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Three coupled pendula via springs

  1. May 1, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    See attached link

    3. The attempt at a solution

    I am fine up to the point where I have to express in matrix form. My method was, given the equation for the energy, to put this in matrix form and then recover the equations of motion from there (by differentiating and setting the result to zero since total energy constant) . When I divide by ##m/2## in the matrix I recover some terms, but in particular the middle entry I have a missing factor of 1/2 (reason is that the mass of the middle pendulum has a mass of ##2m## - so I don't understand why the solution says otherwise). Also missing factors of 1/2 on entries (1,2) and (3,2). All I have done is put the given expression into matrix form. Is this a viable method? It seems so. ( I would have posted my matrices but when I tried the latex was very ugly and I haven't found a way to get the matrices to be all together rather than one every line)

    Many thanks.
     

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  3. May 1, 2013 #2
    $$
    E = \dot{x}^T K \dot{x} + x^T M x

    \\

    \Rightarrow

    \\

    0 = \dot{E} = \ddot{x}^T K \dot{x}+ \dot{x}^T K \ddot{x} + \dot{x}^T M x + x^T M \dot{x}

    \\

    = \dot{x}^T K^T \ddot{x}+ \dot{x}^T K \ddot{x} + \dot{x}^T M x + \dot{x}^T M^T x

    \\

    = \dot{x}^T (K^T + K) \ddot{x} + \dot{x}^T (M + M^T) x

    = \dot{x}^T (P \ddot{x} + Q x)

    \\

    \Rightarrow

    \\

    (P \ddot{x} + Q x) = 0

    \Rightarrow

    \\

    \ddot{x} + P^{-1}Q x = 0

    $$ if ## P^{-1} ## exists, which it should in this case. So your method is sound, but you must have made some errors along the way.
     
  4. May 1, 2013 #3
    Correction: P is symmetric by construction, so its inverse always exists, not just in this case.
     
  5. May 1, 2013 #4
    Correction to correction. A symmetric matrix is not necessarily invertible. I do not know what I was thinking :)
     
  6. May 1, 2013 #5

    CAF123

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    Hi voko,

    My matrix for ##M^{-1}## was $$\begin{pmatrix} \frac{2}{m}&0&0\\0&\frac{2}{m}&0\\0&0&\frac{2}{m} \end{pmatrix}$$ and that for ##K## was $$\begin{pmatrix} \frac{mg}{2a} + \frac{k}{2} & -\frac{k}{2}&0\\-\frac{k}{2}&\frac{mg}{a} + \frac{k}{2}&-\frac{k}{2}\\0&-\frac{k}{2}&\frac{mg}{2a} + \frac{k}{2} \end{pmatrix}. $$ When I multiply these together, I get some of the terms but not all of them.
     
  7. May 1, 2013 #6
    Since both M ## M ## and ## M^{-1} ## are diagonal, their corresponding elements should multiply to 1; but that is impossible with your ## M^{-1} ## because its diagonal elements are all equal to each other, which is not the case in ## M ##. And I think the middle term in ## K ## is wrong, too.
     
  8. May 2, 2013 #7

    CAF123

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    Yes, you're right, the matrix I have for M is $$\begin{pmatrix} m/2&0&0\\0&m&0\\0&0&m/2 \end{pmatrix}$$ from which I can easily attain the inverse: $$\begin{pmatrix} 2/m&0&0\\0&1/m&0\\0&0&2/m \end{pmatrix}$$ The entries don't multiply to 1 though - what is this property you mentioned?

    EDIT: I think I understand - i thought initially you meant when you multiply all the elements on a diagonal matrix together you get 1, when in fact you meant (I think) that when you mulitply the entries in M and M inverse together you get 1's on the diagonal. (as expected MM^{-1} = I)

    However, my K remains the same. Shouldn't the middle term be different in K (from the other entries on the diagonal) since it has double the mass of the other two pendula?
     
    Last edited: May 2, 2013
  9. May 2, 2013 #8
    I think we are using K and M in the exactly opposite ways. Anyway, the middle term I think I got yesterday was $$\frac{mg}{a} + k$$
     
  10. May 2, 2013 #9

    CAF123

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    Hmm, with that correction for M I now recover all the terms in A apart from the one in the middle where I get ##\frac{g}{a} + \frac{k}{2m}## instead of ## \frac{g}{a} + \frac{k}{m}##. (This can't be a coincidence can it - pendulum B has double the mass and so it's potential term is simply ##(mg/a) x_1^2##. Add this to the spring term in ##x_1##: ##(k/2) x_1^2## . That's where I think this (1/2)k/m rather than just k/m comes from. Yet the question says differently?

    Actually what you get for the middle term for K gives me what I want, so I must have made an error somehwere, but I can't see it right now.
     
  11. May 2, 2013 #10
    The last two energy terms yield ##kx_2^2##, which results in ##k##, rather than ##k/2## in the middle of the potential energy matrix.
     
  12. May 2, 2013 #11

    CAF123

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    Thanks voko - this was my error.
    Two further questions:
    1) If instead I differentiated the energy first without putting the it into matrix form first, I end up with all sorts of coupled ##\dot{x_i} ## and ##\ddot{x_i}## terms, (by the chain rule) which makes it seem like I wouldn't be able to put it in the desired form if I went with this approach.

    2) In my expansion for energy I get terms like ##kx_1 x_2##. Why can I not simply neglect these terms? For example, it is given that angle theta deviation is small so if ##x_1## and ##x_2## are abritarily small then isn't their product?

    Thank you.
     
  13. May 2, 2013 #12
    See the derivation in #2. If the matrix equation obtained right after differentiation and before simplification gets exapnded, it will result in the exact same mess you are worried about. Without knowning where it came from it will be difficult, but still not impossible, to press it into a sane expression.

    Note that will make your system decoupled. Generally, if you retain qudartic (or higher degree) terms, you should retain all of them, otherwise you approximate your system non-uniformly.
     
  14. May 2, 2013 #13

    CAF123

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    Ok, thanks

    Yes, when I tried this first, I neglected them and then got decoupled matrices so then went back and included these terms (because clearly the system is coupled)
    Thanks again voko.
     
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