1
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The last time we have driven damped oscillations and we did steady state solutions. 2
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And the steady state solutions have no adjustable constants. 3
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That is strange. 4
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Because at time t equal zero, I can put the object at the certain position. 5
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I can give it certain velocity. And so I have two free choices. 6
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And yet those three choices did not show up in our steady state solution. 7
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Another word, the system has lost its memory of what happened at time t equal zero. 8
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And therefore I discussed with you that you have to wait to go into steady state. 9
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And that is what I will address today. 10
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There must be something missing what we did. 11
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So that the time t equal zero adjustables do show up. 12
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So let's return to our spring system. 13
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So we have here a spring, spring constant k, mass m and this object can be driven. 14
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You can do that with force directly on it here or you can shake on the left side. 15
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ω0 squared equals k/m. I've called that that way. 16
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And γ equals b/m. That's the damping. 17
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Now let us assume that I don't drive it, that I just give it a kick. 18
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Time t equals zero, I let it do its own thing. 19
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We discuss that two lectures ago, so we get an undriven situation. 20
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Not driven. And then the solution, you should remember, 21
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has this exponential decay in it, is that x equals some value X that follows 22
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from the initial conditions times (see the blackboard). 23
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And this the x and the α can be found, 24
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if you know at t equal zero what x is and what x dot is. 25
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ω'(prime) is a frequency which is (see the blackboard). 26
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Just a hair under ω0, but that depends of course on γ. 27
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So this is the situation when we are not driving. 28
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Let's now take a situation that we do drive. 29
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So we drive for instance with a force that we apply directly to this object 30
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which is one way of driving it. And then that force is F0 times cosωt. 31
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This ω is my ω. I force that ω on that system. 32
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I don't give a damped ω' is. I determine what the ω is. 33
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And that's tough luck for this object. 34
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And we know that in the steady state solution only this ω survives. 35
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That one will go. And the solution that we have seen 36
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that we derived last time is (see the blackboard). 37
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That is my ω. That is none negotiable. 38
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This is the steady state. And we derived that the tangent of δ was 39
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(see the blackboard) 40
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And we found for A this monstrous solution, remember? (see the blackboard). 41
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And we spent the entire lecture on dealing with this A. 42
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We changed the ω and we evaluated the ω equal zero, ω at resonance, 43
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ω at very high values. 44
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There is no adjustable constant in this solution. 45
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But there was one in that solution. 46
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Now let us look at the differential equations that we were solving. 47
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And so first we go to the undriven system. 48
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Undriven means at T equals zero, 49
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you give it a kick, you've just let it do its own thing. 50
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The differential equation that we then had was (see the blackboard). 51
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That's the one we have. And we solved it. 52
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And we found that solution there. 53
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Then we were driving it. So now we drive it. 54
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What now was the differential equation? 55
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Well, we had (see the blackboard) and now we had here this driving term. 56
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In the case that we directly put the force on the object, 57
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then we had here F0 divided by m times cosωt. 58
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Because the m comes in because you divide by m, right? 59
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You have Newton's Second Law has ma. 60
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But in the case that you shake the left side with your hands. 61
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You know, with this η0cosωt. 62
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That's another way of effectively driving the object. 63
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We do work that out, then we have here (see the blackboard). 64
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But in any case, you see here a driving term and we solve our equations. 65
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And if you take this case, then this is the solution. 66
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If you take this case, then this takes the place of that. 67
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Suppose now I take this solution, 68
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and I substitute that solution in this differential equation. 69
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Then I get the result is zero, because, look, forget this, so I put it in this part. 70
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I get zero because it fits this differential equation. 71
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So what is wrong with adding zero. 72
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Therefore if I add this two solutions, 73
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it must be a solution to this differential equation. 74
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Because you just add zero. 75
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And if you take 18.03, then they use very nice terms, they say, 76
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yes, of course, if you have this special solution which is this one, 77
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you have to add the homogeneous solution. 78
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And this what homogeneous means that you put a zero here. 79
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And so the general solution is really the sum of the two. 80
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By adding this one, you effectively add zero. 81
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So you add nothing, but you get something in return. 82
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And what you get in return are your two adjustable constants. 83
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So now you can deal with the situation that at T equal zero, 84
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you know exactly where that object is and what its velocity is. 85
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And so I will write down now the general solution, 86
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which is the one that governs a major part of the lecture today. 87
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So x now function of time is the steady state solution (see the blackboard) 88
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This is my ω. It is my will. That Walter Lewin's ω. 89
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Plus (see the blackboard). 90
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And this is the will of the oscillator, and this is my will. 91
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They are two different ωs. 92
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And now you can see what happens. 93
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You see that this term will never die out. It will last forever and ever and ever. 94
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But this one is going to die. It's 1/e decay time of 2/γ. 95
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And so if 2/γ happens to be 10 hours. 96
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Then you have to wait 10 hours for this one to be down by a factor of e. 97
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But if 2/γ happens to be one milliseconds, 98
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then all you have to do is what wait one milliseconds 99
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for that term to go down by a factor of e. 100
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So this is the one that will die out. That's why we have to wait. 101
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And so this is called the transient. And it will die out faster the higher γ is. 102
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And this is called then the steady state solution which ultimately survives. 103
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Suppose I told you that at t equal zero, x equal zero and x dot also equal zero. 104
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And I was so nasty to say "Oh, by the way, why don't you solve for x and α?" 105
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Would be a very nice thing to do, but will take you 50 minutes of grinding 106
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not so fast because remember if you have a x dot, 107
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you have to take the time derivative of this entire function. 108
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And you a t here, and you have a t there, and you have to substitute in there. 109
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Time t equal to zero and you have to make that equal zero. 110
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And it takes you 50 minutes and out pops indeed a value for x and a value for α. 111
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You haven't learned much physics when you do it, but you have learned some algebra. 112
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So I've decided not to spend my time on doing that, but in principle you must agree with me now 113
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that If I specify the initial conditions, I don't have to call this zero, I can call this x0, 114
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I can do anything I want to, I can get x dot any value I want to, 115
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then I get unique values for x and for α, 116
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and that is ultimately then what the solution is. 117
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I must add the two. 118
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The bottom line and that's really where the physics is 119
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and that has to do with Problem 2.5 that you have this week on your plate, 120
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is then the following. 121
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If I make a plot of x as a function of time, then the solution is really the sum of these two. 122
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This is the steady state solution which has a amplitude A, which is none negotiable, 123
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nothing to do with initial conditions. 124
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It has its own ω, Walter Lewin's ω. 125
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So let's assume that this is the period T. 126
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And so out of that pops, a nice cosinusoidal or sinusoidal curve, let me put it in here, 127
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and never changes. It's goes on forever and ever and ever. 128
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And this here, this time T is 2π/ω, my ω. 129
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However, there is also this one, but this one dies out. 130
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So I will now put in some kind of an exponential decay. 131
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Something like this. It has its own frequecny ω'. I can choose any thing. 132
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I can make ω larger, ω' larger than ω. I can make it smaller. 133
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So I just bit one. And let's suppose that the zero crossings are here and here and here... 134
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And so for instance the curve then has to be add it, should be something like this. 135
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And this time is T' is 2π/ω'. 136
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And so you see at the sum of the two, which I will not try to sketch, 137
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is the solution, but it is the pink one that's going to die out. 138
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That's why you have to be, sometimes you have to be patient 139
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except if it is one milliseconds 2/γ, you don't have to be very patient. 140
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Now let us look at the situation 141
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that and I can arrange that and I am going to arrange that. 142
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That ω and ω' are close together. I can choose that. 143
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The system cannot choose ω'. System is dark. 144
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This is ω'. System has no choice, but I have a choice. 145
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I can make the ω any value I want to, so I can make it very close to ω'. 146
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What do you think is going to happen now? 147
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When I turn this systme on, all of a sudden the driver. 148
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And my ω is very close to that ω'. 149
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It's true that the transient will die out, 150
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but let's say we take a system with a pretty high Q. 151
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So it doesn't die out so fast. 152
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What you expect you are going to see? 153
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Excellent, you are going to see beats. 154
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Because now you have two harmonic oscillations which have to be added. 155
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But the frequencies are a little different. 156
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And if this one survives long enough there comes a time that they are inphase 157
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there comes a time that they are out of phase. 158
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And when they are out of phase, you see very low amplitude. 159
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And so you are going to see a beat phenomenon. 160
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And that is exactly what I want you to see. 161
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For that we need a system presumably with a high Q, 162
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and then the driving frequency has to be close to the ω'. 163
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And for that, I have chosen this system here. 164
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Is this an air track? 165
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And we can make the damping very low. 166
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Unpleasantly low, believe me. 167
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A very low value of γ. When I turn on the air flow, 168
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So here is a spring, spring constant k here. 169
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Here is one spring constant k. 170
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And here is the mass. 171
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And there is very little friction. 172
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And now I am going to drive it, just a little bit off resonance. 173
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A little bit below the resonance frequency. 174
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And what you are going to see now is the sum of these two. 175
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But since the γ is so low, it will take a long time for this transient to die out. 176
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And that is exactly what I want you to see. 177
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In addition, you are going to beats. 178
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And as long as it is beating, you know that you haven't reach the steady state solution yet. 179
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And if you are patient, and I am patient, we probably will see it go into steady state, 180
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but may take several minutes. So you're ready for that. I'm going to drive it here. 181
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And I start driving now. So relax and look at the amplitude of this object, 182
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and see what happens. 183
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Hey, Hey! The amplitude is going down. 184
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Hey, Hey, Hey! I call that beat. 185
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You see that? If you see the amplitude go down, 186
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the two frequencies were beating against each other. 187
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Now it's picking up again. 188
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It's nowhere near in steady state. 189
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Very low γ, very high Q. There we go again. 190
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Amplitude way down, picks up again. 191
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Just be patient. Let's be patient and see 192
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whether we have the privliege of seeing it go into steady state. 193
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It's a very high Q system. Since I'm just below resonance. 194
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The driver and the car will be inphase, 195
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when I below resonance in steady state solution. 196
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So this δ will be very close to zero below resonance. Above resonance, 197
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180 degrees out of phase. But I'm just below resonance. 198
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So when we going to steady state, we also will see that we are very close to a δ of zero. 199
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Now let's see what the amplitude is now and whether the amplitude is changing. 200
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Well, we're getting there. It pays off to be patient. 201
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Later today I will do an experiment while 2/γ is two milliseconds. 202
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So all you have to do is wait for milliseconds. 203
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I make up for the fact of now you have to wait so long. 204
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So let's take a look at this now. 205
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I think it looks terrific. It looks terrific. 206
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Inphase, I don't see much beating any more. 207
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It looks alike the amplitude is constant. 208
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I think we are now... We killed this one. 209
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And I think this one has survived. 210
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If you make, if you increase the damping, 211
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This would happen of course earlier that you are going to the steady state solution. 212
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Looks great! I don't see any change any more in the amplitude. 213
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So that's the A that you have there, capital A. 214
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And they are nicely inphase. 215
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So very high Q system. 216
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So the change to go from δ zero to δ π/2 at resonance 217
00:20:20,480 --> 00:20:24,610
takes place over an extremely narrow range of frequency. 218
00:20:24,610 --> 00:20:28,920
So they're still inphase. 219
00:20:28,920 --> 00:20:35,510
All right. 220
00:20:35,510 --> 00:20:44,230
If we are driving, this system with a force say directly on the object, 221
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F zero coswt. 222
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Then in steady state there is energy dissipation. 223
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Because there is friction. 224
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And where there is friction, there is heat. 225
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And that means energy, that means I have to do work to provide that energy. 226
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So it's a steady state situation. 227
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So as the thing is never changing its A just going on forever and ever. 228
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While that happens I must put in energy which comes out in the form of heat. 229
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So let us return to the good old days of 8.01. 230
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And I want to remind you that work is the dot product between a force and a 231
00:21:31,950 --> 00:21:34,830
displacement dx. 232
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It's a dot product. It's a scaler work. 233
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Little bit work is done by this force if it moves over a distance dx. 234
00:21:44,280 --> 00:21:48,880
If the two are perpendicular to each other, then no work is done. 235
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Satellites into circular orbit around the earth no work is done. 236
00:21:58,140 --> 00:22:02,470
And so now I can calculate what the power is. 237
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Because the power is dw/dt how many Joules per second I have to put into the system. 238
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And so if I take the derivative, time derivative. 239
00:22:16,170 --> 00:22:23,810
Then I get F dotted with V because the x/dt is simply the velocity. 240
00:22:23,810 --> 00:22:27,490
Now if I have a one dimensional system, and what I mean by that is that 241
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the force is either in this direction or in this direction. 242
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And the velocity is either in this direction or this direction. 243
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That's what I mean by one dimensional system. 244
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Then I can delete the dot and then the signs will automatically take care of the direction. 245
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And minus v is then this and plus v is that and same for force. 246
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So I can kill the dot. 247
00:22:52,190 --> 00:22:59,400
So now I have to know what the velocity is in the steady state solution. 248
00:22:59,400 --> 00:23:01,400
Oh, that's easy. 249
00:23:01,400 --> 00:23:06,990
Because I go to the steady state solution here and I calculate what x dot is. 250
00:23:06,990 --> 00:23:08,660
So I put that here. 251
00:23:08,660 --> 00:23:12,190
x dot which is v. 252
00:23:12,190 --> 00:23:18,380
So the derivative of coswt is -ωsinωt. 253
00:23:18,380 --> 00:23:26,620
So I got -ωAsin(ωt-δ). 254
00:23:26,620 --> 00:23:28,840
That's the velocity. 255
00:23:28,840 --> 00:23:32,850
But I know what the force is, it's F zero coswt. 256
00:23:32,850 --> 00:23:38,540
So let we go.(see the blackboard) 257
00:23:38,540 --> 00:23:43,580
But I'm going to put these in also now, so I get ωA, 258
00:23:43,580 --> 00:23:47,390
and then I put in the cosωt. 259
00:23:47,390 --> 00:23:55,190
And I'm going to put in this one, sin(ωt-δ). 260
00:23:55,190 --> 00:23:57,860
So take a deep breath. 261
00:23:57,860 --> 00:24:02,030
F zero coswt is the force. 262
00:24:02,030 --> 00:24:03,770
See that there. 263
00:24:03,770 --> 00:24:10,560
And the velocity has been derived from the first derivative of the steady state solution 264
00:24:10,560 --> 00:24:17,400
Give me (see the blackboard) 265
00:24:17,400 --> 00:24:34,380
Now this, what we have between brackets, can be written as the (see the blackboard) 266
00:24:34,380 --> 00:24:40,620
So what you see here between brackets is the same what you see there. 267
00:24:40,620 --> 00:24:48,800
And I'm really interested in knowing every moment in time what exactly the power is. 268
00:24:48,800 --> 00:24:55,290
Not really, most of the time, I'm really interested in knowing what the average power is. 269
00:24:55,290 --> 00:24:58,020
That is required to keep the thing going. 270
00:24:58,020 --> 00:25:01,900
Average over one oscillation or over a hundred of oscillations. 271
00:25:01,900 --> 00:25:04,220
That's really what I'm interested in. 272
00:25:04,220 --> 00:25:07,730
Not necessarilly the instantaneous power. 273
00:25:07,730 --> 00:25:20,130
In other words, most of the time, my interest is really in what this is. That is the time average value. 274
00:25:20,130 --> 00:25:28,340
And think of it as being one oscillation that is fine but you can think of it also as daze. 275
00:25:28,340 --> 00:25:34,120
Now I need some experts in the audience. 276
00:25:34,120 --> 00:25:41,060
I see here with coswt and I see here the sinwt. 277
00:25:41,060 --> 00:25:47,540
What is the time average value of that product? 278
00:25:47,540 --> 00:25:51,580
The time average value would be the sinwt and the coswt 279
00:25:51,580 --> 00:25:58,660
if I average it over one cycle or two cycles or three cycles. 280
00:25:58,660 --> 00:26:00,470
Come on. 281
00:26:00,470 --> 00:26:03,420
Highschool? Yeah! 282
00:26:03,420 --> 00:26:07,430
What is the time average of sinwt times coswt. 283
00:26:07,430 --> 00:26:10,910
Time average of one period. 284
00:26:10,910 --> 00:26:13,040
Zero. 285
00:26:13,040 --> 00:26:17,650
This one times this one time average gives me a zero. 286
00:26:17,650 --> 00:26:19,090
You don't believe it? 287
00:26:19,090 --> 00:26:20,930
Go back to high school. 288
00:26:20,930 --> 00:26:22,230
Ask your high school teacher. 289
00:26:22,230 --> 00:26:24,920
He will agree with me. 290
00:26:24,920 --> 00:26:29,600
Here I see a coswt and I see a coswt there. 291
00:26:29,600 --> 00:26:34,750
What is the time average of cosine square ωt? 292
00:26:34,750 --> 00:26:36,570
Ah...you guys are waking up. 293
00:26:36,570 --> 00:26:38,420
It's one half. 294
00:26:38,420 --> 00:26:43,680
So therefore, for this product, sorry, for this product, 295
00:26:43,680 --> 00:26:45,630
I can write one a half. 296
00:26:45,630 --> 00:26:49,610
And so now we get...Notice that it's a minus here. 297
00:26:49,610 --> 00:26:53,740
And so this minus picks of this minus so it becomes a plus. 298
00:26:53,740 --> 00:26:59,070
So I get F zero. I get the half that is here. 299
00:26:59,070 --> 00:27:06,660
Then I get ω, I get my A, and I get the sinδ. 300
00:27:06,660 --> 00:27:08,390
Are we happy with that? 301
00:27:08,390 --> 00:27:15,900
So you see it collapes into something that is relatively simple. 302
00:27:15,900 --> 00:27:18,440
What is sinδ? 303
00:27:18,440 --> 00:27:24,140
Well, I remember what tanδ is. 304
00:27:24,140 --> 00:27:28,420
And if I know tanδ can I then calculate sinδ? 305
00:27:28,420 --> 00:27:31,590
And the answer is yes of course. 306
00:27:31,590 --> 00:27:37,850
If this is δ, and if this is ωγ. 307
00:27:37,850 --> 00:27:42,490
And this is ω zero squared minus ω squared. 308
00:27:42,490 --> 00:27:46,050
Then it only takes Pythagoras to calculate what this is. 309
00:27:46,050 --> 00:28:08,420
And so I know that the sinδ must be (see the blackboard) 310
00:28:08,420 --> 00:28:14,930
So yes, I do know what the sinδ is. 311
00:28:14,930 --> 00:28:22,300
So now I can come to a close by substituting it in here what the sinδ is. 312
00:28:22,300 --> 00:28:25,650
And I can substitute it in here what A is. 313
00:28:25,650 --> 00:28:29,270
Here is A. 314
00:28:29,270 --> 00:28:30,370
You know what is nice? 315
00:28:30,370 --> 00:28:36,880
Look at the downstairs here, the same as the downstairs there. 316
00:28:36,880 --> 00:28:42,290
So if I multiply them, square root goes away. 317
00:28:42,290 --> 00:28:49,290
So if I can write down now what P average is 318
00:28:49,290 --> 00:29:01,940
Let's go slowly. So we get F0 divided by two, I get an ω. 319
00:29:01,940 --> 00:29:08,330
I will go to the A very shortly. I will first pick ωγ here which is my sinδ. 320
00:29:08,330 --> 00:29:12,060
So that makes this is a square and I get γ. 321
00:29:12,060 --> 00:29:17,260
And now I turn to the A which is F0/m 322
00:29:17,260 --> 00:29:19,430
So I get a square here. 323
00:29:19,430 --> 00:29:21,720
And I get an m here. 324
00:29:21,720 --> 00:29:27,830
And now, I get this downstairs, times this downstairs. 325
00:29:27,830 --> 00:29:43,240
The square root disappears and so I get(see the blackboard) 326
00:29:43,240 --> 00:29:51,670
Almost end of story. I'm going to rewrite it a little as Tony French. 327
00:29:51,670 --> 00:29:53,050
In his book rewrites it in again a different way. 328
00:29:53,050 --> 00:29:56,450
He loves to work with Qs. He puts the Qs in there. 329
00:29:56,450 --> 00:30:01,800
I'm going to divide upstairs and downstairs by ω squared. 330
00:30:01,800 --> 00:30:29,070
And when I do that I get (see the blackboard) 331
00:30:29,070 --> 00:30:31,110
That's what I get. That's not the only way you can write it. 332
00:30:31,110 --> 00:30:32,440
But that's one way you can write it. 333
00:30:32,440 --> 00:30:33,590
Let me check that. 334
00:30:33,590 --> 00:30:42,130
I have F0 squared, I have γ, two m here and I have this downstairs. 335
00:30:42,130 --> 00:30:45,710
So the time has come now. 336
00:30:45,710 --> 00:30:48,990
To try to see through this equation. 337
00:30:48,990 --> 00:30:53,580
Remember last lecture we spend the whole lecture not to look at this done equation 338
00:30:53,580 --> 00:30:55,540
but to see through it. 339
00:30:55,540 --> 00:30:57,700
And we will able to see through it. 340
00:30:57,700 --> 00:31:00,430
See all its idiosyncrasies. 341
00:31:00,430 --> 00:31:10,990
Let's look at the idiosyncrasies of this average power over one circle or a multiple of circles. 342
00:31:10,990 --> 00:31:18,630
Well, let us first make γ infinitely large. 343
00:31:18,630 --> 00:31:20,990
Don't look even at the equation. 344
00:31:20,990 --> 00:31:24,380
γ, infinitely large. 345
00:31:24,380 --> 00:31:27,520
That means there is an infinite amount of friction. 346
00:31:27,520 --> 00:31:29,700
System never gets going. 347
00:31:29,700 --> 00:31:32,140
There is no way that it'll ever move. 348
00:31:32,140 --> 00:31:37,930
So clearly the average power must go to zero. 349
00:31:37,930 --> 00:31:41,350
And indeed, you see that the γ is downstairs. 350
00:31:41,350 --> 00:31:46,840
If γ goes to infinity, power is zero. 351
00:31:46,840 --> 00:31:49,830
Let's make the mass infinitely large. 352
00:31:49,830 --> 00:31:53,240
If the mass is infinitely large you have infinitely high inertia. 353
00:31:53,240 --> 00:31:54,920
Nothing will ever get going. 354
00:31:54,920 --> 00:31:57,580
No force will ever get the mass going. 355
00:31:57,580 --> 00:32:02,480
Well, that means, you expect that the power goes to zero. 356
00:32:02,480 --> 00:32:09,390
And indeed you see here if m goes to infinity the power goes to zero. 357
00:32:09,390 --> 00:32:13,410
Let's say the force goes to zero. 358
00:32:13,410 --> 00:32:14,510
We are not even driving it. 359
00:32:14,510 --> 00:32:16,780
Well if we are not even driving it, 360
00:32:16,780 --> 00:32:20,350
I hope you will agree with me you don't have to put in any work, right? 361
00:32:20,350 --> 00:32:21,570
Nothing gets going. 362
00:32:21,570 --> 00:32:25,880
So clearly you expect then that the power will go to zero. 363
00:32:25,880 --> 00:32:31,890
And indeed if F zero goes to zero, you see that the power is zero. 364
00:32:31,890 --> 00:32:36,040
Suppose you make ω0. 365
00:32:36,040 --> 00:32:38,920
So that means there will never be any velocity. 366
00:32:38,920 --> 00:32:41,060
You never pick up any velocity. 367
00:32:41,060 --> 00:32:46,210
Takes infinitely long. ω is zero. 368
00:32:46,210 --> 00:32:52,540
So clearly, if nothing ever moves, the power will go to zero. 369
00:32:52,540 --> 00:32:57,960
Now look if ω makes zero, then this is zero and this one goes to infinity. 370
00:32:57,960 --> 00:33:00,650
And therefore the power goes to zero. 371
00:33:00,650 --> 00:33:09,780
So you need no equations for that, just common sense to immediately conclude that this has to be the case. 372
00:33:09,780 --> 00:33:14,710
Suppose you go to infinity with ω. 373
00:33:14,710 --> 00:33:16,770
Very very fast. 374
00:33:16,770 --> 00:33:24,420
Well, if you go infinitely fast, because of the inertia of the system it can never react, can never get going. 375
00:33:24,420 --> 00:33:29,520
So I will predict that then the power must go to zero. 376
00:33:29,520 --> 00:33:34,750
And if you put ω infinity in here, this goes to zero, this goes to infinity 377
00:33:34,750 --> 00:33:37,370
and so the power goes to zero. 378
00:33:37,370 --> 00:33:40,040
So all these is complete common sense. 379
00:33:40,040 --> 00:33:43,320
All of these you could have predicted without that equation. 380
00:33:43,320 --> 00:33:44,800
But isn't it nice? 381
00:33:44,800 --> 00:33:48,850
That equation supports my intuition. 382
00:33:48,850 --> 00:33:54,870
So now comes the question if ω goes to ω0, 383
00:33:54,870 --> 00:33:58,350
and that is the reason why I wrote it in this way. 384
00:33:58,350 --> 00:34:05,860
Notice when ω goes to zero, when ω goes to ω0, this one goes to zero. 385
00:34:05,860 --> 00:34:16,380
And therefore, that is the frequency at which the average power is the maximum ever, can never be any higher. 386
00:34:16,380 --> 00:34:19,460
Because it's independent of γ, right, my ω? 387
00:34:19,460 --> 00:34:27,120
So having ω equals ω zero, we reach the maximum value of power. 388
00:34:27,120 --> 00:34:31,290
It can never go any higher, it's exactly at ω0. 389
00:34:31,290 --> 00:34:45,650
And this is zero, you lose one γ, and so you find that this value then becomes F0 squared divided by two m γ. 390
00:34:45,650 --> 00:34:52,780
And you can write that, rewrite that a little bit, I do that because tony franchy likes Qs. 391
00:34:52,780 --> 00:34:55,320
And so I write it with a Q in there. 392
00:34:55,320 --> 00:35:02,300
Q is, remember that? Q is ω0/γ. 393
00:35:02,300 --> 00:35:14,850
So I can rewrite this as Q times F0 squared upstairs so I get the Q in there which is nice divided by 2mω0. 394
00:35:14,850 --> 00:35:21,880
So this is the same thing. 395
00:35:21,880 --> 00:35:37,460
So if I now plot make a curve for you of P average not P maximum but P average as a function of frequency. 396
00:35:37,460 --> 00:35:45,510
So here is ω and here is P average and here is ω0 397
00:35:45,510 --> 00:35:48,870
It goes through a maximum exactly at ω0. 398
00:35:48,870 --> 00:35:54,620
It starts at zero, you see? 399
00:35:54,620 --> 00:36:04,000
And it ends at zero, and it sweeps up to a maximum and then it goes down again. 400
00:36:04,000 --> 00:36:10,220
And for reasonable values of Q these curves look extremely symmetric. 401
00:36:10,220 --> 00:36:25,060
And so this value here is then P average max which is that value, this value. 402
00:36:25,060 --> 00:36:36,160
If we look at the width of this curve at half maximum of the power. 403
00:36:36,160 --> 00:36:39,410
So this is one half times P maximum. 404
00:36:39,410 --> 00:36:43,720
Then you can show and this is not so difficult algebraically. 405
00:36:43,720 --> 00:36:46,000
But I will not attempt it. 406
00:36:46,000 --> 00:36:55,550
You can show that the width at half maximum is very close to γ. 407
00:36:55,550 --> 00:36:57,510
Remember γ and ω have the same units. 408
00:36:57,510 --> 00:37:00,830
One divided by a second. 409
00:37:00,830 --> 00:37:10,260
In other words, if you go to half maximum, this point here is ω0-γ/2. 410
00:37:10,260 --> 00:37:17,880
And this point here is ω0+γ/2. 411
00:37:17,880 --> 00:37:21,700
And so you see immediately which of course make sense. 412
00:37:21,700 --> 00:37:28,210
That if γ is a very small that the peak gets very narrow. 413
00:37:28,210 --> 00:37:32,710
And if γ is very high, the peak gets very broad. 414
00:37:32,710 --> 00:37:35,080
That's intuitively quite pleasing. 415
00:37:35,080 --> 00:37:39,030
High Q systems very narrow peaks. 416
00:37:39,030 --> 00:37:48,960
And that's the way that Tony French likes to plot his data and I'll show you that on the overhead here. 417
00:37:48,960 --> 00:37:55,370
This is just a picture from your book. 418
00:37:55,370 --> 00:38:02,830
Oh what Tony does here he plots not ω here, but he plots ω divided by ω0. 419
00:38:02,830 --> 00:38:07,180
So that means the resonance is at one. 420
00:38:07,180 --> 00:38:10,450
And he doesn't plot the average value for P here. 421
00:38:10,450 --> 00:38:13,390
But he plots it into strange units. 422
00:38:13,390 --> 00:38:17,980
Into the unit F zero squared divided by 2mω0. 423
00:38:17,980 --> 00:38:23,510
So now he affectively can compare the verticle axis with the Q value 424
00:38:23,510 --> 00:38:27,660
because he likes the fact that is Q times higher than something. 425
00:38:27,660 --> 00:38:32,410
And he has plotted this in terms of that something. 426
00:38:32,410 --> 00:38:37,330
And so if you take the curve for Q equals ten which has the peak here in power, 427
00:38:37,330 --> 00:38:44,100
you see indeed that he finds that very close on this scale to ten. 428
00:38:44,100 --> 00:38:47,530
Notice also the nice symmetry. 429
00:38:47,530 --> 00:38:54,190
And you see for lower values of Q which are curves here that indeed the peak get broader. 430
00:38:54,190 --> 00:38:58,110
The width of this peak is 1/Q. 431
00:38:58,110 --> 00:39:01,810
Because you remember this axis is ω divided by ω0. 432
00:39:01,810 --> 00:39:09,040
So if the width is γ, on that peak, it is now γ/ω0 in this plot. 433
00:39:09,040 --> 00:39:12,550
And γ/ω0 is 1/Q. 434
00:39:12,550 --> 00:39:17,890
So here the width in this presentation is directly inverse the proportional of the Q. 435
00:39:17,890 --> 00:39:22,460
So if Q is ten, then the width there is one tenth. 436
00:39:22,460 --> 00:39:26,960
He then shows you another plot whereby he does what I did there, 437
00:39:26,960 --> 00:39:28,990
He plots it as a function of ω. 438
00:39:28,990 --> 00:39:32,560
Not as a function of ω divided by ω0. 439
00:39:32,560 --> 00:39:37,430
And then he emphasizes the fact that the width here is that γ that I mentioned. 440
00:39:37,430 --> 00:39:45,300
And half the maximum power you get here the width of γ. 441
00:39:45,300 --> 00:39:55,040
And this is a picture that I choose forbidden from your book. 442
00:39:55,130 --> 00:40:00,900
This is the best moment for the break that means the mini quiz. 443
00:40:00,900 --> 00:40:02,410
I realize it's a bit early; 444
00:40:02,410 --> 00:40:04,000
we are only 40 minutes into the lecture. 445
00:40:04,000 --> 00:40:05,530
But it's a natural point; 446
00:40:05,530 --> 00:40:09,370
you'll see what comes afterwards. 447
00:40:09,370 --> 00:40:11,940
It’s better that we make the break now. 448
00:40:11,940 --> 00:40:15,360
So therefore I need some help from people 449
00:40:15,360 --> 00:40:24,530
we are willing to hand out the mini quiz. 450
00:40:24,530 --> 00:40:27,780
It will be nice if I can find the mini quizzes. 451
00:40:27,780 --> 00:40:28,650
I have them here, 452
00:40:28,650 --> 00:40:29,730
but someone took them...Oh no, 453
00:40:29,730 --> 00:40:33,990
they're still here...A nice conspiracy. 454
00:40:33,990 --> 00:40:37,470
Afterwards, after the break, 455
00:40:37,470 --> 00:40:39,920
we will collect them this time in some boxes, 456
00:40:39,920 --> 00:40:42,490
so that it's a little bit more organized. 457
00:40:44,150 --> 00:40:50,390
So I'm returning to an RLC circuit which we discussed earlier. 458
00:40:50,390 --> 00:40:55,290
Good old days of 8.02. 459
00:40:55,290 --> 00:40:59,700
I' m going to drive it now, not with a battery, 460
00:40:59,700 --> 00:41:16,300
but with an alternating power supply Vocosωt. 461
00:41:18,440 --> 00:41:23,830
Yeah... Put it in here. Thank you. 462
00:41:23,830 --> 00:41:29,890
So here is the circuit, resistor R, 463
00:41:29,890 --> 00:41:37,260
self-inductance L, capacitor C. 464
00:41:37,260 --> 00:41:44,290
And I have to write down now the differential equation, 465
00:41:44,290 --> 00:41:47,800
I will adopt a positive current in this direction. 466
00:41:47,800 --> 00:41:54,130
That will be my positive current. 467
00:41:54,130 --> 00:41:59,820
The charge here on this right plate I will call q, 468
00:41:59,820 --> 00:42:09,500
and therefore by that definition I is then dq/dt, sign sensitive. 469
00:42:09,500 --> 00:42:12,530
I call the potential difference over this capacitor 470
00:42:12,530 --> 00:42:16,480
in going from the right side to the left side. 471
00:42:16,480 --> 00:42:23,170
I call that V of C, that then is q/C. 472
00:42:23,170 --> 00:42:27,750
All of that is sign sensitive. 473
00:42:28,760 --> 00:42:30,830
I go around the circuit, 474
00:42:30,830 --> 00:42:35,150
and I want to calculate the close loop integral of E dot dl. 475
00:42:35,150 --> 00:42:41,730
And that close loop integral of E dot dl is not zero, 476
00:42:41,730 --> 00:42:45,210
which many books tell you, 477
00:42:45,210 --> 00:42:46,740
even many professors tell you. 478
00:42:46,740 --> 00:42:55,260
It is not zero, but it is -dΦ/dt. This is Faraday's law. 479
00:42:55,260 --> 00:42:59,300
And this runs our economy, 480
00:42:59,300 --> 00:43:03,870
because of the magnetic flux change in close loops, 481
00:43:03,870 --> 00:43:06,510
we can generate induced EMFs, 482
00:43:06,510 --> 00:43:10,960
which run our economy. Look at the lights. 483
00:43:10,960 --> 00:43:16,160
Luckily this is not zero. 484
00:43:16,160 --> 00:43:20,110
This Φ is the magnetic flux that goes through a surface, 485
00:43:20,110 --> 00:43:25,810
any surface that you can attach to this close loop. 486
00:43:25,810 --> 00:43:29,050
So I've done this before, so I can do it a little faster. 487
00:43:29,050 --> 00:43:34,120
I go from here to here, so that is IR. 488
00:43:34,120 --> 00:43:37,360
There is no electric field inside this ideal self-inductor 489
00:43:37,360 --> 00:43:41,240
because super conducting wire can not be an E field. 490
00:43:41,240 --> 00:43:44,990
So that is zero going from here to there. 491
00:43:44,990 --> 00:43:46,900
When I go over the capacitor, 492
00:43:46,900 --> 00:43:54,070
I get my Vc, and here depending upon the phase, 493
00:43:54,070 --> 00:43:56,470
if I assume this plus and this minus, 494
00:43:56,470 --> 00:43:58,540
but you can reverse that, 495
00:43:58,540 --> 00:44:01,870
then I will get when I walking through this direction, 496
00:44:01,870 --> 00:44:07,040
I would get minus Vocosωt, 497
00:44:07,040 --> 00:44:08,820
but if you feel like reversing it, 498
00:44:08,820 --> 00:44:10,540
I have no problem with that. 499
00:44:10,540 --> 00:44:13,050
That's just a matter of 180 degrees phase. 500
00:44:13,050 --> 00:44:14,750
It's no different physics. 501
00:44:14,750 --> 00:44:19,020
And this now equals -dΦ/dt, 502
00:44:19,020 --> 00:44:21,360
the only thing when you apply Faraday's Law, 503
00:44:21,360 --> 00:44:24,120
you should always integrate in the direction 504
00:44:24,120 --> 00:44:26,970
that you have your current assumed. 505
00:44:26,970 --> 00:44:29,030
That it is -LdI/dt, 506
00:44:29,030 --> 00:44:31,140
if you do it in the opposite direction, 507
00:44:31,140 --> 00:44:33,130
then it is +LdI/dt. 508
00:44:33,130 --> 00:44:36,510
I have learned certain discipline in my life, 509
00:44:36,510 --> 00:44:38,040
took me many years, 510
00:44:38,040 --> 00:44:39,290
so you have a long way to go 511
00:44:39,290 --> 00:44:41,590
and I always go in the direction of I, 512
00:44:41,590 --> 00:44:43,250
so I never have to think. 513
00:44:43,250 --> 00:44:47,450
So this is -LdI/dt. 514
00:44:47,450 --> 00:44:51,050
This now covers the -dΦ/dt. 515
00:44:51,050 --> 00:44:53,680
So now what I do, I bring the L in 516
00:44:53,680 --> 00:44:57,480
and I take one more time derivative. 517
00:44:57,480 --> 00:45:06,640
And so I get (see the blackboard), 518
00:45:06,640 --> 00:45:08,190
but I take the time derivative, 519
00:45:08,190 --> 00:45:14,340
so the q dot becomes I, so I get I/C. 520
00:45:14,340 --> 00:45:19,240
And that now becomes the time derivative of this function. 521
00:45:19,240 --> 00:45:23,680
But it goes to the right side, which makes the minus sign plus. 522
00:45:23,680 --> 00:45:29,220
But when I take the derivative of the cosωt, I get -ω out. 523
00:45:29,220 --> 00:45:38,680
So I get here -Voωsinωt. 524
00:45:38,680 --> 00:45:42,340
This is the differential equation that has to be solved 525
00:45:42,340 --> 00:45:48,580
and I will divide this out by L. I will take... 526
00:45:48,580 --> 00:45:50,830
divide everything by L. 527
00:45:50,830 --> 00:45:59,080
I will put the C a little higher, and so with R/L, γ 528
00:45:59,080 --> 00:46:03,760
and with ωo squared equals 1/LC. 529
00:46:03,760 --> 00:46:08,410
This becomes then...(see the picture) 530
00:46:29,850 --> 00:46:35,020
Here you see a differential equation, 531
00:46:35,020 --> 00:46:43,500
and that differential equation looks amazingly similar to this one. 532
00:46:43,500 --> 00:46:46,450
And so you should be able to solve that. 533
00:46:46,450 --> 00:46:48,870
In fact, you wouldn't even want to solve 534
00:46:48,870 --> 00:46:52,020
but you can write down immediately the answer. 535
00:46:52,020 --> 00:46:56,810
You're going to get then I which is an Io, 536
00:46:56,810 --> 00:46:59,660
which is... It takes the place of that A there. 537
00:46:59,660 --> 00:47:01,190
This is steady state solution. 538
00:47:01,190 --> 00:47:10,880
Now I only go for steady state solution. Times the sin(ωt-δ). 539
00:47:10,880 --> 00:47:13,230
No adjustable constants. 540
00:47:13,230 --> 00:47:19,410
It's a steady state solution that I have...Steady state. 541
00:47:19,410 --> 00:47:22,910
And I will leave you to find me Io, 542
00:47:22,910 --> 00:47:25,870
and you can work out what δ is. 543
00:47:25,870 --> 00:47:28,970
That is part of your problem set anyhow. 544
00:47:28,970 --> 00:47:32,190
But with the knowledge that you have here, 545
00:47:32,190 --> 00:47:38,120
you could write it down in the matter of seconds. 546
00:47:38,120 --> 00:47:41,990
So without my telling you what Io is, 547
00:47:41,990 --> 00:47:49,370
at least working it out algeraically. We can talk 8.02 548
00:47:49,370 --> 00:47:51,720
and then we can make all kinds of predictions 549
00:47:51,720 --> 00:47:54,050
without even looking at the equation. 550
00:47:54,050 --> 00:47:55,210
So that's interesting. 551
00:47:55,210 --> 00:47:56,610
So everything that I'm going to tell you now, 552
00:47:56,610 --> 00:47:59,410
I do without knowing what Io is, 553
00:47:59,410 --> 00:48:03,780
and it's better work out that way. 554
00:48:03,780 --> 00:48:12,890
Suppose I make ω go to zero. Remember 8.02? 555
00:48:12,890 --> 00:48:15,880
Remember the word "reactance"? 556
00:48:15,880 --> 00:48:19,730
That the capacitor has a certain reactance 557
00:48:19,730 --> 00:48:26,230
which is 1/ωC, which has unit of Ωs? 558
00:48:26,230 --> 00:48:30,880
If ω goes to zero, this reactance goes to infinity. 559
00:48:30,880 --> 00:48:33,080
No current can ever flow. 560
00:48:33,080 --> 00:48:40,170
So I predict that Io will go to zero. 561
00:48:40,170 --> 00:48:46,270
Suppose my ω goes to ω0, 562
00:48:46,270 --> 00:48:49,710
now your memory may fail you here on 8.02. 563
00:48:49,710 --> 00:48:55,590
but we have a wonderful demonstration in 8.02 564
00:48:55,590 --> 00:49:01,310
that at resonance, 1/ωC, 565
00:49:01,310 --> 00:49:04,870
which is the reactance of the capacitor -ωL 566
00:49:04,870 --> 00:49:11,160
which is the reactance of the inductor is zero. 567
00:49:11,160 --> 00:49:16,010
That determines actually the resonance. 568
00:49:16,010 --> 00:49:18,690
And when this is the case, 569
00:49:18,690 --> 00:49:19,670
perhaps you remember 570
00:49:19,670 --> 00:49:24,060
that the system doesn't even know there is a capacitor 571
00:49:24,060 --> 00:49:26,460
and doesn't even know there is a self-inductor. 572
00:49:26,460 --> 00:49:32,400
The two at all moments in time exactly cancel each other. 573
00:49:32,400 --> 00:49:36,330
And therefore Ohm's Law holds. 574
00:49:36,330 --> 00:49:38,610
There is no L, there is no C, 575
00:49:38,610 --> 00:49:43,460
there is only the power supply and the resistor. 576
00:49:43,460 --> 00:49:46,780
And so since Ohm's Law says V=IR, 577
00:49:46,780 --> 00:49:52,570
you must get Io = Vo/R, 578
00:49:52,570 --> 00:49:55,590
that is what you must get at resonance. 579
00:49:55,590 --> 00:50:01,440
It's none negotiable. 580
00:50:02,150 --> 00:50:07,460
Now if ω goes to infinity, 581
00:50:07,460 --> 00:50:15,800
ωL says " Haha...Yeah...over my dead body. No current ever!" 582
00:50:15,800 --> 00:50:20,390
Imagine a very fast changing signal, 583
00:50:20,390 --> 00:50:24,050
that was the whole self-inductance, it's about, 584
00:50:24,050 --> 00:50:28,380
doesn't want any changes; it's conservative like you and me, 585
00:50:28,380 --> 00:50:30,240
and so the self-inductor says 586
00:50:30,240 --> 00:50:34,120
" sorry, the reactance is infinitely high. No current." 587
00:50:34,120 --> 00:50:38,600
And so Io goes to zero. 588
00:50:38,600 --> 00:50:43,560
And so I make these predictions, 589
00:50:43,560 --> 00:50:47,260
and that's always nice that you can use some knowledge 590
00:50:47,260 --> 00:50:50,550
to make predictions without even having ever looked 591
00:50:50,550 --> 00:50:51,880
at this differential equation. 592
00:50:51,880 --> 00:50:54,650
Any of these predictions I made did not come out of 593
00:50:54,650 --> 00:50:59,900
my knowledge of that differential equation. 594
00:50:59,900 --> 00:51:05,340
So if we make now a plot of the current. 595
00:51:05,340 --> 00:51:07,070
Io that is not the current, 596
00:51:07,070 --> 00:51:10,000
but that is the maximum possible current, 597
00:51:10,000 --> 00:51:15,700
and it is that Io that you see here without having solved it. 598
00:51:15,700 --> 00:51:19,430
I can look now what is going to do. 599
00:51:19,430 --> 00:51:24,240
Exactly at ωo, it will go to a maximum; 600
00:51:24,240 --> 00:51:30,210
it will start at zero and go to the maximum 601
00:51:30,210 --> 00:51:32,660
and then it will fall off a zero. 602
00:51:32,660 --> 00:51:38,780
And this value here is Vo/R. 603
00:51:38,780 --> 00:51:44,960
This is not power. I have now plotted current. 604
00:51:44,960 --> 00:51:48,530
Power of course would go as I squared R, 605
00:51:48,530 --> 00:51:51,370
that is the heat that you dissipate in the resistor, 606
00:51:51,370 --> 00:51:53,060
so that would go as I squared. 607
00:51:53,060 --> 00:51:59,350
I plotted here I as a function of...not t... 608
00:51:59,350 --> 00:52:06,130
but this is I as a function of ω. 609
00:52:07,110 --> 00:52:08,740
You know when you catch an error that I make, 610
00:52:08,740 --> 00:52:10,820
you get partial credit for this course. 611
00:52:10,820 --> 00:52:15,950
So please when you see me make a mistake, scream. 612
00:52:15,950 --> 00:52:22,140
So this is ω. Here is zero. 613
00:52:22,140 --> 00:52:26,000
And this is what I want to demonstrate now. 614
00:52:26,000 --> 00:52:32,820
I'm not going to show you Io only, 615
00:52:32,820 --> 00:52:36,360
but what I'm going to do is the following. 616
00:52:36,360 --> 00:52:43,240
I'm going to show you what I is as a function of ω. 617
00:52:43,240 --> 00:52:47,370
Here being the resonance. 618
00:52:47,370 --> 00:52:55,690
Let us suppose I pick this ω. Well, then this is my solution. 619
00:52:55,690 --> 00:53:01,110
So yes, the amplitude is Io, but here is the sin(ωt-δ). 620
00:53:01,110 --> 00:53:10,000
So all you will see then is this... you plus zero, minus plus zero... 621
00:53:10,000 --> 00:53:15,250
if I do it here, that it go... 622
00:53:15,250 --> 00:53:19,150
and the demonstration that we have prepared for you is one 623
00:53:19,150 --> 00:53:25,280
whereby we will sweep ω from zero to a value 624
00:53:25,280 --> 00:53:29,810
which I remember in terms of Hz, is about 2000Hz. 625
00:53:29,810 --> 00:53:32,430
And we'll do that in 1/6 of a second. 626
00:53:32,430 --> 00:53:37,090
And so what you see is you see this as en envelope 627
00:53:37,090 --> 00:53:49,220
which is the Io envelope, but you will see this....... 628
00:53:49,220 --> 00:53:51,940
and then it sweeps back. 629
00:53:51,940 --> 00:53:58,250
And so you see two things, you'll see the sinωt term..... 630
00:53:58,250 --> 00:54:01,650
but as ω changes you will see it go through resonance 631
00:54:01,650 --> 00:54:06,070
and then you will see it go over resonance. 632
00:54:06,070 --> 00:54:10,980
So I will give you the values that we have chosen. 633
00:54:10,980 --> 00:54:16,030
(see the picture) 634
00:54:27,060 --> 00:54:32,370
which is substantially higher than what we did before. 635
00:54:32,370 --> 00:54:37,470
And we chose C so high because we want a low Q system. 636
00:54:37,470 --> 00:54:43,720
(see the picture) 637
00:54:55,540 --> 00:54:59,070
And we are going to sweep it from zero over a thousand, 638
00:54:59,070 --> 00:55:01,200
which is resonance to about 2000, 639
00:55:01,200 --> 00:55:02,760
and then we sweep it back. 640
00:55:02,760 --> 00:55:06,510
And so the Q of this system (see the picture) 641
00:55:13,690 --> 00:55:16,970
And what Vo is of that circuit is not so important, 642
00:55:16,970 --> 00:55:18,230
but it's just for a volt, 643
00:55:18,230 --> 00:55:19,870
but that's not so important. 644
00:55:19,870 --> 00:55:23,940
And what we're going to show you, 645
00:55:23,940 --> 00:55:27,560
we measured the potential difference over a very small resistor 646
00:55:27,560 --> 00:55:31,620
which is somewhere in that circuit, 1.7Ω I believe, 647
00:55:31,620 --> 00:55:37,180
and so that potential difference over that resistor is IR, 648
00:55:37,180 --> 00:55:40,090
and R is a constant, and that's what we are going to show you. 649
00:55:40,090 --> 00:55:41,790
So we are going to show you something 650
00:55:41,790 --> 00:55:47,780
that is directly linear proportional with I. 651
00:55:47,780 --> 00:55:51,800
It's not power. It's I. 652
00:55:51,800 --> 00:55:55,810
You want to know power, you have to square it. 653
00:55:55,810 --> 00:56:01,620
I squared R is the power. 654
00:56:01,620 --> 00:56:03,260
And we are going to sweep it, 655
00:56:03,260 --> 00:56:09,530
1/6 of a second this way and 1/6 of a second back. 656
00:56:09,530 --> 00:56:15,130
Why did I only take into account the steady state solution? 657
00:56:15,130 --> 00:56:20,200
Why don't I have to also include the transient solution, 658
00:56:20,200 --> 00:56:26,240
which with this experiment took us five minutes to 659
00:56:26,240 --> 00:56:30,550
finally arrived at the steady state solution? 660
00:56:30,550 --> 00:56:34,660
Why am I leaving it out? 661
00:56:34,660 --> 00:56:38,370
I can't hear you. Where is the sound coming from? 662
00:56:38,370 --> 00:56:43,230
Yes, what is 2/γ which is the decay time? 663
00:56:43,230 --> 00:56:46,760
1/e decay time... what is 2/γ? 664
00:56:46,760 --> 00:56:48,600
Oh, I didn't write down what γ is. 665
00:56:48,600 --> 00:56:53,310
γ is a thousand. γ is R/L, right? 666
00:56:53,310 --> 00:56:57,490
(see the picture) 667
00:57:05,600 --> 00:57:08,600
Or another way of putting it 668
00:57:08,600 --> 00:57:19,170
is that in about 2 oscillations I am down by a factor of e. 669
00:57:19,170 --> 00:57:20,680
Remember? 670
00:57:20,680 --> 00:57:28,110
Q/π oscillations will reduce the amplitude by a factor of e. 671
00:57:28,110 --> 00:57:30,180
So in 2 oscillations, 672
00:57:30,180 --> 00:57:35,650
already the transient phenomenon is effectively killed 673
00:57:35,650 --> 00:57:37,750
and so I don't have to take into account. 674
00:57:37,750 --> 00:57:39,710
You won't even notice it. 675
00:57:39,710 --> 00:57:44,230
I told you earlier when the decay time here was very long, 676
00:57:44,230 --> 00:57:45,910
I will show you another experiment 677
00:57:45,910 --> 00:57:52,910
that the decay time is extremely short. 678
00:57:52,910 --> 00:57:57,990
While I met it, experient is set up there and you will see it here. 679
00:57:57,990 --> 00:58:01,910
I am going to make this 100Ω, 680
00:58:01,910 --> 00:58:04,280
I am going to double it to show you 681
00:58:04,280 --> 00:58:08,250
that this point will exactly go down by a factor of two. 682
00:58:08,250 --> 00:58:12,820
Because remember the peak is Vo/R, 683
00:58:12,820 --> 00:58:15,680
and I am not changing Vo. 684
00:58:15,680 --> 00:58:17,530
And so by when I double R, 685
00:58:17,530 --> 00:58:19,420
you will see it come down to here. 686
00:58:19,420 --> 00:58:22,550
And you will see it get broader. 687
00:58:22,550 --> 00:58:29,060
And then I will go to 150Ω. 688
00:58:29,060 --> 00:58:30,980
it's easy to do for us. 689
00:58:30,980 --> 00:58:34,680
So we go to 150Ω. 690
00:58:34,680 --> 00:58:43,110
So it's 100Ω. The Q is 3.15. And 150Ω, 691
00:58:43,110 --> 00:58:46,500
the Q is 3 times smaller. It's about 2.1. 692
00:58:46,500 --> 00:58:51,920
So at 150Ω, it would be 1/3 of this height, 693
00:58:51,920 --> 00:58:55,290
somewhere here, and the peak will be broader. 694
00:58:55,290 --> 00:59:04,050
And all of that we get to they very easily by changing the resistance. 695
00:59:04,050 --> 00:59:06,740
Again this is not power. 696
00:59:06,740 --> 00:59:09,260
Power is I square times R. 697
00:59:09,260 --> 00:59:13,690
This is simply the curve. 698
00:59:13,690 --> 00:59:14,290
Alright. 699
00:59:14,290 --> 00:59:18,570
I will give you the corrects. 700
00:59:18,570 --> 00:59:23,560
Light setting because we have to make it a little darker. 701
00:59:23,560 --> 00:59:31,450
And then we...There it is. 702
00:59:31,450 --> 00:59:34,450
Notice the vertical oscillation goes so fast 703
00:59:34,450 --> 00:59:36,730
that your eyes can not even follow them, 704
00:59:36,730 --> 00:59:41,460
that's the sinωt, and then when it sweeps over resonance, 705
00:59:41,460 --> 00:59:43,460
you see very dramatically, 706
00:59:43,460 --> 00:59:48,030
this value which is Vo/R. 707
00:59:48,030 --> 00:59:52,610
Now notice on the scale here, 708
00:59:52,610 --> 01:00:01,410
That macros has set the Io at resonance at 3 scale units. 709
01:00:01,410 --> 01:00:05,650
So it is nice when we go from R 50Ω to 100Ω, 710
01:00:05,650 --> 01:00:07,580
it's should go down by a factor of two. 711
01:00:07,580 --> 01:00:09,090
You can check that. 712
01:00:09,090 --> 01:00:11,290
And when we go to 150Ω, 713
01:00:11,290 --> 01:00:14,580
it is should go down from 3 units to 1 unit. 714
01:00:14,580 --> 01:00:16,540
So you can quantitate if we check this, 715
01:00:16,540 --> 01:00:20,350
and you will see that the curve gets broader 716
01:00:20,350 --> 01:00:23,140
because it has a lower Q. 717
01:00:23,140 --> 01:00:27,730
So now I will make it 100Ω, 718
01:00:27,730 --> 01:00:30,740
see it's down by a factor of two, 719
01:00:30,740 --> 01:00:33,090
from here to here. 720
01:00:33,090 --> 01:00:35,480
And when you may have noticed that it also gets broader. 721
01:00:35,480 --> 01:00:37,880
And now I will go to 150Ω, 722
01:00:37,880 --> 01:00:41,190
and you see it's down by a factor of three. 723
01:00:41,190 --> 01:00:45,760
And again, it is broader. 724
01:00:45,760 --> 01:00:49,410
So this is an amazing way how with RLC circuit, 725
01:00:49,410 --> 01:00:50,810
you can do wonderful things 726
01:00:50,810 --> 01:00:52,190
because you can manipulate 727
01:00:52,190 --> 01:00:55,470
ω0 very easily by changing L and C. 728
01:00:55,470 --> 01:01:04,840
And you can manipulate the driving frequency also very easily. 729
01:01:04,840 --> 01:01:10,150
So it is clear that systems respond strongly 730
01:01:10,150 --> 01:01:12,860
when they are exposed to their resonance frequency. 731
01:01:12,860 --> 01:01:17,450
We've seen that for pendulums. We've seen that for springs. 732
01:01:17,450 --> 01:01:19,970
We've seen it for a wine glass, 733
01:01:19,970 --> 01:01:24,420
and we've seen this now for an RLC circuit. 734
01:01:24,420 --> 01:01:29,960
So the systems at resonance absorb a large amount of energy 735
01:01:29,960 --> 01:01:36,460
for unit time out of the driver. 736
01:01:36,460 --> 01:01:47,180
I have here two tuning forks, which have an extremely high Q. 737
01:01:47,180 --> 01:01:48,390
My attempt to measure it, 738
01:01:48,390 --> 01:01:51,840
I conclude it is way larger than even a thousand. 739
01:01:51,840 --> 01:01:57,800
And they have exactly the same frequency, both 256Hz. 740
01:01:57,800 --> 01:02:02,950
This one and this one. 741
01:02:02,950 --> 01:02:05,800
Both 256 Hz. That's the wave I designed. 742
01:02:05,800 --> 01:02:07,780
To a high degree of accuracy, 743
01:02:07,780 --> 01:02:11,790
to better than a fraction of one Hz, 744
01:02:11,790 --> 01:02:18,700
but the Qs are so high that if you work to plot 745
01:02:18,700 --> 01:02:22,820
you drive these tuning forks and you work to plot here. 746
01:02:22,820 --> 01:02:27,420
this average power as function of ω, 747
01:02:27,420 --> 01:02:33,190
then you will get something like this. 748
01:02:33,190 --> 01:02:36,900
It means you have to drive it exactly at the right frequency. 749
01:02:36,900 --> 01:02:42,310
Otherwise it will not go to resonance. 750
01:02:42,310 --> 01:02:44,570
Well, we know how to get this going. 751
01:02:44,570 --> 01:02:47,320
You just bang it that means you dump a whole spectrum on it. 752
01:02:47,320 --> 01:02:50,480
It picks out the frequency that it likes. 753
01:02:50,480 --> 01:02:52,940
Now I am going to show you something remarkable. 754
01:02:52,940 --> 01:02:56,930
When this one generates 256 pressure waves, 755
01:02:56,930 --> 01:03:00,140
this one feels those pressure waves. 756
01:03:00,140 --> 01:03:03,560
And it loves it. 757
01:03:03,560 --> 01:03:07,040
Because it is just at the right frequency. 758
01:03:07,040 --> 01:03:10,360
And so it starts to oscillate. 759
01:03:10,360 --> 01:03:14,150
So when I stop this one, you will hear this one. 760
01:03:14,150 --> 01:03:19,680
And that's called resonance absorption. 761
01:03:19,680 --> 01:03:23,720
Let's do that first. 762
01:03:23,720 --> 01:03:28,890
Now you must understand that the sound waves go from here to there. 763
01:03:28,890 --> 01:03:32,100
Not much power reaches that point. 764
01:03:32,100 --> 01:03:35,250
So when I stop this one, you hear sound, 765
01:03:35,250 --> 01:03:37,860
but it's not overwhelming. 766
01:03:37,860 --> 01:03:41,950
So you have to be very quiet. 767
01:03:48,480 --> 01:03:49,080
Hear it? 768
01:03:53,790 --> 01:03:55,350
And I can do the same by hitting this one 769
01:03:55,350 --> 01:04:07,410
and then this one goes to start to resonate. 770
01:04:07,410 --> 01:04:16,410
Now, if the driving frequency is off by a fraction of a Hz. 771
01:04:16,410 --> 01:04:18,640
One Hz is enough. 772
01:04:18,640 --> 01:04:20,640
One Hz difference. 773
01:04:20,640 --> 01:04:22,390
Because the Q is so high, 774
01:04:22,390 --> 01:04:26,990
then this system will not be able to get this one going. 775
01:04:26,990 --> 01:04:31,790
And I can make this frequency a little lower than 256 Hz. 776
01:04:31,790 --> 01:04:35,440
I'm putting this weight on here. 777
01:04:35,440 --> 01:04:40,220
And we've measured the frequency at this loaded weight. 778
01:04:40,220 --> 01:04:45,400
It's roughly 255 Hz. 779
01:04:45,400 --> 01:04:48,490
And you're somewhere here because it's so narrow. 780
01:04:48,490 --> 01:04:50,540
The resonance absorption peak is very sharp. 781
01:04:50,540 --> 01:04:53,840
By the way, this is power. 782
01:04:53,840 --> 01:04:57,970
Because what reaches here is Joules per second, 783
01:04:57,970 --> 01:05:00,140
that's what gets it going, energy per second. 784
01:05:00,140 --> 01:05:02,870
So it's really a power transferred. 785
01:05:02,870 --> 01:05:07,890
So now I've changed the frequency. There we go. 786
01:05:07,890 --> 01:05:10,750
Nothing. 787
01:05:14,750 --> 01:05:19,660
I just change this by one Hz and you hear nothing. 788
01:05:19,660 --> 01:05:20,850
Then. 789
01:05:20,850 --> 01:05:25,090
So now you see you get some respect for high Qs. 790
01:05:25,090 --> 01:05:29,110
If you want to get resonance absorption in a high Q system, 791
01:05:29,110 --> 01:05:34,770
you gonna be dead on that frequency. 792
01:05:34,770 --> 01:05:40,030
So if you, for instance, banged all the keys on the piano, 793
01:05:40,030 --> 01:05:42,050
and this one will be nearby, 794
01:05:42,050 --> 01:05:43,860
it would only start to resonate 795
01:05:43,860 --> 01:05:50,270
if some or one of those strings would produce exactly 256Hz. 796
01:05:50,270 --> 01:05:52,840
Otherwise it would not. 797
01:05:52,840 --> 01:05:54,380
Ignores everything. 798
01:05:54,380 --> 01:06:02,120
It's only sensitive to that resonance frequency. 799
01:06:02,120 --> 01:06:08,080
You probably in high school have learned a little bit about atomic physics. 800
01:06:08,080 --> 01:06:13,280
You probably know that electrons have discrete energy levels PAGE2
| Name | Version | Size | Date | User |
| 04_d.srt | 1 | 81765 | 2/17/06 4:16 AM | OOPSSJTU |
Last Modified 2/21/06 8:38 AM
|
