. 00:06:29.980 --> 00:06:38.050 Cosine omega damped times time minus a little phase angle. 00:06:38.050 --> 00:06:50.830 And there's a second term here, v0 over omega d sine omega dt. 00:06:50.830 --> 00:06:58.560 And the whole thing times e to the minus zeta omega n t. 00:06:58.560 --> 00:07:02.790 So that's our response to an initial deflection 00:07:02.790 --> 00:07:07.260 x0 or an initial velocity v0. 00:07:07.260 --> 00:07:11.000 That's the full kind of messy expression. 00:07:11.000 --> 00:07:13.645 There's another way of writing that, which I'll show you. 00:07:19.760 --> 00:07:34.060 Another way of saying is that it's in x0 cosine omega dt 00:07:34.060 --> 00:07:53.140 plus v0 plus zeta omega n x0, all over omega d sine 00:07:53.140 --> 00:07:58.110 omega d times t e to the minus zeta omega 00:07:58.110 --> 00:08:01.330 nt, the same exponent. 00:08:01.330 --> 00:08:03.260 This is your decaying exponential 00:08:03.260 --> 00:08:04.950 that makes it die out. 00:08:04.950 --> 00:08:07.020 And so I just rearranged some of these things. 00:08:07.020 --> 00:08:08.978 There's another little phase angle in here now. 00:08:08.978 --> 00:08:13.560 So you have just a cosine term, this proportional x0, 00:08:13.560 --> 00:08:18.070 and a sine term, which has both v0 and x0 in it. 00:08:18.070 --> 00:08:19.970 The x0 term, if damping is small, 00:08:19.970 --> 00:08:25.540 this term is pretty small because it's x0 times zeta. 00:08:25.540 --> 00:08:28.450 When you divide by omega d, which is almost omega n, 00:08:28.450 --> 00:08:30.130 that goes away. 00:08:30.130 --> 00:08:34.039 So this term, the scale of it is zeta x0. 00:08:34.039 --> 00:08:38.360 So if this is 1% or 2%, that's a very small number. 00:08:38.360 --> 00:08:40.190 I gave you an approximation, which 00:08:40.190 --> 00:08:48.870 for almost all practical examples that you 00:08:48.870 --> 00:08:55.650 might want to do, make it approximately sine here. 00:08:55.650 --> 00:09:14.080 So this is x0 cosine omega dt plus v0 over omega d sine omega 00:09:14.080 --> 00:09:21.340 dte to the minus zeta omega nt. 00:09:21.340 --> 00:09:23.230 And this is the practical one. 00:09:23.230 --> 00:09:27.160 For any reasonable system that has relatively low damping, 00:09:27.160 --> 00:09:31.940 even 10% or 15% damping, you get part of the transient decay 00:09:31.940 --> 00:09:38.930 comes from x0 cosine, the other part v0 over omega d sine. 00:09:38.930 --> 00:09:42.280 That's what I can remember in my head when I'm trying to do it. 00:09:42.280 --> 00:09:44.930 Now the question, the thing I want to address today 00:09:44.930 --> 00:09:48.140 is what's this useful for? 00:09:48.140 --> 00:09:50.410 My approach to teaching you vibration 00:09:50.410 --> 00:09:54.160 is I want you to go away with a few simple, 00:09:54.160 --> 00:09:57.010 practical understandings so that you can actually 00:09:57.010 --> 00:10:01.820 solve some vibration problems, and one of them 00:10:01.820 --> 00:10:05.412 is just knowing this allows you to do a couple things, 00:10:05.412 --> 00:10:07.370 and we'll do a couple of examples this morning. 00:10:11.120 --> 00:10:23.365 By the way, this form, this is A cosine plus B sine expression. 00:10:26.570 --> 00:10:36.770 And I label them A and B. A and B, 00:10:36.770 --> 00:10:44.830 they're both of the form A1 cosine omega 00:10:44.830 --> 00:10:50.090 t plus B1 sine omega t. 00:10:50.090 --> 00:10:53.020 And you can always add a sine and a cosine 00:10:53.020 --> 00:10:56.290 at the same frequency. 00:10:56.290 --> 00:10:57.670 If I put just any frequency here, 00:10:57.670 --> 00:10:59.620 they just have to be the same. 00:10:59.620 --> 00:11:02.570 You can always take an expression like that 00:11:02.570 --> 00:11:07.520 and rewrite it as some magnitude cosine omega 00:11:07.520 --> 00:11:11.090 t minus a phase angle. 00:11:11.090 --> 00:11:14.100 And the magnitude is just a square root 00:11:14.100 --> 00:11:21.060 of the sum of the squares-- A1 squared plus B1 squared. 00:11:21.060 --> 00:11:25.830 And the phase angle is the tangent inverse 00:11:25.830 --> 00:11:28.600 of the sine term over the cosine term. 00:11:28.600 --> 00:11:32.000 So you can always rewrite sine plus cosine 00:11:32.000 --> 00:11:36.080 as a cosine omega t minus v. We use that a lot, 00:11:36.080 --> 00:11:37.950 and that'll be used a lot in this course. 00:11:43.629 --> 00:11:45.420 And then, of course, if this whole thing is 00:11:45.420 --> 00:11:50.010 multiplied by an e to the minus 8 omega mt, then so is this. 00:11:55.010 --> 00:11:57.420 OK. 00:11:57.420 --> 00:11:59.340 OK, what are these things useful for? 00:12:03.480 --> 00:12:08.470 And we've derived this all for a mass spring system. 00:12:08.470 --> 00:12:12.810 Is that equation applicable to a pendulum? 00:12:12.810 --> 00:12:20.280 So this expression is applicable to any single degree of freedom 00:12:20.280 --> 00:12:23.430 system that oscillates. 00:12:23.430 --> 00:12:25.861 You just have to exchange a couple things. 00:12:25.861 --> 00:12:27.485 So let's think about a simple pendulum. 00:12:43.050 --> 00:12:49.770 So our common massless string and a bob on the end, 00:12:49.770 --> 00:12:58.215 some length L, equation of motion. 00:13:03.460 --> 00:13:05.810 And this is point A up here. 00:13:05.810 --> 00:13:20.930 IZZ with respect to A. Theta double dot plus MgL sine theta 00:13:20.930 --> 00:13:22.580 equals 0. 00:13:22.580 --> 00:13:23.830 That's the equation of motion. 00:13:23.830 --> 00:13:25.329 With no damping, that's the equation 00:13:25.329 --> 00:13:27.080 of motion in this system. 00:13:27.080 --> 00:13:30.060 Is it a linear differential equation? 00:13:33.500 --> 00:13:35.790 And to do the things that we want 00:13:35.790 --> 00:13:40.210 to be able to do in this course, like vibration 00:13:40.210 --> 00:13:43.240 with harmonic inputs and so forth, we 00:13:43.240 --> 00:13:46.040 want to deal with linear equations. 00:13:46.040 --> 00:13:48.940 So one of the topics for today is linearization. 00:13:48.940 --> 00:13:52.550 So this is one of the simplest examples of linearization. 00:13:52.550 --> 00:13:57.470 We need a linearized equation, and we 00:13:57.470 --> 00:14:01.790 need to remember in a couple of approximations. 00:14:01.790 --> 00:14:06.410 So sine of theta, you can do Taylor series expansion. 00:14:06.410 --> 00:14:13.780 It's theta minus theta cubed over 3 factorial plus theta 00:14:13.780 --> 00:14:19.220 to the fifth over 5 factorial plus minus and so forth. 00:14:19.220 --> 00:14:26.970 And cosine of theta is 1 minus theta 00:14:26.970 --> 00:14:32.030 squared over 2 factorial plus theta to the fourth over 4 00:14:32.030 --> 00:14:35.620 factorial, and so forth. 00:14:39.110 --> 00:14:42.480 So what we say, what do we mean when we linearize something? 00:14:42.480 --> 00:14:44.410 So linearization means that we're essentially 00:14:44.410 --> 00:14:51.940 assuming the variable that we're working with is small enough 00:14:51.940 --> 00:14:55.710 that the right hand side, an adequate approximation 00:14:55.710 --> 00:15:00.200 of this function, is to keep only up to the linear terms 00:15:00.200 --> 00:15:03.220 on the right hand side. 00:15:03.220 --> 00:15:08.580 For sine, the term raised to the 1 power, 00:15:08.580 --> 00:15:09.850 that's the linear term. 00:15:09.850 --> 00:15:12.720 This is a cubic term, a fifth order term. 00:15:12.720 --> 00:15:15.790 We're going to throw those away and say, this is close enough. 00:15:15.790 --> 00:15:18.320 For cosine, it's 1 minus theta squared. 00:15:18.320 --> 00:15:19.810 We throw these away. 00:15:19.810 --> 00:15:25.440 The small angle approximation for cosine as it's equal to 1. 00:15:25.440 --> 00:15:27.940 That's the simplest example of linearization, 00:15:27.940 --> 00:15:32.350 of a non-linear term. 00:15:32.350 --> 00:15:37.120 So when you linearize this equation of motion, 00:15:37.120 --> 00:15:47.150 we end up with IZZ with respect to A theta double dot plus MGL 00:15:47.150 --> 00:15:51.660 theta equals 0, and we get our familiar natural frequency 00:15:51.660 --> 00:15:54.820 for Bob as square root of g over L. 00:15:54.820 --> 00:15:57.745 So we need linearization to be able to do pendulum problems. 00:16:04.865 --> 00:16:05.365 Hmm. 00:16:08.790 --> 00:16:10.690 OK. 00:16:10.690 --> 00:16:12.530 Or maybe let's do an example here. 00:16:12.530 --> 00:16:15.060 I've got a pendulum that we'll do an experiment 00:16:15.060 --> 00:16:16.230 with this morning. 00:16:16.230 --> 00:16:20.820 But 30 degrees is about like that. 00:16:20.820 --> 00:16:23.220 17 degrees is about like that. 00:16:23.220 --> 00:16:25.380 That's quite a bit of angle. 00:16:25.380 --> 00:16:32.720 Is that small in the sense that I'm linearizing this equation? 00:16:32.720 --> 00:16:39.520 So 17 degrees happens to be-- I'll have to use this here. 00:16:47.918 --> 00:16:54.550 That's actually 17.2 degrees equals 0.3 radians. 00:16:58.750 --> 00:17:10.140 Sine of 0.3 is 0.2955. 00:17:10.140 --> 00:17:14.240 And if we fill out and look at these terms, 00:17:14.240 --> 00:17:22.220 the lead term here is 0.3-- so plug in the 0.3-- minus-- 00:17:22.220 --> 00:17:29.530 and the second term when you cube 0.3 and divide it by 6, 00:17:29.530 --> 00:17:40.750 the second term is minus-- what is my number here-- 0.0045. 00:17:40.750 --> 00:17:44.130 And if you subtract that from this, you get exactly this. 00:17:44.130 --> 00:17:48.110 So to four decimal places, you only 00:17:48.110 --> 00:17:51.210 need two terms in this series to get exactly the right answer. 00:17:51.210 --> 00:17:54.750 This thing out here, this fifth order term, is really tiny. 00:17:54.750 --> 00:17:58.180 But the approximation, if we say, OK, let's skip this, 00:17:58.180 --> 00:18:02.620 we're saying that 0.3 is approximately 0.2955. 00:18:02.620 --> 00:18:05.670 Pretty close. 00:18:05.670 --> 00:18:08.872 So up to 17 degrees, 0.3 radians, 00:18:08.872 --> 00:18:10.080 that's a great approximation. 00:18:14.060 --> 00:18:17.871 So it's a little high by about, I think it's about 1 and 1/2% 00:18:17.871 --> 00:18:18.370 high. 00:18:26.250 --> 00:18:30.070 So for pretty large angles for pendula, 00:18:30.070 --> 00:18:32.040 that simple linearization works just fine. 00:18:34.700 --> 00:18:40.370 OK, once we get it linearized, that equation of motion 00:18:40.370 --> 00:18:45.420 is of exactly the same form as the one up there. 00:18:45.420 --> 00:18:47.380 We don't have any damping in it. 00:18:47.380 --> 00:18:50.510 We could add some damping. 00:18:50.510 --> 00:18:54.510 We can put a damping in here with a torsional damper-- ct 00:18:54.510 --> 00:18:56.110 theta dot. 00:18:56.110 --> 00:18:59.570 And now that equation is of exactly the same form 00:18:59.570 --> 00:19:04.690 as the linear oscillator, linear meaning translational 00:19:04.690 --> 00:19:06.270 oscillator. 00:19:06.270 --> 00:19:09.480 Have that inertia term, a damping term, a stiffness term. 00:19:09.480 --> 00:19:12.700 It's a second order linear differential equation, 00:19:12.700 --> 00:19:14.499 homogeneous linear differential equation, 00:19:14.499 --> 00:19:15.790 nothing on the right hand side. 00:19:22.730 --> 00:19:25.400 Because they're exactly the same form, 00:19:25.400 --> 00:19:31.940 then the solution for decay, transient decay 00:19:31.940 --> 00:19:35.670 from initial conditions, takes on exactly the same form 00:19:35.670 --> 00:19:40.590 except that it has an initial angle, theta 0, 00:19:40.590 --> 00:19:42.430 and I use the approximation here. 00:19:42.430 --> 00:19:50.940 Cosine omega dt plus theta 0 dot, the initial velocity, 00:19:50.940 --> 00:20:01.070 over omega d sine omega dt all times e to the minus zeta omega 00:20:01.070 --> 00:20:02.960 nt. 00:20:02.960 --> 00:20:06.030 So that's the exact same transient decay equation, 00:20:06.030 --> 00:20:08.850 but now cast in angular terms. 00:20:15.310 --> 00:20:20.460 And if you wanted to express it as a cosine omega t minus v, 00:20:20.460 --> 00:20:24.080 then A would be this squared plus this squared square root 00:20:24.080 --> 00:20:30.570 and the phi would be a similar calculation as we 00:20:30.570 --> 00:20:32.690 have up there someone here. 00:20:35.330 --> 00:20:38.900 Just the B term over the A term, tangent numbers. 00:20:38.900 --> 00:20:41.790 All right. 00:20:41.790 --> 00:20:42.610 What's it good for? 00:20:42.610 --> 00:20:45.350 So I use this, this equation gets used quite a lot. 00:20:45.350 --> 00:20:46.695 It has some practical uses. 00:20:51.140 --> 00:20:57.000 Let's do an example, a little more complicated pendulum. 00:20:57.000 --> 00:20:57.840 Draw a stick maybe. 00:21:01.470 --> 00:21:02.580 Center of mass there. 00:21:07.295 --> 00:21:15.020 A, IZZ with respect to A. We'll call it 00:21:15.020 --> 00:21:18.155 ML cubed over 3 for a slender rod. 00:21:23.310 --> 00:21:26.340 And now, what I want to do is I have 00:21:26.340 --> 00:21:34.490 coming along here a mass, a bullet, that has mass m. 00:21:34.490 --> 00:21:39.940 Has velocity vi for initial here, 00:21:39.940 --> 00:21:46.990 and that's its linear momentum, p initial. 00:21:49.540 --> 00:21:52.532 And it's going to hit this stick and bed in it. 00:21:52.532 --> 00:21:53.990 So you've done this problem before. 00:21:53.990 --> 00:21:54.490 Yeah. 00:21:54.490 --> 00:21:56.720 AUDIENCE: [INAUDIBLE]. 00:21:56.720 --> 00:21:57.470 PROFESSOR: Pardon? 00:21:57.470 --> 00:21:59.730 AUDIENCE: ML cubed [INAUDIBLE]. 00:21:59.730 --> 00:22:01.450 PROFESSOR: ML cubed over 3. 00:22:05.850 --> 00:22:06.470 Good. 00:22:06.470 --> 00:22:08.620 I don't know why I was thinking cubed this morning. 00:22:08.620 --> 00:22:10.090 ML squared over 3. 00:22:10.090 --> 00:22:10.990 Good catch there. 00:22:10.990 --> 00:22:12.360 OK, so we have it. 00:22:12.360 --> 00:22:14.250 This is mass moment of inertia. 00:22:14.250 --> 00:22:15.764 This is a pendulum. 00:22:15.764 --> 00:22:17.180 This bullet's going to come along. 00:22:20.870 --> 00:22:22.550 So this is exactly what I've got here. 00:22:22.550 --> 00:22:24.100 I'll get it in the picture. 00:22:24.100 --> 00:22:25.240 Yeah. 00:22:25.240 --> 00:22:27.140 So it's initially at rest. 00:22:27.140 --> 00:22:29.690 Coming along, this paper, this clip here, 00:22:29.690 --> 00:22:31.005 it represents the bullet. 00:22:31.005 --> 00:22:33.530 So it's swimming along. 00:22:33.530 --> 00:22:35.800 Hits this thing, sticks to it, and when it hits it, 00:22:35.800 --> 00:22:37.870 it does that. 00:22:37.870 --> 00:22:40.900 So then this thing after it hits swings back and forth. 00:22:40.900 --> 00:22:43.420 So what's the response of this system 00:22:43.420 --> 00:22:44.915 to being hit by the bullet? 00:22:44.915 --> 00:22:49.330 Well, I claim you can do it entirely by evaluating response 00:22:49.330 --> 00:22:51.620 to initial conditions. 00:22:51.620 --> 00:22:55.625 But we need to use one conservation law to get there. 00:22:59.790 --> 00:23:03.260 So what's conserved on impact? 00:23:03.260 --> 00:23:05.640 Is linear momentum conserved on impact? 00:23:09.930 --> 00:23:12.580 How many think yes? 00:23:12.580 --> 00:23:14.270 Linear momentum conserved. 00:23:14.270 --> 00:23:16.330 How many think angular momentum's conserved? 00:23:16.330 --> 00:23:16.830 Hm. 00:23:16.830 --> 00:23:19.780 Good. you guys have learned something this year. 00:23:19.780 --> 00:23:20.600 That's great. 00:23:20.600 --> 00:23:22.255 Why is linear momentum not conserved? 00:23:27.560 --> 00:23:31.130 Because are there any possible other external forces 00:23:31.130 --> 00:23:33.320 on the system? 00:23:33.320 --> 00:23:34.270 At the pin joint. 00:23:34.270 --> 00:23:36.860 You can have reaction forces here and there. 00:23:36.860 --> 00:23:38.910 You have no control of them. 00:23:38.910 --> 00:23:41.350 But the moments about this point, 00:23:41.350 --> 00:23:44.640 are there any external moments about that point 00:23:44.640 --> 00:23:45.780 during the impact? 00:23:45.780 --> 00:23:46.280 No. 00:23:46.280 --> 00:23:48.405 They're reaction forces, but there's no moment arm. 00:23:48.405 --> 00:23:49.830 So there's no moments. 00:23:49.830 --> 00:23:52.780 So you can use conservation of angular momentum. 00:23:52.780 --> 00:23:56.620 So H1 I'll call it here with respect to A 00:23:56.620 --> 00:24:01.410 is just R cross Pi. 00:24:01.410 --> 00:24:07.030 And the R is the length in the I direction. 00:24:07.030 --> 00:24:10.730 P is in the j direction, so the momentum is in the k. 00:24:10.730 --> 00:24:17.970 So this should be mv initial times L, 00:24:17.970 --> 00:24:20.830 and its direction is in the k hat direction. 00:24:20.830 --> 00:24:23.960 So that's the initial angular momentum 00:24:23.960 --> 00:24:25.760 of the system with respect to this. 00:24:25.760 --> 00:24:27.520 This has no initial angular momentum 00:24:27.520 --> 00:24:29.550 because it's motionless. 00:24:29.550 --> 00:24:32.010 And since angular momentum is conserved, 00:24:32.010 --> 00:24:35.785 that H2 we'll call it with respect to A 00:24:35.785 --> 00:24:40.910 has got to be equal to H1 with respect to A, 00:24:40.910 --> 00:24:49.880 and that will then be IZZ with respect to A theta dot. 00:24:49.880 --> 00:24:54.040 But I need to account for the mass, this thing. 00:24:54.040 --> 00:24:58.215 So the total mass moment of inertia with respect to A 00:24:58.215 --> 00:25:01.780 is IZZ with respect to A plus M-- what? 00:25:06.580 --> 00:25:09.250 Now I've got the total mass moment of inertia with respect 00:25:09.250 --> 00:25:13.335 to this point, that of the stick plus that of the initial mass 00:25:13.335 --> 00:25:15.140 that I've stuck on there. 00:25:15.140 --> 00:25:18.380 And this must be equal to theta dot. 00:25:18.380 --> 00:25:19.980 And I put a not down here because I'm 00:25:19.980 --> 00:25:23.540 looking for my equivalent initial condition. 00:25:23.540 --> 00:25:31.810 And this then is mv initial times 00:25:31.810 --> 00:25:37.730 L. Then I can solve for theta dot 0, 00:25:37.730 --> 00:25:50.520 and that looks like mv initial L over IZZ A plus mL squared. 00:25:50.520 --> 00:25:55.081 And everything on the right hand side you know. 00:25:55.081 --> 00:25:57.330 You know the initial velocity, the mass of the bullet, 00:25:57.330 --> 00:26:01.470 the length of the distance from the pivot, mass moment 00:26:01.470 --> 00:26:05.240 of inertia, and the additional mass moment of inertia. 00:26:05.240 --> 00:26:06.892 These are all numbers you plug in, 00:26:06.892 --> 00:26:08.100 and you get a value for this. 00:26:08.100 --> 00:26:11.970 And once you have a value for this, you can use that. 00:26:14.750 --> 00:26:17.110 In this problem, what's theta 0? 00:26:21.320 --> 00:26:25.460 The initial angular deflection at time t0 00:26:25.460 --> 00:26:29.550 plus right after the bullets hit it. 00:26:29.550 --> 00:26:35.600 And it hasn't moved because it hasn't had time to move yet. 00:26:35.600 --> 00:26:37.710 At some velocity, it takes finite time 00:26:37.710 --> 00:26:39.500 to get a deflection. 00:26:39.500 --> 00:26:42.700 So there's zero initial angular deflection, 00:26:42.700 --> 00:26:48.020 but you get a step up in initial angular velocity. 00:26:48.020 --> 00:26:51.180 And so the response of this system 00:26:51.180 --> 00:27:05.210 is theta t is theta 0 dot over omega d sine omega dt. 00:27:05.210 --> 00:27:06.060 So what's omega d? 00:27:15.950 --> 00:27:18.930 Remember, I'll define a few things. 00:27:18.930 --> 00:27:25.880 In this case, this is ct over 2 IZZ 00:27:25.880 --> 00:27:32.675 z A plus little ml squared-- we have 00:27:32.675 --> 00:27:37.120 to deal with all the quantities after the collision-- 2 times 00:27:37.120 --> 00:27:38.690 omega n. 00:27:38.690 --> 00:27:47.080 That's the damping ratio for this torsional pendulum, 00:27:47.080 --> 00:27:48.170 with this pendulum. 00:27:48.170 --> 00:27:49.960 It's the damping constant. 00:27:49.960 --> 00:27:53.170 2 times the mass, the inertial quantity, 00:27:53.170 --> 00:27:57.300 times omega n for a translational system 00:27:57.300 --> 00:28:04.260 at c over 2 m omega n. 00:28:04.260 --> 00:28:07.840 For a pendulum system, it's the torsional 00:28:07.840 --> 00:28:13.040 damping over 2 times the mass moment of inertia times omega 00:28:13.040 --> 00:28:14.490 n. 00:28:14.490 --> 00:28:22.630 And omega n, well, it is going to calculate 00:28:22.630 --> 00:28:23.930 the natural frequency. 00:28:23.930 --> 00:28:27.730 It's just MgL divided by IZZ plus this. 00:28:27.730 --> 00:28:29.540 Maybe you ought to write that down. 00:28:35.790 --> 00:28:38.630 So always for a sample singular [INAUDIBLE] oscillator, 00:28:38.630 --> 00:28:40.340 you want the undamped natural frequency. 00:28:40.340 --> 00:28:42.050 Ignore the damping term. 00:28:42.050 --> 00:28:45.880 Take the stiffness term coefficient here and divide it 00:28:45.880 --> 00:28:47.570 by the inertial coefficient. 00:28:47.570 --> 00:28:51.510 But we care about the natural frequency after the impact, 00:28:51.510 --> 00:28:58.055 so this is going to be-- ah. 00:28:58.055 --> 00:29:01.740 The trouble is here I don't know for this system, 00:29:01.740 --> 00:29:03.910 I haven't worked out yet, what this term looks like. 00:29:03.910 --> 00:29:04.698 What is it? 00:29:07.710 --> 00:29:11.920 This result right here is for the simple Bob. 00:29:11.920 --> 00:29:18.880 For this stick, it's MgL over 2 plus the little m times l. 00:29:18.880 --> 00:29:20.690 Little more messy. 00:29:20.690 --> 00:29:27.060 So MgL over 2 plus little ml. 00:29:27.060 --> 00:29:30.880 That'll be the-- come from the potential energy in this system 00:29:30.880 --> 00:29:37.490 all over IZZ A plus mL squared. 00:29:37.490 --> 00:29:39.230 So you get your natural frequency out 00:29:39.230 --> 00:29:42.146 of that expression. 00:29:42.146 --> 00:29:42.645 OK. 00:29:49.450 --> 00:29:53.970 So you do this problem sometimes before when 00:29:53.970 --> 00:29:56.660 you do, say, somebody asks you how high does it swing. 00:29:56.660 --> 00:29:57.250 AND so forth. 00:29:57.250 --> 00:30:00.030 Well, you can do it by conservation 00:30:00.030 --> 00:30:02.050 of energy, et cetera. 00:30:02.050 --> 00:30:04.520 But now, you have actually exact expression 00:30:04.520 --> 00:30:10.130 for the time history of the thing after the impact, 00:30:10.130 --> 00:30:12.420 including the effects of damping. 00:30:12.420 --> 00:30:19.870 And if you were to draw the result of this function 00:30:19.870 --> 00:30:23.970 of theta as a function of time, this one 00:30:23.970 --> 00:30:30.020 starts with no initial displacement but a velocity 00:30:30.020 --> 00:30:30.685 and does this. 00:30:37.600 --> 00:30:39.830 And that's your exponential decay envelope, 00:30:39.830 --> 00:30:42.150 and this is time. 00:30:45.910 --> 00:30:47.970 Now, what-- yeah. 00:30:47.970 --> 00:30:48.470 Excuse me. 00:30:51.970 --> 00:30:52.970 AUDIENCE: [INAUDIBLE]. 00:30:52.970 --> 00:30:54.465 PROFESSOR: Why this one? 00:30:54.465 --> 00:30:57.860 AUDIENCE: Yes. [INAUDIBLE]. 00:30:57.860 --> 00:30:59.170 PROFESSOR: Excuse me. 00:30:59.170 --> 00:30:59.870 Forgot the g. 00:30:59.870 --> 00:31:02.996 I mean, it accounts for the restoring moment, 00:31:02.996 --> 00:31:04.870 the additional little bit of restoring moment 00:31:04.870 --> 00:31:08.040 that you get from having added the mass of this thing 00:31:08.040 --> 00:31:09.250 to it, right. 00:31:09.250 --> 00:31:12.810 So it has by itself MgL sine theta, 00:31:12.810 --> 00:31:13.970 and we linearize that too. 00:31:13.970 --> 00:31:18.650 So it's MgL theta, and those two terms would add together. 00:31:18.650 --> 00:31:22.380 So you just have a second term here 00:31:22.380 --> 00:31:27.030 that has MgL like behavior. 00:31:27.030 --> 00:31:31.470 How I most often personally make use of expressions 00:31:31.470 --> 00:31:34.560 like this, or the one for translation, 00:31:34.560 --> 00:31:39.140 is because experimentally, if I'm 00:31:39.140 --> 00:31:42.440 trying to predict the vibration behavior of a system, 00:31:42.440 --> 00:31:46.360 one of the things you want to know is the damping. 00:31:46.360 --> 00:31:48.790 And one of the simplest ways to measure damping 00:31:48.790 --> 00:31:53.460 is to give a system an initial deflection or initial velocity 00:31:53.460 --> 00:31:56.420 and measure its decay, and from its decay, 00:31:56.420 --> 00:31:58.240 calculate the damping. 00:31:58.240 --> 00:32:06.470 So the last time I gave you an expression for doing that, 00:32:06.470 --> 00:32:13.720 and that was a damping ratio 1 over 2 pi 00:32:13.720 --> 00:32:18.110 times the number of cycles that you count, that you watch it, 00:32:18.110 --> 00:32:27.750 times the natural log of x of t over x of t plus n 00:32:27.750 --> 00:32:31.620 periods of vibration. 00:32:31.620 --> 00:32:32.550 This has a name. 00:32:32.550 --> 00:32:35.940 It's called the logarithmic decrement, this thing. 00:32:35.940 --> 00:32:38.250 So if somebody says log decrement, 00:32:38.250 --> 00:32:40.990 that's where they're referring to this expression. 00:32:51.265 --> 00:32:54.340 A comment about this. 00:32:54.340 --> 00:33:07.380 In this expression, x of t must be zero means 00:33:07.380 --> 00:33:12.080 if your measurement-- we have x of t here, or theta of t-- 00:33:12.080 --> 00:33:14.100 they must be zero mean. 00:33:14.100 --> 00:33:16.670 There must be oscillations around zero 00:33:16.670 --> 00:33:19.460 or you have to have subtracted the mean to get it there 00:33:19.460 --> 00:33:21.730 because if this is displaced and is oscillating 00:33:21.730 --> 00:33:25.480 around some offset, then this calculates and will get really 00:33:25.480 --> 00:33:25.980 messed up. 00:33:25.980 --> 00:33:28.250 It's got an offset plus an offset here 00:33:28.250 --> 00:33:29.160 plus an offset there. 00:33:29.160 --> 00:33:31.500 It means it's totally meaningless. 00:33:31.500 --> 00:33:36.970 So you must remove the mean value from any time history 00:33:36.970 --> 00:33:39.360 that you go to do this. 00:33:39.360 --> 00:33:42.790 So there's an easier way the same expression-- and this is, 00:33:42.790 --> 00:33:45.130 in fact, the way I use this. 00:33:45.130 --> 00:33:48.030 A plot out like just your data acquisition 00:33:48.030 --> 00:33:50.420 grabs it, plots it for you. 00:33:50.420 --> 00:33:53.410 I take this value from here to here, 00:33:53.410 --> 00:33:57.510 and this is my peak to peak amplitude. 00:33:57.510 --> 00:34:01.640 And then I go out n cycles later and find the peak 00:34:01.640 --> 00:34:03.940 to peak amplitude. 00:34:03.940 --> 00:34:06.920 And so this is perfectly, this is just the same as 1 00:34:06.920 --> 00:34:15.530 over 2 pi n, but now you do natural log of x peak 00:34:15.530 --> 00:34:22.969 to peak t over x peak to peak at t plus n periods. 00:34:25.670 --> 00:34:27.844 And that now, peak to peak measurement, 00:34:27.844 --> 00:34:29.010 you totally ignore the mean. 00:34:29.010 --> 00:34:30.870 Doesn't matter where you are. 00:34:30.870 --> 00:34:35.010 You want the here to here, here to here, plug it in there, 00:34:35.010 --> 00:34:37.320 and you're done. 00:34:37.320 --> 00:34:49.219 OK, so let's-- I got 1, 2, 3, 4. 00:34:49.219 --> 00:35:01.060 Let's let n equal 4, and let's assume this expression here-- 00:35:01.060 --> 00:35:05.830 n is 1 over 2 pi times 4. 00:35:05.830 --> 00:35:10.090 And let's assume that in these four periods 00:35:10.090 --> 00:35:15.230 from-- that's 1 period, 2, 3, 4 getting out here 00:35:15.230 --> 00:35:21.710 to this fourth, four periods away, that this is one fifth 00:35:21.710 --> 00:35:23.610 the initial. 00:35:23.610 --> 00:35:25.945 So this would be the natural log of 5. 00:35:28.980 --> 00:35:32.290 So 1 over 2 pi times 4, natural log of 5, 00:35:32.290 --> 00:35:36.895 and you run the numbers, you get 0.064, 00:35:36.895 --> 00:35:42.575 or what we call 6.4% damping. 00:35:42.575 --> 00:35:43.450 That's how you do it. 00:35:43.450 --> 00:35:46.630 That's the way you do a calculation like that. 00:35:46.630 --> 00:35:52.150 Now, I gave you a quick rule of thumb for estimating damping, 00:35:52.150 --> 00:35:54.910 and this is what I-- I can't work. 00:35:54.910 --> 00:35:58.220 I don't do logs in my head, but I 00:35:58.220 --> 00:36:01.050 can do damping estimates without that 00:36:01.050 --> 00:36:06.400 because I know that zeta is also-- 00:36:06.400 --> 00:36:09.040 if I just plug in some numbers here 00:36:09.040 --> 00:36:13.680 and run them all in advance is 0.11 00:36:13.680 --> 00:36:16.695 divided by the number of cycles to decay 50%. 00:36:20.132 --> 00:36:21.590 So we're going to do an experiment. 00:36:26.990 --> 00:36:29.620 And I guess it can be seen with the camera. 00:36:29.620 --> 00:36:33.000 So here's my pendulum. 00:36:33.000 --> 00:36:38.030 This is my initial amplitude, and this is about half. 00:36:38.030 --> 00:36:45.190 So if I take this thing over here, like that, then let go, 00:36:45.190 --> 00:36:49.780 and count the cycles that it takes to decay halfway, 00:36:49.780 --> 00:36:50.889 we can do this experiment. 00:36:50.889 --> 00:36:51.930 So let's do it carefully. 00:36:51.930 --> 00:36:54.160 So line it up like that, and you're 00:36:54.160 --> 00:36:57.480 going to help me tell-- you count how many cycles it 00:36:57.480 --> 00:36:59.250 takes till it gets to here. 00:36:59.250 --> 00:37:15.330 So 1, 2, 3, 4, 5, 6, 7, 8. 00:37:15.330 --> 00:37:16.920 About eight cycles. 00:37:16.920 --> 00:37:20.170 So it decayed halfway in eight cycles. 00:37:20.170 --> 00:37:27.016 So zeta is approximately 0.11/8. 00:37:27.016 --> 00:37:30.240 It's 1 and 1/2%, 1.4%, something like that. 00:37:33.042 --> 00:37:34.500 Perfectly good estimate of damping. 00:37:34.500 --> 00:37:44.780 Now, the stopwatch here, we can do this experiment again. 00:37:47.660 --> 00:37:49.100 I want you to count. 00:37:49.100 --> 00:37:52.000 You're doing the counting. 00:37:52.000 --> 00:37:57.390 And I'm going to say start, and I want you to count cycles 00:37:57.390 --> 00:37:59.570 until I say stop. 00:37:59.570 --> 00:38:04.040 Now, I'll probably stop on 10 to make the calculation easy, 00:38:04.040 --> 00:38:08.090 so quietly to yourself count the number of cycles 00:38:08.090 --> 00:38:11.950 from the time I release it until the time I stop. 00:38:11.950 --> 00:38:13.060 Come back here. 00:38:19.110 --> 00:38:22.150 So this time, the backdrop doesn't matter. 00:38:22.150 --> 00:38:23.870 I just want you to count cycles. 00:38:27.200 --> 00:38:29.220 And I'll start-- I'll let it get going, 00:38:29.220 --> 00:38:30.730 and when it comes back to me is when 00:38:30.730 --> 00:38:33.705 I'm going to start the stopwatch because I 00:38:33.705 --> 00:38:35.580 have a hard time doing both at the same time. 00:38:35.580 --> 00:38:38.190 So start. 00:38:57.810 --> 00:38:59.320 How many cycles? 00:38:59.320 --> 00:39:01.320 AUDIENCE: [INAUDIBLE]. 00:39:01.320 --> 00:39:11.390 PROFESSOR: So I got 17.84 for 10. 00:39:11.390 --> 00:39:25.010 So 10 divided by 10 is 1.784 seconds per cycle. 00:39:25.010 --> 00:39:26.960 Can't write like that. 00:39:26.960 --> 00:39:29.450 1.784 seconds per cycle. 00:39:29.450 --> 00:39:33.159 The frequency would be 1 over that, right. 00:39:33.159 --> 00:39:34.950 The thing you have to be careful about when 00:39:34.950 --> 00:39:44.400 you're counting cycles is if I start here, that's 0, 1, 2. 00:39:44.400 --> 00:39:47.160 A very common human mistake is when you're counting something 00:39:47.160 --> 00:39:49.380 like this is to say one when you start, 00:39:49.380 --> 00:39:52.610 and then you're going to be off by one count. 00:39:52.610 --> 00:39:54.020 Follow me? 00:39:54.020 --> 00:39:59.320 If I start 0, 1, 2. 00:39:59.320 --> 00:40:01.960 So I start the clock on zero, but the first cycle 00:40:01.960 --> 00:40:04.030 isn't completed till one whole cycle later. 00:40:04.030 --> 00:40:06.985 So be careful how you count. 00:40:06.985 --> 00:40:07.485 OK. 00:40:23.670 --> 00:40:35.380 Now we're going to shift gears and take on a new topic, 00:40:35.380 --> 00:40:40.450 and that's the response to a harmonic input, 00:40:40.450 --> 00:40:43.823 some cosine omega t excitation. 00:40:43.823 --> 00:40:44.769 Yeah. 00:40:44.769 --> 00:40:46.065 AUDIENCE: [INAUDIBLE]. 00:40:46.065 --> 00:40:47.190 PROFESSOR: What is omega d? 00:41:02.340 --> 00:41:05.224 So it's the damped natural frequency. 00:41:05.224 --> 00:41:06.390 That's how it's referred to. 00:41:06.390 --> 00:41:09.560 It is the frequency you observe when you do 00:41:09.560 --> 00:41:11.782 an experiment like we just did. 00:41:11.782 --> 00:41:14.430 The actual oscillation frequency when 00:41:14.430 --> 00:41:16.760 it's responding to initial conditions 00:41:16.760 --> 00:41:18.930 is slightly different from the undamped, 00:41:18.930 --> 00:41:23.590 but if you have light damping, if you have even 10% damping, 00:41:23.590 --> 00:41:27.550 0.1 squared is 0.01. 00:41:27.550 --> 00:41:29.370 That's 0.99 square root. 00:41:29.370 --> 00:41:32.090 It's 0.995. 00:41:32.090 --> 00:41:33.660 So you're only off by half. 00:41:33.660 --> 00:41:35.600 They're only half a percent difference. 00:41:35.600 --> 00:41:38.450 So for lightly damped systems, for all intents and purposes, 00:41:38.450 --> 00:41:41.250 mega n and mega d are almost exactly the same. 00:41:43.920 --> 00:41:44.420 OK. 00:41:47.020 --> 00:41:53.890 We now want to think about-- we have a linear system putting 00:41:53.890 --> 00:41:56.360 a force into it. 00:41:56.360 --> 00:42:01.080 It looks like some F0 cosine omega t. 00:42:01.080 --> 00:42:05.430 And out of that system, we measure a response, x of t. 00:42:09.380 --> 00:42:14.300 And inside this box here is my system transfer function. 00:42:14.300 --> 00:42:17.545 It's the mathematics that tells me I can take F of t 00:42:17.545 --> 00:42:21.687 in and predict what x of t out is. 00:42:21.687 --> 00:42:23.270 So I need to know the information that 00:42:23.270 --> 00:42:25.720 goes into this box, and of course, the real system-- this 00:42:25.720 --> 00:42:27.280 is just the mechanical system. 00:42:27.280 --> 00:42:29.880 Force in, measured output out. 00:42:29.880 --> 00:42:33.660 This is what we call a single input single output 00:42:33.660 --> 00:42:37.820 system, SISO, single input single output linear system. 00:42:42.515 --> 00:42:44.140 And there's all sorts of linear systems 00:42:44.140 --> 00:42:46.265 that you're going to study as mechanical engineers, 00:42:46.265 --> 00:42:49.380 and you've already begun, I'm sure, studying some of them. 00:42:49.380 --> 00:42:51.720 One of the properties of a linear system 00:42:51.720 --> 00:42:57.220 is that you put a force in, F1, and measure a response out, x1. 00:42:57.220 --> 00:42:58.970 And then you try a different force, F2, 00:42:58.970 --> 00:43:01.540 and you measure a response out, x2. 00:43:01.540 --> 00:43:03.600 What's the response if you put them both in 00:43:03.600 --> 00:43:05.300 at the same time, F1 and F2? 00:43:09.550 --> 00:43:13.050 You just add the responses to them individually. 00:43:13.050 --> 00:43:15.360 So F1 gives you x1. 00:43:15.360 --> 00:43:16.690 F2 give you x2. 00:43:16.690 --> 00:43:20.010 F1 plus F2 gives you x1 plus x2, and that's 00:43:20.010 --> 00:43:25.020 one of the characteristics of a linear system. 00:43:25.020 --> 00:43:30.240 We use that concept to be able to separate the response. 00:43:30.240 --> 00:43:33.330 Our calculation's about the response of a system, 00:43:33.330 --> 00:43:37.100 like our oscillator here, separate its response 00:43:37.100 --> 00:43:42.070 to transient effects, transience being initial conditions. 00:43:42.070 --> 00:43:45.750 They die out over time-- that's why we call them transients-- 00:43:45.750 --> 00:43:48.690 and steady state effects. 00:43:48.690 --> 00:43:50.450 So cosine omega t, you can leave it 00:43:50.450 --> 00:43:52.950 running for a long, long time, and pretty soon, the system 00:43:52.950 --> 00:43:58.930 will settle down to responding just to that cosine omega t. 00:43:58.930 --> 00:44:00.900 And that we call steady state. 00:44:00.900 --> 00:44:02.800 And we use them separately. 00:44:02.800 --> 00:44:04.640 So we've done initial conditions. 00:44:04.640 --> 00:44:06.920 Now we're going to look at the steady state 00:44:06.920 --> 00:44:13.490 response of a-- say our oscillator, our mass spring 00:44:13.490 --> 00:44:17.920 dashpot, to a harmonic input, F0 cosine omega t. 00:44:20.850 --> 00:44:22.260 Another brief word. 00:44:22.260 --> 00:44:33.080 If I have a force, F0 cosine omega t would look like that. 00:44:35.810 --> 00:44:42.770 And the response that I measure to start off with my-- 00:44:42.770 --> 00:44:47.090 it's sitting here at zero when you turn this on. 00:44:47.090 --> 00:44:52.780 And it's going to do some odd things initially, and then 00:44:52.780 --> 00:45:02.850 eventually settle down to some long term steady response. 00:45:02.850 --> 00:45:04.900 The amplitude stays constant. 00:45:04.900 --> 00:45:08.360 It stays angle with respect to the input 00:45:08.360 --> 00:45:09.840 isn't necessarily the same. 00:45:09.840 --> 00:45:18.547 There's some possibly phase shift. 00:45:18.547 --> 00:45:20.880 And that's so the two, if you're plotting them together, 00:45:20.880 --> 00:45:21.671 they won't line up. 00:45:21.671 --> 00:45:24.090 But see this messy stuff at the beginning? 00:45:24.090 --> 00:45:27.600 When you first turn this on, it jumps from here 00:45:27.600 --> 00:45:31.740 to here, that force, and it gives it a kick to begin with. 00:45:31.740 --> 00:45:36.390 And this will have some response initially due 00:45:36.390 --> 00:45:38.750 to that transient start up. 00:45:38.750 --> 00:45:42.910 And this response is all modeled by the response 00:45:42.910 --> 00:45:44.770 to initial conditions. 00:45:44.770 --> 00:45:47.310 And it'll die out after a while, this messy stuff. 00:45:47.310 --> 00:45:49.280 What's the frequency? 00:45:49.280 --> 00:45:54.950 What frequency do you expect this initial, erratic looking 00:45:54.950 --> 00:45:55.817 stuff to be at? 00:45:59.400 --> 00:46:03.670 Its response to initial conditions. 00:46:03.670 --> 00:46:08.250 What is the model for a response to initial conditions? 00:46:08.250 --> 00:46:09.962 What's the frequency of the response 00:46:09.962 --> 00:46:12.170 to initial conditions of the single degree of freedom 00:46:12.170 --> 00:46:14.470 system? 00:46:14.470 --> 00:46:16.580 We have an equation over here, right? 00:46:20.300 --> 00:46:22.640 The top has a cosine term and a sine term. 00:46:22.640 --> 00:46:24.690 Part of it's a response to initial displacement. 00:46:24.690 --> 00:46:27.560 Part of it's a response to the initial velocity. 00:46:27.560 --> 00:46:31.880 Any of this start up stuff can be cast as initial conditions, 00:46:31.880 --> 00:46:33.620 and the response to initial conditions 00:46:33.620 --> 00:46:37.420 is always at the natural frequency period. 00:46:37.420 --> 00:46:41.040 No other frequencies for same degree of freedom systems. 00:46:41.040 --> 00:46:43.270 So you get a behavior that's oscillating 00:46:43.270 --> 00:46:45.010 at its natural frequency. 00:46:45.010 --> 00:46:49.640 Mixed in there is a response at the excitation frequency. 00:46:49.640 --> 00:46:52.090 And after a long time, the response 00:46:52.090 --> 00:46:54.180 is only excitation frequency. 00:46:54.180 --> 00:46:55.150 This is now out here. 00:46:55.150 --> 00:46:56.590 This is omega. 00:46:56.590 --> 00:47:04.280 In here, you have omega and omega d going on. 00:47:04.280 --> 00:47:06.430 So this is messy. 00:47:06.430 --> 00:47:08.330 Usually isn't important, but it is. 00:47:08.330 --> 00:47:11.690 There are ways of getting the exact solution, but mostly, 00:47:11.690 --> 00:47:15.910 vibration engineers, you're interested in the long term 00:47:15.910 --> 00:47:21.905 steady state response to what we call a harmonic input. 00:47:21.905 --> 00:47:22.405 OK. 00:47:40.510 --> 00:47:44.980 So we'll work a classic single degree 00:47:44.980 --> 00:47:52.360 of freedom oscillator problem-- excited by F0 cosine omega t. 00:47:52.360 --> 00:47:55.800 You've done this in 1803, but now we'll 00:47:55.800 --> 00:48:00.230 do it using engineering terminology. 00:48:00.230 --> 00:48:04.150 We'll look at it the way a person studying vibration 00:48:04.150 --> 00:48:06.160 would think about this. 00:48:06.160 --> 00:48:07.535 We know the equation of motion. 00:48:21.070 --> 00:48:24.640 And I'm interested in the steady state response. 00:48:24.640 --> 00:48:28.470 So this is x, and I'll do-- you just write it once like this. 00:48:28.470 --> 00:48:29.650 SS, steady state. 00:48:35.750 --> 00:48:38.520 I'm only interested in its-- after those transients 00:48:38.520 --> 00:48:39.400 have died out. 00:48:43.680 --> 00:48:46.130 And that steady state response I know 00:48:46.130 --> 00:48:51.100 is going to be some amplitude X0 cosine omega 00:48:51.100 --> 00:48:54.320 t minus some phase angle that I don't necessarily 00:48:54.320 --> 00:48:55.810 know to begin with. 00:48:55.810 --> 00:48:58.280 But that's my input. 00:48:58.280 --> 00:48:59.810 This is my output. 00:48:59.810 --> 00:49:04.820 I plug it into here and turn the crank and see what falls out. 00:49:11.510 --> 00:49:16.230 So you plug both of those in, and you 00:49:16.230 --> 00:49:20.480 get two-- you get-- this is going 00:49:20.480 --> 00:49:23.360 to be a little writing intensive for a few minutes. 00:49:38.220 --> 00:49:46.760 So you plug the X0 cosine omega t into all of these terms. 00:49:46.760 --> 00:49:50.070 The m term gives you minus m omega squared, 00:49:50.070 --> 00:49:55.920 the k term gives you a k, and the damping term, 00:49:55.920 --> 00:50:11.950 minus c omega sine omega t minus v. All of that 00:50:11.950 --> 00:50:18.620 equals the right hand side-- F0 cosine omega t. 00:50:18.620 --> 00:50:20.441 So this just purely from substitution 00:50:20.441 --> 00:50:22.065 and then gathering some terms together. 00:50:36.070 --> 00:50:40.450 I'm going to divide through by k, by k. 00:50:40.450 --> 00:50:42.347 If I divide through by k, k divided by-- this 00:50:42.347 --> 00:50:43.700 gives me a one. 00:50:43.700 --> 00:50:46.100 This gives me an m over k, which is 00:50:46.100 --> 00:50:49.425 1 over the natural frequency squared, for example. 00:50:49.425 --> 00:50:54.400 And I'm going to put this into a form that 00:50:54.400 --> 00:50:58.470 is the standard form for discussing vibration problems. 00:50:58.470 --> 00:51:04.770 So this equation can be rewritten in this form. 00:51:04.770 --> 00:51:15.320 1 minus omega squared over omega n squared cosine omega 00:51:15.320 --> 00:51:26.990 t minus v minus 2 zeta omega over omega n sine omega t 00:51:26.990 --> 00:51:34.740 minus v. All that's still equal to F0 cosine omega t. 00:51:34.740 --> 00:51:36.910 So this is getting into kind of more standard form. 00:51:36.910 --> 00:51:38.270 So there's 1 minus omega. 00:51:38.270 --> 00:51:40.100 This now, this omega over omega n, 00:51:40.100 --> 00:51:45.210 is called the frequency ratio, and you see a lot of that. 00:51:45.210 --> 00:51:48.670 And I've substituted n here. c omega over k 00:51:48.670 --> 00:51:53.230 turns out to be 2 zeta omega over omega n. 00:51:53.230 --> 00:51:55.640 So this frequency ratio appears a lot. 00:51:55.640 --> 00:52:02.000 in our-- let's see here. 00:52:10.220 --> 00:52:16.590 You need a couple of trig identities-- cosine omega t 00:52:16.590 --> 00:52:31.510 minus v. Cosine omega t cosine phi plus sine omega t sine phi, 00:52:31.510 --> 00:52:44.880 and sine omega t minus phi gives you [INAUDIBLE] sine. 00:52:44.880 --> 00:52:57.310 Sine omega t cosine phi minus cosine omega t sine phi. 00:52:57.310 --> 00:52:59.820 So that's a trig identity you actually use quite a bit 00:52:59.820 --> 00:53:01.990 doing vibration problems. 00:53:01.990 --> 00:53:07.290 We need them, so we take these, plug them 00:53:07.290 --> 00:53:11.980 in in all these places, and do quite a bit of cranking. 00:53:11.980 --> 00:53:13.190 Yep. 00:53:13.190 --> 00:53:15.690 AUDIENCE: [INAUDIBLE]. 00:53:15.690 --> 00:53:17.280 PROFESSOR: Yeah. 00:53:17.280 --> 00:53:19.850 Thank you. 00:53:19.850 --> 00:53:22.670 And that's called, that f over k, 00:53:22.670 --> 00:53:26.442 is how much the spring would move statically, 00:53:26.442 --> 00:53:28.750 at which the point would move statically. 00:53:28.750 --> 00:53:31.570 We'll need that term also. 00:53:31.570 --> 00:53:32.850 OK. 00:53:32.850 --> 00:53:36.120 You do all of this. 00:53:36.120 --> 00:53:37.375 Here, I'll call these-- 00:53:59.678 --> 00:54:00.680 OK. 00:54:00.680 --> 00:54:12.188 So this is C. That's expression C. 00:54:12.188 --> 00:54:13.470 I can't see that probably. 00:54:13.470 --> 00:54:24.960 Call this D, this E. So you plug D and E into C 00:54:24.960 --> 00:54:27.105 and work it through, you get two equations. 00:55:03.650 --> 00:55:05.750 You break it into two parts because one 00:55:05.750 --> 00:55:08.350 is a function of cosine omega t, and then 00:55:08.350 --> 00:55:10.890 you have another part after this substitution that's 00:55:10.890 --> 00:55:14.880 a function of sine omega t, and you can separate them. 00:55:38.230 --> 00:55:40.120 But there's no sine omega t force. 00:55:40.120 --> 00:55:41.710 On the right hand side, you get zero. 00:55:41.710 --> 00:55:43.410 There are two equations here. 00:55:43.410 --> 00:55:47.800 How many unknowns do we have? 00:55:47.800 --> 00:55:55.110 All we know when we start this thing is the input, 00:55:55.110 --> 00:56:04.700 and we have unknown response amplitude, 00:56:04.700 --> 00:56:07.850 and we have an unknown phase that we're looking for. 00:56:07.850 --> 00:56:08.730 How many equations? 00:56:08.730 --> 00:56:09.500 How many unknowns? 00:56:09.500 --> 00:56:11.440 Two and two. 00:56:11.440 --> 00:56:13.170 So you can do a lot of cranking, which 00:56:13.170 --> 00:56:14.600 I have no intention of doing here, 00:56:14.600 --> 00:56:20.885 and solve for the amplitude of the response and the phase. 00:56:40.690 --> 00:56:45.480 And every textbook-- the Williams textbook does this. 00:56:45.480 --> 00:56:50.920 There are two readings posted on Stellar 00:56:50.920 --> 00:56:54.790 by [? Row. ?] Every textbook goes through these derivations 00:56:54.790 --> 00:56:56.090 that I've just done. 00:56:56.090 --> 00:56:57.443 Nick, you've got a question. 00:56:57.443 --> 00:56:58.359 AUDIENCE: [INAUDIBLE]. 00:57:00.900 --> 00:57:01.740 PROFESSOR: Pardon? 00:57:01.740 --> 00:57:03.370 AUDIENCE: [INAUDIBLE]. 00:57:03.370 --> 00:57:06.215 PROFESSOR: Yeah, I keep forgetting it. 00:57:06.215 --> 00:57:07.050 You're right. 00:57:07.050 --> 00:57:08.650 So we got a k here. 00:57:08.650 --> 00:57:13.480 And notice, this equation, we throw away that for now. 00:57:13.480 --> 00:57:15.270 We get rid of this for now. 00:57:15.270 --> 00:57:17.420 We have these two equations and two unknowns 00:57:17.420 --> 00:57:22.400 are just algebraic equations There's not time dependent. 00:57:22.400 --> 00:57:24.210 We can get rid of that part. 00:57:24.210 --> 00:57:26.940 So we've now reduced this to algebra, 00:57:26.940 --> 00:57:31.330 and the answer is plotted up there. 00:57:31.330 --> 00:57:33.290 You've probably seen it before. 00:57:33.290 --> 00:57:40.780 It says that x0 is F0 over k-- I can get it right this time 00:57:40.780 --> 00:57:45.030 from the get go-- over a denominator, which 00:57:45.030 --> 00:57:49.420 appears again and again and again in vibration. 00:57:49.420 --> 00:57:54.530 Omega squared over omega n squared 00:57:54.530 --> 00:58:02.570 squared plus 2 zeta omega over omega n 00:58:02.570 --> 00:58:11.161 squared square root, the whole thing, and an expression 00:58:11.161 --> 00:58:11.660 for phi. 00:58:16.050 --> 00:58:23.630 Tangent inverse of 2 zeta omega over omega n, 1 00:58:23.630 --> 00:58:30.000 minus omega squared over omega n squared. 00:58:30.000 --> 00:58:32.310 So you can solve all that-- this mess over here 00:58:32.310 --> 00:58:34.490 for these two quantities. 00:58:34.490 --> 00:58:36.690 Do you need to remember this? 00:58:36.690 --> 00:58:39.770 You ever going to be asked this on a quiz? 00:58:39.770 --> 00:58:42.550 Not by me. 00:58:42.550 --> 00:58:46.320 You ever going to have to use this on a quiz and in homework? 00:58:46.320 --> 00:58:48.820 Absolutely. 00:58:48.820 --> 00:58:52.470 So the takeaway is today know how 00:58:52.470 --> 00:58:56.200 to use those response to initial condition formulas 00:58:56.200 --> 00:58:59.950 and damping and these two. 00:58:59.950 --> 00:59:04.560 So when you plot, when you plot these, 00:59:04.560 --> 00:59:07.320 you get this picture up there. 00:59:07.320 --> 00:59:09.700 And we need to talk about the properties of this. 00:59:13.380 --> 00:59:15.125 Remember, this omega over omega n 00:59:15.125 --> 00:59:16.940 is the same called the frequency ratio. 00:59:16.940 --> 00:59:19.930 It's just the ratio of the excitation frequency 00:59:19.930 --> 00:59:22.790 to the natural frequency of the system. 00:59:22.790 --> 00:59:27.090 And when they're equal, for example, this ratio is one. 00:59:27.090 --> 00:59:30.820 This whole thing in parentheses goes to zero. 00:59:30.820 --> 00:59:36.090 This expression over here goes to 2 zeta, because that's one. 00:59:36.090 --> 00:59:39.210 2 zeta squared square root is just 2 zeta. 00:59:39.210 --> 00:59:41.440 When omega equals omega n, this whole expression 00:59:41.440 --> 00:59:45.400 is F0 over k divided by 2 zeta, for example. 00:59:45.400 --> 00:59:47.570 And that's called resonance, and that's 00:59:47.570 --> 00:59:50.795 when you're right at where that peak goes to its maximum. 00:59:55.840 --> 00:59:59.290 Let's talk about this expression for a moment. 00:59:59.290 --> 01:00:02.790 If we have our cart, our mass-spring 01:00:02.790 --> 01:00:04.930 dashpot we started here. 01:00:04.930 --> 01:00:06.860 If you apply a force, a static force 01:00:06.860 --> 01:00:14.640 F0 and stretch the spring by an amount F0 over k. 01:00:14.640 --> 01:00:21.350 So x-- what we'll call x static is just F0 over k. 01:00:25.390 --> 01:00:46.710 And if I want to plot, I want to-- this has a name. 01:00:46.710 --> 01:00:49.210 This is called, this ratio here, this gives you 01:00:49.210 --> 01:00:52.230 the magnitude of the response. 01:00:52.230 --> 01:00:53.690 It goes by a variety of names. 01:00:53.690 --> 01:00:56.890 Some people call it a transfer function. 01:00:56.890 --> 01:01:00.190 Some people call it a frequency response function. 01:01:04.230 --> 01:01:05.930 I write it intentionally this way. 01:01:05.930 --> 01:01:12.310 This is I put output over input because this expression has 01:01:12.310 --> 01:01:15.170 units of output over input. 01:01:15.170 --> 01:01:18.440 So I just write it like this, remind myself 01:01:18.440 --> 01:01:20.560 what this transfer function is about. 01:01:20.560 --> 01:01:24.850 The input is force, the output is displacement. 01:01:24.850 --> 01:01:27.510 This expression has units of force, 01:01:27.510 --> 01:01:29.024 force per unit displacement. 01:01:38.710 --> 01:01:46.900 If I go to here, if I try to plot this-- let me start over. 01:01:46.900 --> 01:01:49.340 If I try to plot this, it's going 01:01:49.340 --> 01:01:52.960 to be depending on the exact value of the spring constant 01:01:52.960 --> 01:01:55.960 and the exact value of the force every time. 01:01:55.960 --> 01:01:59.840 I have to get a unique plot every time I go to do this. 01:01:59.840 --> 01:02:02.750 So textbooks and engineers, I don't want 01:02:02.750 --> 01:02:04.960 to have to remember this part. 01:02:04.960 --> 01:02:10.140 This is where all of the content is in is in this denominator, 01:02:10.140 --> 01:02:11.320 and it's dimensionless. 01:02:11.320 --> 01:02:17.000 So what I'd really like to plot is x0 over x static. 01:02:19.620 --> 01:02:27.040 And if I do that, that is x0 over the quantity F0 over k. 01:02:27.040 --> 01:02:30.090 If I just divide-- this is x static-- it would bring this 01:02:30.090 --> 01:02:35.800 to this side, then this expression, this is just 1 01:02:35.800 --> 01:02:40.320 over that denominator. 01:02:40.320 --> 01:02:43.810 And sometimes, I think in the handout by [? Row ?], 01:02:43.810 --> 01:02:46.155 they just call this h of omega. 01:02:48.700 --> 01:02:52.810 It's dimensionless, frequency over frequency, 01:02:52.810 --> 01:02:56.690 and that's actually what's plotted up there. 01:02:56.690 --> 01:03:01.950 And this is called-- has different names also. 01:03:01.950 --> 01:03:06.670 Magnification factor, dynamic amplification factor, 01:03:06.670 --> 01:03:10.360 because the ratio of x to x static 01:03:10.360 --> 01:03:20.760 if this is the dynamic effects magnify the response compared 01:03:20.760 --> 01:03:21.780 to the static response. 01:03:21.780 --> 01:03:24.370 So it might be this over this might be 10. 01:03:24.370 --> 01:03:30.020 I mean, the dynamic response is 10 times the static response. 01:03:30.020 --> 01:03:32.140 OK, how do you-- to sum this up-- 01:03:32.140 --> 01:03:37.380 and we'll be kind of getting close to the end. 01:03:37.380 --> 01:03:39.770 We want to talk just about the properties of this. 01:03:39.770 --> 01:03:40.940 How do we use this? 01:03:52.830 --> 01:03:57.730 So in practical use, you have an input specified, 01:03:57.730 --> 01:04:00.260 some force cosine omega t. 01:04:00.260 --> 01:04:02.900 You know you have a single degree of freedom oscillator 01:04:02.900 --> 01:04:09.160 that is governed by equations like that one, 01:04:09.160 --> 01:04:10.940 and you want to predict the response. 01:04:10.940 --> 01:04:18.450 Well, you say x of t is equal to the magnitude 01:04:18.450 --> 01:04:28.437 of the force times the-- and you divide that by-- we 01:04:28.437 --> 01:04:29.311 could do it this way. 01:04:34.230 --> 01:04:36.820 The magnitude of the force divided by k, 01:04:36.820 --> 01:04:38.180 which is the static response. 01:04:40.710 --> 01:04:43.720 To predict x0, we just have to predict this quantity, 01:04:43.720 --> 01:04:46.490 multiply it by F0 over k. 01:04:46.490 --> 01:04:48.180 So you know this. 01:04:48.180 --> 01:04:50.110 You better know that about your system, 01:04:50.110 --> 01:04:55.530 and you multiply it by this quantity magnitude 01:04:55.530 --> 01:04:57.260 of h of omega. 01:04:57.260 --> 01:05:05.400 And the time dependent part is times cosine omega t 01:05:05.400 --> 01:05:07.110 minus the phase angle. 01:05:07.110 --> 01:05:10.610 And the phase angle, you get either off the plot 01:05:10.610 --> 01:05:15.214 or from the-- have I written it down? 01:05:20.540 --> 01:05:22.573 I haven't written the phase angle down yet. 01:05:34.180 --> 01:05:35.770 It's kind of a messy expression too. 01:05:35.770 --> 01:05:39.200 That's why we plot it. 01:05:39.200 --> 01:05:45.330 2 zeta omega over omega n over 1 minus omega squared 01:05:45.330 --> 01:05:48.210 over omega n squared. 01:05:48.210 --> 01:05:52.150 But by knowing just this plot, what you just 01:05:52.150 --> 01:05:54.500 put in every textbook about vibration in the world, 01:05:54.500 --> 01:05:57.770 by knowing this magnification factor, 01:05:57.770 --> 01:06:01.960 calculating the static response, multiplying the two together, 01:06:01.960 --> 01:06:04.820 you have the amplitude of the response, 01:06:04.820 --> 01:06:08.490 and its time dependence is cosine omega t minus the phase 01:06:08.490 --> 01:06:09.880 angle. 01:06:09.880 --> 01:06:11.785 And I've got a little example here. 01:06:21.460 --> 01:06:23.110 Actually, rather than the example, 01:06:23.110 --> 01:06:25.550 I've gone to all the trouble of setting this up. 01:06:43.110 --> 01:06:44.460 All right. 01:06:44.460 --> 01:06:48.210 This is just a beam. 01:06:48.210 --> 01:06:49.860 Where's my other little beam? 01:06:56.910 --> 01:06:59.225 And a beam is just a spring. 01:07:02.420 --> 01:07:04.230 Put a mass on the and. 01:07:04.230 --> 01:07:07.720 This is basically a single degree of freedom system. 01:07:07.720 --> 01:07:09.500 It has a natural frequency. 01:07:09.500 --> 01:07:13.130 The beam has a certain stiffness. 01:07:13.130 --> 01:07:16.520 And now, in this case, we're interested in response 01:07:16.520 --> 01:07:18.750 to some harmonic input. 01:07:18.750 --> 01:07:23.230 So any of you know what a squiggle pen is. 01:07:23.230 --> 01:07:25.045 This is a kid's toy. 01:07:25.045 --> 01:07:26.035 AUDIENCE: Excuse me. 01:07:26.035 --> 01:07:29.005 Can you move the camera a tad to the left so the [INAUDIBLE]? 01:07:32.794 --> 01:07:33.460 PROFESSOR: Yeah. 01:07:36.450 --> 01:07:38.920 So all throughout the term, we've 01:07:38.920 --> 01:07:41.920 studied rotating masses quite a bit, right. 01:07:41.920 --> 01:07:44.730 This thing has a rotating mass that you can see in the end. 01:07:44.730 --> 01:07:46.420 I mean, when you leave it, you can come down and take a look. 01:07:46.420 --> 01:07:47.544 It has a low rotating mass. 01:07:47.544 --> 01:07:50.030 It's actually a pen, but it's a kid's toy. 01:07:50.030 --> 01:07:50.980 Shakes like crazy. 01:07:55.860 --> 01:08:00.599 And now we need the lights down. 01:08:06.830 --> 01:08:14.550 And it happens that the-- I've got a strobe light here, 01:08:14.550 --> 01:08:17.149 and I've kind of preset the frequency 01:08:17.149 --> 01:08:21.870 so it's very close to the frequency of vibration 01:08:21.870 --> 01:08:23.090 of this beam. 01:08:23.090 --> 01:08:28.939 So there's a rotating mass in this pen going round and round, 01:08:28.939 --> 01:08:30.810 and it puts a force into the system 01:08:30.810 --> 01:08:36.000 that looks like F0 cosine omega t in the vertical direction. 01:08:36.000 --> 01:08:38.000 Also does it in the horizontal, but vertical 01:08:38.000 --> 01:08:40.899 is our response direction. 01:08:40.899 --> 01:08:43.129 So it's putting in a force, and I 01:08:43.129 --> 01:08:48.220 have set the length of this beam so that the natural frequency 01:08:48.220 --> 01:08:50.180 of this beam with this mass on the end 01:08:50.180 --> 01:08:54.090 is exactly very close to being the frequency 01:08:54.090 --> 01:08:54.840 of the excitation. 01:08:57.540 --> 01:09:01.430 And the flash rate is slightly different than the vibration 01:09:01.430 --> 01:09:04.542 rate, so you see it illuminated at many positions 01:09:04.542 --> 01:09:05.750 as it goes through the cycle. 01:09:05.750 --> 01:09:09.720 So you see it going up and down. 01:09:09.720 --> 01:09:13.090 So if I mismatch it quite a bit, then you see it going. 01:09:13.090 --> 01:09:15.200 And actually, if you look at the very right end, 01:09:15.200 --> 01:09:18.446 you can see a white thing going up and down. 01:09:18.446 --> 01:09:20.029 That's the mass where you can actually 01:09:20.029 --> 01:09:24.950 see the mass in the very end of the-- right there. 01:09:24.950 --> 01:09:28.220 You can see something going around and round. 01:09:28.220 --> 01:09:31.529 There, that's the rotating mass. 01:09:31.529 --> 01:09:33.580 So the beam is going up and down, 01:09:33.580 --> 01:09:37.710 and I've got this, the vibration frequency of that mass going 01:09:37.710 --> 01:09:40.771 round and round equal to the natural frequency of the beam 01:09:40.771 --> 01:09:41.729 of the mass in the end. 01:09:47.920 --> 01:09:56.480 And it's moving quite a bit, and I'll loosen my clamp, 01:09:56.480 --> 01:09:59.300 and I'm going to change the length of the beam. 01:09:59.300 --> 01:10:00.399 I've shortened it. 01:10:08.890 --> 01:10:11.540 And now it's still moving up and down 01:10:11.540 --> 01:10:17.160 but not as much because the frequency of the rotation 01:10:17.160 --> 01:10:19.250 of the eccentric mass is no longer 01:10:19.250 --> 01:10:21.260 close to the natural frequency of the system. 01:10:24.210 --> 01:10:27.300 In fact, I've made the natural frequency 01:10:27.300 --> 01:10:34.510 of the system-- you can bring the lights back up-- 01:10:34.510 --> 01:10:37.030 I've made the natural frequency of the system. 01:10:37.030 --> 01:10:40.310 By making the beam shorter, I've made it stiffer. 01:10:40.310 --> 01:10:43.235 So the natural frequency has gone up. 01:10:43.235 --> 01:10:46.290 The frequency of the rotation of the eccentric mass 01:10:46.290 --> 01:10:47.920 has stayed about the same. 01:10:47.920 --> 01:10:49.530 So what's happened to that frequency 01:10:49.530 --> 01:10:52.970 ratio, omega over omega n? 01:10:52.970 --> 01:10:54.620 So less than one or greater than one. 01:10:57.230 --> 01:10:59.630 So the omega n has gone up. 01:10:59.630 --> 01:11:00.585 Omega stayed the same. 01:11:03.690 --> 01:11:06.740 The frequency ratio when you shorten this beam 01:11:06.740 --> 01:11:14.570 is less than one, and the properties of this transfer 01:11:14.570 --> 01:11:17.400 function, we call it-- this magnification factor 01:11:17.400 --> 01:11:19.050 looks like this. 01:11:19.050 --> 01:11:22.900 When we're exciting it right at one-- this 01:11:22.900 --> 01:11:27.580 is omega over omega n-- you write it resonance. 01:11:27.580 --> 01:11:31.800 When you excite it at a frequency ratio less than one, 01:11:31.800 --> 01:11:34.750 you start dropping off this backside, 01:11:34.750 --> 01:11:38.700 and the response goes down. 01:11:38.700 --> 01:11:43.370 And if you excite it at frequencies much greater 01:11:43.370 --> 01:11:46.420 than the natural frequency, you end up way out here. 01:11:46.420 --> 01:11:47.350 I can do that too. 01:11:59.820 --> 01:12:05.390 So how much you think it will vibrate now? 01:12:05.390 --> 01:12:07.690 A lot? 01:12:07.690 --> 01:12:08.456 A little? 01:12:16.560 --> 01:12:17.289 Hardly-- oops. 01:12:17.289 --> 01:12:19.330 Oh, I've brought it out so much you can't see it. 01:12:24.780 --> 01:12:32.450 Hardly moving at all, and that's because in terms 01:12:32.450 --> 01:12:36.640 of this terminology of magnification factors, transfer 01:12:36.640 --> 01:12:41.720 functions, this is a plot of x over x static. 01:12:41.720 --> 01:12:43.530 It goes right here. 01:12:43.530 --> 01:12:46.300 When you go to zero frequency, you are at static, 01:12:46.300 --> 01:12:51.052 so the response at very, very low frequency 01:12:51.052 --> 01:12:53.010 goes to being the same as the static frequency. 01:12:53.010 --> 01:12:55.530 So in this plot, it goes to one. 01:12:55.530 --> 01:12:59.650 At resonance, you put in one here. 01:12:59.650 --> 01:13:00.910 This goes to zero. 01:13:00.910 --> 01:13:04.060 That becomes a one 2 zeta squared. 01:13:04.060 --> 01:13:12.350 This height here is 1 over 2 zeta. 01:13:12.350 --> 01:13:14.770 So the dynamic amplification at resonance 01:13:14.770 --> 01:13:16.665 is just 1 over 2 times the damping ratio. 01:13:16.665 --> 01:13:19.280 You have 1% damping. 01:13:19.280 --> 01:13:21.322 Twice that is o2. 01:13:21.322 --> 01:13:24.720 1 over o2 is 50. 01:13:24.720 --> 01:13:29.100 So if you only had 1% damping, the dynamic amplification, 01:13:29.100 --> 01:13:31.600 the amount that this vibrates greater 01:13:31.600 --> 01:13:35.190 than its static response, is a factor of 50 greater. 01:13:35.190 --> 01:13:39.420 But then as you go higher in this omega over omega n, 01:13:39.420 --> 01:13:42.460 and you get way out here, and you get almost no vibration 01:13:42.460 --> 01:13:43.530 at all. 01:13:43.530 --> 01:13:46.296 And that's what's happened when I've lengthened this. 01:13:50.180 --> 01:13:50.680 OK. 01:13:55.930 --> 01:14:02.090 So there's your introduction to linear systems. 01:14:02.090 --> 01:14:05.090 In this case, a single degree of freedom system 01:14:05.090 --> 01:14:10.570 that vibrates an oscillator, we're 01:14:10.570 --> 01:14:15.230 talking about steady state response, not the part 01:14:15.230 --> 01:14:18.440 of the solution, the mathematical solution, 01:14:18.440 --> 01:14:20.850 to the initial conditions. 01:14:20.850 --> 01:14:24.740 So all of this has been about steady state response 01:14:24.740 --> 01:14:27.240 of our simple oscillator to what we call 01:14:27.240 --> 01:14:32.880 a harmonic input, F0 cosine omega t. 01:14:32.880 --> 01:14:34.720 So what's important that you need 01:14:34.720 --> 01:14:36.990 to remember and be able to use? 01:14:39.560 --> 01:14:43.520 This concept here, this idea of a transfer function. 01:14:43.520 --> 01:14:44.720 That's really important. 01:14:44.720 --> 01:14:46.570 You might want to remember it though 01:14:46.570 --> 01:14:51.960 as this dimensionless quantity x over x static. 01:14:51.960 --> 01:14:55.010 Just remember the shape of this transfer function. 01:14:55.010 --> 01:14:59.170 Magnitude of the amplification, 1 over 2 zeta at resonance. 01:14:59.170 --> 01:15:03.380 And it goes to one at low frequency. 01:15:03.380 --> 01:15:06.770 The high frequency, it drops away off. 01:15:06.770 --> 01:15:10.540 Next time, we're going to pick up 01:15:10.540 --> 01:15:14.120 the topic of what we call vibration isolation. 01:15:14.120 --> 01:15:16.540 The practical thing to know as engineers 01:15:16.540 --> 01:15:20.784 is when you have a significant vibration 01:15:20.784 --> 01:15:22.200 problem, like in a lab, and you're 01:15:22.200 --> 01:15:24.750 looking through your microscope, and the floor vibration is 01:15:24.750 --> 01:15:28.510 causing trouble with your microscope, 01:15:28.510 --> 01:15:32.820 and you can't move the subway, what can you 01:15:32.820 --> 01:15:35.989 do to solve that problem? 01:15:35.989 --> 01:15:37.530 Well, you might be able-- what if you 01:15:37.530 --> 01:15:40.620 put some kind of a flexible pad under the microscope? 01:15:40.620 --> 01:15:43.120 You might be able to reduce the vibration of the microscope. 01:15:45.860 --> 01:15:47.110 Things like that. 01:15:47.110 --> 01:15:49.750 That is the topic of vibration isolation, 01:15:49.750 --> 01:15:52.690 so we're going to get into that next time. 01:15:52.690 --> 01:15:54.024 Thanks.