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KIM VANDIVER: Today's lecture is not mathematically hard, 9 00:00:26,150 --> 00:00:30,932 but it's really important to establish vocabulary today. 10 00:00:30,932 --> 00:00:32,390 We're going to talk about vibration 11 00:00:32,390 --> 00:00:34,690 for the rest of the term. 12 00:00:34,690 --> 00:00:40,110 And vibration is essentially applied dynamics. 13 00:00:40,110 --> 00:00:44,740 So up until now, we've been finding equations of motion, 14 00:00:44,740 --> 00:00:46,330 but not solving them. 15 00:00:46,330 --> 00:00:48,190 Did you notice that? 16 00:00:48,190 --> 00:00:52,660 I've almost never asked you to solve the equation of motion 17 00:00:52,660 --> 00:00:55,890 that you've just discovered using Lagrange or whatever. 18 00:00:55,890 --> 00:00:57,390 The rest of the term, we're actually 19 00:00:57,390 --> 00:01:00,980 going to be talking mostly about the resulting motion. 20 00:01:00,980 --> 00:01:03,170 The equations of motion are pretty easy to find. 21 00:01:03,170 --> 00:01:07,020 You have all the techniques that you need to know for finding. 22 00:01:07,020 --> 00:01:11,140 And now, we're going to talk about how things vibrate. 23 00:01:11,140 --> 00:01:14,130 So why do we choose vibration? 24 00:01:14,130 --> 00:01:20,690 Vibration, one, is an incredibly common phenomenon. 25 00:01:20,690 --> 00:01:23,010 We wouldn't have speech without vibration. 26 00:01:23,010 --> 00:01:26,980 You wouldn't have musical instruments without vibration. 27 00:01:26,980 --> 00:01:30,640 It's a positive thing when it's making good music. 28 00:01:30,640 --> 00:01:33,190 It's a negative thing when it's keeping you awake at night 29 00:01:33,190 --> 00:01:35,620 because the air conditioner in the next room 30 00:01:35,620 --> 00:01:38,130 is causing something to rattle in the room 31 00:01:38,130 --> 00:01:40,190 and it's driving you nuts. 32 00:01:40,190 --> 00:01:44,000 So you can want it, it can be desirable, 33 00:01:44,000 --> 00:01:45,880 and you cannot want it. 34 00:01:45,880 --> 00:01:49,060 And you need to know ways of getting rid of it. 35 00:01:49,060 --> 00:01:54,340 And so we're going to talk about vibration, about making 36 00:01:54,340 --> 00:01:56,870 vibration, about suppressing vibration, 37 00:01:56,870 --> 00:02:01,160 about isolating sensitive instruments from the vibration 38 00:02:01,160 --> 00:02:03,240 of the floor, things like that. 39 00:02:03,240 --> 00:02:05,570 So that's the topic of the rest of the term. 40 00:02:05,570 --> 00:02:09,190 And today, we're going to talk about single degree of freedom 41 00:02:09,190 --> 00:02:11,490 systems. 42 00:02:11,490 --> 00:02:17,460 And you might think that we're spending an awful lot of time 43 00:02:17,460 --> 00:02:19,720 on single degree of freedom systems. 44 00:02:19,720 --> 00:02:24,500 But actually, there's a reason for that. 45 00:02:24,500 --> 00:02:27,980 Lots of things in real life, like-- this 46 00:02:27,980 --> 00:02:30,000 is just an aluminum rod. 47 00:02:30,000 --> 00:02:32,400 This will vibrate. 48 00:02:32,400 --> 00:02:35,470 And continuous systems, which this is, 49 00:02:35,470 --> 00:02:40,390 have a theoretically infinite number of degrees of freedom. 50 00:02:40,390 --> 00:02:44,950 Yet when it comes to talking about its vibration, 51 00:02:44,950 --> 00:02:50,320 it is conceptually easy to think about the vibration 52 00:02:50,320 --> 00:02:54,740 of an object like this, one natural frequency, 53 00:02:54,740 --> 00:02:57,280 one natural mode at a time. 54 00:02:57,280 --> 00:02:59,950 And in fact, you can model that natural mode 55 00:02:59,950 --> 00:03:03,600 with its single degree of freedom equivalent. 56 00:03:03,600 --> 00:03:06,300 And that's the way I approach vibration. 57 00:03:06,300 --> 00:03:08,600 So if you can isolate one particular mode, 58 00:03:08,600 --> 00:03:11,560 you can literally model it as a Mass-Spring-Dashpot. 59 00:03:11,560 --> 00:03:15,950 So you need to understand the Mass-Spring-Dashpot behavior 60 00:03:15,950 --> 00:03:21,590 inside and out, because it's the vocabulary we use to do 61 00:03:21,590 --> 00:03:23,470 much more complicated things. 62 00:03:23,470 --> 00:03:26,680 So a single degree of freedom system, 63 00:03:26,680 --> 00:03:32,962 like the simple pendulum, has a natural frequency. 64 00:03:32,962 --> 00:03:34,295 In this case, it has mode shape. 65 00:03:38,710 --> 00:03:44,170 Here's another one, kind of fun, single degree of freedom. 66 00:03:44,170 --> 00:03:46,196 This obviously involves rotation. 67 00:03:58,356 --> 00:03:59,730 And you can figure that out using 68 00:03:59,730 --> 00:04:02,500 Lagrange or whatever, single degree of freedom systems. 69 00:04:02,500 --> 00:04:05,365 But now, I'm going to excite one mode of vibration of this. 70 00:04:08,638 --> 00:04:09,634 [CLANG] 71 00:04:09,634 --> 00:04:11,630 [HIGH-PITCHED TONE] 72 00:04:11,630 --> 00:04:13,225 Hear the real high pitch? 73 00:04:16,730 --> 00:04:19,079 I'll get it down here by the mic so that people at home 74 00:04:19,079 --> 00:04:24,430 can hear it-- about a kilohertz, way up there. 75 00:04:24,430 --> 00:04:27,300 And that's one natural mode of this thing 76 00:04:27,300 --> 00:04:28,620 in longitudinal vibration. 77 00:04:28,620 --> 00:04:32,511 When I thump it sideways-- 78 00:04:32,511 --> 00:04:33,010 [CLANG] 79 00:04:33,010 --> 00:04:33,900 [LOWER TONE] 80 00:04:33,900 --> 00:04:34,870 you hear a lower tone. 81 00:04:34,870 --> 00:04:35,750 Hear that? 82 00:04:35,750 --> 00:04:36,760 [HUMS LOW] 83 00:04:36,760 --> 00:04:37,647 rather than-- 84 00:04:37,647 --> 00:04:38,581 [HUMS HIGH] 85 00:04:38,581 --> 00:04:40,920 [CLANG] 86 00:04:40,920 --> 00:04:43,250 That's bending vibration of this thing. 87 00:04:43,250 --> 00:04:45,040 But each mode of vibration I can think 88 00:04:45,040 --> 00:04:47,930 of in terms of its equivalent single degree of freedom 89 00:04:47,930 --> 00:04:48,900 oscillate. 90 00:04:48,900 --> 00:04:51,430 So we'll get to talking about these things 91 00:04:51,430 --> 00:04:54,140 a little bit-- continuous systems-- in the last couple 92 00:04:54,140 --> 00:04:55,760 lectures of the term. 93 00:04:55,760 --> 00:04:57,940 But for today then, we're really going 94 00:04:57,940 --> 00:05:01,090 to develop this vocabulary around the vibration 95 00:05:01,090 --> 00:05:04,620 of single degree of freedom systems. 96 00:05:04,620 --> 00:05:05,486 So let's start. 97 00:05:24,643 --> 00:05:27,640 All right. 98 00:05:27,640 --> 00:05:31,310 So to keep it from being totally boring, 99 00:05:31,310 --> 00:05:35,280 I'm going to start with a little Mass-Spring-Dashpot that 100 00:05:35,280 --> 00:05:36,305 has two springs. 101 00:05:45,280 --> 00:05:49,210 And they're of such a length that unstretched, 102 00:05:49,210 --> 00:05:52,780 they just meet in the middle. 103 00:05:52,780 --> 00:06:00,050 And then, I'm going to take a mass 104 00:06:00,050 --> 00:06:04,170 and I'm going to squeeze it in between these two springs-- 105 00:06:04,170 --> 00:06:08,170 I can't draw a spring very well today-- and this is k1 106 00:06:08,170 --> 00:06:11,580 and this is k2 and here's m. 107 00:06:11,580 --> 00:06:14,150 And we'll put it on a roller so it's obviously constrained 108 00:06:14,150 --> 00:06:17,910 to motion in one direction. 109 00:06:17,910 --> 00:06:24,830 And I'll pick this point here as the place 110 00:06:24,830 --> 00:06:28,244 I'm going to put my inertial coordinate. 111 00:06:28,244 --> 00:06:30,160 So my inertial coordinate's just measured from 112 00:06:30,160 --> 00:06:32,740 or happens to be where the endpoints of these two springs 113 00:06:32,740 --> 00:06:34,080 were. 114 00:06:34,080 --> 00:06:36,520 Now, to squeeze the spring in here, 115 00:06:36,520 --> 00:06:40,720 I have this clearly pre-compression 116 00:06:40,720 --> 00:06:41,990 in these springs. 117 00:06:41,990 --> 00:06:49,050 So we are no longer in a zero-force state, right? 118 00:06:49,050 --> 00:06:51,660 So and I want to get the equations of motion in this. 119 00:06:51,660 --> 00:06:53,772 And moreover, I want to predict-- 120 00:06:53,772 --> 00:06:55,730 I want to find out what's the natural frequency 121 00:06:55,730 --> 00:06:58,320 of this spring. 122 00:06:58,320 --> 00:07:00,610 So let's check your intuition. 123 00:07:00,610 --> 00:07:02,690 So write down on your piece of paper 124 00:07:02,690 --> 00:07:08,120 whether or not the natural frequency will be different 125 00:07:08,120 --> 00:07:11,090 because there's pre-compression, or whether or not 126 00:07:11,090 --> 00:07:12,810 that pre-compression in the springs 127 00:07:12,810 --> 00:07:16,880 has nothing to do with the natural frequency. 128 00:07:16,880 --> 00:07:19,640 So write down on your paper "natural frequency 129 00:07:19,640 --> 00:07:25,442 is different" or "natural frequency is the same." 130 00:07:25,442 --> 00:07:26,650 Let's have a prediction here. 131 00:07:30,379 --> 00:07:32,170 And then, we'll set about figuring this out 132 00:07:32,170 --> 00:07:33,669 and in the course of doing it, we'll 133 00:07:33,669 --> 00:07:35,680 develop a little vocabulary. 134 00:07:35,680 --> 00:07:37,610 All through the course so far, when 135 00:07:37,610 --> 00:07:39,160 we've done equations of motion, we've 136 00:07:39,160 --> 00:07:42,590 usually picked the zero-spring-force position. 137 00:07:42,590 --> 00:07:46,310 And we sort of led you down this rosy path that suggests 138 00:07:46,310 --> 00:07:47,500 that's the way we do it. 139 00:07:47,500 --> 00:07:50,340 But there are other ways that you're 140 00:07:50,340 --> 00:07:52,800 going to find that are preferable to that, sometimes. 141 00:07:52,800 --> 00:07:56,150 So that's one of the reasons I'm doing this example. 142 00:07:56,150 --> 00:07:57,665 So let's do a free body diagram. 143 00:08:06,030 --> 00:08:08,520 And if I held this mass, for example, 144 00:08:08,520 --> 00:08:10,970 right at the center when I put the springs in, 145 00:08:10,970 --> 00:08:13,590 it's obvious that this spring gets compressed 146 00:08:13,590 --> 00:08:16,200 by half of the length of the mass 147 00:08:16,200 --> 00:08:18,010 and this spring gets compressed by half 148 00:08:18,010 --> 00:08:20,160 of the length of the mass, right? 149 00:08:20,160 --> 00:08:23,980 So this is going to be L long. 150 00:08:23,980 --> 00:08:26,340 So if I held it right in the middle, 151 00:08:26,340 --> 00:08:28,715 it would compress L/2 and L/2. 152 00:08:28,715 --> 00:08:30,590 But then, when I release it, if these springs 153 00:08:30,590 --> 00:08:32,580 are a different spring constant, it's 154 00:08:32,580 --> 00:08:33,840 going to move a little bit. 155 00:08:33,840 --> 00:08:37,260 So the force on this side pushing back 156 00:08:37,260 --> 00:08:43,330 is sum k1 times L/2 minus the distance 157 00:08:43,330 --> 00:08:46,660 that I move in that direction, which would relieve it. 158 00:08:46,660 --> 00:08:49,690 And the force on this side also pushes back. 159 00:08:49,690 --> 00:08:56,840 It's k2 times L/2 over 2 plus x, because when 160 00:08:56,840 --> 00:09:00,550 I go in that direction, I'm compressing it even further. 161 00:09:00,550 --> 00:09:04,360 And those are the total forces in the x-direction 162 00:09:04,360 --> 00:09:05,220 on this body. 163 00:09:05,220 --> 00:09:08,040 There's an N and an mg, which we know 164 00:09:08,040 --> 00:09:09,930 we don't have to deal with because we're only 165 00:09:09,930 --> 00:09:12,750 interested in motion left and right. 166 00:09:12,750 --> 00:09:13,250 All right? 167 00:09:13,250 --> 00:09:18,980 So we can say sum of the forces in the x-direction, 168 00:09:18,980 --> 00:09:22,810 mass times the acceleration. 169 00:09:22,810 --> 00:09:39,130 And those forces are k1 L/2 minus x minus k2 L/2 plus x. 170 00:09:39,130 --> 00:09:44,840 And that's the complete equation of motion for this problem. 171 00:09:44,840 --> 00:09:47,060 And rearrange it so that I get the functions 172 00:09:47,060 --> 00:09:50,006 of x together here. 173 00:09:50,006 --> 00:10:00,160 So mx double dot plus k1 plus k2 times x 174 00:10:00,160 --> 00:10:10,690 equals L/2 times k1 minus k2. 175 00:10:10,690 --> 00:10:13,410 And that's your equation of motion. 176 00:10:13,410 --> 00:10:15,050 It's non-homogeneous. 177 00:10:15,050 --> 00:10:19,620 This is all constants on the right-hand side. 178 00:10:19,620 --> 00:10:25,672 And on the left-hand side are the functions of x, right? 179 00:10:25,672 --> 00:10:27,930 So what's the natural frequency of the system? 180 00:10:34,888 --> 00:10:37,380 AUDIENCE: Square root of k1 plus k2 over m. 181 00:10:37,380 --> 00:10:40,530 J. KIM VANDIVER: I hear a square root of the quantity k1 182 00:10:40,530 --> 00:10:44,300 plus k2, the stiffness, divided by m, k over m, 183 00:10:44,300 --> 00:10:46,899 a usual Mass-Spring-Dashpot system. 184 00:10:46,899 --> 00:10:48,440 Did the pre-compression have anything 185 00:10:48,440 --> 00:10:50,490 to do with the natural frequency? 186 00:10:50,490 --> 00:10:52,960 I won't ask you to embarrass yourselves, but a few of you 187 00:10:52,960 --> 00:10:56,030 probably got that wrong, all right? 188 00:10:56,030 --> 00:11:00,440 So there's a lesson in this that I want you to go away with. 189 00:11:08,090 --> 00:11:09,740 and I'll say it once. 190 00:11:09,740 --> 00:11:17,090 And that is when an external force 191 00:11:17,090 --> 00:11:23,000 has nothing to do with the motion coordinates 192 00:11:23,000 --> 00:11:24,739 in the problem. 193 00:11:24,739 --> 00:11:26,405 It doesn't affect the natural frequency. 194 00:11:29,900 --> 00:11:31,320 These come from external forces. 195 00:11:31,320 --> 00:11:34,340 These are these pre-compressions, right? 196 00:11:34,340 --> 00:11:39,000 And I can separate them out and they are not functions of x. 197 00:11:39,000 --> 00:11:41,215 The stuff on the right-hand side of the equation, 198 00:11:41,215 --> 00:11:44,010 that's not a function of the motion variable-- 199 00:11:44,010 --> 00:11:47,820 cannot affect the natural frequency. 200 00:11:47,820 --> 00:11:49,550 So I'll give you another one. 201 00:11:52,070 --> 00:11:59,910 This is our common thing hanging from a stick. 202 00:12:02,890 --> 00:12:05,940 I've taken my system I built the other day 203 00:12:05,940 --> 00:12:11,620 for a different purpose, but now, it's just a mass 204 00:12:11,620 --> 00:12:13,910 hanging from a spring. 205 00:12:13,910 --> 00:12:16,420 And it's right now at its equilibrium position 206 00:12:16,420 --> 00:12:21,862 or there's non-zero force in the spring. 207 00:12:21,862 --> 00:12:23,320 It clearly has a natural frequency. 208 00:12:26,070 --> 00:12:29,205 And is that natural frequency a function of gravity? 209 00:12:32,420 --> 00:12:34,870 And so if you go to write the equation to motion 210 00:12:34,870 --> 00:12:48,940 of this system, you would find mx double dot plus kx equals mg 211 00:12:48,940 --> 00:12:49,630 g. 212 00:12:49,630 --> 00:12:51,350 But the mg is not a function of x. 213 00:12:51,350 --> 00:12:53,600 The natural frequency's again, the square root of k/m. 214 00:13:01,470 --> 00:13:04,360 Now, we want to talk about solving this differential 215 00:13:04,360 --> 00:13:05,540 equation. 216 00:13:05,540 --> 00:13:07,380 And because it's got this constant term 217 00:13:07,380 --> 00:13:09,350 in the right-hand side, it's non-homogeneous, 218 00:13:09,350 --> 00:13:11,760 which is kind of a nuisance term in terms of dealing 219 00:13:11,760 --> 00:13:13,390 with a differential equation. 220 00:13:13,390 --> 00:13:19,610 It'd be a lot nicer if the right-hand side were 0. 221 00:13:19,610 --> 00:13:26,010 So I want to make the right-hand side of this one 0. 222 00:13:26,010 --> 00:13:32,180 And draw a use of a conclusion from that. 223 00:13:50,540 --> 00:13:51,920 First thing I need to know is I'd 224 00:13:51,920 --> 00:13:56,060 like to know what is the static equilibrium position of this. 225 00:14:07,740 --> 00:14:09,674 And when you go to compute static equilibrium, 226 00:14:09,674 --> 00:14:11,090 you look at the equation of motion 227 00:14:11,090 --> 00:14:13,630 and you say, make all motion variables things 228 00:14:13,630 --> 00:14:15,560 that are functions of time 0. 229 00:14:15,560 --> 00:14:19,160 So no acceleration-- you're left with this. 230 00:14:19,160 --> 00:14:21,730 So you just solve this for whatever the value of x is 231 00:14:21,730 --> 00:14:24,670 and I'll call it x of s for x-static. 232 00:14:24,670 --> 00:14:26,290 And you'll find that, oh, well, it's 233 00:14:26,290 --> 00:14:34,770 that term divided by k1 plus k2, k1 234 00:14:34,770 --> 00:14:40,100 minus k2 all over k1 plus k2. 235 00:14:40,100 --> 00:14:41,465 And that's the static position. 236 00:14:47,970 --> 00:14:50,910 So now, let's say, ah, well, we started off 237 00:14:50,910 --> 00:14:54,340 with this motion variable that wasn't arbitrarily defined 238 00:14:54,340 --> 00:14:56,540 at the middle. 239 00:14:56,540 --> 00:15:00,380 And let's say that, well, it's made up 240 00:15:00,380 --> 00:15:05,060 of a static component, which is a constant, just 241 00:15:05,060 --> 00:15:09,720 a value, plus a dynamic component I'll 242 00:15:09,720 --> 00:15:11,450 call x of d, which moves. 243 00:15:11,450 --> 00:15:14,310 This is the function of time. 244 00:15:14,310 --> 00:15:15,240 This is a constant. 245 00:15:15,240 --> 00:15:17,660 It's not a function of time. 246 00:15:17,660 --> 00:15:19,480 And that means if I take its derivative, 247 00:15:19,480 --> 00:15:21,835 I might need a value for x dot. 248 00:15:21,835 --> 00:15:23,280 That goes away. 249 00:15:23,280 --> 00:15:25,220 It's just xd dot. 250 00:15:25,220 --> 00:15:31,440 And x double dot is xd double dot. 251 00:15:31,440 --> 00:15:36,230 And let's substitute this into my equation of motion. 252 00:15:36,230 --> 00:15:51,300 So it becomes m xd double dot plus k1 plus k2 times-- 253 00:15:51,300 --> 00:16:00,200 and now, this term has got two pieces now-- times xd plus k1 254 00:16:00,200 --> 00:16:11,420 plus k2 times xs equals L/2 k1 minus k2. 255 00:16:15,418 --> 00:16:17,760 All right? 256 00:16:17,760 --> 00:16:22,340 Now if I say, well, let's examine the static case, 257 00:16:22,340 --> 00:16:24,900 then this goes away. 258 00:16:24,900 --> 00:16:27,740 For the static equilibrium case, this term is 0. 259 00:16:27,740 --> 00:16:31,790 This term is 0 because the dynamic motion 260 00:16:31,790 --> 00:16:35,710 is 0 in the static case. 261 00:16:35,710 --> 00:16:39,460 That xd is motion about the static equilibrium position. 262 00:16:39,460 --> 00:16:41,860 So for static case, these two terms go away 263 00:16:41,860 --> 00:16:44,920 and we know that this equals that. 264 00:16:44,920 --> 00:16:47,110 But if that's true, we can get rid of these. 265 00:16:47,110 --> 00:16:49,530 They cancel one another. 266 00:16:49,530 --> 00:16:55,360 These terms cancel and I'm left with m xd double dot 267 00:16:55,360 --> 00:17:03,460 plus k equivalent, I'll call it, xd equals 0. 268 00:17:03,460 --> 00:17:05,589 So the k equivalent's just the total stiffnesses 269 00:17:05,589 --> 00:17:07,297 in the system, whatever works out, right? 270 00:17:07,297 --> 00:17:10,960 In this case, it's k1 plus k2 and the natural frequency, 271 00:17:10,960 --> 00:17:16,440 omega n, is the square root of k equivalent divided by m. 272 00:17:19,450 --> 00:17:23,709 So most often, if you're interested in vibration, 273 00:17:23,709 --> 00:17:25,500 you're interested in natural frequencies, 274 00:17:25,500 --> 00:17:29,360 you're interested in solving the differential equation, 275 00:17:29,360 --> 00:17:33,570 you will find it advantageous to write your equations of motion 276 00:17:33,570 --> 00:17:37,065 around the static equilibrium position. 277 00:17:40,030 --> 00:17:42,840 So I could have started this problem by saying, 278 00:17:42,840 --> 00:17:44,990 whatever the static equilibrium position is 279 00:17:44,990 --> 00:17:48,190 of this thing, that's what I'm measuring x from. 280 00:17:50,850 --> 00:17:54,874 And then, I would have come to this equation eventually. 281 00:17:54,874 --> 00:17:57,540 You'd have to figure out what is the static equilibrium position 282 00:17:57,540 --> 00:18:00,560 and know what you're doing, but once you know it, 283 00:18:00,560 --> 00:18:02,050 then you have the answer. 284 00:18:02,050 --> 00:18:06,830 Now, the same thing is true of that problem. 285 00:18:06,830 --> 00:18:09,600 That's a non-homogeneous differential equation 286 00:18:09,600 --> 00:18:14,610 for the hanging mass. 287 00:18:14,610 --> 00:18:19,610 And we derive the equations of motion things 288 00:18:19,610 --> 00:18:22,780 for this many different ways this term, all right? 289 00:18:22,780 --> 00:18:25,250 But we usually said, zero-spring force. 290 00:18:25,250 --> 00:18:28,010 But now, if you started from here and said, 291 00:18:28,010 --> 00:18:29,940 this is the static equilibrium position, 292 00:18:29,940 --> 00:18:33,110 what's the motion about this position, 293 00:18:33,110 --> 00:18:40,636 then you'd get the equation with 0 on the right-hand side-- 294 00:18:40,636 --> 00:18:42,260 lots of advantages there to using that. 295 00:19:07,450 --> 00:19:10,620 All single degree of freedom oscillators 296 00:19:10,620 --> 00:19:14,120 will boil down to this equation. 297 00:19:14,120 --> 00:19:16,540 This is one involving translation, 298 00:19:16,540 --> 00:19:19,390 but for a simple pendulum. 299 00:19:19,390 --> 00:19:23,975 This object, for example, is a pendulum, but it's rotational. 300 00:19:23,975 --> 00:19:26,100 So it's a pendulum, but it's one degree of freedom. 301 00:19:28,650 --> 00:19:31,210 All pendulum problems, if you do them 302 00:19:31,210 --> 00:19:33,700 about equilibrium positions, boil 303 00:19:33,700 --> 00:19:39,130 down to some I with respect to the point 304 00:19:39,130 --> 00:19:44,580 that they're rocking about, theta double 305 00:19:44,580 --> 00:19:52,030 dot plus some Kt, torsional spring constant theta, 306 00:19:52,030 --> 00:19:52,990 equals 0. 307 00:19:52,990 --> 00:19:55,300 They take the same form. 308 00:19:55,300 --> 00:19:57,545 So all translational single degree 309 00:19:57,545 --> 00:20:00,020 of freedom systems, all rotational single degree 310 00:20:00,020 --> 00:20:03,070 of freedom systems, it's the same differential equation-- 311 00:20:03,070 --> 00:20:06,000 just this involves mass and linear acceleration. 312 00:20:06,000 --> 00:20:08,370 This involves mass moment of inertia and rotational 313 00:20:08,370 --> 00:20:10,270 acceleration. 314 00:20:10,270 --> 00:20:13,260 So everything that I say about the solution 315 00:20:13,260 --> 00:20:15,710 to single degree of freedom systems 316 00:20:15,710 --> 00:20:19,390 applies to both types of problems. 317 00:20:19,390 --> 00:20:27,900 So let's look into the solution of this equation briefly. 318 00:20:33,040 --> 00:20:35,545 Mostly, I'm doing this to establish some terminology. 319 00:20:38,960 --> 00:20:45,320 So a solution I know or I can show that xd of t, 320 00:20:45,320 --> 00:20:47,590 the solution to this problem-- notice, 321 00:20:47,590 --> 00:20:50,090 are there any external forces, by the way, excitations, 322 00:20:50,090 --> 00:20:51,320 f of t's or anything? 323 00:20:51,320 --> 00:20:52,350 No. 324 00:20:52,350 --> 00:20:55,170 So this thing has no external excitation 325 00:20:55,170 --> 00:20:56,620 that's going to make it move. 326 00:20:56,620 --> 00:21:02,640 So it's only source of vibration or motion is what? 327 00:21:02,640 --> 00:21:08,040 Comes from-- I hear initial conditions, right? 328 00:21:08,040 --> 00:21:10,670 You have to do something to perturb it 329 00:21:10,670 --> 00:21:13,980 and then it will vibrate. 330 00:21:13,980 --> 00:21:15,050 So here it is. 331 00:21:15,050 --> 00:21:18,960 It's about its equilibrium position. 332 00:21:18,960 --> 00:21:23,630 I give it an initial deflection and let go. 333 00:21:23,630 --> 00:21:26,560 Or it's around its initial condition 334 00:21:26,560 --> 00:21:29,540 and I give it an initial velocity. 335 00:21:29,540 --> 00:21:32,130 It also responds to some combination of the two. 336 00:21:32,130 --> 00:21:34,190 So initial conditions are the only things 337 00:21:34,190 --> 00:21:37,300 that account for motion of something 338 00:21:37,300 --> 00:21:39,500 without external excitation. 339 00:21:39,500 --> 00:21:42,430 And that motion, I can write that solution 340 00:21:42,430 --> 00:21:49,360 as A cosine omega t. 341 00:21:49,360 --> 00:21:53,070 You'll find this is a possible solution. 342 00:21:53,070 --> 00:21:59,000 B sine omega t is another possible solution. 343 00:21:59,000 --> 00:22:06,690 Sum A cosine omega t minus phase angle's also a solution. 344 00:22:06,690 --> 00:22:15,210 And sum A e to the i omega t you'll find is also a solution. 345 00:22:15,210 --> 00:22:17,150 Any of those things you could throw in 346 00:22:17,150 --> 00:22:20,350 and the precise values of these things, the A's, the B's, the 347 00:22:20,350 --> 00:22:23,965 phi's, and so forth depend on-- 348 00:22:23,965 --> 00:22:25,340 AUDIENCE: the initial conditions. 349 00:22:25,340 --> 00:22:26,740 J. KIM VANDIVER: The initial conditions. 350 00:22:26,740 --> 00:22:28,040 So let's do this one quickly. 351 00:22:37,950 --> 00:22:40,370 All right. 352 00:22:40,370 --> 00:22:54,600 And I'll choose And I'm going to stop writing the x sub d here. 353 00:22:54,600 --> 00:23:02,356 This is now my position from the equilibrium point. 354 00:23:02,356 --> 00:23:05,660 So x of t-- I'm going to say, let 355 00:23:05,660 --> 00:23:16,690 it be an A1 cosine omega t plus a B1 sine omega t 356 00:23:16,690 --> 00:23:18,460 and plug it in. 357 00:23:18,460 --> 00:23:28,670 When I plug it into the equation of motion, 358 00:23:28,670 --> 00:23:31,120 x double dot requires you to take two derivatives of each 359 00:23:31,120 --> 00:23:31,830 of these terms. 360 00:23:31,830 --> 00:23:34,940 Two derivatives of cosine gives you minus omega cosine. 361 00:23:34,940 --> 00:23:39,690 Two derivative sine minus omega squared cosine minus omega 362 00:23:39,690 --> 00:23:40,850 squared sine. 363 00:23:40,850 --> 00:23:46,390 So the answer comes out minus m omega 364 00:23:46,390 --> 00:23:56,300 squared plus k equivalent here times A1 cosine 365 00:23:56,300 --> 00:24:07,836 plus B1 sine-- omega t's obviously in them-- equals 0. 366 00:24:07,836 --> 00:24:09,710 So I just plugged in that equation of motion. 367 00:24:09,710 --> 00:24:11,790 I get this back. 368 00:24:11,790 --> 00:24:13,250 This is what I started with. 369 00:24:13,250 --> 00:24:15,670 That's x. 370 00:24:15,670 --> 00:24:22,990 In general, it is not equal to 0, can take on all 371 00:24:22,990 --> 00:24:23,730 sorts of values. 372 00:24:23,730 --> 00:24:27,350 So that's not generally 0 and that means this must be. 373 00:24:30,470 --> 00:24:32,600 And from this, then, when we solve this, 374 00:24:32,600 --> 00:24:37,907 we find that omega what we call n squared is k over m. 375 00:24:37,907 --> 00:24:40,490 And that's, of course, where our natural frequency comes from. 376 00:24:40,490 --> 00:24:54,349 This is called the undamped natural frequency, 377 00:24:54,349 --> 00:24:56,390 because there's no dampening in this problem yet. 378 00:24:56,390 --> 00:24:57,806 We get the square root of k over m 379 00:24:57,806 --> 00:25:01,650 is the natural frequency of the system. 380 00:25:01,650 --> 00:25:04,420 Let's find out what are A1 and B1. 381 00:25:04,420 --> 00:25:13,864 Well, let's let x0 be x at t equals 0 here. 382 00:25:16,850 --> 00:25:19,950 And if we just plug that in here, put t equals 0 here, 383 00:25:19,950 --> 00:25:21,270 cosine goes to 1. 384 00:25:21,270 --> 00:25:23,140 This term goes away. 385 00:25:23,140 --> 00:25:30,080 So this implies that A1 equals x0. 386 00:25:30,080 --> 00:25:33,970 So we find out right away that the A1 cosine omega 387 00:25:33,970 --> 00:25:38,100 t takes care of the response to an initial deflection. 388 00:25:38,100 --> 00:25:54,700 And we need a x dot here minus A1 omega sine omega t plus B1 389 00:25:54,700 --> 00:25:58,440 omega cosine omega t. 390 00:25:58,440 --> 00:26:02,230 That's the derivative of x. 391 00:26:02,230 --> 00:26:04,550 You know the solution's that, so its first derivative, 392 00:26:04,550 --> 00:26:06,640 the velocity, must look like this. 393 00:26:06,640 --> 00:26:13,450 And let's let v0 equals x dot at t equals 0. 394 00:26:13,450 --> 00:26:17,180 When we plug that in, this term goes away 395 00:26:17,180 --> 00:26:23,570 and we get B1 omega and cosine is 1. 396 00:26:23,570 --> 00:26:29,750 So therefore, B1 is v0 over omega. 397 00:26:29,750 --> 00:26:33,120 But in fact, the omega's omega n, 398 00:26:33,120 --> 00:26:38,740 because we already found that, that the only frequency that 399 00:26:38,740 --> 00:26:42,920 satisfies the equation of motion when 400 00:26:42,920 --> 00:26:46,530 you have only initial conditions in the system, 401 00:26:46,530 --> 00:26:49,360 the only frequency that is allowed in the answer 402 00:26:49,360 --> 00:26:52,460 is the natural frequency. 403 00:26:52,460 --> 00:27:02,630 So we now know B1 is v0 over omega n and A1 is x0. 404 00:27:02,630 --> 00:27:08,300 So if I give you any combination of initial displacement 405 00:27:08,300 --> 00:27:12,120 and initial velocity, you can write out for me the exact time 406 00:27:12,120 --> 00:27:14,510 history of the motion. 407 00:27:14,510 --> 00:27:19,370 X0 to cosine omega t plus v0 over omega n sine omega t 408 00:27:19,370 --> 00:27:23,740 is the complete solution for a response to initial conditions. 409 00:27:55,780 --> 00:28:05,920 So any translational oscillator one degree 410 00:28:05,920 --> 00:28:13,950 of freedom where you have a translational coordinate 411 00:28:13,950 --> 00:28:18,730 measured from its equilibrium position 412 00:28:18,730 --> 00:28:23,290 has the equation of motion-- actually, 413 00:28:23,290 --> 00:28:24,290 you've done this enough. 414 00:28:24,290 --> 00:28:27,620 But if we added a force here and we added some damping 415 00:28:27,620 --> 00:28:29,680 and I wanted the equation of motion of this, 416 00:28:29,680 --> 00:28:34,830 you know that it's m x double dot plus b x 417 00:28:34,830 --> 00:28:42,430 dot plus kx equals F of t. 418 00:28:42,430 --> 00:28:45,034 And so you're going to be confronted with problems-- 419 00:28:45,034 --> 00:28:46,700 find the equation of motion in a system. 420 00:28:46,700 --> 00:28:47,950 It comes up looking like that and they say, 421 00:28:47,950 --> 00:28:49,585 what's the natural frequency? 422 00:28:49,585 --> 00:28:50,850 And I've been a little sloppy. 423 00:28:50,850 --> 00:28:55,310 I really mean, what's the undamped natural frequency? 424 00:28:55,310 --> 00:28:56,890 And so to find the undamped-- when 425 00:28:56,890 --> 00:28:59,360 one says that, what's the undamped natural frequency, 426 00:28:59,360 --> 00:29:11,500 you just temporarily let b and F be 0, just temporarily, 427 00:29:11,500 --> 00:29:19,932 and solve then for omega n equals square root of k/n. 428 00:29:19,932 --> 00:29:21,180 It's what you do. 429 00:29:21,180 --> 00:29:26,890 And then, so we know this is a parameter that tells us 430 00:29:26,890 --> 00:29:29,700 about the behavior of the system, which we always 431 00:29:29,700 --> 00:29:32,320 want to know for the single degree of freedom systems. 432 00:29:32,320 --> 00:29:34,390 What is the natural frequency of the system? 433 00:29:42,712 --> 00:29:49,050 And we know for b equals 0 and F of 0, 434 00:29:49,050 --> 00:29:52,590 then the response can be only due to initial conditions. 435 00:29:52,590 --> 00:29:54,636 So we have x of t. 436 00:29:54,636 --> 00:29:59,770 We know it's going to be some x0 cosine omega n 437 00:29:59,770 --> 00:30:08,110 t plus v0 over omega n sine omega n t. 438 00:30:08,110 --> 00:30:11,510 And every simple vibration system in the world 439 00:30:11,510 --> 00:30:14,310 behaves basically like this from initial conditions. 440 00:30:14,310 --> 00:30:17,570 It'll be some part responding to the initial displacement, 441 00:30:17,570 --> 00:30:19,830 some part to the initial velocity. 442 00:30:19,830 --> 00:30:23,780 And damping is going to make it a little bit more complex, 443 00:30:23,780 --> 00:30:25,470 but not actually by much. 444 00:30:25,470 --> 00:30:29,180 The same basic terms appear even when you have damping in it. 445 00:30:33,940 --> 00:30:45,640 This can be expressed as sum A cosine omega, in this case, 446 00:30:45,640 --> 00:30:48,110 n t minus the phase angle. 447 00:30:48,110 --> 00:30:53,820 And it's useful to know this trigonometric identity 448 00:30:53,820 --> 00:30:55,980 to be able to put things together 449 00:30:55,980 --> 00:30:57,660 into an expression like that. 450 00:30:57,660 --> 00:31:07,100 And you'll find out that A is just the square root of the two 451 00:31:07,100 --> 00:31:07,630 pieces. 452 00:31:07,630 --> 00:31:09,270 It's a sine and cosine term. 453 00:31:09,270 --> 00:31:16,910 So you have an x0 squared plus a v0 over omega n squared 454 00:31:16,910 --> 00:31:18,970 square root. 455 00:31:18,970 --> 00:31:21,620 Remember, this is any A and B. It's just a square root 456 00:31:21,620 --> 00:31:23,620 of A squared plus B squared. 457 00:31:23,620 --> 00:31:25,100 That's what we're doing here. 458 00:31:25,100 --> 00:31:34,360 And the phase angle, the tangent inverse 459 00:31:34,360 --> 00:31:38,270 of this-- we've been calling this like an A and this 460 00:31:38,270 --> 00:31:39,680 is the B quantity. 461 00:31:39,680 --> 00:31:46,230 So tangent inverse of-- get my signs right-- 462 00:31:46,230 --> 00:31:49,480 B over A, which in this case then 463 00:31:49,480 --> 00:32:12,710 is tangent inverse of v0 over x0 omega n. 464 00:32:12,710 --> 00:32:15,310 That's all there is to it. 465 00:32:15,310 --> 00:32:20,030 And finally, another trig thing that you 466 00:32:20,030 --> 00:32:22,910 need to know-- we're going to use it quite a bit-- 467 00:32:22,910 --> 00:32:29,090 is that if you have an expression A cosine omega 468 00:32:29,090 --> 00:32:46,278 t minus phi, that's equal to the real part of A e to the i omega 469 00:32:46,278 --> 00:32:46,778 t. 470 00:32:50,200 --> 00:32:57,710 And if A is real and-- I don't want to write it that way-- 471 00:32:57,710 --> 00:33:09,550 when A is real, it's A times e to the i omega t minus phi, 472 00:33:09,550 --> 00:33:13,530 because Euler's formula says e to the i theta 473 00:33:13,530 --> 00:33:19,850 equals cosine theta plus i sine of theta. 474 00:33:19,850 --> 00:33:22,590 So if you have an i omega t minus phi 475 00:33:22,590 --> 00:33:26,070 here, you get back a cosine omega 476 00:33:26,070 --> 00:33:30,240 t minus phi and another term, an i sine omega t minus phi. 477 00:33:30,240 --> 00:33:36,130 So you can always express that as the real part of that. 478 00:33:36,130 --> 00:33:38,130 So we're going to need that little trig identity 479 00:33:38,130 --> 00:33:39,230 as we go through the term. 480 00:33:44,320 --> 00:33:54,925 Now, I've found in many years of teaching vibration 481 00:33:54,925 --> 00:33:58,800 that something that many students 482 00:33:58,800 --> 00:34:02,810 find a little confusing is this notion of phase angle. 483 00:34:02,810 --> 00:34:05,400 What does "phase angle" really mean? 484 00:34:05,400 --> 00:34:09,080 So I'll try to explain it to you in a couple different ways. 485 00:34:09,080 --> 00:34:15,159 So let's look at what this vibration 486 00:34:15,159 --> 00:34:18,070 that we're talking about here, x0 cosine omega 487 00:34:18,070 --> 00:34:24,179 t plus v0 over omega n sine-- what's it look like? 488 00:34:24,179 --> 00:34:32,530 So that's-- we've just got our-- and we see what it looks like. 489 00:34:32,530 --> 00:34:37,520 But if you plot the motion of this thing just versus time, 490 00:34:37,520 --> 00:34:42,899 what's it look like and where does phase angle come into it? 491 00:34:42,899 --> 00:34:49,560 So this is now x of t and this is 492 00:34:49,560 --> 00:34:58,894 t equals 0 and this undamped system 493 00:34:58,894 --> 00:35:00,860 is essentially going to look like that. 494 00:35:04,540 --> 00:35:07,730 And this is the value x0, the amplitude, 495 00:35:07,730 --> 00:35:11,900 the initial condition on x that you began with. 496 00:35:11,900 --> 00:35:21,420 And right here, the slope-- v0 is the slope, the initial slope 497 00:35:21,420 --> 00:35:27,140 of this curve, right, because the time derivative is F x dot. 498 00:35:27,140 --> 00:35:32,320 If we were plotting x dot, the initial velocity is omega x0. 499 00:35:32,320 --> 00:35:36,200 And so it's just the slope is v0 here. 500 00:35:36,200 --> 00:35:38,450 So this is your initial velocity. 501 00:35:38,450 --> 00:35:42,540 This is the-- and I didn't-- yeah, that's right. 502 00:35:50,850 --> 00:35:54,570 This is the initial displacement. 503 00:35:54,570 --> 00:35:56,290 The total written out mathematically, 504 00:35:56,290 --> 00:35:58,100 it looks like that. 505 00:35:58,100 --> 00:36:04,040 And I'm plotting this function, A cosine omega t minus phi. 506 00:36:04,040 --> 00:36:05,660 Yeah? 507 00:36:05,660 --> 00:36:06,510 Did I see a hand up? 508 00:36:06,510 --> 00:36:12,310 AUDIENCE: Does x0 at t equals 0 or is it a little bit after? 509 00:36:12,310 --> 00:36:14,050 J. KIM VANDIVER: Well, I was just 510 00:36:14,050 --> 00:36:16,174 looking at it myself and said, this can't be right. 511 00:36:18,160 --> 00:36:22,290 This has got to be the initial condition on x and this 512 00:36:22,290 --> 00:36:24,250 has to be the initial condition on v. 513 00:36:24,250 --> 00:36:27,380 Now, whatever this turns out to be is 514 00:36:27,380 --> 00:36:28,589 whatever it turns out to be. 515 00:36:28,589 --> 00:36:29,880 You have some initial velocity. 516 00:36:29,880 --> 00:36:31,550 You have some initial displacement. 517 00:36:31,550 --> 00:36:34,640 The system can actually peak out sometime later 518 00:36:34,640 --> 00:36:36,550 at a maximum value, right? 519 00:36:36,550 --> 00:36:41,550 And that maximum value is that. 520 00:36:41,550 --> 00:36:46,800 So this over here is the square root 521 00:36:46,800 --> 00:36:57,210 of x0 squared plus v0 over omega n squared square root. 522 00:36:57,210 --> 00:36:58,570 That's what the peak value is. 523 00:36:58,570 --> 00:37:01,680 And this system's undamped, so it just goes on forever. 524 00:37:01,680 --> 00:37:07,120 So the question is, though, what is this gap here 525 00:37:07,120 --> 00:37:11,400 between when it starts and when it meets its maximum? 526 00:37:11,400 --> 00:37:19,680 Well, when we use an expression like-- we 527 00:37:19,680 --> 00:37:24,400 said we can express this as some A cosine omega t minus phi. 528 00:37:24,400 --> 00:37:27,250 It's just the point at which the cosine then 529 00:37:27,250 --> 00:37:28,185 reaches its maximum. 530 00:37:32,290 --> 00:37:40,870 So if this axis here is omega t, if we plot this actually 531 00:37:40,870 --> 00:37:45,350 versus omega t, then one full cycle 532 00:37:45,350 --> 00:37:51,150 here is 2 pi or 360 degrees. 533 00:37:51,150 --> 00:37:55,470 So if you plot it versus omega t, then this gap in here 534 00:37:55,470 --> 00:37:58,510 is just phi. 535 00:37:58,510 --> 00:38:02,370 That's the delay in angle, if you will, 536 00:38:02,370 --> 00:38:04,680 that the system goes through between getting 537 00:38:04,680 --> 00:38:06,740 from the initial conditions to getting 538 00:38:06,740 --> 00:38:08,160 to the peak of the cosine. 539 00:38:14,460 --> 00:38:20,250 And phi must also then be equal to some omega 540 00:38:20,250 --> 00:38:28,080 n times a delta tau, I'll call it, some time delay. 541 00:38:28,080 --> 00:38:35,500 So if this is plotted-- if this axis is time-- 542 00:38:35,500 --> 00:38:50,580 not omega t, but time-- then x the same plot, this delay here, 543 00:38:50,580 --> 00:38:52,840 this is a time delay. 544 00:38:52,840 --> 00:38:55,060 And when you plot it against time, 545 00:38:55,060 --> 00:38:58,640 it's a delay in time to get to the peak. 546 00:38:58,640 --> 00:39:04,750 And omega n delta tau, this delay, 547 00:39:04,750 --> 00:39:07,390 must be equal to the phase angle. 548 00:39:07,390 --> 00:39:13,730 So the delta tau, this time delay, is phi over omega n. 549 00:39:16,560 --> 00:39:20,220 So you can think about this as a delay in time 550 00:39:20,220 --> 00:39:23,890 or as a shift in phase angle, depending on whether or not 551 00:39:23,890 --> 00:39:27,130 you want to plot this thing as a function of omega t 552 00:39:27,130 --> 00:39:29,340 or as a function of time. 553 00:39:29,340 --> 00:39:33,820 But you're going to need this concept of phase angle 554 00:39:33,820 --> 00:39:36,690 the rest of the term. 555 00:39:36,690 --> 00:39:39,035 Want to ask any questions about phase? 556 00:39:45,190 --> 00:39:51,170 Because we're doing vibration for the remainder of the term, 557 00:39:51,170 --> 00:39:55,510 this is an introduction to a topic called linear systems. 558 00:39:55,510 --> 00:39:58,960 And so this is basically the fundamental stuff in which you 559 00:39:58,960 --> 00:40:01,230 then, when you go on to 2004, which 560 00:40:01,230 --> 00:40:03,040 is controls and that sort of thing, 561 00:40:03,040 --> 00:40:06,900 this is the basic intro to it. 562 00:40:06,900 --> 00:40:11,350 And we'll talk more about linear system behavior as we go along. 563 00:40:20,600 --> 00:40:23,210 Now, we're going to do something that you've-- much of this 564 00:40:23,210 --> 00:40:24,920 stuff I know you've seen before. 565 00:40:24,920 --> 00:40:29,060 Some of the new parts is just vocabulary and ways 566 00:40:29,060 --> 00:40:31,020 of thinking about vibration that engineers 567 00:40:31,020 --> 00:40:33,880 do that mathematicians tend not to. 568 00:40:33,880 --> 00:40:36,896 So you have seen most of this stuff before where? 569 00:40:36,896 --> 00:40:37,651 AUDIENCE: 1.803. 570 00:40:37,651 --> 00:40:38,900 J. KIM VANDIVER: 1.803, right? 571 00:40:38,900 --> 00:40:40,340 You've done all this. 572 00:40:40,340 --> 00:40:49,210 And a year ago last May, in May, I taught the 1803 lecture 573 00:40:49,210 --> 00:40:51,120 with Professor Haynes Miller. 574 00:40:51,120 --> 00:40:52,600 Now, if you had 1.803 last spring, 575 00:40:52,600 --> 00:40:54,799 I think you had somebody different. 576 00:40:54,799 --> 00:40:56,090 But he invited me to come here. 577 00:40:56,090 --> 00:40:57,620 It was in the same classroom and we 578 00:40:57,620 --> 00:41:02,210 taught the second-order ordinary differential equation together. 579 00:41:02,210 --> 00:41:03,720 It was really a lot of fun. 580 00:41:03,720 --> 00:41:05,529 He said, well, here's what we do. 581 00:41:05,529 --> 00:41:07,070 And then, I said, oh, well, engineers 582 00:41:07,070 --> 00:41:08,890 look at it the following way. 583 00:41:08,890 --> 00:41:10,940 So what I'm going to show you is what 584 00:41:10,940 --> 00:41:12,270 he and I did in class that day. 585 00:41:12,270 --> 00:41:13,940 You can go back and watch that on video. 586 00:41:13,940 --> 00:41:15,150 It's kind of fun. 587 00:41:15,150 --> 00:41:17,080 But I'll give you my take on it today. 588 00:41:17,080 --> 00:41:18,980 So this is the engineer's view of what 589 00:41:18,980 --> 00:41:21,310 you've already seen in 1.803. 590 00:41:21,310 --> 00:41:25,100 So we have that system and we have that equation of motion. 591 00:41:25,100 --> 00:41:27,270 And the engineers and mathematicians 592 00:41:27,270 --> 00:41:31,840 would more or less agree to that m x double dot plus bx. 593 00:41:31,840 --> 00:41:34,120 But I went and looked at the web page last night. 594 00:41:34,120 --> 00:41:37,240 Last spring, the person used c instead of b. 595 00:41:37,240 --> 00:41:40,010 Haynes Miller the year before used b. 596 00:41:40,010 --> 00:41:45,730 So you can't depend on any absolute consistency. 597 00:41:45,730 --> 00:41:51,020 So let's start off with our homogeneous equation here. 598 00:41:51,020 --> 00:41:55,380 And I'm looking now for the response to initial conditions 599 00:41:55,380 --> 00:41:56,480 with damping. 600 00:41:56,480 --> 00:41:58,470 You've done this in 1.803. 601 00:41:58,470 --> 00:42:02,180 You know that you can solve this by assuming a solution 602 00:42:02,180 --> 00:42:05,310 of a form Ae to the st. 603 00:42:05,310 --> 00:42:10,060 Plugging it in gives you a quadratic equation 604 00:42:10,060 --> 00:42:18,170 that looks like s squared plus sb plus k equals 0. 605 00:42:18,170 --> 00:42:19,050 This has roots. 606 00:42:23,770 --> 00:42:27,700 I left out my m here, so it starts off looking like that. 607 00:42:27,700 --> 00:42:30,210 You divide through by the m. s squared 608 00:42:30,210 --> 00:42:36,875 plus b/m s plus k/m equals 0. 609 00:42:36,875 --> 00:42:38,500 And that's where Haynes would leave it. 610 00:42:38,500 --> 00:42:41,910 And he'd give you the entire answer in terms of b/m and k/m 611 00:42:41,910 --> 00:42:43,650 and that kind of thing. 612 00:42:43,650 --> 00:42:47,870 Engineers, we like to call that the natural frequency squared. 613 00:42:47,870 --> 00:42:50,920 And this term, we'd modify to put it 614 00:42:50,920 --> 00:42:56,060 in a terminology that is more convenient to engineering. 615 00:42:56,060 --> 00:42:59,140 So I'll show you how that works out. 616 00:42:59,140 --> 00:43:01,100 When you solve this quadratic just using 617 00:43:01,100 --> 00:43:04,385 the quadratic equation, you get the following. 618 00:43:15,167 --> 00:43:17,000 You get that the roots, there's two of them. 619 00:43:17,000 --> 00:43:20,060 I'll call them S1 and 2. 620 00:43:20,060 --> 00:43:21,900 The roots to this equation look like 621 00:43:21,900 --> 00:43:34,290 minus b over 2m plus or minus square root of b squared 622 00:43:34,290 --> 00:43:39,490 over 4m squared minus k/m. 623 00:43:42,480 --> 00:43:51,811 And that's what you'd get to do in 1.803. 624 00:43:51,811 --> 00:43:56,680 And an engineer would say, well, let's change that a little bit. 625 00:43:56,680 --> 00:44:03,210 So my roots that I would use for S1 and 2, 626 00:44:03,210 --> 00:44:07,120 I just factor out-- that's omega n squared. 627 00:44:07,120 --> 00:44:10,290 I can factor that out and it becomes omega n on the outside. 628 00:44:10,290 --> 00:44:13,000 And I put an omega n in the numerator and denominator here, 629 00:44:13,000 --> 00:44:14,310 as well. 630 00:44:14,310 --> 00:44:49,030 So I get roots that look like-- so I've just 631 00:44:49,030 --> 00:44:50,610 manipulated that a little bit. 632 00:44:53,834 --> 00:44:57,175 I have a name for this term. 633 00:44:57,175 --> 00:45:04,240 I use the Greek letter zeta is b over 2 omega n 634 00:45:04,240 --> 00:45:07,626 m is the way I remember it in my brain. 635 00:45:07,626 --> 00:45:08,875 It's called the damping ratio. 636 00:45:19,900 --> 00:45:24,320 And if I say that, then the roots, S1 and 2 for this, 637 00:45:24,320 --> 00:45:32,060 look like minus zeta omega n plus or minus 638 00:45:32,060 --> 00:45:38,788 omega n times the square root of zeta squared minus 1. 639 00:45:45,340 --> 00:45:48,810 And those are the roots that a vibration engineer 640 00:45:48,810 --> 00:45:55,160 would use to describe this second-order linear 641 00:45:55,160 --> 00:45:58,135 differential equation solution homogeneous solution. 642 00:46:00,720 --> 00:46:03,870 Those are the roots of the equation. 643 00:46:03,870 --> 00:46:07,970 And when you have no damping, then this term goes away 644 00:46:07,970 --> 00:46:12,010 and you're left with-- and I left an i out of here, I think. 645 00:46:15,300 --> 00:46:16,800 No, I'm fine. 646 00:46:16,800 --> 00:46:17,900 The i comes out of here. 647 00:46:33,730 --> 00:46:37,240 So for one thing to absolutely take away from today 648 00:46:37,240 --> 00:46:38,810 is to remember this. 649 00:46:41,800 --> 00:46:44,850 That's our definition of damping called the damping ratio. 650 00:46:44,850 --> 00:46:47,400 When that's 1, it's a number we call critical damping. 651 00:46:47,400 --> 00:46:50,060 I'll show you what that means in a second. 652 00:46:50,060 --> 00:46:55,100 And when it's greater than 1, the system won't vibrate. 653 00:46:55,100 --> 00:46:57,660 It just has exponential decay. 654 00:46:57,660 --> 00:46:59,630 If it's less than 1, you get vibration. 655 00:46:59,630 --> 00:47:02,320 And that's why we like to use it this way as it's meaningful. 656 00:47:02,320 --> 00:47:04,040 Its value, you instantly know if it's 657 00:47:04,040 --> 00:47:05,080 greater than or less than 1, it's 658 00:47:05,080 --> 00:47:06,830 going to change the behavior of the system 659 00:47:06,830 --> 00:47:10,190 from vibrating to not vibrating. 660 00:47:10,190 --> 00:47:14,580 So now, there's four possible solutions to this. 661 00:47:14,580 --> 00:47:20,690 I'm not going to elaborate on all of them, but zeta equals 0, 662 00:47:20,690 --> 00:47:21,795 we've already done. 663 00:47:21,795 --> 00:47:23,870 We know the answer to that. 664 00:47:23,870 --> 00:47:26,210 Response to initial conditions-- simple. 665 00:47:26,210 --> 00:47:27,840 We know that one. 666 00:47:27,840 --> 00:47:31,960 We have another solution when zeta's greater than 1. 667 00:47:31,960 --> 00:47:34,200 When zeta's greater than 1, this quantity here 668 00:47:34,200 --> 00:47:38,780 is the inside is greater than 1, so it's a real positive number. 669 00:47:38,780 --> 00:47:42,580 And both the roots of this thing are completely real. 670 00:47:46,840 --> 00:47:52,200 And you know that the-- remember the response, 671 00:47:52,200 --> 00:47:54,920 we hypothesize in the beginning that response 672 00:47:54,920 --> 00:47:58,210 looks like some Ae to the st. So now, we just plug back in. 673 00:47:58,210 --> 00:47:59,890 This is our st value. 674 00:47:59,890 --> 00:48:01,690 We can plug them back in and we will get 675 00:48:01,690 --> 00:48:04,880 the motion of the system back. 676 00:48:04,880 --> 00:48:11,420 So for zeta greater than 1, st comes out 677 00:48:11,420 --> 00:48:21,490 looking like minus zeta omega n t plus or minus 678 00:48:21,490 --> 00:48:29,040 square root of zeta squared minus 1 times t. 679 00:48:29,040 --> 00:48:32,590 And you just plug this in, and x is just e to the st. 680 00:48:32,590 --> 00:48:35,750 But these are just pure real values. 681 00:48:35,750 --> 00:48:40,930 And you'll find out that the system from initial conditions 682 00:48:40,930 --> 00:48:43,280 on velocity and displacement just-- 683 00:48:43,280 --> 00:48:44,039 [WINDS DOWN] 684 00:48:44,039 --> 00:48:44,580 and dies out. 685 00:48:48,010 --> 00:48:51,180 Zeta equals to 1. 686 00:48:51,180 --> 00:49:00,940 Then, st is just minus-- you get a double root-- minus omega nt, 687 00:49:00,940 --> 00:49:01,440 twice. 688 00:49:04,340 --> 00:49:07,710 And the solution for this, I can write out the whole thing. 689 00:49:07,710 --> 00:49:16,700 x of t here is just some A1 plus t A2 e 690 00:49:16,700 --> 00:49:21,200 to the minus zeta omega n t. 691 00:49:21,200 --> 00:49:24,320 And again, it looks-- it's just some kind 692 00:49:24,320 --> 00:49:28,740 of damp, not very interesting response, no oscillations. 693 00:49:28,740 --> 00:49:31,590 And then finally, zeta less than 1. 694 00:49:31,590 --> 00:49:34,216 And this is the only one-- this one produces oscillation. 695 00:49:37,670 --> 00:49:52,000 And the solution for st is plus or minus-- minus zeta omega n 696 00:49:52,000 --> 00:50:04,950 t, a real part, plus or minus i omega n t 697 00:50:04,950 --> 00:50:07,710 times the square root of 1 minus zeta squared. 698 00:50:07,710 --> 00:50:13,780 Now, I've turned around this zeta squared minus 1. 699 00:50:13,780 --> 00:50:15,220 This is now a negative number. 700 00:50:15,220 --> 00:50:18,270 A square root of a negative number gives me i. 701 00:50:18,270 --> 00:50:20,220 And now, I turn this around, so this is just 702 00:50:20,220 --> 00:50:22,240 a real positive number. 703 00:50:22,240 --> 00:50:24,510 So when you get i into this answer, 704 00:50:24,510 --> 00:50:29,274 what does it tell you that the solution looks like? 705 00:50:29,274 --> 00:50:30,440 AUDIENCE: Sines and cosines. 706 00:50:30,440 --> 00:50:32,460 J. KIM VANDIVER: Sines and cosines, right? 707 00:50:32,460 --> 00:50:36,490 So now, this gives you sines and cosines with a decay. 708 00:50:36,490 --> 00:50:39,215 This is an exponential to e to the minus zeta omega n 709 00:50:39,215 --> 00:50:42,420 t multiplied by a sine and a cosine. 710 00:50:42,420 --> 00:50:44,410 And so this is the interesting part. 711 00:50:44,410 --> 00:50:47,740 So most of the work of the rest of this term, 712 00:50:47,740 --> 00:50:50,585 we're only interested in this final solution. 713 00:50:53,340 --> 00:51:13,336 And what it looks like for this one-- so for zeta less than 1, 714 00:51:13,336 --> 00:51:20,820 x of t is some Ae to the minus zeta omega n 715 00:51:20,820 --> 00:51:32,310 t times a cosine omega d t-- make it d times 716 00:51:32,310 --> 00:51:36,770 t minus a phase angle-- come out looking like that. 717 00:51:36,770 --> 00:51:41,720 And if you draw it, depends on initial conditions, so again, 718 00:51:41,720 --> 00:51:44,290 a positive velocity and a positive displacement. 719 00:51:44,290 --> 00:51:46,240 It does this, but then it dies out. 720 00:51:49,120 --> 00:51:50,900 It's very similar to the undamped case, 721 00:51:50,900 --> 00:51:53,900 except that it has this damping that 722 00:51:53,900 --> 00:51:55,690 causes it to die out with time. 723 00:51:55,690 --> 00:51:59,160 But this right here, this is still the initial slope is v0 724 00:51:59,160 --> 00:52:01,040 and the initial displacement here is x0. 725 00:52:15,050 --> 00:52:19,750 And I'm now going to give you the exact expressions for this 726 00:52:19,750 --> 00:52:20,780 and we'll talk about it. 727 00:52:27,810 --> 00:52:30,410 Another way of writing this then in terms 728 00:52:30,410 --> 00:52:41,170 of the initial conditions is this looks like x0 cosine omega 729 00:52:41,170 --> 00:52:49,680 d t plus v0 over omega d. 730 00:53:46,110 --> 00:53:49,560 So expanding this out, this result clearly 731 00:53:49,560 --> 00:53:52,360 has to depend on the initial displacement 732 00:53:52,360 --> 00:53:54,470 and on the initial velocity. 733 00:53:54,470 --> 00:53:55,230 Now, what's this? 734 00:53:55,230 --> 00:53:57,470 I keep writing this omega d. 735 00:53:57,470 --> 00:53:59,530 So notice in here in the solution, 736 00:53:59,530 --> 00:54:03,370 it's omega n times the square root of 1 minus zeta squared. 737 00:54:03,370 --> 00:54:06,570 So the frequency that's in here isn't exactly omega n. 738 00:54:06,570 --> 00:54:10,470 It's omega n altered by a bit. 739 00:54:10,470 --> 00:54:23,250 Omega sub d is called the damped natural frequency. 740 00:54:28,050 --> 00:54:32,250 And it's equal to omega n times the square root 741 00:54:32,250 --> 00:54:33,995 of 1 minus theta squared. 742 00:54:36,880 --> 00:54:40,790 The system actually oscillates at a slightly different 743 00:54:40,790 --> 00:54:41,400 frequency. 744 00:54:41,400 --> 00:54:47,140 And for most systems that vibrate at all, 745 00:54:47,140 --> 00:54:49,260 this damping term is quite small. 746 00:54:49,260 --> 00:54:52,090 And when you square it, it gets even smaller. 747 00:54:52,090 --> 00:54:58,310 So this is usually a number that's 0.99, oftentimes, 748 00:54:58,310 --> 00:54:59,930 or even bigger than that. 749 00:54:59,930 --> 00:55:04,960 This is very close to 1 for all small amounts of damping. 750 00:55:04,960 --> 00:55:08,350 But being really careful about this 751 00:55:08,350 --> 00:55:11,190 in including it everywhere, that's 752 00:55:11,190 --> 00:55:13,610 what this result looks like. 753 00:55:13,610 --> 00:55:18,080 And this little thing, psi, this little phase angle here, 754 00:55:18,080 --> 00:55:30,860 is tangent inverse of theta over the square root of 1 755 00:55:30,860 --> 00:55:33,220 minus theta squared. 756 00:55:33,220 --> 00:55:36,550 And this number-- when damping is small, 757 00:55:36,550 --> 00:55:39,530 this is a very small number. 758 00:55:39,530 --> 00:55:43,160 And most of the time of problems that we deal with, 759 00:55:43,160 --> 00:55:44,510 the damping will be small. 760 00:55:44,510 --> 00:55:54,230 So let's say, for small damping-- 761 00:55:54,230 --> 00:55:58,640 and by that, I mean zeta, say, less than 10%, 762 00:55:58,640 --> 00:56:01,990 what we call 10%, 0.1. 763 00:56:01,990 --> 00:56:06,560 And if you have a little more-- you don't care too much 764 00:56:06,560 --> 00:56:09,480 about the precision, it might even be 20%. 765 00:56:09,480 --> 00:56:16,470 Actually, if it were 0.2, squared is 0.04, right? 766 00:56:16,470 --> 00:56:25,090 1 minus 0.04-- 0.96 square root, 0.98. 767 00:56:25,090 --> 00:56:27,890 So even with 20% damping, the difference 768 00:56:27,890 --> 00:56:30,370 between the undamped natural frequency 769 00:56:30,370 --> 00:56:31,972 and the damped natural frequency's 2%. 770 00:56:34,480 --> 00:56:39,740 So for most cases with any kind of small damping at all, 771 00:56:39,740 --> 00:56:42,870 we can write an approximation which is easier to remember. 772 00:56:42,870 --> 00:56:45,410 And it's all I carry around in my head. 773 00:56:45,410 --> 00:56:47,800 I can't remember this, quite frankly. 774 00:56:47,800 --> 00:56:51,290 Don't try to and I would instead express 775 00:56:51,290 --> 00:57:01,160 the answer to this as just x0 cosine omega 776 00:57:01,160 --> 00:57:15,530 d t plus v0 over omega d sine omega damped times time 777 00:57:15,530 --> 00:57:21,260 times e to the minus zeta omega n t. 778 00:57:21,260 --> 00:57:24,630 So why do I bother to carry the omega d's along 779 00:57:24,630 --> 00:57:29,590 if I just said that they're almost exactly the same. 780 00:57:29,590 --> 00:57:33,582 For light damping, then omega n's approximately omega d. 781 00:57:33,582 --> 00:57:36,760 Well, you need to keep this one in here 782 00:57:36,760 --> 00:57:43,840 because even though it's only 2% difference at 20% damping, 783 00:57:43,840 --> 00:57:50,530 if you say the solution is omega n when it's really omega d, 784 00:57:50,530 --> 00:57:56,040 this thing will accumulate a phase error over time. 785 00:57:56,040 --> 00:58:00,380 So it's gets bigger and bigger, this error here, 786 00:58:00,380 --> 00:58:02,720 because you haven't taken care of that little 2%. 787 00:58:02,720 --> 00:58:06,550 That 2% can bite you after you go through enough cycles. 788 00:58:06,550 --> 00:58:10,092 So I keep omega d in the expression here. 789 00:58:10,092 --> 00:58:14,300 But other than that, it's almost exactly the same expression 790 00:58:14,300 --> 00:58:17,280 that we just came up to for the simple response 791 00:58:17,280 --> 00:58:20,100 of an undamped system to initial conditions, 792 00:58:20,100 --> 00:58:25,740 x0 cosine plus v0 over omega n sine. 793 00:58:25,740 --> 00:58:30,000 And now, all we've added to it is put the transient decay 794 00:58:30,000 --> 00:58:33,770 and the fact that it decays into the expression 795 00:58:33,770 --> 00:58:38,030 and changed the frequency it oscillates at to omega 796 00:58:38,030 --> 00:58:39,490 d instead of omega n. 797 00:59:13,770 --> 00:59:17,280 So I'm going to try to impress something on you. 798 00:59:17,280 --> 00:59:33,750 If I took this pendulum and my stopwatch, 799 00:59:33,750 --> 00:59:39,690 measured the natural frequency of this thing, 800 00:59:39,690 --> 00:59:43,080 I could get a very accurate value if I do it carefully. 801 00:59:43,080 --> 00:59:47,010 Then, I take the same object and I dunk it in water 802 00:59:47,010 --> 00:59:49,404 and it goes back and forth. 803 00:59:49,404 --> 00:59:51,070 And it conspicuously goes back and forth 804 00:59:51,070 --> 00:59:52,944 but dies down now after a while, because it's 805 00:59:52,944 --> 00:59:54,890 got that water damping it. 806 00:59:54,890 --> 01:00:00,180 But I measure that frequency and it's 807 01:00:00,180 --> 01:00:05,120 10% different, 20% different. 808 01:00:05,120 --> 01:00:09,670 And I have seen people make this mistake dozens of times. 809 01:00:09,670 --> 01:00:12,150 You say, that's the experiment. 810 01:00:12,150 --> 01:00:14,396 Explain why. 811 01:00:14,396 --> 01:00:21,250 What's the reason that that measured frequency has changed? 812 01:00:21,250 --> 01:00:25,469 Got any ocean engineers in the audience? 813 01:00:25,469 --> 01:00:27,380 All right. 814 01:00:27,380 --> 01:00:30,799 So why does-- if you put the pendulum in water-- 815 01:00:30,799 --> 01:00:32,090 and it's still oscillating now. 816 01:00:32,090 --> 01:00:34,330 So it isn't so damp that it's-- 817 01:00:34,330 --> 01:00:35,029 [BLOWS] 818 01:00:35,029 --> 01:00:36,070 So it's got some damping. 819 01:00:36,070 --> 01:00:38,050 It's dying out and the natural frequency's 820 01:00:38,050 --> 01:00:39,820 changed by 15% or 20%. 821 01:00:39,820 --> 01:00:42,040 What's the explanation? 822 01:00:42,040 --> 01:00:49,020 And the answer you always get from people is, damping. 823 01:00:49,020 --> 01:00:49,520 Why? 824 01:00:49,520 --> 01:00:55,665 Because everybody's been taught this thing, right? 825 01:00:55,665 --> 01:00:58,040 And they all then assume that the change in the frequency 826 01:00:58,040 --> 01:00:58,914 is caused by damping. 827 01:00:58,914 --> 01:01:01,500 But damping couldn't possibly be the reason, 828 01:01:01,500 --> 01:01:03,720 because with 20% damping, this thing'll die out 829 01:01:03,720 --> 01:01:06,750 in about two swings and it's done. 830 01:01:06,750 --> 01:01:08,890 That's a lot of damping, actually, 831 01:01:08,890 --> 01:01:11,730 but it only accounts for 2% change in natural frequency, 832 01:01:11,730 --> 01:01:14,160 not 15%. 833 01:01:14,160 --> 01:01:14,710 Hmmm. 834 01:01:14,710 --> 01:01:16,571 So what causes the change in the frequency? 835 01:01:20,892 --> 01:01:22,350 AUDIENCE: Buoyancy of the pendulum? 836 01:01:22,350 --> 01:01:23,766 J. KIM VANDIVER: No, not buoyancy. 837 01:01:27,220 --> 01:01:30,000 That could actually have an effect. 838 01:01:30,000 --> 01:01:33,957 That's actually-- I should say, yes, you're partly right. 839 01:01:33,957 --> 01:01:34,915 There's another reason. 840 01:01:39,190 --> 01:01:43,230 When the thing is swinging back and forth there in the water, 841 01:01:43,230 --> 01:01:46,110 it actually carries some water with it. 842 01:01:46,110 --> 01:01:48,740 Effectively, the kinetic energy-- you 843 01:01:48,740 --> 01:01:51,546 now know how to do vibration problems. 844 01:01:51,546 --> 01:01:53,170 Find the equations of motion accounting 845 01:01:53,170 --> 01:01:56,420 for the potential energy and the kinetic energy. 846 01:01:56,420 --> 01:01:59,250 The kinetic energy changes, because some water 847 01:01:59,250 --> 01:02:03,010 moves with the object and it's called added mass. 848 01:02:03,010 --> 01:02:05,450 It literally-- there is water moving with the object that 849 01:02:05,450 --> 01:02:07,670 has kinetic energy associated with the motion 850 01:02:07,670 --> 01:02:10,270 and it acts like it's more massive. 851 01:02:10,270 --> 01:02:11,940 It is dynamically more massive. 852 01:02:11,940 --> 01:02:15,400 There's water moving with it. 853 01:02:15,400 --> 01:02:17,820 So trying to impress on you that damping 854 01:02:17,820 --> 01:02:21,920 doesn't cause much of a change in systems 855 01:02:21,920 --> 01:02:22,940 that actually vibrate. 856 01:02:22,940 --> 01:02:24,430 Really observe the vibration. 857 01:02:24,430 --> 01:02:26,200 If you can observe the vibration, 858 01:02:26,200 --> 01:02:29,290 damping cannot possibly account for a very large shift 859 01:02:29,290 --> 01:02:30,830 in frequency. 860 01:02:30,830 --> 01:02:34,252 What's the motion look like? 861 01:02:34,252 --> 01:02:36,130 Let's move on a little bit here. 862 01:02:44,110 --> 01:02:46,160 So that's what this solution looks like. 863 01:02:46,160 --> 01:02:48,870 We know it depends on initial conditions. 864 01:02:48,870 --> 01:02:57,020 The distance from here to here will make this a time axis. 865 01:02:57,020 --> 01:02:59,110 This is one period. 866 01:02:59,110 --> 01:03:00,640 So this is tau d. 867 01:03:00,640 --> 01:03:03,810 That's the damped period of vibration. 868 01:03:07,060 --> 01:03:20,820 And we know that x of t is some Ae to the minus theta omega n t 869 01:03:20,820 --> 01:03:26,740 cosine omega d t minus a phase angle. 870 01:03:26,740 --> 01:03:29,370 We can write that expression like this. 871 01:03:31,880 --> 01:03:39,230 And this term, this is just a cosine. 872 01:03:39,230 --> 01:03:44,780 This term repeats every period, right? 873 01:03:47,560 --> 01:03:52,210 If it's at maximum value here, exactly one period later, 874 01:03:52,210 --> 01:03:54,090 it's again at its maximum. 875 01:03:54,090 --> 01:04:00,020 So the cosine term goes to 1 every 2 pi 876 01:04:00,020 --> 01:04:03,530 or every period of motion, right? 877 01:04:03,530 --> 01:04:11,440 So I want to take-- I'm going to define this as the value 878 01:04:11,440 --> 01:04:16,110 at x at some time t. 879 01:04:16,110 --> 01:04:18,970 I'll call it t0. 880 01:04:18,970 --> 01:04:32,220 And out here is x at t0 plus n tau d, n periods later. 881 01:04:32,220 --> 01:04:34,280 So this is the period, defined as period. 882 01:04:36,960 --> 01:04:44,900 Remember, omega d is the same thing 883 01:04:44,900 --> 01:04:51,090 as 2 pi times the frequency in hertz. 884 01:04:51,090 --> 01:05:01,980 And frequency is 1 over period, 2 pi over the period. 885 01:05:01,980 --> 01:05:04,170 So remember, there's a relationship 886 01:05:04,170 --> 01:05:08,180 that you need to remember now that relates radian 887 01:05:08,180 --> 01:05:12,450 frequency to frequency in cycles per second in hertz 888 01:05:12,450 --> 01:05:15,815 to frequency expressed in period. 889 01:05:15,815 --> 01:05:17,030 All right? 890 01:05:17,030 --> 01:05:21,170 This would be tau d here and this would be an f d. 891 01:05:21,170 --> 01:05:23,910 For any frequency, you can say that. 892 01:05:23,910 --> 01:05:27,652 At omega is 2 pi f is 2 pi over tau. 893 01:05:27,652 --> 01:05:29,110 So you've got to be good with that. 894 01:05:29,110 --> 01:05:35,190 But now, so here we are, two peaks separated by n periods. 895 01:05:35,190 --> 01:05:44,050 And I want to take the ratio of x of t to x of t plus n tau d 896 01:05:44,050 --> 01:05:45,680 here. 897 01:05:45,680 --> 01:05:49,600 And that's just going to be then my-- 898 01:05:49,600 --> 01:05:54,260 when I take that ratio, x of t has cosine omega d t 899 01:05:54,260 --> 01:05:56,150 minus phi in it. 900 01:05:56,150 --> 01:05:59,810 And n periods later, exactly the same thing appears, right? 901 01:05:59,810 --> 01:06:02,300 So the cosine term just cancels out. 902 01:06:02,300 --> 01:06:05,390 This just is e-- and the A's cancel out. 903 01:06:05,390 --> 01:06:07,135 That's the initial conditions. 904 01:06:07,135 --> 01:06:11,660 It's e to the minus zeta omega n t-- 905 01:06:11,660 --> 01:06:20,480 and I guess I called it t0-- over e 906 01:06:20,480 --> 01:06:31,225 to the minus zeta omega n t0 plus n damped periods. 907 01:06:33,950 --> 01:06:36,500 And if I bring this into the numerator, 908 01:06:36,500 --> 01:06:38,540 the exponent becomes positive. 909 01:06:38,540 --> 01:06:45,630 The t0 terms, minus zeta omega and t0 plus, those cancel. 910 01:06:45,630 --> 01:06:53,530 And this expression is just e to the plus zeta omega 911 01:06:53,530 --> 01:06:56,844 n times n td. 912 01:07:02,640 --> 01:07:06,390 And the last step that I want to do to this, what I'm coming up 913 01:07:06,390 --> 01:07:10,570 with is a way of estimating-- purposely 914 01:07:10,570 --> 01:07:20,520 doing this-- is this transient curve we know is controlled 915 01:07:20,520 --> 01:07:22,850 by a damping, by zeta. 916 01:07:22,850 --> 01:07:28,340 I want to have an experimental way to determine what is zeta. 917 01:07:28,340 --> 01:07:31,530 And I do it by computing something called 918 01:07:31,530 --> 01:07:33,300 the logarithmic decrement. 919 01:07:33,300 --> 01:07:41,060 So if I take the natural log of x of t over x of t 920 01:07:41,060 --> 01:07:50,270 plus n periods, it's the natural log of this expression. 921 01:07:50,270 --> 01:07:53,580 So I just get the exponent back. 922 01:07:53,580 --> 01:08:07,526 This then is n zeta omega-- I guess 923 01:08:07,526 --> 01:08:13,530 I better to do it carefully-- omega n n tau d. 924 01:08:13,530 --> 01:08:18,439 The tau d is 2 pi over omega and I get some nice things 925 01:08:18,439 --> 01:08:19,640 to cancel out here. 926 01:08:29,630 --> 01:08:38,689 So this natural log over the ratio-- this is n zeta omega n 927 01:08:38,689 --> 01:08:44,040 and this is 2 pi over omega d, which 928 01:08:44,040 --> 01:08:49,649 is omega n times the square root of 1 minus zeta squared. 929 01:08:49,649 --> 01:08:52,260 Omega n's go away. 930 01:08:52,260 --> 01:09:09,149 And for zeta small, this term's approximately 1, in which case 931 01:09:09,149 --> 01:09:19,120 this then becomes n 2 pi zeta. 932 01:09:19,120 --> 01:09:27,200 And zeta equals 1 over 2 pi n natural log 933 01:09:27,200 --> 01:09:36,729 of this ratio of x of t over x of t plus nt. 934 01:09:44,060 --> 01:09:50,310 So experimentally, if you just go in and measure your-- 935 01:09:50,310 --> 01:09:55,960 if you plot out the response, you measure a peak value, 936 01:09:55,960 --> 01:09:58,990 you measure the peak value n periods later, 937 01:09:58,990 --> 01:10:03,950 compute the log of that ratio, divide by 1 over 2 938 01:10:03,950 --> 01:10:06,570 pi n, the number of periods, you have 939 01:10:06,570 --> 01:10:08,840 an estimate of the natural frequency-- 940 01:10:08,840 --> 01:10:10,752 estimate of the damping ratio, excuse me. 941 01:10:15,380 --> 01:10:20,000 And to give you one quick little rule of thumb here, 942 01:10:20,000 --> 01:10:23,790 so this is an experimental way that very quickly, you 943 01:10:23,790 --> 01:10:29,360 can estimate the damping of a pendulum or whatever 944 01:10:29,360 --> 01:10:31,380 by just doing a quick measurement. 945 01:11:05,070 --> 01:11:08,550 So if it happens that after n periods, 946 01:11:08,550 --> 01:11:12,610 this value is half of the initial value, 947 01:11:12,610 --> 01:11:15,960 then this ratio is 2, right? 948 01:11:15,960 --> 01:11:18,570 So x of t-- some n periods later, this is only half 949 01:11:18,570 --> 01:11:19,930 as big. 950 01:11:19,930 --> 01:11:20,950 This value's 2. 951 01:11:20,950 --> 01:11:24,910 The natural log of 2 is some number you can calculate. 952 01:11:24,910 --> 01:11:27,030 So there's a little rule. 953 01:11:27,030 --> 01:11:29,700 If you just work that out, you find 954 01:11:29,700 --> 01:11:44,300 that zeta equals 1 over 2 pi n 50% times the natural log of 2. 955 01:11:44,300 --> 01:11:49,150 And you end up here was 0-- let me do this 956 01:11:49,150 --> 01:11:54,960 carefully-- 1 over 2 pi, n 50%, natural log of 2. 957 01:11:54,960 --> 01:12:03,450 And that is 0.11 over n 50%. 958 01:12:03,450 --> 01:12:07,130 That's a really handy little engineer tool 959 01:12:07,130 --> 01:12:09,140 to carry around in your head. 960 01:12:09,140 --> 01:12:15,190 So if I have an oscillator, this little end here, 961 01:12:15,190 --> 01:12:17,920 I can do an experiment. 962 01:12:17,920 --> 01:12:22,020 Give it initial deflection and it starts off 963 01:12:22,020 --> 01:12:24,940 at six inches or three inches amplitude. 964 01:12:24,940 --> 01:12:27,040 And you let it oscillate until you see it die down 965 01:12:27,040 --> 01:12:28,800 to half of that value. 966 01:12:28,800 --> 01:12:33,795 So let's say, one, two, about four cycles 967 01:12:33,795 --> 01:12:38,710 this thing decays by about 50%. 968 01:12:38,710 --> 01:12:42,580 Four cycles-- plug 4 into that formula. 969 01:12:42,580 --> 01:12:47,240 You get about 0.025. 970 01:12:47,240 --> 01:12:48,762 Agree? 971 01:12:48,762 --> 01:12:52,490 2 and a 1/2% damping. 972 01:12:52,490 --> 01:12:55,720 Really very convenient little thing 973 01:12:55,720 --> 01:12:58,260 to carry around with you-- measure pendulum, 974 01:12:58,260 --> 01:12:59,510 how much damping does it have? 975 01:12:59,510 --> 01:13:00,960 And now, this is what I'm saying. 976 01:13:00,960 --> 01:13:03,850 Most things that have any substantial amount 977 01:13:03,850 --> 01:13:08,078 of vibration, the damping is going to be way less than 10%. 978 01:13:11,330 --> 01:13:15,950 If it dies, if it takes one cycle for the amplitude 979 01:13:15,950 --> 01:13:20,530 to decrease, one cycle for the amplitude to decrease by 50%, 980 01:13:20,530 --> 01:13:22,222 how much damping does it have? 981 01:13:22,222 --> 01:13:23,550 AUDIENCE: 11%. 982 01:13:23,550 --> 01:13:25,930 J. KIM VANDIVER: 11%. 983 01:13:25,930 --> 01:13:28,320 So 11% damping is a lot of damping. 984 01:13:28,320 --> 01:13:32,189 The thing starts out here and the next cycle, it's half gone, 985 01:13:32,189 --> 01:13:34,230 and the next cycle after that, it's half of that. 986 01:13:34,230 --> 01:13:37,520 And so in about three cycles, it's gone. 987 01:13:37,520 --> 01:13:40,060 So if you see anything that's vibrating any length of time 988 01:13:40,060 --> 01:13:43,990 at all, its damping is way less than 10% 989 01:13:43,990 --> 01:13:50,520 and this notion of small damping is a perfectly good one. 990 01:13:50,520 --> 01:13:52,730 And I'll close by just saying one other thing. 991 01:13:52,730 --> 01:13:56,705 If something vibrates a lot, the damping's small. 992 01:14:01,030 --> 01:14:03,160 You need small damping for things 993 01:14:03,160 --> 01:14:04,890 to actually vibrate very much. 994 01:14:04,890 --> 01:14:06,770 This thing, this is vibrating-- 995 01:14:06,770 --> 01:14:07,630 [HIGH TONE] 996 01:14:07,630 --> 01:14:12,320 that high-pitched one, that's about a kilohertz. 997 01:14:12,320 --> 01:14:14,730 How many cycles do you think it's gone through 998 01:14:14,730 --> 01:14:19,270 to get down to 50% of that initial amplitude 999 01:14:19,270 --> 01:14:21,062 that you could hear? 1000 01:14:21,062 --> 01:14:23,730 A few thousand? 1001 01:14:23,730 --> 01:14:27,900 How much damping do you think this rod has? 1002 01:14:27,900 --> 01:14:32,080 Really tiny, really tiny. 1003 01:14:32,080 --> 01:14:32,640 All right. 1004 01:14:32,640 --> 01:14:34,290 So even though all we talked about today was 1005 01:14:34,290 --> 01:14:35,831 single degree of freedom oscillators, 1006 01:14:35,831 --> 01:14:39,840 I hope you learned a few things that we'll carry now 1007 01:14:39,840 --> 01:14:41,230 through the rest of the term. 1008 01:14:41,230 --> 01:14:44,750 We'll use all these concepts that we did today 1009 01:14:44,750 --> 01:14:47,150 to talk about more complicated vibration. 1010 01:14:47,150 --> 01:14:51,820 Good luck on your 2.001 quiz. 1011 01:14:51,820 --> 01:14:55,500 See you on Tuesday.