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Small Osc. Pendulum+Springs

  1. Jul 16, 2009 #1


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

    Three pendulums hand side-by-side and have there masses connected horizontally via springs. All lengths and masses are equal. Find the Lagrangian and put it in terms of "natural units".

    3. The attempt at a solution

    [tex]T = 1/2 m l^2 (\dot{\theta_1}^2 + \dot{\theta_2}^2 + \dot{\theta_3}^2)[/tex]

    [tex]V = 1/2 k (l \theta_1 - l \theta_2)^2 + 1/2 k (l \theta_2 - l \theta_3)^2[/tex]

    Using natural units [itex]q_i = \sqrt{k} x_i[/itex] and [itex]q' = dq/d\tau[/itex] where [itex]\tau=\omega t[/itex]:

    [tex]L = 1/2(q^{'2}_{1} + q^{'2}_{2} + q^{'2}_{3}) - 1/2(q^2_1 + q^2_2 + q^2_3) + (q_1 q_2 + q_2 q_3 - 1/2 q^2_2)[/tex]

    When I try to find the normal modes I only get 3, but since there are three degrees of freedom shouldn't there be three normal modes? I think my T or V expressions must be wrong somewhere. If they are not I can write out the matrices...

    When there is no (1,3) or (3,1) entry in either the T or V matrix, i.e. this case (using natural units), will there ever be 3 modes when solving characteristic eq. for that matrix?
    Last edited: Jul 16, 2009
  2. jcsd
  3. Jul 16, 2009 #2
    Do you mean you only found one normal mode with [tex]\omega[/tex] = 3?

    Your 3x3 matrix should have 3 eigenvalues and hence 3 normal modes, right?
  4. Jul 16, 2009 #3


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    No I mean I get [itex]\omega^2_1 = 0,~\omega^2_2 = 3/m l^2, ~\omega^2_3 = 1/ m l^2[/itex]


    [tex]-.375 m l^2 \omega+.500 m^2 l^4 \omega^2-.125 m^3 l^6 \omega^3=0[/tex]

    I don't really get the idea of the natural units. Is it just convenience, i.e. solve the eigenvalue problem and then convert back?
    Last edited: Jul 16, 2009
  5. Jul 16, 2009 #4
    To be honest I've already finished this topic but I've not heard of "natural units", but if you're solving for normal modes then what you've done looks correct...you have already found the 3 normal frequencies.
  6. Jul 17, 2009 #5


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