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Coupled Oscillation Questions

  1. Feb 12, 2006 #1
    Hi guys, I'm stuck on a problem that states:

    Two equal masses oscillate in the vertical direction. Show that the frequences of the normal modes of oscillation are given by:

    [tex]\omega^2 = (3 +- \sqrt{5})\frac{s}{2m} [/tex]

    and that in the slower mode the ratio of the amplitude of the upper mass to that of the lower mass is [tex] \frac{1}{2}(\sqrt{5}-1) [/tex] whilst in the faster mode this ratio is [tex] \frac{-1}{2}(\sqrt{5}+1) [/tex]. The figure in the problem is basically:

    ______
    s
    s
    s
    M
    s
    s
    s
    M

    Where s is the spring and M are the masses (both springs have equal stiffness s).


    Basically, I'm not sure if I have the equations of motion down correctly. So far, I have:

    1) [tex]\frac{md^2y_{1}}{dt^2} = -sy_{1} + s(y_{2} - y_{1}) => y_{1}'' = -\omega_{0}^2y_{1} + \omega_{0}^2(y_{2} - y_{1}) [/tex]

    2) [tex]\frac{md^2y_{2}}{dt^2} = -s(y_{2} - y_{1}) => y_{2}'' = -\omega_{0}^2(y_{2} - y_{1})[/tex]

    Where [tex]y_{1}[/tex] and [tex]y_{2}[/tex] are the displacements of the first and second mass, respectively.
    However when I use the solutions [tex] y_{1} = A_{1}\cos{wt}[/tex] and [tex]y_{2} = A_{2}\cos{wt} [/tex], find the derivatives, plug back in, etc, I cannot cleanly solve for the normal modes in terms of [tex]\omega[/tex]. I'm suspecting my equations of motion are incorrect, help?

    Thanks
     
    Last edited: Feb 12, 2006
  2. jcsd
  3. Feb 12, 2006 #2

    jamesrc

    User Avatar
    Science Advisor
    Gold Member

    Can you show some of your work so that we can see where you might be getting hung up? Your equations of motion appear fine.
     
  4. Feb 12, 2006 #3
    1st equation: [tex] -A_{1}w^2\cos{wt} = -w_{0}^2A_{1}\cos{wt} + w_{0}^2(A_{2}\cos{wt} - A_{1}\cos{wt}) [/tex]

    cosines factor out, collect like terms, etc....then I get:

    [tex] A_{1}(2w_{0}^2 - w^2) + A_{2}(-w_{0} = 0 [/tex]

    2nd equation: [tex] -A_{2}w^2\cos{wt} = -w_{0}^2(A_{2}\cos{wt} - A_{1}\cos{wt}) [/tex]

    cosines factor out, collect like terms, etc....then I get:

    [tex] A_{1}(-w_{0}^2) + A_{2}(w_{0}^2 - w^2) = 0 [/tex]

    So finally I end up with a system of equations:

    [tex] A_{1}(2w_{0}^2 - w^2) + A_{2}(-w_{0}) = 0 [/tex]

    [tex] A_{1}(-w_{0}^2) + A_{2}(w_{0}^2 - w^2) = 0 [/tex]

    To tackle this, I set the determinant of the matrix equal to zero:

    [tex] (2w_{0}^2 - w^2)(w_{0}^2 - w^2) - (w_{0}^2)^2 = 0 [/tex]

    From this, I can't isolate [tex] w^2 [/tex] to get the answer.

    Thanks
     
    Last edited: Feb 12, 2006
  5. Feb 12, 2006 #4

    jamesrc

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    Science Advisor
    Gold Member

    Expand your last equation and collect like terms. The equation will then be in the form: [tex]a\omega^4+b\omega^2+c=0[/tex] where a, b, and c are constants. You can solve for [tex]\omega^2[/tex] using the quadratic equation.
     
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