1 00:00:00,080 --> 00:00:02,430 The following content is provided under a Creative 2 00:00:02,430 --> 00:00:03,800 Commons license. 3 00:00:03,800 --> 00:00:06,050 Your support will help MIT OpenCourseWare 4 00:00:06,050 --> 00:00:10,140 continue to offer high quality educational resources for free. 5 00:00:10,140 --> 00:00:12,690 To make a donation or to view additional materials 6 00:00:12,690 --> 00:00:16,600 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,600 --> 00:00:17,263 at ocw.mit.edu. 8 00:00:21,670 --> 00:00:24,010 PROFESSOR: So we'll start quickly 9 00:00:24,010 --> 00:00:28,614 going over the concept questions for the homework this week. 10 00:00:28,614 --> 00:00:34,210 So you have a two rotor system. 11 00:00:37,770 --> 00:00:39,425 It says, if you give it a deflection 12 00:00:39,425 --> 00:00:44,450 at a exactly the mode shape of mode two, 13 00:00:44,450 --> 00:00:46,750 what frequency components do expect 14 00:00:46,750 --> 00:00:50,480 the transient response to have? 15 00:00:50,480 --> 00:00:54,890 So most people said only omega 2, but quite a few people 16 00:00:54,890 --> 00:00:56,450 said both. 17 00:00:56,450 --> 00:01:02,830 So this introduction to vibration 18 00:01:02,830 --> 00:01:04,870 that we have been doing for the last few 19 00:01:04,870 --> 00:01:11,700 lectures-- my goal in it is to have you folks go away 20 00:01:11,700 --> 00:01:14,610 with a pretty good conceptual understanding of the basics 21 00:01:14,610 --> 00:01:17,290 of vibration, and on the final exam, 22 00:01:17,290 --> 00:01:19,620 I'm not going to have you derive anything, 23 00:01:19,620 --> 00:01:24,410 like finding the natural frequencies of a 3 24 00:01:24,410 --> 00:01:29,210 by 3 degree of freedom system, solving a sixth order system 25 00:01:29,210 --> 00:01:30,780 in omega squared. 26 00:01:30,780 --> 00:01:33,980 I won't do that kind of thing to you, but questions like this 27 00:01:33,980 --> 00:01:36,280 are really fair game. 28 00:01:36,280 --> 00:01:40,070 So if we did this business bimodal analysis-- 29 00:01:40,070 --> 00:01:45,020 and we did an example the other day where with the system 30 00:01:45,020 --> 00:01:48,040 we had here, if you deflect it in exactly the shape of one 31 00:01:48,040 --> 00:01:50,330 mode, what kind of response do you get? 32 00:01:53,730 --> 00:01:55,560 Initial conditions that are shaped exactly 33 00:01:55,560 --> 00:01:58,190 like one the mode shapes. 34 00:01:58,190 --> 00:02:01,960 AUDIENCE: It has exactly that natural frequency in it. 35 00:02:01,960 --> 00:02:04,560 PROFESSOR: Well, it'll not only have that natural frequency 36 00:02:04,560 --> 00:02:07,270 in it, but if you deflect it initially in the mode 37 00:02:07,270 --> 00:02:10,100 shape, what will the responding motion look like? 38 00:02:10,100 --> 00:02:11,350 AUDIENCE: Just the mode shape. 39 00:02:11,350 --> 00:02:12,849 PROFESSOR: Just like the mode shape, 40 00:02:12,849 --> 00:02:15,100 and if the motion is in a particular mode shape, 41 00:02:15,100 --> 00:02:18,140 it will be at-- this is transient vibration. 42 00:02:18,140 --> 00:02:19,465 No external force. 43 00:02:19,465 --> 00:02:21,600 You deflect it and let it go. 44 00:02:21,600 --> 00:02:24,920 You will see only motion in that mode 45 00:02:24,920 --> 00:02:28,880 if you give it an initial deflection exactly in that mode 46 00:02:28,880 --> 00:02:30,310 shape. 47 00:02:30,310 --> 00:02:32,900 If you went to the other mode, and deflected it that way, 48 00:02:32,900 --> 00:02:34,960 and let go, it would change frequency, 49 00:02:34,960 --> 00:02:38,610 and it would vibrate only in that shape. 50 00:02:38,610 --> 00:02:40,870 And any other combination of motions 51 00:02:40,870 --> 00:02:45,370 is some linear sum of those two mode shapes. 52 00:02:45,370 --> 00:02:47,740 Any other allowable motion of the system 53 00:02:47,740 --> 00:02:52,060 can be made up of some weighted sum of the two mode shapes, 54 00:02:52,060 --> 00:02:53,340 right? 55 00:02:53,340 --> 00:02:55,480 And that's the amount of each mode 56 00:02:55,480 --> 00:02:57,570 that you get-- is the weighted sum. 57 00:02:57,570 --> 00:02:59,320 What are the weightings? 58 00:02:59,320 --> 00:03:01,350 If the weightings are 1, 0, then it's 59 00:03:01,350 --> 00:03:03,280 all one mode and not of another. 60 00:03:03,280 --> 00:03:06,430 If it takes some of each mode to give you 61 00:03:06,430 --> 00:03:09,110 the initial deflected shape, that's how much of each mode 62 00:03:09,110 --> 00:03:10,090 you get. 63 00:03:10,090 --> 00:03:15,780 So the answer to this question is only mode two. 64 00:03:15,780 --> 00:03:16,400 OK, next. 65 00:03:25,570 --> 00:03:28,590 I forget what you were given. 66 00:03:28,590 --> 00:03:29,340 Apply the concept. 67 00:03:29,340 --> 00:03:32,050 Which mode is likely to dominate the steady state response 68 00:03:32,050 --> 00:03:36,360 for the excitation of part D? 69 00:03:36,360 --> 00:03:39,440 So it was probably being excited harmonically 70 00:03:39,440 --> 00:03:41,770 at the natural frequency of mode two. 71 00:03:41,770 --> 00:03:43,510 So which mode? 72 00:03:43,510 --> 00:03:45,470 So harmonic excitation, which mode 73 00:03:45,470 --> 00:03:49,080 do you expect to respond the most when you excite it at one 74 00:03:49,080 --> 00:03:50,610 of the natural frequencies? 75 00:03:50,610 --> 00:03:51,860 AUDIENCE: I'd say [INAUDIBLE]. 76 00:03:51,860 --> 00:03:53,401 PROFESSOR: At that natural frequency. 77 00:03:53,401 --> 00:03:55,590 Why? 78 00:03:55,590 --> 00:03:56,090 Why? 79 00:03:59,436 --> 00:04:01,695 AUDIENCE: Because that's what the eigenvalue is. 80 00:04:01,695 --> 00:04:03,570 PROFESSOR: Well, there's an eigenvalue there, 81 00:04:03,570 --> 00:04:05,653 but now we're talking about steady state response. 82 00:04:05,653 --> 00:04:07,770 What's the transfer function of a single degree 83 00:04:07,770 --> 00:04:09,900 of freedom system look like? 84 00:04:09,900 --> 00:04:11,270 Just trace it in the air. 85 00:04:14,370 --> 00:04:19,100 This is a response per unit input, and where does it go? 86 00:04:19,100 --> 00:04:19,686 Way up high-- 87 00:04:19,686 --> 00:04:20,810 AUDIENCE: At the resonance. 88 00:04:20,810 --> 00:04:21,610 PROFESSOR: At the resonance. 89 00:04:21,610 --> 00:04:23,693 And so if you have two 2 degree of freedom system, 90 00:04:23,693 --> 00:04:25,676 how many resonances do you have? 91 00:04:25,676 --> 00:04:27,300 What do you think the transfer function 92 00:04:27,300 --> 00:04:29,800 is going to look like for a two degree of freedom system 93 00:04:29,800 --> 00:04:33,002 if you just plotted it just plot as a function of frequency? 94 00:04:33,002 --> 00:04:34,150 AUDIENCE: Double peak. 95 00:04:34,150 --> 00:04:35,650 PROFESSOR: Double peak. 96 00:04:35,650 --> 00:04:37,650 So the transfer function is likely to do 97 00:04:37,650 --> 00:04:41,455 that at one frequency and that at the next natural frequency. 98 00:04:41,455 --> 00:04:43,830 And that's what we're going to talk about today, in fact. 99 00:04:43,830 --> 00:04:46,320 OK so if you excited the second natural frequency 100 00:04:46,320 --> 00:04:50,870 and it's lightly damped, it's likely to be dominated 101 00:04:50,870 --> 00:04:53,990 by that modal response. 102 00:04:53,990 --> 00:04:58,720 In that question, is it guaranteed to be only that mode 103 00:04:58,720 --> 00:05:03,190 responding, even if you derive it at the natural frequency? 104 00:05:03,190 --> 00:05:04,690 Another way to saying this question, 105 00:05:04,690 --> 00:05:07,440 can the first mode have some response 106 00:05:07,440 --> 00:05:10,791 at any frequency you excite the system at? 107 00:05:10,791 --> 00:05:11,290 Sure. 108 00:05:11,290 --> 00:05:12,260 It has transfer function. 109 00:05:12,260 --> 00:05:13,480 It's continuous in frequency. 110 00:05:13,480 --> 00:05:14,840 It just won't be very big. 111 00:05:14,840 --> 00:05:16,060 OK, next. 112 00:05:22,400 --> 00:05:24,260 For what value is omega over omega 113 00:05:24,260 --> 00:05:27,020 n is the force transmitted to the wall going 114 00:05:27,020 --> 00:05:29,990 to be the greatest. 115 00:05:29,990 --> 00:05:33,310 So this is-- when we make this, this 116 00:05:33,310 --> 00:05:38,500 is now we're doing-- it's rotating around. 117 00:05:38,500 --> 00:05:42,420 OK, so this thing is rotating mass now 118 00:05:42,420 --> 00:05:44,220 at a constant frequency. 119 00:05:44,220 --> 00:05:47,890 So it's got a static imbalance. 120 00:05:47,890 --> 00:05:52,719 And at what frequency would you expect the force transmitted 121 00:05:52,719 --> 00:05:53,885 to the wall be the greatest? 122 00:05:59,090 --> 00:06:01,070 AUDIENCE: At resonance. 123 00:06:01,070 --> 00:06:03,994 PROFESSOR: At resonance, because that's when you get the most-- 124 00:06:03,994 --> 00:06:05,160 AUDIENCE: The most response. 125 00:06:05,160 --> 00:06:06,368 PROFESSOR: Greatest response. 126 00:06:06,368 --> 00:06:09,214 And the force transmitted to the wall is through the spring 127 00:06:09,214 --> 00:06:10,880 and through the dash pot, and the bigger 128 00:06:10,880 --> 00:06:13,970 the motions, the bigger those forces are going to be. 129 00:06:13,970 --> 00:06:17,060 OK, next. 130 00:06:17,060 --> 00:06:21,070 And so this is-- now, if you'll account for the spring, 131 00:06:21,070 --> 00:06:23,690 do you expect the counting for the mass of the spring 132 00:06:23,690 --> 00:06:25,950 to increase the predicted natural frequency 133 00:06:25,950 --> 00:06:29,525 or decrease it? 134 00:06:29,525 --> 00:06:30,957 AUDIENCE: Decrease. 135 00:06:30,957 --> 00:06:31,790 PROFESSOR: Decrease. 136 00:06:31,790 --> 00:06:33,820 Natural frequencies behave like square root 137 00:06:33,820 --> 00:06:36,090 of k/m kind of thing. 138 00:06:36,090 --> 00:06:38,950 And if you do anything that drives up the m, 139 00:06:38,950 --> 00:06:40,550 frequency is going to go down. 140 00:06:40,550 --> 00:06:42,190 Another way of asking the same question 141 00:06:42,190 --> 00:06:44,710 is, if you neglect mass when you're 142 00:06:44,710 --> 00:06:47,610 estimating the natural frequency of a system, 143 00:06:47,610 --> 00:06:49,590 which way are you likely to be in error? 144 00:06:52,880 --> 00:06:56,037 This system, normally we just ignore the mass of the springs, 145 00:06:56,037 --> 00:06:57,620 and we calculate the natural frequency 146 00:06:57,620 --> 00:07:00,590 as square root of k/m, and we're almost 147 00:07:00,590 --> 00:07:02,920 certain to be-- predict what? 148 00:07:02,920 --> 00:07:03,950 Too high or too low? 149 00:07:03,950 --> 00:07:04,870 AUDIENCE: Too high. 150 00:07:04,870 --> 00:07:06,620 PROFESSOR: Too high, because we've ignored 151 00:07:06,620 --> 00:07:07,890 some mass in the system. 152 00:07:07,890 --> 00:07:08,830 OK, next. 153 00:07:08,830 --> 00:07:10,990 I think that's it. 154 00:07:10,990 --> 00:07:13,320 I want to get on with today's lecture. 155 00:07:13,320 --> 00:07:17,882 We have done vibration from point 156 00:07:17,882 --> 00:07:19,340 of view of single degree of freedom 157 00:07:19,340 --> 00:07:24,190 systems, transient response, and steady escape response. 158 00:07:24,190 --> 00:07:27,170 And then we've started looking at multiple degree of freedom 159 00:07:27,170 --> 00:07:28,722 systems, but we actually started it 160 00:07:28,722 --> 00:07:30,430 from the point of view of modal analysis, 161 00:07:30,430 --> 00:07:32,730 and we looked at two degree of freedom systems, 162 00:07:32,730 --> 00:07:36,450 and found the modal contributions 163 00:07:36,450 --> 00:07:39,640 of each of the modes, and sort of did them one at a time. 164 00:07:39,640 --> 00:07:43,129 So is there another way-- one of you 165 00:07:43,129 --> 00:07:44,670 asked me after class last time, well, 166 00:07:44,670 --> 00:07:48,000 can't you just solve the differential equations directly 167 00:07:48,000 --> 00:07:51,195 and not bother with separating the modes out and figuring out 168 00:07:51,195 --> 00:07:52,195 the modal contributions. 169 00:07:52,195 --> 00:07:55,690 The answer is yes, and you do it via transfer functions, 170 00:07:55,690 --> 00:07:57,987 except now you will need more than one 171 00:07:57,987 --> 00:07:59,820 if you have more than one degree of freedom. 172 00:08:02,420 --> 00:08:05,385 So that's what we'll focus on today. 173 00:08:09,890 --> 00:08:17,220 And we're going to do it by thinking about the example 174 00:08:17,220 --> 00:08:18,920 actually from the homework. 175 00:08:18,920 --> 00:08:22,950 So in the homework you have this problem 176 00:08:22,950 --> 00:08:33,510 of this cart with this pendulum hanging off of it. 177 00:08:33,510 --> 00:08:36,179 And we-- in problem three, I think it is-- 178 00:08:36,179 --> 00:08:38,059 is say, OK, now the pendulum is going 179 00:08:38,059 --> 00:08:40,000 to go around a constant rate, and it's 180 00:08:40,000 --> 00:08:43,750 going to turn this into a rotating imbalance problem. 181 00:08:43,750 --> 00:08:50,810 And I shorten it up so it's just a little mass spinning around. 182 00:08:50,810 --> 00:09:00,350 You have k1, c1, x1 here. 183 00:09:00,350 --> 00:09:05,410 And I think you have draw in there as theta. 184 00:09:05,410 --> 00:09:07,500 This is point A about which it spins. 185 00:09:10,270 --> 00:09:13,320 And you know the non-linear equation 186 00:09:13,320 --> 00:09:17,460 to motion for treating this as a two degree of freedom system, 187 00:09:17,460 --> 00:09:20,040 just pendulum on a cart. 188 00:09:25,180 --> 00:09:31,515 And then nonlinear equation of motion. 189 00:09:42,610 --> 00:09:44,170 And I'm going to put subscripts here, 190 00:09:44,170 --> 00:09:45,890 and I'm going to call this coordinate x1, 191 00:09:45,890 --> 00:09:51,324 because we're going to need a second coordinate-- 192 00:09:51,324 --> 00:09:52,500 or actually, do we? 193 00:09:52,500 --> 00:09:53,580 What am I saying? 194 00:09:53,580 --> 00:09:56,770 We don't need to do that. 195 00:09:56,770 --> 00:09:57,750 Just x will do. 196 00:10:00,950 --> 00:10:03,560 We do need the m1 and m2, though. 197 00:10:03,560 --> 00:10:07,650 And so that is all-- I'm going to move the remaining 198 00:10:07,650 --> 00:10:09,100 stuff to the right hand side. 199 00:10:09,100 --> 00:10:18,390 So this is your nonlinear m2L theta 200 00:10:18,390 --> 00:10:26,060 double dot cos theta minus theta dot 201 00:10:26,060 --> 00:10:35,760 squared sine theta on the right hand side. 202 00:10:35,760 --> 00:10:39,290 And I think I need a minus if I do that. 203 00:10:39,290 --> 00:10:41,030 Now, what I do in this problem, I 204 00:10:41,030 --> 00:10:45,375 say let's let theta dot equals omega, and that's a constant. 205 00:10:49,640 --> 00:10:54,050 And so that says theta double dot is 0. 206 00:10:54,050 --> 00:10:57,080 When you do that, then this term goes away, 207 00:10:57,080 --> 00:11:02,220 and this becomes omega squared minus times a minus. 208 00:11:02,220 --> 00:11:14,570 This equals 1/2 m2L omega squared sine. 209 00:11:14,570 --> 00:11:16,685 And theta is just omega t. 210 00:11:20,280 --> 00:11:24,870 So by forcing what was previously an unknown variable 211 00:11:24,870 --> 00:11:27,720 that you had the solve in this-- you 212 00:11:27,720 --> 00:11:29,610 had two equations and two unknowns 213 00:11:29,610 --> 00:11:32,730 because you had two generalized coordinates 214 00:11:32,730 --> 00:11:36,970 it took to describe the motion of this thing, x and theta. 215 00:11:36,970 --> 00:11:39,690 Now, once you prescribe theta and it's given, 216 00:11:39,690 --> 00:11:41,530 it's no longer a variable. 217 00:11:41,530 --> 00:11:45,590 It's no longer a coordinate in your generalized-- 218 00:11:45,590 --> 00:11:48,260 in your equations of motion that you have to solve for, 219 00:11:48,260 --> 00:11:52,270 and it reduces this equation to the equation of a single degree 220 00:11:52,270 --> 00:11:54,110 of freedom system. 221 00:11:54,110 --> 00:11:56,280 On the left hand side is the response quantities 222 00:11:56,280 --> 00:12:01,200 just like usual-- mx double dot plus bx dot plus kx. 223 00:12:01,200 --> 00:12:04,370 And on the right hand side is our harmonic excitation, 224 00:12:04,370 --> 00:12:09,620 and that's this unbalanced mass going around and around. 225 00:12:09,620 --> 00:12:11,760 So this has the form here. 226 00:12:11,760 --> 00:12:14,650 This is some-- if you want to think of it this way, 227 00:12:14,650 --> 00:12:22,190 this is some F0, and it looks like sine omega t. 228 00:12:22,190 --> 00:12:24,610 So this is just a single degree of freedom system excited 229 00:12:24,610 --> 00:12:27,120 by harmonic excitation, and you know 230 00:12:27,120 --> 00:12:29,022 it has a transfer function, and you 231 00:12:29,022 --> 00:12:30,730 know that there's going to be a resonance 232 00:12:30,730 --> 00:12:32,930 at the natural frequency of the system. 233 00:12:36,980 --> 00:12:39,660 And if you don't like working with sine omega t, 234 00:12:39,660 --> 00:12:43,750 you could say, well, let's measure omega theta from here 235 00:12:43,750 --> 00:12:47,760 if you want to, and now that's cosine omega t. 236 00:12:47,760 --> 00:12:58,500 OK, so that's sort of the set up. 237 00:13:08,100 --> 00:13:11,740 Let me say one-- if I asked you to solve 238 00:13:11,740 --> 00:13:20,620 for the magnitude of the response x1 here, 239 00:13:20,620 --> 00:13:22,010 how would you do it? 240 00:13:26,530 --> 00:13:29,240 So this is now a single degree of freedom system excited 241 00:13:29,240 --> 00:13:30,990 by harmonic excitation. 242 00:13:30,990 --> 00:13:32,795 Steady state response-- how do you do it? 243 00:13:39,620 --> 00:13:42,910 This you do have to know. 244 00:13:42,910 --> 00:13:44,202 AUDIENCE: Transfer function. 245 00:13:44,202 --> 00:13:45,660 PROFESSOR: Use a transfer function. 246 00:13:45,660 --> 00:13:48,020 So this going to be the magnitude of the force 247 00:13:48,020 --> 00:13:55,392 times the magnitude of the HxF of omega transfer function, 248 00:13:55,392 --> 00:13:57,600 and that's how you get the magnitude of the response, 249 00:13:57,600 --> 00:13:59,170 and it happens to be a resonance. 250 00:13:59,170 --> 00:14:03,050 Then you'd be at the peak, and if you're not a resonance, 251 00:14:03,050 --> 00:14:05,496 wherever you happen to be in frequency 252 00:14:05,496 --> 00:14:07,562 is where you would evaluate this, 253 00:14:07,562 --> 00:14:08,645 and there's your response. 254 00:14:12,920 --> 00:14:22,670 Now, a few years ago-- every year-- twice a year, 255 00:14:22,670 --> 00:14:24,760 in fact-- and they're coming up in January, 256 00:14:24,760 --> 00:14:28,070 we have doctoral exams for students 257 00:14:28,070 --> 00:14:31,130 who want to do PhDs in mechanical engineering, 258 00:14:31,130 --> 00:14:34,010 and most departments at MIT have these also. 259 00:14:34,010 --> 00:14:38,900 And in the dynamics portion, there's an oral exam part. 260 00:14:38,900 --> 00:14:42,990 There's a written exam, also, but in the oral exam 261 00:14:42,990 --> 00:14:48,380 in this particular year, gave a single degree 262 00:14:48,380 --> 00:14:49,310 of freedom system. 263 00:14:59,980 --> 00:15:02,480 And we posed-- and we all know that, 264 00:15:02,480 --> 00:15:07,590 if we excite this thing with a harmonic excitation, 265 00:15:07,590 --> 00:15:15,830 some F1 cosine-- and I will use complex notation, 266 00:15:15,830 --> 00:15:19,750 because we're going to need it in a minute-- some F1 e 267 00:15:19,750 --> 00:15:21,550 to the i omega t. 268 00:15:21,550 --> 00:15:23,200 That's the excitation. 269 00:15:23,200 --> 00:15:26,210 We know what the steady state response of this looks like. 270 00:15:26,210 --> 00:15:29,070 The magnitude looks like that transfer function. 271 00:15:31,670 --> 00:15:35,794 But we posed-- so most would be doctoral students 272 00:15:35,794 --> 00:15:38,210 would know all about single degree of freedom [INAUDIBLE]. 273 00:15:38,210 --> 00:15:39,790 So we posed the following question. 274 00:15:39,790 --> 00:15:42,060 Well, we know that the response of this 275 00:15:42,060 --> 00:15:47,250 is going to do something like that, 276 00:15:47,250 --> 00:15:50,060 and you evaluate this at whatever frequency 277 00:15:50,060 --> 00:15:53,242 you're interested in, including right at resonance. 278 00:15:53,242 --> 00:15:54,700 And the question we asked is, is it 279 00:15:54,700 --> 00:16:06,210 possible to add a second spring and a second mass, m2 and k2, 280 00:16:06,210 --> 00:16:31,720 such that-- so is it possible to pick a k2 and an m2, 281 00:16:31,720 --> 00:16:34,660 such that, with this excitation on here, 282 00:16:34,660 --> 00:16:36,170 the motion of that thing will be 0? 283 00:16:38,940 --> 00:16:40,890 And in order to solve this problem, 284 00:16:40,890 --> 00:16:45,540 as soon as you put this on here and assume it's sliding along-- 285 00:16:45,540 --> 00:16:46,250 it can't fall. 286 00:16:46,250 --> 00:16:49,380 It only has horizontal possible motion, 287 00:16:49,380 --> 00:16:53,394 and we'd give it some coordinate describing its motion x2. 288 00:16:53,394 --> 00:16:55,810 So now how many degrees of freedom does this problem have? 289 00:16:58,520 --> 00:16:59,020 Two. 290 00:16:59,020 --> 00:17:02,830 How many equations of motion do you expect? 291 00:17:02,830 --> 00:17:04,970 How many peaks in a transfer function 292 00:17:04,970 --> 00:17:05,970 would you expect to see? 293 00:17:05,970 --> 00:17:08,040 Two resonances. 294 00:17:08,040 --> 00:17:12,160 So now we need to know how to find 295 00:17:12,160 --> 00:17:14,480 transfer functions for a multiple degree of freedom 296 00:17:14,480 --> 00:17:16,140 system. 297 00:17:16,140 --> 00:17:18,950 enough in everything I've said about two degree of freedom 298 00:17:18,950 --> 00:17:23,990 systems, everything is generalizable to n 299 00:17:23,990 --> 00:17:25,244 degrees of freedom. 300 00:17:25,244 --> 00:17:26,660 Though, what I'm going to show you 301 00:17:26,660 --> 00:17:30,205 now is how to do transfer functions for multi-degree 302 00:17:30,205 --> 00:17:31,580 of freedom systems, and I'm going 303 00:17:31,580 --> 00:17:35,000 to do it by way of example of a two a degree of freedom one, 304 00:17:35,000 --> 00:17:40,210 but just keep in mind that you can completely generalize this. 305 00:17:40,210 --> 00:17:49,090 So there is my system-- two masses, two springs-- and I 306 00:17:49,090 --> 00:17:53,420 could even have dash spots in here-- a c1, 307 00:17:53,420 --> 00:17:54,356 and I might have a c2. 308 00:18:01,300 --> 00:18:05,110 And I could even, in general, additionally have 309 00:18:05,110 --> 00:18:07,220 a force acting on this second mass, which 310 00:18:07,220 --> 00:18:11,580 I'll call F2e to the I omega t. 311 00:18:11,580 --> 00:18:14,118 So that's the completely general problem now. 312 00:18:18,700 --> 00:18:27,820 Now, the equations of motion for this system 313 00:18:27,820 --> 00:18:29,660 you could write down. 314 00:18:29,660 --> 00:18:32,810 We've done it many times now-- this particular system 315 00:18:32,810 --> 00:18:35,630 even-- but we could write them in matrix form 316 00:18:35,630 --> 00:18:45,370 as mx double dot plus cx dot plus k. 317 00:18:45,370 --> 00:18:50,170 Stiffness matrix x equals and excite 318 00:18:50,170 --> 00:18:53,280 the excitation in this two degree-- well, 319 00:18:53,280 --> 00:18:55,030 I'll just keep this is completely general. 320 00:18:55,030 --> 00:19:00,800 This is any n degree of freedom system, some Fe to the i omega 321 00:19:00,800 --> 00:19:02,800 t. 322 00:19:02,800 --> 00:19:04,740 So here's kind of an important point. 323 00:19:08,000 --> 00:19:11,920 We solve for the motion of a multiple degree of freedom 324 00:19:11,920 --> 00:19:14,200 system to a harmonic input. 325 00:19:14,200 --> 00:19:17,140 We do it one frequency at a time. 326 00:19:17,140 --> 00:19:18,820 So this doesn't mean that I could 327 00:19:18,820 --> 00:19:23,090 have force on the first mass at omega 1 328 00:19:23,090 --> 00:19:26,900 or at some omega, and force of the second mass 329 00:19:26,900 --> 00:19:28,090 at some other omega. 330 00:19:28,090 --> 00:19:29,720 I'd never do that at the same time. 331 00:19:29,720 --> 00:19:30,950 It's a linear system. 332 00:19:30,950 --> 00:19:32,690 Superposition holds. 333 00:19:32,690 --> 00:19:35,320 You do things one frequency at a time, get the answer, 334 00:19:35,320 --> 00:19:37,230 and if you have another frequency part, 335 00:19:37,230 --> 00:19:39,980 you do that separately, then add the two answers. 336 00:19:39,980 --> 00:19:41,320 So this is assumed. 337 00:19:41,320 --> 00:19:43,070 All the forces acting on the masses 338 00:19:43,070 --> 00:19:45,146 are assumed to occur at one frequency, 339 00:19:45,146 --> 00:19:46,770 but you can't have different amplitudes 340 00:19:46,770 --> 00:19:49,047 on the different masses. 341 00:19:49,047 --> 00:19:50,130 So that's what that means. 342 00:19:53,060 --> 00:20:03,960 So this then for the two degree of freedom system-- F, 343 00:20:03,960 --> 00:20:11,310 for examplet-- F of t would be some constant F1. 344 00:20:11,310 --> 00:20:16,110 Another constant magnitude, F2e to the i omega t. 345 00:20:16,110 --> 00:20:17,870 That's what we mean by that. 346 00:20:22,650 --> 00:20:25,540 Actually, if you do that, if these are just constants, 347 00:20:25,540 --> 00:20:30,450 some constant force vector applied at a single frequency, 348 00:20:30,450 --> 00:20:32,720 what frequency do you expect to see 349 00:20:32,720 --> 00:20:34,850 in the response of the system? 350 00:20:34,850 --> 00:20:36,540 Steady state response. 351 00:20:36,540 --> 00:20:38,940 AUDIENCE: Driving frequency. 352 00:20:38,940 --> 00:20:39,876 Driving frequency. 353 00:20:39,876 --> 00:20:42,000 PROFESSOR: Right, it's a linear system more or less 354 00:20:42,000 --> 00:20:43,870 since you really want to go away with. 355 00:20:43,870 --> 00:20:45,410 This is an intralinear system. 356 00:20:45,410 --> 00:20:48,040 Steady state response of a linear system-- 357 00:20:48,040 --> 00:20:50,045 the frequency in is the frequency out. 358 00:20:53,730 --> 00:20:55,290 Always true. 359 00:20:55,290 --> 00:20:57,285 So we expect to see a solution. 360 00:21:03,700 --> 00:21:08,360 We're going to get a solution of the form x 361 00:21:08,360 --> 00:21:12,650 equals some magnitude vector times an e 362 00:21:12,650 --> 00:21:14,670 to the i omega t also. 363 00:21:14,670 --> 00:21:16,790 And the magnitude might be complex, 364 00:21:16,790 --> 00:21:20,100 because it there be phase angles there due to damping 365 00:21:20,100 --> 00:21:23,000 and so forth, but nonetheless, they're constants. 366 00:21:23,000 --> 00:21:25,660 This part is a constant vector, and that's 367 00:21:25,660 --> 00:21:27,680 its frequency dependence. 368 00:21:27,680 --> 00:21:29,890 And if we know that is true, then we can say, 369 00:21:29,890 --> 00:21:32,330 well, take the time derivative of that. 370 00:21:32,330 --> 00:21:35,840 The only time dependent part is the e to i omega t. 371 00:21:35,840 --> 00:21:46,030 So x dot becomes i omega xe to the i omega t. 372 00:21:46,030 --> 00:21:54,500 And x double dot becomes minus omega squared xe 373 00:21:54,500 --> 00:21:57,660 to the i omega t. 374 00:21:57,660 --> 00:22:05,203 So we can substitute this, this, and this into this equation. 375 00:22:05,203 --> 00:22:06,786 Of course, these are matrix equations. 376 00:22:16,910 --> 00:22:21,030 And when you do that, minus omega 377 00:22:21,030 --> 00:22:26,870 squared times the mass matrix i omega times the c 378 00:22:26,870 --> 00:22:42,990 matrix plus k xe to the i omega t equals Fe to the i omega t. 379 00:22:42,990 --> 00:22:45,010 You can get rid of these, and now you 380 00:22:45,010 --> 00:22:51,610 have a algebraic equation that's no longer a function of time-- 381 00:22:51,610 --> 00:22:53,340 function of your original mass, damping, 382 00:22:53,340 --> 00:22:55,010 and stiffness matrices. 383 00:22:55,010 --> 00:22:57,050 And it's certainly a function of frequency. 384 00:22:57,050 --> 00:22:58,700 And if you think back, this is how 385 00:22:58,700 --> 00:23:01,570 we derive the transfer function for a single degree of freedom 386 00:23:01,570 --> 00:23:02,770 system. 387 00:23:02,770 --> 00:23:05,314 And this statement is true for single degree of freedom, too. 388 00:23:05,314 --> 00:23:07,480 With single degree of freedom, that's just the mass. 389 00:23:07,480 --> 00:23:10,800 That's just the damping, and that's just the stiffness. 390 00:23:10,800 --> 00:23:14,939 And you could solve directly for the hx/f transfer function, 391 00:23:14,939 --> 00:23:16,480 but with multiple degrees of freedom, 392 00:23:16,480 --> 00:23:20,150 we can't just quite divide this out. 393 00:23:20,150 --> 00:23:25,460 This piece here is known as the-- I'll write 394 00:23:25,460 --> 00:23:27,540 it is z omega-- z of omega. 395 00:23:27,540 --> 00:23:29,670 This is known as the impedance matrix. 396 00:23:38,980 --> 00:23:41,470 And if I want to solve, I'm looking for x. 397 00:23:41,470 --> 00:23:43,370 I want to know my solution x. 398 00:23:43,370 --> 00:23:46,496 So this is essentially of the form z. 399 00:23:46,496 --> 00:23:54,600 It's a matrix times x, a vector, equals F, a vector. 400 00:23:54,600 --> 00:23:57,670 And just using what you know about linear algebra, 401 00:23:57,670 --> 00:24:01,730 to solve for x, you just multiply through by z inverse. 402 00:24:01,730 --> 00:24:14,350 So x equals z inverse F. All right? 403 00:24:14,350 --> 00:24:18,860 And z is a two degree of freedom system. z is 404 00:24:18,860 --> 00:24:23,090 a two by two matrix with frequency and everything in it, 405 00:24:23,090 --> 00:24:25,220 but it's a two by two. 406 00:24:25,220 --> 00:24:29,010 So z inverse will be a two by two. 407 00:24:29,010 --> 00:24:33,170 And for two by two, we can just write down the answer, 408 00:24:33,170 --> 00:24:33,930 but let's see. 409 00:24:33,930 --> 00:24:35,870 I'll write out z of omega here. 410 00:24:47,180 --> 00:24:51,550 So just to be clear about what all this is, 411 00:24:51,550 --> 00:24:56,490 z of omega is this, and in this problem, 412 00:24:56,490 --> 00:25:01,330 that's minus omega squared times a mass matrix, which 413 00:25:01,330 --> 00:25:06,990 looks like m1, 0, 0, m2. 414 00:25:06,990 --> 00:25:14,830 And you add to that a damping matrix, i omega, times c. 415 00:25:14,830 --> 00:25:17,550 The c is from the equations of motion. 416 00:25:17,550 --> 00:25:28,100 c1 plus c2, minus c2, minus c2, c2, and plus our k matrix. 417 00:25:31,110 --> 00:25:41,150 k1 plus k2, minus k2, minus k2 k2. 418 00:25:41,150 --> 00:25:45,680 So that's what the z of omega actually looks like. 419 00:25:45,680 --> 00:25:50,310 And so you would collect-- so it's a two by two matrix, 420 00:25:50,310 --> 00:25:54,670 and its upper left term is minus omega squared m1 plus i 421 00:25:54,670 --> 00:25:58,410 omega times this plus that. 422 00:25:58,410 --> 00:26:01,150 Collect them all together. 423 00:26:01,150 --> 00:26:04,420 And this has a form-- this collects together 424 00:26:04,420 --> 00:26:13,010 in a form we call z11, z12, z21, z22. 425 00:26:16,170 --> 00:26:21,560 And very often z12 and z21 are symmetric. 426 00:26:21,560 --> 00:26:22,800 Not always. 427 00:26:22,800 --> 00:26:25,130 The kind of problems we generally do here, 428 00:26:25,130 --> 00:26:28,270 they will be, and it will be generally symmetric 429 00:26:28,270 --> 00:26:34,260 if your coordinate system is measured 430 00:26:34,260 --> 00:26:37,240 from a static equilibrium position. 431 00:26:37,240 --> 00:26:39,810 So if you've been doing 2001, there's 432 00:26:39,810 --> 00:26:42,460 a thing called Maxwell's reciprocal theorem, 433 00:26:42,460 --> 00:26:46,151 which proves this for the stiffness matrix. 434 00:26:46,151 --> 00:26:47,650 They essentially need to be measured 435 00:26:47,650 --> 00:26:50,060 from static equilibrium positions, 436 00:26:50,060 --> 00:26:52,860 but for today's example, this is indeed symmetric. 437 00:26:52,860 --> 00:26:56,850 Minus c2, minus c2, minus k2, minus k2. 438 00:26:56,850 --> 00:26:57,790 0, 0. 439 00:26:57,790 --> 00:26:59,680 Everything is symmetric. 440 00:26:59,680 --> 00:27:04,550 So we'll take advantage of that, and I'll write out 441 00:27:04,550 --> 00:27:06,800 one of the components here-- z11, 442 00:27:06,800 --> 00:27:13,970 for example, when you collect the terms together, 443 00:27:13,970 --> 00:27:26,260 is in general, it would be minus omega squared m11 plus i 444 00:27:26,260 --> 00:27:36,430 omega c11 plus k11, the corner elements of these three pieces. 445 00:27:36,430 --> 00:27:52,150 And in this problem, that's minus omega squared m1 plus i 446 00:27:52,150 --> 00:28:03,230 omega c1 plus c2 plus k1 plus k2. 447 00:28:03,230 --> 00:28:09,070 You just substitute in this, this, and this. 448 00:28:09,070 --> 00:28:11,850 So that's z11, for example, and the other ones 449 00:28:11,850 --> 00:28:15,010 you can figure out what each of the other terms would be. 450 00:28:28,990 --> 00:28:34,130 All right, so we've said that we want to know what the x's are. 451 00:28:34,130 --> 00:28:41,140 And we know that we can get that by doing z inverse times F. 452 00:28:41,140 --> 00:28:44,030 And this gives us our definition. 453 00:28:44,030 --> 00:28:52,000 This z inverse is our H, our transfer function matrix, 454 00:28:52,000 --> 00:28:57,060 times F. So just by getting z inverse, 455 00:28:57,060 --> 00:29:00,240 we get this little set of four transfer functions 456 00:29:00,240 --> 00:29:03,580 that we're interested in. 457 00:29:03,580 --> 00:29:14,710 So H is an N by N matrix of transfer functions. 458 00:29:24,960 --> 00:29:30,910 So in a two by two-- or for the two by two case, 459 00:29:30,910 --> 00:29:34,170 our N equals 2. 460 00:29:34,170 --> 00:29:39,690 We know we can write out directly what z inverse is. 461 00:29:39,690 --> 00:29:56,530 And z inverse-- z22, minus z12, minus z12 when it's symmetric, 462 00:29:56,530 --> 00:30:04,880 z11, all over-- how should I write it? 463 00:30:04,880 --> 00:30:08,882 All over the determinant of z. 464 00:30:08,882 --> 00:30:10,115 So this is the determinant. 465 00:30:16,120 --> 00:30:17,640 So the determinant of z [INAUDIBLE] 466 00:30:17,640 --> 00:30:18,760 do these double bars. 467 00:30:18,760 --> 00:30:23,580 So this over the determinant is the inverse of that matrix, 468 00:30:23,580 --> 00:30:27,940 and this gives you a new result, which 469 00:30:27,940 --> 00:30:52,760 is H11, H12, H12, H22, where Hij-- this is the response at i 470 00:30:52,760 --> 00:31:00,762 per unit force at j. 471 00:31:00,762 --> 00:31:04,480 And I've made this-- I took advantage of the symmetry here, 472 00:31:04,480 --> 00:31:06,990 but this one would normally be called 21, 473 00:31:06,990 --> 00:31:10,500 so this is response at 1 caused by the force at 1. 474 00:31:10,500 --> 00:31:14,230 So we have forces in our problem. 475 00:31:14,230 --> 00:31:16,610 One is the force on the main mass. 476 00:31:16,610 --> 00:31:20,660 So how much response do you get at the main first mass 477 00:31:20,660 --> 00:31:22,545 due to the force on the first mass? 478 00:31:22,545 --> 00:31:23,920 This is how much response you get 479 00:31:23,920 --> 00:31:27,880 on the first mass due to the excitation in the second mass. 480 00:31:27,880 --> 00:31:31,400 Response on the second mass due to the excitation on the first. 481 00:31:31,400 --> 00:31:34,240 Response in the second mass due to the excitation 482 00:31:34,240 --> 00:31:35,240 on the second mass. 483 00:31:35,240 --> 00:31:38,300 So you get four possible contributions here. 484 00:31:42,790 --> 00:31:50,830 And the determinant of z for a two by two 485 00:31:50,830 --> 00:31:53,100 is also very straightforward. 486 00:31:53,100 --> 00:31:59,520 z11, z22, minus z12 squared. 487 00:32:03,160 --> 00:32:09,020 So we can and will work out exactly what the algebra tells 488 00:32:09,020 --> 00:32:11,040 us for this two by two case. 489 00:32:13,600 --> 00:32:17,076 So we've already been discussing it, but what do you expect. 490 00:32:17,076 --> 00:32:18,450 Let's say let's look at this one. 491 00:32:18,450 --> 00:32:21,660 What do you suppose the response at 1 492 00:32:21,660 --> 00:32:29,470 due to a harmonic excitation at 1-- 493 00:32:29,470 --> 00:32:36,690 what do you think the sketch of the magnitude of H11 of omega 494 00:32:36,690 --> 00:32:39,050 would look like as a function a frequency? 495 00:32:41,900 --> 00:32:45,230 AUDIENCE: Will C12 and C21 always be symmetric? 496 00:32:45,230 --> 00:32:49,510 PROFESSOR: No, but for the kind of problems we do, probably. 497 00:32:49,510 --> 00:32:54,060 And if it's simple beams, or masses connected by springs, 498 00:32:54,060 --> 00:32:58,700 if you choose your generalized coordinates in a way that 499 00:32:58,700 --> 00:33:03,950 is, for example, measuring the displacement of each mass 500 00:33:03,950 --> 00:33:08,280 from an inertial static equilibrium starting point, 501 00:33:08,280 --> 00:33:10,930 it'll be symmetric. 502 00:33:10,930 --> 00:33:13,630 But if you measure the-- even in this two degree of freedom, 503 00:33:13,630 --> 00:33:18,370 I can make it non-symmetric just by choosing the coordinate 504 00:33:18,370 --> 00:33:22,030 for the second mass as to be relative to the first mass. 505 00:33:22,030 --> 00:33:24,110 Soon as you do that, it makes it non-symmetric. 506 00:33:24,110 --> 00:33:26,590 You can still solve for transfer function and everything, 507 00:33:26,590 --> 00:33:29,070 but it gets messier. 508 00:33:29,070 --> 00:33:32,840 Though, notice I picked coordinates-- x1 relative 509 00:33:32,840 --> 00:33:35,970 to the inertial frame, x2 relative to the same inertial 510 00:33:35,970 --> 00:33:38,710 frame-- and they're both measuring displacement 511 00:33:38,710 --> 00:33:41,630 from a static equilibrium position. 512 00:33:41,630 --> 00:33:45,126 Then they are symmetric. 513 00:33:45,126 --> 00:33:46,625 And there's the determinant we need, 514 00:33:46,625 --> 00:33:50,630 and this determinant-- see, this is divided into all four 515 00:33:50,630 --> 00:33:52,560 of these terms. 516 00:33:52,560 --> 00:33:56,845 So these four transfer functions all share the same denominator. 517 00:34:00,430 --> 00:34:02,700 They all have the same denominator. 518 00:34:02,700 --> 00:34:04,884 So I just want you to use your intuition. 519 00:34:04,884 --> 00:34:06,550 Tell me what this is going to look like. 520 00:34:10,870 --> 00:34:15,050 What's it look like for a single degree of freedom system? 521 00:34:15,050 --> 00:34:17,679 It's HxF for a single degree of freedom-- 522 00:34:17,679 --> 00:34:20,679 it's H11 for a single degree of freedom system 523 00:34:20,679 --> 00:34:22,800 as a response of those systems to a harmonic 524 00:34:22,800 --> 00:34:27,810 force on that single mass, and that looks like what? 525 00:34:27,810 --> 00:34:28,971 Peak, right? 526 00:34:28,971 --> 00:34:31,179 This one-- what do you think it's going to look like? 527 00:34:35,780 --> 00:34:37,820 Show me again. 528 00:34:37,820 --> 00:34:39,380 OK, but I see one peak. 529 00:34:39,380 --> 00:34:40,219 How many peaks? 530 00:34:40,219 --> 00:34:40,909 AUDIENCE: Two. 531 00:34:40,909 --> 00:34:41,893 PROFESSOR: Why? 532 00:34:41,893 --> 00:34:42,877 AUDIENCE: Because it's two degrees of freedom. 533 00:34:42,877 --> 00:34:44,820 PROFESSOR: Two different natural frequencies. 534 00:34:44,820 --> 00:34:50,400 This thing is going to look something like this, 535 00:34:50,400 --> 00:34:53,195 and I don't know quite how it behaves in here, 536 00:34:53,195 --> 00:34:55,500 but it's going to do that for sure. 537 00:34:55,500 --> 00:34:57,970 It's going to have two resonances-- this one at omega 538 00:34:57,970 --> 00:35:00,180 1, this one at omega 2. 539 00:35:00,180 --> 00:35:02,800 We already know that because we did it by modal analysis. 540 00:35:02,800 --> 00:35:05,210 We know each mode is going to have a resonance in it. 541 00:35:05,210 --> 00:35:10,400 OK, so let's find out exactly what it looks like. 542 00:35:19,140 --> 00:35:22,150 So for this two degree of freedom system, 543 00:35:22,150 --> 00:35:26,275 there's an F1 and an F2, and we're looking for x1. 544 00:35:28,880 --> 00:35:48,490 So x1 x2 here equals H11, H12, H12, H22 times F1 F2. 545 00:35:48,490 --> 00:36:00,727 So if I want to solve for x1, H11 F1 plus H12 F2. 546 00:36:00,727 --> 00:36:02,310 And for the problem I'm going to solve 547 00:36:02,310 --> 00:36:10,030 today, I'm going to let F2 be 0, just to keep it simple, 548 00:36:10,030 --> 00:36:12,310 but also to address the original question 549 00:36:12,310 --> 00:36:14,390 we ask the doctoral students. 550 00:36:14,390 --> 00:36:16,900 You have a force on the first mass. 551 00:36:16,900 --> 00:36:20,540 Can I add a mass and spring to it in such a way 552 00:36:20,540 --> 00:36:23,860 that I can make the response of the first mass go to 0. 553 00:36:23,860 --> 00:36:26,710 So there is no second force. 554 00:36:26,710 --> 00:36:31,030 There's only a first force, and so this term goes away. 555 00:36:31,030 --> 00:36:37,580 And so this problem, x1, is H11 F1. 556 00:36:37,580 --> 00:36:44,360 Therefore, we need to know what H11 looks like. 557 00:36:48,940 --> 00:36:58,290 So H11 is going to be z22 divided by the determinant. 558 00:36:58,290 --> 00:37:15,530 So x1 is going to be z22 over Z11, Z22 minus Z12 squared 559 00:37:15,530 --> 00:37:16,620 times F1. 560 00:37:30,760 --> 00:37:38,230 And actually I can do this so-- and I'm 561 00:37:38,230 --> 00:37:43,780 going to let, for now, damping be 0, 562 00:37:43,780 --> 00:37:46,620 because it also simplifies it for the purposes 563 00:37:46,620 --> 00:37:47,790 of illustration today. 564 00:37:47,790 --> 00:37:50,480 And that's how I was posed to the doctoral student. 565 00:37:50,480 --> 00:37:51,730 Didn't even show any damping. 566 00:37:51,730 --> 00:37:56,110 It was just a simple system, a string and a mass. 567 00:37:56,110 --> 00:37:56,990 How can you do this? 568 00:37:56,990 --> 00:37:59,300 So we'll let the damping be 0 and that considerably 569 00:37:59,300 --> 00:38:00,600 simplifies things then. 570 00:38:19,260 --> 00:38:32,710 So Z11 is minus omega squared m1 plus k1 plus k2. 571 00:38:36,540 --> 00:38:45,430 z22 minus omega squared m2 plus k2. 572 00:38:49,350 --> 00:39:02,330 And z12 just minus k2. 573 00:39:02,330 --> 00:39:05,340 So now these are pretty simple expressions, 574 00:39:05,340 --> 00:39:07,630 and all I have to do is plug them in here. 575 00:39:07,630 --> 00:39:12,070 And notice what's going to happen. z11 times z22-- that's 576 00:39:12,070 --> 00:39:14,710 something involving omega squared 577 00:39:14,710 --> 00:39:18,170 times another term involving omega squared. 578 00:39:18,170 --> 00:39:20,990 It's going to give you a polynomial 579 00:39:20,990 --> 00:39:22,990 and omega to the fourth. 580 00:39:22,990 --> 00:39:25,190 This is the polynomial that has two roots, 581 00:39:25,190 --> 00:39:26,195 and the two routes are? 582 00:39:29,824 --> 00:39:31,270 AUDIENCE: [INAUDIBLE] frequencies. 583 00:39:31,270 --> 00:39:33,280 PROFESSOR: Yeah, natural frequencies. 584 00:39:33,280 --> 00:39:38,010 The determinant of the z matrix is the same problem 585 00:39:38,010 --> 00:39:41,510 you solve when you've found the roots 586 00:39:41,510 --> 00:39:44,450 of the characteristic equation. 587 00:39:44,450 --> 00:39:46,400 It is the characteristic equation 588 00:39:46,400 --> 00:39:49,250 for the system, the denominator, the determinant 589 00:39:49,250 --> 00:39:50,670 of the z matrix. 590 00:39:50,670 --> 00:39:53,220 Therefore it's going to be a fourth order equation in omega. 591 00:39:53,220 --> 00:39:55,870 It'll give you two routes for omega squared. 592 00:39:55,870 --> 00:39:58,690 They're your two undamped natural frequencies 593 00:39:58,690 --> 00:40:01,390 when we leave damping out of this expression, 594 00:40:01,390 --> 00:40:04,050 and they'd be damped-- the damped expressions-- 595 00:40:04,050 --> 00:40:07,500 they'd have complex stuff if you leave in the damping. 596 00:40:07,500 --> 00:40:11,530 But they give us our two roots, and they go in the denominator. 597 00:40:11,530 --> 00:40:12,501 Yeah? 598 00:40:12,501 --> 00:40:13,914 AUDIENCE: If we set the denominator or the determinant 599 00:40:13,914 --> 00:40:15,562 of the matrix equal to 0 and solve, 600 00:40:15,562 --> 00:40:16,980 you get the modal frequencies? 601 00:40:16,980 --> 00:40:19,450 PROFESSOR: Yeah. 602 00:40:19,450 --> 00:40:21,920 That denominator is the characteristic equation 603 00:40:21,920 --> 00:40:26,260 that we did when we set it up to do the natural frequencies. 604 00:40:26,260 --> 00:40:28,230 Same equation. 605 00:40:28,230 --> 00:40:32,136 So we know that this thing has two roots, one at each 606 00:40:32,136 --> 00:40:33,260 of the natural frequencies. 607 00:40:33,260 --> 00:40:37,440 And when omega is at a natural frequency, what 608 00:40:37,440 --> 00:40:39,870 is the value of this characteristic equation? 609 00:40:43,140 --> 00:40:43,924 Well, no. 610 00:40:43,924 --> 00:40:45,090 The characteristic equation. 611 00:40:45,090 --> 00:40:47,740 Think of the denominator. 612 00:40:47,740 --> 00:40:50,350 Add a root, that expression goes to? 613 00:40:50,350 --> 00:40:50,850 0. 614 00:40:50,850 --> 00:40:52,120 So you're dividing by 0. 615 00:40:52,120 --> 00:40:54,146 That's what causes the peaks. 616 00:40:54,146 --> 00:40:55,520 That's where the peaks come from. 617 00:40:55,520 --> 00:40:59,180 That's where the denominator goes to 0, 618 00:40:59,180 --> 00:41:01,530 and that's at the natural frequencies. 619 00:41:01,530 --> 00:41:14,760 OK, so x1 for this problem equals H11F1, 620 00:41:14,760 --> 00:41:17,450 and now we can just write it out. 621 00:41:17,450 --> 00:41:30,460 H11 is minus omega squared m2 plus k2 times F1. 622 00:41:30,460 --> 00:41:33,960 And the denominator, the characteristic equation, 623 00:41:33,960 --> 00:41:41,570 is minus omega squared m1 plus k1 624 00:41:41,570 --> 00:41:47,610 plus k2 times minus omega squared 625 00:41:47,610 --> 00:41:57,160 m2 plus k2 minus k2 squared. 626 00:41:57,160 --> 00:41:59,240 So this whole thing down here-- there's 627 00:41:59,240 --> 00:42:00,500 your characteristic equation. 628 00:42:00,500 --> 00:42:01,333 You multiply it out. 629 00:42:01,333 --> 00:42:03,620 You get your fourth order polynomial. 630 00:42:03,620 --> 00:42:04,180 Solve it. 631 00:42:04,180 --> 00:42:07,300 You get two roots, but this is now 632 00:42:07,300 --> 00:42:09,990 a total expression for the response we were looking for. 633 00:42:15,900 --> 00:42:19,080 So right away this gives me the answer 634 00:42:19,080 --> 00:42:23,320 to the question that was posed to the doctoral students. 635 00:42:23,320 --> 00:42:29,230 So is there a value-- can you set k2 and m2 such 636 00:42:29,230 --> 00:42:33,480 that you can make the response of the system 0 637 00:42:33,480 --> 00:42:34,510 at a particular omega? 638 00:42:38,880 --> 00:42:43,370 All you have to do is make the numerator go to 0, right? 639 00:42:43,370 --> 00:42:48,030 So if you make this go to 0, x goes to 0. 640 00:42:48,030 --> 00:43:01,750 So x1 equals 0 when minus omega squared m2 plus k2 equals zero, 641 00:43:01,750 --> 00:43:09,200 and that happens when k2 over m2 equals omega squared. 642 00:43:16,870 --> 00:43:23,420 So how many-- once you set k2 and m2, how many frequencies 643 00:43:23,420 --> 00:43:25,342 does this happen at? 644 00:43:25,342 --> 00:43:27,620 At how many different operating points 645 00:43:27,620 --> 00:43:31,430 can you make the response of that main mass go to 0? 646 00:43:36,370 --> 00:43:38,040 Well, just one. 647 00:43:38,040 --> 00:43:42,550 Once you choose k1 and m2, whatever you've chosen to be, 648 00:43:42,550 --> 00:43:45,260 they give you some value of omega. 649 00:43:45,260 --> 00:43:49,500 And when I said I was kind of vague about what happens here, 650 00:43:49,500 --> 00:43:50,040 this is H11. 651 00:43:55,030 --> 00:43:56,030 Looks like that. 652 00:43:56,030 --> 00:44:03,615 Right here is when omega squared equals k2/m2. 653 00:44:12,086 --> 00:44:15,746 AUDIENCE: So when it's at the frequency that 654 00:44:15,746 --> 00:44:20,870 makes the numerator 0, wouldn't it also-- I don't remember. 655 00:44:20,870 --> 00:44:22,830 Would the denominator be 0 also? 656 00:44:22,830 --> 00:44:23,950 PROFESSOR: No. 657 00:44:23,950 --> 00:44:24,480 No. 658 00:44:24,480 --> 00:44:27,269 Because, look at the picture here. 659 00:44:27,269 --> 00:44:29,477 Where are the two natural frequencies of this system? 660 00:44:33,166 --> 00:44:34,790 There's one here, and there's one here. 661 00:44:34,790 --> 00:44:37,520 And that's when the denominator goes to 0. 662 00:44:37,520 --> 00:44:39,360 When the denominator goes to 0, the response 663 00:44:39,360 --> 00:44:41,830 goes to infinity with no damping. 664 00:44:41,830 --> 00:44:44,840 So the system has two natural frequencies-- one 665 00:44:44,840 --> 00:44:46,180 to either side of this point. 666 00:44:49,440 --> 00:44:57,890 So I've picked kind of a particular example 667 00:44:57,890 --> 00:45:01,305 to illustrate the use of a transfer function, 668 00:45:01,305 --> 00:45:02,930 and the fact that you can have transfer 669 00:45:02,930 --> 00:45:05,600 functions for multiple degree of freedom systems, 670 00:45:05,600 --> 00:45:08,940 and they essentially become transfer function matrices. 671 00:45:08,940 --> 00:45:12,850 And you use what you need, and for this problem, 672 00:45:12,850 --> 00:45:15,762 we needed H11. 673 00:45:15,762 --> 00:45:17,350 We worked it out. 674 00:45:17,350 --> 00:45:18,810 There it is. 675 00:45:18,810 --> 00:45:22,060 This is a complete equation that describes, for me, 676 00:45:22,060 --> 00:45:26,410 the behavior of mass 1 per unit input force 677 00:45:26,410 --> 00:45:28,310 as a function of frequency. 678 00:45:28,310 --> 00:45:32,226 And it will always have a point right there that it goes to 0. 679 00:45:35,130 --> 00:45:38,582 And we don't have any damping in here, so intuitively, 680 00:45:38,582 --> 00:45:40,540 what do you think damping will do to this plot? 681 00:45:43,850 --> 00:45:45,720 What's the first thing it does to the peaks? 682 00:45:45,720 --> 00:45:48,470 What? 683 00:45:48,470 --> 00:45:49,334 They become? 684 00:45:49,334 --> 00:45:50,250 AUDIENCE: [INAUDIBLE]. 685 00:45:50,250 --> 00:45:51,270 PROFESSOR: Finite. 686 00:45:51,270 --> 00:45:55,480 And so the small damping, you're going to have very high tops. 687 00:45:55,480 --> 00:45:57,830 More damping, they're going to be lower. 688 00:45:57,830 --> 00:46:00,580 So damping especially affects the peaks. 689 00:46:00,580 --> 00:46:04,020 Damping will also pull this off the bottom a little bit. 690 00:46:04,020 --> 00:46:05,510 It won't perfectly go to 0. 691 00:46:11,400 --> 00:46:14,980 So I want to do a very particular case. 692 00:46:14,980 --> 00:46:19,240 So an obvious one is your original system. 693 00:46:19,240 --> 00:46:24,110 It might be just a mass spring with a rotor in it 694 00:46:24,110 --> 00:46:27,350 that's unbalanced, and it's running near resonance, 695 00:46:27,350 --> 00:46:28,860 and it's vibrating like crazy. 696 00:46:28,860 --> 00:46:32,070 A single degree of freedom system vibrating like crazy. 697 00:46:32,070 --> 00:46:38,290 Can I put on a second mass spring and stop the motion? 698 00:46:38,290 --> 00:46:41,140 Well, we can theoretically. 699 00:46:41,140 --> 00:46:44,520 And we'll just say let's let the problem 700 00:46:44,520 --> 00:46:46,520 we're trying to solve-- that first system had 701 00:46:46,520 --> 00:46:47,380 a natural frequency. 702 00:46:47,380 --> 00:46:55,170 I'll call it cap omega n that was the square root k1 over m1, 703 00:46:55,170 --> 00:47:00,360 and I'm going to let that be equal to square root of k2 704 00:47:00,360 --> 00:47:02,240 over m2. 705 00:47:02,240 --> 00:47:05,260 So I've now chosen k2 and m2 such 706 00:47:05,260 --> 00:47:08,350 they're exactly at the natural frequency 707 00:47:08,350 --> 00:47:11,820 of the original single degree of freedom system. 708 00:47:11,820 --> 00:47:16,450 So this is the original system, k1 m1, 709 00:47:16,450 --> 00:47:19,130 and it has that natural frequency. 710 00:47:19,130 --> 00:47:25,890 And now I'm going to stick on a second mass, k2 m2, 711 00:47:25,890 --> 00:47:29,870 but the value k2/m2 of the second little system-- 712 00:47:29,870 --> 00:47:33,281 by itself, it has the same natural frequency as the first 713 00:47:33,281 --> 00:47:34,780 if you want to think of it that way. 714 00:47:34,780 --> 00:47:37,610 If I make this fixed, what's the natural frequency 715 00:47:37,610 --> 00:47:40,350 of that mass and spring? 716 00:47:40,350 --> 00:47:41,730 Same natural frequency. 717 00:47:41,730 --> 00:47:43,290 I'm just making it that way. 718 00:47:43,290 --> 00:47:46,690 So this is now my system. 719 00:47:46,690 --> 00:47:53,640 That's the parameters, and now we can-- 720 00:47:53,640 --> 00:47:58,300 I'm going to give you a plot of the response 721 00:47:58,300 --> 00:47:59,640 of that particular system. 722 00:48:22,000 --> 00:48:23,850 You know how we non-dimensionalize 723 00:48:23,850 --> 00:48:26,300 the single degree of freedom transfer function? 724 00:48:26,300 --> 00:48:29,415 We set x divided by the static motion. 725 00:48:29,415 --> 00:48:31,490 I want to do that again here. 726 00:48:31,490 --> 00:48:34,651 So let's look at this system. 727 00:48:34,651 --> 00:48:38,600 It's a two degree of freedom system. 728 00:48:38,600 --> 00:48:44,010 It has a force F1 on it, and as the frequency goes to 0 729 00:48:44,010 --> 00:48:52,910 and has displacement x1 and x2-- so at omega equals 0, 730 00:48:52,910 --> 00:48:56,120 x1 equals x1s. 731 00:48:56,120 --> 00:48:57,580 I'll call it x1 static. 732 00:48:57,580 --> 00:48:59,940 And how much is it? 733 00:48:59,940 --> 00:49:01,650 So what's the static displacement 734 00:49:01,650 --> 00:49:04,433 at 0 frequency of this system to that force? 735 00:49:12,381 --> 00:49:12,880 Come on. 736 00:49:12,880 --> 00:49:16,180 It's just a spring and a force. 737 00:49:16,180 --> 00:49:18,850 How much will that spring stretch 738 00:49:18,850 --> 00:49:21,204 when you put a force F1 on it? 739 00:49:21,204 --> 00:49:22,120 AUDIENCE: [INAUDIBLE]. 740 00:49:22,120 --> 00:49:23,400 PROFESSOR: F1/k. 741 00:49:23,400 --> 00:49:27,223 So x1 static equals F1/k1. 742 00:49:31,530 --> 00:49:33,850 And F2 is 0. 743 00:49:33,850 --> 00:49:34,860 There's no F2 force. 744 00:49:34,860 --> 00:49:38,480 So how much is x2 static? 745 00:49:38,480 --> 00:49:41,100 There's no forces on the second mass. 746 00:49:41,100 --> 00:49:44,676 How much will it move? 747 00:49:44,676 --> 00:49:45,638 AUDIENCE: [INAUDIBLE]. 748 00:49:45,638 --> 00:49:48,740 PROFESSOR: It just moves the same amount as a first mass. 749 00:49:48,740 --> 00:49:54,810 So this is x2 static, because the forces on the system 750 00:49:54,810 --> 00:49:56,170 are F1 and 0. 751 00:49:56,170 --> 00:49:57,250 There's no force in this. 752 00:49:57,250 --> 00:49:59,310 It's not going to do anything to that spring. 753 00:49:59,310 --> 00:50:01,430 It'll just move with the whole system. 754 00:50:12,420 --> 00:50:15,925 So I want to plot x1 over x1 static. 755 00:50:19,150 --> 00:50:25,040 In general, x1 is a function of frequency-- is F1 times-- 756 00:50:25,040 --> 00:50:27,250 and I'm going to do the magnitudes here-- 757 00:50:27,250 --> 00:50:29,595 times the magnitude of H11. 758 00:50:32,350 --> 00:50:39,680 And I'm going to divide that by F1 over k1. 759 00:50:39,680 --> 00:50:42,180 So the F1's go away. 760 00:50:42,180 --> 00:50:44,420 k1 comes in the numerator. 761 00:50:44,420 --> 00:50:46,500 This looks like k1 h11. 762 00:50:50,200 --> 00:51:00,340 So a plot of k1 magnitude h11. 763 00:51:00,340 --> 00:51:04,760 We know roughly what it's going to look like. 764 00:51:04,760 --> 00:51:07,441 It has a peak. 765 00:51:07,441 --> 00:51:09,570 It has a 0. 766 00:51:09,570 --> 00:51:10,890 It has another peak. 767 00:51:13,740 --> 00:51:16,960 Over here, this plot goes to 1. 768 00:51:16,960 --> 00:51:22,540 There's x1 over-- this is also x1 over x1 static. 769 00:51:22,540 --> 00:51:23,310 Same thing. 770 00:51:23,310 --> 00:51:24,370 It goes to 1. 771 00:51:27,100 --> 00:51:29,510 Here it goes to 0. 772 00:51:29,510 --> 00:51:33,030 Here is omega 1. 773 00:51:33,030 --> 00:51:35,400 Here is omega 2. 774 00:51:35,400 --> 00:51:40,219 And right here you're at the original cap omega n, 775 00:51:40,219 --> 00:51:42,260 because that's the way we've designed this thing, 776 00:51:42,260 --> 00:51:44,820 and we know this is the way H11 behaves. 777 00:51:44,820 --> 00:51:47,850 It goes to 0 at the value we've chosen 778 00:51:47,850 --> 00:51:51,290 for square root of k2/m2. 779 00:51:51,290 --> 00:51:54,270 So we started out with an original system that 780 00:51:54,270 --> 00:51:56,470 had a natural frequency here. 781 00:51:56,470 --> 00:51:58,760 k1/m1 was right here. 782 00:51:58,760 --> 00:52:01,930 We stuck on this second little mass, 783 00:52:01,930 --> 00:52:04,870 and it made into a two degree of freedom system. 784 00:52:04,870 --> 00:52:07,680 It no longer has a natural frequency there. 785 00:52:07,680 --> 00:52:09,340 It has a natural frequency below it, 786 00:52:09,340 --> 00:52:11,270 and a natural frequency above it, 787 00:52:11,270 --> 00:52:14,110 and a 0 right where the original natural frequency was. 788 00:52:29,127 --> 00:52:30,710 Imagine what's going on in the system. 789 00:52:30,710 --> 00:52:33,060 You've got this force being applied to the first mass, 790 00:52:33,060 --> 00:52:35,192 and the first mass isn't moving. 791 00:52:35,192 --> 00:52:37,392 Do you think the second mass is moving? 792 00:52:41,820 --> 00:52:44,000 Think about the free body diagram of the first mass. 793 00:52:44,000 --> 00:52:46,200 F equals ma. 794 00:52:46,200 --> 00:52:49,980 There is a force on that first mass-- F1 cosine omega t-- 795 00:52:49,980 --> 00:52:52,170 and it's not moving. 796 00:52:52,170 --> 00:52:55,880 Mx1 double dot is 0. 797 00:52:55,880 --> 00:52:58,580 So in order for the thing not to move, there's a force on it. 798 00:52:58,580 --> 00:53:03,190 There must be some other force exactly canceling it. 799 00:53:03,190 --> 00:53:04,366 Where does it come from? 800 00:53:04,366 --> 00:53:05,340 AUDIENCE: Mass two. 801 00:53:05,340 --> 00:53:09,770 PROFESSOR: Mass two through the spring. 802 00:53:09,770 --> 00:53:15,560 OK, so I'd also like to know, what is x2? 803 00:53:18,300 --> 00:53:23,670 Well, x2 from our transfer function matrix 804 00:53:23,670 --> 00:53:30,930 is the response of 2 due to a force at 1 times F1, 805 00:53:30,930 --> 00:53:34,680 plus a response at 2 due to a force at 2 times F2, 806 00:53:34,680 --> 00:53:39,330 but that second term is 0 because F2 is 0. 807 00:53:39,330 --> 00:53:46,250 That's a one term expression, and H21 808 00:53:46,250 --> 00:53:52,690 is z11 over the denominator. 809 00:53:52,690 --> 00:54:08,670 And Z11-- so x2 is going to look like that. 810 00:54:27,560 --> 00:54:30,640 I wrote that wrong. 811 00:54:30,640 --> 00:54:33,590 This is not z11. 812 00:54:33,590 --> 00:54:39,160 It is z minus z12 like that. 813 00:54:44,790 --> 00:54:51,640 And if you work that out, it's F1 times k2 814 00:54:51,640 --> 00:55:00,670 over the same denominator, but let's evaluate-- here's our-- 815 00:55:00,670 --> 00:55:03,320 this is the frequency we've been interested in. 816 00:55:03,320 --> 00:55:09,090 Let's evaluate the response of x2 right at this operating 817 00:55:09,090 --> 00:55:11,500 point. 818 00:55:11,500 --> 00:55:18,110 So let's evaluate this at omega equals cap omega n. 819 00:55:18,110 --> 00:55:25,870 So that's k2 down here minus omega squared 820 00:55:25,870 --> 00:55:30,640 m1 plus k1 plus k2. 821 00:55:30,640 --> 00:55:32,325 It's the same denominator as always. 822 00:55:35,090 --> 00:55:43,760 k2-- I'm not going to write it out. 823 00:55:43,760 --> 00:55:46,260 It's exactly the same thing as before, 824 00:55:46,260 --> 00:55:49,460 but I want to plug in to this denominator the operating 825 00:55:49,460 --> 00:55:51,880 frequency. 826 00:55:51,880 --> 00:55:54,940 We're going to operate at cap omega n, which 827 00:55:54,940 --> 00:55:58,790 is k1/m1 squared-- square root of k1/m1, 828 00:55:58,790 --> 00:56:00,190 or square root of k2/me. 829 00:56:00,190 --> 00:56:03,810 I'm going to plug in in this, and let's see what we get. 830 00:56:56,310 --> 00:57:08,400 And I'm going to let omega equal k1/m1 or k2/m2. 831 00:57:08,400 --> 00:57:11,580 It's the same thing, and I'm plug them in whatever 832 00:57:11,580 --> 00:57:12,570 is convenient. 833 00:57:12,570 --> 00:57:16,314 So here I'm going to put k1/m1, N1 and the m1's are 834 00:57:16,314 --> 00:57:16,980 going to cancel. 835 00:57:16,980 --> 00:57:18,438 I'll be left an k1. 836 00:57:22,330 --> 00:57:27,790 F1 over k1 and down in the denominator, this term 837 00:57:27,790 --> 00:57:36,380 turns into k1/m1, so that's minus k1 plus k1 plus k1. 838 00:57:36,380 --> 00:57:37,600 Those two cancel. 839 00:57:37,600 --> 00:57:39,440 I just get a k2. 840 00:57:39,440 --> 00:57:48,980 Over here, I'll use minus k2/m2 times m2 plus k2, 841 00:57:48,980 --> 00:57:52,030 and that gives me a minus k2 plus k2. 842 00:57:52,030 --> 00:57:59,540 This whole thing goes to 0, and I'm left with minus k2 squared 843 00:57:59,540 --> 00:58:02,510 and in the denominator a k2. 844 00:58:02,510 --> 00:58:04,980 This whole thing turns into F1 over k2. 845 00:58:12,690 --> 00:58:18,460 So the total response of that second mass at this operating 846 00:58:18,460 --> 00:58:21,430 frequency is just F1 over k2. 847 00:58:21,430 --> 00:58:37,330 And I would like to plot it as what does x2 over x1 static-- 848 00:58:37,330 --> 00:58:41,290 and x1 static and x2 static are the same. 849 00:58:45,270 --> 00:58:57,850 Well, that is F1 over k2 over F1 over k1. 850 00:58:57,850 --> 00:59:03,200 The F1's go away, and I get k1/k2. 851 00:59:03,200 --> 00:59:07,300 That also happens to be m1/m2, and it's 852 00:59:07,300 --> 00:59:11,410 a quantity we call 1 over mu. 853 00:59:11,410 --> 00:59:18,270 And we have just designed what's called a dynamic absorber. 854 00:59:18,270 --> 00:59:21,270 And a dynamic absorber is a little device 855 00:59:21,270 --> 00:59:24,240 you can use to stop vibration. 856 00:59:24,240 --> 00:59:26,605 So when we were talking about vibration isolation 857 00:59:26,605 --> 00:59:29,010 and vibration mitigation a few days ago 858 00:59:29,010 --> 00:59:32,320 and I said you've got some rotating imbalance [INAUDIBLE] 859 00:59:32,320 --> 00:59:35,380 or something to shake, well, give me 860 00:59:35,380 --> 00:59:37,575 three ways of solving it. 861 00:59:37,575 --> 00:59:39,330 We said balance the rotor. 862 00:59:39,330 --> 00:59:42,170 What were the other two ways of perhaps reducing the vibration? 863 00:59:48,821 --> 00:59:50,770 AUDIENCE: Adding a [INAUDIBLE] damper? 864 00:59:50,770 --> 00:59:52,410 PROFESSOR: Yeah, but also if you've 865 00:59:52,410 --> 00:59:53,885 got something that's shaking like crazy, 866 00:59:53,885 --> 00:59:55,718 and it's putting fibrillation into the floor 867 00:59:55,718 --> 00:59:57,700 or into the table, you can isolate it 868 00:59:57,700 --> 01:00:01,000 with a mass and a spring, or a microscope over here that's 869 01:00:01,000 --> 01:00:02,860 vibrating like crazy, you can isolate it 870 01:00:02,860 --> 01:00:03,940 with a mass or a spring. 871 01:00:03,940 --> 01:00:06,290 So this is to stop a vibration isolation, which 872 01:00:06,290 --> 01:00:09,820 is guaranteed to be on the final-- 873 01:00:09,820 --> 01:00:12,020 the simple practical applications of single degree 874 01:00:12,020 --> 01:00:13,140 of freedom stuff. 875 01:00:13,140 --> 01:00:15,930 So we have three ways-- fix the rotor, 876 01:00:15,930 --> 01:00:18,020 isolate it with a mass and a spring, 877 01:00:18,020 --> 01:00:20,370 isolate the sensitive instrument mass and a spring. 878 01:00:20,370 --> 01:00:22,192 Now you have a fourth way. 879 01:00:22,192 --> 01:00:23,900 If it's a particular operating frequency, 880 01:00:23,900 --> 01:00:27,790 we can operate right here with this thing 881 01:00:27,790 --> 01:00:29,480 called the dynamic absorber. 882 01:00:29,480 --> 01:00:31,430 This mu quantity is called the mass ratio. 883 01:00:39,880 --> 01:00:42,535 That's m2/m1. 884 01:00:45,570 --> 01:00:46,990 Basically, these things are real. 885 01:00:46,990 --> 01:00:48,620 They're actually used in real machines, 886 01:00:48,620 --> 01:00:51,640 and usually you can't-- this dynamic absorber thing you 887 01:00:51,640 --> 01:00:54,590 stick on there can't be as big as the original system. 888 01:00:54,590 --> 01:00:57,120 It's going to be some small fraction of the original system 889 01:00:57,120 --> 01:01:00,390 size-- 5% or 10% if you're lucky. 890 01:01:00,390 --> 01:01:02,040 So the bigger this thing is, you'll 891 01:01:02,040 --> 01:01:03,250 find out the better it works. 892 01:01:03,250 --> 01:01:07,650 So how much is this little second mass bouncing around? 893 01:01:07,650 --> 01:01:11,560 Well it's bouncing around compared to x1 static 894 01:01:11,560 --> 01:01:17,030 in the ratio of k1 to k2, which is this 1 over mu quantity. 895 01:01:17,030 --> 01:01:19,950 So if mu is 10%-- if you've added the second mass, 896 01:01:19,950 --> 01:01:22,310 it's 10% the size of the first one. 897 01:01:22,310 --> 01:01:25,990 1 over mu is a factor of 10. 898 01:01:25,990 --> 01:01:29,370 So this transfer function basically 899 01:01:29,370 --> 01:01:42,950 looks-- do I have some colored-- so the x2/x1 transfer 900 01:01:42,950 --> 01:01:46,420 function basically behaves the same right here 901 01:01:46,420 --> 01:01:49,410 and pretty much the same way out here. 902 01:01:49,410 --> 01:01:54,900 And in here, it comes down like this and comes back up. 903 01:01:54,900 --> 01:02:03,220 And this height right here is 1 over mu. 904 01:02:03,220 --> 01:02:07,060 So the smaller you make this thing, the tinier you make it, 905 01:02:07,060 --> 01:02:14,000 the more it has to shake to force the first mas to go to 0. 906 01:02:14,000 --> 01:02:16,890 So basically, the reason this thing works-- the free body 907 01:02:16,890 --> 01:02:30,200 diagram-- at the 0 point, the main mass m1, x1 equals 0 here. 908 01:02:30,200 --> 01:02:31,860 It's not moving. 909 01:02:31,860 --> 01:02:37,610 It's got a force acting on it that is some F1 cosine omega t. 910 01:02:37,610 --> 01:02:41,110 It's got a spring force acting on it 911 01:02:41,110 --> 01:02:42,950 from this second mass going. 912 01:02:42,950 --> 01:02:46,640 Here is m2 over here, and it's going back and forth 913 01:02:46,640 --> 01:02:51,080 like crazy putting force through this spring. 914 01:02:51,080 --> 01:02:55,180 And the spring force had better be exactly equal to that. 915 01:02:55,180 --> 01:03:02,390 So the F spring is going to be equal to minus F1, 916 01:03:02,390 --> 01:03:08,490 and that will be equal to x2 times kx2. 917 01:03:08,490 --> 01:03:10,390 The second mass has to move enough 918 01:03:10,390 --> 01:03:13,350 that it'll compress the spring enough that it provides 919 01:03:13,350 --> 01:03:15,830 a force equal and opposite to this one 920 01:03:15,830 --> 01:03:19,830 so that it's in equilibrium and it doesn't move. 921 01:03:19,830 --> 01:03:21,430 All right. 922 01:03:21,430 --> 01:03:21,930 Yeah? 923 01:03:26,286 --> 01:03:27,738 AUDIENCE: [INAUDIBLE]. 924 01:03:27,738 --> 01:03:28,706 PROFESSOR: Yeah. 925 01:03:28,706 --> 01:03:32,473 AUDIENCE: So the building, if you 926 01:03:32,473 --> 01:03:34,514 want the masses to be fairly equal to each other, 927 01:03:34,514 --> 01:03:35,889 how is that ever going to happen? 928 01:03:35,889 --> 01:03:37,090 PROFESSOR: It's not. 929 01:03:37,090 --> 01:03:39,700 Not usually. 930 01:03:39,700 --> 01:03:44,640 Dynamic absorbers are used in real things, 931 01:03:44,640 --> 01:03:46,440 like the Hancock Building. 932 01:03:46,440 --> 01:03:49,230 And they're used in engines. 933 01:03:49,230 --> 01:03:51,150 They're used in all sorts of little devices 934 01:03:51,150 --> 01:03:53,130 that you're not aware they they're there. 935 01:03:55,760 --> 01:03:58,000 And you have to usually hide them 936 01:03:58,000 --> 01:04:00,700 inside the footprint of the original device. 937 01:04:00,700 --> 01:04:02,660 So you don't want them-- for a real thing, 938 01:04:02,660 --> 01:04:07,660 you don't want to have a pump with this huge appendage on it. 939 01:04:07,660 --> 01:04:08,780 It's just not practical. 940 01:04:08,780 --> 01:04:10,440 So they tend to be small. 941 01:04:10,440 --> 01:04:12,677 And so, as a consequence, in order 942 01:04:12,677 --> 01:04:14,635 to make them work at this particular frequency, 943 01:04:14,635 --> 01:04:16,590 you have to vibrate like crazy. 944 01:04:16,590 --> 01:04:19,450 So the design cost of one of these things 945 01:04:19,450 --> 01:04:22,230 is you have to design to be able to allow this second mass 946 01:04:22,230 --> 01:04:25,080 to shake like all get out. 947 01:04:25,080 --> 01:04:29,630 Now, how many frequencies does this work at? 948 01:04:29,630 --> 01:04:32,250 Once you've designed it, it only works at one frequency, 949 01:04:32,250 --> 01:04:33,750 so this is only useful for something 950 01:04:33,750 --> 01:04:36,330 that has a fixed operating frequency, 951 01:04:36,330 --> 01:04:38,810 like a synchronous motor. 952 01:04:38,810 --> 01:04:41,520 This isn't a good thing for something 953 01:04:41,520 --> 01:04:45,240 that has a range of frequencies over which it can work, 954 01:04:45,240 --> 01:04:49,570 but it is a different, more complicated theory. 955 01:04:49,570 --> 01:04:51,120 There's a type of dynamic absorber 956 01:04:51,120 --> 01:04:54,905 that you can make work over a wide range of frequencies. 957 01:04:54,905 --> 01:04:56,670 I need to tell you one other thing. 958 01:04:56,670 --> 01:04:59,820 When we put this thing on, the original natural frequency 959 01:04:59,820 --> 01:05:04,380 of the first system was here, and it created two new ones. 960 01:05:04,380 --> 01:05:06,256 You had a further question. 961 01:05:06,256 --> 01:05:09,488 AUDIENCE: So how does making it vibrate like that help? 962 01:05:09,488 --> 01:05:11,404 It's only going to work at that one frequency, 963 01:05:11,404 --> 01:05:13,063 and you're always going to have that gap. 964 01:05:13,063 --> 01:05:14,438 So making it vibrate really fast, 965 01:05:14,438 --> 01:05:17,420 how does that help at all? 966 01:05:17,420 --> 01:05:19,170 PROFESSOR: If, for some reason, you really 967 01:05:19,170 --> 01:05:21,830 don't want that first thing to shake, 968 01:05:21,830 --> 01:05:25,160 this makes it stop shaking at that frequency. 969 01:05:25,160 --> 01:05:32,170 So for example, one of the widest distributed textbooks 970 01:05:32,170 --> 01:05:35,260 in the world was written by an MIT professor named Den Hartog, 971 01:05:35,260 --> 01:05:36,874 and it's called Mechanical Vibration. 972 01:05:36,874 --> 01:05:38,040 It was written in the 1930s. 973 01:05:38,040 --> 01:05:41,570 It's got a wonderful chapter on dynamic absorbers in it. 974 01:05:41,570 --> 01:05:44,150 And the example that he gives of a real device that 975 01:05:44,150 --> 01:05:50,040 actually uses one was an electric hair clipper 976 01:05:50,040 --> 01:05:53,850 that a barber uses. 977 01:05:53,850 --> 01:05:55,990 The head on a hair clipper-- it has 978 01:05:55,990 --> 01:05:59,510 to go back and forth like this in or to cut the hair. 979 01:05:59,510 --> 01:06:02,340 That is an oscillating mass, and you're going to feel it. 980 01:06:05,620 --> 01:06:08,630 You're going to feel the mass times the acceleration-- ma 981 01:06:08,630 --> 01:06:12,300 omega squared acceleration-- at the frequency it runs that. 982 01:06:12,300 --> 01:06:13,800 So they did a clever thing. 983 01:06:13,800 --> 01:06:16,270 They built inside the case of this thing 984 01:06:16,270 --> 01:06:18,890 a little second mass and spring. 985 01:06:18,890 --> 01:06:21,790 And so that when you're holding the clippers, 986 01:06:21,790 --> 01:06:24,830 it doesn't feel like it's-- OK? 987 01:06:24,830 --> 01:06:26,300 So that was an example. 988 01:06:26,300 --> 01:06:30,590 It was actually a great example in his 1938 textbook. 989 01:06:30,590 --> 01:06:32,870 And just so you know, you really do 990 01:06:32,870 --> 01:06:35,910 have two natural frequencies when you do this, 991 01:06:35,910 --> 01:06:37,260 and I'll write them like this. 992 01:06:37,260 --> 01:06:39,080 Omega 1 squared and omega 2 squared 993 01:06:39,080 --> 01:06:45,270 are the original frequency squared times 1 994 01:06:45,270 --> 01:06:54,550 plus mu over 2 plus or minus mu plus mu squared 995 01:06:54,550 --> 01:06:58,570 over 4 square root. 996 01:06:58,570 --> 01:07:03,040 So as your mass ratio gets bigger, 997 01:07:03,040 --> 01:07:05,500 the two frequencies move further and further apart. 998 01:07:05,500 --> 01:07:08,460 When the mass ratio is 0, this quantity-- 999 01:07:08,460 --> 01:07:10,960 they have two natural frequencies. 1000 01:07:10,960 --> 01:07:12,290 It goes to 1 natural frequency. 1001 01:07:12,290 --> 01:07:13,915 It's a single degree of freedom system, 1002 01:07:13,915 --> 01:07:15,710 and it's at the original value. 1003 01:07:15,710 --> 01:07:19,270 As the mass ratio gets bigger and bigger, these two roots-- 1004 01:07:19,270 --> 01:07:23,860 this one and this one-- as you add mass to that second system, 1005 01:07:23,860 --> 01:07:24,930 they spread apart. 1006 01:07:24,930 --> 01:07:27,320 And the bigger the mass ratio, the further apart the two 1007 01:07:27,320 --> 01:07:30,920 natural frequencies become. 1008 01:07:30,920 --> 01:07:34,070 So we've got 10 minutes left, and I have a demo 1009 01:07:34,070 --> 01:07:37,220 which illustrates this thing. 1010 01:07:37,220 --> 01:07:40,449 It's a little delicate to make it work, 1011 01:07:40,449 --> 01:07:42,240 because actually it was really frustrating. 1012 01:07:42,240 --> 01:07:43,530 I had it all set up. 1013 01:07:43,530 --> 01:07:46,350 It worked great in my office. 1014 01:07:46,350 --> 01:07:49,590 Came in 10 minutes early today to set it up, 1015 01:07:49,590 --> 01:07:52,740 and the table in here is so flexible I can't make 1016 01:07:52,740 --> 01:07:56,020 it behave like a fixed surface. 1017 01:07:56,020 --> 01:07:58,990 So I tweaked it, and I think I've got running now, 1018 01:07:58,990 --> 01:08:03,840 but now what it is is a beam with my pen 1019 01:08:03,840 --> 01:08:05,360 on it, my little squiggle pen. 1020 01:08:05,360 --> 01:08:06,640 And you've seen this before. 1021 01:08:06,640 --> 01:08:08,870 It's a rotating mass. 1022 01:08:08,870 --> 01:08:10,630 It's a static imbalance. 1023 01:08:10,630 --> 01:08:13,500 It shakes this beam. 1024 01:08:13,500 --> 01:08:16,960 And I've added to it a second little beam. 1025 01:08:16,960 --> 01:08:20,180 It's a little blade of steal with a heavy magnet on it. 1026 01:08:20,180 --> 01:08:21,700 That's my m2. 1027 01:08:21,700 --> 01:08:24,700 This is my k2-- is this little beam. 1028 01:08:24,700 --> 01:08:27,345 This is the original system, a mass, 1029 01:08:27,345 --> 01:08:29,229 it's close to a natural frequency, 1030 01:08:29,229 --> 01:08:33,084 and the rotation rate of the eccentric mass 1031 01:08:33,084 --> 01:08:36,439 is right at the natural frequency of this beam. 1032 01:08:36,439 --> 01:08:39,250 So because it's delicate to set up, 1033 01:08:39,250 --> 01:08:43,109 I've got to show you the system first with the second mass 1034 01:08:43,109 --> 01:08:45,819 attached to it, and I think I've got it tweaked 1035 01:08:45,819 --> 01:08:46,920 so that it'll sit there. 1036 01:08:46,920 --> 01:08:48,960 It's right at the operating point, 1037 01:08:48,960 --> 01:08:52,189 and you ought to see the second mass moving like crazy, 1038 01:08:52,189 --> 01:08:56,340 and the original mass, the beam, not moving much at all. 1039 01:08:56,340 --> 01:08:59,005 So let's see if we can make it work. 1040 01:08:59,005 --> 01:09:00,630 We're going to need to kill the lights. 1041 01:09:14,590 --> 01:09:19,529 And you can see in here the rotating mass going around 1042 01:09:19,529 --> 01:09:22,300 and around, and so I have the strobe light 1043 01:09:22,300 --> 01:09:25,790 detuned just a little bit so that you can see the system. 1044 01:09:25,790 --> 01:09:27,080 It's not quite synchronous. 1045 01:09:27,080 --> 01:09:28,939 Watch this little white blob out here. 1046 01:09:28,939 --> 01:09:31,420 You see it going up and down. 1047 01:09:31,420 --> 01:09:36,390 That's your second mass, and is the main beam-- 1048 01:09:36,390 --> 01:09:41,840 it's moving a little bit, but not much. 1049 01:09:41,840 --> 01:09:43,340 It's got a little damping, and so it 1050 01:09:43,340 --> 01:09:47,729 isn't perfectly down at that 0 point, but very close. 1051 01:09:47,729 --> 01:09:52,390 So this is the system operating close to that null point. 1052 01:09:52,390 --> 01:09:57,900 So now if I remove the dynamic absorber, 1053 01:09:57,900 --> 01:10:02,330 the second little mass spring system, then the main mass 1054 01:10:02,330 --> 01:10:04,316 and the beam ought to shake like crazy. 1055 01:10:15,052 --> 01:10:17,940 Just changing the tuning a little here. 1056 01:10:33,300 --> 01:10:40,035 So now I'm going to remove the absorber. 1057 01:10:50,150 --> 01:10:52,180 All right? 1058 01:10:52,180 --> 01:10:55,740 And now it's going like crazy. 1059 01:10:59,440 --> 01:11:01,415 So that's an illustration. 1060 01:11:01,415 --> 01:11:02,950 You can come up with the lights. 1061 01:11:08,840 --> 01:11:10,080 Now, you hear beating? 1062 01:11:13,060 --> 01:11:15,950 This little pen-- this little rotor in here 1063 01:11:15,950 --> 01:11:21,280 is driven by a little DC motor and a single AA battery, 1064 01:11:21,280 --> 01:11:23,210 and it's not feedback controlled. 1065 01:11:23,210 --> 01:11:26,539 Its frequency can vary, and it isn't that powerful. 1066 01:11:26,539 --> 01:11:28,330 So as this thing starts moving up and down, 1067 01:11:28,330 --> 01:11:32,200 it actually takes real torque to make that weight go around 1068 01:11:32,200 --> 01:11:35,340 while this whole system is accelerating up and down. 1069 01:11:35,340 --> 01:11:37,010 And the motor just isn't up to it, 1070 01:11:37,010 --> 01:11:41,280 just isn't powerful enough, so the speed changes. 1071 01:11:41,280 --> 01:11:45,390 This thing can't hold constant frequency. 1072 01:11:45,390 --> 01:11:46,260 So that's the demo. 1073 01:11:46,260 --> 01:11:46,759 Questions? 1074 01:11:49,590 --> 01:11:52,460 So we've kind of embedded in this lecture-- this lecture was 1075 01:11:52,460 --> 01:11:55,240 to introduce you to the idea that you can write a transfer 1076 01:11:55,240 --> 01:11:57,830 function for a multiple degree of freedom system-- 1077 01:11:57,830 --> 01:12:00,140 has embedded in it all the natural frequencies 1078 01:12:00,140 --> 01:12:01,440 of the system. 1079 01:12:01,440 --> 01:12:04,350 When you use a transfer function to calculate 1080 01:12:04,350 --> 01:12:07,900 the response of the system, you are getting the contributions 1081 01:12:07,900 --> 01:12:10,870 of all the modes at once. 1082 01:12:10,870 --> 01:12:15,470 So you essentially solve the equations of motion directly. 1083 01:12:15,470 --> 01:12:17,390 So you now have seen there's two ways 1084 01:12:17,390 --> 01:12:21,020 to go about analyzing multiple degree of freedom systems-- 1085 01:12:21,020 --> 01:12:24,400 the technique known as modal analysis-- one mode at a time 1086 01:12:24,400 --> 01:12:26,780 and then add it together. 1087 01:12:26,780 --> 01:12:29,120 Or like with transfer functions, where you're just 1088 01:12:29,120 --> 01:12:32,040 solving the whole thing at once, and each transfer function 1089 01:12:32,040 --> 01:12:36,710 has in it all of the information about all of the modes 1090 01:12:36,710 --> 01:12:37,519 of the system. 1091 01:12:37,519 --> 01:12:39,310 So if it's a five degree of freedom system, 1092 01:12:39,310 --> 01:12:43,261 you're going to see five peaks out here. 1093 01:12:43,261 --> 01:12:43,760 Questions? 1094 01:12:47,946 --> 01:12:50,838 AUDIENCE: So this is for [INAUDIBLE]? 1095 01:12:50,838 --> 01:12:54,230 But initially, since you have a [INAUDIBLE], 1096 01:12:54,230 --> 01:12:57,885 does it have to start out moving to be an [INAUDIBLE] moving 1097 01:12:57,885 --> 01:13:00,140 and then it stops. 1098 01:13:00,140 --> 01:13:03,210 PROFESSOR: So there's going to be a transient phase, sure. 1099 01:13:03,210 --> 01:13:06,010 Any system, when you start it up, 1100 01:13:06,010 --> 01:13:09,450 is going to have transients, and the transients 1101 01:13:09,450 --> 01:13:12,400 are the equivalent of initial conditions. 1102 01:13:12,400 --> 01:13:16,920 So displacement and velocity-- and the response 1103 01:13:16,920 --> 01:13:22,210 of a linear system that vibrates, 1104 01:13:22,210 --> 01:13:25,200 that has natural frequencies to a set of initial conditions, 1105 01:13:25,200 --> 01:13:26,740 is vibration at what frequencies? 1106 01:13:36,027 --> 01:13:37,652 AUDIENCE: Do you know how they overcome 1107 01:13:37,652 --> 01:13:39,636 this in helicopters? [INAUDIBLE] that we saw 1108 01:13:39,636 --> 01:13:42,120 at the beginning of the year? 1109 01:13:42,120 --> 01:13:43,760 PROFESSOR: Yeah, the helicopter-- so 1110 01:13:43,760 --> 01:13:45,080 I want to answer the question. 1111 01:13:45,080 --> 01:13:47,850 Response to that transience is at the natural frequencies 1112 01:13:47,850 --> 01:13:50,030 of the system, maybe some combination. 1113 01:13:50,030 --> 01:13:52,460 The original displacement requires 1114 01:13:52,460 --> 01:13:54,020 contributions of several modes. 1115 01:13:54,020 --> 01:13:57,250 You'll get decay contributions at each 1116 01:13:57,250 --> 01:14:00,100 of the natural frequencies, and once they've died out, 1117 01:14:00,100 --> 01:14:02,570 this motor has started up. 1118 01:14:02,570 --> 01:14:05,730 The thing will settle into what we call steady state response, 1119 01:14:05,730 --> 01:14:08,355 and it will only be at the excitation frequency. 1120 01:14:08,355 --> 01:14:09,980 And I haven't show you anything about-- 1121 01:14:09,980 --> 01:14:14,440 can you solve the equations of motion from start up at time 0 1122 01:14:14,440 --> 01:14:16,350 through the whole messy transient phase 1123 01:14:16,350 --> 01:14:17,210 to steady state? 1124 01:14:17,210 --> 01:14:17,710 Sure. 1125 01:14:17,710 --> 01:14:20,330 And we haven't talked at all about that. 1126 01:14:20,330 --> 01:14:24,410 And actually, for vibration stuff, it is very important. 1127 01:14:24,410 --> 01:14:25,840 Steady state answer is important. 1128 01:14:25,840 --> 01:14:27,548 The transient answer is pretty important, 1129 01:14:27,548 --> 01:14:29,560 and for you to have a feeling for that, 1130 01:14:29,560 --> 01:14:32,690 you actually know quite a fair amount about the basic concepts 1131 01:14:32,690 --> 01:14:34,810 of how things vibrate now. 1132 01:14:34,810 --> 01:14:37,250 Now, your question about the helicopter-- 1133 01:14:37,250 --> 01:14:39,005 so if you have a uniform helicopter 1134 01:14:39,005 --> 01:14:42,642 blade-- they rigged that to do what it did. 1135 01:14:42,642 --> 01:14:44,600 I don't know what they were experimenting with, 1136 01:14:44,600 --> 01:14:47,590 but if you do-- there's a problem that I've actually 1137 01:14:47,590 --> 01:14:48,940 given in a vibration course. 1138 01:14:48,940 --> 01:14:51,470 If you design a really simple helicopter 1139 01:14:51,470 --> 01:14:53,380 that has a uniform blade. 1140 01:14:53,380 --> 01:14:58,990 It's a uniform rod, and it's pivoted at the center, 1141 01:14:58,990 --> 01:15:00,680 and you spin it up, and if you spin it 1142 01:15:00,680 --> 01:15:02,290 fast enough-- if you spin it slowly, 1143 01:15:02,290 --> 01:15:04,470 it will droop because of gravity, but if you spin it 1144 01:15:04,470 --> 01:15:07,570 fast enough, r omega squared is a lot bigger than g. 1145 01:15:07,570 --> 01:15:09,320 And the thing flattens out, and the blades 1146 01:15:09,320 --> 01:15:11,510 are going around and around. 1147 01:15:11,510 --> 01:15:14,990 The natural frequency of that system 1148 01:15:14,990 --> 01:15:18,762 is exactly the rotation rate. 1149 01:15:18,762 --> 01:15:23,050 A uniform blade pivot at the center, going around fast, 1150 01:15:23,050 --> 01:15:24,740 has a natural frequency like this. 1151 01:15:24,740 --> 01:15:28,340 It happens to be exactly at the rate of rotation. 1152 01:15:28,340 --> 01:15:32,760 So any perturbation will start it doing bad things. 1153 01:15:32,760 --> 01:15:37,290 Helicopters are never ever designed like that. 1154 01:15:37,290 --> 01:15:40,930 The helicopter has a rotor disk in here, 1155 01:15:40,930 --> 01:15:43,250 has a finite radius before the pin, 1156 01:15:43,250 --> 01:15:46,270 and then the blades are out here. 1157 01:15:46,270 --> 01:15:49,160 The pin around which-- and that changes the resonance 1158 01:15:49,160 --> 01:15:50,060 frequency. 1159 01:15:50,060 --> 01:15:52,800 And of course, you can change the mass distribution 1160 01:15:52,800 --> 01:15:56,110 in the blade and so forth. 1161 01:15:56,110 --> 01:15:58,254 So what they had done in that case 1162 01:15:58,254 --> 01:16:00,670 to rig it so the thing beat itself to death, I don't know. 1163 01:16:00,670 --> 01:16:03,770 But there's good design practices with helicopters 1164 01:16:03,770 --> 01:16:08,255 so that the blade frequency is not at the rotation rate, 1165 01:16:08,255 --> 01:16:09,838 because at rotation rate, you're going 1166 01:16:09,838 --> 01:16:13,040 to have a lot of excitation, hitting turbulence in the air 1167 01:16:13,040 --> 01:16:16,460 and so forth. 1168 01:16:16,460 --> 01:16:16,960 Yeah? 1169 01:16:16,960 --> 01:16:19,900 AUDIENCE: So you mentioned that you use something 1170 01:16:19,900 --> 01:16:21,370 like this in the Hancock Building, 1171 01:16:21,370 --> 01:16:23,820 but I would imagine that the wind force on the Hancock 1172 01:16:23,820 --> 01:16:27,740 Building does not have a uniform frequency all the time. 1173 01:16:27,740 --> 01:16:31,660 So is the idea that [INAUDIBLE] control 1174 01:16:31,660 --> 01:16:33,620 based on resonant frequency [INAUDIBLE]. 1175 01:16:38,530 --> 01:16:40,769 PROFESSOR: I've forgotten the correct name now, 1176 01:16:40,769 --> 01:16:42,560 even though I've used these many times now. 1177 01:16:42,560 --> 01:16:47,740 The optimally tuned and damped dynamic absorber 1178 01:16:47,740 --> 01:16:56,610 is sort of the second level of dynamic absorber design. 1179 01:16:56,610 --> 01:16:58,590 Typically, a two degree of freedom system 1180 01:16:58,590 --> 01:17:01,330 looks like that if it's lightly damped, 1181 01:17:01,330 --> 01:17:06,000 but you can design that second mass spring and a damper 1182 01:17:06,000 --> 01:17:11,200 so that you can make the transfer function, 1183 01:17:11,200 --> 01:17:18,950 H11 here, look like this, where you 1184 01:17:18,950 --> 01:17:24,910 get the two peaks are of equal height, and the worst-- 1185 01:17:24,910 --> 01:17:27,910 and you can design them so that the worst case amplitude 1186 01:17:27,910 --> 01:17:29,490 response over here is pretty low. 1187 01:17:33,710 --> 01:17:40,660 x1 static is square root of 1 over 1 1188 01:17:40,660 --> 01:17:43,230 plus mu or something like that. 1189 01:17:43,230 --> 01:17:47,250 So the bigger you make this, the better the performance, 1190 01:17:47,250 --> 01:17:50,410 but you have to have optimum damping, 1191 01:17:50,410 --> 01:17:52,860 and you have to have optimum tuning. 1192 01:17:52,860 --> 01:18:00,410 And tuning means, if the there were-- we 1193 01:18:00,410 --> 01:18:05,010 tune this thing so that k2/m2 was 1194 01:18:05,010 --> 01:18:10,470 exactly equal to a particular frequency, in this case k1/n1. 1195 01:18:10,470 --> 01:18:14,070 But for the optimally tuned and damp dynamic absorber, 1196 01:18:14,070 --> 01:18:17,820 you actually tune things-- it's tuned a little differently. 1197 01:18:17,820 --> 01:18:19,590 You put damping in it, and you make 1198 01:18:19,590 --> 01:18:23,440 it behave pretty well over the entire frequency, 1199 01:18:23,440 --> 01:18:28,980 range rather than almost perfect at one frequency. 1200 01:18:28,980 --> 01:18:35,440 And the Hancock Building had its transfer function, so to speak, 1201 01:18:35,440 --> 01:18:39,720 without the dynamic absorber-- it has many, many peaks, 1202 01:18:39,720 --> 01:18:45,450 but it had two problematic ones that were very close together, 1203 01:18:45,450 --> 01:18:55,089 and they were at about 0.8 radians per second, and 0.81, 1204 01:18:55,089 --> 01:18:55,630 or something. 1205 01:18:55,630 --> 01:18:59,115 And this was bending, and this was torsion. 1206 01:19:03,470 --> 01:19:06,660 And at these two resonances-- and the wind would come along, 1207 01:19:06,660 --> 01:19:08,561 and it would excite both of them. 1208 01:19:08,561 --> 01:19:10,810 And the one resonance was just the same as this thing. 1209 01:19:10,810 --> 01:19:12,890 It just bends. 1210 01:19:12,890 --> 01:19:15,720 And the other resonance was the first mode in torsion. 1211 01:19:15,720 --> 01:19:19,095 So if you were standing on one-- and a building cross section 1212 01:19:19,095 --> 01:19:21,530 is a funny-- the Hancock Building 1213 01:19:21,530 --> 01:19:23,390 looks kind of like that in cross section, 1214 01:19:23,390 --> 01:19:31,660 and so it would rotate in torsion and deflect in bending. 1215 01:19:31,660 --> 01:19:36,090 So it would be rotating and deflecting, 1216 01:19:36,090 --> 01:19:37,900 and if they were in equal amounts 1217 01:19:37,900 --> 01:19:41,910 and they're close in frequency, if you're standing up there, 1218 01:19:41,910 --> 01:19:44,450 what would you feel? 1219 01:19:44,450 --> 01:19:45,700 They would beat. 1220 01:19:45,700 --> 01:19:46,630 The motion would beat. 1221 01:19:46,630 --> 01:19:49,369 You'd get a lot of motion, because if you 1222 01:19:49,369 --> 01:19:50,910 are on one end of the building, you'd 1223 01:19:50,910 --> 01:19:54,040 get a lot of torsional motion out here combined 1224 01:19:54,040 --> 01:19:55,296 with the bending motion. 1225 01:19:55,296 --> 01:19:57,170 And then they would cancel, and then the beat 1226 01:19:57,170 --> 01:19:58,970 would build up again. 1227 01:19:58,970 --> 01:20:02,540 And so that was what was happening in the building. 1228 01:20:02,540 --> 01:20:12,140 And they put in two, so on the 58th floor on each end 1229 01:20:12,140 --> 01:20:20,640 they put in a mass spring system and over here another one so 1230 01:20:20,640 --> 01:20:23,199 that they would resist the torsion-- actually, 1231 01:20:23,199 --> 01:20:24,740 really I lined them up the wrong way. 1232 01:20:24,740 --> 01:20:26,850 They line up this way. 1233 01:20:26,850 --> 01:20:28,570 So it has two of them. 1234 01:20:28,570 --> 01:20:32,590 They can resist both the bending and the torsion. 1235 01:20:32,590 --> 01:20:37,300 They each weigh 300 tons, and they 1236 01:20:37,300 --> 01:20:42,410 it's a box filled up with 50 pound lead bricks. 1237 01:20:42,410 --> 01:20:46,240 And the box is about-- it's been 20 years since I was up there, 1238 01:20:46,240 --> 01:20:50,540 but what I recall is this like 8 by 10 feet yea tall. 1239 01:20:50,540 --> 01:20:55,400 And it slides on a pressurized oil film, and the spring on it 1240 01:20:55,400 --> 01:20:58,360 is a big pneumatic spring, a big air spring, 1241 01:20:58,360 --> 01:20:59,910 and a computer runs the whole system. 1242 01:20:59,910 --> 01:21:02,910 It's shut down until the wind gets above 40 miles an hour, 1243 01:21:02,910 --> 01:21:04,720 and then turns it on. 1244 01:21:04,720 --> 01:21:06,390 And it's actually kind of self tuning. 1245 01:21:06,390 --> 01:21:08,540 It can optimize itself, and then it'll sit there, 1246 01:21:08,540 --> 01:21:11,740 and it's designed to do this, because this building has 1247 01:21:11,740 --> 01:21:14,750 these two problem frequencies. 1248 01:21:14,750 --> 01:21:19,410 And so it has been designed to address both of them. 1249 01:21:22,690 --> 01:21:26,505 All right, I think I have to get out of here for the next class. 1250 01:21:26,505 --> 01:21:27,380 See you next Tuesday. 1251 01:21:27,380 --> 01:21:28,834 Last lecture.