1 00:00:00,060 --> 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 00:00:20,906 --> 00:00:23,590 PROFESSOR: Welcome back, and today I'm 9 00:00:23,590 --> 00:00:28,110 going to do for you three problems all related 10 00:00:28,110 --> 00:00:31,310 to driven harmonic oscillators. 11 00:00:31,310 --> 00:00:32,930 Now one of the things you'll notice 12 00:00:32,930 --> 00:00:36,080 as I'm doing these problems, that it seems 13 00:00:36,080 --> 00:00:39,010 I'm doing the same thing over and over again. 14 00:00:39,010 --> 00:00:41,850 You may have also noticed that in the other problems 15 00:00:41,850 --> 00:00:45,280 I did earlier for you in harmonic oscillators, 16 00:00:45,280 --> 00:00:47,440 et cetera. 17 00:00:47,440 --> 00:00:50,350 So at this stage, I have to remind you about something. 18 00:00:50,350 --> 00:00:54,090 This is no accident that it is like that. 19 00:00:54,090 --> 00:00:56,510 It's to do with the scientific method, 20 00:00:56,510 --> 00:00:59,690 and I would want to briefly remind you 21 00:00:59,690 --> 00:01:03,590 about how we are solving problems. 22 00:01:03,590 --> 00:01:07,270 Here is a diagram of a generic problem. 23 00:01:07,270 --> 00:01:08,440 What does it consists of? 24 00:01:08,440 --> 00:01:14,940 A typical problem is we are told of a physical situation 25 00:01:14,940 --> 00:01:17,710 in words, ordinary language. 26 00:01:17,710 --> 00:01:23,090 And in ordinary language, we are asked questions about given 27 00:01:23,090 --> 00:01:25,330 that physical situation, what will 28 00:01:25,330 --> 00:01:27,470 happen as a function of time? 29 00:01:27,470 --> 00:01:33,870 What kind of emotion it has, questions of that kind. 30 00:01:33,870 --> 00:01:38,680 What problem solving is, doing the scientific method, 31 00:01:38,680 --> 00:01:39,970 is the following. 32 00:01:39,970 --> 00:01:50,300 The first thing we have to do is, by studying this situation 33 00:01:50,300 --> 00:01:57,270 and using the laws of nature, describing 34 00:01:57,270 --> 00:02:01,421 the physical situation in terms of mathematical equations. 35 00:02:01,421 --> 00:02:02,420 This is always possible. 36 00:02:05,610 --> 00:02:09,820 So starting with as I say this description, 37 00:02:09,820 --> 00:02:13,150 using laws of nature, we end up with a description 38 00:02:13,150 --> 00:02:16,260 in terms of mathematical equations. 39 00:02:16,260 --> 00:02:19,310 The next thing we do is, we now forget 40 00:02:19,310 --> 00:02:21,650 that this has anything to do with physics. 41 00:02:21,650 --> 00:02:24,260 We have a set of mathematical equations, 42 00:02:24,260 --> 00:02:25,750 which we have to solve. 43 00:02:25,750 --> 00:02:27,180 This is mathematics. 44 00:02:27,180 --> 00:02:29,560 We use mathematics to solve this problem. 45 00:02:29,560 --> 00:02:30,700 All right? 46 00:02:30,700 --> 00:02:34,810 Once we have found the solution to this mathematical equation, 47 00:02:34,810 --> 00:02:41,380 there's only one thing left, we identify the answer 48 00:02:41,380 --> 00:02:46,360 with the questions posed in the problem. 49 00:02:46,360 --> 00:02:50,930 So understanding of physics comes 50 00:02:50,930 --> 00:02:57,720 in this step, which I call step 1, and in the last one. 51 00:02:57,720 --> 00:03:01,580 The middle one, step 2, is pure mathematics. 52 00:03:01,580 --> 00:03:14,050 Now, it so happens that if you take all driven harmonic 53 00:03:14,050 --> 00:03:18,800 oscillators, the mathematics that you end up with 54 00:03:18,800 --> 00:03:19,710 is identical. 55 00:03:19,710 --> 00:03:24,820 In other words, if I look at the description of the problem 56 00:03:24,820 --> 00:03:30,820 in mathematical terms, I cannot tell you which problem that 57 00:03:30,820 --> 00:03:32,230 description belongs. 58 00:03:32,230 --> 00:03:38,330 That's a generic description of many, many, many situations. 59 00:03:38,330 --> 00:03:42,570 Today I'll take three of them, but all three situations, 60 00:03:42,570 --> 00:03:45,130 all three problems-- although to you, they 61 00:03:45,130 --> 00:03:49,600 may look completely different-- after I've translated them 62 00:03:49,600 --> 00:03:53,020 in terms of mathematical equations, are identical. 63 00:03:53,020 --> 00:03:56,460 And so it's not surprising that from here on when we then 64 00:03:56,460 --> 00:03:59,580 try to solve them, et cetera, we are doing the same problem, 65 00:03:59,580 --> 00:04:03,450 or at least it appears so. 66 00:04:03,450 --> 00:04:05,600 Having said that, let's immediately 67 00:04:05,600 --> 00:04:10,090 dive in and take one of the problems. 68 00:04:10,090 --> 00:04:12,810 So the first problem I'm going to consider 69 00:04:12,810 --> 00:04:16,089 is an idealized-driven RLC circuit. 70 00:04:18,750 --> 00:04:21,000 Why do I see idealized? 71 00:04:21,000 --> 00:04:22,690 Because as I've mentioned before, 72 00:04:22,690 --> 00:04:27,390 in real life situations, things extremely complicated, 73 00:04:27,390 --> 00:04:29,570 and if you take everything into account, 74 00:04:29,570 --> 00:04:32,770 it will take you essentially forever 75 00:04:32,770 --> 00:04:35,010 to get an exact solution. 76 00:04:35,010 --> 00:04:37,680 But we're often not most of the time not interested 77 00:04:37,680 --> 00:04:39,290 in the very exact one. 78 00:04:39,290 --> 00:04:42,740 So we simplify a physical situation 79 00:04:42,740 --> 00:04:48,480 to the bear essential parts and solve that problem. 80 00:04:48,480 --> 00:04:54,690 So this is a problem where you have a circuit consisting 81 00:04:54,690 --> 00:04:58,050 all a capacitance, inductance, and a resistance, 82 00:04:58,050 --> 00:05:03,360 in series, which is driven by an alternating voltage source V0 83 00:05:03,360 --> 00:05:04,670 cosine omega t. 84 00:05:07,330 --> 00:05:12,220 Before time t equals 0, I'm assuming the circuit is open. 85 00:05:12,220 --> 00:05:15,330 In other words, there is no charges anywhere, 86 00:05:15,330 --> 00:05:16,750 no current flowing. 87 00:05:16,750 --> 00:05:20,700 At t equals 0, we close the circuit 88 00:05:20,700 --> 00:05:25,690 so the charge on the capacitor t equals 0, is 0. 89 00:05:25,690 --> 00:05:27,790 The current flowing in the circuit is 0. 90 00:05:27,790 --> 00:05:28,750 All right? 91 00:05:28,750 --> 00:05:31,750 And the question we want to answer 92 00:05:31,750 --> 00:05:36,640 is, what will the current be in the circuit at some time t 93 00:05:36,640 --> 00:05:42,570 a specific one, in 17 seconds or whatever, some specific time. 94 00:05:42,570 --> 00:05:45,550 Since this is completely specified, 95 00:05:45,550 --> 00:05:48,480 I can predict what will happen. 96 00:05:48,480 --> 00:05:51,730 That's what it means in this case solving this problem. 97 00:05:51,730 --> 00:05:52,540 So let's do it. 98 00:06:01,270 --> 00:06:05,670 So we are now entering step 1. 99 00:06:05,670 --> 00:06:09,330 We're going to take the physical situation 100 00:06:09,330 --> 00:06:16,050 and convert it to a description in terms of mathematics. 101 00:06:16,050 --> 00:06:22,980 This is the conceptually hardest step of any problem. 102 00:06:22,980 --> 00:06:28,240 It is the one where you need to understand the physics. 103 00:06:28,240 --> 00:06:31,660 It may not necessarily be the hardest or the longest. 104 00:06:31,660 --> 00:06:35,870 Often the solution of the equations 105 00:06:35,870 --> 00:06:40,900 are incredibly tedious and long and take all your time. 106 00:06:40,900 --> 00:06:43,860 But they are routine mathematics. 107 00:06:43,860 --> 00:06:49,370 Here you've got to conceptualize what's going on 108 00:06:49,370 --> 00:06:54,430 and translate it as I say to the language of mathematics. 109 00:06:54,430 --> 00:07:00,230 So in the case of I've had have to define 110 00:07:00,230 --> 00:07:02,240 coordinates for this problem. 111 00:07:02,240 --> 00:07:08,530 So what I have is, I'll assume that at some instant of time 112 00:07:08,530 --> 00:07:10,590 that in this circuit, it's closed 113 00:07:10,590 --> 00:07:12,770 because it's past equal to 0, there 114 00:07:12,770 --> 00:07:18,230 is a current I flowing that on this capacitor on the left 115 00:07:18,230 --> 00:07:20,830 there is plus charge Q and minus charge 116 00:07:20,830 --> 00:07:23,240 Q on the right side that is this inductance. 117 00:07:26,010 --> 00:07:31,300 And over here, we have an alternating potential applied. 118 00:07:31,300 --> 00:07:34,000 Now I have to be very specific at any instant of time which 119 00:07:34,000 --> 00:07:35,540 side is positive and negative. 120 00:07:35,540 --> 00:07:37,700 I'll assume that t equals 0. 121 00:07:37,700 --> 00:07:40,710 It's 0 over here, and at this point 122 00:07:40,710 --> 00:07:45,340 here, it's plus V0 cosine omega t. 123 00:07:45,340 --> 00:07:50,930 So these are the coordinates of this problem. 124 00:07:50,930 --> 00:07:54,350 And now what drives the current? 125 00:07:54,350 --> 00:07:58,320 What decides what happens here? 126 00:07:58,320 --> 00:08:00,560 What are the law of natures applicable? 127 00:08:00,560 --> 00:08:04,750 In this case, are the Maxwell's equations or to be specific, 128 00:08:04,750 --> 00:08:06,810 the Faraday equation. 129 00:08:06,810 --> 00:08:08,340 All right? 130 00:08:08,340 --> 00:08:15,490 One more thing I should add that I is a current here flowing 131 00:08:15,490 --> 00:08:18,430 charge by conservation of charge. 132 00:08:18,430 --> 00:08:20,630 It's law of physics. 133 00:08:20,630 --> 00:08:23,930 The current at the instant of time, 134 00:08:23,930 --> 00:08:28,700 will be equal to the rate of change of the charge. 135 00:08:28,700 --> 00:08:32,450 So at the instant time, I is, in my language, 136 00:08:32,450 --> 00:08:35,240 Q with a dot on top of it. 137 00:08:35,240 --> 00:08:38,559 This is plus because of the way in which I 138 00:08:38,559 --> 00:08:40,270 define my coordinates. 139 00:08:40,270 --> 00:08:43,640 Beware if you by mistake say suppose 140 00:08:43,640 --> 00:08:47,300 I define my coordinates this being minus and this plus, 141 00:08:47,300 --> 00:08:49,890 then this equation would be a minus here, 142 00:08:49,890 --> 00:08:51,920 conservation of charge. 143 00:08:51,920 --> 00:08:54,350 But with the way I've defined these coordinates, 144 00:08:54,350 --> 00:08:56,720 this is plus. 145 00:08:56,720 --> 00:08:57,850 So that's one. 146 00:08:57,850 --> 00:09:00,670 Next I said what the law of nature 147 00:09:00,670 --> 00:09:03,120 which governs the motion of the charges 148 00:09:03,120 --> 00:09:07,390 is the Faraday's law, which states 149 00:09:07,390 --> 00:09:12,350 that if I take the line integral of the electric field 150 00:09:12,350 --> 00:09:17,320 around a closed loop, that's this, 151 00:09:17,320 --> 00:09:23,480 it equals to minus the rate of change 152 00:09:23,480 --> 00:09:27,160 of total magnetic flux linking this circuit, 153 00:09:27,160 --> 00:09:29,980 going through this circuit. 154 00:09:29,980 --> 00:09:34,580 Now, you know that the magnetic flux through a circuit 155 00:09:34,580 --> 00:09:40,520 is equal to the current times the self-inductance 156 00:09:40,520 --> 00:09:45,130 of that circuit or the total inductance in that circuit. 157 00:09:45,130 --> 00:09:49,770 So taking this law of nature, this law of nature, 158 00:09:49,770 --> 00:09:53,240 I can now translate this into mathematics, 159 00:09:53,240 --> 00:09:54,970 and that's what I'm going to do. 160 00:09:54,970 --> 00:09:58,450 So first I'm going to take the line integral of E dl 161 00:09:58,450 --> 00:10:01,700 around this circuit. 162 00:10:01,700 --> 00:10:06,490 If I go around from here to here, 163 00:10:06,490 --> 00:10:10,300 this is positive relative to this point, 164 00:10:10,300 --> 00:10:13,105 so the electric field from here to here 165 00:10:13,105 --> 00:10:19,300 will be in this direction, and so the line integral of E dl 166 00:10:19,300 --> 00:10:24,250 along here will be minus, and by definition of a voltage, 167 00:10:24,250 --> 00:10:27,100 it will be just minus the voltage drop 168 00:10:27,100 --> 00:10:30,360 across here, which is V0 cosine omega t. 169 00:10:30,360 --> 00:10:33,725 So this is minus V0 cosine omega t. 170 00:10:33,725 --> 00:10:35,910 I continue going around. 171 00:10:35,910 --> 00:10:39,530 From here to here, the electric field is in this direction. 172 00:10:39,530 --> 00:10:42,117 And so that will be the integral V dl 173 00:10:42,117 --> 00:10:46,210 is Q over C, that's from the definition of a capacitance, 174 00:10:46,210 --> 00:10:47,840 of course. 175 00:10:47,840 --> 00:10:55,730 Then as we go around in these idealized diagrams, 176 00:10:55,730 --> 00:11:00,250 all wires have 0 resistance and therefore there cannot be field 177 00:11:00,250 --> 00:11:00,750 in them. 178 00:11:00,750 --> 00:11:03,400 That's why there's no contribution to this 179 00:11:03,400 --> 00:11:06,060 from the field inside the wires. 180 00:11:06,060 --> 00:11:07,800 I continue. 181 00:11:07,800 --> 00:11:11,370 This inductance is just coiled wire, 182 00:11:11,370 --> 00:11:15,320 so there's still no contribution to this in the L. I continue. 183 00:11:15,320 --> 00:11:19,690 Here I know there is a current flowing like this means 184 00:11:19,690 --> 00:11:22,590 the reason the electric field in this direction 185 00:11:22,590 --> 00:11:27,170 and once again knowing the definition of our Ohms law, 186 00:11:27,170 --> 00:11:30,400 I get that the integral of the electric fields from here 187 00:11:30,400 --> 00:11:33,300 to here is I times R. 188 00:11:33,300 --> 00:11:37,160 So this is this quantity, the integral 189 00:11:37,160 --> 00:11:39,800 of the electric field around this closed loop. 190 00:11:39,800 --> 00:11:45,930 That must equal by Faraday's law to minus the rate of change 191 00:11:45,930 --> 00:11:49,260 of magnetic flux, which is minus L dI dt. 192 00:11:53,080 --> 00:11:55,170 We're almost finished. 193 00:11:55,170 --> 00:11:56,690 Now we do algebra. 194 00:11:56,690 --> 00:11:57,660 We're playing around. 195 00:11:57,660 --> 00:12:02,270 I can rewrite this like this knowing that the current is 196 00:12:02,270 --> 00:12:07,490 the rate of change of Q, and the rate of change of current 197 00:12:07,490 --> 00:12:09,200 is the second derivative. 198 00:12:09,200 --> 00:12:12,100 So this is just straightforward algebra. 199 00:12:12,100 --> 00:12:15,510 Now, I'm going to rewrite just moving terms around 200 00:12:15,510 --> 00:12:18,830 to make it look in the way I prefer. 201 00:12:18,830 --> 00:12:24,190 And so I've rewritten this equation now in this form. 202 00:12:24,190 --> 00:12:29,980 And now, I will just redefine some constants 203 00:12:29,980 --> 00:12:32,720 so that I recognize the equation better. 204 00:12:32,720 --> 00:12:38,350 So I will define R over L by a constant gamma. 205 00:12:38,350 --> 00:12:43,430 And I will define 1 over LC by the constant omega 0 squared. 206 00:12:43,430 --> 00:12:47,390 And I define V0 L by the constant f. 207 00:12:47,390 --> 00:12:53,270 And so this equation rewritten looks like that. 208 00:12:53,270 --> 00:12:54,740 Eureka. 209 00:12:54,740 --> 00:12:58,340 This is an equation I've seen millions of times, 210 00:12:58,340 --> 00:12:59,840 and I know how to solve it. 211 00:13:03,950 --> 00:13:08,540 So what we've done is, we've taken this physical situation, 212 00:13:08,540 --> 00:13:13,440 translated it into a mathematical equation. 213 00:13:13,440 --> 00:13:16,530 This is the equation of motion, this just 214 00:13:16,530 --> 00:13:20,490 defines the constants, and this is the boundary conditions. 215 00:13:20,490 --> 00:13:23,870 It tells us that in our particular problem, 216 00:13:23,870 --> 00:13:27,630 we switch that circuit at t equals 0. 217 00:13:27,630 --> 00:13:29,680 At that time, what was the condition 218 00:13:29,680 --> 00:13:31,570 of the current and charge? 219 00:13:31,570 --> 00:13:34,650 So Q at 0 we said was 0. 220 00:13:34,650 --> 00:13:37,090 Q dot at 0 was 0. 221 00:13:37,090 --> 00:13:41,530 This completely, mathematically defines that problem. 222 00:13:41,530 --> 00:13:48,270 This is the mathematical description of this problem. 223 00:13:48,270 --> 00:13:51,740 From now on, I don't have to even remember 224 00:13:51,740 --> 00:13:55,380 that this has anything to do with physics. 225 00:13:55,380 --> 00:13:58,790 This is now a problem in mathematics. 226 00:13:58,790 --> 00:14:02,610 What is the solution of this equation 227 00:14:02,610 --> 00:14:05,250 satisfying these boundary conditions? 228 00:14:05,250 --> 00:14:07,380 We'll do that later. 229 00:14:07,380 --> 00:14:11,150 I'm now going to immediately go to the next problem. 230 00:14:11,150 --> 00:14:12,990 See in each case, I've done the physics. 231 00:14:12,990 --> 00:14:15,780 This is the end of the physics in essence. 232 00:14:15,780 --> 00:14:18,220 From now on, I just have to do mathematics. 233 00:14:18,220 --> 00:14:22,010 So let me take the next problem, a different one. 234 00:14:22,010 --> 00:14:23,450 It seems completely different. 235 00:14:23,450 --> 00:14:27,240 Here we were dealing with charges, currents, circuits, et 236 00:14:27,240 --> 00:14:31,290 cetera, and now I'm going to look at the physical one. 237 00:14:31,290 --> 00:14:34,610 I could even bring you a model of that problem. 238 00:14:34,610 --> 00:14:39,670 What I'm going to consider here is an ideal pendulum, 239 00:14:39,670 --> 00:14:44,560 simple pendulum, ideal, which is a heavy mass, considered 240 00:14:44,560 --> 00:14:49,400 to be a point mass, a string, which is considered 241 00:14:49,400 --> 00:14:54,860 to be always taught but yet massless. 242 00:14:54,860 --> 00:14:57,970 So it's an ideal pendulum. 243 00:14:57,970 --> 00:15:01,360 What I have here is an approximation to an ideal one, 244 00:15:01,360 --> 00:15:05,970 and if I understand what the ideal one does, 245 00:15:05,970 --> 00:15:11,070 this in reality, will you do something pretty close to it. 246 00:15:11,070 --> 00:15:14,010 And so what I'm going to consider, 247 00:15:14,010 --> 00:15:19,290 I am going to be oscillating my hand backwards and forwards 248 00:15:19,290 --> 00:15:24,790 sinusoidally starting this at some definite time, 249 00:15:24,790 --> 00:15:28,480 and I'll try to predict the motion of this. 250 00:15:28,480 --> 00:15:31,770 And notice I can make all sorts of motion out of this. 251 00:15:31,770 --> 00:15:35,580 If I do it slowly, look, my hand and the ball 252 00:15:35,580 --> 00:15:37,820 are going in the same direction. 253 00:15:37,820 --> 00:15:42,720 If I go fast, they're going in opposite direction. 254 00:15:42,720 --> 00:15:48,470 And the way this is oscillating is different, 255 00:15:48,470 --> 00:15:52,870 depends on how I am moving my hand backward and forward. 256 00:15:52,870 --> 00:15:56,450 So let me tell you very precisely what is 257 00:15:56,450 --> 00:16:02,270 this idealized situation which I am trying to solve and predict 258 00:16:02,270 --> 00:16:03,310 what will happen. 259 00:16:03,310 --> 00:16:07,120 So I have this ideal pendulum. 260 00:16:07,120 --> 00:16:11,270 It has a length, L, mass, m. 261 00:16:11,270 --> 00:16:12,590 It's in the vertical plane. 262 00:16:12,590 --> 00:16:15,500 Gravity is down there. 263 00:16:15,500 --> 00:16:21,750 At one end of the string, I will be moving my hand 264 00:16:21,750 --> 00:16:25,740 in perfect sinusoidal fashion. 265 00:16:25,740 --> 00:16:30,340 It will have an amplitude of x0 and angular frequency omega, 266 00:16:30,340 --> 00:16:32,896 so it's x0 sine omega t. 267 00:16:32,896 --> 00:16:36,160 I am making the following assumptions, 268 00:16:36,160 --> 00:16:37,610 that's why it's idealized. 269 00:16:37,610 --> 00:16:40,730 I'm assuming it's a massless string. 270 00:16:40,730 --> 00:16:42,960 I'm assuming it's always taught. 271 00:16:42,960 --> 00:16:48,420 I'm going to assume that it's all making small oscillations 272 00:16:48,420 --> 00:16:53,400 so that if in any calculation, I can take a sine 273 00:16:53,400 --> 00:16:55,360 to be an angle to be equal to the angle. 274 00:16:59,940 --> 00:17:04,210 Furthermore I'm going to assume that there is a bit of drag, 275 00:17:04,210 --> 00:17:09,010 there is friction, and I will assume 276 00:17:09,010 --> 00:17:14,210 that the frictional force on this that mass 277 00:17:14,210 --> 00:17:18,160 is proportional to the velocity, and the constant 278 00:17:18,160 --> 00:17:24,349 of that velocity is B. Now you ask me, why choose that? 279 00:17:24,349 --> 00:17:28,300 Well, the simple answer is, I want 280 00:17:28,300 --> 00:17:31,800 to find this situation when I know I can solve it. 281 00:17:31,800 --> 00:17:34,710 And I know how to solve this problem 282 00:17:34,710 --> 00:17:37,160 if it's proportional to velocity. 283 00:17:37,160 --> 00:17:40,090 If it's something more complicated, 284 00:17:40,090 --> 00:17:43,160 then the equation may be correct at the end, 285 00:17:43,160 --> 00:17:47,300 but I will not be able to just like that on the board, 286 00:17:47,300 --> 00:17:48,290 solve it for you. 287 00:17:48,290 --> 00:17:52,260 I could do it numerically, use a computer, et cetera, 288 00:17:52,260 --> 00:17:56,620 but at the same time, I know that if I have something which 289 00:17:56,620 --> 00:18:00,680 is very close to being what the actual situation is, 290 00:18:00,680 --> 00:18:05,790 the prediction of the real world compared to this idealized one 291 00:18:05,790 --> 00:18:07,490 will not be very different. 292 00:18:07,490 --> 00:18:10,410 Qualitatively, I will get a good understanding 293 00:18:10,410 --> 00:18:11,220 of what's going on. 294 00:18:15,680 --> 00:18:19,870 So this is the assumption, and what is the question? 295 00:18:19,870 --> 00:18:21,690 What do I want to predict? 296 00:18:21,690 --> 00:18:24,590 I want to now take a very specific situation. 297 00:18:24,590 --> 00:18:28,980 I want to take a situation that at t equals 0 initially, 298 00:18:28,980 --> 00:18:36,750 the mass is stationary, and it's exactly below my hand, 299 00:18:36,750 --> 00:18:39,880 but my hand is not stationary, it's moving. 300 00:18:39,880 --> 00:18:45,790 So I am moving this backwards and forwards sinusoidally, 301 00:18:45,790 --> 00:18:46,790 all right? 302 00:18:46,790 --> 00:18:52,450 And in the instant, when my hand is over the ball, 303 00:18:52,450 --> 00:18:55,980 I let go of the ball and then I continue moving it. 304 00:18:55,980 --> 00:19:03,010 So I have specified this exactly what this ball is doing at t 305 00:19:03,010 --> 00:19:05,760 equals 0, what my hand is doing, and I 306 00:19:05,760 --> 00:19:12,480 want to predict what will happen at some arbitrary later time. 307 00:19:12,480 --> 00:19:15,820 So my question is what will happen, 308 00:19:15,820 --> 00:19:18,860 and this time to make it a little more interesting also, 309 00:19:18,860 --> 00:19:26,870 I want to know in qualitative terms what kind of motion 310 00:19:26,870 --> 00:19:31,450 will it have if I move this slowly and quickly 311 00:19:31,450 --> 00:19:34,850 and check against reality at the end. 312 00:19:34,850 --> 00:19:36,760 And [? originally ?] I could tell you 313 00:19:36,760 --> 00:19:39,640 that things are different when I'm doing it slowly. 314 00:19:39,640 --> 00:19:41,620 As I told you before, if I do this slowly 315 00:19:41,620 --> 00:19:45,560 like this, my hand and the ball going in the same direction. 316 00:19:45,560 --> 00:19:49,310 If I go fast, I can make them go in opposite directions. 317 00:19:49,310 --> 00:19:53,830 All that should come out of this. 318 00:19:53,830 --> 00:19:57,670 So the part of the question is to qualitatively understand 319 00:19:57,670 --> 00:20:00,060 this phenomenon. 320 00:20:00,060 --> 00:20:02,780 So that's my problem number 2. 321 00:20:02,780 --> 00:20:07,100 The third problem that I'm going to do. 322 00:20:07,100 --> 00:20:08,930 Sorry. 323 00:20:08,930 --> 00:20:09,860 I haven't done it. 324 00:20:09,860 --> 00:20:13,420 That's what I want to do, and I have to do it. 325 00:20:13,420 --> 00:20:18,660 I wanted to get the exact things over quicker, but no such luck. 326 00:20:21,229 --> 00:20:22,770 Now, I'm going to solve this problem. 327 00:20:22,770 --> 00:20:24,360 So back again. 328 00:20:24,360 --> 00:20:27,370 Look at that diagram and what's the first thing to do. 329 00:20:27,370 --> 00:20:38,150 Step 1 use the laws of nature to restate this problem 330 00:20:38,150 --> 00:20:41,270 in terms of math and language of mathematics. 331 00:20:41,270 --> 00:20:42,740 That's what we have to do. 332 00:20:42,740 --> 00:20:43,240 All right. 333 00:20:43,240 --> 00:20:45,430 Let's go a little fast. 334 00:20:45,430 --> 00:20:48,660 So I have to define some coordinate system 335 00:20:48,660 --> 00:20:52,460 so I redraw my picture and now I'm 336 00:20:52,460 --> 00:20:56,640 going to tell you that I'll take some vertical stationary axis 337 00:20:56,640 --> 00:21:01,630 here, and I'll say it instant t, this mass 338 00:21:01,630 --> 00:21:03,700 is at some position, y. 339 00:21:07,090 --> 00:21:11,600 My hand holding here is at position 340 00:21:11,600 --> 00:21:15,700 is at x0 sine omega t, time t. 341 00:21:15,700 --> 00:21:16,370 Fine. 342 00:21:16,370 --> 00:21:18,420 That this string is taught. 343 00:21:18,420 --> 00:21:21,760 I said the assumption is it's massless 344 00:21:21,760 --> 00:21:24,900 so therefore it has to be taught. 345 00:21:24,900 --> 00:21:30,235 That's one of the reasons why we have to make a massless. 346 00:21:30,235 --> 00:21:32,860 If it wasn't massless, it could start getting kinks and I'll be 347 00:21:32,860 --> 00:21:34,957 doing this, but the massless one cannot, 348 00:21:34,957 --> 00:21:37,540 and that's a good question for you to think why that statement 349 00:21:37,540 --> 00:21:38,340 is correct. 350 00:21:38,340 --> 00:21:39,515 It's nontrivial. 351 00:21:39,515 --> 00:21:40,890 All right. 352 00:21:40,890 --> 00:21:46,230 Then at some instant of time, this angle I'll called theta. 353 00:21:46,230 --> 00:21:51,030 And now this is a problem in dynamics. 354 00:21:51,030 --> 00:21:55,490 Basically, I can draw my free body diagram, a force diagram. 355 00:21:55,490 --> 00:21:57,990 I have this force, mass m. 356 00:21:57,990 --> 00:21:59,470 What forces are acting on it? 357 00:21:59,470 --> 00:22:03,510 Well, there is gravity pulling it down, the force mg. 358 00:22:03,510 --> 00:22:06,170 There is a frictional force which 359 00:22:06,170 --> 00:22:12,360 we assume will be proportional to the velocity, 360 00:22:12,360 --> 00:22:13,920 so this is the frictional force which 361 00:22:13,920 --> 00:22:18,200 will be minus b times the velocity. 362 00:22:18,200 --> 00:22:21,230 There will be a tension in this string pulling it 363 00:22:21,230 --> 00:22:22,670 in that direction. 364 00:22:22,670 --> 00:22:26,760 And the sum of forces, by Newton's laws of motion, 365 00:22:26,760 --> 00:22:29,440 will give rise to the acceleration. 366 00:22:29,440 --> 00:22:32,450 The acceleration will be net force 367 00:22:32,450 --> 00:22:38,280 divided by C inertia, the system which is the mass. 368 00:22:38,280 --> 00:22:39,270 We have now experience. 369 00:22:39,270 --> 00:22:41,390 I'm not going to go slowly. 370 00:22:41,390 --> 00:22:45,380 I'll immediately take the components of forces. 371 00:22:45,380 --> 00:22:47,710 First let me consider vertical motion. 372 00:22:47,710 --> 00:22:50,610 I told you the angle is small. 373 00:22:50,610 --> 00:22:52,800 If I make the angle sufficiently small 374 00:22:52,800 --> 00:22:57,420 and that ball is oscillating, it is not moving up and down. 375 00:22:57,420 --> 00:23:00,770 The distance is tiny insignificant in this problem. 376 00:23:00,770 --> 00:23:03,790 So I can ignore the vertical motion. 377 00:23:03,790 --> 00:23:07,400 So the vertical acceleration is 0. 378 00:23:07,400 --> 00:23:11,880 The net vertical force here is the component 379 00:23:11,880 --> 00:23:14,680 of t in the up direction which is t cosine 380 00:23:14,680 --> 00:23:18,300 theta minus the gravity down, and that's 381 00:23:18,300 --> 00:23:20,250 equal to 0, no acceleration. 382 00:23:20,250 --> 00:23:22,120 Therefore t is mg. 383 00:23:22,120 --> 00:23:24,080 T is the tension in this string. 384 00:23:24,080 --> 00:23:25,650 How about horizontal? 385 00:23:25,650 --> 00:23:29,100 For horizontal motion, I have the horizontal component 386 00:23:29,100 --> 00:23:32,820 of this tension here which is t sine theta. 387 00:23:32,820 --> 00:23:36,430 This is theta, so t sine theta is the horizontal component. 388 00:23:36,430 --> 00:23:41,300 Now, this is moving here, therefore, 389 00:23:41,300 --> 00:23:44,580 there will be a frictional force proportional to that. 390 00:23:44,580 --> 00:23:49,520 So that's minus, this is the velocity, times b, 391 00:23:49,520 --> 00:23:50,860 so this is the frictional force. 392 00:23:50,860 --> 00:23:53,330 It's always minus, it's always in opposition 393 00:23:53,330 --> 00:23:55,280 to the actual velocity. 394 00:23:55,280 --> 00:23:58,800 And that by Newton's law is equal to mass 395 00:23:58,800 --> 00:24:01,900 times the horizontal acceleration. 396 00:24:01,900 --> 00:24:06,480 Beware of my symbols. 397 00:24:06,480 --> 00:24:12,060 This whole problem for me is really in one dimension. 398 00:24:12,060 --> 00:24:18,320 So I called x the horizontal motion of my hand, 399 00:24:18,320 --> 00:24:23,620 and I'm calling y the horizontal motion of the mass. 400 00:24:23,620 --> 00:24:26,400 This has nothing to do with x, y, z, one being 401 00:24:26,400 --> 00:24:27,930 up and the other horizontal. 402 00:24:27,930 --> 00:24:29,650 It's all horizontal. 403 00:24:29,650 --> 00:24:30,440 I had a choice. 404 00:24:30,440 --> 00:24:33,700 I could have called this x1 and this x2. 405 00:24:33,700 --> 00:24:35,940 I chose to call them x and y. 406 00:24:35,940 --> 00:24:38,822 You're smart enough to figure the difference, 407 00:24:38,822 --> 00:24:40,280 to follow that kind of terminology. 408 00:24:43,590 --> 00:24:47,660 So now I can rewrite this. 409 00:24:47,660 --> 00:24:53,570 I know what sine theta is if I look at this diagram. 410 00:24:53,570 --> 00:25:03,650 Sine theta is simply equal to the distance. 411 00:25:03,650 --> 00:25:05,740 It's x minus y. 412 00:25:05,740 --> 00:25:06,875 It's sine theta. 413 00:25:09,570 --> 00:25:13,290 This angle equals, I should draw this here. 414 00:25:13,290 --> 00:25:16,180 This angle equals that one, so sine theta 415 00:25:16,180 --> 00:25:19,750 is this distance divided by that length, 416 00:25:19,750 --> 00:25:23,740 and this distance is x minus y over l. 417 00:25:23,740 --> 00:25:27,530 So this is sine theta minus by equals this. 418 00:25:27,530 --> 00:25:30,140 Quickly, I can play with the algebra, 419 00:25:30,140 --> 00:25:32,040 and I end up with this equation. 420 00:25:32,040 --> 00:25:35,610 And lo and behold, Eureka, once again I 421 00:25:35,610 --> 00:25:40,230 get almost the same equation. 422 00:25:40,230 --> 00:25:43,860 I purposely chose it so it didn't come out exactly right. 423 00:25:43,860 --> 00:25:46,480 Before I ended up with a cosine, now I 424 00:25:46,480 --> 00:25:48,690 have ended up with a sine. 425 00:25:48,690 --> 00:25:49,470 Intention. 426 00:25:49,470 --> 00:25:50,190 All right. 427 00:25:50,190 --> 00:25:59,650 So to show that something which seems different often is not. 428 00:25:59,650 --> 00:26:04,020 So rewriting this in terms of constant, which I like, 429 00:26:04,020 --> 00:26:07,030 I do the same as before. 430 00:26:07,030 --> 00:26:12,310 I define this b over m as gamma, this g over l as omega 431 00:26:12,310 --> 00:26:17,720 squared, this thing here as f, and I end up 432 00:26:17,720 --> 00:26:23,150 with this equation where the constants are here 433 00:26:23,150 --> 00:26:27,940 and the initial conditions are that the position 434 00:26:27,940 --> 00:26:32,934 of this mass at time 0 is 0, and that the velocity was 0. 435 00:26:32,934 --> 00:26:35,100 You remember I considered the mass being stationary. 436 00:26:38,170 --> 00:26:43,010 This is now a set of mathematical equations. 437 00:26:43,010 --> 00:26:50,210 I have described this problem in terms of mathematics. 438 00:26:50,210 --> 00:26:53,720 From now on, I can forget that this 439 00:26:53,720 --> 00:26:57,210 has anything to do with that problem. 440 00:26:57,210 --> 00:26:59,650 I have to solve now a mathematical problem. 441 00:27:03,530 --> 00:27:05,360 Finally. 442 00:27:05,360 --> 00:27:07,070 One more. 443 00:27:07,070 --> 00:27:10,840 Let me consider one more problem, 444 00:27:10,840 --> 00:27:18,010 and once we've done all three, at least part 445 00:27:18,010 --> 00:27:25,820 of step 1 of all three we'll go and solve them all in one go. 446 00:27:25,820 --> 00:27:33,380 And I repeat what I've done here I've done all the physics. 447 00:27:33,380 --> 00:27:38,620 This is the part which causes most difficulty, 448 00:27:38,620 --> 00:27:42,740 needs most understanding of the physical situation, et cetera. 449 00:27:42,740 --> 00:27:46,720 It may or may not take a very long time. 450 00:27:46,720 --> 00:27:50,930 The next step is routine mathematics, 451 00:27:50,930 --> 00:27:54,044 which can be the part which takes you a whole night solving 452 00:27:54,044 --> 00:27:55,460 because you make algebraic errors, 453 00:27:55,460 --> 00:27:58,480 and you sweat and you go back, et cetera. 454 00:27:58,480 --> 00:28:02,080 That's the part you may be cursing when you're doing it, 455 00:28:02,080 --> 00:28:05,530 but it is routine from then on. 456 00:28:05,530 --> 00:28:09,890 It doesn't need an understanding of the physics 457 00:28:09,890 --> 00:28:13,590 of the situation, and what I'm trying to help you understand 458 00:28:13,590 --> 00:28:17,050 is physics, not mathematics at this stage. 459 00:28:17,050 --> 00:28:18,690 So let's take another problem, which 460 00:28:18,690 --> 00:28:22,100 again, seems completely different. 461 00:28:22,100 --> 00:28:23,270 All right. 462 00:28:23,270 --> 00:28:26,810 Here it's a seismograph, in other words, 463 00:28:26,810 --> 00:28:32,470 a device for detecting tremors of the Earth. 464 00:28:32,470 --> 00:28:34,640 And the following would work. 465 00:28:34,640 --> 00:28:39,910 Take attached to the floor a spring. 466 00:28:39,910 --> 00:28:46,470 Put the mass, m, on it, and just look at the mass 467 00:28:46,470 --> 00:28:48,380 how it's behaving. 468 00:28:48,380 --> 00:28:53,550 If the Earth starts oscillating, an earthquake, 469 00:28:53,550 --> 00:28:55,540 the mass will oscillate. 470 00:28:55,540 --> 00:28:58,130 How much it oscillates will give you 471 00:28:58,130 --> 00:29:01,630 a measure of how badly the Earth is oscillating. 472 00:29:01,630 --> 00:29:08,800 So I'm going to here do an idealized seismograph. 473 00:29:08,800 --> 00:29:17,280 I will model it by a mass, m, sitting on top of a massless, 474 00:29:17,280 --> 00:29:25,830 ideal spring, which is attached to the floor. 475 00:29:25,830 --> 00:29:29,190 I will idealize the earthquake and saying 476 00:29:29,190 --> 00:29:31,510 it's oscillating with a single frequency 477 00:29:31,510 --> 00:29:32,710 with a single amplitude. 478 00:29:32,710 --> 00:29:39,092 The floor is oscillating by an amplitude y0 sine omega t. 479 00:29:39,092 --> 00:29:39,800 It's oscillating. 480 00:29:42,550 --> 00:29:46,440 So my assumptions are this spring is ideal massless. 481 00:29:46,440 --> 00:29:49,480 The base, Hooke's law, that's the ideal part. 482 00:29:49,480 --> 00:29:53,570 I will again assume that there is a damping, which 483 00:29:53,570 --> 00:29:57,130 will be proportional to the velocity, for the same reasons 484 00:29:57,130 --> 00:30:00,560 as I took the other one to be proportional to velocity. 485 00:30:00,560 --> 00:30:03,050 And the question this time will be 486 00:30:03,050 --> 00:30:07,510 not to predict where that mass is at a given interval of time, 487 00:30:07,510 --> 00:30:12,330 but the question is, for what values of the mass 488 00:30:12,330 --> 00:30:14,960 and the spring constant, k, will I 489 00:30:14,960 --> 00:30:18,360 get the most sensitive instrument? 490 00:30:18,360 --> 00:30:20,560 So this seems completely different 491 00:30:20,560 --> 00:30:23,780 and the question seems different the other ones et cetera, 492 00:30:23,780 --> 00:30:25,880 but as you will see it all boils down 493 00:30:25,880 --> 00:30:29,170 to one and the same question. 494 00:30:29,170 --> 00:30:31,750 So how do we do this? 495 00:30:31,750 --> 00:30:35,860 So now let me come down to step 1, 496 00:30:35,860 --> 00:30:38,930 understanding the situation, the problem 497 00:30:38,930 --> 00:30:41,310 in terms of mathematics. 498 00:30:41,310 --> 00:30:42,890 All right. 499 00:30:42,890 --> 00:30:47,520 So I've re-drawn here the situation. 500 00:30:47,520 --> 00:30:49,940 This is my coordinate system of the problem. 501 00:30:52,930 --> 00:30:56,580 The surface of the floor, I'll assume it's here. 502 00:30:56,580 --> 00:30:58,080 It's moving. 503 00:30:58,080 --> 00:30:59,670 It is not a good reference place. 504 00:30:59,670 --> 00:31:00,570 It's moving. 505 00:31:00,570 --> 00:31:01,430 All right. 506 00:31:01,430 --> 00:31:03,590 So it's here. 507 00:31:03,590 --> 00:31:06,230 The mass, m, is there attached to the spring 508 00:31:06,230 --> 00:31:09,450 of natural length, l0, spring constant, k. 509 00:31:09,450 --> 00:31:12,420 There is the mass sitting on it. 510 00:31:12,420 --> 00:31:17,420 Now, we've got to take a coordinate system which 511 00:31:17,420 --> 00:31:18,090 isn't moving. 512 00:31:18,090 --> 00:31:20,430 Got to have an inertial one. 513 00:31:20,430 --> 00:31:22,690 The surface of the Earth is no good for this. 514 00:31:22,690 --> 00:31:24,500 It's vibrating. 515 00:31:24,500 --> 00:31:28,220 So the best thing we have are the fixed stars. 516 00:31:28,220 --> 00:31:32,750 So let's take the fixed stars as a reference. 517 00:31:32,750 --> 00:31:37,640 So this is a height in that room, which 518 00:31:37,640 --> 00:31:43,370 is fixed in position relative to the fixed stars. 519 00:31:43,370 --> 00:31:47,350 Now, the surface of the Earth can be moving relative to that, 520 00:31:47,350 --> 00:31:51,280 and I am making the assumption it's moving sinusoidally 521 00:31:51,280 --> 00:31:52,660 with amplitude y0. 522 00:31:52,660 --> 00:31:56,990 So this is the floor, position of the floor at time, t. 523 00:31:56,990 --> 00:32:05,430 The mass relative to this, which I defined as my y equal 0, 524 00:32:05,430 --> 00:32:07,025 is at position y of m. 525 00:32:09,970 --> 00:32:15,440 So this now is again a problem in dynamics, 526 00:32:15,440 --> 00:32:21,320 a mass with forces acting on it, what is the acceleration, 527 00:32:21,320 --> 00:32:24,280 and what is the solution of the dynamic equation, 528 00:32:24,280 --> 00:32:25,820 the equation of motion. 529 00:32:25,820 --> 00:32:30,040 So the force diagram or free body diagram 530 00:32:30,040 --> 00:32:36,120 is a point mass, m, with a force fg due to gravity on it, 531 00:32:36,120 --> 00:32:43,050 a force, f, due to the spring acting on it, 532 00:32:43,050 --> 00:32:45,740 and I forgot to put this. 533 00:32:45,740 --> 00:32:54,560 Of course, there will be a force here due to friction 534 00:32:54,560 --> 00:32:55,795 acting on it. 535 00:32:55,795 --> 00:32:57,670 This is probably not a good place to draw it. 536 00:32:57,670 --> 00:33:00,480 Let me draw it so it doesn't confuse you. 537 00:33:00,480 --> 00:33:06,570 I'll draw it next to this, friction. 538 00:33:06,570 --> 00:33:10,692 So these are in same direction et cetera. 539 00:33:10,692 --> 00:33:11,460 All right. 540 00:33:11,460 --> 00:33:16,110 So now let's use Newton's laws of motion, 541 00:33:16,110 --> 00:33:21,260 f equals ma, and relate the forces 542 00:33:21,260 --> 00:33:24,310 to the resulting acceleration. 543 00:33:24,310 --> 00:33:29,560 So first of all what is the net up-going force? 544 00:33:29,560 --> 00:33:32,980 I'm taking a coordinate system where this is a one dimensional 545 00:33:32,980 --> 00:33:35,320 problem, so I don't have to use vectors. 546 00:33:35,320 --> 00:33:37,380 I'll consider just the magnitude. 547 00:33:37,380 --> 00:33:42,430 My positive force is upwards. 548 00:33:42,430 --> 00:33:46,800 So at any instant of time, this mass 549 00:33:46,800 --> 00:33:50,410 will have a force due to the spring and it will depend. 550 00:33:50,410 --> 00:33:55,790 If this spring is compressed, the force will be up. 551 00:33:55,790 --> 00:34:00,530 If it's elongated, stretched, the force will be down. 552 00:34:00,530 --> 00:34:05,780 And the magnitude would be k times the change of length 553 00:34:05,780 --> 00:34:09,570 from the natural length, so the upward force 554 00:34:09,570 --> 00:34:14,190 will be the difference between those two subtracted 555 00:34:14,190 --> 00:34:16,909 from the natural length. 556 00:34:16,909 --> 00:34:21,449 So k times the natural length minus this quantity, which 557 00:34:21,449 --> 00:34:23,170 is the difference between this and this, 558 00:34:23,170 --> 00:34:28,900 which therefore this is how much the string is compressed, 559 00:34:28,900 --> 00:34:31,139 how much is compressed and is forced upwards. 560 00:34:31,139 --> 00:34:34,230 If this is a positive number, it's a negative one, 561 00:34:34,230 --> 00:34:36,580 it's stretched, it's putting it down. 562 00:34:36,580 --> 00:34:39,810 So this is the force up due to the spring, 563 00:34:39,810 --> 00:34:42,330 this is the force down due to gravity, 564 00:34:42,330 --> 00:34:45,723 and this is the force in opposition to the velocity. 565 00:34:45,723 --> 00:34:47,264 I don't know whether it's up or down. 566 00:34:47,264 --> 00:34:52,190 It depends which way it's moving due to the friction. 567 00:34:52,190 --> 00:34:54,239 This is the net force, and that's 568 00:34:54,239 --> 00:34:56,429 got to be equal to the inertia of the system, which 569 00:34:56,429 --> 00:34:59,220 is the mass times the acceleration of it 570 00:34:59,220 --> 00:35:01,570 to the upward direction. 571 00:35:01,570 --> 00:35:05,500 And I told you this spring is massless, et cetera, 572 00:35:05,500 --> 00:35:07,850 so I don't have to take into account 573 00:35:07,850 --> 00:35:09,020 the motion of the spring. 574 00:35:09,020 --> 00:35:11,395 And that's why I have to take an idealized situation. 575 00:35:13,910 --> 00:35:15,590 Back now to algebra. 576 00:35:15,590 --> 00:35:19,810 From here, I manipulate this, and I end up 577 00:35:19,810 --> 00:35:20,970 with this equation. 578 00:35:20,970 --> 00:35:25,920 You can do the same, and I've replaced 579 00:35:25,920 --> 00:35:30,790 for example, the motion of the floor by y0 sine omega t, 580 00:35:30,790 --> 00:35:33,690 et cetera, from here. 581 00:35:33,690 --> 00:35:35,850 And at first I say, oh, hell. 582 00:35:39,390 --> 00:35:45,220 This looks different than the one equation I've seen before. 583 00:35:45,220 --> 00:35:50,450 But then I don't have to be an Einstein to realize, hold on. 584 00:35:50,450 --> 00:36:00,140 If I redefine this as my variable, in other words, 585 00:36:00,140 --> 00:36:08,460 if I take define y as ym plus mg over k minus l0, this quantity, 586 00:36:08,460 --> 00:36:13,310 the rate of change of y is the rate of change of ym 587 00:36:13,310 --> 00:36:17,350 because these are constants, and the second derivative of y 588 00:36:17,350 --> 00:36:20,710 is the second derivative of this, is a constant. 589 00:36:20,710 --> 00:36:23,360 And therefore, I can rewrite the equation 590 00:36:23,360 --> 00:36:28,350 as well y double dot plus gamma y dot plus omega 0 591 00:36:28,350 --> 00:36:33,750 squared y is equal to a constant and sine omega t 592 00:36:33,750 --> 00:36:40,700 where these constants are written here from this equation 593 00:36:40,700 --> 00:36:43,590 and I redefine my variable. 594 00:36:43,590 --> 00:36:47,720 And now, Eureka, I'm back again to the same problem. 595 00:36:47,720 --> 00:36:50,100 So once, again, what we've done. 596 00:36:50,100 --> 00:36:54,470 We've taken a physical situation and a physical question 597 00:36:54,470 --> 00:36:57,210 from our understanding of the laws of nature 598 00:36:57,210 --> 00:36:59,840 which apply to that situation. 599 00:36:59,840 --> 00:37:04,580 We have translated it to a problem in mathematics. 600 00:37:04,580 --> 00:37:06,200 And here it is. 601 00:37:06,200 --> 00:37:10,110 That's the problem, and I can now, 602 00:37:10,110 --> 00:37:12,580 at least for the next few minutes, 603 00:37:12,580 --> 00:37:16,040 forget that this is physics, treat it 604 00:37:16,040 --> 00:37:18,060 as a problem in mathematics. 605 00:37:18,060 --> 00:37:20,190 And that's what I'll do it now, but first I've 606 00:37:20,190 --> 00:37:22,120 got to make some room for myself. 607 00:37:22,120 --> 00:37:25,570 I have nowhere to write, so I have to erase some things, 608 00:37:25,570 --> 00:37:29,440 and we'll continue in a few minutes. 609 00:37:29,440 --> 00:37:30,270 All right. 610 00:37:30,270 --> 00:37:34,720 We've erased the boards, and we can therefore continue. 611 00:37:34,720 --> 00:37:37,450 Let me remind you what we have shown 612 00:37:37,450 --> 00:37:48,140 is we've taken three completely different systems, described 613 00:37:48,140 --> 00:37:51,440 them in terms of mathematics. 614 00:37:51,440 --> 00:37:53,800 The three systems with an electrical circuit, 615 00:37:53,800 --> 00:37:56,600 an RLC circuit. 616 00:37:56,600 --> 00:38:03,240 We took a pendulum, driven on a pendulum, and a seismograph. 617 00:38:03,240 --> 00:38:11,175 Each one of these ended up with almost identical equation. 618 00:38:11,175 --> 00:38:15,110 In fact, the only difference is one of them 619 00:38:15,110 --> 00:38:20,380 ended up with a cosine here, the other two with a sine. 620 00:38:20,380 --> 00:38:24,520 I will comment about that in a second. 621 00:38:24,520 --> 00:38:28,720 We are now, forgetting about the physics, 622 00:38:28,720 --> 00:38:31,690 we are now in the world of mathematics. 623 00:38:31,690 --> 00:38:36,110 These equations of motion need to be 624 00:38:36,110 --> 00:38:38,460 solved if we are to predict what will 625 00:38:38,460 --> 00:38:41,270 happen in those situations. 626 00:38:41,270 --> 00:38:46,300 Now, there are many ways of solving differential equations. 627 00:38:46,300 --> 00:38:51,810 It is not my job to teach you how 628 00:38:51,810 --> 00:38:55,360 to solve differential equations. 629 00:38:55,360 --> 00:38:58,470 For an example, you can go to the lectures of Professor 630 00:38:58,470 --> 00:39:02,350 Walter Lewin, and he does illustrate 631 00:39:02,350 --> 00:39:05,790 how one can solve these using complex amplitudes. 632 00:39:08,740 --> 00:39:14,060 I am satisfied if I find the solutions. 633 00:39:14,060 --> 00:39:15,530 I don't care how. 634 00:39:15,530 --> 00:39:17,880 I can take them from what he did, 635 00:39:17,880 --> 00:39:21,190 I can take them from a book of mathematics, et cetera, 636 00:39:21,190 --> 00:39:25,070 and what you will find is that if you 637 00:39:25,070 --> 00:39:29,490 have a generic second ordered differential 638 00:39:29,490 --> 00:39:35,960 equation of this form, it has solution 639 00:39:35,960 --> 00:39:40,030 which looks like this, the most general solution that 640 00:39:40,030 --> 00:39:41,800 looks like that. 641 00:39:41,800 --> 00:39:46,000 There is no other solution in the universe which 642 00:39:46,000 --> 00:39:48,690 is not represented by this equation. 643 00:39:48,690 --> 00:39:51,250 You can manipulate it, make it look different, 644 00:39:51,250 --> 00:39:53,720 but it will boil down to this equation. 645 00:39:53,720 --> 00:39:55,500 How do I know that? 646 00:39:55,500 --> 00:39:57,250 Because of something I mentioned earlier, 647 00:39:57,250 --> 00:40:00,180 the so-called uniqueness theorem. 648 00:40:00,180 --> 00:40:05,240 This is a second order differential equation. 649 00:40:05,240 --> 00:40:09,110 This equation satisfies it. 650 00:40:09,110 --> 00:40:11,740 Try it, take it, differentiate it twice this, 651 00:40:11,740 --> 00:40:16,590 et cetera, u [? dissatisfied. ?] Furthermore, 652 00:40:16,590 --> 00:40:19,030 it has two arbitrary constants. 653 00:40:22,650 --> 00:40:27,530 This C can be anything and this phi can be anything. 654 00:40:27,530 --> 00:40:29,210 They are arbitrary. 655 00:40:29,210 --> 00:40:34,560 Everything else in this equation is completely determined. 656 00:40:34,560 --> 00:40:41,490 For example, a of omega depends on omega. 657 00:40:41,490 --> 00:40:48,450 This amplitude, a, is equal to f divided 658 00:40:48,450 --> 00:40:52,460 by the square root of omega 0 squared minus omega squared 659 00:40:52,460 --> 00:40:56,760 0 squared plus gamma omega 0 squared. 660 00:40:56,760 --> 00:41:02,310 All these constants-- omega, gamma, f, omega 0-- 661 00:41:02,310 --> 00:41:04,285 are known in each of the problems. 662 00:41:07,270 --> 00:41:10,630 Let me just as an example take the last problem we 663 00:41:10,630 --> 00:41:15,310 did When we derived the equation of motion, 664 00:41:15,310 --> 00:41:20,090 we furthermore knew exactly what each one of those constants 665 00:41:20,090 --> 00:41:20,630 meant. 666 00:41:20,630 --> 00:41:24,140 They are known, so I don't need to solve for them. 667 00:41:24,140 --> 00:41:28,500 So a of omega is this quantity. 668 00:41:28,500 --> 00:41:33,330 The delta, the phase there is, again, a known quantity. 669 00:41:33,330 --> 00:41:35,800 The time of delta is this. 670 00:41:35,800 --> 00:41:38,965 It is only under these conditions 671 00:41:38,965 --> 00:41:43,090 that this satisfies that equation. 672 00:41:43,090 --> 00:41:46,690 And find the omega prime here also is a known quantity. 673 00:41:46,690 --> 00:41:48,400 Nothing here is unknown. 674 00:41:48,400 --> 00:41:51,660 The only unknowns are the C, and the phi, 675 00:41:51,660 --> 00:41:54,220 and they are arbitrary constants. 676 00:41:54,220 --> 00:41:57,790 This works for any value of C and any value of phi. 677 00:42:02,340 --> 00:42:05,760 Now, one more comment. 678 00:42:05,760 --> 00:42:08,080 These look equations. 679 00:42:08,080 --> 00:42:11,000 One looks like a cosine theta sine. 680 00:42:11,000 --> 00:42:13,310 If you stop and think for a second, 681 00:42:13,310 --> 00:42:16,670 they're actually one and the same equation 682 00:42:16,670 --> 00:42:20,840 where you've redefined what is t 0. 683 00:42:20,840 --> 00:42:26,015 So by just changing the phase of this, you'll get to this. 684 00:42:26,015 --> 00:42:28,040 You've changed it by 90 degrees. 685 00:42:28,040 --> 00:42:31,610 Just redefine what t0 is, your clock, 686 00:42:31,610 --> 00:42:34,780 and you'll get from this to that. 687 00:42:34,780 --> 00:42:39,150 So if this equation I've written here 688 00:42:39,150 --> 00:42:44,380 is a solution to the cosine part, 689 00:42:44,380 --> 00:42:46,850 if my driver has this phase, in other words, 690 00:42:46,850 --> 00:42:51,400 sine omega t, then the answer will be displaced with respect 691 00:42:51,400 --> 00:42:53,290 to this one by that same thing. 692 00:42:53,290 --> 00:42:56,740 Instead of having a cosine here, you have a sign. 693 00:42:56,740 --> 00:43:01,570 So from my perspective, these two are identical equations, 694 00:43:01,570 --> 00:43:06,140 and the solution for them is identical. 695 00:43:06,140 --> 00:43:10,420 It's also worth looking at what the mathematics of this looks 696 00:43:10,420 --> 00:43:11,070 like. 697 00:43:11,070 --> 00:43:15,840 What this looks like is the solution to these equations 698 00:43:15,840 --> 00:43:22,460 is some function with a certain amplitude, which is oscillatory 699 00:43:22,460 --> 00:43:28,350 with the same frequency as the driver. 700 00:43:28,350 --> 00:43:30,760 So this is just an oscillatory function. 701 00:43:30,760 --> 00:43:32,870 It's got a different phase to that. 702 00:43:32,870 --> 00:43:35,330 They lag will lead it, different phase, 703 00:43:35,330 --> 00:43:39,740 but it's just constant amplitude oscillatory function. 704 00:43:39,740 --> 00:43:45,270 Added to it is another oscillatory function 705 00:43:45,270 --> 00:43:49,870 with a slightly different frequency, 706 00:43:49,870 --> 00:43:55,920 but which with time decays, because of this e to the minus 707 00:43:55,920 --> 00:43:57,910 gamma over 2t. 708 00:43:57,910 --> 00:44:03,240 And by the way, one more thing I should just emphasize. 709 00:44:03,240 --> 00:44:07,400 In writing this solution, I've made the assumption 710 00:44:07,400 --> 00:44:09,510 that this oscillator does oscil. 711 00:44:09,510 --> 00:44:12,775 You're driving something which oscillates and is 712 00:44:12,775 --> 00:44:18,230 an overdamped system in which case 713 00:44:18,230 --> 00:44:22,560 this term here would be different and not interesting 714 00:44:22,560 --> 00:44:25,730 when normally you wouldn't drive things and be interested in how 715 00:44:25,730 --> 00:44:30,540 it responds if it's incredibly damped system. 716 00:44:30,540 --> 00:44:33,820 So I am making the assumption that this 717 00:44:33,820 --> 00:44:38,110 is an underdamped system, which means that this quantity here 718 00:44:38,110 --> 00:44:42,420 is greater than this quantity here. 719 00:44:42,420 --> 00:44:46,000 And if you're having difficulty with this part, 720 00:44:46,000 --> 00:44:54,080 I urge you to review what happens with just not driven 721 00:44:54,080 --> 00:44:57,860 oscillators under conditions of being underdamped 722 00:44:57,860 --> 00:45:00,875 or overdamped. 723 00:45:00,875 --> 00:45:01,875 That's the same physics. 724 00:45:04,860 --> 00:45:08,000 So we found the most general solution. 725 00:45:08,000 --> 00:45:12,100 This solution must apply to every problem we've done. 726 00:45:12,100 --> 00:45:15,150 So I'll now take one of them. 727 00:45:15,150 --> 00:45:19,130 For example, let's consider this problem. 728 00:45:19,130 --> 00:45:19,950 We did this. 729 00:45:19,950 --> 00:45:22,170 We have an RLC circuit. 730 00:45:22,170 --> 00:45:24,890 They are an equation of motion. 731 00:45:24,890 --> 00:45:28,300 From here, I can immediately write 732 00:45:28,300 --> 00:45:30,930 what is the solution of this. 733 00:45:30,930 --> 00:45:37,060 As a function of time, the charge Q 734 00:45:37,060 --> 00:45:47,650 will be equal to-- just look at that-- a of omega times 735 00:45:47,650 --> 00:45:58,990 cosine omega t minus delta-- I'm just copying from over there-- 736 00:45:58,990 --> 00:46:09,640 plus some arbitrary constant e to the minus gamma over 2t. 737 00:46:09,640 --> 00:46:16,710 And here we have a cosine, so I took the cosine here, and here 738 00:46:16,710 --> 00:46:26,850 is cosine of omega prime t plus this arbitrary phase phi. 739 00:46:26,850 --> 00:46:31,460 So this is a general equation, which 740 00:46:31,460 --> 00:46:38,820 satisfies that one at all times. 741 00:46:38,820 --> 00:46:41,310 And I'm repeating myself. 742 00:46:41,310 --> 00:46:51,450 This, you know, it will be given by A of omega 743 00:46:51,450 --> 00:46:55,030 will be equal to f, and what is f? 744 00:46:55,030 --> 00:47:02,950 f is v0 over L-- I just read it off there-- divided 745 00:47:02,950 --> 00:47:10,470 by square root of omega 0 squared, 746 00:47:10,470 --> 00:47:22,490 which is 1 over LC minus omega squared 747 00:47:22,490 --> 00:47:28,790 That's just the frequencies with which the voltage is driving 748 00:47:28,790 --> 00:47:37,260 there all squared plus gamma omega. 749 00:47:37,260 --> 00:47:39,277 Gamma is R over L omega all squared. 750 00:47:44,840 --> 00:47:52,530 And time delta is equal to gamma omega. 751 00:47:52,530 --> 00:47:59,780 Gamma is R over L divided by omega 0 752 00:47:59,780 --> 00:48:06,070 squared, which is 1 over LC minus omega squared. 753 00:48:11,560 --> 00:48:17,750 So far we know everything except the C and the phi. 754 00:48:17,750 --> 00:48:22,710 The question actually was, I've now erased, 755 00:48:22,710 --> 00:48:27,030 it but the question was what is the current flowing? 756 00:48:27,030 --> 00:48:33,780 And the current flowing I of t, is 757 00:48:33,780 --> 00:48:39,560 equal to dq dt with the way we define 758 00:48:39,560 --> 00:48:48,100 the signs with a part plus here, which is equal to Q dot, which 759 00:48:48,100 --> 00:48:52,520 I can get from this by differentiating this equation, 760 00:48:52,520 --> 00:48:57,630 and I get differentiate this minus omega 761 00:48:57,630 --> 00:49:08,930 A of omega sine of omega t minus delta. 762 00:49:08,930 --> 00:49:13,720 Go to differentiate this, I get minus gamma 763 00:49:13,720 --> 00:49:25,220 over 2 Ce to the minus gamma over 2t times cosine omega 764 00:49:25,220 --> 00:49:29,090 prime t plus phi. 765 00:49:29,090 --> 00:49:31,350 So I've differentiated this times that. 766 00:49:31,350 --> 00:49:34,220 Now I've got to take this differentiated 767 00:49:34,220 --> 00:49:41,570 by that, which minus Ce e to the minus gamma over 2t times 768 00:49:41,570 --> 00:49:52,291 omega prime times sine omega prime t plus phi. 769 00:49:52,291 --> 00:49:52,790 Sorry. 770 00:49:52,790 --> 00:49:55,430 That's so. 771 00:49:55,430 --> 00:50:00,660 So and again, we know every quantity 772 00:50:00,660 --> 00:50:09,030 here except for C and phi. 773 00:50:09,030 --> 00:50:10,065 Those are unknowns. 774 00:50:14,340 --> 00:50:17,640 So this would be the answer to my question 775 00:50:17,640 --> 00:50:20,970 if I knew C and phi, then everything is known. 776 00:50:20,970 --> 00:50:22,960 Then I could predict exactly what 777 00:50:22,960 --> 00:50:25,320 would happen with any instant of time. 778 00:50:25,320 --> 00:50:27,390 How do I find those two? 779 00:50:27,390 --> 00:50:31,190 This is where the boundary conditions come in. 780 00:50:31,190 --> 00:50:33,820 That's why you need at least two boundary 781 00:50:33,820 --> 00:50:36,850 conditions or initial conditions, 782 00:50:36,850 --> 00:50:41,080 two facts about this system to be able to determine 783 00:50:41,080 --> 00:50:43,300 those constants, and we're always 784 00:50:43,300 --> 00:50:47,610 in the physical situation of this kind, 785 00:50:47,610 --> 00:50:49,340 you will have two bits of information. 786 00:50:49,340 --> 00:50:53,490 For example, in this case, I told you 787 00:50:53,490 --> 00:50:55,650 that we will assume that t equals 788 00:50:55,650 --> 00:50:59,340 0 the charge and the capacitance is 0 789 00:50:59,340 --> 00:51:01,972 and the current in the circuit is 0. 790 00:51:01,972 --> 00:51:04,180 I could have told you something else at the beginning 791 00:51:04,180 --> 00:51:06,350 and the answer would be different. 792 00:51:06,350 --> 00:51:07,170 All right. 793 00:51:07,170 --> 00:51:09,310 And the thing that would be different 794 00:51:09,310 --> 00:51:11,050 would be those constants. 795 00:51:11,050 --> 00:51:13,270 So how do we do it? 796 00:51:13,270 --> 00:51:22,520 Simply by taking this equation and writing it at the time t 797 00:51:22,520 --> 00:51:23,380 equals 0. 798 00:51:23,380 --> 00:51:31,510 So at t equals 0, we know what Q of 0 is, 799 00:51:31,510 --> 00:51:35,170 and we know what Q dot of 0 is. 800 00:51:39,290 --> 00:51:45,340 So I take this equation, set t to 0, 801 00:51:45,340 --> 00:51:49,270 and write down this equal to 0. 802 00:51:53,000 --> 00:51:53,500 Sorry. 803 00:51:53,500 --> 00:51:55,965 I take this equation, the Q. This one set 804 00:51:55,965 --> 00:52:01,740 t to 0 and write 0 equals that with t equals 0. 805 00:52:01,740 --> 00:52:04,670 I write another equation from here. 806 00:52:04,670 --> 00:52:11,790 At t equals 0, Q dot is 0, so I take this one, set t to 0, 807 00:52:11,790 --> 00:52:13,510 and write another equation. 808 00:52:13,510 --> 00:52:17,680 I will end up with two algebraic equations 809 00:52:17,680 --> 00:52:21,430 with two unknowns which you can solve. 810 00:52:21,430 --> 00:52:26,660 Now remember I told you, the hard part conceptually 811 00:52:26,660 --> 00:52:32,280 is getting the equations of motion right. 812 00:52:32,280 --> 00:52:38,710 But the slog, the sweat, and a hard labor 813 00:52:38,710 --> 00:52:44,320 is getting this solution right, and I'm not 814 00:52:44,320 --> 00:52:48,760 going to go here now and embarrass myself making error 815 00:52:48,760 --> 00:52:53,110 after algebraic error solving those equations. 816 00:52:53,110 --> 00:52:55,920 I know you can solve them if you keep your cool 817 00:52:55,920 --> 00:52:59,020 and follow all the constants, et cetera. 818 00:52:59,020 --> 00:53:04,070 You take these two algebraic equations, solve them, 819 00:53:04,070 --> 00:53:10,340 you come out with a value for C and a value for the phase phi, 820 00:53:10,340 --> 00:53:12,250 and then I have the final answer. 821 00:53:12,250 --> 00:53:19,350 I then have this equation I of t and Q of t for any T. 822 00:53:19,350 --> 00:53:24,310 And so I can predict at any time with this will do. 823 00:53:24,310 --> 00:53:27,200 End of story. 824 00:53:27,200 --> 00:53:33,204 Now, let me take another case and to look at this one here, 825 00:53:33,204 --> 00:53:34,370 and I'll go a little faster. 826 00:53:38,110 --> 00:53:45,380 This is the mathematical description of the pendulum. 827 00:53:45,380 --> 00:53:49,880 So again, I'll take this, initially I'll 828 00:53:49,880 --> 00:53:59,110 do the same as I did before, so I will write that y at time t 829 00:53:59,110 --> 00:54:03,650 is equal to-- and I have to once again just follow 830 00:54:03,650 --> 00:54:12,320 the thing-- it's f, which is gx over L divided 831 00:54:12,320 --> 00:54:21,140 by the square root of omega squared, 832 00:54:21,140 --> 00:54:35,160 which is g over L minus omega squared all squared plus gamma, 833 00:54:35,160 --> 00:54:40,460 which is b over m this time gamma omega all squared. 834 00:54:46,100 --> 00:54:49,010 So times. 835 00:54:49,010 --> 00:54:52,210 Now, here I have a sign so it's going 836 00:54:52,210 --> 00:55:08,650 to be the sine omega t minus delta plus-- 837 00:55:08,650 --> 00:55:16,067 and to save time, not rewrite it, the full thing, 838 00:55:16,067 --> 00:55:17,400 but I'll start at the beginning. 839 00:55:17,400 --> 00:55:23,190 There's the Ce to the minus gamma over 2t 840 00:55:23,190 --> 00:55:30,220 here at cosine et cetera. 841 00:55:30,220 --> 00:55:36,550 So what we see here as before, we 842 00:55:36,550 --> 00:55:42,920 can get what this mass will do as a function of time, 843 00:55:42,920 --> 00:55:48,600 and as before, I don't want to just go around and around 844 00:55:48,600 --> 00:55:50,370 repeating the thing is. 845 00:55:50,370 --> 00:55:55,590 In order to solve, everything here is known except the C 846 00:55:55,590 --> 00:56:02,070 and this phase here-- maybe I should at least write that in. 847 00:56:02,070 --> 00:56:07,780 That is omega prime t plus phi. 848 00:56:07,780 --> 00:56:11,350 And in principle, this and that you 849 00:56:11,350 --> 00:56:16,570 can get, again, by using the initial conditions. 850 00:56:16,570 --> 00:56:17,930 I don't want to focus on that. 851 00:56:17,930 --> 00:56:21,630 In this case, I want to focus what the problem said. 852 00:56:21,630 --> 00:56:25,410 What it said it was, discuss what 853 00:56:25,410 --> 00:56:29,030 you would expect the motion to be for that. 854 00:56:29,030 --> 00:56:32,030 In other words, when we took this special, 855 00:56:32,030 --> 00:56:40,530 we took this and took in given initial conditions 856 00:56:40,530 --> 00:56:42,720 and let's go. 857 00:56:42,720 --> 00:56:46,280 This will describe what goes on. 858 00:56:46,280 --> 00:56:51,850 And let's look at this solution. 859 00:56:51,850 --> 00:56:58,660 What we see is that there are two terms here as before. 860 00:56:58,660 --> 00:57:05,890 This is an oscillation of the same frequency as a driver. 861 00:57:05,890 --> 00:57:09,850 So other words, that term is something 862 00:57:09,850 --> 00:57:17,050 which has this moving with the same frequency as my hand. 863 00:57:17,050 --> 00:57:21,910 This is a slightly different frequency, 864 00:57:21,910 --> 00:57:26,200 which is exponentially decaying. 865 00:57:26,200 --> 00:57:30,090 When you add two oscillating functions, 866 00:57:30,090 --> 00:57:33,150 you've learned from Professor Walter Lewin's class, 867 00:57:33,150 --> 00:57:35,130 you get beats. 868 00:57:35,130 --> 00:57:42,820 So at first when this is large, these two wave motions, 869 00:57:42,820 --> 00:57:44,720 these oscillatory motions will beat 870 00:57:44,720 --> 00:57:47,365 minutes one against the other and you 871 00:57:47,365 --> 00:57:50,450 will get some kind of a crazy motion. 872 00:57:50,450 --> 00:57:52,850 You can't make sense of it. 873 00:57:52,850 --> 00:57:57,840 As a function of time, as t becomes bigger and bigger, 874 00:57:57,840 --> 00:58:03,410 this exponential kills this term, and after a long time, 875 00:58:03,410 --> 00:58:07,670 you'll end up with only this term, which 876 00:58:07,670 --> 00:58:09,970 is simple sinusoidal motion. 877 00:58:09,970 --> 00:58:12,550 Try it for yourself. 878 00:58:12,550 --> 00:58:15,900 If I start any old way and I then 879 00:58:15,900 --> 00:58:19,000 move with a given frequency, you'll 880 00:58:19,000 --> 00:58:25,340 find this has a constant frequency, some amplitude given 881 00:58:25,340 --> 00:58:29,860 by this, and it continues doing this forever 882 00:58:29,860 --> 00:58:32,740 consistent with it. 883 00:58:32,740 --> 00:58:35,810 But there is another interesting feature, 884 00:58:35,810 --> 00:58:38,350 which I would like you to focus on. 885 00:58:38,350 --> 00:58:44,630 If the friction is very small, like this, this 886 00:58:44,630 --> 00:58:49,980 will have very little damping, after a long time, 887 00:58:49,980 --> 00:58:54,470 this has gone and this term I can neglect. 888 00:58:54,470 --> 00:59:01,040 And what I see here is that this, 889 00:59:01,040 --> 00:59:07,050 the amplitude of the oscillations of this not only 890 00:59:07,050 --> 00:59:12,680 depends on the amplitude-- and again, I notice the time here; 891 00:59:12,680 --> 00:59:18,150 this is the node quantity here-- not only depends 892 00:59:18,150 --> 00:59:24,140 on the amplitude of how far my head is moving, 893 00:59:24,140 --> 00:59:28,900 but it depends critically on the difference between g 894 00:59:28,900 --> 00:59:31,920 over l and omega squared. 895 00:59:31,920 --> 00:59:37,850 This, if you remember from the simple pendulum-- that's 896 00:59:37,850 --> 00:59:44,150 this omega 0 squared-- gives the frequency 897 00:59:44,150 --> 00:59:49,400 of oscillation of a free pendulum of this length. 898 00:59:49,400 --> 00:59:56,710 So what happens depends critically how different 899 00:59:56,710 --> 00:59:59,950 and of which magnitude is between the frequency 900 00:59:59,950 --> 01:00:02,840 of oscillation of my hand compared 901 01:00:02,840 --> 01:00:06,820 to the natural frequency of oscillation. 902 01:00:06,820 --> 01:00:11,110 If-- and here I won't do the algebra. 903 01:00:11,110 --> 01:00:13,590 I'll let you do this as an exercise-- play 904 01:00:13,590 --> 01:00:19,280 for yourself changing the value of this making 905 01:00:19,280 --> 01:00:22,030 it bigger than that or smaller than that. 906 01:00:24,710 --> 01:00:35,250 And what you will find is that the motion of that and sorry, 907 01:00:35,250 --> 01:00:37,180 there's one more thing I should have 908 01:00:37,180 --> 01:00:40,940 said I didn't, and let me just say it now. 909 01:00:40,940 --> 01:00:44,590 There is also this question of this phase delta, 910 01:00:44,590 --> 01:00:53,920 tan delta is equal to gamma omega 911 01:00:53,920 --> 01:01:01,900 over omega 0 squared minus omega squared. 912 01:01:01,900 --> 01:01:09,350 So the sine of delta depends on whether it's bigger or smaller. 913 01:01:09,350 --> 01:01:13,670 Play for yourself changing this value. 914 01:01:13,670 --> 01:01:16,960 See what happens to the amplitude here 915 01:01:16,960 --> 01:01:21,730 and what happens to this phase, and what you'll see 916 01:01:21,730 --> 01:01:28,880 is that if I go slowly, the phase is such 917 01:01:28,880 --> 01:01:31,830 that the two go in phase together 918 01:01:31,830 --> 01:01:36,570 if my oscillating frequency is lower 919 01:01:36,570 --> 01:01:40,020 than the natural frequency of oscillation of this, 920 01:01:40,020 --> 01:01:44,480 while if I go faster, in other words, higher frequency, 921 01:01:44,480 --> 01:01:49,790 this becomes a negative and the phase gets out 922 01:01:49,790 --> 01:01:53,610 of phase by 180 degrees, the two. 923 01:01:53,610 --> 01:01:57,710 Furthermore, if you look at this amplitude, 924 01:01:57,710 --> 01:02:01,290 you can completely understand it in the following way. 925 01:02:06,130 --> 01:02:10,490 Imagine that I had this string and I move it slowly. 926 01:02:15,100 --> 01:02:18,910 I'm moving this slowly, and what you'll see 927 01:02:18,910 --> 01:02:29,750 is it's going in phase with that, and the angle of this 928 01:02:29,750 --> 01:02:34,890 will be such that this string appears 929 01:02:34,890 --> 01:02:40,180 to be the continuation of a single string, which 930 01:02:40,180 --> 01:02:41,275 has a longer length. 931 01:02:44,060 --> 01:02:48,760 And the string with the longer length has a lower frequency. 932 01:02:48,760 --> 01:02:53,410 So if I move here with the lower frequency 933 01:02:53,410 --> 01:02:56,360 than the natural frequency, this will 934 01:02:56,360 --> 01:03:02,810 appear to be part of a string which has just the right length 935 01:03:02,810 --> 01:03:07,970 to give me the frequency of my hand. 936 01:03:07,970 --> 01:03:12,010 While if I move with the higher frequency, 937 01:03:12,010 --> 01:03:16,700 what the string will do is do this. 938 01:03:16,700 --> 01:03:19,580 Here is my hand moving backwards and forwards. 939 01:03:19,580 --> 01:03:21,290 There is the mass, and the mass is 940 01:03:21,290 --> 01:03:23,060 moving backwards and forwards. 941 01:03:23,060 --> 01:03:24,210 All right. 942 01:03:24,210 --> 01:03:30,200 This string is oscillating as if it 943 01:03:30,200 --> 01:03:35,120 was a free string of a shorter length. 944 01:03:35,120 --> 01:03:39,880 I urge you as an exercise, to play with the numbers 945 01:03:39,880 --> 01:03:45,540 here and consider these two cases. 946 01:03:45,540 --> 01:03:48,790 It is a beautiful, educational tool 947 01:03:48,790 --> 01:03:53,720 to understand why if you have a driven system, 948 01:03:53,720 --> 01:03:57,740 response can be in phase or out of phase 949 01:03:57,740 --> 01:04:01,430 depending on the relative frequency you're driving 950 01:04:01,430 --> 01:04:04,380 the system compared to the natural frequency 951 01:04:04,380 --> 01:04:07,170 of oscillation of the oscillator. 952 01:04:07,170 --> 01:04:11,810 And this example brings everything out. 953 01:04:11,810 --> 01:04:15,010 Take this example, consider very little 954 01:04:15,010 --> 01:04:19,120 damping so that you can do the calculations easily. 955 01:04:19,120 --> 01:04:21,170 See what's the relative amplitude 956 01:04:21,170 --> 01:04:25,010 of the driver compared to the response 957 01:04:25,010 --> 01:04:28,350 and what's the relative phase of the driver compared 958 01:04:28,350 --> 01:04:32,110 to the response, and all this characteristics you'll 959 01:04:32,110 --> 01:04:35,720 see which you can play for yourselves with a little model. 960 01:04:35,720 --> 01:04:38,490 I really urge you to do that. 961 01:04:38,490 --> 01:04:43,570 Finally today, I will just comment. 962 01:04:43,570 --> 01:04:46,410 You see we're now in a world of repeating. 963 01:04:46,410 --> 01:04:49,260 We're doing the same thing over and over again. 964 01:04:49,260 --> 01:05:03,680 I would just like to comment about the seismograph. 965 01:05:03,680 --> 01:05:11,580 Once again, this is an equation of this form. 966 01:05:11,580 --> 01:05:13,620 It's the sine version. 967 01:05:13,620 --> 01:05:18,160 So this is the solution to it with a sine, 968 01:05:18,160 --> 01:05:23,510 and the question is, in this case was, 969 01:05:23,510 --> 01:05:30,290 what the value of k over m makes it most sensitive? 970 01:05:30,290 --> 01:05:34,130 That's equivalent to asking the question, 971 01:05:34,130 --> 01:05:39,460 after a long time-- this earthquake 972 01:05:39,460 --> 01:05:45,930 has lasted a long time-- this so-called transient, 973 01:05:45,930 --> 01:05:48,770 this term dies away. 974 01:05:48,770 --> 01:05:50,290 So this we can forget. 975 01:05:50,290 --> 01:05:56,360 This is the response of the mass to the earthquake which 976 01:05:56,360 --> 01:06:05,540 we assumed it's a uniform frequency, sinusoidal driver. 977 01:06:05,540 --> 01:06:10,640 And so the question is, what is the amplitude of the response? 978 01:06:10,640 --> 01:06:14,080 How big is this term? 979 01:06:14,080 --> 01:06:19,920 This term we know how big it is, and if you 980 01:06:19,920 --> 01:06:26,080 look at this function, you see that this amplitude depends 981 01:06:26,080 --> 01:06:27,570 on omega, on the driver. 982 01:06:30,660 --> 01:06:38,860 Now let's assume the damping is not very big. 983 01:06:38,860 --> 01:06:41,790 This will be small compared to that. 984 01:06:41,790 --> 01:06:45,600 And so if you want a big response, 985 01:06:45,600 --> 01:06:49,330 you want this quantity to be very small 986 01:06:49,330 --> 01:06:53,680 so that f divided by this is a big number. 987 01:06:53,680 --> 01:06:56,610 When will this be small? 988 01:06:56,610 --> 01:06:59,500 When these two are almost equal. 989 01:07:02,180 --> 01:07:07,090 So what you want, if you want to have an instrument which 990 01:07:07,090 --> 01:07:16,930 is very sensitive, you want to study so the dominant frequency 991 01:07:16,930 --> 01:07:20,630 of oscillations of earthquakes in your area 992 01:07:20,630 --> 01:07:27,870 and adjust the natural frequency of oscillation close to that. 993 01:07:27,870 --> 01:07:32,470 And the natural frequency is of course given by k over m. 994 01:07:32,470 --> 01:07:38,490 So what matters is, it doesn't matter what m is, what k is, 995 01:07:38,490 --> 01:07:41,870 but what's the ratio of k over m? 996 01:07:41,870 --> 01:07:46,910 You want that to be close to the natural frequency, 997 01:07:46,910 --> 01:07:49,920 that the natural frequency of this oscillator, that value 998 01:07:49,920 --> 01:07:54,220 k over m, to be close to the frequency of the oscillations 999 01:07:54,220 --> 01:07:56,390 of the earth in the earthquake. 1000 01:08:01,030 --> 01:08:10,500 If this driver really had a definite frequency, 1001 01:08:10,500 --> 01:08:15,060 I could solve this problem to find the maximum. 1002 01:08:15,060 --> 01:08:18,960 If I plotted this as a function of omega, 1003 01:08:18,960 --> 01:08:23,439 I would get a curve something like this. 1004 01:08:23,439 --> 01:08:28,260 This is a function of omega, the amplitude, 1005 01:08:28,260 --> 01:08:31,810 the response of that oscillator. 1006 01:08:31,810 --> 01:08:38,609 if you do that, you'll find that if the natural frequency 1007 01:08:38,609 --> 01:08:55,040 of oscillation is over here, the peak 1008 01:08:55,040 --> 01:08:58,000 is slightly at the lower frequency 1009 01:08:58,000 --> 01:09:01,199 than the natural frequency of oscillation omega 0. 1010 01:09:03,750 --> 01:09:09,819 And so you could do this exactly by plotting it out 1011 01:09:09,819 --> 01:09:16,330 to find out what is the optimum value of the frequency 1012 01:09:16,330 --> 01:09:20,620 that you want to adjust so that you get the best response. 1013 01:09:20,620 --> 01:09:23,910 So much for driven oscillators. 1014 01:09:23,910 --> 01:09:30,130 We will now move to couple oscillators, 1015 01:09:30,130 --> 01:09:32,040 and that we'll do next time, some problems. 1016 01:09:32,040 --> 01:09:33,490 Thank you.