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Math Help - Mass on a spring problem

  1. #1
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    Mass on a spring problem


    Can someone explain how to solve this? Cheers.
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  2. #2
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    You will need to solve the homogeneous problem
     \frac{d^2x}{dt^2} + w^2x = 0
    Then you will find a particular solution you can do this using the method of undetermined coefficients. Then you add the two solutions together.
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  3. #3
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    Can you explain how to find the C.F.? Apparently its x=cos(wt)+sin(wt) ? I don't know how they got that answer though.
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  4. #4
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    Okay, I will go through the particular solution first. I will let x_p denote the particular solution. Since the nonhomogeneous part is a trigonometric function we will let the particular solution be of the form x_p = A\sin \gamma t + B\cos \gamma t. Now we must determine the coefficients A,B. To do this we will plug x_p back into the original DEQ.
    x''_p = -\gamma^2 A \sin \gamma t - \gamma^2 B \cos \gamma t
    and so
    -\gamma^2 A \sin \gamma t - \gamma^2 B \cos \gamma t + w^2 \cdot \left( A\sin \gamma t + B\cos \gamma t \right) = \cos \gamma t
    \left(w^2 -\gamma^2\right) A \sin \gamma t + \left(w^2 - \gamma^2\right) B \cos \gamma t = \cos \gamma t.
    Thus we have
     \left(w^2 - \gamma^2\right)A \sin \gamma t = 0
    and
     \left(w^2 - \gamma^2\right)B \cos\gamma t = \cos \gamma t.
    Therefore, we have
     \left(w^2 - \gamma^2\right)A = 0
    and
     \left(w^2 - \gamma^2\right)B  = 1,
    so
     A = 0
     B = \frac{1}{w^2 - \gamma^2}.
    Finally, we see that our particular solution is
    x_p = \frac{1}{w^2 - \gamma^2} \cos \gamma t.

    Deleted mistake see following post.
    Last edited by lvleph; December 5th 2009 at 08:16 AM.
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  5. #5
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    Sorry I made a mistake. The homogeneous solution is wrong. The solution of the characteristic equation is not \lambda = \pm w. The solution is \lambda = \pm w i.
    Therefore, the homogeneous solution is of the form
     x_h = C_1\cos w t + C_2 \sin w t.
    Finally, the general solution should be
     x(t) = C_1\cos w t + C_2 \sin w t + \frac{1}{w^2 - \gamma^2} \cos \gamma t.
    To determine the constants C_1,C_2 we must have initial conditions.
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