1 00:00:00,060 --> 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 --> 00:00:16,590 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:16,590 --> 00:00:17,260 at ocw.mit.edu. 8 00:00:21,380 --> 00:00:24,680 PROFESSOR: So Professor Gossard gave the lecture last week. 9 00:00:24,680 --> 00:00:26,990 I'm going to pick up where he left off. 10 00:00:26,990 --> 00:00:31,260 But let's talk about the concept questions from the homework 11 00:00:31,260 --> 00:00:33,200 you've been working on. 12 00:00:33,200 --> 00:00:35,850 So the first one is our cart. 13 00:00:35,850 --> 00:00:38,520 You'd expect to be able eliminate the terms involving 14 00:00:38,520 --> 00:00:41,070 gravity in the equations of motion 15 00:00:41,070 --> 00:00:43,090 by choosing coordinates with respect 16 00:00:43,090 --> 00:00:45,180 to the static equilibrium position. 17 00:00:45,180 --> 00:00:47,060 So we've talked about that. 18 00:00:47,060 --> 00:00:53,000 And with this one does the restoring force 19 00:00:53,000 --> 00:00:56,200 on the pendulum, what makes it come back 20 00:00:56,200 --> 00:01:01,510 to zero after it's a damped out and hanging straight down. 21 00:01:01,510 --> 00:01:02,260 AUDIENCE: Gravity. 22 00:01:02,260 --> 00:01:02,926 PROFESSOR: What? 23 00:01:02,926 --> 00:01:04,349 Gravity does. 24 00:01:04,349 --> 00:01:09,760 And that gravity term varies with the torque 25 00:01:09,760 --> 00:01:17,780 that the gravity puts around the pivot is MgL over 2 sine theta. 26 00:01:17,780 --> 00:01:21,930 So the gravity term is involved with the motion variable theta. 27 00:01:21,930 --> 00:01:24,080 So in this case, gravity is going 28 00:01:24,080 --> 00:01:27,270 to be involved in the natural frequency 29 00:01:27,270 --> 00:01:30,420 and in the equations of motion, no matter what, 30 00:01:30,420 --> 00:01:36,090 So you will not be able to eliminate the gravity terms. 31 00:01:36,090 --> 00:01:41,220 Next, this one, this is a vibration isolation question. 32 00:01:41,220 --> 00:01:44,430 Will the addition of damping increase or reduce 33 00:01:44,430 --> 00:01:46,570 the vibration of the table in response 34 00:01:46,570 --> 00:01:49,520 to the floor motion at 30 Hertz? 35 00:01:56,490 --> 00:02:00,480 I guess this depends on what the natural frequency of the system 36 00:02:00,480 --> 00:02:00,980 is. 37 00:02:00,980 --> 00:02:03,300 But we're trying to do vibration isolation. 38 00:02:03,300 --> 00:02:05,310 And presuming, if you read the problem, 39 00:02:05,310 --> 00:02:09,880 you're supposed to find a a stiffness such 40 00:02:09,880 --> 00:02:15,060 that you can reduce the response of the table by 12 dB, 41 00:02:15,060 --> 00:02:19,460 I think it said, from the motion of the floor. 42 00:02:19,460 --> 00:02:23,010 So that's something substantially less than 1. 43 00:02:23,010 --> 00:02:31,240 And you will be-- this one takes a, best described 44 00:02:31,240 --> 00:02:32,040 with a picture. 45 00:02:35,650 --> 00:02:42,760 The transfer function for response of the table 46 00:02:42,760 --> 00:02:44,985 over the motion of the floor, the magnitude 47 00:02:44,985 --> 00:02:46,360 of that transfer function, that's 48 00:02:46,360 --> 00:02:49,350 just the ratio of x to y. 49 00:02:49,350 --> 00:02:55,560 That looks like this, if the damping is zero. 50 00:02:55,560 --> 00:03:01,150 And as you add damping, all points cross right here. 51 00:03:01,150 --> 00:03:03,450 Some damping does that. 52 00:03:03,450 --> 00:03:05,800 More damping because of this. 53 00:03:05,800 --> 00:03:13,260 And in order to accomplish what's been described-- 54 00:03:13,260 --> 00:03:16,380 this is 1.0 here. 55 00:03:16,380 --> 00:03:19,580 If you're trying to make this table respond less 56 00:03:19,580 --> 00:03:22,710 than the floor, you must be somewhere out here 57 00:03:22,710 --> 00:03:24,490 where you're below 1. 58 00:03:24,490 --> 00:03:27,970 So this is omega over omega n. 59 00:03:27,970 --> 00:03:30,720 And right here at resonance, you're at 1.0. 60 00:03:30,720 --> 00:03:34,200 So this is, you know, two or three or four for this value 61 00:03:34,200 --> 00:03:35,210 out here. 62 00:03:35,210 --> 00:03:38,470 Let's say here's where you find the answer to B. 63 00:03:38,470 --> 00:03:40,730 And without damping, you're there. 64 00:03:40,730 --> 00:03:44,570 And that's 12 dB down. 65 00:03:44,570 --> 00:03:47,620 If you add damping, it pushes you up these curves. 66 00:03:47,620 --> 00:03:52,830 Does make the response larger, the undesirable response, 67 00:03:52,830 --> 00:03:55,080 the motion of the table larger or smaller 68 00:03:55,080 --> 00:03:59,885 as you add damping at that operating point? 69 00:03:59,885 --> 00:04:02,070 It increases it, right? 70 00:04:02,070 --> 00:04:05,150 OK, so in this case, will the addition of damping 71 00:04:05,150 --> 00:04:06,540 increase or reduce the vibration? 72 00:04:06,540 --> 00:04:07,630 It'll increase it. 73 00:04:07,630 --> 00:04:10,080 But damping's a necessary evil. 74 00:04:10,080 --> 00:04:12,009 You need some damping in the system. 75 00:04:12,009 --> 00:04:14,550 So if you bump it, it doesn't sit there and oscillate all day 76 00:04:14,550 --> 00:04:15,049 long. 77 00:04:15,049 --> 00:04:16,760 Next. 78 00:04:16,760 --> 00:04:17,960 OK, this is a platform. 79 00:04:17,960 --> 00:04:19,459 Do you think I could actually do it? 80 00:04:19,459 --> 00:04:21,250 Did you read this? 81 00:04:21,250 --> 00:04:24,440 So this is a Coast Guard light station off of Cuttyhunk 82 00:04:24,440 --> 00:04:27,190 down off of Woods Hole-- basically, 83 00:04:27,190 --> 00:04:30,840 I was doing this-- in time with the motion of the platform. 84 00:04:30,840 --> 00:04:32,750 Resonance is a wonderful thing. 85 00:04:32,750 --> 00:04:35,230 If you can make the force be right 86 00:04:35,230 --> 00:04:37,070 at the natural frequency of the structure, 87 00:04:37,070 --> 00:04:39,530 it actually doesn't take a lot of force 88 00:04:39,530 --> 00:04:42,950 to drive the amplitude to pretty large amplitudes, 89 00:04:42,950 --> 00:04:44,690 if the damping is small. 90 00:04:44,690 --> 00:04:48,080 So I think the damping in this cases is about 1%. 91 00:04:48,080 --> 00:04:50,250 And that means the amplification, 92 00:04:50,250 --> 00:04:53,020 the dynamic amplification 1 over 2 zeta is about 50. 93 00:04:53,020 --> 00:04:54,970 So I actually could do this. 94 00:04:54,970 --> 00:04:56,190 This is a true story. 95 00:04:56,190 --> 00:04:58,566 OK, next. 96 00:04:58,566 --> 00:05:00,190 For small motions about the horizontal, 97 00:05:00,190 --> 00:05:04,670 you expect the natural frequency to be a function of gravity. 98 00:05:04,670 --> 00:05:09,330 So this is, oh, some of you, about equal, yes no. 99 00:05:09,330 --> 00:05:13,680 But it's just horizontal, the torque that gravity provides 100 00:05:13,680 --> 00:05:18,170 is some Mg pulling down on the center of mass 101 00:05:18,170 --> 00:05:20,330 somewhere in that body. 102 00:05:20,330 --> 00:05:24,950 Not at the pivot, but let's say some distance A away. 103 00:05:24,950 --> 00:05:27,630 So the torque, the gravity, the restoring torque 104 00:05:27,630 --> 00:05:34,940 is some Mg cross r cross Mg, Mgr if r is the distance. 105 00:05:34,940 --> 00:05:36,330 And that's the torque. 106 00:05:36,330 --> 00:05:39,400 And that length of that moment arm might vary. 107 00:05:39,400 --> 00:05:42,230 It's going to vary like cosine theta. 108 00:05:42,230 --> 00:05:46,790 Around horizontal line, if theta is what's, for small angles, 109 00:05:46,790 --> 00:05:48,612 what's cosine? 110 00:05:48,612 --> 00:05:49,990 It goes to 1. 111 00:05:49,990 --> 00:05:52,550 So you find out that this just looks like Mgr. 112 00:05:52,550 --> 00:05:54,190 It's for small angle of vibration. 113 00:05:54,190 --> 00:05:56,070 And you can in fact get it out of-- it 114 00:05:56,070 --> 00:06:01,672 doesn't enter into the equation for the natural frequency. 115 00:06:01,672 --> 00:06:03,130 So the natural frequency of a thing 116 00:06:03,130 --> 00:06:08,960 won't be a function of gravity because of this small angle 117 00:06:08,960 --> 00:06:12,540 vibration around a horizontal point. 118 00:06:12,540 --> 00:06:13,754 OK. 119 00:06:13,754 --> 00:06:15,420 One, when the acceleration of the system 120 00:06:15,420 --> 00:06:19,220 is one half that required to make the mass slide, what's 121 00:06:19,220 --> 00:06:21,555 the magnitude of the friction force? 122 00:06:24,180 --> 00:06:28,020 So friction is one of those things that is only 123 00:06:28,020 --> 00:06:30,670 as big as you need it to be. 124 00:06:30,670 --> 00:06:34,470 So even the largest friction that this thing can sustain 125 00:06:34,470 --> 00:06:41,640 is in fact mu mg-- answer A. But f equals ma 126 00:06:41,640 --> 00:06:43,910 if the acceleration is half of what 127 00:06:43,910 --> 00:06:47,150 is required to have that thing be just slip. 128 00:06:47,150 --> 00:06:56,980 It will just slip when you are at a force which is mu mg. 129 00:06:56,980 --> 00:07:02,430 And so that force is equal to mass times acceleration. 130 00:07:02,430 --> 00:07:05,190 The acceleration then you can figure out 131 00:07:05,190 --> 00:07:07,100 what that will be just when it slips. 132 00:07:07,100 --> 00:07:09,211 But now if you reduce acceleration to half that, 133 00:07:09,211 --> 00:07:11,710 the friction force required to keep it in place is only half 134 00:07:11,710 --> 00:07:13,854 as big. 135 00:07:13,854 --> 00:07:15,270 And it will be that friction force 136 00:07:15,270 --> 00:07:17,830 and it will be half of mu mg. 137 00:07:17,830 --> 00:07:19,370 So it's actually B. 138 00:07:19,370 --> 00:07:21,110 And next-- is that it? 139 00:07:21,110 --> 00:07:21,910 No. 140 00:07:21,910 --> 00:07:24,350 OK. 141 00:07:24,350 --> 00:07:30,890 This is a simple but actually sometimes 142 00:07:30,890 --> 00:07:33,310 hard to see through question. 143 00:07:33,310 --> 00:07:35,710 What initial conditions will be required? 144 00:07:35,710 --> 00:07:38,060 This problem can be solved by initial conditions. 145 00:07:38,060 --> 00:07:42,100 This mortar launches its shell. 146 00:07:42,100 --> 00:07:45,980 And the trick to this question, the key to this question 147 00:07:45,980 --> 00:07:48,750 is for your mathematical model of the system-- 148 00:07:48,750 --> 00:07:52,070 your equation in motion-- is write the equation of motion 149 00:07:52,070 --> 00:07:52,975 without the shell. 150 00:07:55,500 --> 00:07:58,760 Because once it shoots this thing, the shell's gone. 151 00:07:58,760 --> 00:07:59,620 And it's vibrating. 152 00:07:59,620 --> 00:08:02,840 It's now a system without that 25 kilogram mortar 153 00:08:02,840 --> 00:08:04,210 shell part of the system. 154 00:08:04,210 --> 00:08:05,810 It's gone. 155 00:08:05,810 --> 00:08:07,580 Now it's just the mass of the system 156 00:08:07,580 --> 00:08:09,790 without the mortar shell. 157 00:08:09,790 --> 00:08:11,470 And there are two initial conditions 158 00:08:11,470 --> 00:08:15,672 that then you can say when you shot the mortar shell, that was 159 00:08:15,672 --> 00:08:16,880 a certain amount of momentum. 160 00:08:16,880 --> 00:08:19,150 And from conservation of momentum you can figure out 161 00:08:19,150 --> 00:08:24,400 what the momentum of the main mass has to be-- equal 162 00:08:24,400 --> 00:08:26,850 and opposite to the shell you shot. 163 00:08:26,850 --> 00:08:29,240 So that gives you an initial velocity. 164 00:08:29,240 --> 00:08:32,770 But there's also an initial displacement in this problem. 165 00:08:32,770 --> 00:08:34,510 So that's the key to figuring this out. 166 00:08:34,510 --> 00:08:37,860 So what initial conditions would be required? 167 00:08:37,860 --> 00:08:40,260 And it's C-- both an initial velocity 168 00:08:40,260 --> 00:08:42,010 and an initial displacement. 169 00:08:42,010 --> 00:08:44,940 But the key is to make your mathematical model 170 00:08:44,940 --> 00:08:48,630 about the system without the shell. 171 00:08:48,630 --> 00:08:49,440 OK? 172 00:08:49,440 --> 00:08:50,408 Good. 173 00:08:50,408 --> 00:08:51,344 Is that it? 174 00:08:51,344 --> 00:08:53,690 All right. 175 00:08:53,690 --> 00:08:56,580 So today we're going to pick up where 176 00:08:56,580 --> 00:08:59,200 Professor Gossard left off. 177 00:08:59,200 --> 00:09:03,290 But I'm also going to do a bit of a summary 178 00:09:03,290 --> 00:09:07,390 right now about vibration and modeling 179 00:09:07,390 --> 00:09:10,130 the different kinds of systems that we talk about when 180 00:09:10,130 --> 00:09:11,330 we talk about vibration. 181 00:09:11,330 --> 00:09:17,510 They vary from simple, single degree of freedom oscillators, 182 00:09:17,510 --> 00:09:19,560 like a simple pendulum-- one degree 183 00:09:19,560 --> 00:09:24,910 of freedom-- to continuous systems-- beams and vibrate. 184 00:09:24,910 --> 00:09:26,410 So I'm going to try to give you just 185 00:09:26,410 --> 00:09:29,340 sort of an overview of vibration just 186 00:09:29,340 --> 00:09:32,465 to sort of give you a little map of information. 187 00:09:38,528 --> 00:09:44,200 Kind of to let you know what the body of vibration analysis is 188 00:09:44,200 --> 00:09:48,860 and what part of it we're covering in this course. 189 00:09:48,860 --> 00:09:55,723 So I think I will use a little more board. 190 00:10:18,700 --> 00:10:22,330 So we classify dynamics problems into, 191 00:10:22,330 --> 00:10:33,450 for convenience, rigid bodies-- rigid body dynamics 192 00:10:33,450 --> 00:10:35,282 and flexible bodies. 193 00:10:35,282 --> 00:10:36,240 One way to think of it. 194 00:10:41,360 --> 00:10:44,310 And this course is basically about rigid body dynamics. 195 00:10:44,310 --> 00:10:48,640 And under this we then have two categories 196 00:10:48,640 --> 00:10:53,720 that are convenient-- single degree of freedom systems 197 00:10:53,720 --> 00:10:58,045 and multiple degree of freedom systems. 198 00:11:01,420 --> 00:11:03,670 For the purposes of vendors-- talking about vibration. 199 00:11:06,880 --> 00:11:12,030 Single degree of freedom systems have one equation of motion. 200 00:11:12,030 --> 00:11:18,300 And if they vibrate, they have-- and if I'll put over here-- 201 00:11:18,300 --> 00:11:28,195 if vibration occurs then you have one natural frequency. 202 00:11:33,800 --> 00:11:37,877 And it's sort of silly to talk about a mode 203 00:11:37,877 --> 00:11:39,710 shape for a single degree of freedom system, 204 00:11:39,710 --> 00:11:42,090 because it's only relative to itself. 205 00:11:42,090 --> 00:11:45,920 So one natural frequency and one sort of degenerate mode shape. 206 00:11:45,920 --> 00:11:47,840 Multiple degree of freedom systems 207 00:11:47,840 --> 00:11:52,140 have n equations of motion for the number of degrees 208 00:11:52,140 --> 00:11:53,370 of freedom. 209 00:11:53,370 --> 00:12:02,510 And if they vibrate they have n omega i's, or m values 210 00:12:02,510 --> 00:12:08,650 of omega i for i equals 1 to n. 211 00:12:08,650 --> 00:12:12,640 You get n natural frequencies of the system. 212 00:12:12,640 --> 00:12:18,155 And you will get with it n mode shapes. 213 00:12:20,920 --> 00:12:23,240 So a n degree of freedom, this is 214 00:12:23,240 --> 00:12:27,320 equal to the number of degrees of freedom. 215 00:12:27,320 --> 00:12:28,830 An n degree of freedom system will 216 00:12:28,830 --> 00:12:30,570 have n natural frequencies and n mode 217 00:12:30,570 --> 00:12:33,220 shapes that go along with it. 218 00:12:33,220 --> 00:12:35,820 Now, what about what about flexible bodies? 219 00:12:35,820 --> 00:12:42,500 So a taut string like a guitar string. 220 00:12:47,490 --> 00:12:50,080 And actually I should say over here these rigid body 221 00:12:50,080 --> 00:12:52,920 things-- we have found what kind of equations of motion? 222 00:12:52,920 --> 00:13:06,815 These are ordinary differential equations. 223 00:13:11,230 --> 00:13:13,960 And there's a finite number of them and so forth. 224 00:13:13,960 --> 00:13:17,840 The flexible bodies like taut strings 225 00:13:17,840 --> 00:13:24,440 are described by partial differential equations. 226 00:13:34,050 --> 00:13:36,460 The number of degrees of freedom n 227 00:13:36,460 --> 00:13:40,960 here is the number of degrees of freedom 228 00:13:40,960 --> 00:13:42,635 actually goes to infinity. 229 00:13:45,300 --> 00:13:53,070 And you get an infinite number of omega i's, 230 00:13:53,070 --> 00:14:00,210 the natural frequencies, and an infinite number 231 00:14:00,210 --> 00:14:01,675 of corresponding mode shapes. 232 00:14:10,020 --> 00:14:13,740 So just about everything in the world can be made to vibrate. 233 00:14:19,860 --> 00:14:25,520 So how do you tell if a-- you've got a mechanical system, 234 00:14:25,520 --> 00:14:28,007 rigid bodies, you've got three degrees of freedom. 235 00:14:28,007 --> 00:14:30,215 How do you know whether or not it's going to vibrate? 236 00:14:32,577 --> 00:14:33,660 It will exhibit vibration? 237 00:14:40,500 --> 00:14:43,540 Well, one thing you could do is figure out 238 00:14:43,540 --> 00:14:45,420 all the equations of motion and solve them 239 00:14:45,420 --> 00:14:48,840 and see if cosine omega t is a solution. 240 00:14:48,840 --> 00:14:50,460 Right? 241 00:14:50,460 --> 00:14:51,960 That's the hard way. 242 00:14:51,960 --> 00:14:55,010 The other way is to go up to it and give it a smack 243 00:14:55,010 --> 00:14:57,450 and see if it vibrates. 244 00:14:57,450 --> 00:14:58,499 That's the simple way. 245 00:14:58,499 --> 00:15:00,790 If you have the mechanical system just give it a whack. 246 00:15:00,790 --> 00:15:05,850 And if it oscillates around some stable equilibrium position, 247 00:15:05,850 --> 00:15:07,070 it exhibits vibration. 248 00:15:07,070 --> 00:15:08,430 So this is a flexible system. 249 00:15:08,430 --> 00:15:10,620 You can actually probably see this from there. 250 00:15:10,620 --> 00:15:14,150 Just by giving this frame a smack, it will sit and vibrate. 251 00:15:14,150 --> 00:15:16,860 And it does it at some natural frequency. 252 00:15:16,860 --> 00:15:18,360 But that's a continuous system. 253 00:15:22,060 --> 00:15:25,340 This continuously improving little demo-- 254 00:15:25,340 --> 00:15:28,060 so Professor Gossard for his lecture last weekend 255 00:15:28,060 --> 00:15:31,280 had done this really neat embellishment, 256 00:15:31,280 --> 00:15:33,770 which allows you to figure out and excite the two 257 00:15:33,770 --> 00:15:35,344 different natural modes. 258 00:15:35,344 --> 00:15:37,510 But this system you have equations of motion for it. 259 00:15:37,510 --> 00:15:38,301 You could write it. 260 00:15:38,301 --> 00:15:41,190 And if you come up and give it a whack, it oscillates. 261 00:15:41,190 --> 00:15:44,280 And you could also find out that sure enough cosine omega t sine 262 00:15:44,280 --> 00:15:48,090 omega t are solutions to the equations of motion. 263 00:15:48,090 --> 00:15:52,320 So systems that vibrate are systems 264 00:15:52,320 --> 00:15:57,070 that oscillate about static equilibrium positions. 265 00:15:57,070 --> 00:15:59,020 And another way you can say that is 266 00:15:59,020 --> 00:16:01,050 when mechanical vibration occurs, 267 00:16:01,050 --> 00:16:03,960 there's always an exchange of energy 268 00:16:03,960 --> 00:16:07,860 between kinetic and potential, kinetic and potential. 269 00:16:07,860 --> 00:16:12,650 So our pendulum it goes-- when it 270 00:16:12,650 --> 00:16:16,260 reaches zero velocity up here, it's all potential energy. 271 00:16:16,260 --> 00:16:19,170 It reaches maximum velocity down here, it's all kinetic. 272 00:16:19,170 --> 00:16:20,905 And it goes back and forth. 273 00:16:20,905 --> 00:16:23,540 As it sloshes back and forth the energy system 274 00:16:23,540 --> 00:16:25,550 from kinetic to potential and back again. 275 00:16:25,550 --> 00:16:29,850 All vibration has that property. 276 00:16:29,850 --> 00:16:34,310 So that sets some basic properties of vibration. 277 00:16:34,310 --> 00:16:51,660 And now there's a whole body of knowledge about vibration. 278 00:16:51,660 --> 00:16:56,160 And we choose, or for the purposes of this class, 279 00:16:56,160 --> 00:16:58,510 we choose to break it down into two kinds of vibration. 280 00:17:19,003 --> 00:17:21,694 And one is what we call free vibration. 281 00:17:30,620 --> 00:17:34,490 And that we've learned already is response, only a response. 282 00:17:34,490 --> 00:17:46,970 It's a response to initial conditions 283 00:17:46,970 --> 00:17:49,135 and what we call forced vibrations. 284 00:17:56,700 --> 00:17:58,820 Now, forces can come of all kinds. 285 00:17:58,820 --> 00:18:01,890 And for the purposes of this course 286 00:18:01,890 --> 00:18:05,730 we look at a particular kind of force. 287 00:18:05,730 --> 00:18:11,310 So we focus on harmonic excitation. 288 00:18:18,860 --> 00:18:32,740 So excitation that is of the form cosine omega t, 289 00:18:32,740 --> 00:18:35,260 or e to the i omega t. 290 00:18:35,260 --> 00:18:41,483 These are external excitations. 291 00:18:46,270 --> 00:18:48,700 So we choose to break down the analysis 292 00:18:48,700 --> 00:18:53,550 of the vibration of systems into response 293 00:18:53,550 --> 00:18:55,430 to initial conditions called free vibration, 294 00:18:55,430 --> 00:18:58,400 no external forces, and force vibration. 295 00:18:58,400 --> 00:19:00,940 But we focus on a particular kind-- harmonic. 296 00:19:00,940 --> 00:19:04,120 And we go even one step further and say we're only 297 00:19:04,120 --> 00:19:05,450 going to study steady state. 298 00:19:11,009 --> 00:19:13,050 And steady state means you've waited a long time. 299 00:19:13,050 --> 00:19:14,960 Turned it on, let it shake for quite awhile. 300 00:19:14,960 --> 00:19:18,190 All the initial startup transients 301 00:19:18,190 --> 00:19:19,460 have been damped out. 302 00:19:19,460 --> 00:19:21,930 And you're left with a steady state vibration. 303 00:19:21,930 --> 00:19:24,900 And that leads to things like the transfer functions 304 00:19:24,900 --> 00:19:26,400 for single degree of freedom systems 305 00:19:26,400 --> 00:19:27,500 that we've talked about. 306 00:19:30,670 --> 00:19:36,500 Now, there's one other breakdown or subdivision 307 00:19:36,500 --> 00:19:38,450 that we need to talk about. 308 00:19:38,450 --> 00:19:41,680 And that is whether systems are linear or non-linear. 309 00:19:45,680 --> 00:19:49,350 And this is all set up so you can see it. 310 00:19:49,350 --> 00:19:51,670 This is a double pendulum. 311 00:19:51,670 --> 00:19:55,130 How many degrees of freedom? 312 00:19:55,130 --> 00:19:56,414 Two. 313 00:19:56,414 --> 00:19:58,580 And in general, do you think the equations of motion 314 00:19:58,580 --> 00:20:01,770 of this thing are going to be non-linear? 315 00:20:01,770 --> 00:20:03,610 Right. 316 00:20:03,610 --> 00:20:06,540 Just a simple pendulum is the restoring torque 317 00:20:06,540 --> 00:20:09,320 is Mgl sine theta. 318 00:20:09,320 --> 00:20:11,470 So you know it's got sine theta and that. 319 00:20:11,470 --> 00:20:13,362 And this one gets quite messy. 320 00:20:13,362 --> 00:20:15,320 And especially if you give it large amplitudes. 321 00:20:23,020 --> 00:20:26,060 And that really isn't vibration. 322 00:20:26,060 --> 00:20:26,600 It's not. 323 00:20:26,600 --> 00:20:29,410 It's looping all over itself and then doing other things. 324 00:20:29,410 --> 00:20:32,640 So cosine omega t is not a solution. 325 00:20:32,640 --> 00:20:34,090 It's not a solution to this. 326 00:20:34,090 --> 00:20:36,600 It's got to be more complicated than that. 327 00:20:36,600 --> 00:20:39,550 So when this thing is exhibiting large motions, 328 00:20:39,550 --> 00:20:41,792 the equations of motion are completely non-linear. 329 00:20:41,792 --> 00:20:43,250 And you're going to need a computer 330 00:20:43,250 --> 00:20:46,620 to crank out the full solution to integrate 331 00:20:46,620 --> 00:20:48,770 these non-linear equations of motion. 332 00:20:48,770 --> 00:20:58,400 But as the amplitude settles down to something pretty small, 333 00:20:58,400 --> 00:21:02,155 now it's vibrating about an equilibrium position. 334 00:21:02,155 --> 00:21:05,060 The equilibrium position is straight down. 335 00:21:05,060 --> 00:21:06,990 And the damping of it has made it such 336 00:21:06,990 --> 00:21:09,270 that the only motion left is what's called 337 00:21:09,270 --> 00:21:12,650 its first mode of vibration. 338 00:21:12,650 --> 00:21:16,970 And so if we linearize the equations of motion, 339 00:21:16,970 --> 00:21:21,830 assuming small amplitudes around static equilibrium positions, 340 00:21:21,830 --> 00:21:25,870 then we can find a vibration solution 341 00:21:25,870 --> 00:21:29,220 and work it out by hand probably. 342 00:21:29,220 --> 00:21:30,940 That's first mode for this system. 343 00:21:30,940 --> 00:21:36,110 And if I'm careful-- there's second mode. 344 00:21:39,390 --> 00:21:42,100 And for small oscillations it has 345 00:21:42,100 --> 00:21:45,170 a very clear single frequency that it vibrates at. 346 00:21:45,170 --> 00:21:49,425 The amplitude decays over time because of damping. 347 00:21:49,425 --> 00:21:51,600 And for every natural frequency there 348 00:21:51,600 --> 00:21:54,100 is a particular mode shape that goes along 349 00:21:54,100 --> 00:21:55,800 with that natural frequency. 350 00:21:55,800 --> 00:21:59,040 The first one for this system-- I 351 00:21:59,040 --> 00:22:02,800 have to wait for this thing to damp out. 352 00:22:02,800 --> 00:22:04,410 It's got a little mix of the two, 353 00:22:04,410 --> 00:22:09,250 but as it-- the second natural frequency motion 354 00:22:09,250 --> 00:22:11,010 dies out faster than the first because it 355 00:22:11,010 --> 00:22:13,769 has more cycles per unit time. 356 00:22:13,769 --> 00:22:14,560 So it settles down. 357 00:22:14,560 --> 00:22:16,310 This is now mostly first mode vibration. 358 00:22:16,310 --> 00:22:20,480 And you can see that both move in the same direction, 359 00:22:20,480 --> 00:22:23,330 the bottom one a little more than the top. 360 00:22:23,330 --> 00:22:24,500 And that's the first mode. 361 00:22:24,500 --> 00:22:26,200 It has a unique natural frequency. 362 00:22:26,200 --> 00:22:30,424 And a mode shape that is specific that goes along 363 00:22:30,424 --> 00:22:31,590 with that natural frequency. 364 00:22:48,280 --> 00:23:04,560 So this further break down here, I'll call it, 365 00:23:04,560 --> 00:23:14,215 is basically into non-linear and linearized. 366 00:23:20,900 --> 00:23:24,820 So in our discussions of vibration in this course 367 00:23:24,820 --> 00:23:28,400 we basically only talk about this. 368 00:23:28,400 --> 00:23:49,810 So we're only doing-- so that's quite a breakdown. 369 00:23:49,810 --> 00:23:53,220 You start at all possible vibration systems, 370 00:23:53,220 --> 00:23:54,900 rigid bodies, single degree of freedom, 371 00:23:54,900 --> 00:23:57,920 multidegree of freedom, finite number of degrees of freedom 372 00:23:57,920 --> 00:24:00,490 or continuous. 373 00:24:00,490 --> 00:24:03,330 They can have linear or non-linear equations of motion. 374 00:24:03,330 --> 00:24:06,590 But if we require them to be linear, 375 00:24:06,590 --> 00:24:10,550 and that's what we're going to look at then 376 00:24:10,550 --> 00:24:13,490 you sort of narrow this down what we're looking at to this. 377 00:24:13,490 --> 00:24:15,720 So there's lots of other things possible to look 378 00:24:15,720 --> 00:24:18,650 at it, like that really non-linear motion of that two 379 00:24:18,650 --> 00:24:19,800 dimensional thing. 380 00:24:19,800 --> 00:24:22,796 But our study of vibration is here. 381 00:24:22,796 --> 00:24:25,320 So this is what we're doing in 2003. 382 00:24:25,320 --> 00:24:27,930 But there's a lot of important problems 383 00:24:27,930 --> 00:24:29,100 that are covered by that. 384 00:24:31,760 --> 00:24:33,710 Lots of real things in nature that 385 00:24:33,710 --> 00:24:36,675 are problematic for engineers and problematic for design 386 00:24:36,675 --> 00:24:41,410 can be analyzed with linear equations of motion. 387 00:24:41,410 --> 00:24:44,560 And even if they're not linear, if you do the linear solution 388 00:24:44,560 --> 00:24:46,280 first, it gives you a starting point 389 00:24:46,280 --> 00:24:48,080 to think about what's the behavior 390 00:24:48,080 --> 00:24:50,360 of the non-linear systems. 391 00:24:50,360 --> 00:24:52,680 But this is our study of vibration. 392 00:24:52,680 --> 00:24:56,890 And we're going to do that for-- and what we had 393 00:24:56,890 --> 00:25:03,750 started doing-- in two ways. 394 00:25:03,750 --> 00:25:06,220 We look at the response to initial conditions called 395 00:25:06,220 --> 00:25:07,450 free vibration. 396 00:25:07,450 --> 00:25:09,840 And we look at response, steady state 397 00:25:09,840 --> 00:25:15,720 response of now these linear systems to force vibration. 398 00:25:15,720 --> 00:25:20,630 And last week you were looking for the first time 399 00:25:20,630 --> 00:25:24,850 at Professor Gossard's lecture about the free vibration 400 00:25:24,850 --> 00:25:29,840 response of basically a two degree freedom system. 401 00:25:29,840 --> 00:25:31,550 So why do two degree of freedom systems? 402 00:25:31,550 --> 00:25:35,830 Well, it's the simplest next step up from a single degree. 403 00:25:35,830 --> 00:25:38,050 And they're sort of mathematically tractable. 404 00:25:38,050 --> 00:25:39,210 You can do them on paper. 405 00:25:39,210 --> 00:25:42,084 We emphasize looking at two degree freedom systems 406 00:25:42,084 --> 00:25:43,750 because we can do the math on the board, 407 00:25:43,750 --> 00:25:44,991 do the math on the paper. 408 00:25:44,991 --> 00:25:46,740 But as you get to more degrees of freedom, 409 00:25:46,740 --> 00:25:49,470 you basically are going to have to do-- it's easier 410 00:25:49,470 --> 00:25:51,460 to do it using the computer. 411 00:25:51,460 --> 00:25:54,960 And in order to do that you need to know some linear algebra. 412 00:25:54,960 --> 00:25:57,860 So I'm kind curious. 413 00:25:57,860 --> 00:26:01,130 In terms of linear algebra, like multiplying two matrices 414 00:26:01,130 --> 00:26:04,560 together, or finding the determinant of a matrix, 415 00:26:04,560 --> 00:26:07,780 or inverting a matrix, how many of you 416 00:26:07,780 --> 00:26:11,980 actually have been taught that? 417 00:26:11,980 --> 00:26:14,690 How may perhaps have it? 418 00:26:14,690 --> 00:26:16,770 Do you do that in 1803 now? 419 00:26:16,770 --> 00:26:17,970 Is that where you do it? 420 00:26:17,970 --> 00:26:18,470 OK. 421 00:26:18,470 --> 00:26:18,970 Good. 422 00:26:18,970 --> 00:26:19,830 So that's helpful. 423 00:26:19,830 --> 00:26:21,038 I wasn't sure whether or not. 424 00:26:21,038 --> 00:26:23,670 I can assume that you at least know what 425 00:26:23,670 --> 00:26:27,040 the determinant of a matrix is. 426 00:26:27,040 --> 00:26:27,590 That's great. 427 00:26:27,590 --> 00:26:29,291 That's really helpful. 428 00:26:29,291 --> 00:26:29,790 OK. 429 00:26:32,840 --> 00:26:35,630 Let's talk. 430 00:26:35,630 --> 00:26:38,250 So we want linear equations of motion. 431 00:26:38,250 --> 00:26:44,200 And I've done a little bit about linearization but not much. 432 00:26:44,200 --> 00:26:49,600 So let's talk a little bit about that for a second. 433 00:26:52,530 --> 00:26:59,001 For a pendulum we know the equation of motion for it. 434 00:26:59,001 --> 00:27:01,500 And actually we could make this a more complicated pendulum. 435 00:27:01,500 --> 00:27:05,920 It could be a stick or any rigid body swinging about this point 436 00:27:05,920 --> 00:27:09,140 A. 437 00:27:09,140 --> 00:27:12,530 We know that we can write the equation of motion Izz 438 00:27:12,530 --> 00:27:19,270 with respect to A, theta double dot, plus Mgl-- omega L. 439 00:27:19,270 --> 00:27:32,020 The distance here to wherever g is, Mgl sine theta. 440 00:27:32,020 --> 00:27:34,780 And for free vibration that's all there is to it. 441 00:27:34,780 --> 00:27:38,790 And to linearize this equation we just say, 442 00:27:38,790 --> 00:27:44,660 well, we know that sine theta is equal to theta minus theta 443 00:27:44,660 --> 00:27:48,990 cubed over 3 factorial plus theta to the fifth over 5 444 00:27:48,990 --> 00:27:51,080 factorial. 445 00:27:51,080 --> 00:27:53,890 And the cosine theta-- just to have it available here-- 446 00:27:53,890 --> 00:28:00,930 is 1 minus theta squared over 2 factorial and so forth. 447 00:28:00,930 --> 00:28:02,470 And when we say linearize, we really 448 00:28:02,470 --> 00:28:06,890 mean we want our equations to involve the motion variables 449 00:28:06,890 --> 00:28:09,580 at most to first order. 450 00:28:09,580 --> 00:28:12,280 So the first order term for sine is theta. 451 00:28:12,280 --> 00:28:14,370 There's no first order term for cosine. 452 00:28:17,610 --> 00:28:18,940 Theta squared is non-linear. 453 00:28:18,940 --> 00:28:23,190 So the small angle approximation to cosine 454 00:28:23,190 --> 00:28:24,930 is it's approximately 1. 455 00:28:24,930 --> 00:28:27,480 And to sine is it's approximately 456 00:28:27,480 --> 00:28:30,340 theta for small motions. 457 00:28:30,340 --> 00:28:31,090 Theta-- small. 458 00:28:33,650 --> 00:28:35,540 So when we linearize this equation, 459 00:28:35,540 --> 00:28:39,925 we just substitute in for sine theta its linear approximation. 460 00:28:39,925 --> 00:28:42,200 And we get Mgl theta. 461 00:28:42,200 --> 00:28:43,920 So we've seen that one many times. 462 00:28:50,970 --> 00:28:53,330 And that's your linearized equation of motion. 463 00:28:53,330 --> 00:28:57,122 But on this week's homework you've got a harder problem. 464 00:28:57,122 --> 00:28:57,955 And that's our cart. 465 00:29:08,480 --> 00:29:13,620 And here you have a theta and an x are your two equations. 466 00:29:13,620 --> 00:29:15,710 And you've worked this problem before. 467 00:29:15,710 --> 00:29:20,520 And you know with the previous homeworks you've 468 00:29:20,520 --> 00:29:21,840 gotten the equation of motion. 469 00:29:21,840 --> 00:29:24,670 I'll write one of them down here. 470 00:29:24,670 --> 00:29:26,490 So one of the equations of motion 471 00:29:26,490 --> 00:29:34,480 is-- this is m1, m1, k, b. 472 00:29:39,960 --> 00:29:46,130 And this is a stick, it's l long, g in the middle. 473 00:29:46,130 --> 00:29:54,050 So the equation of motion for this looks like m1 plus m2, x 474 00:29:54,050 --> 00:30:08,130 double dot plus m2 l over 2, theta double dot, 475 00:30:08,130 --> 00:30:18,510 minus m2 l over 2 theta dot squared theta, 476 00:30:18,510 --> 00:30:27,370 and then plus bx dot, plus kx, and equals, 477 00:30:27,370 --> 00:30:32,580 and in fact, this one has a force on it. 478 00:30:32,580 --> 00:30:33,790 It's equal to F of t. 479 00:30:36,680 --> 00:30:42,480 Now, is that a non-linear equation? 480 00:30:54,760 --> 00:30:56,690 So this is the force equation mass 481 00:30:56,690 --> 00:30:58,050 times acceleration or forces. 482 00:30:58,050 --> 00:30:59,450 You know you've got another equation of motion 483 00:30:59,450 --> 00:31:01,150 in here, which is the torque one. 484 00:31:01,150 --> 00:31:02,210 This is just one of them. 485 00:31:02,210 --> 00:31:03,525 So is it linear or non-linear? 486 00:31:07,890 --> 00:31:09,620 How many think it's non-linear? 487 00:31:09,620 --> 00:31:11,060 OK. 488 00:31:11,060 --> 00:31:19,820 If I have number the terms here-- one, two, three, four, 489 00:31:19,820 --> 00:31:20,340 five. 490 00:31:20,340 --> 00:31:22,000 And that's not a motion. 491 00:31:22,000 --> 00:31:23,260 This doesn't involve motion. 492 00:31:23,260 --> 00:31:26,424 So if one through five, which ones have a non-linear term? 493 00:31:26,424 --> 00:31:27,090 AUDIENCE: Three. 494 00:31:27,090 --> 00:31:27,670 J. KIM VANDIVER: Three. 495 00:31:27,670 --> 00:31:28,510 OK. 496 00:31:28,510 --> 00:31:31,080 So how do you linearize that thing? 497 00:31:36,738 --> 00:31:40,240 AUDIENCE: You make it zero, because it's second order. 498 00:31:40,240 --> 00:31:42,480 J. KIM VANDIVER: In fact, it's third order. 499 00:31:42,480 --> 00:31:46,190 So the reason I wanted to mention this today-- 500 00:31:46,190 --> 00:31:49,210 if you haven't thought about being confronted 501 00:31:49,210 --> 00:31:52,100 with linearization problems before-- we're 502 00:31:52,100 --> 00:31:55,740 trying to linearize the system so that we 503 00:31:55,740 --> 00:32:00,900 can by making it linear we can make cosine omega t a solution. 504 00:32:00,900 --> 00:32:01,800 Right? 505 00:32:01,800 --> 00:32:06,120 We want cosine omega t to be a solution to this thing. 506 00:32:06,120 --> 00:32:15,240 So if you let-- so you've got a term that 507 00:32:15,240 --> 00:32:24,860 looks like ml over m2l over 2, theta dot squared theta. 508 00:32:24,860 --> 00:32:28,180 Well, theta in this problem of some function of time 509 00:32:28,180 --> 00:32:30,460 we're hoping-- we want to find a solution that 510 00:32:30,460 --> 00:32:37,260 has some amplitude times, say, a cosine omega t. 511 00:32:37,260 --> 00:32:46,310 And theta dot is minus omega theta naught sine omega t. 512 00:32:46,310 --> 00:32:48,790 And so that expression up there, the magnitude 513 00:32:48,790 --> 00:32:52,540 of that expression or the magnitude of theta dot 514 00:32:52,540 --> 00:32:58,100 squared theta is proportional to-- you get a theta 515 00:32:58,100 --> 00:33:02,740 naught here and you get omega squared 516 00:33:02,740 --> 00:33:08,590 theta naught squared here. 517 00:33:08,590 --> 00:33:12,700 So this term is proportional to theta naught cubed. 518 00:33:12,700 --> 00:33:18,140 And if the angle theta is small, then a small angle cubed 519 00:33:18,140 --> 00:33:20,090 is really small. 520 00:33:20,090 --> 00:33:23,870 And so the way you linearize this equation 521 00:33:23,870 --> 00:33:27,200 is to throw this out. 522 00:33:27,200 --> 00:33:30,635 So when you've done all your tricks you can, 523 00:33:30,635 --> 00:33:35,090 like replacing sine theta with theta and cosine theta 524 00:33:35,090 --> 00:33:39,630 with one, and you still end up with terms have a higher 525 00:33:39,630 --> 00:33:44,170 order than one in the motion variable, theta or x, 526 00:33:44,170 --> 00:33:44,970 you throw it out. 527 00:33:50,510 --> 00:33:52,700 So if you throw that term out then you 528 00:33:52,700 --> 00:33:55,580 end up with a nice linear equation in motion. 529 00:34:09,060 --> 00:34:09,730 OK. 530 00:34:09,730 --> 00:34:12,100 So now for the rest of today we're 531 00:34:12,100 --> 00:34:26,380 going to talk about free vibration solution. 532 00:34:26,380 --> 00:34:29,170 So we're not going to worry for the moment about the force 533 00:34:29,170 --> 00:34:32,230 vibration steady state transfer function stuff. 534 00:34:32,230 --> 00:34:34,790 We're talking just about free vibration. 535 00:34:34,790 --> 00:34:42,130 And this is of linear equations of motion. 536 00:35:00,490 --> 00:35:03,220 So vibration is a pretty big body of knowledge. 537 00:35:03,220 --> 00:35:06,990 And we're doing an introduction to vibration in about half 538 00:35:06,990 --> 00:35:08,930 a dozen lectures here. 539 00:35:08,930 --> 00:35:13,180 So there's lots of things that I'm not 540 00:35:13,180 --> 00:35:15,730 going to have time to teach you, but there 541 00:35:15,730 --> 00:35:19,320 are a few things I really want you to go away with 542 00:35:19,320 --> 00:35:20,800 and understanding. 543 00:35:20,800 --> 00:35:29,160 And one of these key concepts is that the vibration 544 00:35:29,160 --> 00:35:32,950 of a multiple degree of freedom system-- 545 00:35:32,950 --> 00:35:35,950 say this is a two degree of freedom system. 546 00:35:35,950 --> 00:35:38,510 That the vibration of this system, the free vibration, 547 00:35:38,510 --> 00:35:45,670 can be made up of the sum of two parts 548 00:35:45,670 --> 00:35:49,040 at any vibration of this system at all. 549 00:35:49,040 --> 00:35:51,240 So at any arbitrary set of initial conditions 550 00:35:51,240 --> 00:35:53,480 I give it-- I let it go. 551 00:35:53,480 --> 00:35:56,640 The key concept is the response will 552 00:35:56,640 --> 00:36:01,410 be made up of two pieces-- vibration 553 00:36:01,410 --> 00:36:03,940 in each of the two modes. 554 00:36:03,940 --> 00:36:05,845 And if you can solve the vibration that's 555 00:36:05,845 --> 00:36:07,970 in first mode-- first mode is the one where they're 556 00:36:07,970 --> 00:36:09,850 going kind of together, second mode 557 00:36:09,850 --> 00:36:11,640 they're opposite one another. 558 00:36:11,640 --> 00:36:13,750 That the total solution can be made up 559 00:36:13,750 --> 00:36:16,880 of a contribution for mode one and a contribution 560 00:36:16,880 --> 00:36:18,960 from mode two. 561 00:36:18,960 --> 00:36:29,200 So this is this concept called mode superposition. 562 00:36:29,200 --> 00:36:30,650 It's really quite powerful. 563 00:36:30,650 --> 00:36:34,630 So you can figure out the response of the first mode 564 00:36:34,630 --> 00:36:37,670 in the system, figure out the response of the second mode's 565 00:36:37,670 --> 00:36:42,230 contribution, add them together, and that's the total solution. 566 00:36:42,230 --> 00:36:51,480 And this concept works-- there's all sorts of caveats 567 00:36:51,480 --> 00:36:57,040 that one gets into-- but basically this is true 568 00:36:57,040 --> 00:37:11,860 for all lightly damped systems. 569 00:37:11,860 --> 00:37:15,570 You get into heavy damping and strange damping, 570 00:37:15,570 --> 00:37:17,230 you have to make some adjustments. 571 00:37:17,230 --> 00:37:19,770 But for lightly damped systems you'll 572 00:37:19,770 --> 00:37:21,890 find that this concept of mode superposition 573 00:37:21,890 --> 00:37:25,970 works out just fine. 574 00:37:25,970 --> 00:37:31,070 So an illustration of this, a really simple illustration-- 575 00:37:31,070 --> 00:37:33,500 in some ways easier than this one. 576 00:37:33,500 --> 00:37:35,840 I don't know if I can get this where you can see it 577 00:37:35,840 --> 00:37:37,090 in the picture or not. 578 00:37:37,090 --> 00:37:38,210 Maybe not really. 579 00:37:38,210 --> 00:37:41,190 This is just two little lead weights. 580 00:37:41,190 --> 00:37:42,470 This is a double pendulum. 581 00:37:42,470 --> 00:37:46,080 It has two natural frequencies. 582 00:37:46,080 --> 00:37:48,190 One is that one. 583 00:37:48,190 --> 00:37:51,730 You can see the two weights going the same direction. 584 00:37:51,730 --> 00:37:54,740 The bottom weight at a little bit more larger angle 585 00:37:54,740 --> 00:37:57,540 than the top weight. 586 00:37:57,540 --> 00:37:59,960 And it's at a particular frequency. 587 00:37:59,960 --> 00:38:03,940 And that's the mode shape that goes through this frequency. 588 00:38:03,940 --> 00:38:07,790 So another key concept is that for free vibration 589 00:38:07,790 --> 00:38:12,490 the total solution is made up of the free vibration 590 00:38:12,490 --> 00:38:13,420 of each mode. 591 00:38:13,420 --> 00:38:15,775 And each mode has a particular frequency 592 00:38:15,775 --> 00:38:18,740 and a particular shape to it. 593 00:38:18,740 --> 00:38:23,020 So that's the first mode frequency and the first mode 594 00:38:23,020 --> 00:38:23,520 shape. 595 00:38:23,520 --> 00:38:26,440 The second mode-- I have a little harder time getting it 596 00:38:26,440 --> 00:38:29,560 started-- it looks like that. 597 00:38:29,560 --> 00:38:31,930 Masses move in opposite directions. 598 00:38:31,930 --> 00:38:35,225 It's kind of rotating around where you can't see it. 599 00:38:35,225 --> 00:38:38,920 I have to do it in the plane. 600 00:38:38,920 --> 00:38:39,870 It's hard to do here. 601 00:38:39,870 --> 00:38:42,455 It doesn't want to behave like it's confined to a plane. 602 00:38:47,420 --> 00:38:49,570 They're going opposite directions. 603 00:38:49,570 --> 00:38:51,520 And the frequency is higher. 604 00:38:51,520 --> 00:38:53,600 But this motion, that mode shape, 605 00:38:53,600 --> 00:38:58,530 is a fixed feature of this mode of vibration 606 00:38:58,530 --> 00:39:00,470 along with this natural frequency. 607 00:39:00,470 --> 00:39:04,540 So this idea of mode superposition-- 608 00:39:04,540 --> 00:39:27,310 and a second concept here is that for free vibration 609 00:39:27,310 --> 00:40:00,740 of each mode it oscillates at a unique frequency for this two 610 00:40:00,740 --> 00:40:01,970 degree of freedom system. 611 00:40:01,970 --> 00:40:04,060 You have two natural frequencies-- 612 00:40:04,060 --> 00:40:07,450 omega 1 and omega 2. 613 00:40:07,450 --> 00:40:23,220 And at each omega n there is a corresponding mode shape. 614 00:40:25,790 --> 00:40:31,960 So any vibration of a linear system, free vibration of it, 615 00:40:31,960 --> 00:40:36,870 any vibration at all is composed of a superposition of the two 616 00:40:36,870 --> 00:40:37,870 modes. 617 00:40:37,870 --> 00:40:39,930 Part of this motion is in the first mode 618 00:40:39,930 --> 00:40:42,370 at its natural frequency and in its shape. 619 00:40:42,370 --> 00:40:47,440 And part of the motion has a second contribution, 620 00:40:47,440 --> 00:40:50,030 which is at the natural frequency of the second mode 621 00:40:50,030 --> 00:40:51,105 and in its shape. 622 00:40:53,670 --> 00:40:57,360 So I'm going to give you a quick demo 623 00:40:57,360 --> 00:41:02,450 and ask you-- let's if you can use what I just 624 00:41:02,450 --> 00:41:08,676 said to analyze a motion. 625 00:41:16,970 --> 00:41:21,330 So this is just a block on some strings. 626 00:41:21,330 --> 00:41:23,910 And I'm going to show you a motion. 627 00:41:23,910 --> 00:41:27,430 And I want you to tell me whether or not 628 00:41:27,430 --> 00:41:36,580 it could possibly be a natural frequency motion in one mode, 629 00:41:36,580 --> 00:41:40,552 or the other answer is it's a sum of multiple modes. 630 00:41:40,552 --> 00:41:42,010 But I'm going to show you a motion, 631 00:41:42,010 --> 00:41:44,070 and I want you to tell me and argue 632 00:41:44,070 --> 00:41:47,160 on the basis of what I've just told you whether or not 633 00:41:47,160 --> 00:41:49,260 you are seeing a single mode of vibration. 634 00:41:52,810 --> 00:41:54,779 And maybe I'll use the clamp here 635 00:41:54,779 --> 00:41:56,570 so I don't have to stand there and hold it. 636 00:42:25,802 --> 00:42:27,010 I'm going to just place this. 637 00:42:27,010 --> 00:42:28,720 So the way you do free vibration is 638 00:42:28,720 --> 00:42:30,360 you give it an initial displacement, 639 00:42:30,360 --> 00:42:32,680 some initial conditions, and let go. 640 00:42:32,680 --> 00:42:39,975 So I'm going to pull this over and back and let go. 641 00:42:43,935 --> 00:42:46,480 And just watch closely what you see it do. 642 00:43:14,794 --> 00:43:16,710 All right, now it's doing more of what I want. 643 00:43:16,710 --> 00:43:19,224 It's like it's going in a circle right now. 644 00:43:19,224 --> 00:43:21,390 And now it looks like it's just going back and forth 645 00:43:21,390 --> 00:43:23,140 on a diagonal. 646 00:43:23,140 --> 00:43:25,940 And then it's going to start circling the other way. 647 00:43:25,940 --> 00:43:27,650 It's going in a circle. 648 00:43:27,650 --> 00:43:31,320 And now it goes to on the diagonal-- left and right. 649 00:43:31,320 --> 00:43:36,210 And then it starts back into a circle again. 650 00:43:36,210 --> 00:43:40,950 Are you observing a natural mode of vibration? 651 00:43:40,950 --> 00:43:42,780 It looks like it's single frequency, right? 652 00:43:42,780 --> 00:43:45,340 This looks like it's all happening at one frequency. 653 00:43:45,340 --> 00:43:50,340 But is it a natural mode, a unique natural mode? 654 00:43:57,150 --> 00:44:00,070 Who wants to make a case for whether it is or isn't? 655 00:44:00,070 --> 00:44:01,600 How many believe that you're seeing 656 00:44:01,600 --> 00:44:03,240 a natural mode of vibration? 657 00:44:03,240 --> 00:44:03,740 None. 658 00:44:03,740 --> 00:44:05,810 How many think you're not seeing a natural mode of vibration? 659 00:44:05,810 --> 00:44:06,770 Let's see if you're awake. 660 00:44:06,770 --> 00:44:07,270 OK. 661 00:44:07,270 --> 00:44:09,150 So you don't believe that it's natural mode. 662 00:44:09,150 --> 00:44:10,830 Make the case. 663 00:44:10,830 --> 00:44:12,460 Why? 664 00:44:12,460 --> 00:44:16,090 How do you use sort of this definition of a natural mode 665 00:44:16,090 --> 00:44:19,676 to tell me why this can't be? 666 00:44:19,676 --> 00:44:21,652 AUDIENCE: It looks like a superposition 667 00:44:21,652 --> 00:44:25,110 of at least two different kinds of vibration. 668 00:44:25,110 --> 00:44:26,592 J. KIM VANDIVER: OK. 669 00:44:26,592 --> 00:44:29,145 The evidence that you see is because what does it do? 670 00:44:29,145 --> 00:44:30,436 AUDIENCE: It circled sometimes. 671 00:44:30,436 --> 00:44:33,407 And sometimes it goes straight back and forth on a diagonal. 672 00:44:33,407 --> 00:44:34,240 J. KIM VANDIVER: OK. 673 00:44:34,240 --> 00:44:35,490 So it circled around part of the time 674 00:44:35,490 --> 00:44:38,060 and then goes straight back and forth part of the time. 675 00:44:38,060 --> 00:44:44,580 Is it a constant mode shape of vibration? 676 00:44:44,580 --> 00:44:45,220 No. 677 00:44:45,220 --> 00:44:47,680 And that's all you need to observe. 678 00:44:47,680 --> 00:44:52,500 If the thing doesn't keep a constant single shape 679 00:44:52,500 --> 00:44:55,400 at a single frequency, it's not a natural mode. 680 00:44:55,400 --> 00:44:57,230 So let's do a different case. 681 00:44:57,230 --> 00:45:00,690 I'll deflect it just this way. 682 00:45:00,690 --> 00:45:02,510 And ignore that little bit of torsion. 683 00:45:02,510 --> 00:45:05,670 So it's just going back and forth in line. 684 00:45:05,670 --> 00:45:09,900 Other than slowly damping out, that has just one motion to it, 685 00:45:09,900 --> 00:45:12,190 and it's at one natural frequency. 686 00:45:12,190 --> 00:45:14,740 So do you think that's a mode? 687 00:45:14,740 --> 00:45:16,100 That probably is two. 688 00:45:16,100 --> 00:45:17,160 And so is this one. 689 00:45:20,279 --> 00:45:21,570 And ignore that high frequency. 690 00:45:21,570 --> 00:45:22,778 Now it's just back and forth. 691 00:45:22,778 --> 00:45:24,160 It's just a pendulum. 692 00:45:24,160 --> 00:45:27,480 And it just stays in just pendular motion, no circling 693 00:45:27,480 --> 00:45:28,790 around or any of that. 694 00:45:28,790 --> 00:45:31,210 So that's also natural, and it occurs 695 00:45:31,210 --> 00:45:33,530 at a particular frequency. 696 00:45:33,530 --> 00:45:37,610 So these are two individual pendular emotions-- 697 00:45:37,610 --> 00:45:39,330 one this way and one that way. 698 00:45:39,330 --> 00:45:41,310 And what I was doing at the beginning 699 00:45:41,310 --> 00:45:44,580 is I pulled it to the side, which 700 00:45:44,580 --> 00:45:46,470 would start one of those modes. 701 00:45:46,470 --> 00:45:48,340 And I pulled it back, which will put 702 00:45:48,340 --> 00:45:51,670 some energy into the other mode, and let it go. 703 00:45:51,670 --> 00:45:55,540 And now what you have is the sum of these two different motions 704 00:45:55,540 --> 00:45:57,230 adding up. 705 00:45:57,230 --> 00:46:00,020 It goes in circles and then in straight lines. 706 00:46:00,020 --> 00:46:05,700 And the fact that they-- this is a phenomenon called beading. 707 00:46:05,700 --> 00:46:09,240 And it is because these two pendulums, 708 00:46:09,240 --> 00:46:13,410 even though they have strings of the same length, 709 00:46:13,410 --> 00:46:16,654 they actually have slightly different natural frequencies. 710 00:46:16,654 --> 00:46:18,570 They're each single degree of freedom systems. 711 00:46:18,570 --> 00:46:21,040 They're two independent single degree of freedom systems, 712 00:46:21,040 --> 00:46:22,990 each with their own natural frequency. 713 00:46:22,990 --> 00:46:27,789 But if you mix them then they're going to exhibit this motion. 714 00:46:27,789 --> 00:46:29,830 So that's something really important to remember. 715 00:46:29,830 --> 00:46:34,900 A quiz question that I like to ask is-- it's easy to grade 716 00:46:34,900 --> 00:46:38,850 and it's no math required-- is to literally-- I've often 717 00:46:38,850 --> 00:46:40,870 done this in exams-- walk in with something 718 00:46:40,870 --> 00:46:45,330 like that block of wood and say, is this a natural mode? 719 00:46:53,880 --> 00:47:00,910 Time to do one-- let me see here. 720 00:47:29,460 --> 00:47:31,930 So now let's pick up where Professor Gossard left off. 721 00:47:31,930 --> 00:47:34,380 Let's go talking about natural frequencies and mode 722 00:47:34,380 --> 00:47:39,100 shapes of linearized two degree of freedom systems. 723 00:47:39,100 --> 00:47:43,880 But I want to generalize a little bit on what he did. 724 00:47:43,880 --> 00:47:49,400 So he, in his lecture, analyzed this system like this. 725 00:47:52,710 --> 00:47:54,470 I'll just kind of put the highlights here. 726 00:48:09,860 --> 00:48:22,100 And this is now solving for natural frequencies and mode 727 00:48:22,100 --> 00:48:22,600 shapes. 728 00:48:26,460 --> 00:48:28,845 He came up with a set of equations of motion for this. 729 00:48:28,845 --> 00:48:33,605 This was, I guess, M1, M2. 730 00:48:33,605 --> 00:48:40,580 And the equations of motion for this are m1 in matrix form. 731 00:48:49,436 --> 00:48:53,000 Now I'm going to do this to emphasize something. 732 00:48:57,480 --> 00:48:59,040 In general there could be damping 733 00:48:59,040 --> 00:49:01,590 in our linearized system. 734 00:49:01,590 --> 00:49:08,520 And we have a stiffness matrix-- K1 plus K2 minus K2. 735 00:49:19,380 --> 00:49:23,570 And in general there could be forces, 736 00:49:23,570 --> 00:49:28,130 which are functions of time on that system. 737 00:49:28,130 --> 00:49:30,850 Now, if we want to find natural frequencies in mode shapes, 738 00:49:30,850 --> 00:49:33,980 we go looking for what we call with the undamped natural 739 00:49:33,980 --> 00:49:35,700 frequencies in mode shapes. 740 00:49:35,700 --> 00:49:37,710 So this problem doesn't even have dampers in it. 741 00:49:37,710 --> 00:49:40,290 But if it did for the purpose of finding 742 00:49:40,290 --> 00:49:45,280 natural frequencies in mode shapes, you just set to 0. 743 00:49:45,280 --> 00:49:47,130 And with the forces you do the same thing. 744 00:49:52,030 --> 00:49:57,850 And now you have undamped, unforced equations of motion. 745 00:50:01,610 --> 00:50:05,630 And this is then of the form of mass matrix times 746 00:50:05,630 --> 00:50:11,720 an acceleration vector, X1, X2, plus a stiffness matrix, 747 00:50:11,720 --> 00:50:15,815 times a displacement vector equals 0. 748 00:50:15,815 --> 00:50:18,250 So in matrix notation it looks like that. 749 00:50:20,950 --> 00:50:26,670 This is the way you would do any rigid body vibration problem. 750 00:50:26,670 --> 00:50:28,490 This is two degrees of freedom. 751 00:50:28,490 --> 00:50:29,940 But this is the general expression 752 00:50:29,940 --> 00:50:33,681 for an n degree of freedom system. 753 00:50:33,681 --> 00:50:36,770 If we had three masses here, then these would be 3 by 3 754 00:50:36,770 --> 00:50:40,390 matrices instead of 2 by 2's. 755 00:50:40,390 --> 00:50:44,160 So that's the basic formulation. 756 00:50:44,160 --> 00:50:47,660 And you went through last time with 2 by 2. 757 00:50:47,660 --> 00:50:52,890 You can actually go through and find the fourth order equation 758 00:50:52,890 --> 00:50:59,180 in omega and solve for two roots of omega squared. 759 00:50:59,180 --> 00:51:02,370 And you've got the two natural frequencies, plug them back in. 760 00:51:02,370 --> 00:51:05,050 You've got the two mode shapes that go along with them. 761 00:51:05,050 --> 00:51:08,020 So you did it that way by hand so you 762 00:51:08,020 --> 00:51:10,650 can see how you can work out the natural frequencies. 763 00:51:10,650 --> 00:51:14,850 How can you do-- I'm going to show an approach that you'd 764 00:51:14,850 --> 00:51:16,790 more likely use on a computer. 765 00:51:16,790 --> 00:51:20,850 And if you get the larger order n degree of freedom systems, 766 00:51:20,850 --> 00:51:23,630 you're going to want to do this-- instead of by hand-- 767 00:51:23,630 --> 00:51:25,750 have a computer do the work for you. 768 00:51:25,750 --> 00:51:30,495 So this is the generic form. 769 00:51:30,495 --> 00:51:34,580 And let's just assume for a minute 770 00:51:34,580 --> 00:51:37,050 it's an n degree of freedom system. 771 00:51:37,050 --> 00:51:41,470 So these are n by n matrices. 772 00:51:45,750 --> 00:51:48,100 How would we find the natural frequencies and mode 773 00:51:48,100 --> 00:51:52,490 shapes of this general system? 774 00:51:52,490 --> 00:52:05,920 So you assume solutions of the form 775 00:52:05,920 --> 00:52:09,760 of what I've been describing-- a natural mode. 776 00:52:09,760 --> 00:52:15,110 Any natural mode of the system has a particular shape 777 00:52:15,110 --> 00:52:18,519 to it and a particular frequency. 778 00:52:18,519 --> 00:52:19,310 And that's the key. 779 00:52:19,310 --> 00:52:21,150 That's the key assumption here. 780 00:52:21,150 --> 00:52:26,647 You assume solutions of the form that this vector x-- 781 00:52:26,647 --> 00:52:28,480 instead of writing it in brackets like this, 782 00:52:28,480 --> 00:52:30,532 I'm going to make this. 783 00:52:30,532 --> 00:52:37,460 So X here is just with a line underneath it. 784 00:52:37,460 --> 00:52:48,080 So x is of the form X1 of t down to Xn 785 00:52:48,080 --> 00:52:50,950 if you have n degrees of freedom. 786 00:52:50,950 --> 00:52:53,890 You're looking for a solution for that thing. 787 00:52:53,890 --> 00:52:57,900 And it's going to have an amplitude to it-- 788 00:52:57,900 --> 00:53:02,770 A1 down to An. 789 00:53:02,770 --> 00:53:06,590 And this is any one mode. 790 00:53:06,590 --> 00:53:10,710 So any one mode will look like a set of amplitudes 791 00:53:10,710 --> 00:53:14,610 that govern its mode shape. 792 00:53:14,610 --> 00:53:17,110 And it will oscillate. 793 00:53:17,110 --> 00:53:21,110 We can write the oscillation as cosine omega. 794 00:53:21,110 --> 00:53:24,070 And I'll put an i here, it's the i-th natural frequency 795 00:53:24,070 --> 00:53:26,970 minus some phase angle. 796 00:53:26,970 --> 00:53:29,380 So in general, each mode-- assume solutions-- 797 00:53:29,380 --> 00:53:34,560 I'll say here for each mode. 798 00:53:37,520 --> 00:53:41,210 So each mode, any mode, mode I will look like this. 799 00:53:41,210 --> 00:53:43,759 It will have a shape to it governed by this. 800 00:53:43,759 --> 00:53:45,300 And these are basically on constants. 801 00:53:45,300 --> 00:53:47,630 Once determined, this is just a constant. 802 00:53:47,630 --> 00:53:49,510 And here's your time dependence. 803 00:53:49,510 --> 00:53:53,660 And it's going to-- I left out my t-- oscillate 804 00:53:53,660 --> 00:53:54,950 at some natural frequency. 805 00:53:54,950 --> 00:53:57,770 So we know this is what the solution has to look like. 806 00:53:57,770 --> 00:54:01,485 And we can take this and plug it in to this equation. 807 00:54:11,710 --> 00:54:18,230 This vector of responses is some vector of amplitude 808 00:54:18,230 --> 00:54:24,690 times the cosine omega t minus phi. 809 00:54:24,690 --> 00:54:28,180 And just plug that into this set of matrix equations. 810 00:54:28,180 --> 00:54:37,590 Note that x double dot is just a-- you get minus omega squared 811 00:54:37,590 --> 00:54:41,910 a cosine omega t. 812 00:54:45,720 --> 00:54:49,455 And we now substitute these into here. 813 00:54:56,040 --> 00:55:01,570 You get minus omega squared for the first term. 814 00:55:01,570 --> 00:55:04,190 Minus omega squared times the mass 815 00:55:04,190 --> 00:55:12,280 matrix a cosine omega t minus phi, 816 00:55:12,280 --> 00:55:19,410 plus this stiffness matrix a-- better consistent notation 817 00:55:19,410 --> 00:55:26,157 here, excuse me-- a cosine omega t minus phi. 818 00:55:26,157 --> 00:55:27,365 And all that's equal to zero. 819 00:55:30,060 --> 00:55:32,120 So these go away. 820 00:55:32,120 --> 00:55:34,770 You can cancel them out. 821 00:55:34,770 --> 00:55:38,950 And we can factor out this a quantity. 822 00:55:38,950 --> 00:55:56,740 And we have minus omega squared m, plus k times a equals 0. 823 00:55:56,740 --> 00:55:59,067 So you can do this with any linearized n 824 00:55:59,067 --> 00:56:00,650 degree of freedom system that you know 825 00:56:00,650 --> 00:56:02,740 has a vibration solution to it. 826 00:56:09,110 --> 00:56:12,100 These are the unknown mode shapes. 827 00:56:12,100 --> 00:56:15,010 And so in order to satisfy this equation, 828 00:56:15,010 --> 00:56:19,290 either this a has to be 0, which is a trivial solution. 829 00:56:19,290 --> 00:56:21,770 There's no motion, no mode shape-- 830 00:56:21,770 --> 00:56:35,700 or which this is trivial, not too useful. 831 00:56:35,700 --> 00:56:48,520 Either a has to be 0, or the determinant of this quantity 832 00:56:48,520 --> 00:56:49,450 has to be 0. 833 00:56:58,970 --> 00:57:00,660 But the way you do that on a computer-- 834 00:57:00,660 --> 00:57:02,220 so that would be beginning the way 835 00:57:02,220 --> 00:57:05,120 you would analyze this by hand. 836 00:57:05,120 --> 00:57:07,160 You find the determinant of that matrix. 837 00:57:07,160 --> 00:57:12,550 And if it is a two degree of freedom system, 838 00:57:12,550 --> 00:57:16,810 you'll get an equation n omega to the fourth, which has 839 00:57:16,810 --> 00:57:18,180 two roots for omega squared. 840 00:57:18,180 --> 00:57:19,888 If it's a three degree of freedom system, 841 00:57:19,888 --> 00:57:22,930 you'll get an equation of omega to the sixth when you write out 842 00:57:22,930 --> 00:57:24,180 that determinant. 843 00:57:24,180 --> 00:57:27,020 And it has three roots for omega squared. 844 00:57:27,020 --> 00:57:28,740 An n degree of freedom system has 845 00:57:28,740 --> 00:57:32,850 an equation that's of order 2n omega to the 2nth power. 846 00:57:32,850 --> 00:57:36,110 And it'll have n solutions or roots 847 00:57:36,110 --> 00:57:39,684 for the natural frequency for omega squared. 848 00:57:39,684 --> 00:57:41,100 But that would be if you're trying 849 00:57:41,100 --> 00:57:42,290 to grind this out by hand. 850 00:57:46,450 --> 00:57:52,552 The way you do this on a computer-- maybe I 851 00:57:52,552 --> 00:57:54,111 can get a little bit more on here. 852 00:57:56,700 --> 00:57:58,080 Come back here. 853 00:58:15,370 --> 00:58:21,730 So I'll go back to the earlier form I had here plus ka 854 00:58:21,730 --> 00:58:23,850 equals 0. 855 00:58:23,850 --> 00:58:35,110 And I'm going to multiply by m inverse. 856 00:58:35,110 --> 00:58:39,200 So if I invert the mass matrix, if I 857 00:58:39,200 --> 00:58:43,230 multiply a matrix by its inverse, what do you get? 858 00:58:43,230 --> 00:58:45,720 So if I multiply m times m inverse? 859 00:58:45,720 --> 00:58:46,860 AUDIENCE: A unit matrix. 860 00:58:46,860 --> 00:58:48,776 J. KIM VANDIVER: You get a unit matrix, right? 861 00:58:48,776 --> 00:58:50,390 It has ones on the diagonal. 862 00:58:50,390 --> 00:58:53,360 So I'm going to multiply through here by this. 863 00:58:53,360 --> 00:58:56,240 And so this gives me a minus omega squared. 864 00:58:56,240 --> 00:59:00,180 And m times m inverse gives me the unit matrix-- 865 00:59:00,180 --> 00:59:02,460 ones on the diagonal. 866 00:59:02,460 --> 00:59:15,910 Times a plus M inverse times ka equals 0. 867 00:59:15,910 --> 00:59:19,210 And this product is just a matrix product-- m inverse 868 00:59:19,210 --> 00:59:20,110 times k. 869 00:59:20,110 --> 00:59:21,693 And I'm going to call it the a matrix. 870 00:59:25,060 --> 00:59:27,790 And I'm going to move this to the other side. 871 00:59:27,790 --> 00:59:32,440 So I have a linear algebraic expression 872 00:59:32,440 --> 00:59:42,500 of the form a times the vector equals omega squared 873 00:59:42,500 --> 00:59:46,500 times the unit matrix times a. 874 00:59:49,040 --> 00:59:55,300 And I could go ahead and multiply this out. 875 00:59:55,300 --> 00:59:58,770 For example, this times that and I'll get a vector. 876 00:59:58,770 --> 01:00:02,100 So this just looks like omega squared 877 01:00:02,100 --> 01:00:04,250 a if you multiply it out. 878 01:00:06,940 --> 01:00:11,580 The vector times the matrix is a vector on the left side. 879 01:00:11,580 --> 01:00:15,550 A vector times a matrix gives me back a vector. 880 01:00:15,550 --> 01:00:16,720 It's just the unit matrix. 881 01:00:16,720 --> 01:00:19,630 So it gives me back the vector times omega squared. 882 01:00:19,630 --> 01:00:24,140 This is in what is known as standard eigenvalue 883 01:00:24,140 --> 01:00:26,150 formulation. 884 01:00:26,150 --> 01:00:28,760 It's a standard eigenvalue problem now. 885 01:00:28,760 --> 01:00:36,020 It's a problem of the form a times a vector equals 886 01:00:36,020 --> 01:00:40,310 something lambda times a. 887 01:00:40,310 --> 01:00:42,060 A parameter which we know happens 888 01:00:42,060 --> 01:00:44,130 to be the frequency squared. 889 01:00:44,130 --> 01:00:49,580 But this is standard eigenvalue formulation. 890 01:00:49,580 --> 01:00:50,771 Yeah? 891 01:00:50,771 --> 01:00:52,645 AUDIENCE: I was just asking if you wrote down 892 01:00:52,645 --> 01:00:55,340 omega squared a because it's equal to the length up there. 893 01:00:55,340 --> 01:00:58,660 J. KIM VANDIVER: Ah, good. 894 01:00:58,660 --> 01:01:00,320 Omega squared a. 895 01:01:00,320 --> 01:01:04,190 So a times the unit vector you get a back as a vector. 896 01:01:04,190 --> 01:01:06,060 And I got the omega squared in front of it. 897 01:01:06,060 --> 01:01:09,640 And oftentimes in a manual for Matlab or something 898 01:01:09,640 --> 01:01:12,410 they'll describe this as some parameter. 899 01:01:12,410 --> 01:01:15,460 It's a constant times a. 900 01:01:15,460 --> 01:01:20,160 And this is standard what they call eigenvalue formulation. 901 01:01:20,160 --> 01:01:38,720 And in Matlab if you say, for example, E equals EIG of A. 902 01:01:38,720 --> 01:01:49,455 This returns a vector, which is the natural frequency squared. 903 01:01:52,392 --> 01:01:54,080 It will return these lambdas. 904 01:01:54,080 --> 01:01:56,835 And the first one is omega 1 squared down 905 01:01:56,835 --> 01:01:58,060 to omega n squared. 906 01:02:00,780 --> 01:02:02,820 And if you go with this function, 907 01:02:02,820 --> 01:02:12,170 if you go a little further, if you say V comma D is EIG A, 908 01:02:12,170 --> 01:02:16,570 then this gives you two matrices back. 909 01:02:16,570 --> 01:02:21,910 It gives you V. And V is a matrix, which 910 01:02:21,910 --> 01:02:25,310 its columns are the mode shape. 911 01:02:25,310 --> 01:02:36,420 So A1 to An, this is mode 1 over to A1 to An for mode n. 912 01:02:36,420 --> 01:02:38,530 It gives you two matrices. 913 01:02:38,530 --> 01:02:40,070 One that's that. 914 01:02:40,070 --> 01:02:49,330 And another one a D matrix, which has the lambdas-- lambda 915 01:02:49,330 --> 01:02:55,650 1 lambda n on the diagonals. 916 01:02:55,650 --> 01:02:58,210 And it's a diagonal matrix. 917 01:02:58,210 --> 01:03:00,120 So it gives you two matrices back. 918 01:03:00,120 --> 01:03:04,270 One that has the eigenvectors, the mode shapes. 919 01:03:04,270 --> 01:03:06,970 And another matrix whose diagonal elements are 920 01:03:06,970 --> 01:03:10,190 the natural frequency squared. 921 01:03:10,190 --> 01:03:14,620 And that's all there is to it if you do this numerically. 922 01:03:14,620 --> 01:03:17,450 And there's lots of different programs. 923 01:03:17,450 --> 01:03:20,340 There's multiple ways of doing this in Matlab. 924 01:03:20,340 --> 01:03:23,720 When you do it this way it doesn't come out sometimes 925 01:03:23,720 --> 01:03:26,670 nicely ordered and what I call normalized. 926 01:03:26,670 --> 01:03:31,067 But it does produce the eigenvalues. 927 01:03:31,067 --> 01:03:32,900 They're called eigenvalues and eigenvectors. 928 01:03:32,900 --> 01:03:36,210 The eigenvalues are the lambdas, the natural frequencies 929 01:03:36,210 --> 01:03:37,180 squared. 930 01:03:37,180 --> 01:03:40,580 And the eigenvectors are these mode shapes that 931 01:03:40,580 --> 01:03:45,510 go with each natural frequency. 932 01:03:45,510 --> 01:03:47,760 Once you know the natural frequencies and mode shapes, 933 01:03:47,760 --> 01:03:50,310 now we want to get back to talking about solutions. 934 01:03:50,310 --> 01:03:51,950 This idea of mode superposition. 935 01:03:51,950 --> 01:03:55,070 And if you give it a set of initial conditions, what 936 01:03:55,070 --> 01:03:55,950 is the response? 937 01:03:55,950 --> 01:03:59,380 How do you add these two modes together? 938 01:03:59,380 --> 01:04:01,220 So let's go back now. 939 01:04:01,220 --> 01:04:08,670 We'll return to two degree of freedom systems 940 01:04:08,670 --> 01:04:11,415 like this one to do an example. 941 01:04:14,140 --> 01:04:27,840 And we assume that the solution was sum A1, A2, cosine omega 942 01:04:27,840 --> 01:04:31,020 t minus a phase angle. 943 01:04:31,020 --> 01:04:34,750 That each mode would have this character to it. 944 01:04:34,750 --> 01:04:39,645 And I'm going to normalize my mode shapes. 945 01:04:51,790 --> 01:04:54,940 So for each mode shape of the system-- so 946 01:04:54,940 --> 01:04:58,290 this could be for mode one. 947 01:04:58,290 --> 01:04:59,990 This is the mode shape for mode one. 948 01:04:59,990 --> 01:05:03,720 This is natural frequency one and phase angle one. 949 01:05:03,720 --> 01:05:06,610 Each mode shape-- I could write this then 950 01:05:06,610 --> 01:05:10,460 as-- I could factor out the A1. 951 01:05:10,460 --> 01:05:13,090 Just pull out A1, divide each member by A1. 952 01:05:13,090 --> 01:05:22,430 So this can be written as A1 times 1 and A2 over A1. 953 01:05:22,430 --> 01:05:23,720 So I've just factored out. 954 01:05:29,670 --> 01:05:33,920 So for mode one it's normalized mode shape-- 955 01:05:33,920 --> 01:05:41,370 by normalize you just pick some way that you repeatedly use, 956 01:05:41,370 --> 01:05:42,550 consistent in its use. 957 01:05:42,550 --> 01:05:47,100 I often say let's make the top element of the vector 1. 958 01:05:47,100 --> 01:05:49,990 And to make the top one 1 you factor out 959 01:05:49,990 --> 01:05:52,365 whatever its value is that you get back from the computer 960 01:05:52,365 --> 01:05:54,080 or from your calculation. 961 01:05:54,080 --> 01:05:56,930 You factor that out of every member. 962 01:05:56,930 --> 01:05:59,430 Now you have a normalized mode shape whose top element is 1. 963 01:05:59,430 --> 01:06:01,263 There's lots of other normalization schemes. 964 01:06:01,263 --> 01:06:03,380 That's just one way to do it. 965 01:06:03,380 --> 01:06:12,340 And that's one of the mode shapes. 966 01:06:12,340 --> 01:06:26,790 The total solution is X1, X2-- and this 967 01:06:26,790 --> 01:06:29,690 is where the mode superposition part comes in-- 968 01:06:29,690 --> 01:06:38,080 is some undetermined constant A1 times the mode shape 969 01:06:38,080 --> 01:06:47,836 A2 over A1 for mode 1 cosine omega 1 t minus phi 1. 970 01:06:47,836 --> 01:06:49,210 And I'm going to run out of room. 971 01:07:14,455 --> 01:07:17,780 And now the responses to initial conditions-- this 972 01:07:17,780 --> 01:07:20,040 has got another term. 973 01:07:20,040 --> 01:07:21,460 I'm just going to rewrite it here. 974 01:07:21,460 --> 01:07:24,090 So we're looking at-- our total motion response 975 01:07:24,090 --> 01:07:39,515 now by mode superposition will be A1, 1 A2 over A1, mode 1. 976 01:08:01,160 --> 01:08:05,840 So the free vibration response of any two degree 977 01:08:05,840 --> 01:08:09,920 of freedom system, linearized equation, any two 978 01:08:09,920 --> 01:08:12,940 degree of freedom linear system can be made up 979 01:08:12,940 --> 01:08:16,050 of the sum of two terms. 980 01:08:16,050 --> 01:08:20,810 The motion at its first natural frequency in its first mode 981 01:08:20,810 --> 01:08:22,990 shape. 982 01:08:22,990 --> 01:08:26,250 And another term is the motion at a second natural frequency 983 01:08:26,250 --> 01:08:28,220 and its second mode shape. 984 01:08:28,220 --> 01:08:31,740 But now you have two undetermined constants 985 01:08:31,740 --> 01:08:33,350 out here-- A1 and A2. 986 01:08:33,350 --> 01:08:34,775 Where do they come from? 987 01:08:38,340 --> 01:08:43,140 You have to use your initial conditions to get those. 988 01:08:43,140 --> 01:08:44,460 I'll write down-- let's see. 989 01:08:44,460 --> 01:08:48,875 I have just maybe enough time to write this down. 990 01:08:51,560 --> 01:08:52,810 These are functions of time. 991 01:08:55,340 --> 01:09:07,740 So A1 and A2 come from the ICs, the initial conditions. 992 01:09:07,740 --> 01:09:16,310 So at t equals 0, for example, plug in t equals 0 into here. 993 01:09:16,310 --> 01:09:20,380 You get cosine phi. 994 01:09:20,380 --> 01:09:23,890 And over here another phi 1 and another way over here. 995 01:09:23,890 --> 01:09:26,680 Cosine of minus phi 1 is cosine phi 1. 996 01:09:26,680 --> 01:09:31,740 So if you put in t equals 0, you find out the X1 997 01:09:31,740 --> 01:09:39,420 at 0, which I'll write X1 0, and X2 of 0 998 01:09:39,420 --> 01:09:45,069 is equal to-- I'll actually write them out. 999 01:09:45,069 --> 01:09:57,750 This is going to be A1 times 1 cosine phi 1, plus A2 times 1000 01:09:57,750 --> 01:10:03,050 1 cosine phi 2. 1001 01:10:03,050 --> 01:10:08,920 And the second equation that this gives you is A1-- 1002 01:10:08,920 --> 01:10:12,030 and now to keep from writing these many, many times, 1003 01:10:12,030 --> 01:10:21,275 I'm going to let this first A2 over A1 for mode 1 1004 01:10:21,275 --> 01:10:28,130 be R1, and the second one A2 over A1 for mode 2. 1005 01:10:28,130 --> 01:10:30,360 I'll just call it R2. 1006 01:10:30,360 --> 01:10:31,710 And then I can write this out. 1007 01:10:31,710 --> 01:10:40,430 So the second equation, this is R1 cosine phi 1, 1008 01:10:40,430 --> 01:10:48,300 plus A2 R2 cosine phi 2. 1009 01:10:48,300 --> 01:10:51,350 And now I have initial conditions 1010 01:10:51,350 --> 01:10:53,790 that are normally given. 1011 01:10:53,790 --> 01:10:56,220 This is an initial displacement on 1 1012 01:10:56,220 --> 01:11:00,180 and a initial displacement on 2. 1013 01:11:00,180 --> 01:11:01,379 The phis I don't know. 1014 01:11:01,379 --> 01:11:02,420 And the A's I don't know. 1015 01:11:02,420 --> 01:11:05,370 I have four unknowns and two equations. 1016 01:11:05,370 --> 01:11:07,215 How do I get two more equations? 1017 01:11:07,215 --> 01:11:11,520 I take the derivative of this expression to get velocity. 1018 01:11:11,520 --> 01:11:16,620 And I get an A omega here and an A omega here. 1019 01:11:16,620 --> 01:11:21,440 And I plug in t equals 0. 1020 01:11:21,440 --> 01:11:24,310 And I get two more equations. 1021 01:11:24,310 --> 01:11:41,050 So X1 dot and X2 dot equal at t equals 0. 1022 01:11:41,050 --> 01:11:44,700 This gives me two more equations. 1023 01:11:44,700 --> 01:11:53,560 And if I have a place to write them-- for example, 1024 01:11:53,560 --> 01:12:00,750 X1 dot-- or X1 0 dot, the initial condition on velocity. 1025 01:12:00,750 --> 01:12:03,875 X2 dot-- this is two equations. 1026 01:12:06,420 --> 01:12:08,790 I've only got time to write down one of them. 1027 01:12:08,790 --> 01:12:11,680 And you could do the other for exercise. 1028 01:12:11,680 --> 01:12:23,970 You find this is A1 omega 1 sine phi 1, plus A2 omega 1029 01:12:23,970 --> 01:12:27,940 2 sine phi 2. 1030 01:12:27,940 --> 01:12:35,430 And the second equation is you get A1 R1 omega 1 sine phi 1, 1031 01:12:35,430 --> 01:12:43,390 plus A2 R2 omega 2 sine phi 2. 1032 01:12:43,390 --> 01:12:47,390 So now you have one, two. 1033 01:12:47,390 --> 01:12:49,865 And this is the initial conditions on velocity. 1034 01:12:49,865 --> 01:12:52,850 These are initial values of velocity. 1035 01:12:52,850 --> 01:12:55,760 Now you have one, two, three, four equations and one, 1036 01:12:55,760 --> 01:13:02,750 two, three, four unknowns-- the A1, A2, phi 1, phi 2. 1037 01:13:02,750 --> 01:13:13,070 And I guess I will give you the answer, so you have it once. 1038 01:13:30,810 --> 01:13:34,980 So a little tedious but this is sort of in the spirit of we 1039 01:13:34,980 --> 01:13:36,940 do two degree of freedom systems so 1040 01:13:36,940 --> 01:13:39,680 that we can see how it works. 1041 01:13:39,680 --> 01:13:43,360 And then for larger degrees of freedom systems 1042 01:13:43,360 --> 01:13:45,580 you would do this with a computer. 1043 01:13:45,580 --> 01:13:52,555 But the solution for A1 is 1 over R2 minus R1. 1044 01:13:55,485 --> 01:13:57,990 It's all in terms now of things you know. 1045 01:13:57,990 --> 01:14:00,910 These R2's and R1's are part of the mode shape, 1046 01:14:00,910 --> 01:14:06,270 and the rest is initial conditions. 1047 01:14:06,270 --> 01:14:23,020 R2 x1 0, minus X2 0 squared, plus R2 V1 0, 1048 01:14:23,020 --> 01:14:31,550 minus V2 0 quantity squared, over omega 1 1049 01:14:31,550 --> 01:14:36,080 squared, the whole thing square root. 1050 01:14:36,080 --> 01:14:37,630 But now this is all stuff you know. 1051 01:14:37,630 --> 01:14:39,790 The given initial conditions on velocity, 1052 01:14:39,790 --> 01:14:43,530 given initial conditions on displacement. 1053 01:14:43,530 --> 01:14:44,930 You know the natural frequency. 1054 01:14:44,930 --> 01:14:47,020 You know the pieces of the mode shapes. 1055 01:14:47,020 --> 01:14:50,940 Just plug it in, you're going to get a number. 1056 01:14:50,940 --> 01:14:54,350 The magnitude of A1 for the given initial conditions. 1057 01:14:54,350 --> 01:15:10,600 A2-- a very similar expression-- minus R1 X1 0 plus x2 0 1058 01:15:10,600 --> 01:15:35,770 squared And you'd solve for A2 and phi 1059 01:15:35,770 --> 01:15:41,240 1-- it's a little simpler-- minus tangent inverse, 1060 01:15:41,240 --> 01:16:00,540 phi 2 0 minus R2 V1 0, all over omega 1 R2 x1 0, minus x2 0. 1061 01:16:00,540 --> 01:16:02,210 That's one of the phase angles. 1062 01:16:02,210 --> 01:16:06,570 And the other phase angle, a very similar expression. 1063 01:16:06,570 --> 01:16:18,650 Minus tangent inverse, minus R1 V1 0, plus V2 0 omega 1064 01:16:18,650 --> 01:16:25,650 2 R1 x1 0, minus x2 0. 1065 01:16:25,650 --> 01:16:30,100 So just expressions in terms of the initial conditions and you 1066 01:16:30,100 --> 01:16:31,869 can get all four quantities. 1067 01:16:31,869 --> 01:16:33,410 You can also do this on the computer. 1068 01:16:33,410 --> 01:16:37,040 But in the few short lectures that we have 1069 01:16:37,040 --> 01:16:38,967 we're not going to get into that. 1070 01:16:38,967 --> 01:16:40,550 But this just shows you where it goes. 1071 01:16:40,550 --> 01:16:41,590 You could do this now. 1072 01:16:41,590 --> 01:16:44,040 There are straightforward ways of doing it 1073 01:16:44,040 --> 01:16:47,420 with matrix algebra on the computer. 1074 01:16:47,420 --> 01:16:52,470 Next time I'll do maybe just a quick example-- I didn't quite 1075 01:16:52,470 --> 01:16:57,830 get to it today-- of a response to initial conditions problem. 1076 01:16:57,830 --> 01:16:59,410 Plug it in there. 1077 01:16:59,410 --> 01:17:00,715 See what happens. 1078 01:17:00,715 --> 01:17:03,260 But we're out of time. 1079 01:17:03,260 --> 01:17:06,680 See you on Thursday.