1 00:00:00,070 --> 00:00:01,770 The following content is provided 2 00:00:01,770 --> 00:00:04,010 under a Creative Commons license. 3 00:00:04,010 --> 00:00:06,860 Your support will help MIT OpenCourseWare continue 4 00:00:06,860 --> 00:00:10,720 to offer high quality educational resources for free. 5 00:00:10,720 --> 00:00:13,330 To make a donation or view additional materials 6 00:00:13,330 --> 00:00:17,209 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,209 --> 00:00:17,834 at ocw.mit.edu. 8 00:00:20,830 --> 00:00:22,090 PROFESSOR: Well, we're back. 9 00:00:22,090 --> 00:00:25,830 And today we will consider solutions 10 00:00:25,830 --> 00:00:31,950 to harmonic oscillators with damped harmonic oscillators. 11 00:00:31,950 --> 00:00:35,420 And what I thought would be best, in order 12 00:00:35,420 --> 00:00:41,570 to focus on the new part, the question of damping-- 13 00:00:41,570 --> 00:00:43,620 what effect it has on the motion of harmonic 14 00:00:43,620 --> 00:00:47,830 oscillators-- I decided I'll take an oscillator, which 15 00:00:47,830 --> 00:00:50,590 is one of those we considered last time, 16 00:00:50,590 --> 00:00:53,210 I could've taken any one of them, 17 00:00:53,210 --> 00:00:58,180 and add to it a drag force. 18 00:00:58,180 --> 00:01:04,200 So let me discuss what we're going to consider. 19 00:01:07,150 --> 00:01:12,810 Suppose we have a rod, the uniform rod of mass m 20 00:01:12,810 --> 00:01:15,510 suspended from the pivot here. 21 00:01:15,510 --> 00:01:17,780 This is the vertical plane. 22 00:01:17,780 --> 00:01:22,140 So imagine a ruler, for example, with a nail 23 00:01:22,140 --> 00:01:24,880 through the end of it can oscillate 24 00:01:24,880 --> 00:01:27,070 in the plane of this board. 25 00:01:27,070 --> 00:01:33,680 So we at a gain considering rotations in two dimensions. 26 00:01:33,680 --> 00:01:36,750 We will consider rotations, which 27 00:01:36,750 --> 00:01:41,960 are anti-clockwise, like this, to have a positive sign. 28 00:01:41,960 --> 00:01:46,520 And if it's clockwise, that's a negative rotation. 29 00:01:46,520 --> 00:01:51,310 This particular mass rod we assume has a mass m. 30 00:01:51,310 --> 00:01:53,820 It's uniform length l. 31 00:01:53,820 --> 00:01:57,280 And initially, it's just suspended like this. 32 00:01:57,280 --> 00:01:59,145 So at time equals 0. 33 00:02:02,520 --> 00:02:05,700 This is the position, and at that instant, 34 00:02:05,700 --> 00:02:07,650 we hit the bottom. 35 00:02:07,650 --> 00:02:09,830 There's an impulse acting on it. 36 00:02:09,830 --> 00:02:15,180 And that impulse instantaneously gives an angular velocity 37 00:02:15,180 --> 00:02:19,570 to this rod, which is of course the rate of change 38 00:02:19,570 --> 00:02:21,540 of the angle. 39 00:02:21,540 --> 00:02:25,660 And we assume it has some number. 40 00:02:25,660 --> 00:02:29,930 And so that it's easy for you to remember what that number is, 41 00:02:29,930 --> 00:02:35,600 I'm calling it the angular velocity at t equals 0. 42 00:02:35,600 --> 00:02:40,470 But remember, this is simply the initial angular velocity 43 00:02:40,470 --> 00:02:42,330 that this rod has. 44 00:02:42,330 --> 00:02:46,300 At any instant of time later, this rod 45 00:02:46,300 --> 00:02:48,430 will be at some other angle. 46 00:02:48,430 --> 00:02:51,000 I've shown here, theta of t. 47 00:02:51,000 --> 00:02:52,890 This is at time t. 48 00:02:52,890 --> 00:02:55,780 And it may be moving at that time. 49 00:02:55,780 --> 00:03:01,330 And the angular velocity in this direction we call theta dot t. 50 00:03:01,330 --> 00:03:07,330 These are anti-clockwise rotations. 51 00:03:07,330 --> 00:03:09,740 And so these are positive. 52 00:03:09,740 --> 00:03:11,250 This is as that angle. 53 00:03:11,250 --> 00:03:13,480 And if this is a positive number, 54 00:03:13,480 --> 00:03:17,490 this is rotating in that direction. 55 00:03:17,490 --> 00:03:20,910 So this is the same simple harmonic oscillator 56 00:03:20,910 --> 00:03:22,670 we considered last time. 57 00:03:22,670 --> 00:03:26,230 And what I now want to take a slightly more 58 00:03:26,230 --> 00:03:29,160 realistic situation. 59 00:03:29,160 --> 00:03:32,190 All harmonic oscillators in the universe 60 00:03:32,190 --> 00:03:34,770 have some drag, maybe very small, 61 00:03:34,770 --> 00:03:39,630 some mechanism by which they lose kinetic energy. 62 00:03:39,630 --> 00:03:46,460 So we were going to assume that this oscillator has 63 00:03:46,460 --> 00:03:49,540 some drag torque applied to it. 64 00:03:49,540 --> 00:03:52,946 It could be friction around the pivot, it could be the air, 65 00:03:52,946 --> 00:03:57,230 this could be inside the liquid, you name it. 66 00:03:57,230 --> 00:04:01,690 Now, I am going to make the assumption 67 00:04:01,690 --> 00:04:08,780 that the presence of the damping mechanism 68 00:04:08,780 --> 00:04:15,750 exerts a drag torque on this rod. 69 00:04:15,750 --> 00:04:19,690 And I'm going to make the assumption that the drag 70 00:04:19,690 --> 00:04:26,080 force is proportional to the angular velocity. 71 00:04:26,080 --> 00:04:29,670 Since the drag force is in the opposite direction 72 00:04:29,670 --> 00:04:31,980 to the angular velocity obviously, 73 00:04:31,980 --> 00:04:33,890 there is a minus sign here and this 74 00:04:33,890 --> 00:04:36,840 is the constant of proportionality. 75 00:04:36,840 --> 00:04:41,490 Now, you may well ask why did I assume this is the case? 76 00:04:41,490 --> 00:04:43,830 Is that because in nature it's normally like that? 77 00:04:43,830 --> 00:04:44,870 No. 78 00:04:44,870 --> 00:04:48,860 The reason why I chose this form is very simple. 79 00:04:48,860 --> 00:04:55,880 I know from experience that if I use this drag force, 80 00:04:55,880 --> 00:04:58,590 I can analytically solve my equations. 81 00:04:58,590 --> 00:05:03,510 If I take a more realistic version of the force, 82 00:05:03,510 --> 00:05:06,170 then the final solutions I'm going to get it 83 00:05:06,170 --> 00:05:08,579 are not analytically soluble. 84 00:05:08,579 --> 00:05:10,120 It doesn't mean you can't solve them. 85 00:05:10,120 --> 00:05:11,810 You can solve them by a computer. 86 00:05:11,810 --> 00:05:14,950 But for teaching you how to solve problems, 87 00:05:14,950 --> 00:05:16,900 it's not going to be very helpful if I then 88 00:05:16,900 --> 00:05:19,000 say go into a computer and get the computer 89 00:05:19,000 --> 00:05:20,270 to solve it for you. 90 00:05:20,270 --> 00:05:24,600 So the choice of this resistive force 91 00:05:24,600 --> 00:05:31,100 which is proportional to the rate of change of position 92 00:05:31,100 --> 00:05:34,550 of the rod is chosen for that reason. 93 00:05:34,550 --> 00:05:39,820 The other thing I will do, once again-- because of what you 94 00:05:39,820 --> 00:05:43,040 saw last time that happened-- assume 95 00:05:43,040 --> 00:05:46,920 that the displacement of this rod is small. 96 00:05:46,920 --> 00:05:51,600 We're only going to consider small oscillations. 97 00:05:51,600 --> 00:05:58,210 In other words, we'll only assume 98 00:05:58,210 --> 00:06:02,340 it's sufficiently small so that we can approximate sine 99 00:06:02,340 --> 00:06:04,800 of the angle by the angle itself. 100 00:06:04,800 --> 00:06:08,850 Again, the reason why we do that is so that at the end, 101 00:06:08,850 --> 00:06:14,840 we'll be able to solve the equations of motion. 102 00:06:14,840 --> 00:06:18,690 Now, if you ask yourself, hold on. 103 00:06:18,690 --> 00:06:21,090 We're doing all these approximations. 104 00:06:21,090 --> 00:06:23,020 Does this correspond to reality? 105 00:06:23,020 --> 00:06:25,030 And the answer is the following. 106 00:06:27,920 --> 00:06:32,070 In problems that we are doing on the board like this, 107 00:06:32,070 --> 00:06:35,760 we're taking idealized situations. 108 00:06:35,760 --> 00:06:39,590 For those idealized situations, the prediction 109 00:06:39,590 --> 00:06:43,480 of what will happen is exact. 110 00:06:43,480 --> 00:06:49,090 But since the description of the situations 111 00:06:49,090 --> 00:06:52,300 is it not exactly equal to what you'd normally 112 00:06:52,300 --> 00:06:56,320 have in the world, the answers that we get 113 00:06:56,320 --> 00:07:01,970 are only as accurate to the degree of the assumptions. 114 00:07:01,970 --> 00:07:05,830 But certainly from solutions like this, 115 00:07:05,830 --> 00:07:09,080 we can get a sense in understanding 116 00:07:09,080 --> 00:07:13,670 what happens, because for example, the drag 117 00:07:13,670 --> 00:07:17,170 force slightly different will not qualitatively 118 00:07:17,170 --> 00:07:20,840 change what the actual motion is, et cetera. 119 00:07:20,840 --> 00:07:24,530 And this happens throughout so-called problem solving 120 00:07:24,530 --> 00:07:30,830 land in any physics that you consider. 121 00:07:30,830 --> 00:07:34,290 So this is the physical description of the situation. 122 00:07:34,290 --> 00:07:37,290 And what we want to do, as before, 123 00:07:37,290 --> 00:07:43,350 find out what this rod will do qualitatively 124 00:07:43,350 --> 00:07:45,700 as a function of time and also be 125 00:07:45,700 --> 00:07:48,840 able to predict exactly where it will 126 00:07:48,840 --> 00:07:51,930 be at the given instant of time. 127 00:07:51,930 --> 00:07:56,760 So that's what I mean by I'm going to solve this problem. 128 00:07:56,760 --> 00:08:01,310 Now, I chose the same oscillator as I 129 00:08:01,310 --> 00:08:07,040 say so that you don't have to focus so much on the rising 130 00:08:07,040 --> 00:08:14,430 equation of motion, but more on what the frictional torque 131 00:08:14,430 --> 00:08:20,450 does, how that changes what we learned before. 132 00:08:20,450 --> 00:08:26,240 So here is the free body diagrams or the force 133 00:08:26,240 --> 00:08:31,260 diagram, which corresponds to this. 134 00:08:31,260 --> 00:08:32,710 Here is the pivot. 135 00:08:32,710 --> 00:08:35,289 This represents the rod. 136 00:08:35,289 --> 00:08:37,950 What forces act on this rod? 137 00:08:37,950 --> 00:08:41,280 Well, there's going to be the force of gravity. 138 00:08:41,280 --> 00:08:45,410 And you know from the studies of rigid body motion 139 00:08:45,410 --> 00:08:52,330 that we can understand what torque this force exerts 140 00:08:52,330 --> 00:08:55,000 by considering that the force of gravity 141 00:08:55,000 --> 00:08:58,110 acts at the center of mass of this rod. 142 00:08:58,110 --> 00:09:00,050 So if you take the center of mass, 143 00:09:00,050 --> 00:09:03,400 this location, then there'll be a force 144 00:09:03,400 --> 00:09:06,060 of gravity on this mass. 145 00:09:06,060 --> 00:09:07,810 The mass of the rod will be m, so there'll 146 00:09:07,810 --> 00:09:11,530 be an mg force acting downwards there. 147 00:09:11,530 --> 00:09:19,810 This force applied to this rigid body will exert a torque. 148 00:09:19,810 --> 00:09:22,990 In this case, it'll be a clockwise and negative torque. 149 00:09:25,950 --> 00:09:29,390 In addition to that, there will be 150 00:09:29,390 --> 00:09:35,370 a torque acting on this due to the frictional force. 151 00:09:35,370 --> 00:09:44,760 And that, we said, we will call b theta dot. 152 00:09:44,760 --> 00:09:51,400 So what is the equation of motion for this rod? 153 00:09:51,400 --> 00:09:56,070 The acceleration of the angular acceleration 154 00:09:56,070 --> 00:10:05,360 of this rod from tau is equal to i alpha. 155 00:10:05,360 --> 00:10:09,590 The acceleration will be equal to the torque divided 156 00:10:09,590 --> 00:10:12,310 by the moment of inertia. 157 00:10:12,310 --> 00:10:17,480 Since this is rotations of a rigid body in a plane in two 158 00:10:17,480 --> 00:10:22,820 dimensions, I don't have to write any vector quantities. 159 00:10:25,540 --> 00:10:28,950 On writing the equations, this is the component 160 00:10:28,950 --> 00:10:34,760 of the rotations about an axis perpendicular to the board 161 00:10:34,760 --> 00:10:37,230 here. 162 00:10:37,230 --> 00:10:40,360 So the angular acceleration will be 163 00:10:40,360 --> 00:10:43,170 equal to the total torque, which is the torque due 164 00:10:43,170 --> 00:10:47,580 to gravity plus the torque due to the drag divided 165 00:10:47,580 --> 00:10:49,270 by the moment of inertia. 166 00:10:49,270 --> 00:10:51,590 And by the way, the moment of inertia 167 00:10:51,590 --> 00:10:58,840 of a rod like this about one end is 1/3 ml squared. 168 00:10:58,840 --> 00:11:06,010 So this is the torque due to gravity on this. 169 00:11:06,010 --> 00:11:08,670 It's minus because, as I mentioned here, 170 00:11:08,670 --> 00:11:14,390 it's a clockwise torque minus b theta dot. 171 00:11:14,390 --> 00:11:18,230 That's also minus that, because the drag force is always 172 00:11:18,230 --> 00:11:26,010 in opposition to the angular velocity divided by i. 173 00:11:26,010 --> 00:11:31,380 As we did last time for the harmonic oscillator, 174 00:11:31,380 --> 00:11:37,400 in order for me to write fewer numbers or algebraic symbols 175 00:11:37,400 --> 00:11:40,840 on the board, I define some quantities. 176 00:11:40,840 --> 00:11:47,990 I will define omega 0 squared to be equal to 3 g over l. 177 00:11:47,990 --> 00:11:53,010 I'll define gamma to be b over I-- 178 00:11:53,010 --> 00:11:55,820 that constant proportionality divided 179 00:11:55,820 --> 00:11:59,070 by the inertia of this system, the moment of inertia. 180 00:11:59,070 --> 00:12:04,180 And I will, as I said, assume that this angle's always 181 00:12:04,180 --> 00:12:09,930 sufficiently small for the accuracy with which one 182 00:12:09,930 --> 00:12:12,490 I want to predict what will happen, 183 00:12:12,490 --> 00:12:18,690 I can assume that sine delta t is equal to theta t. 184 00:12:18,690 --> 00:12:20,960 Now let me be honest with you. 185 00:12:20,960 --> 00:12:24,310 I defined these quantities not only 186 00:12:24,310 --> 00:12:27,840 to make my life easier in writing things down, 187 00:12:27,840 --> 00:12:33,670 but as you'll see as we go on, it will serve a useful purpose 188 00:12:33,670 --> 00:12:38,210 for comparing how all sorts of harmonic oscillators 189 00:12:38,210 --> 00:12:42,290 behave with or without damping, et cetera. 190 00:12:42,290 --> 00:12:45,480 If I take these definitions and make this assumption 191 00:12:45,480 --> 00:12:50,730 and I take this and play around with the algebra, 192 00:12:50,730 --> 00:12:53,490 I'll end up with this equation. 193 00:12:53,490 --> 00:12:56,530 I end up with an equation that theta double 194 00:12:56,530 --> 00:12:59,570 dot, the second derivative of theta and the angular 195 00:12:59,570 --> 00:13:04,600 acceleration times gamma times the angular velocity plus omega 196 00:13:04,600 --> 00:13:09,840 0 squared times the angular position is equal to 0. 197 00:13:09,840 --> 00:13:13,700 This is the equation of motion for this system. 198 00:13:13,700 --> 00:13:15,900 And the other thing which we know 199 00:13:15,900 --> 00:13:21,670 that initially the theta at 0 is 0 200 00:13:21,670 --> 00:13:25,780 and the initial angular velocity is 201 00:13:25,780 --> 00:13:30,280 this number, angular velocity at t equals 0. 202 00:13:33,390 --> 00:13:41,200 These three equations are absolutely, 203 00:13:41,200 --> 00:13:43,340 from the point of view of physics, 204 00:13:43,340 --> 00:13:48,160 equivalent to the description we had. 205 00:13:48,160 --> 00:13:51,310 Here we had this physical description 206 00:13:51,310 --> 00:13:55,465 of somebody's-- this rod, et cetera. 207 00:14:00,480 --> 00:14:05,803 These mathematical equations describe this situation. 208 00:14:08,840 --> 00:14:15,179 In order to predict what will happen, what we now have to do 209 00:14:15,179 --> 00:14:16,220 is solve these equations. 210 00:14:19,000 --> 00:14:22,090 So that's what I will do now. 211 00:14:22,090 --> 00:14:24,865 So I will solve this equation. 212 00:14:31,020 --> 00:14:36,390 And let me remind you, as I discussed in the past, 213 00:14:36,390 --> 00:14:40,450 I cannot emphasize it enough now. 214 00:14:40,450 --> 00:14:44,580 I am not trying to teach you mathematics. 215 00:14:44,580 --> 00:14:51,805 I am trying to help you learn how to solve physical systems, 216 00:14:51,805 --> 00:14:57,340 how to predict how a given physical system will 217 00:14:57,340 --> 00:14:59,620 behave as a function of time. 218 00:14:59,620 --> 00:15:05,280 So for me, mathematics is a tool. 219 00:15:05,280 --> 00:15:08,020 I need a solution of that equation. 220 00:15:08,020 --> 00:15:11,860 And this is a linear second order differential 221 00:15:11,860 --> 00:15:14,350 equation in time. 222 00:15:14,350 --> 00:15:31,410 And I know from mathematics that if I find an equation which 223 00:15:31,410 --> 00:15:38,320 solves that and if it has 2 arbitrary constants-- 224 00:15:38,320 --> 00:15:42,840 in other words, the solution satisfies 225 00:15:42,840 --> 00:15:46,260 that equation for all values of those arbitrary constants. 226 00:15:46,260 --> 00:15:56,300 Then I have found the one and only solution of that equation. 227 00:15:56,300 --> 00:15:57,970 And so let me tell you. 228 00:16:01,080 --> 00:16:08,760 I will solve this by using this uniqueness theorem. 229 00:16:11,390 --> 00:16:18,830 I know that that equation, unfortunately, 230 00:16:18,830 --> 00:16:22,170 is a little harder to solve than the one we 231 00:16:22,170 --> 00:16:25,650 did last time for simple harmonic motion. 232 00:16:25,650 --> 00:16:28,850 One finds that the solution of that equation 233 00:16:28,850 --> 00:16:39,700 actually depends on these constants. 234 00:16:39,700 --> 00:16:47,950 For the harmonic oscillator, the type of equation, 235 00:16:47,950 --> 00:16:51,620 which was the solution, did not depend on those constants. 236 00:16:51,620 --> 00:16:54,450 Here, it does. 237 00:16:54,450 --> 00:17:01,520 One finds that if you have gamma squared greater 238 00:17:01,520 --> 00:17:08,010 than omega 0 squared-- in other words, 239 00:17:08,010 --> 00:17:12,780 if you have this big constant bigger than a certain amount, 240 00:17:12,780 --> 00:17:16,050 this means you have heavy damping. 241 00:17:16,050 --> 00:17:20,550 You have one kind of a solution, while if it's that other way 242 00:17:20,550 --> 00:17:24,849 about, if this is bigger than that, 243 00:17:24,849 --> 00:17:28,560 you get a different kind of a solution. 244 00:17:28,560 --> 00:17:33,150 It's not obvious, but that's what mathematics tells you. 245 00:17:33,150 --> 00:17:38,050 So let me consider first the case of strong damping. 246 00:17:38,050 --> 00:17:40,500 B is big. 247 00:17:40,500 --> 00:17:42,600 And so we're talking about situation, 248 00:17:42,600 --> 00:17:46,370 you have this oscillator. 249 00:17:46,370 --> 00:17:48,470 But there's lots of friction, so it 250 00:17:48,470 --> 00:17:52,680 doesn't move smoothly at all. 251 00:17:52,680 --> 00:17:54,680 This is greater than that. 252 00:17:54,680 --> 00:17:57,790 Under those conditions, you'll find 253 00:17:57,790 --> 00:18:08,350 that this equation satisfies our equation of motion. 254 00:18:08,350 --> 00:18:16,200 Now actually, Professor Walter Lewin, in his lectures, 255 00:18:16,200 --> 00:18:28,410 did show you how you can prove that this equation satisfies 256 00:18:28,410 --> 00:18:34,370 our equation of motion for these specific conditions. 257 00:18:34,370 --> 00:18:37,720 He used complex amplitude. 258 00:18:37,720 --> 00:18:41,600 And you can review his lectures and you can see that. 259 00:18:41,600 --> 00:18:47,080 Or you can look it up in books on the solution of differential 260 00:18:47,080 --> 00:18:49,300 equations. 261 00:18:49,300 --> 00:18:51,990 For me, it is adequate that I found 262 00:18:51,990 --> 00:19:00,290 a solution which satisfies this equation. 263 00:19:00,290 --> 00:19:07,570 Now, note this is a solution of that equation 264 00:19:07,570 --> 00:19:14,570 where the alpha 1 and alpha 2 are defined here. 265 00:19:14,570 --> 00:19:17,710 It's all defined in known quantities. 266 00:19:17,710 --> 00:19:21,950 Gamma omega 0 are known quantities 267 00:19:21,950 --> 00:19:24,910 for any specific problem in particular for the one 268 00:19:24,910 --> 00:19:26,810 we're solving. 269 00:19:26,810 --> 00:19:29,940 So here, every term is known. 270 00:19:29,940 --> 00:19:34,210 The A and the B are the 2 arbitrary constants. 271 00:19:34,210 --> 00:19:38,650 In other words, this equation satisfies our equation 272 00:19:38,650 --> 00:19:42,470 of motion for all values of A and B. 273 00:19:42,470 --> 00:19:45,110 So again, by the uniqueness theorem, 274 00:19:45,110 --> 00:19:49,090 this is the most general solution to our problem 275 00:19:49,090 --> 00:19:54,470 provided the damping is very high. 276 00:19:54,470 --> 00:19:59,140 If this is the theta of t, then of course 277 00:19:59,140 --> 00:20:04,090 the angular velocity theta dot of t, just differentiate here. 278 00:20:04,090 --> 00:20:07,940 And I get minus alpha 1 A e to the minus alpha 1 t 279 00:20:07,940 --> 00:20:11,940 and minus alpha 2 B to the minus alpha t. 280 00:20:16,580 --> 00:20:22,580 We want to find the solution to our particular problem. 281 00:20:22,580 --> 00:20:26,870 So we found it that satisfies the equation 282 00:20:26,870 --> 00:20:28,680 of motion for our system. 283 00:20:28,680 --> 00:20:33,990 We now want to make sure that we have a solution which 284 00:20:33,990 --> 00:20:36,120 satisfies the boundary conditions. 285 00:20:36,120 --> 00:20:39,460 In other words, which is consistent with what 286 00:20:39,460 --> 00:20:43,480 we said this rod was doing at t equals 0. 287 00:20:43,480 --> 00:20:48,930 At t equals 0, we said the rod was vertically, so theta of 0 288 00:20:48,930 --> 00:20:50,160 is 0. 289 00:20:50,160 --> 00:20:52,760 And we also said that at that instance, 290 00:20:52,760 --> 00:20:54,780 there was an impulse given to the rod. 291 00:20:54,780 --> 00:21:00,740 And so its angular velocity at t equals 0 was this number, 292 00:21:00,740 --> 00:21:03,990 the angular velocity at t equal 0. 293 00:21:03,990 --> 00:21:12,210 So we must now make sure that the A and B, these arbitrary 294 00:21:12,210 --> 00:21:18,250 constants, are consistent with these initial conditions. 295 00:21:18,250 --> 00:21:20,290 And it's obvious how you do that. 296 00:21:20,290 --> 00:21:29,120 You take this equation and I take what is theta at t 297 00:21:29,120 --> 00:21:29,830 equals 0. 298 00:21:29,830 --> 00:21:30,520 Well, it's 0. 299 00:21:30,520 --> 00:21:33,450 So that's 0 equals to t equals 0. 300 00:21:33,450 --> 00:21:38,160 This is one, so it's A, plus when t equals 0, this is B. 301 00:21:38,160 --> 00:21:42,720 So it's A plus B. So we know that A must be equal to minus 302 00:21:42,720 --> 00:21:47,220 B. That's to satisfy the boundary 303 00:21:47,220 --> 00:21:51,040 conditions that the rod was vertical at t equals 0. 304 00:21:51,040 --> 00:21:52,880 How about the other boundary condition? 305 00:21:52,880 --> 00:21:58,130 We know that t equals 0, the angular velocity theta dot 306 00:21:58,130 --> 00:22:02,730 is angular velocity at t equals 0. 307 00:22:02,730 --> 00:22:10,000 And so angular at t equals 0 must be equal to minus alpha 1 308 00:22:10,000 --> 00:22:13,770 times A. And t equals 0, this is 1. 309 00:22:13,770 --> 00:22:20,470 So it's minus alpha 1 A. And here, minus alpha 2 B. OK, 310 00:22:20,470 --> 00:22:29,000 we now have 2 equations with 2 knowns, 2 algebraic equations. 311 00:22:29,000 --> 00:22:31,620 Therefore, we can find A and B. In fact, 312 00:22:31,620 --> 00:22:36,590 we found already that A is equal to minus B from this equation. 313 00:22:36,590 --> 00:22:39,580 And then replacing B by minus A, we 314 00:22:39,580 --> 00:22:50,140 find that A is angular velocity t equals 0 and alpha t over 1. 315 00:22:50,140 --> 00:22:52,320 So what do we find? 316 00:22:52,320 --> 00:23:00,070 We have found that the specific solution to our equation 317 00:23:00,070 --> 00:23:03,780 of motion which satisfies these boundary conditions 318 00:23:03,780 --> 00:23:16,834 is theta of t is equal A e to the minus alpha 1 t. 319 00:23:16,834 --> 00:23:33,400 A is angular velocity at t equals 0 into e 320 00:23:33,400 --> 00:23:38,870 to the minus alpha 1 t. 321 00:23:38,870 --> 00:23:45,447 And B is minus A. So it's minus e to the minus alpha 2t. 322 00:23:50,950 --> 00:24:05,840 So we have solved that problem. 323 00:24:05,840 --> 00:24:11,010 This equation describes the motion 324 00:24:11,010 --> 00:24:22,150 of this compound pendulum, the rod, completely. 325 00:24:22,150 --> 00:24:26,850 If you ask me for a time, I will tell you 326 00:24:26,850 --> 00:24:32,540 exactly the angle at which this will be hanging. 327 00:24:32,540 --> 00:24:36,200 There are no more unknowns in this. 328 00:24:36,200 --> 00:24:44,050 Furthermore, it gives me a sense of what that rod is doing. 329 00:24:44,050 --> 00:24:47,770 And actually, it's a bit of a surprise. 330 00:24:47,770 --> 00:24:51,270 We've considered here an oscillator, 331 00:24:51,270 --> 00:24:54,320 we'd expect an oscillator to oscillate. 332 00:24:54,320 --> 00:24:56,780 This equation doesn't oscillate. 333 00:24:56,780 --> 00:25:00,210 It is the sum of 2 exponential functions. 334 00:25:02,910 --> 00:25:14,060 If I just qualitatively plot this, 335 00:25:14,060 --> 00:25:17,030 if you compare alpha 2 to alpha 1, 336 00:25:17,030 --> 00:25:20,720 you find that alpha 2 is greater than alpha 1. 337 00:25:23,510 --> 00:25:30,380 So this exponential has a bigger exponent. 338 00:25:30,380 --> 00:25:32,960 It will be dropping faster than this one. 339 00:25:32,960 --> 00:25:34,010 So what do we have? 340 00:25:38,090 --> 00:25:41,200 This is my angular velocity. 341 00:25:45,050 --> 00:25:49,395 Velocity at t equals 0. 342 00:25:52,330 --> 00:25:53,630 Let's consider one of them. 343 00:25:53,630 --> 00:25:57,100 This is 1. 344 00:25:57,100 --> 00:25:58,540 It's an exponential function. 345 00:25:58,540 --> 00:26:03,910 Here, I'm plotting the angle as a function of time. 346 00:26:03,910 --> 00:26:12,070 I'll get some exponential function like this, 347 00:26:12,070 --> 00:26:14,130 starting at that value. 348 00:26:14,130 --> 00:26:18,350 And the other function's also an exponential one, 349 00:26:18,350 --> 00:26:21,770 which has a higher value of the exponent. 350 00:26:21,770 --> 00:26:24,810 And therefore, it'll drop more quickly, 351 00:26:24,810 --> 00:26:27,860 starts with the same amplitude, so it'll 352 00:26:27,860 --> 00:26:30,300 go more quickly like this. 353 00:26:35,360 --> 00:26:42,640 The net motion is the difference between those 2. 354 00:26:42,640 --> 00:26:46,620 Take this one minus that one. 355 00:26:46,620 --> 00:26:49,040 What does that look like? 356 00:26:49,040 --> 00:26:52,425 Well, qualitative you can see I start at 0. 357 00:26:56,650 --> 00:27:00,620 Far out here, it'll be 0, because the 2 exponentials 358 00:27:00,620 --> 00:27:01,880 will approach 0. 359 00:27:01,880 --> 00:27:06,430 So out here, and it'll be 0 or approaching 0. 360 00:27:06,430 --> 00:27:09,210 It'll be some magnitude in here. 361 00:27:09,210 --> 00:27:16,880 And so I will get something like this happening. 362 00:27:19,985 --> 00:27:22,110 It's the difference of the 2 exponential functions. 363 00:27:26,320 --> 00:27:28,860 Isn't it amazing? 364 00:27:28,860 --> 00:27:32,230 Let's think for a second what we have shown. 365 00:27:32,230 --> 00:27:37,920 We've taken this harmonic oscillator, 366 00:27:37,920 --> 00:27:43,850 we've put it in a situation with a lot of friction, 367 00:27:43,850 --> 00:27:53,880 we gave it a kick, and we described it 368 00:27:53,880 --> 00:27:58,090 in terms of these mathematical equations. 369 00:27:58,090 --> 00:28:02,250 I then blindly applied mathematics, 370 00:28:02,250 --> 00:28:06,342 solved these equations, and predicted 371 00:28:06,342 --> 00:28:09,040 that this will be the motion. 372 00:28:09,040 --> 00:28:11,790 No oscillations. 373 00:28:11,790 --> 00:28:13,580 Let's stop now for a second. 374 00:28:13,580 --> 00:28:17,780 By common sense, what would you expect? 375 00:28:17,780 --> 00:28:19,940 Imagine you had this rod and something which 376 00:28:19,940 --> 00:28:26,600 is very, very viscous-- very high drag. 377 00:28:26,600 --> 00:28:30,345 You give it a kick, it will move, because of inertia it 378 00:28:30,345 --> 00:28:30,960 will move. 379 00:28:30,960 --> 00:28:32,930 But it will get slowed down. 380 00:28:32,930 --> 00:28:35,610 It will finally come to a halt. 381 00:28:35,610 --> 00:28:39,890 And it'll slowly come back. 382 00:28:39,890 --> 00:28:47,060 But if the resistive torque is so big, it'll never overshoot. 383 00:28:47,060 --> 00:28:49,440 So this is exactly what you'd expect. 384 00:28:52,220 --> 00:28:57,920 And as I say, it is amazing. 385 00:28:57,920 --> 00:29:00,240 I always find it amazing, because this 386 00:29:00,240 --> 00:29:06,620 is a magnificent example that mathematics 387 00:29:06,620 --> 00:29:12,190 is the language that one can use to describe the nature, 388 00:29:12,190 --> 00:29:17,140 and that solving the mathematical equations 389 00:29:17,140 --> 00:29:21,110 predicts what will actually happen in a given situation. 390 00:29:25,920 --> 00:29:28,090 Enough philosophizing. 391 00:29:28,090 --> 00:29:31,880 Let's now consider the other possibility. 392 00:29:31,880 --> 00:29:37,770 So I'll now do a second problem where 393 00:29:37,770 --> 00:29:40,850 I will have very weak damping. 394 00:29:40,850 --> 00:29:43,810 So imagine a more realistic situation. 395 00:29:43,810 --> 00:29:47,490 We take our pendulum, this compound pendulum, 396 00:29:47,490 --> 00:29:54,420 and we include the resistive torque 397 00:29:54,420 --> 00:29:58,880 due to the air, for example, that's there, et cetera. 398 00:29:58,880 --> 00:30:03,070 Again, I am taking the same pendulum 399 00:30:03,070 --> 00:30:09,400 so that we just focus on the effect of the damping. 400 00:30:09,400 --> 00:30:14,273 So I think I have it here. 401 00:30:22,970 --> 00:30:27,550 Everything up to here will be exactly the same. 402 00:30:27,550 --> 00:30:29,980 The only thing we've now changed, 403 00:30:29,980 --> 00:30:39,720 we've said that this little b is big such that the gamma squared 404 00:30:39,720 --> 00:30:45,030 is smaller, it doesn't have to be much smaller, 405 00:30:45,030 --> 00:30:48,250 so that gamma squared is less than 4 omega squared. 406 00:30:48,250 --> 00:30:51,840 In other words, B is less than this. 407 00:30:51,840 --> 00:30:56,090 So it's the weak damping case. 408 00:30:56,090 --> 00:30:59,670 You may well ask, hold on. 409 00:30:59,670 --> 00:31:02,450 Why look for any other solutions? 410 00:31:02,450 --> 00:31:08,700 I have found a solution for this equation before. 411 00:31:08,700 --> 00:31:11,850 And up there was the two exponentials. 412 00:31:11,850 --> 00:31:15,150 Why can't we just take those and say look, 413 00:31:15,150 --> 00:31:22,970 that equation satisfies this equation, that value of theta, 414 00:31:22,970 --> 00:31:25,460 and just repeat this? 415 00:31:25,460 --> 00:31:28,650 Well, try it and you'll see you immediately get 416 00:31:28,650 --> 00:31:29,980 into difficulties. 417 00:31:29,980 --> 00:31:35,800 In this formulation, you see that alpha 1 and alpha 2 418 00:31:35,800 --> 00:31:41,390 include a square root of gamma squared minus 4 omega squared. 419 00:31:41,390 --> 00:31:44,580 We had no problem with that when gamma squared was 420 00:31:44,580 --> 00:31:49,970 greater than 4 omega squared and take the square root of it. 421 00:31:49,970 --> 00:31:54,580 But if gamma squared is less than 4 omega squared, 422 00:31:54,580 --> 00:32:00,810 that square root is the square root of a minus number. 423 00:32:00,810 --> 00:32:06,390 And how is that going to represent a physical situation? 424 00:32:06,390 --> 00:32:10,170 So the method breaks down. 425 00:32:10,170 --> 00:32:13,760 In other words, for this case, we 426 00:32:13,760 --> 00:32:19,640 had not found an equation which satisfies 427 00:32:19,640 --> 00:32:24,320 our equation of motion with these boundary conditions. 428 00:32:24,320 --> 00:32:26,790 So we've got to try again. 429 00:32:26,790 --> 00:32:29,519 And there is an equation that does. 430 00:32:29,519 --> 00:32:30,560 And I've written it here. 431 00:32:33,200 --> 00:32:40,030 One finds that theta of t is equal 432 00:32:40,030 --> 00:32:44,500 to some arbitrary constant A e to the minus 433 00:32:44,500 --> 00:32:50,510 gamma over 2t times cosine, omega prime t plus phi 434 00:32:50,510 --> 00:32:54,632 where omega prime is well defined in terms of omega 0 435 00:32:54,632 --> 00:32:57,480 and gamma is given here. 436 00:32:57,480 --> 00:33:00,630 And there's no problem. 437 00:33:00,630 --> 00:33:03,860 We have now gamma squared is less than that, 438 00:33:03,860 --> 00:33:05,200 so this is a real number. 439 00:33:05,200 --> 00:33:10,250 So it's no problem with that square root plus phi. 440 00:33:10,250 --> 00:33:14,650 So we have 2 arbitrary constants. 441 00:33:14,650 --> 00:33:15,840 Try it. 442 00:33:15,840 --> 00:33:23,020 Take this equation, differentiate it twice, 443 00:33:23,020 --> 00:33:24,450 and calculate it. 444 00:33:24,450 --> 00:33:28,700 Then differentiate it once and multiply it by gamma. 445 00:33:28,700 --> 00:33:29,610 Don't differentiate. 446 00:33:29,610 --> 00:33:34,220 Take that equation and multiply the omega 0 squared. 447 00:33:34,220 --> 00:33:37,780 Add those 3 terms, you'll get 0. 448 00:33:37,780 --> 00:33:39,460 I tried it last night to make sure I 449 00:33:39,460 --> 00:33:42,330 didn't write the wrong thing here. 450 00:33:42,330 --> 00:33:44,730 So that really works. 451 00:33:47,810 --> 00:33:51,890 Again, if you want to know how did I find this equation, 452 00:33:51,890 --> 00:33:54,630 did I take it out of a hat, like a rabbit out of a hat? 453 00:33:54,630 --> 00:33:55,650 No. 454 00:33:55,650 --> 00:33:58,660 You can look it up in books, solutions of equations, 455 00:33:58,660 --> 00:34:01,840 or actually Professor Walter Lewin 456 00:34:01,840 --> 00:34:07,160 did derive this solution also using complex amplitudes. 457 00:34:07,160 --> 00:34:08,909 And you can find it in his lectures. 458 00:34:08,909 --> 00:34:12,469 But from my point of view, I don't care where I got it from. 459 00:34:12,469 --> 00:34:15,270 That works. 460 00:34:15,270 --> 00:34:22,170 So this must describe the motion. 461 00:34:22,170 --> 00:34:28,120 And it has 2 arbitrary constants which 462 00:34:28,120 --> 00:34:33,880 I can find out, because I know the initial conditions 463 00:34:33,880 --> 00:34:36,860 for this pendulum. 464 00:34:36,860 --> 00:34:45,000 So once again, now I will try to find the 2 arbitrary constants. 465 00:34:55,989 --> 00:34:57,540 How did we do it before? 466 00:34:57,540 --> 00:35:03,710 By taking our solution. 467 00:35:03,710 --> 00:35:10,165 Well, we know that theta of 0 is 0. 468 00:35:12,970 --> 00:35:23,310 Therefore, if 0 is equal to A e to the minus gamma over 2t, 469 00:35:23,310 --> 00:35:24,370 that's t 0. 470 00:35:24,370 --> 00:35:31,500 So that's 1 cosine t 0, so I get phi. 471 00:35:31,500 --> 00:35:42,740 So one boundary condition tells me that phi is pi over 2. 472 00:35:42,740 --> 00:35:46,170 Or there are others, minus pi over 2 works, et cetera. 473 00:35:46,170 --> 00:35:47,430 Doesn't matter. 474 00:35:47,430 --> 00:35:54,120 Take any one of them and providing you're consistent, 475 00:35:54,120 --> 00:35:58,090 the other one will make up for it to give you the same answer. 476 00:35:58,090 --> 00:36:01,590 So anything that satisfies this boundary condition 477 00:36:01,590 --> 00:36:05,370 is fine for one of those constants. 478 00:36:05,370 --> 00:36:07,810 Now, the other one tells me something 479 00:36:07,810 --> 00:36:12,490 about the initial angular velocity of the rod. 480 00:36:12,490 --> 00:36:16,930 So I need to calculate the angular velocity of this rod. 481 00:36:16,930 --> 00:36:23,530 So here, I calculate theta dot t at time t 482 00:36:23,530 --> 00:36:25,930 by differentiating theta. 483 00:36:25,930 --> 00:36:28,520 And I get this lower equation. 484 00:36:28,520 --> 00:36:35,110 And now again, theta dot at 0 time 485 00:36:35,110 --> 00:36:39,640 is equal to-- we know it is-- this number, 486 00:36:39,640 --> 00:36:43,760 angular velocity at t equals 0. 487 00:36:43,760 --> 00:36:47,470 So in this equation, I plug that in. 488 00:36:47,470 --> 00:36:51,600 I get that the angular velocity at t 489 00:36:51,600 --> 00:36:56,090 equals 0 will have to equal 2. 490 00:36:56,090 --> 00:37:02,710 A, e to the minus gamma over 2t. 491 00:37:02,710 --> 00:37:05,780 t is 0, so that's 1. 492 00:37:05,780 --> 00:37:17,710 And 2 minus omega prime sine of omega prime t, but t is 0. 493 00:37:17,710 --> 00:37:18,760 So that's 0. 494 00:37:18,760 --> 00:37:23,010 Phi, that's the sine of pi over 2, because we find phi. 495 00:37:23,010 --> 00:37:24,850 That's 1. 496 00:37:24,850 --> 00:37:27,530 So I don't do anything with that. 497 00:37:27,530 --> 00:37:34,210 Minus gamma over 2 cosine of omega prime t. t is 0, 498 00:37:34,210 --> 00:37:36,220 so it's cosine of phi. 499 00:37:36,220 --> 00:37:40,230 Phi is pi over 2, so that is 0. 500 00:37:40,230 --> 00:37:42,810 So that's all right. 501 00:37:42,810 --> 00:37:54,950 Therefore, A is equal to now angular 502 00:37:54,950 --> 00:38:02,580 velocity at t equals 0 divided by omega prime. 503 00:38:06,330 --> 00:38:12,640 And therefore now, let me move over here. 504 00:38:12,640 --> 00:38:16,430 So now I can write the full solution. 505 00:38:16,430 --> 00:38:18,950 I have found everything, and I find 506 00:38:18,950 --> 00:38:23,570 that theta of t from the second line 507 00:38:23,570 --> 00:38:32,300 down is equal to A, which we found, 508 00:38:32,300 --> 00:38:41,470 angular velocity t equals 0 over omega prime minus-- 509 00:38:41,470 --> 00:38:46,790 there's a minus-- and now it's cosine omega prime t 510 00:38:46,790 --> 00:38:48,980 plus 90 degrees. 511 00:38:48,980 --> 00:38:58,640 So that's the same as minus sine of omega prime t. 512 00:38:58,640 --> 00:39:02,680 The 2 minuses cancel, so I get a plus. 513 00:39:02,680 --> 00:39:09,000 And if earlier on you chose, for example here, minus pi over 2, 514 00:39:09,000 --> 00:39:11,960 then both of those signs would have been plus, 515 00:39:11,960 --> 00:39:14,050 and you still get the same answer. 516 00:39:14,050 --> 00:39:15,780 So it doesn't matter. 517 00:39:15,780 --> 00:39:19,060 The algebra looks after itself. 518 00:39:19,060 --> 00:39:31,800 So we have to found the solution of this equation of motion 519 00:39:31,800 --> 00:39:36,460 with these boundary conditions for this situation 520 00:39:36,460 --> 00:39:38,990 where the gamma and the omega 0 squared 521 00:39:38,990 --> 00:39:42,260 is such that this is angle to correspond 522 00:39:42,260 --> 00:39:45,480 to the low damping situation. 523 00:39:45,480 --> 00:39:47,195 What kind of a motion is that? 524 00:39:51,550 --> 00:40:06,710 If I plot this, again as we do as a function of time, 525 00:40:06,710 --> 00:40:11,550 if I had a room full of students here who would correct me, then 526 00:40:11,550 --> 00:40:14,350 I would have caught the mistake. 527 00:40:14,350 --> 00:40:16,700 If you look at the second equation, 528 00:40:16,700 --> 00:40:20,140 it says take theta of t is A e to the minus gamma 529 00:40:20,140 --> 00:40:23,500 2t cosine omega prime t. 530 00:40:23,500 --> 00:40:26,855 When I rewrote it plugging in the value of A, 531 00:40:26,855 --> 00:40:32,603 I forgot the e to the minus gamma over 2t. 532 00:40:37,420 --> 00:40:38,800 Sorry about that. 533 00:40:38,800 --> 00:40:40,190 And we continue. 534 00:40:40,190 --> 00:40:43,140 So now if I plot this, what do I see? 535 00:40:43,140 --> 00:40:47,510 If it wasn't for the e to the minus gamma over 2t, 536 00:40:47,510 --> 00:40:50,760 it would be a sine. 537 00:40:50,760 --> 00:40:53,110 That's what it would look like. 538 00:40:53,110 --> 00:40:58,490 t, and this is the angle theta of t. 539 00:40:58,490 --> 00:41:01,200 This would be the simple harmonic motion 540 00:41:01,200 --> 00:41:07,880 with no drag, no damping. 541 00:41:07,880 --> 00:41:13,250 But this function is multiplied by an exponential, 542 00:41:13,250 --> 00:41:16,210 like this, which is dropping. 543 00:41:16,210 --> 00:41:22,360 So the net result-- and excuse my sloppy drawing-- 544 00:41:22,360 --> 00:41:25,770 is this modulated by that. 545 00:41:25,770 --> 00:41:29,970 So it will be something like this. 546 00:41:34,080 --> 00:41:37,830 So what we get is damped harmonic oscillation. 547 00:41:42,260 --> 00:41:49,860 Now again, I can't resist pointing out. 548 00:41:49,860 --> 00:41:52,900 To me, it seems almost miraculous. 549 00:41:52,900 --> 00:42:01,290 You have the same equation in form with the same boundary 550 00:42:01,290 --> 00:42:09,840 conditions predicting completely different behavior 551 00:42:09,840 --> 00:42:14,690 if some of these constants are bigger than the others or vice 552 00:42:14,690 --> 00:42:15,190 versa. 553 00:42:18,100 --> 00:42:25,680 So the mathematics has it built in to that kind of solution. 554 00:42:25,680 --> 00:42:27,740 Those two kinds of solutions exist. 555 00:42:27,740 --> 00:42:31,310 And that corresponds exactly what 556 00:42:31,310 --> 00:42:36,180 you would expect if you imagine an oscillator with damping. 557 00:42:36,180 --> 00:42:40,670 If you have this oscillator, if it's highly damped, as I've 558 00:42:40,670 --> 00:42:45,360 said before, you would expect this crazy motion reaching 559 00:42:45,360 --> 00:42:49,310 a maximum slowly coming down and coming to a grinding halt. 560 00:42:49,310 --> 00:42:51,830 Or if it's lightly damped, you would 561 00:42:51,830 --> 00:42:56,490 expect this to almost behave like a normal harmonic 562 00:42:56,490 --> 00:43:00,080 oscillator, but slowly losing energy 563 00:43:00,080 --> 00:43:02,750 due to damping and oscillating with smaller 564 00:43:02,750 --> 00:43:05,490 and smaller amplitude, which is exactly 565 00:43:05,490 --> 00:43:07,240 what the mathematics said would happen. 566 00:43:09,860 --> 00:43:12,420 I always like to ask a crazy question, 567 00:43:12,420 --> 00:43:16,975 how does the oscillator know what the mathematical solution 568 00:43:16,975 --> 00:43:17,800 is? 569 00:43:17,800 --> 00:43:19,770 But somehow or other, it does. 570 00:43:19,770 --> 00:43:24,302 It's a crazy way to phrase it, but I always think of that. 571 00:43:31,460 --> 00:43:41,120 So let's put together actually what 572 00:43:41,120 --> 00:43:50,730 we have seen in combination of last time and this time, 573 00:43:50,730 --> 00:43:56,700 my problem solving, my attempts to teach you 574 00:43:56,700 --> 00:43:59,610 how to solve problems. 575 00:43:59,610 --> 00:44:05,000 What we found last time was that we 576 00:44:05,000 --> 00:44:08,770 took three different harmonic oscillators. 577 00:44:08,770 --> 00:44:11,790 One was a mass on a spring. 578 00:44:11,790 --> 00:44:18,020 Another was this rod oscillating like this. 579 00:44:18,020 --> 00:44:22,950 The third was an electrical system, an LC circuit. 580 00:44:22,950 --> 00:44:30,030 In each case, we found that we the equation of motion 581 00:44:30,030 --> 00:44:32,200 looked exactly the same. 582 00:44:32,200 --> 00:44:40,780 The only difference was what was the name of the variable. 583 00:44:40,780 --> 00:44:46,180 For example, in the case of the spring, 584 00:44:46,180 --> 00:44:51,120 the displacement from the equilibrium was a distance. 585 00:44:51,120 --> 00:44:58,350 So this equation was of this form where the size was y. 586 00:45:01,050 --> 00:45:07,060 When we came to this oscillating rod, 587 00:45:07,060 --> 00:45:11,690 the quantity which was displaced from equilibrium was an angle. 588 00:45:11,690 --> 00:45:15,310 So this sine was a theta. 589 00:45:15,310 --> 00:45:21,760 In the case of the LC circuit, the quantity was a charge. 590 00:45:21,760 --> 00:45:27,090 So the charge which was displaced from equilibrium. 591 00:45:27,090 --> 00:45:31,570 But I could have summarized the equations of motion 592 00:45:31,570 --> 00:45:34,240 for the three harmonic oscillators 593 00:45:34,240 --> 00:45:37,530 by this one equation and just tell you, well, this 594 00:45:37,530 --> 00:45:41,190 is the equation of motion for that 595 00:45:41,190 --> 00:45:44,540 and put in the appropriate quantity, which 596 00:45:44,540 --> 00:45:49,510 is displaced and also put in the appropriate value 597 00:45:49,510 --> 00:45:53,310 of this constant, which for the spring was k over n. 598 00:45:53,310 --> 00:45:55,400 For the rod, 3g over 2l. 599 00:45:55,400 --> 00:45:59,510 For the LC circuit is 1 over LC. 600 00:45:59,510 --> 00:46:02,520 And believe me, we could have taken a million harmonic 601 00:46:02,520 --> 00:46:08,670 oscillator with no damping, anyone you can think of, 602 00:46:08,670 --> 00:46:10,770 and I could reduce it to this form. 603 00:46:13,360 --> 00:46:17,850 So here's something new that we observe, 604 00:46:17,850 --> 00:46:22,850 that you can take lots of physical situation 605 00:46:22,850 --> 00:46:26,100 but describe them in terms of the same mathematical 606 00:46:26,100 --> 00:46:27,460 equations. 607 00:46:27,460 --> 00:46:29,980 The same mathematical equation can 608 00:46:29,980 --> 00:46:34,120 describe millions and millions of different kinds 609 00:46:34,120 --> 00:46:36,020 of harmonic oscillators. 610 00:46:36,020 --> 00:46:39,180 So that is another feature of the scientific method 611 00:46:39,180 --> 00:46:40,520 that is very important. 612 00:46:40,520 --> 00:46:45,410 It means now we don't have to understand 613 00:46:45,410 --> 00:46:48,510 every little thing that happens in the universe. 614 00:46:48,510 --> 00:46:55,410 We can reduce it to a smaller subset of description. 615 00:46:58,430 --> 00:47:03,670 Now, we've now add the new feature today. 616 00:47:03,670 --> 00:47:07,900 We added some loss mechanism. 617 00:47:07,900 --> 00:47:13,450 We took a specific kind where the resistive force 618 00:47:13,450 --> 00:47:21,370 or the resist torque is proportional to the velocity 619 00:47:21,370 --> 00:47:24,030 of the analysis. 620 00:47:24,030 --> 00:47:29,530 In our case, to the angular velocity of the rod. 621 00:47:29,530 --> 00:47:33,070 And we derived an equation of motion, 622 00:47:33,070 --> 00:47:40,100 which looks like this, where these psis for the oscillator 623 00:47:40,100 --> 00:47:44,220 we considered was the angular displacement. 624 00:47:44,220 --> 00:47:48,960 I don't have to be an Einstein-- by analogies to what I said 625 00:47:48,960 --> 00:47:52,330 here-- to say look, I could have taken 626 00:47:52,330 --> 00:47:57,220 any one of the oscillators in the universe. 627 00:47:57,220 --> 00:48:03,510 And with damping, if the damping is 628 00:48:03,510 --> 00:48:07,150 proportional to the velocity of the analogous 629 00:48:07,150 --> 00:48:09,160 quantity of velocity, I'll end up 630 00:48:09,160 --> 00:48:10,940 with this equation of motion. 631 00:48:10,940 --> 00:48:13,180 So this is the equation of motion 632 00:48:13,180 --> 00:48:15,060 of simple harmonic motion. 633 00:48:15,060 --> 00:48:20,240 This is the equation of motion of damped harmonic oscillators. 634 00:48:20,240 --> 00:48:21,090 Every one of them. 635 00:48:24,340 --> 00:48:28,850 But there is an interesting difference we saw today. 636 00:48:28,850 --> 00:48:34,770 There is only qualitatively one kind of solution here. 637 00:48:34,770 --> 00:48:38,700 In the case of the damped harmonic oscillator, 638 00:48:38,700 --> 00:48:47,080 there are two qualitatively different solutions to that. 639 00:48:47,080 --> 00:48:51,520 One is the sum of exponentials and the other 640 00:48:51,520 --> 00:48:54,580 is damped harmonic motion. 641 00:48:57,420 --> 00:49:04,330 Both the quality of the solution and the details, 642 00:49:04,330 --> 00:49:07,760 whether it's oscillating fast or slow, et cetera, 643 00:49:07,760 --> 00:49:11,080 depends on those quantities-- the gamma 644 00:49:11,080 --> 00:49:14,870 and the omega 0 squared, which in turn, for example, 645 00:49:14,870 --> 00:49:17,890 in the various cases it's k over n, 3g, et cetera. 646 00:49:22,200 --> 00:49:27,440 So that will describe all oscillators. 647 00:49:27,440 --> 00:49:30,220 And in any particular oscillator, 648 00:49:30,220 --> 00:49:34,070 if you want to know what exactly happens at the given 649 00:49:34,070 --> 00:49:36,570 instant of time, you then need also 650 00:49:36,570 --> 00:49:41,900 to know those two constants, the boundary condition. 651 00:49:41,900 --> 00:49:45,400 If I have some mass on the spring oscillating, 652 00:49:45,400 --> 00:49:48,570 where it will be at some later time 653 00:49:48,570 --> 00:49:52,580 does depend where it is at this very instant. 654 00:49:55,790 --> 00:49:59,052 Finally, there's just one more thing I would like to do today, 655 00:49:59,052 --> 00:50:00,010 and it's the following. 656 00:50:02,580 --> 00:50:07,680 Oscillators are not only nice in order 657 00:50:07,680 --> 00:50:10,830 to give you a lecture on oscillators 658 00:50:10,830 --> 00:50:13,740 or show you how to solve them, they are practical devices. 659 00:50:16,290 --> 00:50:22,860 So sometimes we want to know some property of a given 660 00:50:22,860 --> 00:50:24,200 oscillator. 661 00:50:24,200 --> 00:50:29,510 Well, there are two properties which 662 00:50:29,510 --> 00:50:33,390 describe the quality of a given oscillator 663 00:50:33,390 --> 00:50:37,010 and what its behavior is. 664 00:50:37,010 --> 00:50:41,571 One you've seen already is to do the period or the frequency 665 00:50:41,571 --> 00:50:42,320 of the oscillator. 666 00:50:42,320 --> 00:50:45,880 In other words, you could have one oscillator 667 00:50:45,880 --> 00:50:48,340 which goes very slowly. 668 00:50:48,340 --> 00:50:52,510 You're going to have one which goes very fast. 669 00:50:52,510 --> 00:50:55,450 I have an oscillator and you want 670 00:50:55,450 --> 00:50:59,840 to tell somebody what it's like, a crucial number is what 671 00:50:59,840 --> 00:51:04,550 its period is, the 1 over the period which is the frequency. 672 00:51:04,550 --> 00:51:07,810 Imagine, for example, a bell. 673 00:51:07,810 --> 00:51:09,960 That's an oscillator. 674 00:51:09,960 --> 00:51:15,590 You could have a bell which has a low tone or a very small one 675 00:51:15,590 --> 00:51:18,390 with a very high pitched tone. 676 00:51:18,390 --> 00:51:20,590 So that's one number that's important. 677 00:51:20,590 --> 00:51:23,580 But there is another number that's important. 678 00:51:23,580 --> 00:51:28,250 Whether it's a good or a bad oscillator. 679 00:51:28,250 --> 00:51:34,040 And it's obvious what is it good oscillator. 680 00:51:34,040 --> 00:51:38,020 A good oscillator is one that oscillates for a long time. 681 00:51:38,020 --> 00:51:49,010 A bad one is that doesn't continue 682 00:51:49,010 --> 00:51:52,340 oscillating for a very long time. 683 00:51:52,340 --> 00:51:56,370 But again, you'd I'm sure agree with me 684 00:51:56,370 --> 00:52:03,010 that it isn't just how many oscillations it does, say, 685 00:52:03,010 --> 00:52:06,260 in a given time, like 1 second. 686 00:52:06,260 --> 00:52:09,000 If I take an oscillator that makes 687 00:52:09,000 --> 00:52:16,316 one oscillation per second, then if it survives 1 second, 688 00:52:16,316 --> 00:52:18,190 you wouldn't say it's a very good oscillator. 689 00:52:18,190 --> 00:52:24,040 It barely made one turn, while if the oscillator is making 690 00:52:24,040 --> 00:52:26,850 a million oscillations per second 691 00:52:26,850 --> 00:52:28,710 and that survives for a second, you 692 00:52:28,710 --> 00:52:30,460 would say that's a pretty good oscillator. 693 00:52:30,460 --> 00:52:39,410 It made a million oscillations before it ran out of steam. 694 00:52:39,410 --> 00:52:46,140 So to quantify the quality of an oscillator, we use a term, 695 00:52:46,140 --> 00:52:48,462 and being very imaginative we call 696 00:52:48,462 --> 00:52:52,290 it the quality of the oscillator, or Q value. 697 00:52:52,290 --> 00:52:57,050 So I will finally want to derive for you 698 00:52:57,050 --> 00:53:02,910 for this particular oscillator what is its Q or Q value. 699 00:53:02,910 --> 00:53:06,480 And that's the last thing I will do today. 700 00:53:06,480 --> 00:53:15,520 Now, it only makes sense to talk about the Q of an oscillator 701 00:53:15,520 --> 00:53:19,080 if it's at least a reasonably good oscillator. 702 00:53:19,080 --> 00:53:22,470 This one, this overdamped one who didn't even 703 00:53:22,470 --> 00:53:26,250 make one complete cycle, it doesn't make sense 704 00:53:26,250 --> 00:53:29,490 to talk about its Q. So the only case 705 00:53:29,490 --> 00:53:33,940 where it makes sense to talk about the Q of an oscillator 706 00:53:33,940 --> 00:53:38,050 is if it makes at least a few oscillations. 707 00:53:38,050 --> 00:53:41,440 It's got to be the underdamped case. 708 00:53:41,440 --> 00:53:45,190 So I will now calculate for you what 709 00:53:45,190 --> 00:53:51,120 is the Q of this compound pendulum, this rod which 710 00:53:51,120 --> 00:53:57,230 oscillates on the conditions where the drag force is small 711 00:53:57,230 --> 00:54:03,500 so that it has this undamped motion. 712 00:54:13,090 --> 00:54:17,770 So let me do that. 713 00:54:17,770 --> 00:54:26,940 The Q value of an oscillator is defined 714 00:54:26,940 --> 00:54:40,950 by no good reason of pi multiply by the number of oscillations 715 00:54:40,950 --> 00:54:49,010 the oscillator makes before its amplitude drops by 1 over e. 716 00:54:49,010 --> 00:54:51,710 It's a formal definition. 717 00:54:51,710 --> 00:54:58,655 So take some oscillator, which is a good oscillator. 718 00:54:58,655 --> 00:55:00,990 So it oscillates like this. 719 00:55:00,990 --> 00:55:04,370 It's decaying. 720 00:55:04,370 --> 00:55:06,880 This has an exponential form, as we 721 00:55:06,880 --> 00:55:10,270 saw for a damped harmonic oscillator. 722 00:55:10,270 --> 00:55:13,760 And we ask the question how many oscillations 723 00:55:13,760 --> 00:55:18,380 it makes before the amplitude drops by 1 over e. 724 00:55:21,310 --> 00:55:24,700 And for historic reason, 1 takes pi 725 00:55:24,700 --> 00:55:38,130 multiply by the number of oscillations for amplitude 726 00:55:38,130 --> 00:55:42,480 to go down by amplitude over e. 727 00:55:42,480 --> 00:55:45,300 1 over e. 728 00:55:45,300 --> 00:55:49,000 Because the energy stored in an oscillator 729 00:55:49,000 --> 00:55:52,400 is proportional to the amplitude squared, 730 00:55:52,400 --> 00:56:01,610 this is equivalent to 2 pi times the number of oscillations 731 00:56:01,610 --> 00:56:11,830 it takes for the energy to go to the energy over e. 732 00:56:15,270 --> 00:56:18,910 These are equivalent because 1 is the square of the other. 733 00:56:18,910 --> 00:56:20,770 And in fact, just let me tell you. 734 00:56:20,770 --> 00:56:22,830 I'm digressing here. 735 00:56:22,830 --> 00:56:26,310 Originally this was the definition. 736 00:56:26,310 --> 00:56:33,040 The 2 pi is there because in the definition, 737 00:56:33,040 --> 00:56:38,870 one said what will be the change of phase 738 00:56:38,870 --> 00:56:44,280 of the oscillator for the energy stored in the oscillator 739 00:56:44,280 --> 00:56:46,420 to go down by 1 over e? 740 00:56:46,420 --> 00:56:49,000 If you're talking about the change of phase 741 00:56:49,000 --> 00:56:52,100 rather than number of oscillations, 742 00:56:52,100 --> 00:56:55,420 you have 2 pi there, because the phase 743 00:56:55,420 --> 00:56:59,820 in one oscillation, the phase changes by 2 pi. 744 00:56:59,820 --> 00:57:02,000 So historically, this came in. 745 00:57:02,000 --> 00:57:03,315 That's equivalent to this. 746 00:57:06,180 --> 00:57:13,550 And so we have to calculate that. 747 00:57:13,550 --> 00:57:17,260 Now, for this particular oscillator, 748 00:57:17,260 --> 00:57:23,790 we actually calculated the motion, how the angle changes 749 00:57:23,790 --> 00:57:28,260 with time, and we see that the amplitude drops 750 00:57:28,260 --> 00:57:32,780 like e to the minus gamma over 2. 751 00:57:32,780 --> 00:57:35,010 So we saw that the amplitude drops 752 00:57:35,010 --> 00:57:40,413 like e to the minus gamma over 2t. 753 00:57:44,700 --> 00:57:52,350 So in what time will the amplitude drop by 1 over e? 754 00:57:52,350 --> 00:58:01,350 That time will be when gamma t over 2 is equal to 1. 755 00:58:03,950 --> 00:58:07,040 So you get e to the minus 1. 756 00:58:07,040 --> 00:58:19,600 So it will be in a time t equal to 2 over gamma. 757 00:58:22,710 --> 00:58:30,210 Now, what we are interested in is how many 758 00:58:30,210 --> 00:58:38,620 oscillations will, in this time, the oscillator make. 759 00:58:41,760 --> 00:58:55,860 But you know the period T is equal to 2 pi over omega prime. 760 00:58:55,860 --> 00:59:05,640 I'm using here omega prime because that is the angular 761 00:59:05,640 --> 00:59:09,190 frequency of the damped oscillator. 762 00:59:09,190 --> 00:59:12,710 It was not omega 0, like the undamped one. 763 00:59:12,710 --> 00:59:19,140 And omega prime I defined for you earlier. 764 00:59:19,140 --> 00:59:26,810 But you know that if the damping is weak, 765 00:59:26,810 --> 00:59:33,726 omega prime is essentially equal to omega 0. 766 00:59:33,726 --> 00:59:35,340 Q is a quality thing. 767 00:59:35,340 --> 00:59:37,310 We don't need it exactly. 768 00:59:37,310 --> 00:59:41,540 So it's just to give you a feel whether something 769 00:59:41,540 --> 00:59:43,960 is good or bad oscillator. 770 00:59:43,960 --> 00:59:49,280 So since omega prime is essentially equal to omega 0, 771 00:59:49,280 --> 00:59:52,300 I'll take the period to be approximately 772 00:59:52,300 --> 00:59:54,370 equal to this over omega 0. 773 00:59:58,440 --> 01:00:06,860 So in time t, how many periods will I have? 774 01:00:06,860 --> 01:00:12,900 I have to divide t by the period. 775 01:00:12,900 --> 01:00:16,810 That'll be the number of oscillations I'll make. 776 01:00:16,810 --> 01:00:24,080 So it's 2 over gamma divided by 2 pi over omega 0. 777 01:00:28,520 --> 01:00:33,640 And so this is the number of oscillations. 778 01:00:33,640 --> 01:00:38,580 So the Q will be pi times the number of oscillations 779 01:00:38,580 --> 01:00:41,740 for the amplitude to drop by this. 780 01:00:41,740 --> 01:00:48,230 So the Q will be equal to pi times 781 01:00:48,230 --> 01:00:54,450 this quantity, which is 2 over gamma divided 782 01:00:54,450 --> 01:01:02,570 by 2 pi over omega 0, which is equal to omega 0 over gamma. 783 01:01:02,570 --> 01:01:08,270 So the quality of this particular oscillator 784 01:01:08,270 --> 01:01:11,770 will be equal to omega 0 over gamma. 785 01:01:11,770 --> 01:01:17,270 And I hope I still have it. 786 01:01:17,270 --> 01:01:18,150 Do I have that? 787 01:01:20,850 --> 01:01:31,710 Well, by going back through this video, 788 01:01:31,710 --> 01:01:37,680 omega 0 we defined earlier on and gamma also. 789 01:01:37,680 --> 01:01:40,520 And so you calculated this. 790 01:01:40,520 --> 01:01:44,000 That's as much as I was going to do 791 01:01:44,000 --> 01:01:47,330 about damped oscillations for you. 792 01:01:47,330 --> 01:01:48,880 Thank you.