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I will start today with a very interesting phenomenon,which is known as a beat phenomenon. 2
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Suppose you have two simple harmonic motions. With the same amplitude, 3
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but with different frequencies. So we have X1 (see the blackboard) and you have X2 which is (see the blackboard) 4
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and two are different. Same amplitude. 5
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If you sum them,X1 + X2,you get (see the blackboard) 6
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And when you look at this,you have here a frequency which is high compared to this frequency. 7
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You can best see that if ω1 is very close to ω2 and you could simply replace that by ω. 8
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Then this is simply saying cosωt. 9
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And so if you change from the notation of ω in your heads to frequency in hertz 10
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which is a lot easier to think of (see the blackboard) This is in hertz. 11
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This is in radiance per second. Suppose you have an f1 which is 256Hz 12
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and you have an f2 which is 254Hz. You take these numbers now,as a working example. 13
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Then the period of this oscillation is 1/256 of a second 14
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but the period of this oscillation is only one second. 15
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Because it's the difference between the two divided by two. 16
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And so what happens is that if you make a plot of the,of X as a function of time 17
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I'll first sketch this a very slow term. So this is very slow compared to this one 18
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which is very fast. I'll first make a sketch of the very slow one. 19
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So this is the slow one. And I'll make also dotted line here to guide my pen 20
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when I going to make the plot. So what you're going to see now is that the fast term 21
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go "chi..." and the slow one goes like this and the net result is this. 22
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This is the fast one,fast one. And so you see right here is when the slow one is zero. 23
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So the slow one kills the amplitude. You can think of as an amplitude modulation. 24
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That the A is multiplied by that cosine function. 25
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We define that,it is just a matter of definition,we defined this as the beat period. 26
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When the two are in phase and here they are again in phase. 27
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And two are here 180 degrees out of phase. We define this as the beat period. 28
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For the working example that we have on the blackboard there the 256 Hz and the 254 Hz. 29
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From here to here would be one second. That's the time for the slow one to go 360 degrees. 30
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And so the beat period would in that case be half a second. 31
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So if you want to see in your head what is happening. 32
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Then if one oscillation is like this,then when they are in phase, 33
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each one has an amplitude A,when they are in phase you get 2A, 34
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but when they're out of phase,you have to add this 180 degrees out fo phase 35
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and the net width is also zero. And that's the case here and that's the case there 36
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and that's the case there. This is known in the literature as a beat phenomenon. 37
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And the frequency of the beat,so fbeat is then defined as delta f, 38
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in our case 2Hz,indeed you see that the beat period is half a second. 39
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A very nice way to demonstrate this is to make you listen to it. 40
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I have here a tuning fork of 256Hz and I have another one which is also 256Hz. 41
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But I can change this one by putting a little weight on them. 42
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I can make the frequency a little lower. 43
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I put a weight here. And I'll simultaneously make you listen to them. 44
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And if somehow I managed to give them the same amplitude,your eardrum will 45
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reach this signal which is 256 and then that comes a time that you hear nothing. 46
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That means sound plus sound make silence,a interesting concept for the interference. 47
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You can hear that. So listen carefully! That's the beat. 48
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If I make the frequency different larger,the beat comes faster. 49
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Can all of you hear this? Also in the back? 50
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So now I will make the offset larger. It's unpleasant,you know,make you 51
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feel like slowing up with so,very high beat frequency. 52
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If the amplitude of the two are not the same, 53
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then you don't have to be a rocket scientist to anticipate what will happen. 54
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You just never get a zero here,which will always have some residual, 55
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and so you will get something that looks like this. And you still have an oscillation here. 56
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And I can also make you see that. It's harder to make you hear that because your ears 57
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can not really tell the difference between a little bit of sound and no sound. 58
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You hear "woo woo woo...” That's all you hear. 59
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But I can make you see that. And I'm going to make you see that right there. 60
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When we pick up the signals of these two tuning forks with a microphone 61
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and we'll display it there with a oscilloscope. 62
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So you will hear it and you will see it at the same time. And it is unlikely 63
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when I strike them that the amplitude as received 64
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by the microphone will be exactly the same. 65
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So it's not likely you will see it go through zero. 66
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But I can of course change the distance to the microphone, 67
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and try to get this closest I can to zero. 68
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Not bad. Now it's close to zero. 69
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I'll do it once more. Now I perfectly make the amplitude different, 70
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bring one very close to the microphone. This is what you saw there on the board. 71
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If I make the amplitudes the same,then you get... You get the zero back. 72
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Beat phenomenon are actually quite common. 73
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When you sit in an airplane on the runway and the airplane has two engines. 74
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Perhaps you have noticed something very irritating at times, 75
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that is some crazy humming sound that goes up in volume and it goes down and it goes up. 76
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That it is almost certainly the two engines beating against each other. 77
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There are not rotating at exactly the same frequency but slightly off. 78
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That's my explanation why you have that strangely rising volume and then decaying volume. 79
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It's very common. 80
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When I was a student,I remember there were two faucets in my bedroom. 81
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And no matter how much I tighten the faucets,they were leaking,they were dripping. 82
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And they were dripping in a such way that frequency 83
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reflects the drops came off was almost the same. 84
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So it went like this "clank,clank..." 85
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But the frequency was never exactly the same. 86
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So after a while they will go "clank,clank,clank ..." 87
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And then after a while "clank,clank..." 88
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It drove me crazy. But of course I do pride in knowing that is a beat phenomenon. 89
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except that their sound plus sound never made silence because when they were 90
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180 degrees out of phase you would hear both drops. 91
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Irritating but that's the way it is. 92
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Another example of beat phenomenon will by two oscillators 93
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very closely the same period given enough time they go out of phase. 94
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and they go back into phase. We have made an effort here 95
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with these two pendulums really in effort. 96
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We've 60-centimeter-length. To really make these lengths the same. 97
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So when I give them both certain amplitude,they will go in unison. 98
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They are in phase,but no matter how hard you try they will never be exactly the same. 99
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So as I continued the lecture,you will see that they gradually go out of phase. 100
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And It may take five minutes for them to be out of phase. 101
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That will be then the time from here to here. It will be half a beat period. 102
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So I will show you that. And I am not going to wait for them to be 180 degrees out of phase. 103
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Because it can take a long time,since the periods are so closely the same. 104
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And if you look at it now,and you will say "boy,are they in unison?" 105
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Keep an eye on it,and you will see if a few minutes from now, 106
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they will no longer be exactly in phase. 107
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When you think of a conductor for an orchestra,what conductor is really doing? 108
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He is making sure that all the instruments stay in phase. He's a really a phase locker. 109
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He's a phase keeper. If he didn't do that,then each person will play at his own pace. 110
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Then of course you would end up with chaos. 111
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So now perhaps you know what the expression comes from keep the phase. 112
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Now I go to the heart of this lecture. And the heart of this lecture is damping. 113
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Do you see it's already out of phase a little. Not completely out of phase, 114
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but they're no longer. Keep an eye on them. You will see that. It will grow. 115
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And the heart of this lecture is damping. When we have a spring or we have a pendulum, 116
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and we let it oscillate we all know that the oscillation will come to a halt. 117
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That means there is friction or there is air drag and it will ultimately die out and it will stop. 118
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Just like these will stop. The frictional force always opposes velocity. 119
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And the frictional force,it’s general form as a vector is (see the blackboard) 120
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whereby C1 is a positive number. So you see that it opposes the velocity. minus (see the blackboard) 121
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And I put here the unit vector of V to remind you that is opposing the velocity. 122
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This term has a name that is called the viscous term. 123
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And the other term has a name which is called the pressure term. 124
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One is proportional with the velocity,and the other is proportional with the square of that. 125
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In my lecture NO.12,which is on OCW of my 1999 Anthony mechanic lecture. 126
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I spared an entire lecture on this equation. And I discuss when the viscous term 127
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is dominant and when the square term is dominant. I do demonstrations in both domains 128
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if the velocity is low this term dominates. If the velocity is high,the speed is high,that term dominates. 129
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When you drive your car 10-20 miles an hour,this is the V square term that dominates. 130
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The air drag that slows you down. A range of that force is the V square term that dominates,not this one. 131
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But if you take ball bearings and you drop them in syrup,then the velocity is always very low. 132
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The speed is very low. And then this term dominates. 133
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That's one of the experiments I do during this lecture 134
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to show that indeed the predictions that follow from this part 135
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can really be verified quite beautifully. 136
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So if you have the patience,I’ll strongly advise you to watch that lecture. 137
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To manage the mass,it becomes almost impossible except when you want to do it numerically 138
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to take both terms into account. And so to be nice to you in 8.03 we always ignore this term. 139
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I will only take that one into account. And of course that is only valid if the velocities are low, 140
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which is often the case,but,of course,not always. I'll start with an extremely pedestrian example. 141
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I have here a spring on a horizontal surface. And as it moves,there is friction. 142
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And we assume that the friction has this form. The spring constant is K. 143
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The mass is M. And if I move it the distance x away from equilibrium. 144
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Then there is a spring force that drive it back to equilibrium. 145
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In addition there is this frictional force and I have no idea in which direction it is. 146
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Because if your object moves in this direction,the frictional force is like so. 147
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If the object at this moment moves in this direction,then the frictional force is so. 148
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So I cannot put it in there. I have to know the velocity in order to put in the frictional force. 149
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So I won't do that. 150
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But what I can do,I can write down Newton's Second Law, 151
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which says that mx double dot is -kx, 152
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that is this force,when x is positive,the force is in the negative direction. 153
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So the minus sign takes into account to do dealing with vectors. 154
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And then we have minus this term,now when we deal with 8.03, 155
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we will call this C1 just b. That's a tradition. 156
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We won't call it C1. We call it b. 157
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Another words,we have -b times x dot,x dot is a vector. 158
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It's a velocity,and if the velocity is positive,then the force is in this direction. 159
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If the velocity is negative,then the frictional force is in that direction. 160
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So that now is the differential equation that we have to solve. 161
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I will introduce some short hand notations: 162
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the k over m is ωo squared,remember that is the angular frequency of 163
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a spring oscillating in the absence of any friction. We've seen that last time. 164
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And I will also introduce a symbol b/m = γ. 165
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That's also a tradition. γ has the same unit as ω, 166
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seconds to the power -1,and so if I rewrite this differential equation, 167
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then I get (see the blackboard). 168
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And you see now our task ahead of us 169
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to solve this differential equation,what is x as a function of time? 170
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Without starting the math,you can sort off in your head,see what's going to happen. 171
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An oscillation that goes on forever and ever and ever is going to come to a halt. 172
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So instead of a nice cosine that continues,the cosine will gradually decay away, 173
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and then the thing will stop. 174
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I take a spring,I offset it,starts to oscillate,and it comes to a halt. 175
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So the amplitude must die out,must decay. That's a prediction. 176
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Now let me try your intuition. Let us assume that the system will oscillate with a unique frequency ω. 177
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Do you think that this ω will be larger than ωo,smaller than ωo,or equals to ωo? 178
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Or to put it in the other way,will a whole oscillation under the influence of friction take longer, 179
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or shorter,or the same amount of time than in the absence of friction? 180
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So who thinks that if I have damping, 181
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that a whole periods of a whole oscillation will take longer. 182
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Who thinks it will take shorter? Who thinks it will be the same? 183
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OK,we will what comes out of it. No one thinks that it takes shorter. 184
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Your intuition is very good there. 185
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Now I'm going to give you some advice. 186
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Some fatherly advice. 187
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During the next 18 minutes,take no notes. 188
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Because during the next 18 minutes, 189
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I'm going to solve this differential equation for you. 190
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And I have decided which I not always go to do exact the way the trains does it. 191
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Very elegant. 192
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So if you take notes,you're wasting you time because you have it in your book. 193
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And I would much rather prefer that you follow each step that I make. 194
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So you have more time to concentrate on the methods. 195
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So don't write anything down. That is my recommendation. 196
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18 minutes. 197
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You can time,may if you don't believe me. 198
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We are going to jump through the complex plane. 199
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This is a beautiful example where the complex solutions really come in handy. 200
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And so I'm going to change this equation to Z which is now a complex notation. 201
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(see the blackboard) 202
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And then Z my trial function is some amplitude A 203
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(see the blackboard) 204
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And clearly this has a smell of amplitude. 205
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And this p has a small of frequency. 206
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Anything e to the power jpt has the smell of frequency. 207
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If you'll see there you're in full surprise what happens with this p. 208
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Once we have found the proper solution to Z, 209
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we take the real part,and we're on business then we get back to x. 210
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I'm going to take the second derivative of this equation. 211
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So outcomes jp comes out twice so it is j square times p squared. 212
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And so I'm going to see now minus p squared, 213
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that is the second derivative of this function and Z comes later 214
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Then I have plus γ and then I'll pops out this jp. 215
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Because it is only the first derivative. 216
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jp. Has to be very careful that you can distinguish your js from your γs. 217
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And then you get plus ωo squared 218
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The whole thing times Z and that now be equal to zero. 219
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Well,if you have a equation like this,you may think it's one equation but it's really two equations. 220
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Because if something like this has to be zero,and of course Z is zero is not an acceptable solution. 221
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Because that's the oscillation. 222
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If you have an equation like this,which has a real part and it has an imaginary part. 223
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Then both must independently be zero. 224
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Obvious reasons. 225
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These are apples and these are oranges,and five apples minus five oranges is never zero oranges, 226
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is never zero apples. 227
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You can not subtract apples from coconuts. 228
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So this has to be zero and this has to be zero. 229
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Now,if you make γ zero,you have no problem,there is no damping. 230
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So that's ridiculous. 231
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If you make p zero,then you no longer have any oscillation going on because you smell that pt 232
00:24:34,080 --> 00:24:38,980
you know,that's going to be the oscillatory thing,so that's also unacceptable. 233
00:24:38,980 --> 00:24:47,670
And so this one work,therefore p itself must be complex. 234
00:24:47,670 --> 00:24:51,190
And so we're going one step further now. 235
00:24:51,190 --> 00:25:02,690
And we are going to say,the only way that we can make this work is if we say p=n+js 236
00:25:02,690 --> 00:25:08,660
well,by n and s then,let's hope,are real. 237
00:25:08,660 --> 00:25:17,010
So let's first calculate what p square is. So we go slowly. So we don't get into trouble with the algebra. 238
00:25:17,010 --> 00:25:25,960
So p squared is n squared,plus this one squared,the j squared minus 1 239
00:25:25,960 --> 00:25:34,790
So that is minus s squared,plus 2n times j times s. 240
00:25:34,790 --> 00:25:40,340
So now I can go back to this form and have minus p squared. 241
00:25:40,340 --> 00:25:53,210
So I get minus n squared that is minus p squared plus s squared minus 2nj times s. 242
00:25:53,210 --> 00:25:55,100
So I'm done here. 243
00:25:55,100 --> 00:26:04,640
Now I get γjp plus γ times j 244
00:26:04,640 --> 00:26:15,770
And I'll have to multiply that by p which is n plus γ times j times js 245
00:26:15,770 --> 00:26:25,590
And then I have plus ωo squared and this now has to be zero. 246
00:26:25,590 --> 00:26:31,980
I have here j square. j square is -1 247
00:26:31,980 --> 00:26:38,700
So I can now erase this and replace it by minus γs 248
00:26:38,700 --> 00:26:40,890
You can not do that. 249
00:26:40,890 --> 00:26:47,260
That's the price you pay if you don't take my advice not to take any notes. 250
00:26:47,260 --> 00:26:51,860
So this is minus γs. 251
00:26:51,860 --> 00:26:59,550
And now this has to be zero that means the apples must add up to zero. 252
00:26:59,550 --> 00:27:05,020
These are the apples and the oranges have to be zero. 253
00:27:05,020 --> 00:27:12,780
This is an orange,this is an orange,this is an orange and this is an orange. 254
00:27:12,780 --> 00:27:19,010
For this to be zero,there's a j here,there's a j here,there's an n here there's an n here. 255
00:27:19,010 --> 00:27:26,780
Two s must be γ,so γ/2 is s. 256
00:27:26,780 --> 00:27:30,750
Oh,we found s. 257
00:27:30,750 --> 00:27:42,570
For this to be zero,we find that n squared minus n squared plus γ squared over four, 258
00:27:42,570 --> 00:27:45,240
because we know what s is now. 259
00:27:45,240 --> 00:27:53,510
so plus γ squared over four minus γ over two times minus γ s 260
00:27:53,510 --> 00:27:57,260
which is minus γ squared over two . 261
00:27:57,260 --> 00:28:06,890
Minus γ squared over two plus ωo squared equals zero. 262
00:28:06,890 --> 00:28:16,460
And that means that n squared is ωo squared minus γ squared over four. 263
00:28:16,460 --> 00:28:20,930
So we also have n. 264
00:28:20,930 --> 00:28:26,090
So we now have n and we have s. 265
00:28:26,090 --> 00:28:31,620
And so now I can go back to my trial function. 266
00:28:31,620 --> 00:28:38,890
And I can substitute the value that we have found in that complex function. 267
00:28:38,890 --> 00:28:46,190
So we get Z equals A times e to the power j. 268
00:28:46,190 --> 00:28:51,220
And now we have to put in pt plus α. 269
00:28:51,220 --> 00:28:54,950
But p is n+js. 270
00:28:54,950 --> 00:29:05,200
(see the blackboard) 271
00:29:05,200 --> 00:29:11,390
Yeah,I go slowly. I don't skip too many steps. 272
00:29:11,390 --> 00:29:17,500
But I know that s is γ over two. 273
00:29:17,500 --> 00:29:22,300
But I know that j squared is -1. 274
00:29:22,300 --> 00:29:29,090
So therefore this part here and this part here can be brought out. 275
00:29:29,090 --> 00:29:36,020
So we get A times e to the minus γ over two times t. 276
00:29:36,020 --> 00:29:42,920
Here we have the t. The j squared makes it minus,and s is γ over two. 277
00:29:42,920 --> 00:29:52,220
And then I get e to the power j times nt plus α. 278
00:29:52,220 --> 00:29:56,620
The first thing that you see here which is amazing. 279
00:29:56,620 --> 00:30:05,210
This term here is an exponential decay in time,nothing oscillatory about it. 280
00:30:05,210 --> 00:30:14,820
It means that this A which is the original amplitude is gradually dying out with a decay time. 281
00:30:14,820 --> 00:30:21,570
τ,which is 2/γ,I remember γ has unit second -1. 282
00:30:21,570 --> 00:30:24,170
So this really has unit time. 283
00:30:24,170 --> 00:30:32,620
So this one is going to be responsible for that decay that we discussed earlier. 284
00:30:32,620 --> 00:30:35,480
What is this now? 285
00:30:35,480 --> 00:30:38,450
e to the power jnt 286
00:30:38,450 --> 00:30:41,650
but n is obviously a frequency. 287
00:30:41,650 --> 00:30:48,560
This is striking example in the complex plane of a rotating vector. 288
00:30:48,560 --> 00:30:54,760
And the angular frequency is n but we know what n is. 289
00:30:54,760 --> 00:30:58,490
And so this n really is nothing but ω squared. 290
00:30:58,490 --> 00:31:02,990
n squared is really ω squared. 291
00:31:02,990 --> 00:31:11,110
And so I can replace this n now by whatever you have here. 292
00:31:11,110 --> 00:31:22,930
And so if I write this now (see the blackboard) 293
00:31:22,930 --> 00:31:26,010
And I know exactly now what that ω is. 294
00:31:26,010 --> 00:31:31,850
That is the square root of this number. 295
00:31:31,850 --> 00:31:40,240
And therefore ω squared (see the blackboard) 296
00:31:40,240 --> 00:31:45,850
So those of you who'd been audience who intuitively sense that frequency would go down 297
00:31:45,850 --> 00:31:48,280
because of the friction they were right. 298
00:31:48,280 --> 00:31:52,490
See ω is lower than ωo. 299
00:31:52,490 --> 00:31:56,870
Now of course if γ is very low the two could be close together. 300
00:31:56,870 --> 00:32:04,460
Then so γ that's obvious if you have a lot of friction,then ω will be a lot of less than ωo. 301
00:32:04,460 --> 00:32:09,650
If a very little friction,well then they will be very closely the same. 302
00:32:09,650 --> 00:32:15,550
And so we can now write down in all its glory. 303
00:32:15,550 --> 00:32:35,330
The real part of this function which then becomes (see the blackboard) 304
00:32:35,330 --> 00:32:40,620
woo,look what I did? 305
00:32:40,620 --> 00:32:44,520
I have an equation in the complex plane. 306
00:32:44,520 --> 00:32:47,100
That's perfectly okay. 307
00:32:47,100 --> 00:32:54,620
And then I went back to the real world going from Z,the complex plane,to the real part of Z. 308
00:32:54,620 --> 00:32:56,570
And what did I write down? 309
00:32:56,570 --> 00:33:02,850
This is fine,and then I write e to the power minus jωt plus α. 310
00:33:02,850 --> 00:33:08,370
First of all,there should not be a minus sign in front of the j. That's not a major problem. 311
00:33:08,370 --> 00:33:11,930
But I do not take it out of the complex plane. 312
00:33:11,930 --> 00:33:14,160
I leave it into the complex plane. 313
00:33:14,160 --> 00:33:28,660
So clearly what I should have written,which is what I will do now,(see the blackboard) 314
00:33:28,660 --> 00:33:32,820
Now we are in the real world. 315
00:33:32,820 --> 00:33:39,010
You're going to see this equation at least for the next five or ten minutes on the tape. 316
00:33:39,010 --> 00:33:42,620
It is all in my side,but there is nothing I can do about it. 317
00:33:42,620 --> 00:33:44,430
You just have to live with it. 318
00:33:44,430 --> 00:33:48,880
And think of it has been this. 319
00:33:48,880 --> 00:33:54,640
And there're two adjustable constants which depend entirely on the initial condition. 320
00:33:54,640 --> 00:33:58,750
A time t equal zero there are two things you can do to the object. 321
00:33:58,750 --> 00:34:03,790
You can bring it to a certain point away from equilibrium and you can give it a kick. 322
00:34:03,790 --> 00:34:05,110
We call that the velocity. 323
00:34:05,110 --> 00:34:08,300
And you are free to choose any way you what to do that. 324
00:34:08,300 --> 00:34:13,940
And so it's clear that with those two choices that you have,that in your final solution, 325
00:34:13,940 --> 00:34:20,250
you end up with two adjustable constants which depend on the initial condition. 326
00:34:20,250 --> 00:34:27,580
And so you see that this amplitude is going to die out with 1/e decay time of 2/γ. 327
00:34:27,580 --> 00:34:38,460
Seconds,and that the frequency is lower than the frequency of the undamped system. 328
00:34:38,460 --> 00:34:43,520
I try to make a plot. 329
00:34:43,520 --> 00:34:48,730
So here is t and here is x 330
00:34:48,730 --> 00:34:59,130
And to guide my hands I'm going to first put in this exponential. 331
00:34:59,130 --> 00:35:05,630
And now I'm going to put in that oscillatory term here. 332
00:35:05,630 --> 00:35:10,570
Whereby the frequency is uniquely determined. ω is uniquely determined. 333
00:35:10,570 --> 00:35:20,160
And the period t is that ω/2pi. 334
00:35:20,160 --> 00:35:24,100
Oops,how could I? 335
00:35:24,100 --> 00:35:25,700
Look what I wrote. 336
00:35:25,700 --> 00:35:31,820
I wrote t = ω/2pi 337
00:35:31,820 --> 00:35:34,120
It was not my day. 338
00:35:34,120 --> 00:35:39,900
Clearly it should be t = 2pi/ω. 339
00:35:39,900 --> 00:35:45,650
Now,fortunately for me,very shortly afterwards I erase this from the blackboard. 340
00:35:45,650 --> 00:35:47,600
So you won't see it for very long. 341
00:35:47,600 --> 00:35:52,220
And any case this is what it should be. 342
00:35:52,220 --> 00:35:56,460
And therefore I'm going to put in here. 343
00:35:56,460 --> 00:36:03,630
Zero crossing's to guide myself so that I don't make the mistake which you often see. 344
00:36:03,630 --> 00:36:07,360
The people think that this period is slowly getting shorter in time. 345
00:36:07,360 --> 00:36:09,860
That is not true. 346
00:36:09,860 --> 00:36:13,080
But it is the amplitude that decays away. 347
00:36:13,080 --> 00:36:21,600
And so if we now try to put in the oscillation. 348
00:36:21,600 --> 00:36:24,180
And this is what going to happen. 349
00:36:24,180 --> 00:36:32,040
And so you see the oscillation dies away in time but the period t is uniquely determined. 350
00:36:32,040 --> 00:36:39,560
And that depends how much friction there is. 351
00:36:39,560 --> 00:36:52,820
It is not uncommon to introduce a quality factor Q that is high if there is a little damping 352
00:36:52,820 --> 00:36:58,120
and that is low if there is a lot of damping. 353
00:36:58,120 --> 00:37:04,930
And that Q which is dimensionless is ωo/γ 354
00:37:04,930 --> 00:37:08,740
You see immediately if γ is high,then Q is low. 355
00:37:08,740 --> 00:37:12,920
So low quality oscillator. 356
00:37:12,920 --> 00:37:18,190
If you introduce that,you can go back to your ω square equation. 357
00:37:18,190 --> 00:37:26,350
And then ω squared becomes (see the blackboard) 358
00:37:26,350 --> 00:37:28,660
Is this any different from that? 359
00:37:28,660 --> 00:37:34,310
No. Just a different way of writing it because you introduce this. 360
00:37:34,310 --> 00:37:40,380
What it tell you is that Q is about 10. 361
00:37:40,380 --> 00:37:44,360
And I bet the Qs are much higher for these pendulums. 362
00:37:44,360 --> 00:37:46,990
Then you have here 1/400. 363
00:37:46,990 --> 00:37:49,290
That is one quarter of a percents. 364
00:37:49,290 --> 00:37:54,680
But since ω is the square root of that,it's only one-eighth of a percent. 365
00:37:54,680 --> 00:38:03,730
So for Q of 10,ω is only one-eighth of a percent lower than ωo. 366
00:38:03,730 --> 00:38:13,560
Even if you make the Q as low as 2,the frequencies are only off by about 3.2% 367
00:38:13,560 --> 00:38:28,740
So I want you to appreciate that most of the time,but not always,is ω very close to ωo. 368
00:38:28,740 --> 00:38:37,210
We can look at the decay in a different way. 369
00:38:37,210 --> 00:38:40,100
We can not do that here. 370
00:38:40,100 --> 00:38:47,400
I will do that here on the center board because we don't need that anymore. 371
00:38:47,400 --> 00:38:57,830
Instead of saying I have to wait 2/γ seconds for the amplitude to go down by a factor of e. 372
00:38:57,830 --> 00:39:09,910
I can ask myself how many oscillations do I have to wait for the amplitude to go down by a factor of e. 373
00:39:09,910 --> 00:39:13,030
How many oscillations? 374
00:39:13,030 --> 00:39:18,370
Well,N oscillations will take this long. 375
00:39:18,370 --> 00:39:21,550
T being the period of one oscillation. 376
00:39:21,550 --> 00:39:30,300
But for reasonable values of Q,T and T0 are the same like ω and ωo are closely the same. 377
00:39:30,300 --> 00:39:35,450
So I can write for this,and this is approximately N times T0 . 378
00:39:35,450 --> 00:39:44,040
So that's approximately N times 2pi/ωo. 379
00:39:44,040 --> 00:39:48,790
I can now substitute this time in this t. 380
00:39:48,790 --> 00:39:51,370
And only concentrate on that decay potion. 381
00:39:51,370 --> 00:39:53,470
That early part. 382
00:39:53,470 --> 00:40:07,930
So what I find then that A after N oscillations is (see the black board) times this time after N oscillations 383
00:40:07,930 --> 00:40:14,600
and divided by 2 pi,divided by ωo. 384
00:40:14,600 --> 00:40:17,610
And you lose a two. 385
00:40:17,610 --> 00:40:20,910
But ωo divided by γ is Q. 386
00:40:20,910 --> 00:40:31,850
And so now you have it in the form which is minus N times pi over Q. 387
00:40:31,980 --> 00:40:37,160
Or no! Here we say, 388
00:40:37,160 --> 00:40:42,540
I have to wait 2/γ seconds for the amplitude to go down by a factor of e. 389
00:40:42,540 --> 00:40:48,650
Here we say,if N is Q/pi, 390
00:40:48,650 --> 00:40:52,180
then the amplitude goes down by 1/e. 391
00:40:52,180 --> 00:40:56,770
So in one case,I ask myself how many seconds do I have to wait? 392
00:40:56,770 --> 00:41:01,680
In the other case,I ask myself how many oscillations do I have to wait? 393
00:41:01,680 --> 00:41:06,150
And so if Q is ten,it tells you that 394
00:41:06,150 --> 00:41:09,340
you have to wait about three oscillations roughly 395
00:41:09,340 --> 00:41:15,690
for the amplitude to go down by a factor of e. 396
00:41:15,690 --> 00:41:17,400
And if Q is a hundred, 397
00:41:17,400 --> 00:41:26,060
you have to wait more like 32 oscillations for the amplitude to go down by a factor of e. 398
00:41:26,060 --> 00:41:29,090
Eighteen minutes are up. 399
00:41:29,090 --> 00:41:33,420
So now you can start taking notes again. 400
00:41:33,420 --> 00:41:37,330
I have here two pendulums, 401
00:41:37,330 --> 00:41:40,800
and the pendulums have about the same length, 402
00:41:40,800 --> 00:41:46,670
and the objects have about the same radius. 403
00:41:46,670 --> 00:41:54,430
That means the b which is this coefficient 404
00:41:54,430 --> 00:42:01,340
in front of the velocity is about the same. 405
00:42:01,340 --> 00:42:04,230
But γ is not the same, 406
00:42:04,230 --> 00:42:05,910
because it has the same b, 407
00:42:05,910 --> 00:42:09,060
but if the mass of the two objects is very different, 408
00:42:09,060 --> 00:42:12,490
this is a Styrofoam and this is a billiard ball, 409
00:42:12,490 --> 00:42:15,390
there is huge difference in γ. 410
00:42:15,390 --> 00:42:18,190
And since the period of the two pendulums is very closely the same. 411
00:42:18,190 --> 00:42:20,200
They have the same length. 412
00:42:20,200 --> 00:42:25,460
You see that the Q of the two systems must be very different. 413
00:42:25,460 --> 00:42:30,470
Because if b is the same,m is very much higher of the billiard ball, 414
00:42:30,470 --> 00:42:35,900
then the γ is much lower and therefore the Q of this system 415
00:42:35,900 --> 00:42:38,630
is weighed higher than the Q of this system. 416
00:42:38,630 --> 00:42:43,020
In fact,if I want it to wait how many oscillations it will take 417
00:42:43,020 --> 00:42:46,640
for that amplitude to go down by a factor of e, 418
00:42:46,640 --> 00:42:50,880
I may have to wait 5 or 10 minutes. 419
00:42:50,880 --> 00:42:52,360
So I will not attend that, 420
00:42:52,360 --> 00:42:59,580
but I will attend that to measure Q with this pendulum. 421
00:42:59,580 --> 00:43:06,410
If I bring this here,the separation from equilibrium is 27cm 422
00:43:06,410 --> 00:43:10,200
and by the time that it has decayed to this,it's 10 cm. 423
00:43:10,200 --> 00:43:13,590
So that's about a factor of e. 424
00:43:13,590 --> 00:43:15,380
And I want the students in the audience 425
00:43:15,380 --> 00:43:19,300
who are sitting here who can really see it ahead on, 426
00:43:19,300 --> 00:43:22,080
I want them to say "stop". 427
00:43:22,080 --> 00:43:27,750
Scream the word "stop" when the 27 has decayed to 10. 428
00:43:27,750 --> 00:43:31,520
And the mean time,we count the number of oscillations. 429
00:43:31,520 --> 00:43:35,760
So we have then counted the number of oscillations 430
00:43:35,760 --> 00:43:38,480
and therefore we know what Q is 431
00:43:38,480 --> 00:43:41,290
because we multiply the number of oscillations by pi. 432
00:43:41,290 --> 00:43:43,540
So we have measured Q. 433
00:43:43,540 --> 00:43:45,650
We could also have done it through a time measurement 434
00:43:45,650 --> 00:43:48,200
but I prefer to do it this way. 435
00:43:48,200 --> 00:43:49,930
Now you guys over there keep your mouths shut 436
00:43:49,930 --> 00:43:51,750
because there is no way you can see it. 437
00:43:51,750 --> 00:43:53,110
Right? You just can't. 438
00:43:53,110 --> 00:43:54,600
You have a projection effect which is awful. 439
00:43:54,600 --> 00:43:56,540
So only want to hear from you. 440
00:43:56,540 --> 00:43:59,740
So you ready for that? Count and say "stop". 441
00:43:59,740 --> 00:44:02,540
Not all of you may say stop at the say moment. 442
00:44:02,540 --> 00:44:04,880
Some of you may say "stop" after 10 oscillations. 443
00:44:04,880 --> 00:44:07,320
Others may say "stop" after 11 oscillations. 444
00:44:07,320 --> 00:44:08,530
That's fine. 445
00:44:08,530 --> 00:44:11,830
That is part of the uncertainty in the measurement. 446
00:44:11,830 --> 00:44:18,880
Ready? 27. There we go. 447
00:44:18,880 --> 00:44:39,380
1...2...You see the decay already...3...4...5...6...7...8...9...10 448
00:44:39,380 --> 00:44:43,610
My goodness! You guys crazy? 449
00:44:43,610 --> 00:44:44,930
You're paying a lot of tuition, 450
00:44:44,930 --> 00:44:48,130
but you can't even pay attention to a demonstration. 451
00:44:48,130 --> 00:44:51,300
But...at nine...I think it was already a ten. 452
00:44:51,300 --> 00:44:54,320
When it is here,then it's ten. 453
00:44:54,320 --> 00:44:58,300
I will give you a second chance. 454
00:44:58,300 --> 00:45:04,600
You have to scream "stop". 455
00:45:07,480 --> 00:45:32,750
1...2...3...4...5...6...7...8...9...10...11...12 456
00:45:32,750 --> 00:45:35,600
Okay,I did it somewhere around ten,eleven 457
00:45:35,600 --> 00:45:37,750
and I don't know why you didn't scream "stop", 458
00:45:37,750 --> 00:45:45,620
but that's your problem. So N is about 11,maybe plus or minus 1,10% uncertainty. 459
00:45:45,620 --> 00:45:49,510
And so Q then is about 5 times higher. 460
00:45:49,510 --> 00:45:55,630
It's a crude measurement but very roughly you'll get then... 461
00:45:55,630 --> 00:45:59,870
you get about thirty-five. 462
00:45:59,870 --> 00:46:04,350
Thirty-five that means the frequency damped 463
00:46:04,350 --> 00:46:07,230
is almost identical to the frequency undamped 464
00:46:07,230 --> 00:46:12,880
because,remember,this equation,the 1/4. 465
00:46:12,880 --> 00:46:17,260
I erased it,perhaps,but this is....yeah,you have it up there. 466
00:46:17,260 --> 00:46:26,730
If Q were thirty-five,you can see how close ω is to ωo. 467
00:46:26,730 --> 00:46:33,020
On problem set II your very first task is to do a take-home experiment 468
00:46:33,020 --> 00:46:35,540
with something similar to what I did today. 469
00:46:35,540 --> 00:46:38,770
Make sure you pick up a kit today. 470
00:46:38,770 --> 00:46:43,710
You can do that between 11.1.3 to 5 in room 4335 471
00:46:43,710 --> 00:46:46,670
or you can do it tomorrow between 2 and 5. 472
00:46:46,670 --> 00:46:48,830
You share one kit so you choose your partner 473
00:46:48,830 --> 00:46:53,920
and you can do all these experiments together with your partner. 474
00:46:53,920 --> 00:46:57,050
I want to see in your solutions uncertainties. 475
00:46:57,050 --> 00:46:59,630
Any timing measurements that you have to do 476
00:46:59,630 --> 00:47:03,240
has to come with an estimate of your uncertainty. 477
00:47:03,240 --> 00:47:07,460
And all the conclusions that you draw must carry on these uncertainties 478
00:47:07,460 --> 00:47:12,390
just as we did last time when we were exploring the possibility 479
00:47:12,390 --> 00:47:16,320
whether the equation of the spring was indeed accurate. 480
00:47:16,320 --> 00:47:20,740
We will only able to come to a conclusion that it was not accurate 481
00:47:20,740 --> 00:47:23,630
because we had our proper uncertainties in there. 482
00:47:23,630 --> 00:47:29,200
So I want to see uncertainties in there. 483
00:47:29,200 --> 00:47:34,620
Now comes the mini quiz. Five-minute break. It's a bit early. 484
00:47:34,620 --> 00:47:37,930
But that's the best point today to do. 485
00:47:37,930 --> 00:47:42,560
So if some of you can help me handing it out, 486
00:47:42,560 --> 00:47:45,800
and also at the end of the five minutes bringing it back to me. 487
00:47:45,800 --> 00:47:49,620
Shouldn't it take you more than 1 minute 488
00:47:49,620 --> 00:47:51,640
that leaves you is still 4 minutes to stretch your leg. 489
00:47:53,830 --> 00:47:59,340
I will now take you back to the good old days of 8.02, 490
00:47:59,340 --> 00:48:03,990
and I will take you back to an RLC circuit. 491
00:48:03,990 --> 00:48:07,860
Whereby we have a battery, 492
00:48:07,860 --> 00:48:11,030
this is the plus side and this is the minus side. 493
00:48:11,030 --> 00:48:19,040
Here I have a resistor R. I have here a pure self-inductor L 494
00:48:19,040 --> 00:48:22,780
and here I have a capacitor C. 495
00:48:22,780 --> 00:48:27,370
And then I have a switch here and I can throw that switch. 496
00:48:27,370 --> 00:48:31,430
And then the battery is going to charge up the capacitor 497
00:48:31,430 --> 00:48:34,280
and then you'll get oscillations. 498
00:48:34,280 --> 00:48:41,840
This is a wonderful example of damped oscillations. 499
00:48:41,840 --> 00:48:50,520
The way that I solve these problems is the strict discipline. 500
00:48:50,520 --> 00:48:59,170
I assume that when I start it that there's a current going in this direction I. 501
00:48:59,170 --> 00:49:05,800
That current will make this side of the capacitor more positive than that side. 502
00:49:05,800 --> 00:49:16,080
So I equals dq/dt,q being the charge on this side of the capacitor. 503
00:49:16,080 --> 00:49:23,650
And now I have to do the close-loop integral of E·dl. 504
00:49:23,650 --> 00:49:26,680
Kirchhoff's Law does not hold. 505
00:49:26,680 --> 00:49:30,850
The close-loop integral is not zero 506
00:49:30,850 --> 00:49:39,840
because we have a magnetic flux going through a surface attached to this close loop. 507
00:49:39,840 --> 00:49:46,570
So the close-loop integral of E·dl is -dΦ/dt, 508
00:49:46,570 --> 00:49:50,530
Φ being that magnetic flux going through that surface. 509
00:49:50,530 --> 00:49:58,130
And so I want to know what the e factors are in each one of these components 510
00:49:58,130 --> 00:50:00,710
if the current is in this direction. 511
00:50:00,710 --> 00:50:05,380
Then the electric field in the resistor is in this direction. 512
00:50:05,380 --> 00:50:09,460
The self-inductor is made of super-conducting wire. 513
00:50:09,460 --> 00:50:12,100
So there is never any E field inside the self-inductor. 514
00:50:12,100 --> 00:50:15,280
Unlike what many EE people think, 515
00:50:15,280 --> 00:50:19,980
there is never any electric field inside a self-inductor. 516
00:50:19,980 --> 00:50:23,620
So the E here is zero. 517
00:50:23,620 --> 00:50:25,220
This is plus and this is minus, 518
00:50:25,220 --> 00:50:28,580
so the potential,the electric field is in this direction. 519
00:50:28,580 --> 00:50:30,380
This is plus and this is minus, 520
00:50:30,380 --> 00:50:35,780
and so the electric field is opposing me if I go around clockwise. 521
00:50:35,780 --> 00:50:38,470
So if now I go around clockwise, 522
00:50:38,470 --> 00:50:44,440
and I call the integral E·dl from this side to this side of the capacitor, 523
00:50:44,440 --> 00:50:49,890
if I call that Vc,which is q/C, 524
00:50:49,890 --> 00:50:53,210
that's the definition of capacitor, 525
00:50:53,210 --> 00:50:54,700
if you're ready for this 526
00:50:54,700 --> 00:50:58,680
and I start here and I go clockwise around, 527
00:50:58,680 --> 00:51:05,760
then from here to here I get plus IR going through the self-inductor, 528
00:51:05,760 --> 00:51:09,640
the integral E·dl from here to here is zero 529
00:51:09,640 --> 00:51:13,820
because there is no electric field inside the self-inductor. 530
00:51:13,820 --> 00:51:18,190
Going from here to here,I get +Vc. 531
00:51:18,190 --> 00:51:22,130
That is what we called the potential difference over the capacitor. 532
00:51:22,130 --> 00:51:24,020
Here the electric field is opposing me. 533
00:51:24,020 --> 00:51:25,620
I walk into the electric field, 534
00:51:25,620 --> 00:51:31,630
so E?dl is negative,and this is -Vo. 535
00:51:31,630 --> 00:51:35,240
And now I go to Mr. Maxwell to Faraday 536
00:51:35,240 --> 00:51:40,000
and he says this now is -LdI/dt, 537
00:51:40,000 --> 00:51:45,180
Kirchhoff,-LdI/dt. 538
00:51:45,180 --> 00:51:48,480
and now I have my equation correct. 539
00:51:48,480 --> 00:51:55,500
So now I'm going to replace I by dq/dt and I'm bring L to the left. 540
00:51:55,500 --> 00:52:04,910
So I get (see the blackboard). 541
00:52:04,910 --> 00:52:14,030
Plus Vc which is q/C and that now equals Vo, 542
00:52:14,030 --> 00:52:31,640
and I'm going to divide it by L (see the blackboard). 543
00:52:31,640 --> 00:52:37,050
I'm going to replace R/L by γ. 544
00:52:37,050 --> 00:52:39,320
You will see very shortly why we do that. 545
00:52:39,320 --> 00:52:45,470
And 1/LC is ωo squared. 546
00:52:45,470 --> 00:52:49,490
And when we do that,we get an equation which is almost identical 547
00:52:49,490 --> 00:52:54,100
to the one we had on the blackboard for the spring. 548
00:52:54,100 --> 00:53:17,250
We get (see the blackboard) and now we have to solve this differential equation. 549
00:53:17,250 --> 00:53:22,100
You recognize the γ is the damping. 550
00:53:22,100 --> 00:53:29,930
R/L has the same function as the damping had in the case of the spring. 551
00:53:29,930 --> 00:53:33,650
In the case of the spring,it was b/m. Here it is R/L. 552
00:53:33,650 --> 00:53:37,300
The larger R is,the more heat dissipation there is. 553
00:53:37,300 --> 00:53:41,320
Heat dissipation goes in terms of I square R, 554
00:53:41,320 --> 00:53:44,270
and clearly that means you take energy out of the system. 555
00:53:44,270 --> 00:53:47,720
So that means there is damping. 556
00:53:47,720 --> 00:53:54,910
And this is then the natural frequency of the RLC circuit,if there is no R, 557
00:53:54,910 --> 00:54:02,730
if there is only an L and a C.This is the square of the frequency. 558
00:54:02,730 --> 00:54:06,490
If only this were zero,then I know the solution 559
00:54:06,490 --> 00:54:08,410
because we had it on the board there. 560
00:54:08,410 --> 00:54:10,970
We still have the complex notation there. 561
00:54:10,970 --> 00:54:40,980
Then q would be some q1,which we have an a there. (see the blackboard) 562
00:54:40,980 --> 00:54:46,720
That would be the solution if this were zero. 563
00:54:46,720 --> 00:54:48,940
Now it's not a zero, 564
00:54:48,940 --> 00:54:54,290
so you can take 1803 and try to solve it if it's not zero. 565
00:54:54,290 --> 00:55:03,610
Or you can think like a physicist and say I don't need 1803. 566
00:55:03,610 --> 00:55:06,810
What is the difference between this differential equation 567
00:55:06,810 --> 00:55:10,110
and the one before with the spring that we had a zero here. 568
00:55:10,110 --> 00:55:17,050
In the case of zero,it means that the oscillation would end up at x equal zero. 569
00:55:17,050 --> 00:55:23,420
Well,as here the oscillation in the charge will end up if we wait long enough 570
00:55:23,420 --> 00:55:28,680
with a fully charged capacitor coz when the oscillation has died out, 571
00:55:28,680 --> 00:55:31,830
the capacitor has been fully charged. 572
00:55:31,830 --> 00:55:40,730
And so clearly if you add here this charge on that capacitor,then you must be okay. 573
00:55:40,730 --> 00:55:43,970
That must take into account that it is not zero, 574
00:55:43,970 --> 00:55:50,980
and this qmax then is simply Vo times C, 575
00:55:50,980 --> 00:55:55,060
which means the capacitor is fully charged. 576
00:55:55,060 --> 00:55:58,720
So do it your 1803 way or do it with brains 577
00:55:58,720 --> 00:56:06,370
and then you'll immediately agree to this solution to the differential equation. 578
00:56:06,370 --> 00:56:14,120
So you see here the decay. You see the oscillation and then when you end up 579
00:56:14,120 --> 00:56:22,830
you must end up there with fully charged capacitor. 580
00:56:22,830 --> 00:56:28,260
You now have to solve,depending upon your initial conditions,for α 581
00:56:28,260 --> 00:56:31,380
and you have to solve for q1. 582
00:56:31,380 --> 00:56:34,870
So that means you have to know what're your initial conditions. 583
00:56:34,870 --> 00:56:38,440
And initial conditions of this problem for instance would be that 584
00:56:38,440 --> 00:56:44,110
t = 0 when I throw the switch that there is no charge on the capacitor. 585
00:56:44,110 --> 00:56:45,680
That is one way I could do it. 586
00:56:45,680 --> 00:56:51,030
And there is no current flowing that could be my initial condition. 587
00:56:51,030 --> 00:56:56,830
Not the only one possible but that is a possible initial condition. 588
00:56:56,830 --> 00:56:59,650
And now I will leave you with a little bit work. 589
00:56:59,650 --> 00:57:06,540
First of all,you substitute in this equation,q=0 when t=0. 590
00:57:06,540 --> 00:57:10,810
So you still have q1 and you still have the cosine of α. 591
00:57:10,810 --> 00:57:15,820
Then you have to take the derivative of this equation which gives you the current. 592
00:57:15,820 --> 00:57:17,920
And then you have to make that zero. 593
00:57:17,920 --> 00:57:20,280
Now you get two terms when you do the derivative. 594
00:57:20,280 --> 00:57:22,780
So be careful. You get the derivative of this one times this 595
00:57:22,780 --> 00:57:25,900
and the derivative of this one times this. 596
00:57:25,900 --> 00:57:38,550
So you have to do it slowly. When you do that,you'll find that q1 equals -qmax 597
00:57:38,550 --> 00:57:43,140
divided by the cosine of α. 598
00:57:43,140 --> 00:57:55,610
And you'll find that the tangent of α equals -γ/2ω. 599
00:57:55,610 --> 00:57:59,820
I don't have a very good feeling for these numbers 600
00:57:59,820 --> 00:58:02,060
but what I do had a good feeling for is that PAGE2
| Name | Version | Size | Date | User |
| 02_b.srt | 1 | 79087 | 2/17/06 4:15 AM | OOPSSJTU |
Last Modified 2/21/06 8:32 AM
|
