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 --> 00:00:06,040 Your support will help MIT OpenCourseWare 4 00:00:06,040 --> 00:00:10,130 continue to offer high quality educational resources for free. 5 00:00:10,130 --> 00:00:12,690 To make a donation or to view additional materials 6 00:00:12,690 --> 00:00:16,590 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,590 --> 00:00:17,297 at ocw.mit.edu. 8 00:00:21,343 --> 00:00:22,218 PROFESSOR: All right. 9 00:00:22,218 --> 00:00:25,270 I'm a little slow getting started today, 10 00:00:25,270 --> 00:00:26,100 better get going. 11 00:00:32,030 --> 00:00:38,590 What we're going to talk about today is a technique-- 12 00:00:38,590 --> 00:00:40,130 you guys done? 13 00:00:40,130 --> 00:00:41,550 OK, thanks. 14 00:00:41,550 --> 00:00:43,760 We're going to talk about today a technique known 15 00:00:43,760 --> 00:00:47,840 as modal analysis, and it's a way 16 00:00:47,840 --> 00:00:52,290 of analyzing things that vibrate, essentially thinking 17 00:00:52,290 --> 00:00:54,860 about them one mode and a time. 18 00:00:54,860 --> 00:00:58,220 Though you might not make a lot of use 19 00:00:58,220 --> 00:01:01,280 of the actual calculations, doing the math, 20 00:01:01,280 --> 00:01:04,959 throughout your careers, I think if you understand it 21 00:01:04,959 --> 00:01:09,370 conceptually it'll help you just have a better 22 00:01:09,370 --> 00:01:12,170 understanding of what vibration is all about, 23 00:01:12,170 --> 00:01:14,640 just give you some insight to it that you otherwise 24 00:01:14,640 --> 00:01:16,480 wouldn't have. 25 00:01:16,480 --> 00:01:18,960 So the basic concept is that you can model just 26 00:01:18,960 --> 00:01:23,080 about any structural vibration as the summation 27 00:01:23,080 --> 00:01:25,970 of the individual contributions of each 28 00:01:25,970 --> 00:01:28,370 what we call natural mode. 29 00:01:28,370 --> 00:01:30,440 So what we mean by that is, let's 30 00:01:30,440 --> 00:01:32,800 start by thinking-- actually, let 31 00:01:32,800 --> 00:01:36,150 me say that this applies to both continuous systems 32 00:01:36,150 --> 00:01:41,010 like vibrating strings or beams or buildings as it 33 00:01:41,010 --> 00:01:45,510 does to finite degree of freedom rigid body systems. 34 00:01:45,510 --> 00:01:47,450 We haven't talked about continuous systems. 35 00:01:47,450 --> 00:01:51,060 I'll do a lecture on it as the last lecture 36 00:01:51,060 --> 00:01:54,230 of the term, just kind of an enrichment sort of lecture. 37 00:01:54,230 --> 00:01:57,590 But everything I say about finite degree of freedom 38 00:01:57,590 --> 00:02:01,360 systems can be extended to continuous systems. 39 00:02:01,360 --> 00:02:04,610 But since we've been studying rigid bodies and systems 40 00:02:04,610 --> 00:02:07,270 with finite numbers of degrees of freedom, 41 00:02:07,270 --> 00:02:10,430 I'll explain-- I'll go through this analysis 42 00:02:10,430 --> 00:02:14,330 in the context of rigid body finite degree of freedom 43 00:02:14,330 --> 00:02:16,700 systems. 44 00:02:16,700 --> 00:02:23,020 So in general, we can write the equations of motion 45 00:02:23,020 --> 00:02:28,380 for finite degree of freedom systems as a mass matrix. 46 00:02:28,380 --> 00:02:30,020 And to keep the kind of writing down, 47 00:02:30,020 --> 00:02:35,080 I'm just going to underline matrices and a squiggle 48 00:02:35,080 --> 00:02:36,525 under vectors so we have them. 49 00:02:36,525 --> 00:02:39,300 In general, we can write the equations of motion 50 00:02:39,300 --> 00:02:44,880 as a mass matrix times an acceleration vector 51 00:02:44,880 --> 00:02:50,030 plus a damping matrix times a velocity vector, 52 00:02:50,030 --> 00:02:54,430 stiffness matrix times a displacement vector, 53 00:02:54,430 --> 00:03:00,970 all equal to some external vector of excitations. 54 00:03:00,970 --> 00:03:03,100 And I'm writing these as if these are translations, 55 00:03:03,100 --> 00:03:07,570 but you know, like from doing the pendulum on the cart 56 00:03:07,570 --> 00:03:09,860 problem, that the equations of motion 57 00:03:09,860 --> 00:03:12,700 might involve rotations and displacements. 58 00:03:12,700 --> 00:03:16,250 And we let them-- they mix together here however they 59 00:03:16,250 --> 00:03:17,440 fall out. 60 00:03:17,440 --> 00:03:20,000 But just to write them symbolically, 61 00:03:20,000 --> 00:03:23,990 I'm just going to refer to all of those coordinates 62 00:03:23,990 --> 00:03:26,650 with an x vector. 63 00:03:26,650 --> 00:03:33,880 OK, now the basic premise of modal analysis 64 00:03:33,880 --> 00:03:43,825 is a thing called the modal expansion theorem. 65 00:03:48,040 --> 00:03:51,330 It's basically the assertion that you 66 00:03:51,330 --> 00:03:56,460 can represent any motion set of vectors-- 67 00:03:56,460 --> 00:04:03,420 I'll write them kind of as a vector here for a moment-- x, 68 00:04:03,420 --> 00:04:10,270 as the superposition of each contributing mode. 69 00:04:10,270 --> 00:04:14,100 Now each mode has a mode shape to it, 70 00:04:14,100 --> 00:04:19,490 which I'm going to call u, and up here I'll 71 00:04:19,490 --> 00:04:21,315 put a superscript for what mode it 72 00:04:21,315 --> 00:04:27,980 is, the first mode, times its time-dependent behavior. 73 00:04:27,980 --> 00:04:31,010 And this is called, what they call in textbooks, 74 00:04:31,010 --> 00:04:33,700 the natural coordinates. 75 00:04:33,700 --> 00:04:35,430 And we'll see what those are in a second. 76 00:04:35,430 --> 00:04:37,640 So mode shape one. 77 00:04:37,640 --> 00:04:41,430 This is the time-dependent and amplitude part 78 00:04:41,430 --> 00:04:45,630 that says how much the contribution of mode one 79 00:04:45,630 --> 00:04:51,000 is to this motion and what its time dependence is, this is. 80 00:04:51,000 --> 00:04:57,200 And then you'd have mode two's contribution, q2. 81 00:05:00,460 --> 00:05:10,000 And this goes out to the nth mode's contribution, qn of t. 82 00:05:10,000 --> 00:05:12,150 And that's the proposition, that you 83 00:05:12,150 --> 00:05:15,300 can represent the total response of the system 84 00:05:15,300 --> 00:05:18,230 as a superposition of the response of each 85 00:05:18,230 --> 00:05:20,250 of the natural modes of the system. 86 00:05:20,250 --> 00:05:22,790 And if it's an n degree of freedom system, 87 00:05:22,790 --> 00:05:24,590 there will be n natural modes, so. 88 00:05:42,850 --> 00:05:45,710 Now something I didn't say here. 89 00:05:45,710 --> 00:05:50,250 This all assumes that the system vibrates. 90 00:05:50,250 --> 00:05:52,820 So this is all in the discussion of things 91 00:05:52,820 --> 00:05:55,160 that exhibit vibratory motion. 92 00:05:55,160 --> 00:06:08,243 So this is all, it should say here, of vibrating systems, OK? 93 00:06:14,470 --> 00:06:17,190 So this kind of a long and cumbersome way 94 00:06:17,190 --> 00:06:18,830 of writing this out. 95 00:06:18,830 --> 00:06:24,475 So if you notice, each one of these is the mode shape vector. 96 00:06:24,475 --> 00:06:31,010 And if I put them together in a matrix just side by side, 97 00:06:31,010 --> 00:06:41,920 here's a u1 over to un and multiply it by this vector, 98 00:06:41,920 --> 00:06:45,815 q1 of t down to qn. 99 00:06:48,750 --> 00:06:50,800 That's the same statement but said 100 00:06:50,800 --> 00:06:53,670 in a much more compact way. 101 00:06:53,670 --> 00:06:56,070 So this statement, this modal expansion theorem, 102 00:06:56,070 --> 00:07:00,230 basically says the vector of-- these 103 00:07:00,230 --> 00:07:09,720 are your generalized coordinates, which 104 00:07:09,720 --> 00:07:11,269 we've been using all term long. 105 00:07:11,269 --> 00:07:12,810 These are the generalized coordinates 106 00:07:12,810 --> 00:07:15,820 that you choose to derive the equations of motion around. 107 00:07:15,820 --> 00:07:18,300 The vector of generalized coordinates 108 00:07:18,300 --> 00:07:20,320 can be written as uq. 109 00:07:23,990 --> 00:07:32,190 And these are often called the modal coordinates 110 00:07:32,190 --> 00:07:40,450 or sometimes called the natural coordinates, OK? 111 00:07:51,690 --> 00:07:58,820 So if we can say that x is uq, then x dot, 112 00:07:58,820 --> 00:08:02,260 you take the derivative of each one of those expressions. 113 00:08:02,260 --> 00:08:07,450 You'll find that's going to be uq dot. 114 00:08:07,450 --> 00:08:12,800 And x double dot equals uq double 115 00:08:12,800 --> 00:08:15,000 dot because these are just constants. 116 00:08:15,000 --> 00:08:18,270 The mode shape vectors are just a fixed set of numbers 117 00:08:18,270 --> 00:08:21,620 that represent the mode shape to the system. 118 00:08:21,620 --> 00:08:27,132 Now just to-- I think maybe this is a good time to do this. 119 00:08:27,132 --> 00:08:28,070 You grab one end. 120 00:08:31,430 --> 00:08:34,960 So this is a-- and it's hard to see black against black. 121 00:08:34,960 --> 00:08:36,919 My apologies for that. 122 00:08:36,919 --> 00:08:42,770 So this is a guitar string or any stringed instrument. 123 00:08:42,770 --> 00:08:48,200 In fact, any long, slender thing under tension will vibrate. 124 00:08:48,200 --> 00:08:51,620 And it has, if I do this carefully, 125 00:08:51,620 --> 00:08:56,290 that's called the first mode of vibration. 126 00:08:56,290 --> 00:08:59,240 And that's when you pluck your guitar string or violin 127 00:08:59,240 --> 00:09:00,060 in the middle. 128 00:09:00,060 --> 00:09:01,280 You mostly hear that. 129 00:09:04,150 --> 00:09:09,300 But at twice the frequency, if I can get it going here, 130 00:09:09,300 --> 00:09:11,400 there's a second mode of vibration. 131 00:09:11,400 --> 00:09:13,460 And for a taut string, it happens 132 00:09:13,460 --> 00:09:15,900 to be at twice the frequency of the first. 133 00:09:15,900 --> 00:09:19,530 And if my hand is well calibrated-- 134 00:09:19,530 --> 00:09:22,060 it may be easier if it's a little longer-- if I 135 00:09:22,060 --> 00:09:32,580 get this going right, there's a third mode, OK? 136 00:09:32,580 --> 00:09:38,020 So that's what we're calling-- oh, 137 00:09:38,020 --> 00:09:39,660 what I meant to say when I was doing 138 00:09:39,660 --> 00:09:44,070 this is these shapes for the vibrating string, 139 00:09:44,070 --> 00:09:47,933 that second mode shape happens to be one full sine wave. 140 00:09:47,933 --> 00:09:54,160 And the mode shape has the form sine n pi x over l, 141 00:09:54,160 --> 00:09:57,290 where l's the length of the string. 142 00:09:57,290 --> 00:09:58,430 n's the mode number. 143 00:09:58,430 --> 00:09:59,540 So first mode. 144 00:09:59,540 --> 00:10:02,060 Second mode is this, when n is 2. 145 00:10:02,060 --> 00:10:04,530 First mode, n is 1. 146 00:10:04,530 --> 00:10:08,300 nth mode or something high, you get higher modes like that. 147 00:10:08,300 --> 00:10:11,780 So these are the mode shapes for a vibrating string. 148 00:10:14,900 --> 00:10:15,930 That's good for now. 149 00:10:18,530 --> 00:10:23,114 This two degree of freedom system with the two lump 150 00:10:23,114 --> 00:10:24,530 masses-- and it's going to show up 151 00:10:24,530 --> 00:10:28,570 there, yeah-- this is basically two lump masses. 152 00:10:28,570 --> 00:10:31,960 And we idealize the springs as being massless, 153 00:10:31,960 --> 00:10:34,610 but it's a pretty good approximation. 154 00:10:34,610 --> 00:10:36,920 This has two modes of vibration. 155 00:10:36,920 --> 00:10:46,060 And Professor Gossard made these neat little things that 156 00:10:46,060 --> 00:10:48,810 can make it so-- and I'm going to come back to this, 157 00:10:48,810 --> 00:10:52,130 but there's mode one. 158 00:10:52,130 --> 00:10:54,160 And the mode shape is as this goes down 159 00:10:54,160 --> 00:10:58,310 one unit, that goes down about two times as much. 160 00:10:58,310 --> 00:11:00,340 I'll give you the exact numbers in a minute. 161 00:11:00,340 --> 00:11:10,504 And the other mode shape of the system-- 162 00:11:10,504 --> 00:11:12,920 we're going to talk about this today and why this happens. 163 00:11:12,920 --> 00:11:15,010 But if I give it the right initial conditions, 164 00:11:15,010 --> 00:11:18,360 I can make it vibrate only in the second mode shape. 165 00:11:18,360 --> 00:11:21,780 And so it's now deflected with the right conditions 166 00:11:21,780 --> 00:11:23,922 so that it'll respond only in second mode. 167 00:11:26,520 --> 00:11:28,190 This mass goes up and down a lot. 168 00:11:28,190 --> 00:11:31,130 That mass goes up in that little, opposite to it, 169 00:11:31,130 --> 00:11:31,710 actually. 170 00:11:31,710 --> 00:11:34,420 The frequency is different from the first. 171 00:11:34,420 --> 00:11:37,640 But if this is moving one unit, then this down here 172 00:11:37,640 --> 00:11:40,220 is moving minus 0.3 or something like that. 173 00:11:40,220 --> 00:11:42,740 And that ratio is constant. 174 00:11:42,740 --> 00:11:45,660 And that's called the mode shape. 175 00:11:45,660 --> 00:11:49,740 So if you just pick one of them and say, let its motion be one, 176 00:11:49,740 --> 00:11:52,090 then all of the other masses in the system 177 00:11:52,090 --> 00:11:56,760 will move in a particular ratio to the motion of that one 178 00:11:56,760 --> 00:11:58,830 that you arbitrarily set to one. 179 00:11:58,830 --> 00:12:00,620 So this is what we mean by mode shapes 180 00:12:00,620 --> 00:12:04,150 and their natural frequencies. 181 00:12:04,150 --> 00:12:05,900 There's the natural frequency associated 182 00:12:05,900 --> 00:12:07,030 with that first mode. 183 00:12:07,030 --> 00:12:09,360 And we can solve these things mathematically, 184 00:12:09,360 --> 00:12:11,499 and we've been doing that a little bit 185 00:12:11,499 --> 00:12:12,790 in the last couple of lectures. 186 00:12:15,500 --> 00:12:16,040 All right. 187 00:12:18,830 --> 00:12:21,670 So this is the relationship between these things, 188 00:12:21,670 --> 00:12:25,200 the generalized coordinates and the modal coordinates. 189 00:12:25,200 --> 00:12:34,140 And we now need to see how we're going to use these. 190 00:12:40,330 --> 00:12:42,710 So in general, we have our equations of motion. 191 00:12:57,950 --> 00:13:01,310 And I'm going to substitute for x, x dot, 192 00:13:01,310 --> 00:13:27,870 and x double dot, these and pre multiply by u transpose. 193 00:13:27,870 --> 00:13:29,545 Remember the transpose of a matrix. 194 00:13:29,545 --> 00:13:32,410 You just take the first column, make it the first row. 195 00:13:32,410 --> 00:13:34,580 Second column, make it the second row. 196 00:13:34,580 --> 00:13:38,540 So if I plug in these up here, I get 197 00:13:38,540 --> 00:13:47,213 muq-- I'm going to leave some space here 198 00:13:47,213 --> 00:13:55,980 because I'm going to pre multiply in a second-- plus cuq 199 00:13:55,980 --> 00:14:10,740 dot plus kuq equals the external exciting forces. 200 00:14:10,740 --> 00:14:25,150 Now I'm going to pre multiply by u transpose, OK? 201 00:14:25,150 --> 00:14:30,300 Now a remarkable thing happens. 202 00:14:30,300 --> 00:14:35,350 It happens that when you do this calculation, when you multiply 203 00:14:35,350 --> 00:14:39,640 this matrix times that, one row at a time-- so this 204 00:14:39,640 --> 00:14:42,430 has vectors in it, 1 through n. 205 00:14:42,430 --> 00:14:44,860 I'm going to pick vector r, the rth one. 206 00:14:44,860 --> 00:14:49,160 If I take that rth vector and multiply it one 207 00:14:49,160 --> 00:14:53,700 at a time by row by row by row, then I 208 00:14:53,700 --> 00:15:00,350 get a new vector that results, which 209 00:15:00,350 --> 00:15:03,240 I'm going to multiply by this. 210 00:15:03,240 --> 00:15:07,190 And so if I'm going to pick out one of the vectors, 211 00:15:07,190 --> 00:15:13,850 multiply it through times one of the rows 212 00:15:13,850 --> 00:15:18,290 here-- when you transpose them, the rows are now the vectors. 213 00:15:18,290 --> 00:15:19,230 So I'm going to pick. 214 00:15:19,230 --> 00:15:26,640 If I do the calculation-- lost my right piece of paper here. 215 00:15:41,690 --> 00:15:46,670 So I'm going to just pull out one of the calculations 216 00:15:46,670 --> 00:15:51,600 that you end up doing if you do this whole triple matrix 217 00:15:51,600 --> 00:16:00,610 multiplication, you need to know the following fact. 218 00:16:00,610 --> 00:16:04,280 So for the mode s transpose-- that's one 219 00:16:04,280 --> 00:16:10,410 of the rows out of here-- times m times 220 00:16:10,410 --> 00:16:17,340 one of the columns, the rth one from here, 221 00:16:17,340 --> 00:16:25,165 and I do this calculation, this is 0 for r not equal to s. 222 00:16:27,670 --> 00:16:32,240 What that statement says is, the only non-zero result from this 223 00:16:32,240 --> 00:16:36,810 is when you multiply-- when you take the rth column from here 224 00:16:36,810 --> 00:16:39,380 and you use the rth from here. 225 00:16:39,380 --> 00:16:42,780 All the other combinations of this thing go to 0. 226 00:16:42,780 --> 00:16:45,580 And the net result of that is that this 227 00:16:45,580 --> 00:16:53,820 implies that u transpose mu always 228 00:16:53,820 --> 00:17:01,080 equals to a diagonal vector, which I'll call this like that. 229 00:17:01,080 --> 00:17:03,590 Sometimes a mass matrix is diagonal to start with. 230 00:17:03,590 --> 00:17:06,050 But even if it isn't, you do this calculation, 231 00:17:06,050 --> 00:17:08,310 it will produce a diagonal matrix. 232 00:17:08,310 --> 00:17:12,680 And that's because these multiplications are always 233 00:17:12,680 --> 00:17:17,160 0 unless r is the same as s. 234 00:17:17,160 --> 00:17:23,619 And the same is true for u transpose Ku will give you 235 00:17:23,619 --> 00:17:31,090 a K matrix that is diagonal. 236 00:17:31,090 --> 00:17:32,640 And you know, normally the stiffness 237 00:17:32,640 --> 00:17:34,930 matrix we've come up with, they've 238 00:17:34,930 --> 00:17:37,330 generally been full matrices oftentimes. 239 00:17:37,330 --> 00:17:42,640 But you do uKu, you will get a diagonal stiffness matrix. 240 00:17:45,600 --> 00:17:48,430 And there the little problem comes 241 00:17:48,430 --> 00:18:04,510 because u transpose cu, well, sometimes, this one is diagonal 242 00:18:04,510 --> 00:18:14,110 only for ideal conditions of damping. 243 00:18:14,110 --> 00:18:16,205 So that's something you just have to address. 244 00:18:29,110 --> 00:18:31,470 So only for ideal conditions, and that's 245 00:18:31,470 --> 00:18:33,250 just something you have to deal with. 246 00:18:40,180 --> 00:18:42,720 So why is this? 247 00:18:42,720 --> 00:18:44,720 Why is there this special, wonderful thing? 248 00:18:44,720 --> 00:18:49,050 The natural modes of a system-- this one is 249 00:18:49,050 --> 00:18:54,510 a two degree of freedom system-- form a complete and independent 250 00:18:54,510 --> 00:18:58,430 set of vectors. 251 00:18:58,430 --> 00:19:01,810 And in this case of this two degree of freedom system, 252 00:19:01,810 --> 00:19:07,440 I can pick any kinematically allowable position, 253 00:19:07,440 --> 00:19:13,000 like this-- stationary, static is one of the solutions, 254 00:19:13,000 --> 00:19:15,150 right, to this two degree of freedom system-- 255 00:19:15,150 --> 00:19:19,720 so any possible allowable position of these two things, 256 00:19:19,720 --> 00:19:23,270 static or moving, can be described 257 00:19:23,270 --> 00:19:28,950 by a linear combination of the mode shapes of the system, 258 00:19:28,950 --> 00:19:32,041 a weighted sum of the mode shapes of the system. 259 00:19:32,041 --> 00:19:33,040 And that's all it takes. 260 00:19:33,040 --> 00:19:34,990 So this one has two mode shapes, one 261 00:19:34,990 --> 00:19:37,050 that looks like that, one that looks 262 00:19:37,050 --> 00:19:40,490 like this one's going down, this one's going up, 263 00:19:40,490 --> 00:19:44,990 their particular ratios. 264 00:19:44,990 --> 00:19:49,000 And I can take a weighted amount of that first mode, so much 265 00:19:49,000 --> 00:19:51,000 of it, and a weighted amount of the second mode 266 00:19:51,000 --> 00:19:55,500 and add them together and describe any possible position 267 00:19:55,500 --> 00:19:56,670 of the system. 268 00:19:56,670 --> 00:19:59,590 The same thing is true of that string. 269 00:19:59,590 --> 00:20:01,790 It has mode shapes that are sine waves, 270 00:20:01,790 --> 00:20:08,100 but they're sine 1 pi x, sine pi 2x, over and so forth. 271 00:20:08,100 --> 00:20:11,190 Any possible allowable shape of that guitar string 272 00:20:11,190 --> 00:20:13,980 can be made up of a weighted sum of the mode 273 00:20:13,980 --> 00:20:15,670 shapes of the system. 274 00:20:15,670 --> 00:20:19,940 And moreover, the mode shapes, the reason 275 00:20:19,940 --> 00:20:34,500 this works is because the mode shapes 276 00:20:34,500 --> 00:20:36,540 are orthogonal to one another. 277 00:20:45,812 --> 00:20:50,860 Now, you know that if you take 2 sine waves like that string 278 00:20:50,860 --> 00:20:55,140 and you take first mode sine pi x over l, 279 00:20:55,140 --> 00:20:57,560 and second mode say sine 2 pi x over l 280 00:20:57,560 --> 00:20:59,984 and you multiply them together and integrate from 0 281 00:20:59,984 --> 00:21:00,900 to l, what do you get? 282 00:21:05,040 --> 00:21:08,950 You'll always get 0 if the two sines are-- 283 00:21:08,950 --> 00:21:11,530 if they're full wavelengths, they go to nodes at the end, 284 00:21:11,530 --> 00:21:15,890 you will always get 0 if the wavelengths are different, 285 00:21:15,890 --> 00:21:18,610 always, right? 286 00:21:18,610 --> 00:21:20,510 That's a statement of orthogonality 287 00:21:20,510 --> 00:21:23,500 of sine functions. 288 00:21:23,500 --> 00:21:24,000 All right. 289 00:21:24,000 --> 00:21:25,958 The same thing is true of these simple vectors. 290 00:21:25,958 --> 00:21:28,140 They are orthogonal to one another such 291 00:21:28,140 --> 00:21:30,940 that if you do this multiplication, 292 00:21:30,940 --> 00:21:34,930 you transpose mu, you only get contributions 293 00:21:34,930 --> 00:21:39,475 when you are using mode r transpose m mode r. 294 00:21:39,475 --> 00:21:41,602 You only get a contribution of each of those. 295 00:21:41,602 --> 00:21:42,810 That gives you the diagonals. 296 00:21:42,810 --> 00:21:45,600 The same is true when you do u transpose ku. 297 00:21:45,600 --> 00:21:48,620 Because of orthogonality, you only get a diagonal matrix 298 00:21:48,620 --> 00:21:49,610 at the end. 299 00:21:49,610 --> 00:21:52,450 And under the right conditions, u transpose cu 300 00:21:52,450 --> 00:21:55,950 gives you a diagonal matrix. 301 00:21:55,950 --> 00:21:57,370 So what's that good for? 302 00:22:11,130 --> 00:22:13,470 Well, here was the set of equations 303 00:22:13,470 --> 00:22:16,240 that we get when we make that substitution. 304 00:22:16,240 --> 00:22:23,730 This is going to give us a diagonal mass 305 00:22:23,730 --> 00:22:29,590 matrix times q double dot plus, when conditions are right, 306 00:22:29,590 --> 00:22:34,518 a diagonal damping matrix times q dot, 307 00:22:34,518 --> 00:22:41,990 plus a diagonal stiffness matrix times q 308 00:22:41,990 --> 00:22:49,970 equals u transpose F, which as a vector times a matrix 309 00:22:49,970 --> 00:22:52,390 gives you back a vector, which we call capital 310 00:22:52,390 --> 00:22:54,795 Q. It's a function of time. 311 00:22:54,795 --> 00:22:56,465 And this is called the modal force. 312 00:23:02,140 --> 00:23:04,240 But if you look carefully at these, 313 00:23:04,240 --> 00:23:13,150 if I pick the rth one, mode r out of this whole thing-- 314 00:23:13,150 --> 00:23:16,540 if I just pick any mode out of this, any part of this vector, 315 00:23:16,540 --> 00:23:18,100 and complete this multiplication, 316 00:23:18,100 --> 00:23:22,750 I will find that I get an Mr, which is the rth entry here. 317 00:23:22,750 --> 00:23:26,125 And now I'm going to refer to these as the modal masses, 318 00:23:26,125 --> 00:23:28,080 and I'll write them with capitals 319 00:23:28,080 --> 00:23:30,690 and I'll give a subscript to tell you what the mode is. 320 00:23:30,690 --> 00:23:32,250 This is a number. 321 00:23:32,250 --> 00:23:35,140 This is the modal mass for mode r. 322 00:23:35,140 --> 00:23:37,070 This gives me an equation that looks 323 00:23:37,070 --> 00:23:50,161 like Mrqr double dot plus crqr dot plus Krqr equals Qr of t. 324 00:23:50,161 --> 00:23:55,410 And what does that remind you of that we've done a lot of work 325 00:23:55,410 --> 00:23:56,517 with? 326 00:23:56,517 --> 00:23:58,271 AUDIENCE: [INAUDIBLE] 327 00:23:58,271 --> 00:24:00,145 PROFESSOR: How many degree of freedom system? 328 00:24:03,490 --> 00:24:07,150 That's the equation of motion, the generic equation of motion, 329 00:24:07,150 --> 00:24:09,130 of a single degree of freedom oscillator. 330 00:24:09,130 --> 00:24:11,450 And you know how to calculate the response 331 00:24:11,450 --> 00:24:13,160 to initial conditions for that. 332 00:24:13,160 --> 00:24:14,920 You know how to calculate the steady state 333 00:24:14,920 --> 00:24:18,600 response for that when you have a harmonic input, right? 334 00:24:18,600 --> 00:24:23,360 What I said at the beginning of the discussion about vibration 335 00:24:23,360 --> 00:24:25,540 is it's really important to understand 336 00:24:25,540 --> 00:24:27,410 the single degree of freedom oscillator 337 00:24:27,410 --> 00:24:31,290 because it'll give you insight as to the behavior 338 00:24:31,290 --> 00:24:33,920 of complicated multiple degree of freedom systems. 339 00:24:33,920 --> 00:24:36,723 And here's the proof of this. 340 00:24:40,347 --> 00:24:47,430 This is now n uncoupled single degree of freedom systems. 341 00:24:47,430 --> 00:25:01,000 This is n independent single, one degree of freedom systems 342 00:25:01,000 --> 00:25:02,600 which you can solve one at a time. 343 00:25:19,100 --> 00:25:25,610 Now, lots of times a vibrating system, a complicated one, 344 00:25:25,610 --> 00:25:28,440 might be this thing. 345 00:25:28,440 --> 00:25:30,110 If I hit this, it's vibrating. 346 00:25:30,110 --> 00:25:31,930 And actually, it's pretty much vibrating 347 00:25:31,930 --> 00:25:32,846 at a single frequency. 348 00:25:36,260 --> 00:25:41,320 And once I've hit it, are there any external forces driving it? 349 00:25:41,320 --> 00:25:43,060 So what kind of response are you seeing? 350 00:25:45,950 --> 00:25:47,032 Response to? 351 00:25:47,032 --> 00:25:48,240 AUDIENCE: Initial conditions. 352 00:25:48,240 --> 00:25:50,230 PROFESSOR: Initial conditions, right? 353 00:25:50,230 --> 00:25:54,720 Now in general, each one of the natural modes of a system 354 00:25:54,720 --> 00:25:58,730 has a different natural frequency, right? 355 00:25:58,730 --> 00:26:00,979 So if I hit this thing and I look at it, 356 00:26:00,979 --> 00:26:03,020 really, I can just see it wiggling back and forth 357 00:26:03,020 --> 00:26:06,140 basically at one frequency So if you 358 00:26:06,140 --> 00:26:09,580 wanted to come up with a simple model of this system, 359 00:26:09,580 --> 00:26:11,630 how many natural modes you think you'd 360 00:26:11,630 --> 00:26:15,320 have to include to describe the motion of this system? 361 00:26:15,320 --> 00:26:16,220 AUDIENCE: One. 362 00:26:16,220 --> 00:26:17,150 PROFESSOR: One. 363 00:26:17,150 --> 00:26:19,570 Now is that a lot easier than having 364 00:26:19,570 --> 00:26:22,120 to do the full general equation of motion 365 00:26:22,120 --> 00:26:25,780 for all the possible modes that this thing has? 366 00:26:25,780 --> 00:26:27,654 And it turns out a lot, right, you 367 00:26:27,654 --> 00:26:29,320 have to deal with the equation of motion 368 00:26:29,320 --> 00:26:31,890 of a single degree of freedom system to describe this. 369 00:26:31,890 --> 00:26:34,050 And that's the real point. 370 00:26:34,050 --> 00:26:36,440 You know they built the Hancock building 371 00:26:36,440 --> 00:26:40,350 across the river 35 years ago. 372 00:26:40,350 --> 00:26:42,510 It was losing windows like crazy. 373 00:26:42,510 --> 00:26:44,250 It was a brand new building. 374 00:26:44,250 --> 00:26:47,800 And when the wind would get above 40 miles an hour, 375 00:26:47,800 --> 00:26:49,180 the windows started falling out. 376 00:26:49,180 --> 00:26:53,900 60 stories high, 60, 61 stories high, and the wind was blowing. 377 00:26:53,900 --> 00:26:56,440 Where do you suppose the windows fall out? 378 00:26:56,440 --> 00:26:58,072 What part of the building? 379 00:26:58,072 --> 00:26:59,425 AUDIENCE: [INAUDIBLE] 380 00:26:59,425 --> 00:27:00,359 PROFESSOR: Huh? 381 00:27:00,359 --> 00:27:02,400 I mean, you'd think that when the wind is blowing 382 00:27:02,400 --> 00:27:03,650 it get stronger as it goes up. 383 00:27:03,650 --> 00:27:06,860 It was probably blowing out the windows at the top, right? 384 00:27:06,860 --> 00:27:10,010 But the windows were breaking-- as time went by, 385 00:27:10,010 --> 00:27:13,530 every time a window would break, they replaced this five foot 386 00:27:13,530 --> 00:27:16,580 by nine foot sheet of glass with a piece of plywood. 387 00:27:16,580 --> 00:27:18,250 And so you get this statistical sampling 388 00:27:18,250 --> 00:27:20,190 after a while of where the breakage was. 389 00:27:20,190 --> 00:27:24,060 So you had no windows broken at the top and a few 390 00:27:24,060 --> 00:27:27,380 as you got further down and more and lots of them broken out 391 00:27:27,380 --> 00:27:29,860 at the bottom. 392 00:27:29,860 --> 00:27:34,750 It turns out that that building was vibrating mostly 393 00:27:34,750 --> 00:27:35,925 in its first bending mode. 394 00:27:35,925 --> 00:27:39,315 It was going back and forth like this. 395 00:27:39,315 --> 00:27:40,940 Also happened to have a torsional mode. 396 00:27:40,940 --> 00:27:43,120 Its first torsional mode was kind of twisting around 397 00:27:43,120 --> 00:27:44,400 the base like that. 398 00:27:44,400 --> 00:27:46,415 So in fact the moment when the wind 399 00:27:46,415 --> 00:27:47,790 would get about 40 miles an hour, 400 00:27:47,790 --> 00:27:50,270 this building would start rocking and rolling, 401 00:27:50,270 --> 00:27:54,590 mostly like this with a little of this thrown in, OK? 402 00:27:54,590 --> 00:27:57,420 But you can basically model that complicated building that 403 00:27:57,420 --> 00:28:01,200 has millions of possible natural modes in it by one 404 00:28:01,200 --> 00:28:05,520 or possibly two single degree of freedom oscillators. 405 00:28:05,520 --> 00:28:07,240 So that's the power of modal analysis. 406 00:28:07,240 --> 00:28:09,580 But I think the real power of understanding 407 00:28:09,580 --> 00:28:12,080 that you can do this is that it gives you 408 00:28:12,080 --> 00:28:13,779 this immediate insight as to what 409 00:28:13,779 --> 00:28:15,070 might be going on in something. 410 00:28:15,070 --> 00:28:16,090 So I look at this. 411 00:28:16,090 --> 00:28:17,860 I don't see a complicated thing that I 412 00:28:17,860 --> 00:28:19,900 have to model with a big finite element model. 413 00:28:19,900 --> 00:28:22,570 I see something that's vibrating at one frequency. 414 00:28:22,570 --> 00:28:26,260 And I know it has a little decay. 415 00:28:26,260 --> 00:28:27,390 It'll have damping. 416 00:28:27,390 --> 00:28:29,480 It'll have a natural frequency. 417 00:28:29,480 --> 00:28:31,970 And I get immediate insight as to its behavior 418 00:28:31,970 --> 00:28:34,238 by knowing this, OK. 419 00:28:34,238 --> 00:28:38,060 And that's the real reason why I wanted to show this to you. 420 00:28:38,060 --> 00:28:46,595 So today we'll do-- there are sort of two directions we 421 00:28:46,595 --> 00:28:47,640 can go with this. 422 00:28:47,640 --> 00:28:51,340 One is to talk about response to initial conditions, 423 00:28:51,340 --> 00:28:53,190 and the other is to talk about the response 424 00:28:53,190 --> 00:28:54,700 to force excitation. 425 00:28:54,700 --> 00:29:08,810 So we're going to begin by doing response to ICs, OK? 426 00:29:08,810 --> 00:29:11,350 And then we'll move on probably next time 427 00:29:11,350 --> 00:29:13,960 and talk about response to harmonic excitations. 428 00:29:17,420 --> 00:29:21,310 And we're going to use that as the example. 429 00:29:21,310 --> 00:29:25,920 Before I go there, once we have broken the system down 430 00:29:25,920 --> 00:29:29,637 and analyzed this way, how do we get back 431 00:29:29,637 --> 00:29:34,811 to the motion of the system in our generalized coordinates, 432 00:29:34,811 --> 00:29:36,560 which are the ones we're comfortable with? 433 00:29:36,560 --> 00:29:39,930 Because I don't know where to take a ruler and go 434 00:29:39,930 --> 00:29:45,090 measure this natural coordinate. 435 00:29:45,090 --> 00:29:46,850 So somehow I have to get back to putting 436 00:29:46,850 --> 00:29:49,570 in the real physical measurements 437 00:29:49,570 --> 00:29:50,590 that I can relate to. 438 00:29:50,590 --> 00:29:53,620 Well, that's easy because where did we start with this? 439 00:29:53,620 --> 00:29:56,960 We started by saying this whole thing began right here. 440 00:30:03,760 --> 00:30:09,180 And so at the end we just come back and say, oh, well, 441 00:30:09,180 --> 00:30:12,420 x here, our generalized coordinates, 442 00:30:12,420 --> 00:30:24,180 is this summation of the mode shapes ui here, summed 443 00:30:24,180 --> 00:30:30,880 over i of qi of t. 444 00:30:30,880 --> 00:30:35,750 Now the reason I wrote it here as a summation 445 00:30:35,750 --> 00:30:37,735 is to remind you that you do this. 446 00:30:50,272 --> 00:30:51,730 That's the beauty of this thing, is 447 00:30:51,730 --> 00:30:54,570 you only have to do it over the modes that matter. 448 00:30:54,570 --> 00:30:56,830 So if you've decided to approximate the motion 449 00:30:56,830 --> 00:31:00,020 of this complicated system, by just a couple 450 00:31:00,020 --> 00:31:01,550 of motile contributions because you 451 00:31:01,550 --> 00:31:04,210 know they're the important ones, this 452 00:31:04,210 --> 00:31:06,450 is a pretty short summation. 453 00:31:06,450 --> 00:31:09,770 This is how you get back to your original modal coordinates. 454 00:31:09,770 --> 00:31:15,630 Just take the modal amplitude, multiply it by the mode shape. 455 00:31:15,630 --> 00:31:18,360 And when you do that, it says, if this turns out 456 00:31:18,360 --> 00:31:23,130 to be, say, sum a sine omega t, when you multiply by the mode 457 00:31:23,130 --> 00:31:25,440 shape it basically tells you how much 458 00:31:25,440 --> 00:31:29,150 each generalized coordinate gets of the motion. 459 00:31:29,150 --> 00:31:31,570 The mode shape distributes thing answer 460 00:31:31,570 --> 00:31:34,360 out proportionally in the correct amount. 461 00:31:34,360 --> 00:31:37,270 So this is how you get back to the original. 462 00:31:37,270 --> 00:31:43,710 So let's think about that system and we'll 463 00:31:43,710 --> 00:31:46,190 do an initial conditions kind of problem. 464 00:31:46,190 --> 00:31:48,120 So I think Professor Gossard-- I think 465 00:31:48,120 --> 00:31:51,070 in class you sort of figured out what the approximate ks 466 00:31:51,070 --> 00:31:53,500 and ms and things were for that system. 467 00:31:53,500 --> 00:31:56,830 So I actually took it apart, weighed it, 468 00:31:56,830 --> 00:31:58,470 measured some natural frequencies, 469 00:31:58,470 --> 00:32:06,120 and have come up with a pretty good model, 470 00:32:06,120 --> 00:32:08,800 or at least pretty good set of numbers, 471 00:32:08,800 --> 00:32:11,690 characterizing this two degree of freedom system. 472 00:32:22,250 --> 00:32:36,210 So c1, k1, m1, k2, c2, x1, x2. 473 00:32:36,210 --> 00:32:39,094 So these are my generalized coordinates, measured probably 474 00:32:39,094 --> 00:32:39,885 from what position? 475 00:32:45,140 --> 00:32:46,690 Static equilibrium, right? 476 00:32:46,690 --> 00:32:49,980 So I don't have to mess with gravity in this. 477 00:32:49,980 --> 00:32:51,960 Measured from static equilibrium. 478 00:32:51,960 --> 00:32:58,780 And to try to help keep things understandable, 479 00:32:58,780 --> 00:33:01,840 I tried to write the parameters of the system 480 00:33:01,840 --> 00:33:04,690 as lowercase k1s, k2s, k3s because I 481 00:33:04,690 --> 00:33:09,490 want to write modal stiffness for mode one as a capital K1, 482 00:33:09,490 --> 00:33:12,200 so I try to be consistent about that. 483 00:33:12,200 --> 00:33:16,000 And notice where I put the dampers in the system. 484 00:33:16,000 --> 00:33:18,010 That's because most of the damping in this thing 485 00:33:18,010 --> 00:33:23,400 comes from the upper mass rubs against a stationary object, 486 00:33:23,400 --> 00:33:25,930 which is the bar here. 487 00:33:25,930 --> 00:33:28,610 The lower mass rubs against a stationary object. 488 00:33:28,610 --> 00:33:32,010 So I'm going to model that as a dashpot between each mass 489 00:33:32,010 --> 00:33:36,230 and the fixed reference frame because the bar doesn't move. 490 00:33:36,230 --> 00:33:38,680 So it's an approximate model of the damping. 491 00:33:38,680 --> 00:33:43,180 And so if we do our sum of forces on each of these masses, 492 00:33:43,180 --> 00:33:45,440 just do Newton's laws on the mass, 493 00:33:45,440 --> 00:33:48,530 we can come up with our two equations of motion. 494 00:33:48,530 --> 00:33:50,140 We get two equations of motion. 495 00:33:54,550 --> 00:33:55,740 And let's see. 496 00:33:59,010 --> 00:34:01,746 I think I'll give you some information here first. 497 00:34:01,746 --> 00:34:02,246 m1. 498 00:34:43,420 --> 00:34:45,150 And I really don't know the damping, 499 00:34:45,150 --> 00:34:49,000 but we can get that by just counting how many cycles it 500 00:34:49,000 --> 00:34:51,110 takes to decay and so forth. 501 00:34:51,110 --> 00:34:53,850 So that's basically what I come into this problem knowing. 502 00:34:57,580 --> 00:35:00,210 And I'm going to write my equations of motion 503 00:35:00,210 --> 00:35:02,300 in matrix form. 504 00:35:02,300 --> 00:35:04,560 So it's going to end up looking like m1. 505 00:35:18,790 --> 00:35:22,320 Now notice the damping in this one, the damping force, 506 00:35:22,320 --> 00:35:26,190 is only proportion-- it'll be c1 x1. 507 00:35:26,190 --> 00:35:28,640 Doesn't involve the motion of the other object. 508 00:35:28,640 --> 00:35:30,270 In this one, the damping force only 509 00:35:30,270 --> 00:35:31,990 involves the second motion. 510 00:35:31,990 --> 00:35:37,680 So this one happens to look like a c1, 0, c2. 511 00:35:45,950 --> 00:35:55,170 And the stiffness matrix, well, that's k1 plus k2, minus k2, 512 00:35:55,170 --> 00:36:05,520 minus k2, and k2, x1, x2. 513 00:36:05,520 --> 00:36:11,586 And for no external forces, this starts off this one 514 00:36:11,586 --> 00:36:14,380 has nothing on the right hand side. it's equal to 0. 515 00:36:14,380 --> 00:36:16,130 So those are my equations of motion. 516 00:36:16,130 --> 00:36:17,730 And you know if you multiply these out 517 00:36:17,730 --> 00:36:19,620 you'd get two equations. 518 00:36:19,620 --> 00:36:22,030 And each one would be this result 519 00:36:22,030 --> 00:36:25,380 that you get by apply Newton's law to mass one 520 00:36:25,380 --> 00:36:27,239 and Newton's law to mass two. 521 00:36:27,239 --> 00:36:28,780 But you we've done that enough times. 522 00:36:28,780 --> 00:36:31,070 I'm not going to go through that part of it. 523 00:36:34,014 --> 00:36:34,514 OK. 524 00:37:03,580 --> 00:37:17,115 And putting it in real numbers, that's our mass matrix. 525 00:37:20,670 --> 00:37:22,760 I don't know this. 526 00:37:22,760 --> 00:37:24,180 My stiffness matrix. 527 00:37:47,080 --> 00:37:50,160 So this is my K matrix here. 528 00:37:50,160 --> 00:37:55,635 And stiffness matrices, they're always symmetric. 529 00:37:58,190 --> 00:38:00,130 Although this one happened to be diagonal, 530 00:38:00,130 --> 00:38:02,660 you'll find that mass matrices and even the damping matrices 531 00:38:02,660 --> 00:38:05,430 for our linear systems are symmetric. 532 00:38:05,430 --> 00:38:06,960 So here's my stiffness matrix. 533 00:38:06,960 --> 00:38:09,770 Here's my mass matrix, OK? 534 00:38:12,280 --> 00:38:14,690 And also in this case here's my damping matrix, 535 00:38:14,690 --> 00:38:16,981 but I'm going to leave that because it's the one that's 536 00:38:16,981 --> 00:38:19,420 a little bit troublesome. 537 00:38:19,420 --> 00:38:24,810 So what do I need to do to this to carry out my modal analysis? 538 00:38:24,810 --> 00:38:31,250 So I need to go find the results of computing u transpose m 539 00:38:31,250 --> 00:38:34,912 and u and transpose Ku. 540 00:38:34,912 --> 00:38:36,130 And let's see what we get. 541 00:38:43,390 --> 00:38:45,630 So we need to know a couple things about this system. 542 00:38:50,460 --> 00:38:52,920 We need to know natural frequencies and mode shapes. 543 00:38:52,920 --> 00:38:55,210 So if we have this mass matrix and we 544 00:38:55,210 --> 00:39:00,710 have the stiffness matrix, then we know we can cast this. 545 00:39:00,710 --> 00:39:09,555 We want the undamped natural frequencies and our mode 546 00:39:09,555 --> 00:39:10,055 shapes. 547 00:39:28,510 --> 00:39:32,440 And we know that we can transform 548 00:39:32,440 --> 00:39:35,700 the equations of motion into an algebraic problem where 549 00:39:35,700 --> 00:39:38,390 we solve for the natural frequencies and mode shapes. 550 00:39:38,390 --> 00:39:43,340 So we have, just to remind you really quickly of that, 551 00:39:43,340 --> 00:39:46,550 remember our equations look like this undamped. 552 00:39:49,670 --> 00:39:58,240 And you assume that x is some form u in fact e to the i omega 553 00:39:58,240 --> 00:39:59,990 t. 554 00:39:59,990 --> 00:40:12,210 Plug it in, you get minus omega squared m plus K u e 555 00:40:12,210 --> 00:40:15,155 to the i omega t equals 0. 556 00:40:17,740 --> 00:40:20,160 And this now is your algebraic problem. 557 00:40:20,160 --> 00:40:24,140 e to this unknown set of amplitudes is 0. 558 00:40:24,140 --> 00:40:26,850 These are going to turn out to be the mode shapes. 559 00:40:26,850 --> 00:40:30,660 And they're not generally 0 so that means this has to be 0. 560 00:40:30,660 --> 00:40:33,460 That means we know the determinant of this matrix. 561 00:40:41,740 --> 00:40:44,670 And that'll give you in this case the two 562 00:40:44,670 --> 00:40:45,680 natural frequencies. 563 00:40:45,680 --> 00:40:48,490 This gives of you the omega ns of the system. 564 00:40:48,490 --> 00:40:51,190 Omega n squareds is what you solve for, OK? 565 00:40:51,190 --> 00:40:57,830 And then you go back and you get the mode shapes out of it. 566 00:40:57,830 --> 00:40:59,640 But this you can do on the computer too. 567 00:40:59,640 --> 00:41:03,190 You can either crank out-- for a two degree of freedom system, 568 00:41:03,190 --> 00:41:06,140 this gives you a quadratic omega squared. 569 00:41:06,140 --> 00:41:07,507 You solve it. 570 00:41:07,507 --> 00:41:09,340 You plug it back in and get the mode shapes. 571 00:41:09,340 --> 00:41:10,839 I'm not going to take the time to do 572 00:41:10,839 --> 00:41:13,820 that today because I want to emphasize the modal analysis 573 00:41:13,820 --> 00:41:14,527 part. 574 00:41:14,527 --> 00:41:15,735 So I'll give you the answers. 575 00:41:22,930 --> 00:41:23,910 Where are we here? 576 00:41:32,240 --> 00:41:38,250 So you get omega 1 is 5.6546. 577 00:41:38,250 --> 00:41:42,080 And I seem to be keeping a lot of significant digits, 578 00:41:42,080 --> 00:41:43,330 and there's a reason for that. 579 00:41:53,790 --> 00:41:56,040 In both mode shapes and natural frequencies 580 00:41:56,040 --> 00:41:58,020 you need to carry a lot of significant digits 581 00:41:58,020 --> 00:42:01,070 or modal analysis doesn't work, or at least you 582 00:42:01,070 --> 00:42:02,710 don't get the clean results you expect. 583 00:42:02,710 --> 00:42:06,150 If you're sloppy about the number of significant digits 584 00:42:06,150 --> 00:42:09,440 and you compute u transpose mu, then the 585 00:42:09,440 --> 00:42:12,750 [? off ?] diagonal terms won't quite go to 0. 586 00:42:12,750 --> 00:42:17,040 And it's just because you're not carrying enough precision. 587 00:42:17,040 --> 00:42:22,040 OK, now that's the two natural frequencies. 588 00:42:22,040 --> 00:42:25,320 Now the u matrix, the mode shapes for this system 589 00:42:25,320 --> 00:42:27,270 that goes with that. 590 00:42:27,270 --> 00:42:40,089 u comes out to be 1.0 and 2.2667. 591 00:42:40,089 --> 00:42:40,755 And that's mode. 592 00:42:40,755 --> 00:42:43,230 I'll do this to help you. 593 00:42:43,230 --> 00:42:46,040 The columns are the mode shapes. 594 00:42:46,040 --> 00:42:47,925 That's the first mode shape. 595 00:42:47,925 --> 00:43:01,100 And the second mode shape is 1 and minus 0.2236. 596 00:43:01,100 --> 00:43:04,010 So those are the mode shapes for the first and second mode 597 00:43:04,010 --> 00:43:06,740 that go with these two natural frequencies. 598 00:43:06,740 --> 00:43:08,310 So that's for this system. 599 00:43:08,310 --> 00:43:09,810 The top one moves one unit. 600 00:43:09,810 --> 00:43:15,380 The bottom one moves 2.27 times that, same direction, 601 00:43:15,380 --> 00:43:16,750 positive, positive. 602 00:43:16,750 --> 00:43:18,396 So the upper one moves one unit. 603 00:43:18,396 --> 00:43:20,270 The bottom one moves the opposite direction-- 604 00:43:20,270 --> 00:43:22,186 that's the minus signs-- equivalent to a phase 605 00:43:22,186 --> 00:43:26,740 angle of 180 degrees minus 22% of the amount 606 00:43:26,740 --> 00:43:28,570 that the upper one moves. 607 00:43:28,570 --> 00:43:31,760 So first one moves one unit. 608 00:43:31,760 --> 00:43:35,820 The bottom one moves 2.2 times that. 609 00:43:35,820 --> 00:43:39,100 And then the second mode, which is much harder to get going. 610 00:43:42,765 --> 00:43:44,140 Guess the only way I can do it is 611 00:43:44,140 --> 00:43:46,470 to do it the way Professor Gossard intended here. 612 00:43:51,910 --> 00:43:58,510 One unit up and down, minus 0.2236, going the opposite way. 613 00:43:58,510 --> 00:43:59,890 So those are our mode shapes. 614 00:43:59,890 --> 00:44:01,570 These are the natural frequencies. 615 00:44:01,570 --> 00:44:04,240 I calculated this one and measured it with a stopwatch. 616 00:44:04,240 --> 00:44:06,650 This one I can do watching it with a stopwatch. 617 00:44:06,650 --> 00:44:12,856 And I came within better than 1% of getting the same number. 618 00:44:12,856 --> 00:44:13,355 OK. 619 00:44:36,360 --> 00:44:38,510 So I want my model mass matrix. 620 00:44:38,510 --> 00:44:40,680 I carry out this calculation. 621 00:44:40,680 --> 00:44:42,680 And for this system, remember, it's 622 00:44:42,680 --> 00:44:48,540 going to give me back a diagonal matrix looking like this. 623 00:44:48,540 --> 00:45:05,120 And in fact, the numbers are 3.5562, 0, 0, and 0.3508. 624 00:45:05,120 --> 00:45:10,430 And when I calculate u transpose Ku, 625 00:45:10,430 --> 00:45:12,690 gives me a diagonal stiffness matrix. 626 00:45:16,940 --> 00:45:38,670 And I get the numbers 113.71 and 0, 0, 109.839. 627 00:45:38,670 --> 00:45:43,480 And that's my diagonalized stiffness matrix. 628 00:45:43,480 --> 00:45:46,090 Now something had better be true. 629 00:45:46,090 --> 00:45:47,820 I'm saying that this is now going 630 00:45:47,820 --> 00:45:50,230 to give me my two independent single degree of freedom 631 00:45:50,230 --> 00:45:51,640 equations of motion, right? 632 00:46:03,219 --> 00:46:07,380 So what I'm seeking here, I want to get two equations, 633 00:46:07,380 --> 00:46:17,550 one that looks like m1q1 double dot plus c1q1 dot plus K1q1 634 00:46:17,550 --> 00:46:20,137 equals 0 for no external force. 635 00:46:20,137 --> 00:46:21,720 That's one of the equations I'm after. 636 00:46:21,720 --> 00:46:30,496 And the other one will look like m2q2 double dot plus c2q2 dot 637 00:46:30,496 --> 00:46:30,995 K2q2. 638 00:46:33,930 --> 00:46:36,730 Now one way to check that you've gotten the right thing 639 00:46:36,730 --> 00:46:39,700 is now these are two independent single degree of freedom 640 00:46:39,700 --> 00:46:40,340 systems. 641 00:46:40,340 --> 00:46:42,173 What's the natural frequency of this system? 642 00:46:52,750 --> 00:46:53,250 Yeah? 643 00:46:53,250 --> 00:46:55,870 Actually, I heard somebody say square root of K1 over m1. 644 00:46:55,870 --> 00:46:56,870 That had better be true. 645 00:46:56,870 --> 00:46:58,650 But numerically what's the number? 646 00:46:58,650 --> 00:47:02,010 What had it better be? 647 00:47:02,010 --> 00:47:04,510 It better be the omega 1 of the system, right? 648 00:47:04,510 --> 00:47:07,230 And so a check that you can perform 649 00:47:07,230 --> 00:47:18,560 is to check to see if the omega 1 squared equals K1/m1. 650 00:47:18,560 --> 00:47:20,590 You found two numbers. 651 00:47:20,590 --> 00:47:27,950 You've got, up here, K1 is 113.7. 652 00:47:27,950 --> 00:47:29,990 m1 is 3.55. 653 00:47:29,990 --> 00:47:33,850 Take K1/m1, and take its square root. 654 00:47:33,850 --> 00:47:39,530 So K1/m1, that's about 30 something. 655 00:47:39,530 --> 00:47:43,946 Square root of 30 something is a little less than 6. 656 00:47:43,946 --> 00:47:46,660 Omega 1 is 5.65. 657 00:47:46,660 --> 00:47:49,980 And same thing, omega 2 had better 658 00:47:49,980 --> 00:47:54,130 be equal to the square root of K2/m2. 659 00:47:54,130 --> 00:47:55,640 So one of the things you can always 660 00:47:55,640 --> 00:47:57,100 do when you do your modal analysis, 661 00:47:57,100 --> 00:48:01,480 you do your calculations, u transpose mu, u transpose Ku. 662 00:48:01,480 --> 00:48:04,326 If you calculate the ratios of each one of these things, 663 00:48:04,326 --> 00:48:05,742 you can go back and check that you 664 00:48:05,742 --> 00:48:07,408 can see that the natural frequencies are 665 00:48:07,408 --> 00:48:09,020 the ones that you started with. 666 00:48:09,020 --> 00:48:11,730 If they are not, then you've messed up in your arithmetic. 667 00:48:11,730 --> 00:48:14,550 So now we've got our two independent equations. 668 00:48:14,550 --> 00:48:18,310 And the natural frequencies check out. 669 00:48:18,310 --> 00:48:21,880 But we still have a couple of things to deal with. 670 00:48:21,880 --> 00:48:24,030 We have to figure out how to calculate 671 00:48:24,030 --> 00:48:26,469 the initial conditions, and we have to figure out 672 00:48:26,469 --> 00:48:27,510 how to deal with damping. 673 00:48:30,980 --> 00:48:34,280 Let's do ICs first. 674 00:48:53,690 --> 00:48:56,020 So those of you who were here last time, 675 00:48:56,020 --> 00:48:59,930 I ended kind of right at the end. 676 00:48:59,930 --> 00:49:02,550 We kind of worked our way through figuring out 677 00:49:02,550 --> 00:49:06,214 the initial conditions for a two degree of freedom system 678 00:49:06,214 --> 00:49:07,130 doing it the hard way. 679 00:49:07,130 --> 00:49:09,840 You end up with four equations and four unknowns 680 00:49:09,840 --> 00:49:12,450 for the a1, a2, phi 1, phi two. 681 00:49:12,450 --> 00:49:13,070 Remember that? 682 00:49:13,070 --> 00:49:16,150 I mean, it's really painful. 683 00:49:16,150 --> 00:49:18,510 This is incredibly easier. 684 00:49:18,510 --> 00:49:23,160 We're going to do the same thing, but extremely easily. 685 00:49:23,160 --> 00:49:27,000 So I would never go myself given the choice of grinding out 686 00:49:27,000 --> 00:49:29,850 all those phase angles and amplitudes 687 00:49:29,850 --> 00:49:31,260 in simultaneous equations. 688 00:49:31,260 --> 00:49:33,740 I'd do the following. 689 00:49:33,740 --> 00:49:35,960 Generally now I know the initial conditions are 690 00:49:35,960 --> 00:49:38,790 going to be specified not in q coordinates 691 00:49:38,790 --> 00:49:40,050 but in what coordinate system? 692 00:49:44,310 --> 00:49:47,200 In your original generalized coordinates, right? 693 00:49:47,200 --> 00:49:49,687 You know, your x, this one. 694 00:49:49,687 --> 00:49:51,520 If I'm going to set initial conditions here, 695 00:49:51,520 --> 00:49:54,640 I'm not going to say q1 is equal to something. 696 00:49:54,640 --> 00:49:57,420 I'm going to put this one down one unit and this one 697 00:49:57,420 --> 00:49:59,110 down two units and let go. 698 00:49:59,110 --> 00:50:01,070 This is in x1 and x2 coordinates. 699 00:50:01,070 --> 00:50:07,085 But the beautiful thing here is that we know that x equals uq. 700 00:50:10,130 --> 00:50:13,340 So if I know the initial conditions on I'll 701 00:50:13,340 --> 00:50:18,840 call it x0 here, if I know the initial deflections 702 00:50:18,840 --> 00:50:21,540 of the system, they're going to be u 703 00:50:21,540 --> 00:50:27,190 times the initial values of q. 704 00:50:27,190 --> 00:50:33,070 And if I know a vector of initial velocities at time 0, 705 00:50:33,070 --> 00:50:35,780 they're going to be uq0 dot. 706 00:50:39,570 --> 00:50:44,070 So if I told you values of x0 and you 707 00:50:44,070 --> 00:50:48,050 know that this equation's true, what we need is the q0s. 708 00:50:48,050 --> 00:50:50,730 We need the initial conditions in the modal 709 00:50:50,730 --> 00:50:53,249 coordinates in order to finish this problem. 710 00:50:53,249 --> 00:50:55,290 If I told you this, how would you solve for that? 711 00:50:59,510 --> 00:51:01,160 Just a little linear algebra here. 712 00:51:04,520 --> 00:51:05,960 AUDIENCE: Inverse matrix of u? 713 00:51:05,960 --> 00:51:07,400 PROFESSOR: Yeah, do what with it? 714 00:51:07,400 --> 00:51:08,900 AUDIENCE: Then you multiply x by it. 715 00:51:08,900 --> 00:51:13,290 PROFESSOR: Multiply it by u inverse, right? 716 00:51:13,290 --> 00:51:24,440 OK, so this implies that q0-- well, 717 00:51:24,440 --> 00:51:26,480 I'll write it out a little more fully here. 718 00:51:26,480 --> 00:51:37,590 So if I do u inverse x0, that's going 719 00:51:37,590 --> 00:51:40,885 to be equal to u inverse uq0. 720 00:51:45,210 --> 00:51:48,390 u inverse times u gives you? 721 00:51:48,390 --> 00:51:50,010 1, basically, right? 722 00:51:50,010 --> 00:51:58,337 And so if I do u inverse x0, I get q0. 723 00:51:58,337 --> 00:51:59,420 That's all there is to it. 724 00:51:59,420 --> 00:51:59,920 Yeah? 725 00:51:59,920 --> 00:52:02,510 AUDIENCE: [INAUDIBLE] initial conditions, what about c? 726 00:52:02,510 --> 00:52:04,820 PROFESSOR: All right, c's a problem, OK, 727 00:52:04,820 --> 00:52:07,050 and I'm leaving it to the end. 728 00:52:07,050 --> 00:52:09,530 We're going to deal with it as the last step. 729 00:52:09,530 --> 00:52:15,690 And if I have initial velocities u inverse times x 730 00:52:15,690 --> 00:52:20,440 initial velocities vector, I get the initial velocities vector 731 00:52:20,440 --> 00:52:22,700 in the natural coordinates. 732 00:52:22,700 --> 00:52:25,600 So that's how simple it is to get the initial conditions 733 00:52:25,600 --> 00:52:26,650 in modal coordinates. 734 00:52:26,650 --> 00:52:29,030 Boom, OK? 735 00:52:34,280 --> 00:52:36,830 And we'll do a numerical example in a second here. 736 00:52:43,290 --> 00:52:51,089 We're seeking a solution of the form to do response 737 00:52:51,089 --> 00:52:52,005 to initial conditions. 738 00:52:55,290 --> 00:53:05,094 We seek equations that we know are 739 00:53:05,094 --> 00:53:07,260 right for a single degree of freedom system response 740 00:53:07,260 --> 00:53:08,380 to initial conditions. 741 00:53:08,380 --> 00:53:16,430 So we know that for a single degree of freedom system, x 742 00:53:16,430 --> 00:53:33,940 of t-- this is for SDOF system here-- 743 00:53:33,940 --> 00:53:38,660 we worked out before is equal to some e to the minus zeta 744 00:53:38,660 --> 00:53:40,570 omega nt. 745 00:53:40,570 --> 00:53:50,640 This is just a transient decay problem of x0 cosine omega dt 746 00:53:50,640 --> 00:54:07,860 plus v0 plus zeta omega n x0 all over omega d sine omega dt. 747 00:54:07,860 --> 00:54:10,680 We know that that's what the response of a single degree 748 00:54:10,680 --> 00:54:16,000 of freedom system looks like to initial conditions x0 and v0. 749 00:54:16,000 --> 00:54:18,980 And for light damping, for small damping, 750 00:54:18,980 --> 00:54:21,280 you can usually even ignore this term. 751 00:54:21,280 --> 00:54:24,300 So it's just even simpler. 752 00:54:24,300 --> 00:54:28,700 This term is small compared to that, all right? 753 00:54:28,700 --> 00:54:33,040 This term, contribution from x0, is small compared to this term. 754 00:54:33,040 --> 00:54:36,340 So it's basically dominated by an x0 cosine and a v0 755 00:54:36,340 --> 00:54:38,070 over omega d sine. 756 00:54:38,070 --> 00:54:40,770 But we know that's the exact response 757 00:54:40,770 --> 00:54:43,900 for a single degree of freedom system to initial conditions. 758 00:54:43,900 --> 00:54:46,120 So just by analogy to that, we're 759 00:54:46,120 --> 00:54:49,890 looking for mode one in modal coordinates. 760 00:54:49,890 --> 00:54:52,360 It's going to look exactly the same way. 761 00:54:52,360 --> 00:55:03,120 e to the minus zeta omega nt, q0 cosine omega 1 762 00:55:03,120 --> 00:55:17,960 d-- this is omega dt-- plus q0 dot plus zeta 1 omega 1 q0. 763 00:55:17,960 --> 00:55:24,080 I guess I need to do q10 like that. 764 00:55:24,080 --> 00:55:27,710 This is the first mode's equation, zeta 1. 765 00:55:27,710 --> 00:55:29,507 And I'll call this omega 1. 766 00:55:29,507 --> 00:55:31,840 But now that you get multiple degree of freedom systems, 767 00:55:31,840 --> 00:55:34,173 you got to keep track of what mode you're talking about. 768 00:55:34,173 --> 00:55:36,140 Mode one, damping ratio mode one, 769 00:55:36,140 --> 00:55:40,800 natural frequency mode one, initial displacement mode one, 770 00:55:40,800 --> 00:55:48,810 initial velocity mode one, omega 1 d like that. 771 00:55:48,810 --> 00:55:51,670 And mode two is going to be exactly analogous. 772 00:55:51,670 --> 00:55:56,650 q2 equals, and it's exactly similar, 773 00:55:56,650 --> 00:55:59,210 except you update it with a 2 instead of a 1. 774 00:56:02,550 --> 00:56:05,510 And if you plug in the initial-- you over here 775 00:56:05,510 --> 00:56:14,990 have found the initial values for q10 and q20 and q1 dot 0 776 00:56:14,990 --> 00:56:15,650 and so forth. 777 00:56:15,650 --> 00:56:19,410 You found the initial values that plug into that equation 778 00:56:19,410 --> 00:56:20,560 by just doing this. 779 00:56:23,150 --> 00:56:25,950 And once we have this, then we can go back to saying, 780 00:56:25,950 --> 00:56:29,570 how do you get to the final answer? 781 00:56:29,570 --> 00:56:31,350 Well, you just multiply the result 782 00:56:31,350 --> 00:56:33,300 for q times the mode shape and add them up. 783 00:56:33,300 --> 00:56:35,435 And you have the answer. 784 00:56:35,435 --> 00:56:37,560 But we still have to deal with the damping problem. 785 00:56:41,224 --> 00:56:43,687 We're going to do that one next. 786 00:56:43,687 --> 00:56:45,770 But I see a bunch of hands and some puzzled looks, 787 00:56:45,770 --> 00:56:49,060 so it means it's a good time to stop and talk for a second. 788 00:56:49,060 --> 00:56:50,354 Yeah? 789 00:56:50,354 --> 00:56:53,040 AUDIENCE: [INAUDIBLE] 790 00:56:53,040 --> 00:56:53,872 PROFESSOR: Pardon? 791 00:56:53,872 --> 00:56:55,080 AUDIENCE: What if you [INAUDIBLE] 792 00:56:55,080 --> 00:56:56,329 PROFESSOR: I can't quite hear. 793 00:56:56,329 --> 00:56:58,489 AUDIENCE: The sine theta and the sine rate of g. 794 00:56:58,489 --> 00:56:59,780 PROFESSOR: Yeah, what about it? 795 00:56:59,780 --> 00:57:00,800 AUDIENCE: Why do we lose it? 796 00:57:00,800 --> 00:57:01,660 PROFESSOR: Why do we use it? 797 00:57:01,660 --> 00:57:02,409 AUDIENCE: Lose it. 798 00:57:02,409 --> 00:57:03,860 PROFESSOR: Oh, you don't lose it. 799 00:57:03,860 --> 00:57:07,450 I was saying, you see this bit, it's like that. 800 00:57:07,450 --> 00:57:11,860 These are two pieces that behave like sine. 801 00:57:11,860 --> 00:57:14,630 And see, this one depends on initial displacement 802 00:57:14,630 --> 00:57:17,002 but is multiplied by the damping ratio. 803 00:57:17,002 --> 00:57:19,210 And the damping ratio for things that are interesting 804 00:57:19,210 --> 00:57:21,040 is usually pretty small. 805 00:57:21,040 --> 00:57:29,480 So here you have a term that's x0 cosine omega 1 damped, 806 00:57:29,480 --> 00:57:34,030 and here you have a contribution that's x0 small times 807 00:57:34,030 --> 00:57:39,600 x0 sine omega 1 damped. 808 00:57:39,600 --> 00:57:40,910 So you multiply the same. 809 00:57:40,910 --> 00:57:42,620 They're operating on the same frequency. 810 00:57:42,620 --> 00:57:44,920 Two terms at the same frequency, you add them together, 811 00:57:44,920 --> 00:57:49,620 it's like a cosine omega t minus some phase angle. 812 00:57:49,620 --> 00:57:52,197 If this little term is small, that phase angle's almost 0. 813 00:57:59,560 --> 00:58:03,940 x0 cosine plus something x0 sine, 814 00:58:03,940 --> 00:58:07,930 it gives you a cosine term that is shifted a little bit 815 00:58:07,930 --> 00:58:11,562 and its magnitude is different by this little amount. 816 00:58:11,562 --> 00:58:13,270 I'm just saying oftentimes this is small. 817 00:58:13,270 --> 00:58:14,820 But if you don't want to make that approximation, 818 00:58:14,820 --> 00:58:15,660 just carry it along. 819 00:58:15,660 --> 00:58:16,321 Just do it. 820 00:58:16,321 --> 00:58:17,196 AUDIENCE: [INAUDIBLE] 821 00:58:19,540 --> 00:58:20,290 PROFESSOR: Mm-hmm. 822 00:58:20,290 --> 00:58:21,410 AUDIENCE: [INAUDIBLE] 823 00:58:21,410 --> 00:58:22,790 PROFESSOR: Oh, no. 824 00:58:22,790 --> 00:58:25,950 I'm just saying you can throw out this piece usually. 825 00:58:25,950 --> 00:58:31,930 And it makes-- I keep in my mind-- let me see. 826 00:58:31,930 --> 00:58:32,430 OK, now. 827 00:58:32,430 --> 00:58:35,990 Vibration engineering is full of lots of approximations 828 00:58:35,990 --> 00:58:38,570 because it's very hard oftentimes 829 00:58:38,570 --> 00:58:43,990 to get detailed quantitative numbers on exactly 830 00:58:43,990 --> 00:58:45,310 everything you need to know. 831 00:58:45,310 --> 00:58:47,120 So I carry around little approximations 832 00:58:47,120 --> 00:58:49,390 that I know is the way the world mostly behaves. 833 00:58:49,390 --> 00:58:50,890 And the way the world mostly behaves 834 00:58:50,890 --> 00:58:52,431 for a single degree of freedom system 835 00:58:52,431 --> 00:58:55,610 is the response to initial conditions looks like this. 836 00:58:58,320 --> 00:59:03,690 And this initial value here is always approximately x0. 837 00:59:03,690 --> 00:59:09,920 And this initial slope here is always approximately v0. 838 00:59:12,842 --> 00:59:13,550 That's the slope. 839 00:59:16,120 --> 00:59:20,590 Now, it turns out that this thing is shifted just slightly. 840 00:59:20,590 --> 00:59:21,090 Why? 841 00:59:21,090 --> 00:59:23,600 Because of this term, OK? 842 00:59:23,600 --> 00:59:25,430 But honest, to tell you the truth, 843 00:59:25,430 --> 00:59:27,420 it really rarely matters. 844 00:59:27,420 --> 00:59:30,590 So as a vibration engineer, I just 845 00:59:30,590 --> 00:59:32,860 remember I have an x0 cosine. 846 00:59:32,860 --> 00:59:34,650 I have a v0 over omega d sine. 847 00:59:34,650 --> 00:59:37,090 And the whole thing decays like that. 848 00:59:37,090 --> 00:59:40,080 But if you like to be mathematically precise, 849 00:59:40,080 --> 00:59:41,630 you carry along that a little bit. 850 00:59:41,630 --> 00:59:42,130 Yeah? 851 00:59:42,130 --> 00:59:44,510 AUDIENCE: Don't you lose [INAUDIBLE] sine wave? 852 00:59:44,510 --> 00:59:46,476 PROFESSOR: You're not going to lose the sine. 853 00:59:46,476 --> 00:59:48,474 AUDIENCE: [INAUDIBLE] 854 00:59:48,474 --> 00:59:49,390 PROFESSOR: Oh, oh, oh. 855 00:59:49,390 --> 00:59:50,500 Wait a minute. 856 00:59:50,500 --> 00:59:51,455 I just left it out. 857 00:59:54,660 --> 00:59:57,310 You guys are-- well, I'm glad you're awake. 858 00:59:57,310 --> 00:59:58,060 This is good. 859 01:00:01,990 --> 01:00:03,680 Now how's that? 860 01:00:03,680 --> 01:00:04,560 Ah, good. 861 01:00:04,560 --> 01:00:07,390 Now I know why I had so many puzzled looks. 862 01:00:07,390 --> 01:00:08,860 Anybody have a different question? 863 01:00:08,860 --> 01:00:11,677 Just anything now about this whole modal analysis thing? 864 01:00:11,677 --> 01:00:14,010 Because then we have to deal with this awkward part that 865 01:00:14,010 --> 01:00:16,620 has to do with the damping. 866 01:00:16,620 --> 01:00:19,326 And I've got to finish on time, OK? 867 01:00:23,821 --> 01:00:24,320 All right. 868 01:00:27,540 --> 01:00:28,760 So damping. 869 01:00:28,760 --> 01:00:30,460 I've gotten this far. 870 01:00:30,460 --> 01:00:37,080 What I need is I need estimates for the damping for mode one 871 01:00:37,080 --> 01:00:39,400 and damping for mode two, right? 872 01:00:54,030 --> 01:01:12,110 So the problem is that utcu does not always 873 01:01:12,110 --> 01:01:17,210 equal some nice diagonalized matrix. 874 01:01:17,210 --> 01:01:29,350 You sometimes get these are not always 0, OK? 875 01:01:29,350 --> 01:01:31,330 The orthogonality principle just doesn't 876 01:01:31,330 --> 01:01:34,879 apply to the damping terms. 877 01:01:34,879 --> 01:01:35,420 Just doesn't. 878 01:01:38,000 --> 01:01:40,780 But this actually doesn't hurt you a lot. 879 01:01:40,780 --> 01:01:43,210 You just got to know that this is going to be a problem. 880 01:01:43,210 --> 01:01:45,170 And when the systems are lightly damped, 881 01:01:45,170 --> 01:01:49,050 the approximation, even if your true damping in the system 882 01:01:49,050 --> 01:01:51,700 gives you some non-zero elements here, 883 01:01:51,700 --> 01:01:53,974 the first order behavior of the system 884 01:01:53,974 --> 01:01:55,890 is basically going to be-- you can just ignore 885 01:01:55,890 --> 01:01:57,800 the off-diagonal elements. 886 01:01:57,800 --> 01:01:59,520 What practical consequence do you 887 01:01:59,520 --> 01:02:06,520 think it has if you have some actual non-zero numbers here? 888 01:02:06,520 --> 01:02:08,080 Go back and look at the equations 889 01:02:08,080 --> 01:02:11,170 that you're trying to derive. 890 01:02:11,170 --> 01:02:14,810 These were the equations that we were trying to come up with. 891 01:02:14,810 --> 01:02:20,970 And we wanted them to be n individual single degree 892 01:02:20,970 --> 01:02:22,570 of freedom systems. 893 01:02:22,570 --> 01:02:26,060 But if this has non-zero off-diagonal terms, 894 01:02:26,060 --> 01:02:29,230 you're going to find popping up in this single degree 895 01:02:29,230 --> 01:02:34,200 of freedom equation another term that couples it through damping 896 01:02:34,200 --> 01:02:35,950 to the other modes. 897 01:02:35,950 --> 01:02:39,160 It provides a little bit of coupling to other modes. 898 01:02:39,160 --> 01:02:43,170 They can talk to one another, all right? 899 01:02:43,170 --> 01:02:47,340 And what that means is if I-- this may be a good time 900 01:02:47,340 --> 01:02:50,590 to do the demonstration. 901 01:03:03,485 --> 01:03:04,970 How do I want to say this? 902 01:03:18,870 --> 01:03:22,210 If the initial displacement of the system 903 01:03:22,210 --> 01:03:32,260 is in the shape of one of the natural modes-- 904 01:03:32,260 --> 01:03:36,810 so if this is some u, this is exactly shaped like mode r. 905 01:03:36,810 --> 01:03:38,605 So this looks like the ur vector. 906 01:03:43,970 --> 01:03:47,250 When I carry out this multiplication, 907 01:03:47,250 --> 01:03:50,152 what do you think will happen? 908 01:03:56,260 --> 01:03:58,350 If this is shaped like mode r, because 909 01:03:58,350 --> 01:04:01,870 of orthogonality when I do u inverse, which 910 01:04:01,870 --> 01:04:04,450 is all about the mode shapes information and the mode 911 01:04:04,450 --> 01:04:08,890 shapes are these orthogonal set of independent orthogonal 912 01:04:08,890 --> 01:04:12,970 vectors, if this is exactly one mode 913 01:04:12,970 --> 01:04:17,160 and I do u inverse times that, I will get 0 over here 914 01:04:17,160 --> 01:04:22,200 on the right hand side for every mode except the mode 915 01:04:22,200 --> 01:04:24,650 that that's shaped like. 916 01:04:24,650 --> 01:04:28,210 So if this is shaped like a particular mode, 917 01:04:28,210 --> 01:04:31,310 then over here all the modal initial conditions 918 01:04:31,310 --> 01:04:35,140 are 0 except that mode. 919 01:04:35,140 --> 01:04:39,530 That means if I set this, give its initial conditions are 920 01:04:39,530 --> 01:04:45,370 equal to the shape exactly of mode one, 921 01:04:45,370 --> 01:04:49,490 it only responds in mode one. 922 01:04:49,490 --> 01:04:56,180 And if I give it initial conditions that are exactly 923 01:04:56,180 --> 01:05:01,090 shaped like that of mode two, then it only 924 01:05:01,090 --> 01:05:03,670 responds in mode two. 925 01:05:03,670 --> 01:05:10,290 And if I give it anything else, like I move just the top one 926 01:05:10,290 --> 01:05:13,550 but not the bottom and let it go, 927 01:05:13,550 --> 01:05:17,180 then there's-- maybe I better do the other one. 928 01:05:17,180 --> 01:05:19,610 That one had too much of one and not the other. 929 01:05:19,610 --> 01:05:21,590 If I hold this one, here's its reference. 930 01:05:21,590 --> 01:05:23,256 I'm going to hold it right there and I'm 931 01:05:23,256 --> 01:05:26,352 going to give this one a unit deflection and let go. 932 01:05:26,352 --> 01:05:30,240 Now you see a get some of both, all right? 933 01:05:30,240 --> 01:05:46,520 So if when I do this first one, say first mode, 934 01:05:46,520 --> 01:05:48,660 I could sit here and measure how many cycles 935 01:05:48,660 --> 01:05:51,400 it takes to decay halfway and estimate the damping ratio 936 01:05:51,400 --> 01:05:52,680 for that mode. 937 01:05:52,680 --> 01:05:54,630 If it's only moving in this mode, 938 01:05:54,630 --> 01:05:58,300 I can estimate its damping directly for that mode 939 01:05:58,300 --> 01:05:59,375 and get zeta 1. 940 01:05:59,375 --> 01:06:00,563 You agree? 941 01:06:00,563 --> 01:06:01,430 OK. 942 01:06:01,430 --> 01:06:04,460 And I did the same thing with mode two, it's too fast for me 943 01:06:04,460 --> 01:06:07,680 to catch it with a stopwatch, but I 944 01:06:07,680 --> 01:06:08,980 could measure its damping. 945 01:06:08,980 --> 01:06:11,930 And as it decays, I could get an estimate for zeta 2, 946 01:06:11,930 --> 01:06:15,090 for the damping ratio for mode two, all right? 947 01:06:15,090 --> 01:06:16,370 All right. 948 01:06:16,370 --> 01:06:18,710 But somehow I have to get damping 949 01:06:18,710 --> 01:06:21,300 ratio for mode one, zeta 1, and damping ratio 950 01:06:21,300 --> 01:06:22,590 for mode two, zeta 2. 951 01:06:22,590 --> 01:06:26,260 I have to somehow get it out of this. 952 01:06:26,260 --> 01:06:30,030 I have to model it somehow with these damping coefficients that 953 01:06:30,030 --> 01:06:35,210 come from computing this u transpose cu, OK. 954 01:06:38,580 --> 01:06:41,160 So I'm going to show you kind of damping 955 01:06:41,160 --> 01:06:53,830 called Rayleigh damping, OK? 956 01:06:53,830 --> 01:06:56,630 Lord Rayleigh, who did lots of things in science 957 01:06:56,630 --> 01:07:00,590 that you've probably run into, proposed 958 01:07:00,590 --> 01:07:07,940 that if you model your damping, the c matrix as-- this 959 01:07:07,940 --> 01:07:11,210 is just now the system damping matrix 960 01:07:11,210 --> 01:07:20,150 that you start with-- some alpha times the mass matrix 961 01:07:20,150 --> 01:07:24,520 plus beta times the stiffness matrix-- these are now 962 01:07:24,520 --> 01:07:27,400 the original ones in your generalized coordinates, 963 01:07:27,400 --> 01:07:30,740 just your mass and stiffness matrices. 964 01:07:30,740 --> 01:07:33,340 If you say, I'm going to approximate my damping model 965 01:07:33,340 --> 01:07:50,265 like this, then I want to compute u transpose cu. 966 01:07:52,900 --> 01:07:57,000 I'm going to get alpha u transpose 967 01:07:57,000 --> 01:08:02,390 mu plus beta u transpose Ku. 968 01:08:05,160 --> 01:08:10,020 But we know that this gives you the diagonalized mass matrix, 969 01:08:10,020 --> 01:08:11,630 known as the modal mass matrix. 970 01:08:11,630 --> 01:08:14,960 This gives you the diagonalized stiffness matrix. 971 01:08:14,960 --> 01:08:17,810 And so this damping model, this is 972 01:08:17,810 --> 01:08:23,560 guaranteed to give you a diagonalized damping matrix 973 01:08:23,560 --> 01:08:27,899 which we'll call, somehow, some capital C2, 0, 974 01:08:27,899 --> 01:08:31,920 0, C2, all right? 975 01:08:31,920 --> 01:08:40,810 And it's going to be alpha times the modal mass matrix 976 01:08:40,810 --> 01:08:46,319 plus a beta times the modal stiffness matrix. 977 01:08:46,319 --> 01:08:47,979 And those alphas and betas you adjust. 978 01:08:47,979 --> 01:08:49,395 They're just parameters you adjust 979 01:08:49,395 --> 01:08:54,380 to get the amount of damping you need, OK? 980 01:08:54,380 --> 01:09:09,450 So for a two degree of freedom system, 981 01:09:09,450 --> 01:09:23,240 C1 here is alpha m1 plus beta K1. 982 01:09:23,240 --> 01:09:25,620 Modal mass, alpha times the modal mass 983 01:09:25,620 --> 01:09:27,310 plus beta times the modal stiffness. 984 01:09:27,310 --> 01:09:29,715 That's what you get for the first one. 985 01:09:29,715 --> 01:09:39,288 And C2 is alpha m2 plus beta K2, OK? 986 01:09:46,120 --> 01:09:48,510 And the alphas and betas give you two free parameters 987 01:09:48,510 --> 01:09:49,950 you can play with. 988 01:09:49,950 --> 01:09:52,520 And for a two degree of freedom system, 989 01:09:52,520 --> 01:09:55,680 I can manipulate alpha and beta to get the damping 990 01:09:55,680 --> 01:09:58,020 that I measure. 991 01:09:58,020 --> 01:10:01,900 And I forced my equations of motion a couple. 992 01:10:01,900 --> 01:10:04,160 Now, Mother Nature may say, you know, Vandiver, 993 01:10:04,160 --> 01:10:06,227 they don't uncouple, and there's going 994 01:10:06,227 --> 01:10:07,810 to be a little crosstalk between them. 995 01:10:07,810 --> 01:10:10,580 But I say, yeah, but to first order 996 01:10:10,580 --> 01:10:13,580 I'm going to get a pretty good model of the system. 997 01:10:13,580 --> 01:10:15,280 So let's do that in this case. 998 01:10:15,280 --> 01:10:19,650 Let's maybe just to keep it-- I've got numbers here, 999 01:10:19,650 --> 01:10:24,840 so let my notes so I don't get completely lost here. 1000 01:10:24,840 --> 01:10:28,020 So I'm going to just pick one for now. 1001 01:10:28,020 --> 01:10:29,890 I'm going to model my damping with just 1002 01:10:29,890 --> 01:10:54,640 beta K, beta times my diagonal, my stiffness matrix. 1003 01:10:54,640 --> 01:10:57,257 And let's see what happens here. 1004 01:11:01,550 --> 01:11:07,240 So that says my modal damping is going to be some, 1005 01:11:07,240 --> 01:11:12,110 for mode one, beta K1. 1006 01:11:12,110 --> 01:11:13,790 Now what's damping ratio? 1007 01:11:13,790 --> 01:11:18,850 Zeta 1 for a single degree of freedom system 1008 01:11:18,850 --> 01:11:27,540 is the damping constant for the system over 2 omega 1 m1. 1009 01:11:27,540 --> 01:11:37,610 But that's going to be beta K1 over 2 omega 1 m1. 1010 01:11:37,610 --> 01:11:41,980 But m1/K1 is omega 1 squared. 1011 01:11:41,980 --> 01:11:45,720 So I get an omega 1 squared in the numerator. 1012 01:11:45,720 --> 01:11:54,344 Beta omega 1 squared over 2 omega 1. 1013 01:11:54,344 --> 01:11:56,920 Remember, the K over the m gave me the omega one squared, 1014 01:11:56,920 --> 01:11:58,560 so the ms are gone. 1015 01:11:58,560 --> 01:12:00,200 You can cancel one of these. 1016 01:12:00,200 --> 01:12:04,010 This gives me beta omega 1 over 2. 1017 01:12:06,760 --> 01:12:14,390 So this now gives me a way I can fit one of the dampings. 1018 01:12:14,390 --> 01:12:17,450 I can get exactly what I want, say, for mode one 1019 01:12:17,450 --> 01:12:21,910 if I pick beta to be the right number. 1020 01:12:21,910 --> 01:12:27,820 OK, so in this case, I actually did some numbers. 1021 01:12:33,772 --> 01:12:36,715 Pardon? 1022 01:12:36,715 --> 01:12:37,340 Can't hear you. 1023 01:12:37,340 --> 01:12:38,980 AUDIENCE: [INAUDIBLE] 1024 01:12:38,980 --> 01:12:40,690 PROFESSOR: No. 1025 01:12:40,690 --> 01:12:43,790 K1/m1 is omega 1 squared. 1026 01:12:43,790 --> 01:12:45,730 Omega 1 squared takes care of the m1. 1027 01:12:45,730 --> 01:12:47,490 I get rid of one of the omega 1s. 1028 01:12:47,490 --> 01:12:51,531 I'm left with this, OK? 1029 01:12:51,531 --> 01:12:52,030 OK. 1030 01:12:54,690 --> 01:13:00,970 So let's just let beta equal 0.01. 1031 01:13:00,970 --> 01:13:03,250 And if you let beta equal to 0.01, 1032 01:13:03,250 --> 01:13:09,420 then zeta 1 equals 0.01 omega 1 over 2. 1033 01:13:09,420 --> 01:13:13,250 We know omega 1 is 5.65. 1034 01:13:13,250 --> 01:13:15,940 This when you work it out then gives you 1035 01:13:15,940 --> 01:13:24,230 a number of 0.0283, about 3%. 1036 01:13:24,230 --> 01:13:28,290 And that would say that this system when it vibrates 1037 01:13:28,290 --> 01:13:30,960 in mode one is going to damp out up to 50% 1038 01:13:30,960 --> 01:13:34,020 in about three cycles. 1039 01:13:34,020 --> 01:13:35,282 Not bad approximations. 1040 01:13:35,282 --> 01:13:36,740 I'm just guessing about what it is. 1041 01:13:36,740 --> 01:13:39,910 That's a reasonable amount of damping for mode one. 1042 01:13:39,910 --> 01:13:45,300 Now the problem is when I only use just beta K as my model. 1043 01:13:45,300 --> 01:13:48,220 Now I'm stuck with whatever happens for mode two 1044 01:13:48,220 --> 01:13:53,450 once I pick beta because zeta 2 is going 1045 01:13:53,450 --> 01:13:57,840 to be beta omega 2 over 2. 1046 01:13:57,840 --> 01:14:00,350 And omega 2 is quite a bit larger, 1047 01:14:00,350 --> 01:14:05,770 so now I'm stuck with a greater value for the second mode. 1048 01:14:05,770 --> 01:14:11,800 In this case, it's 0.0885. 1049 01:14:11,800 --> 01:14:16,010 So if I just pick a one parameter model for my damping, 1050 01:14:16,010 --> 01:14:18,160 I can make one perfect. 1051 01:14:18,160 --> 01:14:20,050 I can match it perfectly, but then I'm 1052 01:14:20,050 --> 01:14:22,440 stuck with whatever the other one is. 1053 01:14:22,440 --> 01:14:25,230 So I did this because I could do it simply with one. 1054 01:14:25,230 --> 01:14:30,710 But if I'd kept the full two-parameter model, 1055 01:14:30,710 --> 01:14:33,260 with manipulating both alpha and beta 1056 01:14:33,260 --> 01:14:37,050 I could actually get both of the two measured dampings 1057 01:14:37,050 --> 01:14:38,569 exactly right. 1058 01:14:38,569 --> 01:14:40,110 But if I have an n degree-- if I have 1059 01:14:40,110 --> 01:14:42,220 three degree of freedom system, I only have two parameters. 1060 01:14:42,220 --> 01:14:43,800 I can fit two of the damping ratios, 1061 01:14:43,800 --> 01:14:46,133 but then I'm going to be stuck with whatever it gives me 1062 01:14:46,133 --> 01:14:46,750 for the third. 1063 01:14:46,750 --> 01:14:50,000 But oftentimes it's just one mode you really care about. 1064 01:14:50,000 --> 01:14:50,937 It's the problem mode. 1065 01:14:50,937 --> 01:14:52,270 You're at its natural frequency. 1066 01:14:52,270 --> 01:14:54,670 It's going like crazy. 1067 01:14:54,670 --> 01:14:58,320 Initial conditions make it vibrate a lot in that mode. 1068 01:14:58,320 --> 01:15:01,440 But this is what Rayleigh damping allows you to do. 1069 01:15:01,440 --> 01:15:05,260 It guarantees you that you will have a diagonalized set 1070 01:15:05,260 --> 01:15:06,754 of equations of motion. 1071 01:15:06,754 --> 01:15:08,420 And it gives you two parameters that you 1072 01:15:08,420 --> 01:15:12,260 can play with to fit the damping model however you want. 1073 01:15:12,260 --> 01:15:17,900 Once you have damping, now you have the complete solution 1074 01:15:17,900 --> 01:15:19,394 for decay from initial conditions. 1075 01:15:23,950 --> 01:15:25,130 And there's your two models. 1076 01:15:25,130 --> 01:15:28,710 You can solve for q1, transient decay given initial conditions. 1077 01:15:28,710 --> 01:15:33,100 You can solve for q2, transient K of the second mode. 1078 01:15:33,100 --> 01:15:37,130 And then to get back to the initial 1079 01:15:37,130 --> 01:15:39,900 to the response in terms of your modal coordinates, 1080 01:15:39,900 --> 01:15:44,342 you just add the two together, OK? 1081 01:15:48,100 --> 01:15:50,345 I got some numbers here which are just instructive. 1082 01:15:52,950 --> 01:15:54,072 u Inverse. 1083 01:15:54,072 --> 01:15:55,780 In order to get these initial conditions, 1084 01:15:55,780 --> 01:15:56,988 you've got to know u inverse. 1085 01:15:56,988 --> 01:15:58,300 Do we know u? 1086 01:15:58,300 --> 01:15:59,430 I gave us u. 1087 01:15:59,430 --> 01:16:03,080 Here's our set of mode shape vectors. 1088 01:16:03,080 --> 01:16:05,530 And I've run out of boards. 1089 01:16:30,410 --> 01:16:32,710 So we have the u matrix. 1090 01:16:32,710 --> 01:16:36,020 We need u inverse, so u inverse for this problem. 1091 01:16:51,030 --> 01:16:53,110 And we're going to quickly do some examples. 1092 01:16:53,110 --> 01:16:59,320 Let's let the v0s be 0. 1093 01:16:59,320 --> 01:17:02,260 No initial conditions on velocities. 1094 01:17:02,260 --> 01:17:12,450 And let's do x0, the initial displacements, be 1 and 0. 1095 01:17:12,450 --> 01:17:14,690 So the 1 and 0, what we're saying 1096 01:17:14,690 --> 01:17:18,945 is the bottom one doesn't move, unit deflection here, let 1097 01:17:18,945 --> 01:17:20,000 it go. 1098 01:17:20,000 --> 01:17:23,530 What are you're going to get for the initial conditions? 1099 01:17:23,530 --> 01:17:35,290 x0 equals 1 and 0, well, that implies that the qs are 1100 01:17:35,290 --> 01:17:38,460 going to be u inverse x0. 1101 01:17:38,460 --> 01:17:40,370 So by the way, if this is true, this 1102 01:17:40,370 --> 01:17:47,820 implies that all q dot initial conditions equal 0, right? 1103 01:17:47,820 --> 01:17:49,730 No initial velocities in generalized 1104 01:17:49,730 --> 01:17:52,770 coordinates, no initial velocities in modal 1105 01:17:52,770 --> 01:17:54,462 coordinates. 1106 01:17:54,462 --> 01:17:56,420 But we are going to have an initial deflection. 1107 01:17:59,360 --> 01:18:06,280 I want to then compute u inverse x0 and see what I get. 1108 01:18:06,280 --> 01:18:20,320 And what I get back when I do this one is 0.0898 and 0.9102. 1109 01:18:20,320 --> 01:18:24,280 Remember, this is q10, q20. 1110 01:18:27,480 --> 01:18:33,640 So for that case, it says I'm going to get 0.08 or 0.09 1111 01:18:33,640 --> 01:18:37,770 equal to q1 and 0.9 of q2. 1112 01:18:37,770 --> 01:18:41,120 And I go back over here to my transient decay. 1113 01:18:41,120 --> 01:18:42,850 There's no velocity. 1114 01:18:42,850 --> 01:18:51,720 So it's basically going to look like q10 cosine omega 1115 01:18:51,720 --> 01:18:58,190 dt, e the minus zeta omega t, decaying, cosine. 1116 01:18:58,190 --> 01:19:03,510 But for mode one, its initial amplitude's less than 0.1. 1117 01:19:03,510 --> 01:19:06,580 And mode two, it's got a lot of mode two. 1118 01:19:06,580 --> 01:19:09,910 So what happens? 1119 01:19:09,910 --> 01:19:15,650 So unit deflection here, in fact it's mostly mode two. 1120 01:19:15,650 --> 01:19:18,820 And just quickly I'll do one other. 1121 01:19:18,820 --> 01:19:22,990 x0 is 0, 1. 1122 01:19:22,990 --> 01:19:30,400 That implies that q0 that you get from that 1123 01:19:30,400 --> 01:19:40,830 is 0.4016 and minus 0.4016. 1124 01:19:40,830 --> 01:19:44,690 Says you get about equal amounts. 1125 01:19:44,690 --> 01:19:46,100 So that's this one. 1126 01:19:46,100 --> 01:19:49,110 I don't move this one, but I give this one 1127 01:19:49,110 --> 01:19:52,390 a unit deflection, let go. 1128 01:19:52,390 --> 01:19:55,190 I get about equal amounts of each one. 1129 01:19:55,190 --> 01:19:59,120 And of course I've told you the answer to this one. 1130 01:19:59,120 --> 01:20:04,250 If I let x equal mode one's mode shape, 1131 01:20:04,250 --> 01:20:16,250 1 and 2.266, that implies that q1 equals 1 and q2, when 1132 01:20:16,250 --> 01:20:18,780 you multiply it out, is zero 0. 1133 01:20:18,780 --> 01:20:23,260 If I deflect it in the shape of mode one and I do u inverse x0, 1134 01:20:23,260 --> 01:20:25,620 I will get back 0 and 1. 1135 01:20:25,620 --> 01:20:27,960 And if I make this the shape of mode two, 1136 01:20:27,960 --> 01:20:31,676 I will get back 0 for mode one and 1 for mode two. 1137 01:20:36,760 --> 01:20:41,780 I've out of time, but that's your intro to modal analysis. 1138 01:20:41,780 --> 01:20:43,966 So I think it's conceptually powerful. 1139 01:20:43,966 --> 01:20:44,466 Yes. 1140 01:20:44,466 --> 01:20:47,868 AUDIENCE: How did you get from the 0.898 value 1141 01:20:47,868 --> 01:20:56,616 to the 0.0898 value? 1142 01:20:56,616 --> 01:21:00,018 AUDIENCE: The inverse should be 0.0898. 1143 01:21:00,018 --> 01:21:01,910 PROFESSOR: Oh, is this 0.08? 1144 01:21:01,910 --> 01:21:03,416 Yeah, OK. 1145 01:21:03,416 --> 01:21:04,730 I may have written that down. 1146 01:21:10,720 --> 01:21:11,870 Yeah. 1147 01:21:11,870 --> 01:21:12,970 I'll double check that. 1148 01:21:12,970 --> 01:21:14,040 But yeah, question? 1149 01:21:14,040 --> 01:21:16,550 AUDIENCE: Why is it for here that we picked c 1150 01:21:16,550 --> 01:21:18,840 to be only a function of-- 1151 01:21:18,840 --> 01:21:19,575 PROFESSOR: Beta? 1152 01:21:19,575 --> 01:21:19,830 AUDIENCE: A. 1153 01:21:19,830 --> 01:21:22,205 PROFESSOR: Because I want to get done by the end of the-- 1154 01:21:22,205 --> 01:21:22,760 AUDIENCE: OK. 1155 01:21:22,760 --> 01:21:24,620 PROFESSOR: --60 minutes, 80 minutes. 1156 01:21:24,620 --> 01:21:26,550 I could have put both of them in, 1157 01:21:26,550 --> 01:21:29,120 manipulated both parameters as two equation with [? two ?] 1158 01:21:29,120 --> 01:21:31,130 modes, two target values of dampings. 1159 01:21:31,130 --> 01:21:33,070 I'd find an alpha and a beta that would 1160 01:21:33,070 --> 01:21:35,410 make both work exactly right. 1161 01:21:35,410 --> 01:21:38,870 Actually, just this one model is pretty good for this case. 1162 01:21:38,870 --> 01:21:40,840 The damping for second mode is greater 1163 01:21:40,840 --> 01:21:42,890 than the first mode just happens to be. 1164 01:21:42,890 --> 01:21:44,032 This model's not bad. 1165 01:21:44,032 --> 01:21:45,740 AUDIENCE: All right, so you try the three 1166 01:21:45,740 --> 01:21:47,690 and see what gets you the best results? 1167 01:21:47,690 --> 01:21:49,240 PROFESSOR: Yeah.