1
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To avoid possible misunderstandings, 2
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my lectures start at 9:30 eastern standard time 3
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which is different from Lowell Seven(?) time as you may have noticed. 4
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The clock in Lowell Seven(?) is 7 minutes slow. 5
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Today we are going to cover coupled oscillators. 6
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It is a big part of 8.03. 7
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So we leave our damping in order to avoid a major complication. 8
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Imaging that I have a pendulum length l mass m. 9
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And I have another pendulum also mass m length l. 10
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And I connect a spring between them as you see there. 11
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Spring constant k. 12
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Imaging now at time t equal zero that I give this object which I call object number one 13
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and this is object number two. 14
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Give it a certain position and give it a certain velocity. 15
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So I have four choices and I let the system go. 16
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Which you're going to see is something extremely chaotic 17
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and our task today is to predict what position of this one is at any moment in time 18
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and what position, the position of that one is at any moment in time. 19
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And to show you how chaotic that motion is, 20
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I will just show you this. 21
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So I take this one and just displace it from equilibrium. 22
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I displace this one from equilibrium. 23
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At t time equal zero I will give the one, my left hand just a certain velocity. 24
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And if you now look at the position of the individual objects, 25
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it would see nearly impossible to come with an analytic solution 26
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which tells you what these motions are. 27
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You will see it the amplitudes build up, 28
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of certain one's amplitude goes down. 29
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This one is hardly moving at all now. 30
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Now it is picking up again. 31
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And so our task today is to work on that. 32
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What is by no means obvious but I will show that to you 33
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that any motion, no matter how you start it off 34
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is going to be the superposition of two normal mode solutions. 35
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Anyway you start it can always be written as the superposition of two normal mode solutions. 36
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What is a normal mode? 37
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A normal mode is in this case that both objects have exactly the same frequency. 38
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That is fundamental to normal mode. 39
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And that they are either inphase with each other or out of phase with each other. 40
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Nothing in between. 41
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Because it is no damping so it's either inphase or it's out of phase. 42
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That is a normal mode. 43
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In other words, if I call one of those frequencies ω-, 44
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minus means it is the lowest frequency. 45
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There has two frequencies in this system because there are two objects. 46
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ω- I call that the lowest frequency 47
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Then it would mean if they are inphase with each other the two. 48
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That they come to a halt at the same moment in time. 49
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That means inphase and in the same direction. 50
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If they come to a halt at the same moment in time, 51
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in the same direction and they have the same frequency. 52
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If then I have another frequency which is higher. 53
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There are two normal modes because we have two objects. 54
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If we have three objects there are three normal modes. 55
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If we go to the higher normal mode the higher frequency. 56
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They have the same frequency but they are 180 degrees out of phase. 57
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So when one come to a halt here, the other one comes to a halt there. 58
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That is what means 180 degrees out of phase. 59
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I can excite and I will excite this system into its normal modes only. 60
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I can excite this normal mode alone and this normal mode alone, 61
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if I choose the correct initial conditions. 62
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So for any randomly chosen initial condition, 63
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the motion of each object can be written as a linear combination of these two modes. 64
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If you take my words for that for now but of course I will demonstrate to you 65
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and I will prove that to you. 66
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It will mean then that x1 as a function of time can then be written as having some kind of an amplitude xo 67
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I give it a minus sign because it's related to that normal mode frequency 68
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(see the blackboard) 69
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plus some other amplitude which I call xo plus which is going to be related to this frequency 70
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(see the blackboard) 71
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Let's look at this. Let's try to see through this what this means. 72
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It means that if I know my initial conditions, 73
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that I can determine this amplitude, I can determine this phase, 74
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I can determine this amplitude, and I can determine this phase. 75
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There have to be four adjustable constant. 76
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Because I have the choice between the positions at to and the velocity. 77
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So there must be four. 78
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What you see is that ω- and ω+ 79
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which are these normal mode frequencies are independent of the initial conditions. 80
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So now I am going to write down the position in time for object number two. 81
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And I know that in this mode it must has the same frequency. 82
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So there must be here a cosine minus t plus the same, exactly the same phase, 83
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because I have told you the normal mode means same frequency in phase 84
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or same frequency out of phase. 85
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And this is the next one which is going to be out of phase, 86
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but you will see that shortly for this term must become ω+ t 87
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for now I will say plus phi, 88
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plus, which you are going to see how the out of phase comes in very shortly. 89
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Notice this ω- must be that ω-, 90
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otherwise it wouldn't be a normal mode. 91
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This ω+ must be this ω+, 92
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otherwise it wouldn't be a normal mode. 93
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These phis are the same to allow them to be in phase. 94
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These phis are the same even though they are out of phase, 95
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but you will see very shortly that I will get a minus sign here which will take care of the 180 degrees. 96
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I can with this system excite the lowest mode. 97
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So it'd only oscillate in that lowest mode. 98
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So this term is not there and this term is not there. 99
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I can do that. 100
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I can always oscillate something in one of the normal mode, 101
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but if I set it off at random that it is going to be a superposition. 102
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Even though it is early for you, 103
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can you show me using your hands and your legs, whatever you want to. 104
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When I set this off just a right initial conditions, 105
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what will that lowest normal mode look like, how will these pendulums oscillate. 106
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Very good, very good. So we have a good instinct. 107
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Very good. Let's do that. Let's set them off in the same direction and let them go. 108
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Those are ideal initial conditions for this mode 109
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and you will see that this is a normal mode. 110
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Namely they have the same frequency and they are in phase with each other, 111
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they come to a halt at the same moment in time. 112
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So what you see already is that this value here must also be xo minus. 113
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That's some us...you've just seen it. 114
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In this case what it means is that you might as well remove this sprint. 115
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The spring is not doing anything. The spring never gets any longer and never gets any shorter. 116
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What do you think will the pendulums do when I excite the other mode 117
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which has a higher frequency. 118
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Very good. Can I see some other hands or heads or legs? Good, that's this. 119
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Symmetry. You see there is a symmetry in this system 120
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and that's why you people are so smart and see immediately the connection. 121
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So let me make here the drawing of the ω- and then the amplitudes are the same 122
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and here I make the drawing of the ω+ 123
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higher frequency and the amplitude are now the same but 180 degrees out of phase. 124
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So it is easy to excite that. There we go. 125
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Notice they oscillate with the same frequency and they are coming to a halt at the same moment in time. 126
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But if one is here, the other is there. 127
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And so therefore this one here must be minus xo plus. 128
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That minus sign now gives me the 180 degrees so I could still keep that phi plus there. 129
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It is utterly trivial without any work to calculate or to tell you what ω- is. 130
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That's just the frequency of a single pendulum. 131
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So that's easy.(See blackboard) 132
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There is no spring, right? 133
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Each one is doing its own thing and the spring is never pushing and pulling. 134
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But we should be able to calculate ω+. So let's do that. 135
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And I will only make a drawing of one of those pendulums. I don't need them both. 136
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So here is one of those pendulums. And let this separation in distance be x from equilibrium. 137
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And I call this angle theta. And so here is that spring. 138
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And the other one is on the other side. 139
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Let's put in all the forces that we can think of. 140
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So there is gravity mg, there is tension T, but there is more. 141
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What else is there? Spring. How much longer is the spring than it wants to be? 142
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2x exactly. In another words, there is a force here, I call that the spring force Fs 143
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which is bringing it also back to equilibrium. And that one is 2kx. 144
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Now, the tension is very close to mg, we have discuss that several times. 145
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I'm going to introduce the short hand notation (see blackboard) 146
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And now I have to set up the differential equation. I have to apply Newton's Second Law. 147
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And remind you that the magnitude of the spring force is 2kx. 148
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So Newton's Second Law. m x double dots equals. 149
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Now there are two forces that I have to take into account. 150
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First, the spring force which is -2kx. 151
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And then I need the horizontal component of the tension. 152
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Remember that the only one that is important here which is Tsinθ. 153
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But T is mg, and sinθ is x divided by l. l is the length of the pendulum. 154
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So we get (see blackboard) And so this is the differential equation. 155
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I'm going to divide m out, and I am going to bring everything to one side. 156
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So I'm going to get (see blackboard) 157
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And our task now is to solve this differential equation. 158
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And that of course you can do in three seconds. 159
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As you recognize this differential equation, it is x double dots plus something times x. 160
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And so the, the new frequency which I will call ω+. 161
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That new frequency ω+ is (see blackboard) 162
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And what you see is something that I already anticipated which was consistent with your intuition. 163
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It is larger than omega zero because this is the effect of the spring. 164
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So omega zero is the one here. And ω+ now is (see blackboard) 165
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And omega s square is to find this way. 166
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So if I now turn to my general solution 167
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if for now you accept the fact that that is the general solution 168
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the superposition of the normal modes. 169
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Then, the key point is that independent of your initial conditions is ω- 170
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that was the square root of g over l. I never put in any initial condition. 171
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Independent of all those initial conditions is ω+ 172
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that this one I never put in any initial conditions. 173
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Independent of the initial conditions is this ratio which is +1. 174
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And independent of the initial conditions is this ratio that is -1. 175
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Another words, if I didn't tell you the initial conditions which I haven't. 176
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I can predict that this ratio is +1 and I can predict that this ratio is -1. 177
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If I give this one...If I make this 3 and this is -3. 178
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If I make this 5 and this is -5. The ratio is -1. 179
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So the ratios are independent of the initial conditions 180
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and the frequencies are independent of the initial conditions. 181
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If I tell you the initial conditions then of course you can also calculate the individual amplitudes. 182
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Suppose now I start this system off in some way at time t equal zero which I can choose. 183
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And I choose the following. 184
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At t equal zero I make x1 C. 185
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Just some number C that we can choose. 186
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3 cm whatever you can choose but I make v1 zero. 187
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So as I release it, no speed. 188
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And suppose I make x2 zero and I make v2 zero 189
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but if you don't know what that means because I am going to demonstrate it. 190
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It means that this one is here, the other one I offset. That's all I do and I let this one go. 191
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Pull my hands off, and then I want to know what is going to happen. 192
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Right? Because I release this one with zero speed. 193
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That's what you see in there. That is one of the infinite number of initial conditions that you may choose. 194
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So this now has to be substituted into my general solution, and so you have to take the derivative of x_1. 195
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So you have to take x1 dot, and put that equal to zero. I leave that in your competent hands. 196
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Then you have to take x2 you have to take x2 dot and you have to make that equal to zero. 197
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And I leave that into your competent hands. That's easy. 198
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And you will find then that in this particular case for this particular example 199
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phi minus equals zero and phi2 and phi plus equal zero. 200
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It will not take you more than a few minutes to find that. 201
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I didn't want to waste your time on taking a derivative of such a simple function. 202
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So if we take this right now then I will substitute these results in the equation. 203
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So you did the hard work, you did the velocities I will do the positions. 204
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So I am going to substitute in that equation t equal zero I know already that the phis are not there. 205
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And so then I get that C at time t equals zero is going to be x0 minus. 206
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t is zero so this is one, t is zero so this is one plus x0 plus. 207
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That's this initial condition. Then I go to the second initial condition 208
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that the zero is x0 minus minus x0 plus.Minus sign. 209
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So what do I find I have solve now the general solution? 210
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I will find that x0 minus is one half C and x0 plus is also one half C 211
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which of course should not come as a surprise to you. 212
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So let me write down now the general solution that we have. 213
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For this, the specific initial condition. So we know everything. 214
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We know x01, we know x01-, x0- not 1, x0-, we know x0+, 215
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we know φs, we know everything. 216
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So I'm going to write it down here. 217
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For x1, it's going to be c/2. Remember? Found that is c/2, 218
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times the cosine of ωt, because φ is 0, plus c/2 times, 219
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this is minus by the way cos(ω-*t), cos(ω+*t), that is x1. 220
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And x2 (see the blackboard). 221
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Take a deep breath, substitute it in that t equal 0. 222
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And you see immediately that x is c, and substitute t equal 0 in the second, 223
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and then you see indeed that x2 is 0. No surprise, because that's my initial condition. 224
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Now I remember for my high school days that the cosα plus the cosβ 225
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is twice the cosine of half the sum times the cosine of half the difference. 226
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So I can write down x1 as twice the cosine of half the sum times the cosine of the difference. 227
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So that two that I get, each of these one halves. 228
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So I get (see the blackboard). 229
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I've just rewriten it in a different form. 230
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And x2 as a function of time, I now have the cosine of α minus the cosine of β. 231
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That is twice the sine half the sum times the sine of half the difference. 232
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So now I get here (see the blackboard). 233
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Notice when t is 0, that the sine of this one is 0. 234
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Consistent with what our initial conditions:x2 is zero, remember? 235
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All there, initial conditions are all there, just written in the different form. 236
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Now, the image now in your mind that these two frequencies are not too far apart. 237
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Then these equations have the smell of what? Beats. 238
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See it is this here is the fast term. This one and that one. 239
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And this one is the slow one. 240
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And so if these two are close, then what you'll see is something quite remarkable. 241
00:23:55,830 --> 00:24:05,470
At t equal 0, this one stands still, and this cosine term is going to be 1, because t is 1(t is 0). 242
00:24:05,470 --> 00:24:08,980
And this is going to oscillate happily with this frequency. 243
00:24:08,980 --> 00:24:16,700
But this cosine term is very gradually going to 0, and as that cosine term gradually goes to 0, 244
00:24:16,700 --> 00:24:22,340
this one will stop oscillating, but this sine term becomes +1. 245
00:24:22,340 --> 00:24:24,930
And so the other one will start to oscillate. 246
00:24:24,930 --> 00:24:30,700
And in a little later in time, the cosine term will become -1, so it starts to oscillate. 247
00:24:30,700 --> 00:24:34,690
But when that happens, this sine term is 0 again, so it stops. 248
00:24:34,690 --> 00:24:41,060
So you see a beautiful beat phenomenon. One, the first one will gradually come to a halt, 249
00:24:41,060 --> 00:24:44,440
and the other one will pick up, and then the other one will come to a halt, 250
00:24:44,440 --> 00:24:48,790
and then it transfers in the wave energy. 251
00:24:48,790 --> 00:24:56,240
It is of course consistent with the conservation of mechanical energy. 252
00:24:56,240 --> 00:24:58,130
And I want to demonstrate that. 253
00:25:03,410 --> 00:25:06,220
So we have this here. 254
00:25:06,220 --> 00:25:12,020
So all I have to do is offset x1 over a distance C that we can choose. 255
00:25:12,020 --> 00:25:16,870
And then we release this one at 0 speed, and then we will just watch it. 256
00:25:16,870 --> 00:25:21,460
And then you should see that strange beat phenomenon. 257
00:25:21,460 --> 00:25:23,270
You're ready for this? 258
00:25:23,270 --> 00:25:27,860
This one have to hold inphase. 259
00:25:27,860 --> 00:25:34,620
Three, two, one, zero. 260
00:25:34,620 --> 00:25:39,260
Look! This one is standing still. 261
00:25:39,260 --> 00:25:46,430
Look! This one is standing still. 262
00:25:46,430 --> 00:25:51,260
That's beating, and you see that energy is transfered from one to the other. 263
00:25:51,260 --> 00:25:56,190
And that is a beat phenomenon that follows immediately from this. 264
00:25:56,190 --> 00:26:01,700
What would happen if I move the spring up. 265
00:26:01,700 --> 00:26:08,950
Suppose I move the spring here. Higher, say halfway. 266
00:26:08,950 --> 00:26:17,010
So anyone without looking too much at the blackboard for the few sure intuition, 267
00:26:17,010 --> 00:26:20,380
what would happened? Would the same phenomenon happened? 268
00:26:25,170 --> 00:26:30,040
Yeah, yeah. 269
00:26:30,040 --> 00:26:31,830
The spring what? 270
00:26:31,830 --> 00:26:36,000
Yes? Would the same phenomenon happened? 271
00:26:36,000 --> 00:26:38,280
But what would happened with the beat periods? 272
00:26:44,300 --> 00:26:47,430
Now you may look at the blackboard. 273
00:26:47,430 --> 00:26:52,320
Do you think this had a certain separation, right? 274
00:26:52,320 --> 00:26:56,040
You make that separation smaller, you make the effect of the spring smaller. 275
00:26:56,040 --> 00:26:59,590
So your ω- becomes even closer to the ω+. 276
00:26:59,590 --> 00:27:04,230
So the beat period will increase, will take longer for one to stop. 277
00:27:04,230 --> 00:27:08,700
Let's find that. 278
00:27:08,700 --> 00:27:14,750
So we're going to move this up. 279
00:27:14,750 --> 00:27:17,050
I put it roughly halfway. 280
00:27:22,800 --> 00:27:24,690
OK, ready for this? 281
00:27:24,690 --> 00:27:27,030
And so we're doing it again. 282
00:27:27,030 --> 00:27:28,910
And then you'll see the same phenomenon 283
00:27:28,910 --> 00:27:34,490
except it will take longer for the first one to come to a halt. 284
00:27:34,490 --> 00:27:36,820
So I have decreased the coupling. 285
00:27:48,360 --> 00:27:53,540
Is it still swinging happily? 286
00:27:53,540 --> 00:27:57,920
Now it's beginning to change its mind, you see? 287
00:27:57,920 --> 00:27:59,770
And now it's standing still. 288
00:27:59,770 --> 00:28:05,180
Took longer, do we agree? 289
00:28:05,180 --> 00:28:11,160
Now there are something mind bothering, something that you really want to see. 290
00:28:11,160 --> 00:28:17,010
Suppose I bring this all the way to the top. 291
00:28:17,010 --> 00:28:23,340
That's interesting because then ω- is exactly the same as ω+, 292
00:28:23,340 --> 00:28:27,840
because look, this term goes away. 293
00:28:27,840 --> 00:28:31,660
What hell will happen? 294
00:28:31,660 --> 00:28:35,800
It's almost this cutting out the spring. There is no spring anymore. 295
00:28:35,800 --> 00:28:38,640
What now would happen if I start one here. 296
00:28:38,640 --> 00:28:40,170
Three, two, one, zero. 297
00:28:45,610 --> 00:28:48,810
What do you think? Yeah? 298
00:28:48,810 --> 00:28:51,510
Ah, brilliant! Brilliant! 299
00:28:51,510 --> 00:28:54,090
One would swing and the other wouldn't. 300
00:28:54,090 --> 00:28:57,130
Isn't that shocking? Isn't that shocking? 301
00:28:57,130 --> 00:29:00,790
You have two pendulums and they are not even connected anymore. 302
00:29:00,790 --> 00:29:05,320
And you start this one swinging and the other one will never start to swing. 303
00:29:05,320 --> 00:29:08,040
Is that amazing for thirty thousand dollars tuition? 304
00:29:08,040 --> 00:29:11,430
You learned something fantastic. 305
00:29:11,430 --> 00:29:16,790
Except that the beat period is infinitely long. 306
00:29:16,790 --> 00:29:19,750
So with a bit of patience. 307
00:29:19,750 --> 00:29:26,320
So let's demonstrate that. So let's demonstrate this. 308
00:29:26,320 --> 00:29:31,380
So there we go. 309
00:29:31,380 --> 00:29:37,650
Unbelievable. Physics works. 310
00:29:37,650 --> 00:29:40,300
And look at this equation. 311
00:29:40,300 --> 00:29:46,400
When ω- is ω+, this term is always 0. 312
00:29:46,400 --> 00:29:50,980
So you see x2 remains 0. Hahaha, that's what you see. 313
00:29:50,980 --> 00:29:56,350
And then ω- is ω+, this cosine term is always +1. 314
00:29:56,350 --> 00:29:59,830
Hahaha, that's what you see. 315
00:29:59,830 --> 00:30:06,470
Isn't it amazing the power of physics? 316
00:30:06,470 --> 00:30:15,320
So it is truely remarkable that we can describe for any initial condition 317
00:30:15,320 --> 00:30:22,340
the motion in terms of the linear superposition of the two normal modes. 318
00:30:22,340 --> 00:30:27,170
And so what originally look like an impossibility. When I start it, 319
00:30:27,170 --> 00:30:30,730
the first 30 seconds of my lecture with that demonstration, 320
00:30:30,730 --> 00:30:37,180
when I started them off in a random way, it looks so chaotic that it looks almost unimaginable. 321
00:30:37,180 --> 00:30:40,310
that we would be able to sort that out 322
00:30:40,310 --> 00:30:48,210
and be able to predict the motion of each one individually as a function of time. 323
00:30:48,210 --> 00:30:53,800
I want you to appreciate that we have two coupled oscillators here 324
00:30:53,800 --> 00:30:55,810
that give you two normal modes. 325
00:30:55,810 --> 00:30:58,460
If you have three coupled oscillators, we will get back to that. 326
00:30:58,460 --> 00:31:00,500
Then there are three normal modes. 327
00:31:00,500 --> 00:31:06,530
And if you have four, then you have four normal modes. 328
00:31:06,530 --> 00:31:12,830
Now clearly what we need is a general recipe. 329
00:31:12,830 --> 00:31:18,410
In this case, the problem was so beautifully symmetric. We could do ω- just a split seconds, 330
00:31:18,410 --> 00:31:21,190
ω+ we could see exactly the motion, 331
00:31:21,190 --> 00:31:26,780
so we know time when you get that ratio of the amplitude was +1 and was -1. 332
00:31:26,780 --> 00:31:31,350
But let me tell you if I break the symmetry. 333
00:31:31,350 --> 00:31:34,740
Oh, man, all hell break its rules. 334
00:31:34,740 --> 00:31:37,380
For instance, if you change the ratio of the masses. 335
00:31:37,380 --> 00:31:39,840
You make one longer than the other. 336
00:31:39,840 --> 00:31:44,840
You have three beats connected with springs. 337
00:31:44,840 --> 00:31:48,920
You break the symmetry. It is very very hard work. 338
00:31:48,920 --> 00:31:52,790
And for that we need a general recipe. 339
00:31:52,790 --> 00:31:55,540
And I'm going to give you the general recipe. 340
00:31:55,540 --> 00:32:00,570
And then I'm going to apply that general recipe to this case. 341
00:32:00,570 --> 00:32:05,640
So everything that you have seen will then come out after a lot of algebra, 342
00:32:05,640 --> 00:32:11,860
but at least you know that the recipe is working. 343
00:32:11,860 --> 00:32:14,430
So the first thing that you do. 344
00:32:14,430 --> 00:32:29,830
That's NO.1 you give each object a displacement from equilibrium. 345
00:32:29,830 --> 00:32:35,970
Even though you're free to choose the direction, I always in matter of discipline 346
00:32:35,970 --> 00:32:38,640
set them all off in the same direction. 347
00:32:38,640 --> 00:32:44,630
That's not a must, but that reduces the chance of mistakes. 348
00:32:44,630 --> 00:32:48,750
But you give them a displacement, and I always do them always in the same direction. 349
00:32:48,750 --> 00:32:57,740
That's NO.1. Once you have done that, you write down Newton's Second Law for each. 350
00:33:03,940 --> 00:33:13,980
Which means if you have two objects, you have three unknowns. 351
00:33:13,980 --> 00:33:15,330
What are the three unknowns? 352
00:33:23,870 --> 00:33:30,100
So you want to find one of the normal mode frequencies. 353
00:33:30,100 --> 00:33:33,000
One of the normal mode frequencies, do you want to find this one 354
00:33:33,000 --> 00:33:38,110
and independently do you want to find that one which has to come out of the recipe. 355
00:33:38,110 --> 00:33:43,730
What are in principle the three unknowns? 356
00:33:43,730 --> 00:33:48,790
This ratio +1 holds here, but it doesn't hold in other cases. 357
00:33:48,790 --> 00:33:53,920
And so in principle you have as an uncertainty, 358
00:33:53,920 --> 00:33:58,290
this amplitude, this amplitude and ω. 359
00:33:58,290 --> 00:34:04,050
So you always end up with three equations, two equations with three unknowns. 360
00:34:04,050 --> 00:34:10,480
So If I say two objects, then you have three unknowns, and you have two equations. 361
00:34:10,480 --> 00:34:13,380
So that looks like a problem. 362
00:34:13,380 --> 00:34:20,590
NO.3 you are going to put in the condition for normal modes 363
00:34:20,590 --> 00:34:31,560
which means that x1 is (see the blackboard), but x2 can be some other amplitude, 364
00:34:31,560 --> 00:34:38,960
which we have to calculate also times the same ωt. 365
00:34:38,960 --> 00:34:43,120
I don't have to put a phase in there, because it's either inphase or out of phase. 366
00:34:43,120 --> 00:34:46,840
And out of phase means that you get minus signs. 367
00:34:46,840 --> 00:34:52,010
And so you see immediately that if you substitute this in Newton's Second Law 368
00:34:52,010 --> 00:34:56,330
that you get for two equations, three unknowns. C1 is unknown, C2 is unknown 369
00:34:56,330 --> 00:35:06,990
and ω is unknown. So you substitute this in here, and then comes the problem. 370
00:35:06,990 --> 00:35:13,390
How are you going to deal with two equations with three unknowns? 371
00:35:13,390 --> 00:35:17,680
Well, think about what I told you earlier. 372
00:35:17,680 --> 00:35:22,630
If you don't know the initial conditions, you can never find x0-, 373
00:35:22,630 --> 00:35:29,620
so you can never find c1. But the ratio is independent of the initial conditions. 374
00:35:29,620 --> 00:35:36,510
So that means the ratio c1/c2 cannot be dependent on the initial conditions, 375
00:35:36,510 --> 00:35:45,690
in other words, you can always solve these equations in terms of c1/c2 and ω. 376
00:35:45,690 --> 00:35:51,230
And I will show you that that works. 377
00:35:51,230 --> 00:35:57,060
So you follow this recipe, and we will do that religiously together. 378
00:35:57,060 --> 00:36:05,870
We will find then in this case two values for ω, we will find an ω-, 379
00:36:05,870 --> 00:36:10,570
which has an associated value of c1/c2, I will put a minus there. 380
00:36:10,570 --> 00:36:17,450
And we will find that ω+ with has its own ratio associated with it. 381
00:36:17,450 --> 00:36:23,870
In the case of our simple symmetric system, this will be +1 and this will be -1. 382
00:36:23,870 --> 00:36:31,790
But that is not the case when you break symmetry. 383
00:36:31,790 --> 00:36:41,560
So my task is now to apply this recipe in its most general form to this system. 384
00:36:41,560 --> 00:36:46,260
It will be sixteen minutes of grinding, we wil go through each step. 385
00:36:46,260 --> 00:36:49,060
And outcome something that we already know. 386
00:36:49,060 --> 00:36:53,620
But at least you will see that I have to make no assumption 387
00:36:53,620 --> 00:36:56,010
which I did there about symmetry. 388
00:36:56,010 --> 00:37:00,150
Therefore this is an ideal moment for you to have your five minutes break. 389
00:37:00,150 --> 00:37:07,410
So you take a deep breath to get ready for this sixteen minutes marathon. 390
00:37:07,410 --> 00:37:10,800
All right, general recipe. 391
00:37:10,800 --> 00:37:15,350
Someone asked me during the intermittion whether this ω is the same as that one. 392
00:37:15,350 --> 00:37:18,390
Yes, of course. Otherwise it wouldn't be in normal mode. 393
00:37:18,390 --> 00:37:21,430
We are going to search for the normal modes. 394
00:37:21,430 --> 00:37:27,680
And in the normal mode, the frequencies must be the same of all objects 395
00:37:27,680 --> 00:37:31,150
whether you have five or six or two or three. 396
00:37:31,150 --> 00:37:34,980
But the ratios is different. That's the different issue. 397
00:37:34,980 --> 00:37:42,170
However, they are inphase or out of phase. So the ratios can be negative or positive. 398
00:37:42,170 --> 00:37:46,200
But you never have any phase angles other than 180 degrees or 0. 399
00:37:46,200 --> 00:37:49,280
Because we have no damping. We have taken the damping out. 400
00:37:49,280 --> 00:37:53,410
So it is essential that when you substitute that in Newton's Second Law, 401
00:37:53,410 --> 00:38:01,570
that these ωs are the same, because they will then give you the normal modes. 402
00:38:01,570 --> 00:38:02,800
OK, you are ready? 403
00:38:08,510 --> 00:38:17,410
I'm going to offset the first one, over a distance x1. 404
00:38:17,410 --> 00:38:22,440
So they have mass m, they have length l to remind you. 405
00:38:22,440 --> 00:38:32,910
And we are going to have (see the blackboard). 406
00:38:32,910 --> 00:38:39,540
And this one I offset over a distance x2. 407
00:38:39,540 --> 00:38:47,790
Notice you don't have to do that. I always offset them in the same direction. 408
00:38:47,790 --> 00:38:52,620
And there is the spring that connects them. 409
00:38:52,620 --> 00:39:00,790
I will just make it very ?? otherwise the picture becomes a little bit too complicated. 410
00:39:00,790 --> 00:39:07,900
So let this angle be θ1, and let this angle be θ2. 411
00:39:07,900 --> 00:39:14,890
And so we have here mg, we've here mg. 412
00:39:16,400 --> 00:39:18,260
We have here the tension; 413
00:39:18,260 --> 00:39:20,340
it is roughly mg for small angles. 414
00:39:20,340 --> 00:39:24,730
And we have here the tension, which is roughly mg. 415
00:39:24,730 --> 00:39:26,940
A little bit too big the way I drew it, 416
00:39:26,940 --> 00:39:29,470
but that's not so important now. 417
00:39:29,470 --> 00:39:34,220
And now there is another force, that x on both, 418
00:39:34,220 --> 00:39:36,580
and which force is that? 419
00:39:36,580 --> 00:39:38,840
That's the spring force. 420
00:39:38,840 --> 00:39:46,470
Now do we agree what the magnitude of this spring force is? 421
00:39:46,470 --> 00:39:52,480
The magnitude, then we will argue about the direction. 422
00:39:52,480 --> 00:39:55,930
Nico, magnitude, look very closely, 423
00:39:55,930 --> 00:40:04,600
I offset one by x2 and I offset the other by x1. 424
00:40:05,340 --> 00:40:07,270
I was asking you a question. 425
00:40:07,270 --> 00:40:08,970
You can't answer it with a question. 426
00:40:08,970 --> 00:40:14,760
So it's going to be ... Very good. 427
00:40:14,760 --> 00:40:23,010
The magnitude of that force is k(x2-x1), 428
00:40:23,010 --> 00:40:25,810
non-negotiable, right? 429
00:40:25,810 --> 00:40:29,070
Now if x2 is larger than x1, 430
00:40:29,070 --> 00:40:35,290
do we agree that this spring force is in this direction? 431
00:40:35,290 --> 00:40:40,590
And do we agree then that this spring force must be in this direction, 432
00:40:40,590 --> 00:40:43,130
if x2 is larger than x1. 433
00:40:43,130 --> 00:40:47,620
Well, that means I can leave everything the way it is now. 434
00:40:47,620 --> 00:40:50,460
As long as I give this a minus sign and this a plus sign, 435
00:40:50,460 --> 00:40:54,530
I'm ok because if x2 is smaller than x1, 436
00:40:54,530 --> 00:40:59,220
it will automatically flip over, and my algebra is fine. 437
00:40:59,220 --> 00:41:02,810
So therefore, in my head, 438
00:41:02,810 --> 00:41:05,510
I can just think of x2 being larger than x1, 439
00:41:05,510 --> 00:41:07,190
set up the differential equations, 440
00:41:07,190 --> 00:41:10,220
and I no longer have to worry about the fact 441
00:41:10,220 --> 00:41:16,370
that maybe x2 at certain moment in time is not larger than x1. 442
00:41:16,370 --> 00:41:21,860
So now I set up the differential equation mx1 double dot. 443
00:41:21,860 --> 00:41:28,360
So that's this one. Let's first do the pendulum 444
00:41:28,360 --> 00:41:34,500
that is minus T times the sinθ1 and T is mg. 445
00:41:34,500 --> 00:41:40,520
So this is -mgx1/l. 446
00:41:40,520 --> 00:41:44,390
Do we agree this at horizontal component here? 447
00:41:44,390 --> 00:41:46,100
The horizontal component of T is 448
00:41:46,100 --> 00:41:49,530
driving it back to equilibrium minus sign. 449
00:41:49,530 --> 00:41:53,730
Fs, the spring force, is driving it away from equilibrium. 450
00:41:53,730 --> 00:42:03,300
So it's going to get +k (x2-x1). 451
00:42:03,300 --> 00:42:12,010
Look at this, science are very important now, very important. 452
00:42:12,010 --> 00:42:19,070
And the next one, mx2 double dot. It has again, 453
00:42:19,070 --> 00:42:28,450
the restoring force due to the pendulum, which is -mgx2/l, 454
00:42:28,450 --> 00:42:32,340
cos it's the sinθ2 now that comes in. 455
00:42:32,340 --> 00:42:36,280
And now it has the spring force, which now is a negative sign, 456
00:42:36,280 --> 00:42:44,080
and so now we get -k *(x2-x1). 457
00:42:44,080 --> 00:42:49,460
And when you have this done, on your next exam, 458
00:42:49,460 --> 00:42:51,090
you take a deep breath, 459
00:42:51,090 --> 00:42:54,170
and you go over each individual step 460
00:42:54,170 --> 00:42:58,100
to make absolutely sure that this is correct, 461
00:42:58,100 --> 00:43:01,710
if there is anything wrong with this, you dead in the waters, 462
00:43:01,710 --> 00:43:03,260
the whole problem fall apart, 463
00:43:03,260 --> 00:43:06,070
it may not even be a simple harmonic oscillator, 464
00:43:06,070 --> 00:43:08,540
you get something ridiculous. 465
00:43:08,540 --> 00:43:10,190
Science are crucial here. 466
00:43:10,190 --> 00:43:16,120
So therefore, look at this again, take a pause. 467
00:43:16,120 --> 00:43:19,710
This one, if I give this a positive offset, 468
00:43:19,710 --> 00:43:25,570
is indeed in the negative direction, the component of Tsinθ. 469
00:43:25,570 --> 00:43:29,930
This one, if x2 is larger than x1, is indeed in that direction. 470
00:43:29,930 --> 00:43:32,150
And you know this if x2 becomes smaller than x1, 471
00:43:32,150 --> 00:43:34,790
well then this becomes automatically negative. 472
00:43:34,790 --> 00:43:37,980
So are you okay? So this is fine. 473
00:43:37,980 --> 00:43:42,940
This one has a restoring force due to the pendulum, 474
00:43:42,940 --> 00:43:47,320
which is Tsinθ2 negative sign, 475
00:43:47,320 --> 00:43:51,790
and the restoring force of this spring always opposite this one, 476
00:43:51,790 --> 00:43:57,570
I am happy. I am happy. 477
00:43:57,570 --> 00:44:00,620
So now we are going to rearrange it a little bit, 478
00:44:00,620 --> 00:44:06,520
and we're going to introduce the ωo square notation. 479
00:44:06,520 --> 00:44:09,560
So I'm going to divide m out, 480
00:44:09,560 --> 00:44:15,230
and so I'm going to get x1 double dot, 481
00:44:15,230 --> 00:44:19,660
and then I'm going to bring the x1s to the left 482
00:44:19,660 --> 00:44:32,370
and so I get +ω0 squared times x1, 483
00:44:32,370 --> 00:44:37,170
but you have a -kx1. You divide it by m, 484
00:44:37,170 --> 00:44:45,290
so you get plus ωs squared and that both times x1. 485
00:44:45,290 --> 00:44:49,190
You notice that? They both have an x1 term. 486
00:44:49,190 --> 00:44:51,840
And now the x2 comes out, 487
00:44:51,840 --> 00:44:54,490
the x2 has a plus sign on the right side, 488
00:44:54,490 --> 00:45:01,380
so it gets a minus sign, so I get minus ωs squared times x2, 489
00:45:01,380 --> 00:45:07,110
and that is zero. 490
00:45:07,110 --> 00:45:09,900
And then we do the next one, 491
00:45:09,900 --> 00:45:12,400
we get x2 double prime...double dot, 492
00:45:12,400 --> 00:45:16,900
not double prime, x2 double dot. 493
00:45:16,900 --> 00:45:21,740
Ahhh, I get the same terms. I get an ω0 squared. 494
00:45:21,740 --> 00:45:28,190
And I get an ωs squared times x2, 495
00:45:28,190 --> 00:45:35,990
and then I get this plus, minus times minus is plus. 496
00:45:35,990 --> 00:45:43,510
So I get -ωs squared times x1 equals zero. 497
00:45:43,510 --> 00:45:47,810
When you have reached this point, 498
00:45:47,810 --> 00:45:50,060
in your exam, you take a deep breath, 499
00:45:50,060 --> 00:45:52,620
and you go each over each little term to make sure 500
00:45:52,620 --> 00:45:55,940
that you haven't accidentally slipped on a minus sign, 501
00:45:55,940 --> 00:45:57,610
because if you slip on one minus sign, 502
00:45:57,610 --> 00:45:58,990
you dead in the waters, 503
00:45:58,990 --> 00:46:02,670
it may not even become a simple harmonic oscillation. 504
00:46:02,670 --> 00:46:05,120
So let me do that, are you agree with that? 505
00:46:05,120 --> 00:46:06,330
Are you agree with this? 506
00:46:06,330 --> 00:46:08,200
Are you agree with that? Are you agree with that? 507
00:46:08,200 --> 00:46:11,980
Differential equation is fine. 508
00:46:11,980 --> 00:46:16,110
Notice that I already did for number 1, 509
00:46:16,110 --> 00:46:18,230
and that already I did even number 2, 510
00:46:18,230 --> 00:46:22,080
I've set up the differential equations, Newton's Second Law. 511
00:46:22,080 --> 00:46:24,590
So now comes No.3. 512
00:46:24,590 --> 00:46:28,200
And No.3 means I'm going to substitute in 513
00:46:28,200 --> 00:46:42,370
there x1 is C1cosωt and x2 is C2cosωt. 514
00:46:44,120 --> 00:46:48,650
We go slowly. I want you to follow each step. 515
00:46:48,650 --> 00:46:52,850
And I'm going to work on here so that you can see it high. 516
00:46:52,850 --> 00:46:55,100
So I'm going to substitute this in here, 517
00:46:55,100 --> 00:47:00,690
so the x1 double dots will give me minus ω squared, 518
00:47:00,690 --> 00:47:03,860
right? Because you get twice ω out, 519
00:47:03,860 --> 00:47:10,790
minus ω squared times C1, and the hell was cosωt. 520
00:47:10,790 --> 00:47:21,030
Why the hell was cosωt? Because every term will have cosωt, 521
00:47:21,030 --> 00:47:23,010
so minus will divide out right away. 522
00:47:23,010 --> 00:47:23,590
Right? 523
00:47:23,590 --> 00:47:26,160
Every term that you're going to have in here 524
00:47:26,160 --> 00:47:27,960
will have a cosωt in it, 525
00:47:27,960 --> 00:47:31,220
so I'll leave the cosωt already out now. 526
00:47:31,220 --> 00:47:55,320
So I get minus ω squared C1...(see the blackboard) 527
00:47:55,320 --> 00:48:01,220
Because remember x2 has the C2. 528
00:48:01,220 --> 00:48:05,390
Yeah, am I going too fast? Beautiful. 529
00:48:05,390 --> 00:48:07,040
So I go to the second one. 530
00:48:07,040 --> 00:48:08,820
The second one is x2 double dot, 531
00:48:08,820 --> 00:48:13,700
so I also get minus ω squared times C2. 532
00:48:14,740 --> 00:48:33,350
(see the blackboard) 533
00:48:33,350 --> 00:48:37,010
And when you have reached this point in your exam, 534
00:48:37,010 --> 00:48:39,010
you take a deep breath, 535
00:48:39,010 --> 00:48:42,730
and you make sure that all your signs are correct. 536
00:48:42,730 --> 00:48:49,090
If not, you dead in the waters, and the problem will fall apart. 537
00:48:49,090 --> 00:48:54,230
So let's do that. Minus ω squared C1, I can live with that, 538
00:48:54,230 --> 00:48:56,930
plus that term with a C1, yes. 539
00:48:56,930 --> 00:49:01,490
Minus ωs squared C2, even if you make a little slip of the pen, 540
00:49:01,490 --> 00:49:04,700
and you change this into a one, and this into a two, 541
00:49:04,700 --> 00:49:07,190
it's all over of course. 542
00:49:07,190 --> 00:49:10,240
And then the next equation, minus ω squared C2, 543
00:49:10,240 --> 00:49:12,370
then I have this times C2, 544
00:49:12,370 --> 00:49:14,950
and then it minus that times C1. 545
00:49:14,950 --> 00:49:15,520
We're all most there 546
00:49:15,520 --> 00:49:18,870
even though it doesn't look that way, does it? 547
00:49:18,870 --> 00:49:20,360
Remember what I said? 548
00:49:20,360 --> 00:49:26,240
you cannot solve 2 equations with 3 unknowns. 549
00:49:26,240 --> 00:49:30,820
There is no way that you can solve for C1, C2 and for this ω, 550
00:49:30,820 --> 00:49:33,100
which is.. which you're really after, 551
00:49:33,100 --> 00:49:35,970
this is the ω you're after. 552
00:49:35,970 --> 00:49:41,590
But you can solve for C1/C2 and ω. Knowing that already, 553
00:49:41,590 --> 00:49:46,950
I'm going to eliminate C1/C2 and you will see how I do that. 554
00:49:46,950 --> 00:49:54,470
C1 divided by C2, I go to the first equation, 555
00:49:54,470 --> 00:49:56,120
this is not so hard what I'm doing. 556
00:49:56,120 --> 00:49:59,360
I put in my mind just on the right side, 557
00:49:59,360 --> 00:50:03,240
so I get plus ωs squared times C2. 558
00:50:03,240 --> 00:50:06,190
And then I divide C1 by C2, 559
00:50:06,190 --> 00:50:12,880
and so what I get then is the following (see the blackboard) 560
00:50:23,500 --> 00:50:25,170
Do we agree? That's the first equation, 561
00:50:25,170 --> 00:50:28,900
I've simply written it as C1 divided by C2, 562
00:50:28,900 --> 00:50:30,140
I can always do that, right? 563
00:50:30,140 --> 00:50:33,180
Bring the C1s to one side, the C2s to one side, 564
00:50:33,180 --> 00:50:36,260
and divide them. That's this. 565
00:50:36,260 --> 00:50:40,060
And I'm going to do the same with the next equation. 566
00:50:40,060 --> 00:50:41,800
So I bring this one to the right side, 567
00:50:41,800 --> 00:50:45,980
then I have C1s there, and C2s here. 568
00:50:45,980 --> 00:50:49,000
And now this becomes upstairs. 569
00:50:49,000 --> 00:51:03,410
(see the blackboard) 570
00:51:03,410 --> 00:51:06,960
And now I have eliminated C1 and C2, 571
00:51:06,960 --> 00:51:10,640
because if I solve this equation, 572
00:51:10,640 --> 00:51:12,590
I get my solutions for ω. 573
00:51:12,590 --> 00:51:13,770
This is one equation. 574
00:51:13,770 --> 00:51:17,490
There is one unknown. That's ω. We're gonna admit. 575
00:51:17,490 --> 00:51:19,540
And there're better be two solutions, 576
00:51:19,540 --> 00:51:21,290
there're better be an ω-, 577
00:51:21,290 --> 00:51:25,270
and there're better be an ω+ coming out of this. 578
00:51:25,270 --> 00:51:26,890
Untouched by young human hands, 579
00:51:26,890 --> 00:51:32,190
I must find two solutions. 580
00:51:32,190 --> 00:51:36,580
Let's first make sure that this is correct. 581
00:51:36,580 --> 00:51:39,880
And the answer is yes. 582
00:51:39,880 --> 00:51:42,010
So what will I do now, well, 583
00:51:42,010 --> 00:51:44,600
I am going to simplify it one step further, 584
00:51:44,600 --> 00:51:47,260
which is now very easy. 585
00:51:47,260 --> 00:51:49,870
I multiply this by this, 586
00:51:49,870 --> 00:51:59,720
and this with this, so we get ...(see the blackboard) 587
00:52:08,220 --> 00:52:15,130
One equation with one unknown. ω is the only unknown. 588
00:52:15,130 --> 00:52:18,440
Take this square root left and right, 589
00:52:18,440 --> 00:52:32,770
so I get...(see the blackboard) 590
00:52:32,770 --> 00:52:35,370
Do not forget the plus or minus! 591
00:52:35,370 --> 00:52:40,270
Because the square root of ωs to the fourth 592
00:52:40,270 --> 00:52:45,970
is plus or minus ωs squared. 593
00:52:45,970 --> 00:52:49,600
And now we're going to see the light of the day, 594
00:52:49,600 --> 00:52:51,110
you won't believe this. 595
00:52:51,110 --> 00:52:53,820
I bring ω squared to the other side, 596
00:52:53,820 --> 00:53:03,510
and I get (see the blackboard) 597
00:53:03,510 --> 00:53:06,350
and the simplicity is overpowering, 598
00:53:06,350 --> 00:53:09,730
I feel it over my whole body. 599
00:53:09,730 --> 00:53:12,100
It is unbelievable. 600
00:53:12,100 --> 00:53:14,020
When they have a minus sign here, PAGE2
| Name | Version | Size | Date | User |
| 05_e.srt | 1 | 85872 | 2/17/06 4:17 AM | OOPSSJTU |
Last Modified 2/22/06 2:18 AM
|
