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Math Help - Damped spring - general solution of IVP

  1. #1
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    Damped spring - general solution of IVP

    Hi,

    I'm working on a 2nd order DE which describes the motion of a damped spring

    mx'' + bx' + kx = 0

    I have found the general solution:

    x(t) = A cos(\omega t + \phi)e^{-\alpha t}

    Where:

    \alpha = b/{2 m} and \omega=\sqrt{k/m - b^2/{4 m^2}}

    which I believe is correct. I now have to find the general solution of A and \phi using x(0) = x0 and x'(0) = v0. I'm not really sure where to go with this, so could someone point me in the right direction please?

    Thanks!
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  2. #2
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    Quote Originally Posted by CorruptioN View Post
    Hi,

    I'm working on a 2nd order DE which describes the motion of a damped spring

    mx'' + bx' + kx = 0

    I have found the general solution:

    x(t) = A cos(\omega t + \phi)e^{-\alpha t}

    Where:

    \alpha = b/{2 m} and \omega=\sqrt{k/m - b^2/{4 m^2}}

    which I believe is correct. I now have to find the general solution of A and \phi using x(0) = x0 and x'(0) = v0. I'm not really sure where to go with this, so could someone point me in the right direction please?

    Thanks!
    Okay, what you have done looks good. Now put in the given values:
    x(0)= A cos(\phi)= x_0

    From x(t)= A cos(\omega t+ \phi)e^{-\alpha t}, x'(t)= -\omega Asin(\omega t+ \phi)e^{-\alpha t}- \alpha A cos(\omega t+ \phi)e^{-\alpha t}
    so that x'(0)= \omega A sin(\phi)- \alpha A cos(\phi)= v_0.

    Solve the two equations x(0)= A cos(\phi)= x_0 and x'(0)= \omega A sin(\phi)- \alpha A cos(\phi)= v_0 for A and \phi in terms of \omega, \alpha, x_0, and v_0.

    You can eliminate A, getting an equation for \phi only, by dividing one equation by the other.
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  3. #3
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    Thanks for the reply! That's kind of where I got to, except I found v_0 = \frac{ \omega A sin(\phi)- \alpha A cos(\phi)}{\alpha^2 + \omega^2}. Am I missing something?
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