1
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I will start today calculating for you the normal mode frequencies 2
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of the double pendulum and then I will drive that double pendulum, 3
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and then we'll see very dramatic things that I change it. 4
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So let us start here with the double pendulum. 5
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This is the equilibrium position if it hangs straight, 6
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length L, mass is M, this angle be θ1, and this angle θ2, 7
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I call this position X1, and I call this position X2. 8
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I'm going to introduce a short hand notation that (see the blackboard) 9
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I want to remind you that (see the blackboard) 10
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For small angular approximation, 11
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we do know the tensions, 12
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here we have the tension which I will call T2. 13
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That is here of course mg. 14
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So there is also a tension T2 here. 15
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Action equals minus reaction. 16
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And then there is here the tension T1. 17
00:02:14,130 --> 00:02:20,250
And here is mg, for that is the only force on these objects. 18
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At small angles I do know that T1 must be approximately 2mg 19
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coz it's carrying both low, so to speak, 20
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and we know that T2 is very close to mg. 21
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And we are going to use that in our approximations. 22
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So let me first start with the bottom one that has only 23
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two forces on it so that may be the easiest (see the blackboard). 24
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It's all the only force that it's driving it back to equilibrium in the 25
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small angular approximation is then the horizontal component of the T2, 26
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so that equals (see the blackboard). 27
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And for that I can write (see the blackboard). 28
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We devide out m and we use our short hand notation, 29
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until we get that X2 double dot and now we have here(see the blackboard), 30
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So this is my first differential equation for the second object. 31
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So now I am going to my first object. 32
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So now we get (see the blackboard). 33
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And now there is one force that is driving it back to 34
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equilibrium that is the horizontal component of T1. 35
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But the horizontal component of T2 is driving it away from equilibrium 36
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in the drawing that I have made here. 37
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So you get (see the blackboard). 38
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And so I am going to substitute in there now. 39
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The signs that I am going to substitute in there T1=2mg and T2=mg. 40
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So this becomes equal to (see the blackboard). 41
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I am going to divide by m and I am going to use 42
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my short hand notation and when I do that, 43
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I will come out here X1 double dot is this one. 44
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Here the angles g over l becomes ωo squared. 45
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But now look closely. 46
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There is here a 2 times X1 and here is 1 times X1. 47
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And both have a minus sign. 48
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So when they come out, I get three ωo squared. 49
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So I get (see the blackboard). 50
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And then I have to bring X2 out that becomes (see the blackboard) . 51
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And so we have to solve now this differential equation coupled 52
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with X1 and X2 together with this one. 53
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My solution has to satisfy both. 54
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I want to go over that one to make sure that this is correct. 55
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It what we pays off because one sign wrong and you hang. 56
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The whole thing falls apart. 57
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You dead in the waters so it pays off to think about it again. 58
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So we have the X2 double dot, I can live with that, plus ωo squared X2, 59
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that has the right smell for me, minus ωo squared X1. 60
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I can live with that differential equation. 61
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I go to this one of course the three is well known in the system 62
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like this that you get the three I have (see the blackboard). 63
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I think we are in go check. 64
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And so now we are going to put in our trial solutions. 65
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(see the blackboard) 66
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I will going to search for the frequencies. 67
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So this ω we have to solve for this ω. 68
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This is not a given. 69
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Since we are looking for normal mode solutions, these two ωs must be the same. 70
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Cannot write down ω1 and ω2. 71
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And I don't have to worry about any phase angles because since we have no damping. 72
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Either they are inphase or they are 180 degrees out of phase. 73
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180 degrees out of phase simply means a minus sign. 74
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So that's the great thing about this. 75
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The signs will take care then of the possible phase angles. 76
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So now we are going to substitute this solution in these two differential expressions. 77
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And I am going to write it down in the forms that 78
00:08:16,640 --> 00:08:22,820
I put C1s to the left and C2s to the right. 79
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And so I am first going to this one that is my object number one. 80
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I take the second derivative. 81
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So I get (see the blackboard) 82
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Because the second derivative here gives me a minus ω squared. 83
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I ditch the cosωt term because I'm going to have a cosωt everywhere. 84
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So I'm not going to write down the cosωt. 85
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I have here (see the blackboard). 86
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Notice I put the C1s here and C2 there. 87
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I am going to the other equation, then I am going to put C1s on the left. 88
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So we get (see the blackboard). 89
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That's the term you have here. 90
00:09:21,260 --> 00:09:31,780
And then we have (see the blackboard). 91
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And this minus ω squared comes from the second derivative of this one. 92
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Second derivative of this one, minus ω squared. 93
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Right. You get ω squared comes out 94
00:09:43,410 --> 00:09:48,040
and then the cosine goes to the sine that if you add a minus sign. 95
00:09:48,040 --> 00:09:53,940
So now we have here two equations with three unknowns. 96
00:09:53,940 --> 00:09:59,900
That is typical for normal mode solutions of a system with two oscillations. 97
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We don't know C1, we don't know C2, and we don't know ω. 98
00:10:04,870 --> 00:10:10,790
And so we remember from last time that you can always solve for the ratio C1 over C2. 99
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And you can solve for ω. 100
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You can only find C1 if you also know the initial conditions 101
00:10:16,530 --> 00:10:19,690
which I have not given. 102
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Instead of solving it in the high school way as I did last week, 103
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the simple way, the fast way. 104
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I am going to do now using Cramer's Rule. 105
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And the reason why I want to do this once even though 106
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in this case is really is not necessary. 107
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When you have three coupled oscillators or four, 108
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there is just no way that the high school methods will do it for you, 109
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and you have to have a more general approach. 110
00:10:43,040 --> 00:10:45,130
I have sent this to you by emailing all of you 111
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and I also assume that you have work on this a little bit 112
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since I request that you prepare for that. 113
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So I’m first going to write down what D is which is the determinant 114
00:11:02,350 --> 00:11:05,830
of these columns that you see that a, b and c of course 115
00:11:05,830 --> 00:11:12,560
we only have here two columns C1's and we have the C2's. 116
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So let me here write down what D is. 117
00:11:16,880 --> 00:11:24,640
So I'm going to have(see the blackboard). 118
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That's the C1. 119
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(see the blackboard). 120
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That's the C2. 121
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And then I get (see the blackboard). 122
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And here I get (see the blackboard). 123
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And the determinant of this, that's D. 124
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And I presume you all know how to get the determinant of this very simple matrix. 125
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we will do that together of course. 126
00:11:57,620 --> 00:12:00,520
So following the Cramer's Rule then. 127
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My C1 which is x there that's the one I want to solve for. 128
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So my C1, I now have to take this 0. 129
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This is also a zero by the way I forgot to mention that this is a 0. 130
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Right? This equation this is a 0. 131
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So this column now zero-zero has to come first. 132
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(see the blackboard) 133
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And then the second column. 134
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This is the same as it was here. 135
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(see the blackboard) 136
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So we now know that is C1. 137
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And then we go to C2. 138
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So now the zero-zero column shifts towards the right. 139
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Both here and the first column is unchanged. 140
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So we have here (see the blackboard) 141
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You got to admit that the upstairs here, the determinant of 142
00:13:21,530 --> 00:13:24,950
this matrix, is 0 because you have two 0s here 143
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and it's also 0 for this one. 144
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But clearly zero solution to C1 and C2 are meaningless. 145
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They are not in correct because you will get in this 146
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differential equation that zero is zero which is while obvious. 147
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So we don't want the solutions that C1 and C2 are 0. 148
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And the only way we can avoid that is to demand that D becomes 0. 149
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Because now you get 0 divided by 0 that's a whole different story. 150
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That's not necessarily zero. 151
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And that then is the idea behind getting the solutions 152
00:13:58,500 --> 00:14:03,340
to the search for normal mode frequencies. 153
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So we must demand in this case that this becomes 0 154
00:14:06,530 --> 00:14:11,510
otherwise you get trivial solutions which are of no interest. 155
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So then we make D equal 0. 156
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We do get that the determinant that matrix becomes (see the blackboard). 157
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We do get that the determinant (see the blackboard). 158
00:14:50,860 --> 00:14:54,860
And this equation is the equation in ω to the fourth. 159
00:14:54,860 --> 00:14:59,950
You can solve that. You can solve that for ω squared. 160
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I will leave you with that solution. This is utterly trivial and outcomes then 161
00:15:06,200 --> 00:15:20,460
that ω- squared which is the lowest frequency of the two is (see the blackboard) 162
00:15:20,460 --> 00:15:29,320
And ω+ squared is (see the blackboard) 163
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So this step for you will take you no more than may be half a minute. 164
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But be careful because you can easily slip up of course. 165
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So we now have a solution that ω- is approximately... 166
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If I calculate 2-sqrt(2), it's approximately 0.77. 167
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0.76 ωo. 168
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Not intuitive it also. It's lower than the resonance frequency of a single pendulum. 169
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And then I can substitute this value for ω either back to my equations 170
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or I substitute it back in this if you want to. 171
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You get 0 divided by 0 which is not going to be 0. 172
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And that you are going to find then that C2 divided by C1 in that minus mode 173
00:16:29,490 --> 00:16:42,730
in that lowest possible mode, you will find that that is (see the blackboard). 174
00:16:42,730 --> 00:16:50,450
And the ω+ solution gives you a frequency which is about 0.85 ωo. 175
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That solution you can put back into your differential equations here 176
00:16:54,290 --> 00:16:57,210
or not the differential equations I mean into this one 177
00:16:57,210 --> 00:17:00,080
or you put it back into this if you want to. 178
00:17:00,080 --> 00:17:07,880
And you will find now that C2 over C1 for that plus mode, I call this a plus mode, 179
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that's my short hand notation, is going to be 180
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(see the blackboard). 181
00:17:26,660 --> 00:17:31,530
So those are the formal solutions for the normal modes. 182
00:17:31,530 --> 00:17:36,940
And if I give you the initial conditions, then of course you would calculate C1 183
00:17:36,940 --> 00:17:40,470
and then you know everything because you know the ratios C2 over C1. 184
00:17:40,470 --> 00:17:45,470
But without the initial conditions you can not do that. 185
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Each of these normal mode solutions satisfies both differential equations. 186
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So therefore 187
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a linear superposition of the two normal mode solutions is the general solution. 188
00:17:59,210 --> 00:18:03,500
So when you start that system off at time t equal 0, 189
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you specify X1, you specify the velocity of object one, you specify X2, 190
00:18:09,850 --> 00:18:13,830
and you specify the velocity of that object. 191
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Then it's going to oscillate in the superposition, 192
00:18:17,640 --> 00:18:21,050
the linear superposition of two normal mode solutions, 193
00:18:21,050 --> 00:18:26,900
One you see here as ω- and this will be the ratio of the amplitudes. 194
00:18:26,900 --> 00:18:30,810
And this is the ω+, and that will be the ratio of the amplitudes. 195
00:18:30,810 --> 00:18:33,280
That is non-negotiable. 196
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That what the system will do. 197
00:18:37,680 --> 00:18:41,890
Now we are going to make a dramatic change. 198
00:18:41,890 --> 00:18:46,330
Now we are going to drive this system. 199
00:18:46,330 --> 00:18:53,870
So now this is the equilibrium position now of the whole system. 200
00:18:53,870 --> 00:18:59,080
And I, I'm going to drive it back and forth holded in my hand like this. 201
00:18:59,080 --> 00:19:01,270
And I'm going to shake it. 202
00:19:01,270 --> 00:19:09,490
I call the position of my hand η equals ηocosωt. 203
00:19:09,490 --> 00:19:15,510
So ηo is the amplitude of my motion of my hand. 204
00:19:15,510 --> 00:19:19,600
That is what I'm going to do. 205
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And so when I look now at this equation, at this figure not the equation, 206
00:19:24,800 --> 00:19:28,430
but that the figure, I call this now X1. 207
00:19:28,430 --> 00:19:34,460
I call this now X2 because you should always call X1 and X2 the distance from equilibrium. 208
00:19:34,460 --> 00:19:35,980
And this is now equilibrium. 209
00:19:35,980 --> 00:19:38,370
If I don't shake at all I have the pendulum here. 210
00:19:38,370 --> 00:19:41,300
And so it's hanging straight down so this is now equilibrium. 211
00:19:41,300 --> 00:19:48,080
And this location here at a random moment in time is now η. 212
00:19:48,080 --> 00:19:51,310
What has changed? 213
00:19:51,310 --> 00:19:54,720
Well at first sight, very little has changed. 214
00:19:54,720 --> 00:19:57,030
The only thing that has changed now is that 215
00:19:57,030 --> 00:20:04,130
the sine of θ1 is now (x1-η)/l. 216
00:20:04,130 --> 00:20:08,200
So I have here a -η. 217
00:20:08,200 --> 00:20:16,110
It looks rather innocent. But the consequencies will be unbelievable. 218
00:20:16,110 --> 00:20:19,770
So I can leave everything on the blackboard the way I have it. 219
00:20:19,770 --> 00:20:27,180
All I have to do is to put instead of the sinθ, x1/l. 220
00:20:27,180 --> 00:20:31,870
I have to put in (x1-η)/l. 221
00:20:31,870 --> 00:20:34,840
The nature was very kind to me it already left some space there. 222
00:20:34,840 --> 00:20:36,140
Do you see that? 223
00:20:36,140 --> 00:20:39,840
It's anticipated that I am going to do that. 224
00:20:39,840 --> 00:20:43,290
So you have -η there. 225
00:20:43,290 --> 00:20:51,870
Well, if now you divide m out and you are going to use the short hand notation, 226
00:20:51,870 --> 00:20:58,850
then leaving η on the right side, which is nice to do, we get -η. 227
00:20:58,850 --> 00:21:16,800
So this 0 now becomes two ωo squared times ηo times the cosine of ωt. 228
00:21:16,800 --> 00:21:21,620
It becomes two ωo squared times η. 229
00:21:21,620 --> 00:21:29,370
You see it do. So it's no longer a 0. 230
00:21:29,370 --> 00:21:35,930
And so when we now substitute in there these trial functions. 231
00:21:35,930 --> 00:21:39,480
They now have a completely different meaning. 232
00:21:39,480 --> 00:21:46,460
ω is no longer negotiable, ω is set by me is a driver. 233
00:21:46,460 --> 00:21:51,600
So we are not going to solve for ω. ω is a given. 234
00:21:51,600 --> 00:21:57,080
That means if ω is a given that we are going to end up not with two equations 235
00:21:57,080 --> 00:22:00,640
with three unknowns C1, C2 and ω, 236
00:22:00,640 --> 00:22:03,320
but we are going to end up with two equations with two unknowns, 237
00:22:03,320 --> 00:22:06,800
only C1 and C2, because ω is a given. 238
00:22:06,800 --> 00:22:10,700
And so in the steady state solution you get a number for C1 239
00:22:10,700 --> 00:22:15,140
and a number for C2 in terms of ηo of course. 240
00:22:15,140 --> 00:22:17,380
And you will see how that works. 241
00:22:17,380 --> 00:22:22,370
So when we put in these functions now these ωs are fixed 242
00:22:22,370 --> 00:22:26,840
are no longer something that we search for. 243
00:22:26,840 --> 00:22:32,590
It is my ω. I can make it 0. I can make it infinite. I can make it anything I want to. 244
00:22:32,590 --> 00:22:36,370
That is what we want to study. 245
00:22:36,370 --> 00:22:39,610
So if we want to go back now to these two equations. 246
00:22:39,610 --> 00:22:44,010
What changes here since we have put in this trial function. 247
00:22:44,010 --> 00:22:47,430
The only thing that disappeares is this cosine ωt. 248
00:22:47,430 --> 00:22:54,350
So we end up here with two ωo squared times ηo. 249
00:22:54,350 --> 00:22:57,060
And nothing else changes. 250
00:22:57,060 --> 00:23:05,560
But now we're looking now at our solution for C1 251
00:23:05,560 --> 00:23:14,860
and we are going to look at our solution for C2 in steady state. 252
00:23:14,860 --> 00:23:20,630
Ah... It's easy, right? We apply Cramer's Rule. 253
00:23:20,630 --> 00:23:24,230
And the only thing that changes is this one which 254
00:23:24,230 --> 00:23:32,390
has to be replaced by this two ωo squared times ηo. 255
00:23:32,390 --> 00:23:37,340
ηo remember is the amplitude of my hand. 256
00:23:37,340 --> 00:23:43,940
η is the displacement of my hand at any moment in time. ηo, the amplitude. 257
00:23:43,940 --> 00:23:57,420
And so here we get then also (see the blackboard) 258
00:23:57,420 --> 00:24:01,790
And so now we can solve for C1 and C2. 259
00:24:01,790 --> 00:24:03,950
We get an answer. 260
00:24:03,950 --> 00:24:06,960
Not just a ratio only, we get an answer. 261
00:24:06,960 --> 00:24:08,990
We know exactly what C1 is going to be 262
00:24:08,990 --> 00:24:11,240
and we know exactly what C2 is going to be. 263
00:24:11,240 --> 00:24:14,790
Because ω is none negotiable. 264
00:24:14,790 --> 00:24:16,920
ω is now unknown. 265
00:24:16,920 --> 00:24:21,640
When we solve this, we were searching for the ω and outcame these ω. 266
00:24:21,640 --> 00:24:23,190
That's not the case anymore. 267
00:24:23,190 --> 00:24:28,960
We know ω. It's called ω. That's it. 268
00:24:28,960 --> 00:24:37,050
So I can write down now C1. So I take the determinant there of the upstairs. 269
00:24:37,050 --> 00:24:52,430
So that gives me (see the blackboard) 270
00:24:52,430 --> 00:24:57,750
that is this diagonal and this one is 0. 271
00:24:57,750 --> 00:25:01,830
And I have to divide that by D. 272
00:25:01,830 --> 00:25:06,330
Now I could write D in that form and I could do that. 273
00:25:06,330 --> 00:25:08,900
There is nothing wrong with that but I can write it in the form 274
00:25:08,900 --> 00:25:12,310
which is a little bit more transparent. 275
00:25:12,310 --> 00:25:20,140
We do know that ω- and ω+ will make D zero. 276
00:25:20,140 --> 00:25:31,080
So you can write this then in the following way you can here (see the blackboard) 277
00:25:31,080 --> 00:25:33,320
That must be the same as what I have there 278
00:25:33,320 --> 00:25:37,430
because you see with this one becomes ω- then this goes to be 0 279
00:25:37,430 --> 00:25:39,720
and when this one becomes ω+, that's also 0. 280
00:25:39,720 --> 00:25:43,270
So it's different way of writing give a little bit more insight. 281
00:25:43,270 --> 00:25:47,160
It reminds you that the downstairs will go to 0 282
00:25:47,160 --> 00:25:51,020
when you hit those resonance frequencies. 283
00:25:51,020 --> 00:25:56,190
And so I will also write down C2 then. 284
00:25:56,190 --> 00:26:00,590
This one is 0, I get minus this one. 285
00:26:00,590 --> 00:26:11,310
So I get (see the blackboard) 286
00:26:11,310 --> 00:26:12,600
So you can write this for it. 287
00:26:15,600 --> 00:26:18,130
So I will check that see whether I am happy with that?. 288
00:26:18,130 --> 00:26:23,330
Two ωo squared ηo ωo squared divided by ω squared that looks good. 289
00:26:23,330 --> 00:26:27,450
This looks good. This looks good. And I have here I am happy. 290
00:26:34,130 --> 00:26:43,590
Our task now is not to look at these equations but to see through them. 291
00:26:43,590 --> 00:26:49,280
And they are by no means trivial for they are going to do as a function of ω. 292
00:26:49,280 --> 00:26:57,150
It's extraordinary complicated dependence on ω. 293
00:26:57,150 --> 00:27:02,680
And I have plotted for you these values of C1 and C2 294
00:27:02,680 --> 00:27:04,050
for which you get an answer now. 295
00:27:04,050 --> 00:27:06,380
You also know C1 over C2 of course 296
00:27:06,380 --> 00:27:09,630
because you know C1 and you know C2. 297
00:27:09,630 --> 00:27:12,700
And I am going to show you this as a function of ω 298
00:27:12,700 --> 00:27:16,010
and then we will try to digest that together. 299
00:27:16,010 --> 00:27:17,740
And this plot, 300
00:27:17,740 --> 00:27:24,050
like the other plots that I will show you later today will be on the 8.03 website. 301
00:27:24,050 --> 00:27:26,480
They will be part of lecture notes. 302
00:27:26,480 --> 00:27:30,580
So you have to click on lecture notes and then you will see these plots. 303
00:27:30,580 --> 00:27:35,400
This is the first one which is the double pendulum. 304
00:27:35,400 --> 00:27:41,160
What is plotted here horizontally is ω divided by ωo. 305
00:27:41,160 --> 00:27:43,580
That's the ωo that we mention there. 306
00:27:43,580 --> 00:27:50,930
And so you see the first resonance here is about at 0.76 ωo. 307
00:27:50,930 --> 00:27:55,980
And the second resonance here is about at 1.85 ωo. 308
00:27:55,980 --> 00:28:01,950
And what we plot here is the amplitude C divided by ηo. 309
00:28:01,950 --> 00:28:07,590
Because obviously if you make ηo larger you expect that has an effect on the C of course. 310
00:28:07,590 --> 00:28:10,050
You know if you drive it to a large amplitude 311
00:28:10,050 --> 00:28:13,580
of course the object will also respond accordingly. 312
00:28:13,580 --> 00:28:18,330
And so therefore we have it as a function of ηo. You see the ηo here? 313
00:28:18,330 --> 00:28:23,100
C1 is linearly proportional with ηo. C2 is linearly proportional with ηo. 314
00:28:23,100 --> 00:28:27,790
That's no surprise. So we divided by ηo. 315
00:28:27,790 --> 00:28:32,980
When we plot it upstairs it means that 316
00:28:32,980 --> 00:28:38,440
the amplitude is in phase with the driver that's the way we have written it. 317
00:28:38,440 --> 00:28:41,550
And if it is below the zero line it means that 318
00:28:41,550 --> 00:28:46,790
the amplitude of the object is out of phase with the driver. 319
00:28:46,790 --> 00:28:51,610
That's all it means. That's the science convention. 320
00:28:51,610 --> 00:29:02,800
So let us now look at this and try to digest it and use, to some degree, our intuition 321
00:29:02,800 --> 00:29:07,200
and see whether that agrees with our intuition. 322
00:29:07,200 --> 00:29:10,390
And let us start when ω goes to 0. 323
00:29:10,390 --> 00:29:13,000
And don't look now at the solutions. 324
00:29:13,000 --> 00:29:17,640
If ω is 0, I have a double pendulum in my hand 325
00:29:17,640 --> 00:29:20,550
and I am going to move it to the left 326
00:29:20,550 --> 00:29:24,480
and twenty five years from now I am going to go back and I move it to the left again. 327
00:29:24,480 --> 00:29:27,310
So the pendulum is always straight. 328
00:29:27,310 --> 00:29:33,500
And it is clear that C1 and C2 both mut be ηo. 329
00:29:33,500 --> 00:29:39,310
You'd better believe it that if you substitute in there ωo that's what you will find. 330
00:29:39,310 --> 00:29:47,390
So C1 must be C2 and must be ηo and it must even be +ηo. 331
00:29:47,390 --> 00:29:50,650
It must be in phase with the driver. 332
00:29:50,650 --> 00:29:54,950
And you see that the ratio, that not the ratio which is C divided by ηo, 333
00:29:54,950 --> 00:29:58,810
that is a ratio, that is plotted at +1? 334
00:29:58,810 --> 00:30:02,510
And so both are in phase with the driver. 335
00:30:02,510 --> 00:30:04,520
So that's the trivial result. 336
00:30:04,520 --> 00:30:11,300
It means that the pendulum which is here. Twenty five years from now is here. 337
00:30:11,300 --> 00:30:14,920
That's ωo, almost 0. 338
00:30:14,920 --> 00:30:19,880
So notice that we now know the ratio C2 over C1 which is +1. 339
00:30:19,880 --> 00:30:30,020
Here are the ratio C1 over C2 or C2 over C1 and now entirely dictated by this. 340
00:30:30,020 --> 00:30:31,930
Nothig to do with normal modes any more. 341
00:30:31,930 --> 00:30:36,740
Don't be surprised that C1 over C2 is now +1. 342
00:30:36,740 --> 00:30:41,860
Now look what happens I am going to increase my ω 343
00:30:41,860 --> 00:30:44,580
and what you see is that the red curve 344
00:30:44,580 --> 00:30:48,840
which is the second object is the lowest one of the two 345
00:30:48,840 --> 00:30:52,730
is going to have a large amplitude than the top one. 346
00:30:52,730 --> 00:30:56,650
You see it already begins to grow very rapidly. 347
00:30:56,650 --> 00:31:03,140
And when you reach resonance the ratio is going to be +2.4 of course. 348
00:31:03,140 --> 00:31:04,940
That's obvious. 349
00:31:04,940 --> 00:31:09,390
Now at resonance you get an infinite amplitude for each. 350
00:31:09,390 --> 00:31:12,990
That's because it's nonsence that has physically no meaning. 351
00:31:12,990 --> 00:31:16,180
So you shouldn't really think of it as going to infinity. 352
00:31:16,180 --> 00:31:20,480
For one thing, if C1 became infinity 353
00:31:20,480 --> 00:31:24,590
that means if this points here ends up there in a halt. 354
00:31:24,590 --> 00:31:27,330
It's hard to argue that there is a amall angle, right? 355
00:31:27,330 --> 00:31:30,930
So in any case the solution wouldn't even halt apart from the fact 356
00:31:30,930 --> 00:31:32,850
of course there is always some damping. 357
00:31:32,850 --> 00:31:36,990
And so we never go completely off scale, so to speak. 358
00:31:36,990 --> 00:31:41,930
However you will see that when you drive the system the double pendulum 359
00:31:41,930 --> 00:31:45,120
and you will approach resonance that you will very quickly see 360
00:31:45,120 --> 00:31:51,240
that the ratio of the amplitude of the second one over the top one 361
00:31:51,240 --> 00:31:56,890
will grow, will become one and a half, will become two and then in the limited case, 362
00:31:56,890 --> 00:32:00,080
you will get the +2.4. 363
00:32:00,080 --> 00:32:09,060
So as ω goes up, you are going to see that C2 over C1 going to be large than one, 364
00:32:09,060 --> 00:32:14,480
right there this one and then ultimately it will reach that 2.4. 365
00:32:14,480 --> 00:32:20,300
But that is that extreme case of resonance. 366
00:32:20,300 --> 00:32:26,320
And when you look at the motion of the pendulum if you drive it someway here. 367
00:32:26,320 --> 00:32:32,370
Notice that they are both in phase with the driver and C2 is larger than C1. 368
00:32:32,370 --> 00:32:39,140
So what you will see is this is C1 and then this is C2. 369
00:32:39,140 --> 00:32:45,150
So C2 is larger than C1 and they are in phase with the driver. 370
00:32:45,150 --> 00:32:49,360
And so when they return you will see the pendulum like this. 371
00:32:49,360 --> 00:32:56,100
That's the sweeping that you will see. 372
00:32:56,100 --> 00:33:00,870
Now there is something truely bizarre. 373
00:33:00,870 --> 00:33:06,490
I go a little higher in frequency I go over the resonance and I see here a point 374
00:33:06,490 --> 00:33:11,710
whereby the frequency happens to be exactly the frequency of a single pendulum. 375
00:33:11,710 --> 00:33:14,600
ω divided by ωo is one. 376
00:33:14,600 --> 00:33:19,040
And the top one refuses to move. 377
00:33:19,040 --> 00:33:25,810
But the bottom one does. The bottom one here has an amplitude 378
00:33:25,810 --> 00:33:30,470
which is twice the amplitude of the driver. Look, you see it two here? 379
00:33:30,470 --> 00:33:34,420
And it is out of phase with the driver. 380
00:33:34,420 --> 00:33:40,330
That is unimaginable. It's unimaginable which you are going to see. 381
00:33:40,330 --> 00:33:51,580
So when ω is ωo, C1 become 0, C2 becomes -2ηo. 382
00:33:51,580 --> 00:33:56,870
And so this pendulum is going to look then as follows. 383
00:33:56,870 --> 00:34:00,800
I will make a drawing here have a little bit more space. 384
00:34:00,800 --> 00:34:05,820
So if my hand is here at ηo， 385
00:34:05,820 --> 00:34:11,640
then number one will stand still, won't do anything. 386
00:34:11,640 --> 00:34:20,130
But number two will have an amplitude which is twice ηo but is on the other side. 387
00:34:20,130 --> 00:34:26,130
So this is 2ηo. 388
00:34:26,130 --> 00:34:34,130
And then if you look half a period later it will look like this. 389
00:34:34,130 --> 00:34:43,500
So this is ηo and this is 2ηo. And this one doesn't move. 390
00:34:43,500 --> 00:34:48,180
That's what it tells you. 391
00:34:48,180 --> 00:34:56,850
How on the earth can the lower pendulum oscillate if the upper one stands still. 392
00:34:56,850 --> 00:35:00,340
I want you to think about that. 393
00:35:00,340 --> 00:35:06,960
I am still having sleepless nights about it. And so maybe you will have some, too. 394
00:35:06,960 --> 00:35:10,760
And if you have some clever ideas come to see me. 395
00:35:10,760 --> 00:35:15,940
But the logical consequence of what we did is that this one will stand still, 396
00:35:15,940 --> 00:35:22,600
and the other one will still oscillate and be driven by my hands, 397
00:35:22,600 --> 00:35:25,550
I have to keep moving this otherwise this will go to peaces? 398
00:35:25,550 --> 00:35:29,720
I must keep doing this all the time at that frequency 399
00:35:29,720 --> 00:35:34,930
which is exactly the resonance frequency of the single pendulum which is that ωo. 400
00:35:34,930 --> 00:35:38,300
I must keep doing this and this one does nothing 401
00:35:38,300 --> 00:35:44,200
and this one has double the swing of this and is out of the phase. 402
00:35:44,200 --> 00:35:52,810
If then you go even higher then you will see that the two object will go out of phase. 403
00:35:52,810 --> 00:35:55,370
The upper one will be in phase with the driver 404
00:35:55,370 --> 00:35:57,540
and the lower one will be out of phase with the driver. 405
00:35:57,540 --> 00:36:02,370
and then you hit the second resonance when things will get out of hands 406
00:36:02,370 --> 00:36:11,820
and you get that ratio -0.42 back of course. 407
00:36:11,820 --> 00:36:16,900
I will want to demonstrate to you this situation here 408
00:36:16,900 --> 00:36:22,910
and this situation to see whether they make sense. 409
00:36:22,910 --> 00:36:27,680
And so now I am going to use a double pendulum 410
00:36:27,680 --> 00:36:32,990
and drive it with my own frequency which I determine. 411
00:36:32,990 --> 00:36:38,040
I am the boss. I determine ω. Not looking for normal mode solutions. 412
00:36:38,040 --> 00:36:46,720
I determine ω and I am going to first drive it with ωo which ω is about 0. 413
00:36:46,720 --> 00:36:53,860
In another words, I am going to driver it here. 414
00:36:53,860 --> 00:37:03,700
And what you going to see is of course fantastic. Absolutely fantastic. 415
00:37:03,700 --> 00:37:11,020
It's hanging straight down now. And I am moving it over a distance of one foot. 416
00:37:11,020 --> 00:37:18,890
So ηo is one foot and C1 is one foot and C2 is one foot and they are inphase with the driver. 417
00:37:18,890 --> 00:37:23,030
Physics works. 418
00:37:23,030 --> 00:37:25,670
If I go a little higher up here, 419
00:37:25,670 --> 00:37:29,370
so I go somewhere here approaching resonance, 420
00:37:29,370 --> 00:37:35,180
then you will see that C2 becomes larger than C1. This is C2 and this is C1. 421
00:37:35,180 --> 00:37:40,100
And you get to see a picture which is very much like this. 422
00:37:40,100 --> 00:37:45,580
Now try that. 423
00:37:45,580 --> 00:37:53,450
So I drive it below resonance but not too far below. And then you see it. 424
00:37:53,450 --> 00:37:59,160
You see C1 is smaller than C2. No longer +1. 425
00:37:59,160 --> 00:38:05,390
If this was 1. You will see one over C2 is 1. That's no longer the case now. 426
00:38:05,390 --> 00:38:07,630
You really see that C2 is getting ahead. 427
00:38:07,630 --> 00:38:09,620
The ahead not in terms of a phase ahead 428
00:38:09,620 --> 00:38:14,720
but in terms of amplitude. Very clear. 429
00:38:14,720 --> 00:38:18,050
Now I am going to attemp to do the impossible. 430
00:38:18,050 --> 00:38:22,780
And the impossible is to try to hit this point. 431
00:38:22,780 --> 00:38:25,710
The point whereby the upper one stands still 432
00:38:25,710 --> 00:38:29,490
and whereby the lower one will have an amplitude 433
00:38:29,490 --> 00:38:35,280
which is twice that of my hands but out of phase of my hands. 434
00:38:35,280 --> 00:38:39,200
How on earth can I ever drive the system 435
00:38:39,200 --> 00:38:45,720
with that frequency ωo which is the frequency of a single pendulum? 436
00:38:45,720 --> 00:38:49,020
Well, maybe I can't. 437
00:38:49,020 --> 00:38:54,010
But I will try and the way that I am going to try it is the following. 438
00:38:54,010 --> 00:39:00,750
I know what the resonance frequency of a single pendulum is. That's this. 439
00:39:00,750 --> 00:39:02,950
Yeah, I can feel it in my hands. I can feel it in my stomach. 440
00:39:02,950 --> 00:39:07,840
I can feel it in my brain. I feel it all over my body. 441
00:39:07,840 --> 00:39:15,350
I can burn the frequency into my chips here and then I can close my eyes. 442
00:39:15,350 --> 00:39:19,050
But you are looking and generate that frequency 443
00:39:19,050 --> 00:39:23,710
which is burns here and drive the system as a double pendulum. 444
00:39:23,710 --> 00:39:27,010
If I succeeds, you will see the upper one stands still 445
00:39:27,010 --> 00:39:30,250
and the bottom one will have twice the amplitude of my hands. 446
00:39:30,250 --> 00:39:34,800
So the success of this depends exclusively on how accurately 447
00:39:34,800 --> 00:39:40,370
I can burn this frequency into my chips. 448
00:39:40,370 --> 00:39:42,520
So you have to be quiet. So I am going to count. 449
00:39:42,520 --> 00:39:46,920
One, two, three. I am burning now. 450
00:39:46,920 --> 00:39:52,630
One, two, three, four. 451
00:39:52,630 --> 00:39:57,890
One, two, three, four. 452
00:39:57,890 --> 00:39:58,850
Got closing my eyes. 453
00:39:58,850 --> 00:40:04,680
One, two, three, four. 454
00:40:04,680 --> 00:40:10,620
One, two, three, four. 455
00:40:10,620 --> 00:40:16,730
One, two, three, four. 456
00:40:16,730 --> 00:40:22,740
One, two, three, four. 457
00:40:22,740 --> 00:40:24,500
One... 458
00:40:24,500 --> 00:40:29,730
You are not saying anything? Didn't you see this one standing still? 459
00:40:29,730 --> 00:40:36,500
Yes you see it. Did you also see the other one have twice the amplitude of my hands, 460
00:40:36,500 --> 00:40:39,110
and out of phase with my hand? You didn't see that, right? 461
00:40:39,110 --> 00:40:41,020
Admit it you didn't see it. 462
00:40:41,020 --> 00:40:46,780
Because you were not looking for it. Specially for you, I'll do it again. 463
00:40:46,780 --> 00:40:52,240
So you really have to see No.1 that this one will practically stand still, 464
00:40:52,240 --> 00:40:59,200
No.2 that this one has doubled the amplitude of my hand but out of phase. 465
00:40:59,200 --> 00:41:05,310
The decay time of burning is only one minute. So I have to burn it again. 466
00:41:05,310 --> 00:41:11,870
One, two, three, four, five. 467
00:41:11,870 --> 00:41:19,570
One, two, three, four, five. 468
00:41:19,570 --> 00:41:23,440
One... Uu~Uu~ 469
00:41:23,440 --> 00:41:28,170
With things happened and you have to start all over with the burning. 470
00:41:28,170 --> 00:41:30,240
There we go. 471
00:41:30,240 --> 00:41:37,500
One, two, three, four, five. 472
00:41:37,500 --> 00:41:44,790
One, two, three, four, five. 473
00:41:44,790 --> 00:41:51,880
One, two, three, four, five. 474
00:41:51,880 --> 00:41:55,840
One, two, three... 475
00:41:55,840 --> 00:41:58,340
Are you seeing it? 476
00:41:58,340 --> 00:42:07,310
Yeah!(students) All right. 477
00:42:07,310 --> 00:42:15,230
This is an ideal moment for the break. Going to handout the mini quiz. 478
00:42:15,230 --> 00:42:18,720
And we will reconvene. I will give you six minutes this time. 479
00:42:18,720 --> 00:42:25,280
So you can even stretch your legs and I would like some help handing this out 480
00:42:25,280 --> 00:42:28,830
and then you bring it back, and I will put some boxes out there. 481
00:42:28,830 --> 00:42:34,840
So if you can help me handing it out here You can start right away. 482
00:42:34,840 --> 00:42:39,780
Hand this out here. For those of you have no seats, come forward and get some seats. 483
00:42:39,780 --> 00:42:44,190
Nico, we are still friends, right? So why don't you hand that out? 484
00:42:44,190 --> 00:42:46,590
And why don't you hand this out here? 485
00:42:46,590 --> 00:42:55,280
You can also give it to people here. Here, you do that. You can start right way. 486
00:42:55,280 --> 00:43:07,520
I'm now going to do something perhaps even more ambitious. 487
00:43:07,520 --> 00:43:12,610
And I'm going to now couple three oscillators not pendulums yet, 488
00:43:12,610 --> 00:43:18,720
but I'm to couple three oscillators which are connected with four springs, 489
00:43:18,720 --> 00:43:23,590
I'm going to work on this, three masses equal masses, 490
00:43:23,590 --> 00:43:26,120
four springs, spring constant K. 491
00:43:26,120 --> 00:43:28,480
And the spring constant are the same. 492
00:43:28,480 --> 00:43:31,550
And I'm going to drive that system. 493
00:43:31,550 --> 00:43:42,060
One, two, three, four, and this is the end. 494
00:43:42,060 --> 00:43:50,590
In other words, I have here a spring and here is the first mass, second mass, 495
00:43:50,590 --> 00:43:55,240
third mass, and here it is fixed. 496
00:43:55,240 --> 00:44:07,550
And I'm going to drive it here with a displacement η which is ηocosωt. 497
00:44:07,550 --> 00:44:15,740
So at random moment in time, this is where my hand will be, so this is η. 498
00:44:15,740 --> 00:44:18,270
This is where the first mass will be. 499
00:44:18,270 --> 00:44:22,850
Remember you always called the displacement x1 from its equilibrium, 500
00:44:22,850 --> 00:44:27,730
that is its dotted line. So here is the spring. 501
00:44:27,730 --> 00:44:33,180
This one is here, so I call this x2. 502
00:44:33,180 --> 00:44:41,240
Here is the spring, and this one is here, so this is x3. 503
00:44:41,240 --> 00:44:44,870
And here is the spring, and here is the spring. 504
00:44:44,870 --> 00:44:51,560
You may have noticed more than once now that I have the certian discipline 505
00:44:51,560 --> 00:44:56,180
that I always offset them in the same direction, do you have to do that? 506
00:44:56,180 --> 00:45:02,240
No. If you don't do it, your chance over mistake on the sign slip is much larger than 507
00:45:02,240 --> 00:45:06,100
if you always set them off in the same direction, you will see shortly why. 508
00:45:06,100 --> 00:45:11,520
So that's certainly someting that is not must, but it's the smart thing to do. 509
00:45:11,520 --> 00:45:17,270
I defined this as my positive direction but that of course is free free choice. 510
00:45:17,270 --> 00:45:27,630
Now let's at this situation and at this moment in time, that x1 be larger than η, 511
00:45:27,630 --> 00:45:36,350
let x2 be larger than x1and let x3 be larger than x2. 512
00:45:36,350 --> 00:45:41,730
And this assumption will have no consequences for what follows 513
00:45:41,730 --> 00:45:50,010
at least not for the differential equations. If x1 is larger than η, 514
00:45:50,010 --> 00:45:55,030
that first spring is longer than it wants to be 515
00:45:55,030 --> 00:45:58,030
because I've assumed that x1 is larger than η. 516
00:45:58,030 --> 00:46:00,920
And so that means there will be a force in this direction, 517
00:46:00,920 --> 00:46:05,180
because this spring is longer than it wants to be. 518
00:46:05,180 --> 00:46:09,540
If x2 is larger than x1, the spring is also longer than it wants to be. 519
00:46:09,540 --> 00:46:15,400
So I would want to contract, so that is a force in this direction. 520
00:46:15,400 --> 00:46:21,620
And so I can write down now the differential equation for my first object. 521
00:46:21,620 --> 00:46:36,660
So that's going to be m x1(double dot). That equals -k(x1-η), 522
00:46:36,660 --> 00:46:40,550
because this the amount by which it is longer than it wants to be. 523
00:46:40,550 --> 00:46:47,500
So times x1-η. That is this force. 524
00:46:47,500 --> 00:46:55,790
And this force is now in the plus direction, is +k times... 525
00:46:55,790 --> 00:47:04,870
This spring here is longer than it wants to be by an amount of x2-x1. 526
00:47:04,870 --> 00:47:10,660
Not ω1, but x1. 527
00:47:10,660 --> 00:47:16,380
That's my differential equation for the first object. 528
00:47:16,380 --> 00:47:21,580
And this one is always correct, even if x1 is not larger than η. 529
00:47:21,580 --> 00:47:26,770
Because if x1 is not larger than η, then this force flips over. 530
00:47:26,770 --> 00:47:28,970
Well, this will also flip over. 531
00:47:28,970 --> 00:47:31,830
So that's why it's always causal and advisable 532
00:47:31,830 --> 00:47:34,770
to make that assumption to start with because again 533
00:47:34,770 --> 00:47:38,330
it reduces the probability of making mistakes. 534
00:47:38,330 --> 00:47:43,970
That's all. It's nothing else to do it, just reduce the chance of flipping up. 535
00:47:43,970 --> 00:47:49,870
So let's now go to this object. If this spring is larger, longer than it wants to be, 536
00:47:49,870 --> 00:47:54,750
it wants to contract, so this object will see a force to the left. 537
00:47:54,750 --> 00:47:59,400
But if this spring is longer than it wants to be because x3 is larger than x2, 538
00:47:59,400 --> 00:48:03,390
it will experience a force to the right. 539
00:48:03,390 --> 00:48:08,230
So I can write down now the differential equation for object No.2. 540
00:48:08,230 --> 00:48:14,750
m x2(double dot), 541
00:48:14,750 --> 00:48:20,170
Notice that the one that is here to the left is the same one that is here to the right. 542
00:48:20,170 --> 00:48:22,310
Right? Action equals minus reaction. 543
00:48:22,310 --> 00:48:24,380
This pull is the same as this pull. 544
00:48:24,380 --> 00:48:27,090
So it is going to be this term which is now has a minus sign. 545
00:48:27,090 --> 00:48:29,730
You also see that in coupled oscillators. 546
00:48:29,730 --> 00:48:36,620
That was a plus here, you're going to come out here with a minus sign. 547
00:48:36,620 --> 00:48:38,610
You see that? Comes out nicely. 548
00:48:38,610 --> 00:48:43,600
So this spring is longer than it once to be by the amount x2-x1, 549
00:48:43,600 --> 00:48:45,630
and the force is in the minus direction. 550
00:48:45,630 --> 00:48:55,050
And this one is now going to be +k(x3-x2). 551
00:48:55,050 --> 00:49:00,940
So now I go to the next spring, to the next object. 552
00:49:00,940 --> 00:49:05,910
So this object here will experience a force to the left. 553
00:49:05,910 --> 00:49:08,460
Because this spring is longer than it wants to be. 554
00:49:08,460 --> 00:49:13,590
So it wants to contract, but this one is pushing. 555
00:49:13,590 --> 00:49:18,530
So therefore the force due to this spring is now also in this direction. 556
00:49:18,530 --> 00:49:21,290
Because the end is fixed. 557
00:49:21,290 --> 00:49:23,460
So we get for the third object 558
00:49:23,460 --> 00:49:33,190
(see the blackboard) 559
00:49:33,190 --> 00:49:36,010
which is this term to switch a sign. 560
00:49:36,010 --> 00:49:43,080
And then in addition, I get -kx3. 561
00:49:43,080 --> 00:49:49,920
When you reach this point over an exam, you pause to take a deep breath. 562
00:49:49,920 --> 00:49:53,670
And you go over every term and every sign. 563
00:49:53,670 --> 00:49:58,810
If you slip up on one sign, one casual mistake that you just... 564
00:49:58,810 --> 00:50:03,530
Even though you know it, you casually write here, for instance, x1 instead of 3, 565
00:50:03,530 --> 00:50:08,160
it's all over. You're dead in the waters and problem will fall apart 566
00:50:08,160 --> 00:50:11,870
and it may not even oscillate in the simple harmonic way. 567
00:50:11,870 --> 00:50:15,060
So therefore let's look at it. 568
00:50:15,060 --> 00:50:22,240
mx1(double dot) x1 is larger than η, therefore force is in this direction. I love it. 569
00:50:22,240 --> 00:50:26,570
The other one is in this direction, perfect, x2-x1. 570
00:50:26,570 --> 00:50:29,730
That same force here is going to pull on the second one. 571
00:50:29,730 --> 00:50:32,630
So that's is this correct. This is also correct. 572
00:50:32,630 --> 00:50:38,220
This one is driving it away from equilibrium. x3-x2 gotta be right. 573
00:50:38,220 --> 00:50:41,920
This term shows up here with a minus sign can go along there, 574
00:50:41,920 --> 00:50:46,700
and since this spring is always shorter here, if push to the right. 575
00:50:46,700 --> 00:50:50,260
I am happy with my differential equations. 576
00:50:50,260 --> 00:50:55,080
So now we are going to substitute it in here. 577
00:50:55,080 --> 00:51:01,280
x1 is C1cosωt, trial functions. 578
00:51:01,280 --> 00:51:11,160
x2 is C2cosωt, and x3 is C3cosωt. 579
00:51:11,160 --> 00:51:16,010
I'll looking for ωs. Oh, no, oh, no! 580
00:51:16,010 --> 00:51:21,020
ω is given by me, I'm telling you what ω is. 581
00:51:21,020 --> 00:51:23,100
You're not going to negotiate that with me. 582
00:51:23,100 --> 00:51:25,880
We only solving for C1, C2 and C3, 583
00:51:25,880 --> 00:51:30,680
and in the steady state you will be able to do that because ω is nonnegotiable. 584
00:51:30,680 --> 00:51:36,830
You're going to get three equation with three unknowns, C1, C2 and C3. 585
00:51:36,830 --> 00:51:41,990
You don't have to settle, to only calculate the ratios of the amplitude. 586
00:51:41,990 --> 00:51:44,350
No, you're going to get the real answer. 587
00:51:44,350 --> 00:51:50,140
For C1, for C2 and for C3, which of course really depend on ηo. 588
00:51:50,140 --> 00:51:53,150
Sure, if you know ηo. 589
00:51:53,150 --> 00:51:58,080
Do we worry about phase, angles here? No, because there's no damping. 590
00:51:58,080 --> 00:52:02,220
And if there is no damping, either the objects are inphase or they are out of phase. 591
00:52:02,220 --> 00:52:05,100
Because it is damping that gives this phase angle in between. 592
00:52:05,100 --> 00:52:07,940
And 180 degree out of phase is a minus sign. 593
00:52:07,940 --> 00:52:12,270
So we have the power to introduce 180 degree phase changes, 594
00:52:12,270 --> 00:52:19,630
and 0 phase for that we have plus a minus sign. 595
00:52:19,630 --> 00:52:22,920
Now you are going to do some grinding. 596
00:52:22,920 --> 00:52:27,580
I did all the grinding in every detail on the double pendulums. 597
00:52:27,580 --> 00:52:29,860
Now you are going to do the grinding, 598
00:52:29,860 --> 00:52:36,350
however I want to make sure that if you go through that effort to make the grinding. 599
00:52:36,350 --> 00:52:40,930
That you indeed end up with the right solution. 600
00:52:40,930 --> 00:52:45,270
So in that sense I'm going to help you a little bit by giving you the D, PAGE2
| Name | Version | Size | Date | User |
| 06_f.srt | 1 | 87046 | 2/17/06 4:18 AM | OOPSSJTU |
Last Modified 2/22/06 6:43 AM
|
