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Using Matrices to Solve Systems of Masses and Springs

  1. May 26, 2014 #1
    I am going through Mary Boas' "Mathematical Methods in the Physical Sciences 3rd Ed". I finished the chapter 3 section 12 problem set, but I do not understand how she gets eq. 12.39. These don't seem obviously equal to each other. Here is the equation:

    $$\lambda Tr=Vr$$ Where T is a matrix representing the kinetic energy, and V is a matrix representing the stiffness of the springs (potential energy).

    Any help would be appreciated.

    Thanks,
    Chris Maness
     
  2. jcsd
  3. May 26, 2014 #2

    UltrafastPED

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    Ideal springs transform potential to kinetic energy, and back again.

    When averaged over a full cycle the kinetic & potential energy each equal 1/2 of the total energy.

    The problem description will provide further detail for the details of the expression.
     
  4. May 26, 2014 #3
    Ironically, I only see the equation pop into the text ex nihilo. Would this be an equivalent expression?:

    $$k\left< { r }|{ V }|{ r } \right> =m\left< { \dot { r } }|{ T }|{ \dot { r } } \right> \quad if\quad r={ e }^{ i\omega t } $$

    This seems to be equivalent to the original equation in my mind. I used the BRA/KET just to reflect the transpose since it seems to be an inner product with a the matrices sandwiched in between. The equation above does not seem to work. That is why I am scratching my head.

    Thanks,
    Chris Maness
     
    Last edited: May 26, 2014
  5. May 27, 2014 #4

    AlephZero

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    Your equation isn't quite right. Using matrix notation (I'm an engineer not a physicist!) the potential energy is ##\frac 1 2 r^T V r## and the kinetic energy is ##\frac 1 2 \dot r^T T \dot r##. The T matrix includes the mass so you don't need another ##m## term.

    If you assume harmonic motion ##r = e^{i\omega t}##, then ##\dot r = i\omega r## so ## T = -\omega^2 \frac 1 2 r^T T r##, or ## T = -\lambda \frac 1 2 r^T T r##.

    By conservation of energy, ##-\lambda \frac 1 2 r^T T r + \frac 1 2 r^T V r## is constant.

    If you consider this as a function of ##r##, the physical solutions are when the energy does not change for small perturbations of ##r##, i.e. when ##\frac {\partial}{\partial r}(-\lambda \frac 1 2 r^T T r + \frac 1 2 r^T V r) = 0## which gives your equation ##\lambda Tr = Vr##.

    You probably want a longer explanation of the last step of taking ##\frac {\partial}{\partial r}##, but you should be able to find that in a textbook. I don't have Boas so I can't give you a page reference.
     
  6. May 27, 2014 #5
    Ah, I now see where I get my omega squared. There are two r dots there.

    Chris
     
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