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 --> 00:00:16.600 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:16.600 --> 00:00:17.263 at ocw.mit.edu. 00:00:21.670 --> 00:00:24.010 PROFESSOR: So we'll start quickly 00:00:24.010 --> 00:00:28.614 going over the concept questions for the homework this week. 00:00:28.614 --> 00:00:34.210 So you have a two rotor system. 00:00:37.770 --> 00:00:39.425 It says, if you give it a deflection 00:00:39.425 --> 00:00:44.450 at a exactly the mode shape of mode two, 00:00:44.450 --> 00:00:46.750 what frequency components do expect 00:00:46.750 --> 00:00:50.480 the transient response to have? 00:00:50.480 --> 00:00:54.890 So most people said only omega 2, but quite a few people 00:00:54.890 --> 00:00:56.450 said both. 00:00:56.450 --> 00:01:02.830 So this introduction to vibration 00:01:02.830 --> 00:01:04.870 that we have been doing for the last few 00:01:04.870 --> 00:01:11.700 lectures-- my goal in it is to have you folks go away 00:01:11.700 --> 00:01:14.610 with a pretty good conceptual understanding of the basics 00:01:14.610 --> 00:01:17.290 of vibration, and on the final exam, 00:01:17.290 --> 00:01:19.620 I'm not going to have you derive anything, 00:01:19.620 --> 00:01:24.410 like finding the natural frequencies of a 3 00:01:24.410 --> 00:01:29.210 by 3 degree of freedom system, solving a sixth order system 00:01:29.210 --> 00:01:30.780 in omega squared. 00:01:30.780 --> 00:01:33.980 I won't do that kind of thing to you, but questions like this 00:01:33.980 --> 00:01:36.280 are really fair game. 00:01:36.280 --> 00:01:40.070 So if we did this business bimodal analysis-- 00:01:40.070 --> 00:01:45.020 and we did an example the other day where with the system 00:01:45.020 --> 00:01:48.040 we had here, if you deflect it in exactly the shape of one 00:01:48.040 --> 00:01:50.330 mode, what kind of response do you get? 00:01:53.730 --> 00:01:55.560 Initial conditions that are shaped exactly 00:01:55.560 --> 00:01:58.190 like one the mode shapes. 00:01:58.190 --> 00:02:01.960 AUDIENCE: It has exactly that natural frequency in it. 00:02:01.960 --> 00:02:04.560 PROFESSOR: Well, it'll not only have that natural frequency 00:02:04.560 --> 00:02:07.270 in it, but if you deflect it initially in the mode 00:02:07.270 --> 00:02:10.100 shape, what will the responding motion look like? 00:02:10.100 --> 00:02:11.350 AUDIENCE: Just the mode shape. 00:02:11.350 --> 00:02:12.849 PROFESSOR: Just like the mode shape, 00:02:12.849 --> 00:02:15.100 and if the motion is in a particular mode shape, 00:02:15.100 --> 00:02:18.140 it will be at-- this is transient vibration. 00:02:18.140 --> 00:02:19.465 No external force. 00:02:19.465 --> 00:02:21.600 You deflect it and let it go. 00:02:21.600 --> 00:02:24.920 You will see only motion in that mode 00:02:24.920 --> 00:02:28.880 if you give it an initial deflection exactly in that mode 00:02:28.880 --> 00:02:30.310 shape. 00:02:30.310 --> 00:02:32.900 If you went to the other mode, and deflected it that way, 00:02:32.900 --> 00:02:34.960 and let go, it would change frequency, 00:02:34.960 --> 00:02:38.610 and it would vibrate only in that shape. 00:02:38.610 --> 00:02:40.870 And any other combination of motions 00:02:40.870 --> 00:02:45.370 is some linear sum of those two mode shapes. 00:02:45.370 --> 00:02:47.740 Any other allowable motion of the system 00:02:47.740 --> 00:02:52.060 can be made up of some weighted sum of the two mode shapes, 00:02:52.060 --> 00:02:53.340 right? 00:02:53.340 --> 00:02:55.480 And that's the amount of each mode 00:02:55.480 --> 00:02:57.570 that you get-- is the weighted sum. 00:02:57.570 --> 00:02:59.320 What are the weightings? 00:02:59.320 --> 00:03:01.350 If the weightings are 1, 0, then it's 00:03:01.350 --> 00:03:03.280 all one mode and not of another. 00:03:03.280 --> 00:03:06.430 If it takes some of each mode to give you 00:03:06.430 --> 00:03:09.110 the initial deflected shape, that's how much of each mode 00:03:09.110 --> 00:03:10.090 you get. 00:03:10.090 --> 00:03:15.780 So the answer to this question is only mode two. 00:03:15.780 --> 00:03:16.400 OK, next. 00:03:25.570 --> 00:03:28.590 I forget what you were given. 00:03:28.590 --> 00:03:29.340 Apply the concept. 00:03:29.340 --> 00:03:32.050 Which mode is likely to dominate the steady state response 00:03:32.050 --> 00:03:36.360 for the excitation of part D? 00:03:36.360 --> 00:03:39.440 So it was probably being excited harmonically 00:03:39.440 --> 00:03:41.770 at the natural frequency of mode two. 00:03:41.770 --> 00:03:43.510 So which mode? 00:03:43.510 --> 00:03:45.470 So harmonic excitation, which mode 00:03:45.470 --> 00:03:49.080 do you expect to respond the most when you excite it at one 00:03:49.080 --> 00:03:50.610 of the natural frequencies? 00:03:50.610 --> 00:03:51.860 AUDIENCE: I'd say [INAUDIBLE]. 00:03:51.860 --> 00:03:53.401 PROFESSOR: At that natural frequency. 00:03:53.401 --> 00:03:55.590 Why? 00:03:55.590 --> 00:03:56.090 Why? 00:03:59.436 --> 00:04:01.695 AUDIENCE: Because that's what the eigenvalue is. 00:04:01.695 --> 00:04:03.570 PROFESSOR: Well, there's an eigenvalue there, 00:04:03.570 --> 00:04:05.653 but now we're talking about steady state response. 00:04:05.653 --> 00:04:07.770 What's the transfer function of a single degree 00:04:07.770 --> 00:04:09.900 of freedom system look like? 00:04:09.900 --> 00:04:11.270 Just trace it in the air. 00:04:14.370 --> 00:04:19.100 This is a response per unit input, and where does it go? 00:04:19.100 --> 00:04:19.686 Way up high-- 00:04:19.686 --> 00:04:20.810 AUDIENCE: At the resonance. 00:04:20.810 --> 00:04:21.610 PROFESSOR: At the resonance. 00:04:21.610 --> 00:04:23.693 And so if you have two 2 degree of freedom system, 00:04:23.693 --> 00:04:25.676 how many resonances do you have? 00:04:25.676 --> 00:04:27.300 What do you think the transfer function 00:04:27.300 --> 00:04:29.800 is going to look like for a two degree of freedom system 00:04:29.800 --> 00:04:33.002 if you just plotted it just plot as a function of frequency? 00:04:33.002 --> 00:04:34.150 AUDIENCE: Double peak. 00:04:34.150 --> 00:04:35.650 PROFESSOR: Double peak. 00:04:35.650 --> 00:04:37.650 So the transfer function is likely to do 00:04:37.650 --> 00:04:41.455 that at one frequency and that at the next natural frequency. 00:04:41.455 --> 00:04:43.830 And that's what we're going to talk about today, in fact. 00:04:43.830 --> 00:04:46.320 OK so if you excited the second natural frequency 00:04:46.320 --> 00:04:50.870 and it's lightly damped, it's likely to be dominated 00:04:50.870 --> 00:04:53.990 by that modal response. 00:04:53.990 --> 00:04:58.720 In that question, is it guaranteed to be only that mode 00:04:58.720 --> 00:05:03.190 responding, even if you derive it at the natural frequency? 00:05:03.190 --> 00:05:04.690 Another way to saying this question, 00:05:04.690 --> 00:05:07.440 can the first mode have some response 00:05:07.440 --> 00:05:10.791 at any frequency you excite the system at? 00:05:10.791 --> 00:05:11.290 Sure. 00:05:11.290 --> 00:05:12.260 It has transfer function. 00:05:12.260 --> 00:05:13.480 It's continuous in frequency. 00:05:13.480 --> 00:05:14.840 It just won't be very big. 00:05:14.840 --> 00:05:16.060 OK, next. 00:05:22.400 --> 00:05:24.260 For what value is omega over omega 00:05:24.260 --> 00:05:27.020 n is the force transmitted to the wall going 00:05:27.020 --> 00:05:29.990 to be the greatest. 00:05:29.990 --> 00:05:33.310 So this is-- when we make this, this 00:05:33.310 --> 00:05:38.500 is now we're doing-- it's rotating around. 00:05:38.500 --> 00:05:42.420 OK, so this thing is rotating mass now 00:05:42.420 --> 00:05:44.220 at a constant frequency. 00:05:44.220 --> 00:05:47.890 So it's got a static imbalance. 00:05:47.890 --> 00:05:52.719 And at what frequency would you expect the force transmitted 00:05:52.719 --> 00:05:53.885 to the wall be the greatest? 00:05:59.090 --> 00:06:01.070 AUDIENCE: At resonance. 00:06:01.070 --> 00:06:03.994 PROFESSOR: At resonance, because that's when you get the most-- 00:06:03.994 --> 00:06:05.160 AUDIENCE: The most response. 00:06:05.160 --> 00:06:06.368 PROFESSOR: Greatest response. 00:06:06.368 --> 00:06:09.214 And the force transmitted to the wall is through the spring 00:06:09.214 --> 00:06:10.880 and through the dash pot, and the bigger 00:06:10.880 --> 00:06:13.970 the motions, the bigger those forces are going to be. 00:06:13.970 --> 00:06:17.060 OK, next. 00:06:17.060 --> 00:06:21.070 And so this is-- now, if you'll account for the spring, 00:06:21.070 --> 00:06:23.690 do you expect the counting for the mass of the spring 00:06:23.690 --> 00:06:25.950 to increase the predicted natural frequency 00:06:25.950 --> 00:06:29.525 or decrease it? 00:06:29.525 --> 00:06:30.957 AUDIENCE: Decrease. 00:06:30.957 --> 00:06:31.790 PROFESSOR: Decrease. 00:06:31.790 --> 00:06:33.820 Natural frequencies behave like square root 00:06:33.820 --> 00:06:36.090 of k/m kind of thing. 00:06:36.090 --> 00:06:38.950 And if you do anything that drives up the m, 00:06:38.950 --> 00:06:40.550 frequency is going to go down. 00:06:40.550 --> 00:06:42.190 Another way of asking the same question 00:06:42.190 --> 00:06:44.710 is, if you neglect mass when you're 00:06:44.710 --> 00:06:47.610 estimating the natural frequency of a system, 00:06:47.610 --> 00:06:49.590 which way are you likely to be in error? 00:06:52.880 --> 00:06:56.037 This system, normally we just ignore the mass of the springs, 00:06:56.037 --> 00:06:57.620 and we calculate the natural frequency 00:06:57.620 --> 00:07:00.590 as square root of k/m, and we're almost 00:07:00.590 --> 00:07:02.920 certain to be-- predict what? 00:07:02.920 --> 00:07:03.950 Too high or too low? 00:07:03.950 --> 00:07:04.870 AUDIENCE: Too high. 00:07:04.870 --> 00:07:06.620 PROFESSOR: Too high, because we've ignored 00:07:06.620 --> 00:07:07.890 some mass in the system. 00:07:07.890 --> 00:07:08.830 OK, next. 00:07:08.830 --> 00:07:10.990 I think that's it. 00:07:10.990 --> 00:07:13.320 I want to get on with today's lecture. 00:07:13.320 --> 00:07:17.882 We have done vibration from point 00:07:17.882 --> 00:07:19.340 of view of single degree of freedom 00:07:19.340 --> 00:07:24.190 systems, transient response, and steady escape response. 00:07:24.190 --> 00:07:27.170 And then we've started looking at multiple degree of freedom 00:07:27.170 --> 00:07:28.722 systems, but we actually started it 00:07:28.722 --> 00:07:30.430 from the point of view of modal analysis, 00:07:30.430 --> 00:07:32.730 and we looked at two degree of freedom systems, 00:07:32.730 --> 00:07:36.450 and found the modal contributions 00:07:36.450 --> 00:07:39.640 of each of the modes, and sort of did them one at a time. 00:07:39.640 --> 00:07:43.129 So is there another way-- one of you 00:07:43.129 --> 00:07:44.670 asked me after class last time, well, 00:07:44.670 --> 00:07:48.000 can't you just solve the differential equations directly 00:07:48.000 --> 00:07:51.195 and not bother with separating the modes out and figuring out 00:07:51.195 --> 00:07:52.195 the modal contributions. 00:07:52.195 --> 00:07:55.690 The answer is yes, and you do it via transfer functions, 00:07:55.690 --> 00:07:57.987 except now you will need more than one 00:07:57.987 --> 00:07:59.820 if you have more than one degree of freedom. 00:08:02.420 --> 00:08:05.385 So that's what we'll focus on today. 00:08:09.890 --> 00:08:17.220 And we're going to do it by thinking about the example 00:08:17.220 --> 00:08:18.920 actually from the homework. 00:08:18.920 --> 00:08:22.950 So in the homework you have this problem 00:08:22.950 --> 00:08:33.510 of this cart with this pendulum hanging off of it. 00:08:33.510 --> 00:08:36.179 And we-- in problem three, I think it is-- 00:08:36.179 --> 00:08:38.059 is say, OK, now the pendulum is going 00:08:38.059 --> 00:08:40.000 to go around a constant rate, and it's 00:08:40.000 --> 00:08:43.750 going to turn this into a rotating imbalance problem. 00:08:43.750 --> 00:08:50.810 And I shorten it up so it's just a little mass spinning around. 00:08:50.810 --> 00:09:00.350 You have k1, c1, x1 here. 00:09:00.350 --> 00:09:05.410 And I think you have draw in there as theta. 00:09:05.410 --> 00:09:07.500 This is point A about which it spins. 00:09:10.270 --> 00:09:13.320 And you know the non-linear equation 00:09:13.320 --> 00:09:17.460 to motion for treating this as a two degree of freedom system, 00:09:17.460 --> 00:09:20.040 just pendulum on a cart. 00:09:25.180 --> 00:09:31.515 And then nonlinear equation of motion. 00:09:42.610 --> 00:09:44.170 And I'm going to put subscripts here, 00:09:44.170 --> 00:09:45.890 and I'm going to call this coordinate x1, 00:09:45.890 --> 00:09:51.324 because we're going to need a second coordinate-- 00:09:51.324 --> 00:09:52.500 or actually, do we? 00:09:52.500 --> 00:09:53.580 What am I saying? 00:09:53.580 --> 00:09:56.770 We don't need to do that. 00:09:56.770 --> 00:09:57.750 Just x will do. 00:10:00.950 --> 00:10:03.560 We do need the m1 and m2, though. 00:10:03.560 --> 00:10:07.650 And so that is all-- I'm going to move the remaining 00:10:07.650 --> 00:10:09.100 stuff to the right hand side. 00:10:09.100 --> 00:10:18.390 So this is your nonlinear m2L theta 00:10:18.390 --> 00:10:26.060 double dot cos theta minus theta dot 00:10:26.060 --> 00:10:35.760 squared sine theta on the right hand side. 00:10:35.760 --> 00:10:39.290 And I think I need a minus if I do that. 00:10:39.290 --> 00:10:41.030 Now, what I do in this problem, I 00:10:41.030 --> 00:10:45.375 say let's let theta dot equals omega, and that's a constant. 00:10:49.640 --> 00:10:54.050 And so that says theta double dot is 0. 00:10:54.050 --> 00:10:57.080 When you do that, then this term goes away, 00:10:57.080 --> 00:11:02.220 and this becomes omega squared minus times a minus. 00:11:02.220 --> 00:11:14.570 This equals 1/2 m2L omega squared sine. 00:11:14.570 --> 00:11:16.685 And theta is just omega t. 00:11:20.280 --> 00:11:24.870 So by forcing what was previously an unknown variable 00:11:24.870 --> 00:11:27.720 that you had the solve in this-- you 00:11:27.720 --> 00:11:29.610 had two equations and two unknowns 00:11:29.610 --> 00:11:32.730 because you had two generalized coordinates 00:11:32.730 --> 00:11:36.970 it took to describe the motion of this thing, x and theta. 00:11:36.970 --> 00:11:39.690 Now, once you prescribe theta and it's given, 00:11:39.690 --> 00:11:41.530 it's no longer a variable. 00:11:41.530 --> 00:11:45.590 It's no longer a coordinate in your generalized-- 00:11:45.590 --> 00:11:48.260 in your equations of motion that you have to solve for, 00:11:48.260 --> 00:11:52.270 and it reduces this equation to the equation of a single degree 00:11:52.270 --> 00:11:54.110 of freedom system. 00:11:54.110 --> 00:11:56.280 On the left hand side is the response quantities 00:11:56.280 --> 00:12:01.200 just like usual-- mx double dot plus bx dot plus kx. 00:12:01.200 --> 00:12:04.370 And on the right hand side is our harmonic excitation, 00:12:04.370 --> 00:12:09.620 and that's this unbalanced mass going around and around. 00:12:09.620 --> 00:12:11.760 So this has the form here. 00:12:11.760 --> 00:12:14.650 This is some-- if you want to think of it this way, 00:12:14.650 --> 00:12:22.190 this is some F0, and it looks like sine omega t. 00:12:22.190 --> 00:12:24.610 So this is just a single degree of freedom system excited 00:12:24.610 --> 00:12:27.120 by harmonic excitation, and you know 00:12:27.120 --> 00:12:29.022 it has a transfer function, and you 00:12:29.022 --> 00:12:30.730 know that there's going to be a resonance 00:12:30.730 --> 00:12:32.930 at the natural frequency of the system. 00:12:36.980 --> 00:12:39.660 And if you don't like working with sine omega t, 00:12:39.660 --> 00:12:43.750 you could say, well, let's measure omega theta from here 00:12:43.750 --> 00:12:47.760 if you want to, and now that's cosine omega t. 00:12:47.760 --> 00:12:58.500 OK, so that's sort of the set up. 00:13:08.100 --> 00:13:11.740 Let me say one-- if I asked you to solve 00:13:11.740 --> 00:13:20.620 for the magnitude of the response x1 here, 00:13:20.620 --> 00:13:22.010 how would you do it? 00:13:26.530 --> 00:13:29.240 So this is now a single degree of freedom system excited 00:13:29.240 --> 00:13:30.990 by harmonic excitation. 00:13:30.990 --> 00:13:32.795 Steady state response-- how do you do it? 00:13:39.620 --> 00:13:42.910 This you do have to know. 00:13:42.910 --> 00:13:44.202 AUDIENCE: Transfer function. 00:13:44.202 --> 00:13:45.660 PROFESSOR: Use a transfer function. 00:13:45.660 --> 00:13:48.020 So this going to be the magnitude of the force 00:13:48.020 --> 00:13:55.392 times the magnitude of the HxF of omega transfer function, 00:13:55.392 --> 00:13:57.600 and that's how you get the magnitude of the response, 00:13:57.600 --> 00:13:59.170 and it happens to be a resonance. 00:13:59.170 --> 00:14:03.050 Then you'd be at the peak, and if you're not a resonance, 00:14:03.050 --> 00:14:05.496 wherever you happen to be in frequency 00:14:05.496 --> 00:14:07.562 is where you would evaluate this, 00:14:07.562 --> 00:14:08.645 and there's your response. 00:14:12.920 --> 00:14:22.670 Now, a few years ago-- every year-- twice a year, 00:14:22.670 --> 00:14:24.760 in fact-- and they're coming up in January, 00:14:24.760 --> 00:14:28.070 we have doctoral exams for students 00:14:28.070 --> 00:14:31.130 who want to do PhDs in mechanical engineering, 00:14:31.130 --> 00:14:34.010 and most departments at MIT have these also. 00:14:34.010 --> 00:14:38.900 And in the dynamics portion, there's an oral exam part. 00:14:38.900 --> 00:14:42.990 There's a written exam, also, but in the oral exam 00:14:42.990 --> 00:14:48.380 in this particular year, gave a single degree 00:14:48.380 --> 00:14:49.310 of freedom system. 00:14:59.980 --> 00:15:02.480 And we posed-- and we all know that, 00:15:02.480 --> 00:15:07.590 if we excite this thing with a harmonic excitation, 00:15:07.590 --> 00:15:15.830 some F1 cosine-- and I will use complex notation, 00:15:15.830 --> 00:15:19.750 because we're going to need it in a minute-- some F1 e 00:15:19.750 --> 00:15:21.550 to the i omega t. 00:15:21.550 --> 00:15:23.200 That's the excitation. 00:15:23.200 --> 00:15:26.210 We know what the steady state response of this looks like. 00:15:26.210 --> 00:15:29.070 The magnitude looks like that transfer function. 00:15:31.670 --> 00:15:35.794 But we posed-- so most would be doctoral students 00:15:35.794 --> 00:15:38.210 would know all about single degree of freedom [INAUDIBLE]. 00:15:38.210 --> 00:15:39.790 So we posed the following question. 00:15:39.790 --> 00:15:42.060 Well, we know that the response of this 00:15:42.060 --> 00:15:47.250 is going to do something like that, 00:15:47.250 --> 00:15:50.060 and you evaluate this at whatever frequency 00:15:50.060 --> 00:15:53.242 you're interested in, including right at resonance. 00:15:53.242 --> 00:15:54.700 And the question we asked is, is it 00:15:54.700 --> 00:16:06.210 possible to add a second spring and a second mass, m2 and k2, 00:16:06.210 --> 00:16:31.720 such that-- so is it possible to pick a k2 and an m2, 00:16:31.720 --> 00:16:34.660 such that, with this excitation on here, 00:16:34.660 --> 00:16:36.170 the motion of that thing will be 0? 00:16:38.940 --> 00:16:40.890 And in order to solve this problem, 00:16:40.890 --> 00:16:45.540 as soon as you put this on here and assume it's sliding along-- 00:16:45.540 --> 00:16:46.250 it can't fall. 00:16:46.250 --> 00:16:49.380 It only has horizontal possible motion, 00:16:49.380 --> 00:16:53.394 and we'd give it some coordinate describing its motion x2. 00:16:53.394 --> 00:16:55.810 So now how many degrees of freedom does this problem have? 00:16:58.520 --> 00:16:59.020 Two. 00:16:59.020 --> 00:17:02.830 How many equations of motion do you expect? 00:17:02.830 --> 00:17:04.970 How many peaks in a transfer function 00:17:04.970 --> 00:17:05.970 would you expect to see? 00:17:05.970 --> 00:17:08.040 Two resonances. 00:17:08.040 --> 00:17:12.160 So now we need to know how to find 00:17:12.160 --> 00:17:14.480 transfer functions for a multiple degree of freedom 00:17:14.480 --> 00:17:16.140 system. 00:17:16.140 --> 00:17:18.950 enough in everything I've said about two degree of freedom 00:17:18.950 --> 00:17:23.990 systems, everything is generalizable to n 00:17:23.990 --> 00:17:25.244 degrees of freedom. 00:17:25.244 --> 00:17:26.660 Though, what I'm going to show you 00:17:26.660 --> 00:17:30.205 now is how to do transfer functions for multi-degree 00:17:30.205 --> 00:17:31.580 of freedom systems, and I'm going 00:17:31.580 --> 00:17:35.000 to do it by way of example of a two a degree of freedom one, 00:17:35.000 --> 00:17:40.210 but just keep in mind that you can completely generalize this. 00:17:40.210 --> 00:17:49.090 So there is my system-- two masses, two springs-- and I 00:17:49.090 --> 00:17:53.420 could even have dash spots in here-- a c1, 00:17:53.420 --> 00:17:54.356 and I might have a c2. 00:18:01.300 --> 00:18:05.110 And I could even, in general, additionally have 00:18:05.110 --> 00:18:07.220 a force acting on this second mass, which 00:18:07.220 --> 00:18:11.580 I'll call F2e to the I omega t. 00:18:11.580 --> 00:18:14.118 So that's the completely general problem now. 00:18:18.700 --> 00:18:27.820 Now, the equations of motion for this system 00:18:27.820 --> 00:18:29.660 you could write down. 00:18:29.660 --> 00:18:32.810 We've done it many times now-- this particular system 00:18:32.810 --> 00:18:35.630 even-- but we could write them in matrix form 00:18:35.630 --> 00:18:45.370 as mx double dot plus cx dot plus k. 00:18:45.370 --> 00:18:50.170 Stiffness matrix x equals and excite 00:18:50.170 --> 00:18:53.280 the excitation in this two degree-- well, 00:18:53.280 --> 00:18:55.030 I'll just keep this is completely general. 00:18:55.030 --> 00:19:00.800 This is any n degree of freedom system, some Fe to the i omega 00:19:00.800 --> 00:19:02.800 t. 00:19:02.800 --> 00:19:04.740 So here's kind of an important point. 00:19:08.000 --> 00:19:11.920 We solve for the motion of a multiple degree of freedom 00:19:11.920 --> 00:19:14.200 system to a harmonic input. 00:19:14.200 --> 00:19:17.140 We do it one frequency at a time. 00:19:17.140 --> 00:19:18.820 So this doesn't mean that I could 00:19:18.820 --> 00:19:23.090 have force on the first mass at omega 1 00:19:23.090 --> 00:19:26.900 or at some omega, and force of the second mass 00:19:26.900 --> 00:19:28.090 at some other omega. 00:19:28.090 --> 00:19:29.720 I'd never do that at the same time. 00:19:29.720 --> 00:19:30.950 It's a linear system. 00:19:30.950 --> 00:19:32.690 Superposition holds. 00:19:32.690 --> 00:19:35.320 You do things one frequency at a time, get the answer, 00:19:35.320 --> 00:19:37.230 and if you have another frequency part, 00:19:37.230 --> 00:19:39.980 you do that separately, then add the two answers. 00:19:39.980 --> 00:19:41.320 So this is assumed. 00:19:41.320 --> 00:19:43.070 All the forces acting on the masses 00:19:43.070 --> 00:19:45.146 are assumed to occur at one frequency, 00:19:45.146 --> 00:19:46.770 but you can't have different amplitudes 00:19:46.770 --> 00:19:49.047 on the different masses. 00:19:49.047 --> 00:19:50.130 So that's what that means. 00:19:53.060 --> 00:20:03.960 So this then for the two degree of freedom system-- F, 00:20:03.960 --> 00:20:11.310 for examplet-- F of t would be some constant F1. 00:20:11.310 --> 00:20:16.110 Another constant magnitude, F2e to the i omega t. 00:20:16.110 --> 00:20:17.870 That's what we mean by that. 00:20:22.650 --> 00:20:25.540 Actually, if you do that, if these are just constants, 00:20:25.540 --> 00:20:30.450 some constant force vector applied at a single frequency, 00:20:30.450 --> 00:20:32.720 what frequency do you expect to see 00:20:32.720 --> 00:20:34.850 in the response of the system? 00:20:34.850 --> 00:20:36.540 Steady state response. 00:20:36.540 --> 00:20:38.940 AUDIENCE: Driving frequency. 00:20:38.940 --> 00:20:39.876 Driving frequency. 00:20:39.876 --> 00:20:42.000 PROFESSOR: Right, it's a linear system more or less 00:20:42.000 --> 00:20:43.870 since you really want to go away with. 00:20:43.870 --> 00:20:45.410 This is an intralinear system. 00:20:45.410 --> 00:20:48.040 Steady state response of a linear system-- 00:20:48.040 --> 00:20:50.045 the frequency in is the frequency out. 00:20:53.730 --> 00:20:55.290 Always true. 00:20:55.290 --> 00:20:57.285 So we expect to see a solution. 00:21:03.700 --> 00:21:08.360 We're going to get a solution of the form x 00:21:08.360 --> 00:21:12.650 equals some magnitude vector times an e 00:21:12.650 --> 00:21:14.670 to the i omega t also. 00:21:14.670 --> 00:21:16.790 And the magnitude might be complex, 00:21:16.790 --> 00:21:20.100 because it there be phase angles there due to damping 00:21:20.100 --> 00:21:23.000 and so forth, but nonetheless, they're constants. 00:21:23.000 --> 00:21:25.660 This part is a constant vector, and that's 00:21:25.660 --> 00:21:27.680 its frequency dependence. 00:21:27.680 --> 00:21:29.890 And if we know that is true, then we can say, 00:21:29.890 --> 00:21:32.330 well, take the time derivative of that. 00:21:32.330 --> 00:21:35.840 The only time dependent part is the e to i omega t. 00:21:35.840 --> 00:21:46.030 So x dot becomes i omega xe to the i omega t. 00:21:46.030 --> 00:21:54.500 And x double dot becomes minus omega squared xe 00:21:54.500 --> 00:21:57.660 to the i omega t. 00:21:57.660 --> 00:22:05.203 So we can substitute this, this, and this into this equation. 00:22:05.203 --> 00:22:06.786 Of course, these are matrix equations. 00:22:16.910 --> 00:22:21.030 And when you do that, minus omega 00:22:21.030 --> 00:22:26.870 squared times the mass matrix i omega times the c 00:22:26.870 --> 00:22:42.990 matrix plus k xe to the i omega t equals Fe to the i omega t. 00:22:42.990 --> 00:22:45.010 You can get rid of these, and now you 00:22:45.010 --> 00:22:51.610 have a algebraic equation that's no longer a function of time-- 00:22:51.610 --> 00:22:53.340 function of your original mass, damping, 00:22:53.340 --> 00:22:55.010 and stiffness matrices. 00:22:55.010 --> 00:22:57.050 And it's certainly a function of frequency. 00:22:57.050 --> 00:22:58.700 And if you think back, this is how 00:22:58.700 --> 00:23:01.570 we derive the transfer function for a single degree of freedom 00:23:01.570 --> 00:23:02.770 system. 00:23:02.770 --> 00:23:05.314 And this statement is true for single degree of freedom, too. 00:23:05.314 --> 00:23:07.480 With single degree of freedom, that's just the mass. 00:23:07.480 --> 00:23:10.800 That's just the damping, and that's just the stiffness. 00:23:10.800 --> 00:23:14.939 And you could solve directly for the hx/f transfer function, 00:23:14.939 --> 00:23:16.480 but with multiple degrees of freedom, 00:23:16.480 --> 00:23:20.150 we can't just quite divide this out. 00:23:20.150 --> 00:23:25.460 This piece here is known as the-- I'll write 00:23:25.460 --> 00:23:27.540 it is z omega-- z of omega. 00:23:27.540 --> 00:23:29.670 This is known as the impedance matrix. 00:23:38.980 --> 00:23:41.470 And if I want to solve, I'm looking for x. 00:23:41.470 --> 00:23:43.370 I want to know my solution x. 00:23:43.370 --> 00:23:46.496 So this is essentially of the form z. 00:23:46.496 --> 00:23:54.600 It's a matrix times x, a vector, equals F, a vector. 00:23:54.600 --> 00:23:57.670 And just using what you know about linear algebra, 00:23:57.670 --> 00:24:01.730 to solve for x, you just multiply through by z inverse. 00:24:01.730 --> 00:24:14.350 So x equals z inverse F. All right? 00:24:14.350 --> 00:24:18.860 And z is a two degree of freedom system. z is 00:24:18.860 --> 00:24:23.090 a two by two matrix with frequency and everything in it, 00:24:23.090 --> 00:24:25.220 but it's a two by two. 00:24:25.220 --> 00:24:29.010 So z inverse will be a two by two. 00:24:29.010 --> 00:24:33.170 And for two by two, we can just write down the answer, 00:24:33.170 --> 00:24:33.930 but let's see. 00:24:33.930 --> 00:24:35.870 I'll write out z of omega here. 00:24:47.180 --> 00:24:51.550 So just to be clear about what all this is, 00:24:51.550 --> 00:24:56.490 z of omega is this, and in this problem, 00:24:56.490 --> 00:25:01.330 that's minus omega squared times a mass matrix, which 00:25:01.330 --> 00:25:06.990 looks like m1, 0, 0, m2. 00:25:06.990 --> 00:25:14.830 And you add to that a damping matrix, i omega, times c. 00:25:14.830 --> 00:25:17.550 The c is from the equations of motion. 00:25:17.550 --> 00:25:28.100 c1 plus c2, minus c2, minus c2, c2, and plus our k matrix. 00:25:31.110 --> 00:25:41.150 k1 plus k2, minus k2, minus k2 k2. 00:25:41.150 --> 00:25:45.680 So that's what the z of omega actually looks like. 00:25:45.680 --> 00:25:50.310 And so you would collect-- so it's a two by two matrix, 00:25:50.310 --> 00:25:54.670 and its upper left term is minus omega squared m1 plus i 00:25:54.670 --> 00:25:58.410 omega times this plus that. 00:25:58.410 --> 00:26:01.150 Collect them all together. 00:26:01.150 --> 00:26:04.420 And this has a form-- this collects together 00:26:04.420 --> 00:26:13.010 in a form we call z11, z12, z21, z22. 00:26:16.170 --> 00:26:21.560 And very often z12 and z21 are symmetric. 00:26:21.560 --> 00:26:22.800 Not always. 00:26:22.800 --> 00:26:25.130 The kind of problems we generally do here, 00:26:25.130 --> 00:26:28.270 they will be, and it will be generally symmetric 00:26:28.270 --> 00:26:34.260 if your coordinate system is measured 00:26:34.260 --> 00:26:37.240 from a static equilibrium position. 00:26:37.240 --> 00:26:39.810 So if you've been doing 2001, there's 00:26:39.810 --> 00:26:42.460 a thing called Maxwell's reciprocal theorem, 00:26:42.460 --> 00:26:46.151 which proves this for the stiffness matrix. 00:26:46.151 --> 00:26:47.650 They essentially need to be measured 00:26:47.650 --> 00:26:50.060 from static equilibrium positions, 00:26:50.060 --> 00:26:52.860 but for today's example, this is indeed symmetric. 00:26:52.860 --> 00:26:56.850 Minus c2, minus c2, minus k2, minus k2. 00:26:56.850 --> 00:26:57.790 0, 0. 00:26:57.790 --> 00:26:59.680 Everything is symmetric. 00:26:59.680 --> 00:27:04.550 So we'll take advantage of that, and I'll write out 00:27:04.550 --> 00:27:06.800 one of the components here-- z11, 00:27:06.800 --> 00:27:13.970 for example, when you collect the terms together, 00:27:13.970 --> 00:27:26.260 is in general, it would be minus omega squared m11 plus i 00:27:26.260 --> 00:27:36.430 omega c11 plus k11, the corner elements of these three pieces. 00:27:36.430 --> 00:27:52.150 And in this problem, that's minus omega squared m1 plus i 00:27:52.150 --> 00:28:03.230 omega c1 plus c2 plus k1 plus k2. 00:28:03.230 --> 00:28:09.070 You just substitute in this, this, and this. 00:28:09.070 --> 00:28:11.850 So that's z11, for example, and the other ones 00:28:11.850 --> 00:28:15.010 you can figure out what each of the other terms would be. 00:28:28.990 --> 00:28:34.130 All right, so we've said that we want to know what the x's are. 00:28:34.130 --> 00:28:41.140 And we know that we can get that by doing z inverse times F. 00:28:41.140 --> 00:28:44.030 And this gives us our definition. 00:28:44.030 --> 00:28:52.000 This z inverse is our H, our transfer function matrix, 00:28:52.000 --> 00:28:57.060 times F. So just by getting z inverse, 00:28:57.060 --> 00:29:00.240 we get this little set of four transfer functions 00:29:00.240 --> 00:29:03.580 that we're interested in. 00:29:03.580 --> 00:29:14.710 So H is an N by N matrix of transfer functions. 00:29:24.960 --> 00:29:30.910 So in a two by two-- or for the two by two case, 00:29:30.910 --> 00:29:34.170 our N equals 2. 00:29:34.170 --> 00:29:39.690 We know we can write out directly what z inverse is. 00:29:39.690 --> 00:29:56.530 And z inverse-- z22, minus z12, minus z12 when it's symmetric, 00:29:56.530 --> 00:30:04.880 z11, all over-- how should I write it? 00:30:04.880 --> 00:30:08.882 All over the determinant of z. 00:30:08.882 --> 00:30:10.115 So this is the determinant. 00:30:16.120 --> 00:30:17.640 So the determinant of z [INAUDIBLE] 00:30:17.640 --> 00:30:18.760 do these double bars. 00:30:18.760 --> 00:30:23.580 So this over the determinant is the inverse of that matrix, 00:30:23.580 --> 00:30:27.940 and this gives you a new result, which 00:30:27.940 --> 00:30:52.760 is H11, H12, H12, H22, where Hij-- this is the response at i 00:30:52.760 --> 00:31:00.762 per unit force at j. 00:31:00.762 --> 00:31:04.480 And I've made this-- I took advantage of the symmetry here, 00:31:04.480 --> 00:31:06.990 but this one would normally be called 21, 00:31:06.990 --> 00:31:10.500 so this is response at 1 caused by the force at 1. 00:31:10.500 --> 00:31:14.230 So we have forces in our problem. 00:31:14.230 --> 00:31:16.610 One is the force on the main mass. 00:31:16.610 --> 00:31:20.660 So how much response do you get at the main first mass 00:31:20.660 --> 00:31:22.545 due to the force on the first mass? 00:31:22.545 --> 00:31:23.920 This is how much response you get 00:31:23.920 --> 00:31:27.880 on the first mass due to the excitation in the second mass. 00:31:27.880 --> 00:31:31.400 Response on the second mass due to the excitation on the first. 00:31:31.400 --> 00:31:34.240 Response in the second mass due to the excitation 00:31:34.240 --> 00:31:35.240 on the second mass. 00:31:35.240 --> 00:31:38.300 So you get four possible contributions here. 00:31:42.790 --> 00:31:50.830 And the determinant of z for a two by two 00:31:50.830 --> 00:31:53.100 is also very straightforward. 00:31:53.100 --> 00:31:59.520 z11, z22, minus z12 squared. 00:32:03.160 --> 00:32:09.020 So we can and will work out exactly what the algebra tells 00:32:09.020 --> 00:32:11.040 us for this two by two case. 00:32:13.600 --> 00:32:17.076 So we've already been discussing it, but what do you expect. 00:32:17.076 --> 00:32:18.450 Let's say let's look at this one. 00:32:18.450 --> 00:32:21.660 What do you suppose the response at 1 00:32:21.660 --> 00:32:29.470 due to a harmonic excitation at 1-- 00:32:29.470 --> 00:32:36.690 what do you think the sketch of the magnitude of H11 of omega 00:32:36.690 --> 00:32:39.050 would look like as a function a frequency? 00:32:41.900 --> 00:32:45.230 AUDIENCE: Will C12 and C21 always be symmetric? 00:32:45.230 --> 00:32:49.510 PROFESSOR: No, but for the kind of problems we do, probably. 00:32:49.510 --> 00:32:54.060 And if it's simple beams, or masses connected by springs, 00:32:54.060 --> 00:32:58.700 if you choose your generalized coordinates in a way that 00:32:58.700 --> 00:33:03.950 is, for example, measuring the displacement of each mass 00:33:03.950 --> 00:33:08.280 from an inertial static equilibrium starting point, 00:33:08.280 --> 00:33:10.930 it'll be symmetric. 00:33:10.930 --> 00:33:13.630 But if you measure the-- even in this two degree of freedom, 00:33:13.630 --> 00:33:18.370 I can make it non-symmetric just by choosing the coordinate 00:33:18.370 --> 00:33:22.030 for the second mass as to be relative to the first mass. 00:33:22.030 --> 00:33:24.110 Soon as you do that, it makes it non-symmetric. 00:33:24.110 --> 00:33:26.590 You can still solve for transfer function and everything, 00:33:26.590 --> 00:33:29.070 but it gets messier. 00:33:29.070 --> 00:33:32.840 Though, notice I picked coordinates-- x1 relative 00:33:32.840 --> 00:33:35.970 to the inertial frame, x2 relative to the same inertial 00:33:35.970 --> 00:33:38.710 frame-- and they're both measuring displacement 00:33:38.710 --> 00:33:41.630 from a static equilibrium position. 00:33:41.630 --> 00:33:45.126 Then they are symmetric. 00:33:45.126 --> 00:33:46.625 And there's the determinant we need, 00:33:46.625 --> 00:33:50.630 and this determinant-- see, this is divided into all four 00:33:50.630 --> 00:33:52.560 of these terms. 00:33:52.560 --> 00:33:56.845 So these four transfer functions all share the same denominator. 00:34:00.430 --> 00:34:02.700 They all have the same denominator. 00:34:02.700 --> 00:34:04.884 So I just want you to use your intuition. 00:34:04.884 --> 00:34:06.550 Tell me what this is going to look like. 00:34:10.870 --> 00:34:15.050 What's it look like for a single degree of freedom system? 00:34:15.050 --> 00:34:17.679 It's HxF for a single degree of freedom-- 00:34:17.679 --> 00:34:20.679 it's H11 for a single degree of freedom system 00:34:20.679 --> 00:34:22.800 as a response of those systems to a harmonic 00:34:22.800 --> 00:34:27.810 force on that single mass, and that looks like what? 00:34:27.810 --> 00:34:28.971 Peak, right? 00:34:28.971 --> 00:34:31.179 This one-- what do you think it's going to look like? 00:34:35.780 --> 00:34:37.820 Show me again. 00:34:37.820 --> 00:34:39.380 OK, but I see one peak. 00:34:39.380 --> 00:34:40.219 How many peaks? 00:34:40.219 --> 00:34:40.909 AUDIENCE: Two. 00:34:40.909 --> 00:34:41.893 PROFESSOR: Why? 00:34:41.893 --> 00:34:42.877 AUDIENCE: Because it's two degrees of freedom. 00:34:42.877 --> 00:34:44.820 PROFESSOR: Two different natural frequencies. 00:34:44.820 --> 00:34:50.400 This thing is going to look something like this, 00:34:50.400 --> 00:34:53.195 and I don't know quite how it behaves in here, 00:34:53.195 --> 00:34:55.500 but it's going to do that for sure. 00:34:55.500 --> 00:34:57.970 It's going to have two resonances-- this one at omega 00:34:57.970 --> 00:35:00.180 1, this one at omega 2. 00:35:00.180 --> 00:35:02.800 We already know that because we did it by modal analysis. 00:35:02.800 --> 00:35:05.210 We know each mode is going to have a resonance in it. 00:35:05.210 --> 00:35:10.400 OK, so let's find out exactly what it looks like. 00:35:19.140 --> 00:35:22.150 So for this two degree of freedom system, 00:35:22.150 --> 00:35:26.275 there's an F1 and an F2, and we're looking for x1. 00:35:28.880 --> 00:35:48.490 So x1 x2 here equals H11, H12, H12, H22 times F1 F2. 00:35:48.490 --> 00:36:00.727 So if I want to solve for x1, H11 F1 plus H12 F2. 00:36:00.727 --> 00:36:02.310 And for the problem I'm going to solve 00:36:02.310 --> 00:36:10.030 today, I'm going to let F2 be 0, just to keep it simple, 00:36:10.030 --> 00:36:12.310 but also to address the original question 00:36:12.310 --> 00:36:14.390 we ask the doctoral students. 00:36:14.390 --> 00:36:16.900 You have a force on the first mass. 00:36:16.900 --> 00:36:20.540 Can I add a mass and spring to it in such a way 00:36:20.540 --> 00:36:23.860 that I can make the response of the first mass go to 0. 00:36:23.860 --> 00:36:26.710 So there is no second force. 00:36:26.710 --> 00:36:31.030 There's only a first force, and so this term goes away. 00:36:31.030 --> 00:36:37.580 And so this problem, x1, is H11 F1. 00:36:37.580 --> 00:36:44.360 Therefore, we need to know what H11 looks like. 00:36:48.940 --> 00:36:58.290 So H11 is going to be z22 divided by the determinant. 00:36:58.290 --> 00:37:15.530 So x1 is going to be z22 over Z11, Z22 minus Z12 squared 00:37:15.530 --> 00:37:16.620 times F1. 00:37:30.760 --> 00:37:38.230 And actually I can do this so-- and I'm 00:37:38.230 --> 00:37:43.780 going to let, for now, damping be 0, 00:37:43.780 --> 00:37:46.620 because it also simplifies it for the purposes 00:37:46.620 --> 00:37:47.790 of illustration today. 00:37:47.790 --> 00:37:50.480 And that's how I was posed to the doctoral student. 00:37:50.480 --> 00:37:51.730 Didn't even show any damping. 00:37:51.730 --> 00:37:56.110 It was just a simple system, a string and a mass. 00:37:56.110 --> 00:37:56.990 How can you do this? 00:37:56.990 --> 00:37:59.300 So we'll let the damping be 0 and that considerably 00:37:59.300 --> 00:38:00.600 simplifies things then. 00:38:19.260 --> 00:38:32.710 So Z11 is minus omega squared m1 plus k1 plus k2. 00:38:36.540 --> 00:38:45.430 z22 minus omega squared m2 plus k2. 00:38:49.350 --> 00:39:02.330 And z12 just minus k2. 00:39:02.330 --> 00:39:05.340 So now these are pretty simple expressions, 00:39:05.340 --> 00:39:07.630 and all I have to do is plug them in here. 00:39:07.630 --> 00:39:12.070 And notice what's going to happen. z11 times z22-- that's 00:39:12.070 --> 00:39:14.710 something involving omega squared 00:39:14.710 --> 00:39:18.170 times another term involving omega squared. 00:39:18.170 --> 00:39:20.990 It's going to give you a polynomial 00:39:20.990 --> 00:39:22.990 and omega to the fourth. 00:39:22.990 --> 00:39:25.190 This is the polynomial that has two roots, 00:39:25.190 --> 00:39:26.195 and the two routes are? 00:39:29.824 --> 00:39:31.270 AUDIENCE: [INAUDIBLE] frequencies. 00:39:31.270 --> 00:39:33.280 PROFESSOR: Yeah, natural frequencies. 00:39:33.280 --> 00:39:38.010 The determinant of the z matrix is the same problem 00:39:38.010 --> 00:39:41.510 you solve when you've found the roots 00:39:41.510 --> 00:39:44.450 of the characteristic equation. 00:39:44.450 --> 00:39:46.400 It is the characteristic equation 00:39:46.400 --> 00:39:49.250 for the system, the denominator, the determinant 00:39:49.250 --> 00:39:50.670 of the z matrix. 00:39:50.670 --> 00:39:53.220 Therefore it's going to be a fourth order equation in omega. 00:39:53.220 --> 00:39:55.870 It'll give you two routes for omega squared. 00:39:55.870 --> 00:39:58.690 They're your two undamped natural frequencies 00:39:58.690 --> 00:40:01.390 when we leave damping out of this expression, 00:40:01.390 --> 00:40:04.050 and they'd be damped-- the damped expressions-- 00:40:04.050 --> 00:40:07.500 they'd have complex stuff if you leave in the damping. 00:40:07.500 --> 00:40:11.530 But they give us our two roots, and they go in the denominator. 00:40:11.530 --> 00:40:12.501 Yeah? 00:40:12.501 --> 00:40:13.914 AUDIENCE: If we set the denominator or the determinant 00:40:13.914 --> 00:40:15.562 of the matrix equal to 0 and solve, 00:40:15.562 --> 00:40:16.980 you get the modal frequencies? 00:40:16.980 --> 00:40:19.450 PROFESSOR: Yeah. 00:40:19.450 --> 00:40:21.920 That denominator is the characteristic equation 00:40:21.920 --> 00:40:26.260 that we did when we set it up to do the natural frequencies. 00:40:26.260 --> 00:40:28.230 Same equation. 00:40:28.230 --> 00:40:32.136 So we know that this thing has two roots, one at each 00:40:32.136 --> 00:40:33.260 of the natural frequencies. 00:40:33.260 --> 00:40:37.440 And when omega is at a natural frequency, what 00:40:37.440 --> 00:40:39.870 is the value of this characteristic equation? 00:40:43.140 --> 00:40:43.924 Well, no. 00:40:43.924 --> 00:40:45.090 The characteristic equation. 00:40:45.090 --> 00:40:47.740 Think of the denominator. 00:40:47.740 --> 00:40:50.350 Add a root, that expression goes to? 00:40:50.350 --> 00:40:50.850 0. 00:40:50.850 --> 00:40:52.120 So you're dividing by 0. 00:40:52.120 --> 00:40:54.146 That's what causes the peaks. 00:40:54.146 --> 00:40:55.520 That's where the peaks come from. 00:40:55.520 --> 00:40:59.180 That's where the denominator goes to 0, 00:40:59.180 --> 00:41:01.530 and that's at the natural frequencies. 00:41:01.530 --> 00:41:14.760 OK, so x1 for this problem equals H11F1, 00:41:14.760 --> 00:41:17.450 and now we can just write it out. 00:41:17.450 --> 00:41:30.460 H11 is minus omega squared m2 plus k2 times F1. 00:41:30.460 --> 00:41:33.960 And the denominator, the characteristic equation, 00:41:33.960 --> 00:41:41.570 is minus omega squared m1 plus k1 00:41:41.570 --> 00:41:47.610 plus k2 times minus omega squared 00:41:47.610 --> 00:41:57.160 m2 plus k2 minus k2 squared. 00:41:57.160 --> 00:41:59.240 So this whole thing down here-- there's 00:41:59.240 --> 00:42:00.500 your characteristic equation. 00:42:00.500 --> 00:42:01.333 You multiply it out. 00:42:01.333 --> 00:42:03.620 You get your fourth order polynomial. 00:42:03.620 --> 00:42:04.180 Solve it. 00:42:04.180 --> 00:42:07.300 You get two roots, but this is now 00:42:07.300 --> 00:42:09.990 a total expression for the response we were looking for. 00:42:15.900 --> 00:42:19.080 So right away this gives me the answer 00:42:19.080 --> 00:42:23.320 to the question that was posed to the doctoral students. 00:42:23.320 --> 00:42:29.230 So is there a value-- can you set k2 and m2 such 00:42:29.230 --> 00:42:33.480 that you can make the response of the system 0 00:42:33.480 --> 00:42:34.510 at a particular omega? 00:42:38.880 --> 00:42:43.370 All you have to do is make the numerator go to 0, right? 00:42:43.370 --> 00:42:48.030 So if you make this go to 0, x goes to 0. 00:42:48.030 --> 00:43:01.750 So x1 equals 0 when minus omega squared m2 plus k2 equals zero, 00:43:01.750 --> 00:43:09.200 and that happens when k2 over m2 equals omega squared. 00:43:16.870 --> 00:43:23.420 So how many-- once you set k2 and m2, how many frequencies 00:43:23.420 --> 00:43:25.342 does this happen at? 00:43:25.342 --> 00:43:27.620 At how many different operating points 00:43:27.620 --> 00:43:31.430 can you make the response of that main mass go to 0? 00:43:36.370 --> 00:43:38.040 Well, just one. 00:43:38.040 --> 00:43:42.550 Once you choose k1 and m2, whatever you've chosen to be, 00:43:42.550 --> 00:43:45.260 they give you some value of omega. 00:43:45.260 --> 00:43:49.500 And when I said I was kind of vague about what happens here, 00:43:49.500 --> 00:43:50.040 this is H11. 00:43:55.030 --> 00:43:56.030 Looks like that. 00:43:56.030 --> 00:44:03.615 Right here is when omega squared equals k2/m2. 00:44:12.086 --> 00:44:15.746 AUDIENCE: So when it's at the frequency that 00:44:15.746 --> 00:44:20.870 makes the numerator 0, wouldn't it also-- I don't remember. 00:44:20.870 --> 00:44:22.830 Would the denominator be 0 also? 00:44:22.830 --> 00:44:23.950 PROFESSOR: No. 00:44:23.950 --> 00:44:24.480 No. 00:44:24.480 --> 00:44:27.269 Because, look at the picture here. 00:44:27.269 --> 00:44:29.477 Where are the two natural frequencies of this system? 00:44:33.166 --> 00:44:34.790 There's one here, and there's one here. 00:44:34.790 --> 00:44:37.520 And that's when the denominator goes to 0. 00:44:37.520 --> 00:44:39.360 When the denominator goes to 0, the response 00:44:39.360 --> 00:44:41.830 goes to infinity with no damping. 00:44:41.830 --> 00:44:44.840 So the system has two natural frequencies-- one 00:44:44.840 --> 00:44:46.180 to either side of this point. 00:44:49.440 --> 00:44:57.890 So I've picked kind of a particular example 00:44:57.890 --> 00:45:01.305 to illustrate the use of a transfer function, 00:45:01.305 --> 00:45:02.930 and the fact that you can have transfer 00:45:02.930 --> 00:45:05.600 functions for multiple degree of freedom systems, 00:45:05.600 --> 00:45:08.940 and they essentially become transfer function matrices. 00:45:08.940 --> 00:45:12.850 And you use what you need, and for this problem, 00:45:12.850 --> 00:45:15.762 we needed H11. 00:45:15.762 --> 00:45:17.350 We worked it out. 00:45:17.350 --> 00:45:18.810 There it is. 00:45:18.810 --> 00:45:22.060 This is a complete equation that describes, for me, 00:45:22.060 --> 00:45:26.410 the behavior of mass 1 per unit input force 00:45:26.410 --> 00:45:28.310 as a function of frequency. 00:45:28.310 --> 00:45:32.226 And it will always have a point right there that it goes to 0. 00:45:35.130 --> 00:45:38.582 And we don't have any damping in here, so intuitively, 00:45:38.582 --> 00:45:40.540 what do you think damping will do to this plot? 00:45:43.850 --> 00:45:45.720 What's the first thing it does to the peaks? 00:45:45.720 --> 00:45:48.470 What? 00:45:48.470 --> 00:45:49.334 They become? 00:45:49.334 --> 00:45:50.250 AUDIENCE: [INAUDIBLE]. 00:45:50.250 --> 00:45:51.270 PROFESSOR: Finite. 00:45:51.270 --> 00:45:55.480 And so the small damping, you're going to have very high tops. 00:45:55.480 --> 00:45:57.830 More damping, they're going to be lower. 00:45:57.830 --> 00:46:00.580 So damping especially affects the peaks. 00:46:00.580 --> 00:46:04.020 Damping will also pull this off the bottom a little bit. 00:46:04.020 --> 00:46:05.510 It won't perfectly go to 0. 00:46:11.400 --> 00:46:14.980 So I want to do a very particular case. 00:46:14.980 --> 00:46:19.240 So an obvious one is your original system. 00:46:19.240 --> 00:46:24.110 It might be just a mass spring with a rotor in it 00:46:24.110 --> 00:46:27.350 that's unbalanced, and it's running near resonance, 00:46:27.350 --> 00:46:28.860 and it's vibrating like crazy. 00:46:28.860 --> 00:46:32.070 A single degree of freedom system vibrating like crazy. 00:46:32.070 --> 00:46:38.290 Can I put on a second mass spring and stop the motion? 00:46:38.290 --> 00:46:41.140 Well, we can theoretically. 00:46:41.140 --> 00:46:44.520 And we'll just say let's let the problem 00:46:44.520 --> 00:46:46.520 we're trying to solve-- that first system had 00:46:46.520 --> 00:46:47.380 a natural frequency. 00:46:47.380 --> 00:46:55.170 I'll call it cap omega n that was the square root k1 over m1, 00:46:55.170 --> 00:47:00.360 and I'm going to let that be equal to square root of k2 00:47:00.360 --> 00:47:02.240 over m2. 00:47:02.240 --> 00:47:05.260 So I've now chosen k2 and m2 such 00:47:05.260 --> 00:47:08.350 they're exactly at the natural frequency 00:47:08.350 --> 00:47:11.820 of the original single degree of freedom system. 00:47:11.820 --> 00:47:16.450 So this is the original system, k1 m1, 00:47:16.450 --> 00:47:19.130 and it has that natural frequency. 00:47:19.130 --> 00:47:25.890 And now I'm going to stick on a second mass, k2 m2, 00:47:25.890 --> 00:47:29.870 but the value k2/m2 of the second little system-- 00:47:29.870 --> 00:47:33.281 by itself, it has the same natural frequency as the first 00:47:33.281 --> 00:47:34.780 if you want to think of it that way. 00:47:34.780 --> 00:47:37.610 If I make this fixed, what's the natural frequency 00:47:37.610 --> 00:47:40.350 of that mass and spring? 00:47:40.350 --> 00:47:41.730 Same natural frequency. 00:47:41.730 --> 00:47:43.290 I'm just making it that way. 00:47:43.290 --> 00:47:46.690 So this is now my system. 00:47:46.690 --> 00:47:53.640 That's the parameters, and now we can-- 00:47:53.640 --> 00:47:58.300 I'm going to give you a plot of the response 00:47:58.300 --> 00:47:59.640 of that particular system. 00:48:22.000 --> 00:48:23.850 You know how we non-dimensionalize 00:48:23.850 --> 00:48:26.300 the single degree of freedom transfer function? 00:48:26.300 --> 00:48:29.415 We set x divided by the static motion. 00:48:29.415 --> 00:48:31.490 I want to do that again here. 00:48:31.490 --> 00:48:34.651 So let's look at this system. 00:48:34.651 --> 00:48:38.600 It's a two degree of freedom system. 00:48:38.600 --> 00:48:44.010 It has a force F1 on it, and as the frequency goes to 0 00:48:44.010 --> 00:48:52.910 and has displacement x1 and x2-- so at omega equals 0, 00:48:52.910 --> 00:48:56.120 x1 equals x1s. 00:48:56.120 --> 00:48:57.580 I'll call it x1 static. 00:48:57.580 --> 00:48:59.940 And how much is it? 00:48:59.940 --> 00:49:01.650 So what's the static displacement 00:49:01.650 --> 00:49:04.433 at 0 frequency of this system to that force? 00:49:12.381 --> 00:49:12.880 Come on. 00:49:12.880 --> 00:49:16.180 It's just a spring and a force. 00:49:16.180 --> 00:49:18.850 How much will that spring stretch 00:49:18.850 --> 00:49:21.204 when you put a force F1 on it? 00:49:21.204 --> 00:49:22.120 AUDIENCE: [INAUDIBLE]. 00:49:22.120 --> 00:49:23.400 PROFESSOR: F1/k. 00:49:23.400 --> 00:49:27.223 So x1 static equals F1/k1. 00:49:31.530 --> 00:49:33.850 And F2 is 0. 00:49:33.850 --> 00:49:34.860 There's no F2 force. 00:49:34.860 --> 00:49:38.480 So how much is x2 static? 00:49:38.480 --> 00:49:41.100 There's no forces on the second mass. 00:49:41.100 --> 00:49:44.676 How much will it move? 00:49:44.676 --> 00:49:45.638 AUDIENCE: [INAUDIBLE]. 00:49:45.638 --> 00:49:48.740 PROFESSOR: It just moves the same amount as a first mass. 00:49:48.740 --> 00:49:54.810 So this is x2 static, because the forces on the system 00:49:54.810 --> 00:49:56.170 are F1 and 0. 00:49:56.170 --> 00:49:57.250 There's no force in this. 00:49:57.250 --> 00:49:59.310 It's not going to do anything to that spring. 00:49:59.310 --> 00:50:01.430 It'll just move with the whole system. 00:50:12.420 --> 00:50:15.925 So I want to plot x1 over x1 static. 00:50:19.150 --> 00:50:25.040 In general, x1 is a function of frequency-- is F1 times-- 00:50:25.040 --> 00:50:27.250 and I'm going to do the magnitudes here-- 00:50:27.250 --> 00:50:29.595 times the magnitude of H11. 00:50:32.350 --> 00:50:39.680 And I'm going to divide that by F1 over k1. 00:50:39.680 --> 00:50:42.180 So the F1's go away. 00:50:42.180 --> 00:50:44.420 k1 comes in the numerator. 00:50:44.420 --> 00:50:46.500 This looks like k1 h11. 00:50:50.200 --> 00:51:00.340 So a plot of k1 magnitude h11. 00:51:00.340 --> 00:51:04.760 We know roughly what it's going to look like. 00:51:04.760 --> 00:51:07.441 It has a peak. 00:51:07.441 --> 00:51:09.570 It has a 0. 00:51:09.570 --> 00:51:10.890 It has another peak. 00:51:13.740 --> 00:51:16.960 Over here, this plot goes to 1. 00:51:16.960 --> 00:51:22.540 There's x1 over-- this is also x1 over x1 static. 00:51:22.540 --> 00:51:23.310 Same thing. 00:51:23.310 --> 00:51:24.370 It goes to 1. 00:51:27.100 --> 00:51:29.510 Here it goes to 0. 00:51:29.510 --> 00:51:33.030 Here is omega 1. 00:51:33.030 --> 00:51:35.400 Here is omega 2. 00:51:35.400 --> 00:51:40.219 And right here you're at the original cap omega n, 00:51:40.219 --> 00:51:42.260 because that's the way we've designed this thing, 00:51:42.260 --> 00:51:44.820 and we know this is the way H11 behaves. 00:51:44.820 --> 00:51:47.850 It goes to 0 at the value we've chosen 00:51:47.850 --> 00:51:51.290 for square root of k2/m2. 00:51:51.290 --> 00:51:54.270 So we started out with an original system that 00:51:54.270 --> 00:51:56.470 had a natural frequency here. 00:51:56.470 --> 00:51:58.760 k1/m1 was right here. 00:51:58.760 --> 00:52:01.930 We stuck on this second little mass, 00:52:01.930 --> 00:52:04.870 and it made into a two degree of freedom system. 00:52:04.870 --> 00:52:07.680 It no longer has a natural frequency there. 00:52:07.680 --> 00:52:09.340 It has a natural frequency below it, 00:52:09.340 --> 00:52:11.270 and a natural frequency above it, 00:52:11.270 --> 00:52:14.110 and a 0 right where the original natural frequency was. 00:52:29.127 --> 00:52:30.710 Imagine what's going on in the system. 00:52:30.710 --> 00:52:33.060 You've got this force being applied to the first mass, 00:52:33.060 --> 00:52:35.192 and the first mass isn't moving. 00:52:35.192 --> 00:52:37.392 Do you think the second mass is moving? 00:52:41.820 --> 00:52:44.000 Think about the free body diagram of the first mass. 00:52:44.000 --> 00:52:46.200 F equals ma. 00:52:46.200 --> 00:52:49.980 There is a force on that first mass-- F1 cosine omega t-- 00:52:49.980 --> 00:52:52.170 and it's not moving. 00:52:52.170 --> 00:52:55.880 Mx1 double dot is 0. 00:52:55.880 --> 00:52:58.580 So in order for the thing not to move, there's a force on it. 00:52:58.580 --> 00:53:03.190 There must be some other force exactly canceling it. 00:53:03.190 --> 00:53:04.366 Where does it come from? 00:53:04.366 --> 00:53:05.340 AUDIENCE: Mass two. 00:53:05.340 --> 00:53:09.770 PROFESSOR: Mass two through the spring. 00:53:09.770 --> 00:53:15.560 OK, so I'd also like to know, what is x2? 00:53:18.300 --> 00:53:23.670 Well, x2 from our transfer function matrix 00:53:23.670 --> 00:53:30.930 is the response of 2 due to a force at 1 times F1, 00:53:30.930 --> 00:53:34.680 plus a response at 2 due to a force at 2 times F2, 00:53:34.680 --> 00:53:39.330 but that second term is 0 because F2 is 0. 00:53:39.330 --> 00:53:46.250 That's a one term expression, and H21 00:53:46.250 --> 00:53:52.690 is z11 over the denominator. 00:53:52.690 --> 00:54:08.670 And Z11-- so x2 is going to look like that. 00:54:27.560 --> 00:54:30.640 I wrote that wrong. 00:54:30.640 --> 00:54:33.590 This is not z11. 00:54:33.590 --> 00:54:39.160 It is z minus z12 like that. 00:54:44.790 --> 00:54:51.640 And if you work that out, it's F1 times k2 00:54:51.640 --> 00:55:00.670 over the same denominator, but let's evaluate-- here's our-- 00:55:00.670 --> 00:55:03.320 this is the frequency we've been interested in. 00:55:03.320 --> 00:55:09.090 Let's evaluate the response of x2 right at this operating 00:55:09.090 --> 00:55:11.500 point. 00:55:11.500 --> 00:55:18.110 So let's evaluate this at omega equals cap omega n. 00:55:18.110 --> 00:55:25.870 So that's k2 down here minus omega squared 00:55:25.870 --> 00:55:30.640 m1 plus k1 plus k2. 00:55:30.640 --> 00:55:32.325 It's the same denominator as always. 00:55:35.090 --> 00:55:43.760 k2-- I'm not going to write it out. 00:55:43.760 --> 00:55:46.260 It's exactly the same thing as before, 00:55:46.260 --> 00:55:49.460 but I want to plug in to this denominator the operating 00:55:49.460 --> 00:55:51.880 frequency. 00:55:51.880 --> 00:55:54.940 We're going to operate at cap omega n, which 00:55:54.940 --> 00:55:58.790 is k1/m1 squared-- square root of k1/m1, 00:55:58.790 --> 00:56:00.190 or square root of k2/me. 00:56:00.190 --> 00:56:03.810 I'm going to plug in in this, and let's see what we get. 00:56:56.310 --> 00:57:08.400 And I'm going to let omega equal k1/m1 or k2/m2. 00:57:08.400 --> 00:57:11.580 It's the same thing, and I'm plug them in whatever 00:57:11.580 --> 00:57:12.570 is convenient. 00:57:12.570 --> 00:57:16.314 So here I'm going to put k1/m1, N1 and the m1's are 00:57:16.314 --> 00:57:16.980 going to cancel. 00:57:16.980 --> 00:57:18.438 I'll be left an k1. 00:57:22.330 --> 00:57:27.790 F1 over k1 and down in the denominator, this term 00:57:27.790 --> 00:57:36.380 turns into k1/m1, so that's minus k1 plus k1 plus k1. 00:57:36.380 --> 00:57:37.600 Those two cancel. 00:57:37.600 --> 00:57:39.440 I just get a k2. 00:57:39.440 --> 00:57:48.980 Over here, I'll use minus k2/m2 times m2 plus k2, 00:57:48.980 --> 00:57:52.030 and that gives me a minus k2 plus k2. 00:57:52.030 --> 00:57:59.540 This whole thing goes to 0, and I'm left with minus k2 squared 00:57:59.540 --> 00:58:02.510 and in the denominator a k2. 00:58:02.510 --> 00:58:04.980 This whole thing turns into F1 over k2. 00:58:12.690 --> 00:58:18.460 So the total response of that second mass at this operating 00:58:18.460 --> 00:58:21.430 frequency is just F1 over k2. 00:58:21.430 --> 00:58:37.330 And I would like to plot it as what does x2 over x1 static-- 00:58:37.330 --> 00:58:41.290 and x1 static and x2 static are the same. 00:58:45.270 --> 00:58:57.850 Well, that is F1 over k2 over F1 over k1. 00:58:57.850 --> 00:59:03.200 The F1's go away, and I get k1/k2. 00:59:03.200 --> 00:59:07.300 That also happens to be m1/m2, and it's 00:59:07.300 --> 00:59:11.410 a quantity we call 1 over mu. 00:59:11.410 --> 00:59:18.270 And we have just designed what's called a dynamic absorber. 00:59:18.270 --> 00:59:21.270 And a dynamic absorber is a little device 00:59:21.270 --> 00:59:24.240 you can use to stop vibration. 00:59:24.240 --> 00:59:26.605 So when we were talking about vibration isolation 00:59:26.605 --> 00:59:29.010 and vibration mitigation a few days ago 00:59:29.010 --> 00:59:32.320 and I said you've got some rotating imbalance [INAUDIBLE] 00:59:32.320 --> 00:59:35.380 or something to shake, well, give me 00:59:35.380 --> 00:59:37.575 three ways of solving it. 00:59:37.575 --> 00:59:39.330 We said balance the rotor. 00:59:39.330 --> 00:59:42.170 What were the other two ways of perhaps reducing the vibration? 00:59:48.821 --> 00:59:50.770 AUDIENCE: Adding a [INAUDIBLE] damper? 00:59:50.770 --> 00:59:52.410 PROFESSOR: Yeah, but also if you've 00:59:52.410 --> 00:59:53.885 got something that's shaking like crazy, 00:59:53.885 --> 00:59:55.718 and it's putting fibrillation into the floor 00:59:55.718 --> 00:59:57.700 or into the table, you can isolate it 00:59:57.700 --> 01:00:01.000 with a mass and a spring, or a microscope over here that's 01:00:01.000 --> 01:00:02.860 vibrating like crazy, you can isolate it 01:00:02.860 --> 01:00:03.940 with a mass or a spring. 01:00:03.940 --> 01:00:06.290 So this is to stop a vibration isolation, which 01:00:06.290 --> 01:00:09.820 is guaranteed to be on the final-- 01:00:09.820 --> 01:00:12.020 the simple practical applications of single degree 01:00:12.020 --> 01:00:13.140 of freedom stuff. 01:00:13.140 --> 01:00:15.930 So we have three ways-- fix the rotor, 01:00:15.930 --> 01:00:18.020 isolate it with a mass and a spring, 01:00:18.020 --> 01:00:20.370 isolate the sensitive instrument mass and a spring. 01:00:20.370 --> 01:00:22.192 Now you have a fourth way. 01:00:22.192 --> 01:00:23.900 If it's a particular operating frequency, 01:00:23.900 --> 01:00:27.790 we can operate right here with this thing 01:00:27.790 --> 01:00:29.480 called the dynamic absorber. 01:00:29.480 --> 01:00:31.430 This mu quantity is called the mass ratio. 01:00:39.880 --> 01:00:42.535 That's m2/m1. 01:00:45.570 --> 01:00:46.990 Basically, these things are real. 01:00:46.990 --> 01:00:48.620 They're actually used in real machines, 01:00:48.620 --> 01:00:51.640 and usually you can't-- this dynamic absorber thing you 01:00:51.640 --> 01:00:54.590 stick on there can't be as big as the original system. 01:00:54.590 --> 01:00:57.120 It's going to be some small fraction of the original system 01:00:57.120 --> 01:01:00.390 size-- 5% or 10% if you're lucky. 01:01:00.390 --> 01:01:02.040 So the bigger this thing is, you'll 01:01:02.040 --> 01:01:03.250 find out the better it works. 01:01:03.250 --> 01:01:07.650 So how much is this little second mass bouncing around? 01:01:07.650 --> 01:01:11.560 Well it's bouncing around compared to x1 static 01:01:11.560 --> 01:01:17.030 in the ratio of k1 to k2, which is this 1 over mu quantity. 01:01:17.030 --> 01:01:19.950 So if mu is 10%-- if you've added the second mass, 01:01:19.950 --> 01:01:22.310 it's 10% the size of the first one. 01:01:22.310 --> 01:01:25.990 1 over mu is a factor of 10. 01:01:25.990 --> 01:01:29.370 So this transfer function basically 01:01:29.370 --> 01:01:42.950 looks-- do I have some colored-- so the x2/x1 transfer 01:01:42.950 --> 01:01:46.420 function basically behaves the same right here 01:01:46.420 --> 01:01:49.410 and pretty much the same way out here. 01:01:49.410 --> 01:01:54.900 And in here, it comes down like this and comes back up. 01:01:54.900 --> 01:02:03.220 And this height right here is 1 over mu. 01:02:03.220 --> 01:02:07.060 So the smaller you make this thing, the tinier you make it, 01:02:07.060 --> 01:02:14.000 the more it has to shake to force the first mas to go to 0. 01:02:14.000 --> 01:02:16.890 So basically, the reason this thing works-- the free body 01:02:16.890 --> 01:02:30.200 diagram-- at the 0 point, the main mass m1, x1 equals 0 here. 01:02:30.200 --> 01:02:31.860 It's not moving. 01:02:31.860 --> 01:02:37.610 It's got a force acting on it that is some F1 cosine omega t. 01:02:37.610 --> 01:02:41.110 It's got a spring force acting on it 01:02:41.110 --> 01:02:42.950 from this second mass going. 01:02:42.950 --> 01:02:46.640 Here is m2 over here, and it's going back and forth 01:02:46.640 --> 01:02:51.080 like crazy putting force through this spring. 01:02:51.080 --> 01:02:55.180 And the spring force had better be exactly equal to that. 01:02:55.180 --> 01:03:02.390 So the F spring is going to be equal to minus F1, 01:03:02.390 --> 01:03:08.490 and that will be equal to x2 times kx2. 01:03:08.490 --> 01:03:10.390 The second mass has to move enough 01:03:10.390 --> 01:03:13.350 that it'll compress the spring enough that it provides 01:03:13.350 --> 01:03:15.830 a force equal and opposite to this one 01:03:15.830 --> 01:03:19.830 so that it's in equilibrium and it doesn't move. 01:03:19.830 --> 01:03:21.430 All right. 01:03:21.430 --> 01:03:21.930 Yeah? 01:03:26.286 --> 01:03:27.738 AUDIENCE: [INAUDIBLE]. 01:03:27.738 --> 01:03:28.706 PROFESSOR: Yeah. 01:03:28.706 --> 01:03:32.473 AUDIENCE: So the building, if you 01:03:32.473 --> 01:03:34.514 want the masses to be fairly equal to each other, 01:03:34.514 --> 01:03:35.889 how is that ever going to happen? 01:03:35.889 --> 01:03:37.090 PROFESSOR: It's not. 01:03:37.090 --> 01:03:39.700 Not usually. 01:03:39.700 --> 01:03:44.640 Dynamic absorbers are used in real things, 01:03:44.640 --> 01:03:46.440 like the Hancock Building. 01:03:46.440 --> 01:03:49.230 And they're used in engines. 01:03:49.230 --> 01:03:51.150 They're used in all sorts of little devices 01:03:51.150 --> 01:03:53.130 that you're not aware they they're there. 01:03:55.760 --> 01:03:58.000 And you have to usually hide them 01:03:58.000 --> 01:04:00.700 inside the footprint of the original device. 01:04:00.700 --> 01:04:02.660 So you don't want them-- for a real thing, 01:04:02.660 --> 01:04:07.660 you don't want to have a pump with this huge appendage on it. 01:04:07.660 --> 01:04:08.780 It's just not practical. 01:04:08.780 --> 01:04:10.440 So they tend to be small. 01:04:10.440 --> 01:04:12.677 And so, as a consequence, in order 01:04:12.677 --> 01:04:14.635 to make them work at this particular frequency, 01:04:14.635 --> 01:04:16.590 you have to vibrate like crazy. 01:04:16.590 --> 01:04:19.450 So the design cost of one of these things 01:04:19.450 --> 01:04:22.230 is you have to design to be able to allow this second mass 01:04:22.230 --> 01:04:25.080 to shake like all get out. 01:04:25.080 --> 01:04:29.630 Now, how many frequencies does this work at? 01:04:29.630 --> 01:04:32.250 Once you've designed it, it only works at one frequency, 01:04:32.250 --> 01:04:33.750 so this is only useful for something 01:04:33.750 --> 01:04:36.330 that has a fixed operating frequency, 01:04:36.330 --> 01:04:38.810 like a synchronous motor. 01:04:38.810 --> 01:04:41.520 This isn't a good thing for something 01:04:41.520 --> 01:04:45.240 that has a range of frequencies over which it can work, 01:04:45.240 --> 01:04:49.570 but it is a different, more complicated theory. 01:04:49.570 --> 01:04:51.120 There's a type of dynamic absorber 01:04:51.120 --> 01:04:54.905 that you can make work over a wide range of frequencies. 01:04:54.905 --> 01:04:56.670 I need to tell you one other thing. 01:04:56.670 --> 01:04:59.820 When we put this thing on, the original natural frequency 01:04:59.820 --> 01:05:04.380 of the first system was here, and it created two new ones. 01:05:04.380 --> 01:05:06.256 You had a further question. 01:05:06.256 --> 01:05:09.488 AUDIENCE: So how does making it vibrate like that help? 01:05:09.488 --> 01:05:11.404 It's only going to work at that one frequency, 01:05:11.404 --> 01:05:13.063 and you're always going to have that gap. 01:05:13.063 --> 01:05:14.438 So making it vibrate really fast, 01:05:14.438 --> 01:05:17.420 how does that help at all? 01:05:17.420 --> 01:05:19.170 PROFESSOR: If, for some reason, you really 01:05:19.170 --> 01:05:21.830 don't want that first thing to shake, 01:05:21.830 --> 01:05:25.160 this makes it stop shaking at that frequency. 01:05:25.160 --> 01:05:32.170 So for example, one of the widest distributed textbooks 01:05:32.170 --> 01:05:35.260 in the world was written by an MIT professor named Den Hartog, 01:05:35.260 --> 01:05:36.874 and it's called Mechanical Vibration. 01:05:36.874 --> 01:05:38.040 It was written in the 1930s. 01:05:38.040 --> 01:05:41.570 It's got a wonderful chapter on dynamic absorbers in it. 01:05:41.570 --> 01:05:44.150 And the example that he gives of a real device that 01:05:44.150 --> 01:05:50.040 actually uses one was an electric hair clipper 01:05:50.040 --> 01:05:53.850 that a barber uses. 01:05:53.850 --> 01:05:55.990 The head on a hair clipper-- it has 01:05:55.990 --> 01:05:59.510 to go back and forth like this in or to cut the hair. 01:05:59.510 --> 01:06:02.340 That is an oscillating mass, and you're going to feel it. 01:06:05.620 --> 01:06:08.630 You're going to feel the mass times the acceleration-- ma 01:06:08.630 --> 01:06:12.300 omega squared acceleration-- at the frequency it runs that. 01:06:12.300 --> 01:06:13.800 So they did a clever thing. 01:06:13.800 --> 01:06:16.270 They built inside the case of this thing 01:06:16.270 --> 01:06:18.890 a little second mass and spring. 01:06:18.890 --> 01:06:21.790 And so that when you're holding the clippers, 01:06:21.790 --> 01:06:24.830 it doesn't feel like it's-- OK? 01:06:24.830 --> 01:06:26.300 So that was an example. 01:06:26.300 --> 01:06:30.590 It was actually a great example in his 1938 textbook. 01:06:30.590 --> 01:06:32.870 And just so you know, you really do 01:06:32.870 --> 01:06:35.910 have two natural frequencies when you do this, 01:06:35.910 --> 01:06:37.260 and I'll write them like this. 01:06:37.260 --> 01:06:39.080 Omega 1 squared and omega 2 squared 01:06:39.080 --> 01:06:45.270 are the original frequency squared times 1 01:06:45.270 --> 01:06:54.550 plus mu over 2 plus or minus mu plus mu squared 01:06:54.550 --> 01:06:58.570 over 4 square root. 01:06:58.570 --> 01:07:03.040 So as your mass ratio gets bigger, 01:07:03.040 --> 01:07:05.500 the two frequencies move further and further apart. 01:07:05.500 --> 01:07:08.460 When the mass ratio is 0, this quantity-- 01:07:08.460 --> 01:07:10.960 they have two natural frequencies. 01:07:10.960 --> 01:07:12.290 It goes to 1 natural frequency. 01:07:12.290 --> 01:07:13.915 It's a single degree of freedom system, 01:07:13.915 --> 01:07:15.710 and it's at the original value. 01:07:15.710 --> 01:07:19.270 As the mass ratio gets bigger and bigger, these two roots-- 01:07:19.270 --> 01:07:23.860 this one and this one-- as you add mass to that second system, 01:07:23.860 --> 01:07:24.930 they spread apart. 01:07:24.930 --> 01:07:27.320 And the bigger the mass ratio, the further apart the two 01:07:27.320 --> 01:07:30.920 natural frequencies become. 01:07:30.920 --> 01:07:34.070 So we've got 10 minutes left, and I have a demo 01:07:34.070 --> 01:07:37.220 which illustrates this thing. 01:07:37.220 --> 01:07:40.449 It's a little delicate to make it work, 01:07:40.449 --> 01:07:42.240 because actually it was really frustrating. 01:07:42.240 --> 01:07:43.530 I had it all set up. 01:07:43.530 --> 01:07:46.350 It worked great in my office. 01:07:46.350 --> 01:07:49.590 Came in 10 minutes early today to set it up, 01:07:49.590 --> 01:07:52.740 and the table in here is so flexible I can't make 01:07:52.740 --> 01:07:56.020 it behave like a fixed surface. 01:07:56.020 --> 01:07:58.990 So I tweaked it, and I think I've got running now, 01:07:58.990 --> 01:08:03.840 but now what it is is a beam with my pen 01:08:03.840 --> 01:08:05.360 on it, my little squiggle pen. 01:08:05.360 --> 01:08:06.640 And you've seen this before. 01:08:06.640 --> 01:08:08.870 It's a rotating mass. 01:08:08.870 --> 01:08:10.630 It's a static imbalance. 01:08:10.630 --> 01:08:13.500 It shakes this beam. 01:08:13.500 --> 01:08:16.960 And I've added to it a second little beam. 01:08:16.960 --> 01:08:20.180 It's a little blade of steal with a heavy magnet on it. 01:08:20.180 --> 01:08:21.700 That's my m2. 01:08:21.700 --> 01:08:24.700 This is my k2-- is this little beam. 01:08:24.700 --> 01:08:27.345 This is the original system, a mass, 01:08:27.345 --> 01:08:29.229 it's close to a natural frequency, 01:08:29.229 --> 01:08:33.084 and the rotation rate of the eccentric mass 01:08:33.084 --> 01:08:36.439 is right at the natural frequency of this beam. 01:08:36.439 --> 01:08:39.250 So because it's delicate to set up, 01:08:39.250 --> 01:08:43.109 I've got to show you the system first with the second mass 01:08:43.109 --> 01:08:45.819 attached to it, and I think I've got it tweaked 01:08:45.819 --> 01:08:46.920 so that it'll sit there. 01:08:46.920 --> 01:08:48.960 It's right at the operating point, 01:08:48.960 --> 01:08:52.189 and you ought to see the second mass moving like crazy, 01:08:52.189 --> 01:08:56.340 and the original mass, the beam, not moving much at all. 01:08:56.340 --> 01:08:59.005 So let's see if we can make it work. 01:08:59.005 --> 01:09:00.630 We're going to need to kill the lights. 01:09:14.590 --> 01:09:19.529 And you can see in here the rotating mass going around 01:09:19.529 --> 01:09:22.300 and around, and so I have the strobe light 01:09:22.300 --> 01:09:25.790 detuned just a little bit so that you can see the system. 01:09:25.790 --> 01:09:27.080 It's not quite synchronous. 01:09:27.080 --> 01:09:28.939 Watch this little white blob out here. 01:09:28.939 --> 01:09:31.420 You see it going up and down. 01:09:31.420 --> 01:09:36.390 That's your second mass, and is the main beam-- 01:09:36.390 --> 01:09:41.840 it's moving a little bit, but not much. 01:09:41.840 --> 01:09:43.340 It's got a little damping, and so it 01:09:43.340 --> 01:09:47.729 isn't perfectly down at that 0 point, but very close. 01:09:47.729 --> 01:09:52.390 So this is the system operating close to that null point. 01:09:52.390 --> 01:09:57.900 So now if I remove the dynamic absorber, 01:09:57.900 --> 01:10:02.330 the second little mass spring system, then the main mass 01:10:02.330 --> 01:10:04.316 and the beam ought to shake like crazy. 01:10:15.052 --> 01:10:17.940 Just changing the tuning a little here. 01:10:33.300 --> 01:10:40.035 So now I'm going to remove the absorber. 01:10:50.150 --> 01:10:52.180 All right? 01:10:52.180 --> 01:10:55.740 And now it's going like crazy. 01:10:59.440 --> 01:11:01.415 So that's an illustration. 01:11:01.415 --> 01:11:02.950 You can come up with the lights. 01:11:08.840 --> 01:11:10.080 Now, you hear beating? 01:11:13.060 --> 01:11:15.950 This little pen-- this little rotor in here 01:11:15.950 --> 01:11:21.280 is driven by a little DC motor and a single AA battery, 01:11:21.280 --> 01:11:23.210 and it's not feedback controlled. 01:11:23.210 --> 01:11:26.539 Its frequency can vary, and it isn't that powerful. 01:11:26.539 --> 01:11:28.330 So as this thing starts moving up and down, 01:11:28.330 --> 01:11:32.200 it actually takes real torque to make that weight go around 01:11:32.200 --> 01:11:35.340 while this whole system is accelerating up and down. 01:11:35.340 --> 01:11:37.010 And the motor just isn't up to it, 01:11:37.010 --> 01:11:41.280 just isn't powerful enough, so the speed changes. 01:11:41.280 --> 01:11:45.390 This thing can't hold constant frequency. 01:11:45.390 --> 01:11:46.260 So that's the demo. 01:11:46.260 --> 01:11:46.759 Questions? 01:11:49.590 --> 01:11:52.460 So we've kind of embedded in this lecture-- this lecture was 01:11:52.460 --> 01:11:55.240 to introduce you to the idea that you can write a transfer 01:11:55.240 --> 01:11:57.830 function for a multiple degree of freedom system-- 01:11:57.830 --> 01:12:00.140 has embedded in it all the natural frequencies 01:12:00.140 --> 01:12:01.440 of the system. 01:12:01.440 --> 01:12:04.350 When you use a transfer function to calculate 01:12:04.350 --> 01:12:07.900 the response of the system, you are getting the contributions 01:12:07.900 --> 01:12:10.870 of all the modes at once. 01:12:10.870 --> 01:12:15.470 So you essentially solve the equations of motion directly. 01:12:15.470 --> 01:12:17.390 So you now have seen there's two ways 01:12:17.390 --> 01:12:21.020 to go about analyzing multiple degree of freedom systems-- 01:12:21.020 --> 01:12:24.400 the technique known as modal analysis-- one mode at a time 01:12:24.400 --> 01:12:26.780 and then add it together. 01:12:26.780 --> 01:12:29.120 Or like with transfer functions, where you're just 01:12:29.120 --> 01:12:32.040 solving the whole thing at once, and each transfer function 01:12:32.040 --> 01:12:36.710 has in it all of the information about all of the modes 01:12:36.710 --> 01:12:37.519 of the system. 01:12:37.519 --> 01:12:39.310 So if it's a five degree of freedom system, 01:12:39.310 --> 01:12:43.261 you're going to see five peaks out here. 01:12:43.261 --> 01:12:43.760 Questions? 01:12:47.946 --> 01:12:50.838 AUDIENCE: So this is for [INAUDIBLE]? 01:12:50.838 --> 01:12:54.230 But initially, since you have a [INAUDIBLE], 01:12:54.230 --> 01:12:57.885 does it have to start out moving to be an [INAUDIBLE] moving 01:12:57.885 --> 01:13:00.140 and then it stops. 01:13:00.140 --> 01:13:03.210 PROFESSOR: So there's going to be a transient phase, sure. 01:13:03.210 --> 01:13:06.010 Any system, when you start it up, 01:13:06.010 --> 01:13:09.450 is going to have transients, and the transients 01:13:09.450 --> 01:13:12.400 are the equivalent of initial conditions. 01:13:12.400 --> 01:13:16.920 So displacement and velocity-- and the response 01:13:16.920 --> 01:13:22.210 of a linear system that vibrates, 01:13:22.210 --> 01:13:25.200 that has natural frequencies to a set of initial conditions, 01:13:25.200 --> 01:13:26.740 is vibration at what frequencies? 01:13:36.027 --> 01:13:37.652 AUDIENCE: Do you know how they overcome 01:13:37.652 --> 01:13:39.636 this in helicopters? [INAUDIBLE] that we saw 01:13:39.636 --> 01:13:42.120 at the beginning of the year? 01:13:42.120 --> 01:13:43.760 PROFESSOR: Yeah, the helicopter-- so 01:13:43.760 --> 01:13:45.080 I want to answer the question. 01:13:45.080 --> 01:13:47.850 Response to that transience is at the natural frequencies 01:13:47.850 --> 01:13:50.030 of the system, maybe some combination. 01:13:50.030 --> 01:13:52.460 The original displacement requires 01:13:52.460 --> 01:13:54.020 contributions of several modes. 01:13:54.020 --> 01:13:57.250 You'll get decay contributions at each 01:13:57.250 --> 01:14:00.100 of the natural frequencies, and once they've died out, 01:14:00.100 --> 01:14:02.570 this motor has started up. 01:14:02.570 --> 01:14:05.730 The thing will settle into what we call steady state response, 01:14:05.730 --> 01:14:08.355 and it will only be at the excitation frequency. 01:14:08.355 --> 01:14:09.980 And I haven't show you anything about-- 01:14:09.980 --> 01:14:14.440 can you solve the equations of motion from start up at time 0 01:14:14.440 --> 01:14:16.350 through the whole messy transient phase 01:14:16.350 --> 01:14:17.210 to steady state? 01:14:17.210 --> 01:14:17.710 Sure. 01:14:17.710 --> 01:14:20.330 And we haven't talked at all about that. 01:14:20.330 --> 01:14:24.410 And actually, for vibration stuff, it is very important. 01:14:24.410 --> 01:14:25.840 Steady state answer is important. 01:14:25.840 --> 01:14:27.548 The transient answer is pretty important, 01:14:27.548 --> 01:14:29.560 and for you to have a feeling for that, 01:14:29.560 --> 01:14:32.690 you actually know quite a fair amount about the basic concepts 01:14:32.690 --> 01:14:34.810 of how things vibrate now. 01:14:34.810 --> 01:14:37.250 Now, your question about the helicopter-- 01:14:37.250 --> 01:14:39.005 so if you have a uniform helicopter 01:14:39.005 --> 01:14:42.642 blade-- they rigged that to do what it did. 01:14:42.642 --> 01:14:44.600 I don't know what they were experimenting with, 01:14:44.600 --> 01:14:47.590 but if you do-- there's a problem that I've actually 01:14:47.590 --> 01:14:48.940 given in a vibration course. 01:14:48.940 --> 01:14:51.470 If you design a really simple helicopter 01:14:51.470 --> 01:14:53.380 that has a uniform blade. 01:14:53.380 --> 01:14:58.990 It's a uniform rod, and it's pivoted at the center, 01:14:58.990 --> 01:15:00.680 and you spin it up, and if you spin it 01:15:00.680 --> 01:15:02.290 fast enough-- if you spin it slowly, 01:15:02.290 --> 01:15:04.470 it will droop because of gravity, but if you spin it 01:15:04.470 --> 01:15:07.570 fast enough, r omega squared is a lot bigger than g. 01:15:07.570 --> 01:15:09.320 And the thing flattens out, and the blades 01:15:09.320 --> 01:15:11.510 are going around and around. 01:15:11.510 --> 01:15:14.990 The natural frequency of that system 01:15:14.990 --> 01:15:18.762 is exactly the rotation rate. 01:15:18.762 --> 01:15:23.050 A uniform blade pivot at the center, going around fast, 01:15:23.050 --> 01:15:24.740 has a natural frequency like this. 01:15:24.740 --> 01:15:28.340 It happens to be exactly at the rate of rotation. 01:15:28.340 --> 01:15:32.760 So any perturbation will start it doing bad things. 01:15:32.760 --> 01:15:37.290 Helicopters are never ever designed like that. 01:15:37.290 --> 01:15:40.930 The helicopter has a rotor disk in here, 01:15:40.930 --> 01:15:43.250 has a finite radius before the pin, 01:15:43.250 --> 01:15:46.270 and then the blades are out here. 01:15:46.270 --> 01:15:49.160 The pin around which-- and that changes the resonance 01:15:49.160 --> 01:15:50.060 frequency. 01:15:50.060 --> 01:15:52.800 And of course, you can change the mass distribution 01:15:52.800 --> 01:15:56.110 in the blade and so forth. 01:15:56.110 --> 01:15:58.254 So what they had done in that case 01:15:58.254 --> 01:16:00.670 to rig it so the thing beat itself to death, I don't know. 01:16:00.670 --> 01:16:03.770 But there's good design practices with helicopters 01:16:03.770 --> 01:16:08.255 so that the blade frequency is not at the rotation rate, 01:16:08.255 --> 01:16:09.838 because at rotation rate, you're going 01:16:09.838 --> 01:16:13.040 to have a lot of excitation, hitting turbulence in the air 01:16:13.040 --> 01:16:16.460 and so forth. 01:16:16.460 --> 01:16:16.960 Yeah? 01:16:16.960 --> 01:16:19.900 AUDIENCE: So you mentioned that you use something 01:16:19.900 --> 01:16:21.370 like this in the Hancock Building, 01:16:21.370 --> 01:16:23.820 but I would imagine that the wind force on the Hancock 01:16:23.820 --> 01:16:27.740 Building does not have a uniform frequency all the time. 01:16:27.740 --> 01:16:31.660 So is the idea that [INAUDIBLE] control 01:16:31.660 --> 01:16:33.620 based on resonant frequency [INAUDIBLE]. 01:16:38.530 --> 01:16:40.769 PROFESSOR: I've forgotten the correct name now, 01:16:40.769 --> 01:16:42.560 even though I've used these many times now. 01:16:42.560 --> 01:16:47.740 The optimally tuned and damped dynamic absorber 01:16:47.740 --> 01:16:56.610 is sort of the second level of dynamic absorber design. 01:16:56.610 --> 01:16:58.590 Typically, a two degree of freedom system 01:16:58.590 --> 01:17:01.330 looks like that if it's lightly damped, 01:17:01.330 --> 01:17:06.000 but you can design that second mass spring and a damper 01:17:06.000 --> 01:17:11.200 so that you can make the transfer function, 01:17:11.200 --> 01:17:18.950 H11 here, look like this, where you 01:17:18.950 --> 01:17:24.910 get the two peaks are of equal height, and the worst-- 01:17:24.910 --> 01:17:27.910 and you can design them so that the worst case amplitude 01:17:27.910 --> 01:17:29.490 response over here is pretty low. 01:17:33.710 --> 01:17:40.660 x1 static is square root of 1 over 1 01:17:40.660 --> 01:17:43.230 plus mu or something like that. 01:17:43.230 --> 01:17:47.250 So the bigger you make this, the better the performance, 01:17:47.250 --> 01:17:50.410 but you have to have optimum damping, 01:17:50.410 --> 01:17:52.860 and you have to have optimum tuning. 01:17:52.860 --> 01:18:00.410 And tuning means, if the there were-- we 01:18:00.410 --> 01:18:05.010 tune this thing so that k2/m2 was 01:18:05.010 --> 01:18:10.470 exactly equal to a particular frequency, in this case k1/n1. 01:18:10.470 --> 01:18:14.070 But for the optimally tuned and damp dynamic absorber, 01:18:14.070 --> 01:18:17.820 you actually tune things-- it's tuned a little differently. 01:18:17.820 --> 01:18:19.590 You put damping in it, and you make 01:18:19.590 --> 01:18:23.440 it behave pretty well over the entire frequency, 01:18:23.440 --> 01:18:28.980 range rather than almost perfect at one frequency. 01:18:28.980 --> 01:18:35.440 And the Hancock Building had its transfer function, so to speak, 01:18:35.440 --> 01:18:39.720 without the dynamic absorber-- it has many, many peaks, 01:18:39.720 --> 01:18:45.450 but it had two problematic ones that were very close together, 01:18:45.450 --> 01:18:55.089 and they were at about 0.8 radians per second, and 0.81, 01:18:55.089 --> 01:18:55.630 or something. 01:18:55.630 --> 01:18:59.115 And this was bending, and this was torsion. 01:19:03.470 --> 01:19:06.660 And at these two resonances-- and the wind would come along, 01:19:06.660 --> 01:19:08.561 and it would excite both of them. 01:19:08.561 --> 01:19:10.810 And the one resonance was just the same as this thing. 01:19:10.810 --> 01:19:12.890 It just bends. 01:19:12.890 --> 01:19:15.720 And the other resonance was the first mode in torsion. 01:19:15.720 --> 01:19:19.095 So if you were standing on one-- and a building cross section 01:19:19.095 --> 01:19:21.530 is a funny-- the Hancock Building 01:19:21.530 --> 01:19:23.390 looks kind of like that in cross section, 01:19:23.390 --> 01:19:31.660 and so it would rotate in torsion and deflect in bending. 01:19:31.660 --> 01:19:36.090 So it would be rotating and deflecting, 01:19:36.090 --> 01:19:37.900 and if they were in equal amounts 01:19:37.900 --> 01:19:41.910 and they're close in frequency, if you're standing up there, 01:19:41.910 --> 01:19:44.450 what would you feel? 01:19:44.450 --> 01:19:45.700 They would beat. 01:19:45.700 --> 01:19:46.630 The motion would beat. 01:19:46.630 --> 01:19:49.369 You'd get a lot of motion, because if you 01:19:49.369 --> 01:19:50.910 are on one end of the building, you'd 01:19:50.910 --> 01:19:54.040 get a lot of torsional motion out here combined 01:19:54.040 --> 01:19:55.296 with the bending motion. 01:19:55.296 --> 01:19:57.170 And then they would cancel, and then the beat 01:19:57.170 --> 01:19:58.970 would build up again. 01:19:58.970 --> 01:20:02.540 And so that was what was happening in the building. 01:20:02.540 --> 01:20:12.140 And they put in two, so on the 58th floor on each end 01:20:12.140 --> 01:20:20.640 they put in a mass spring system and over here another one so 01:20:20.640 --> 01:20:23.199 that they would resist the torsion-- actually, 01:20:23.199 --> 01:20:24.740 really I lined them up the wrong way. 01:20:24.740 --> 01:20:26.850 They line up this way. 01:20:26.850 --> 01:20:28.570 So it has two of them. 01:20:28.570 --> 01:20:32.590 They can resist both the bending and the torsion. 01:20:32.590 --> 01:20:37.300 They each weigh 300 tons, and they 01:20:37.300 --> 01:20:42.410 it's a box filled up with 50 pound lead bricks. 01:20:42.410 --> 01:20:46.240 And the box is about-- it's been 20 years since I was up there, 01:20:46.240 --> 01:20:50.540 but what I recall is this like 8 by 10 feet yea tall. 01:20:50.540 --> 01:20:55.400 And it slides on a pressurized oil film, and the spring on it 01:20:55.400 --> 01:20:58.360 is a big pneumatic spring, a big air spring, 01:20:58.360 --> 01:20:59.910 and a computer runs the whole system. 01:20:59.910 --> 01:21:02.910 It's shut down until the wind gets above 40 miles an hour, 01:21:02.910 --> 01:21:04.720 and then turns it on. 01:21:04.720 --> 01:21:06.390 And it's actually kind of self tuning. 01:21:06.390 --> 01:21:08.540 It can optimize itself, and then it'll sit there, 01:21:08.540 --> 01:21:11.740 and it's designed to do this, because this building has 01:21:11.740 --> 01:21:14.750 these two problem frequencies. 01:21:14.750 --> 01:21:19.410 And so it has been designed to address both of them. 01:21:22.690 --> 01:21:26.505 All right, I think I have to get out of here for the next class. 01:21:26.505 --> 01:21:27.380 See you next Tuesday. 01:21:27.380 --> 01:21:28.834 Last lecture.