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Homework Help: 2nd order ordinary differential equation for damped harmonic motion

  1. Apr 22, 2007 #1
    1. The problem statement, all variables and given/known data
    the equation of motion for a damped harmonic oscillator is

    d^2x/dt^2 + 2(gamma)dx/dt +[(omega0)^2]x =0


    show that

    x(t) = Ae^(mt) + Be^(pt)


    m= -(gamma) + [(gamma)^2 - (omega0)^2 ]^1/2
    p =-(gamma) - [(gamma)^2 - (omega0)^2 ]^1/2

    If x=x0 and dx/dt =v0 at t=0. show that

    A= v0 - px0
    m - p ​

    B =mx0 - v0
    m - p​

    In the case of very strong damping (i.e gamma >> omega0) show that

    p (approx)= -2(gamma)


    m (approx)= -(omega0)^2

    Hence show that if v0 = 0, the displacement of the oscillator is given approximately by

    x(t) = x0 e^(q)

    where q = (t(omega0)^2) / 2(gamma)

    2. Relevant equations

    3. The attempt at a solution

    Im okay up until the asterixes. Dont understand how you get the approximation for p in the limit gamma>>omega0. It just tends towards 0 as far as i can see. I tried l'hopitals, isolating the dominant term etc, but couldnt get anywhere

    Dont quite get the last bit either, but that might be because i dont get the step before.

    (PS im gonna learn latex soon :biggrin: sorry for all the mess)

    EDIT: Yes, youre right Aleph. thankyou
    Last edited: Apr 22, 2007
  2. jcsd
  3. Apr 22, 2007 #2


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    I think there is typo in your post (or the original question).

    m and p should not be the same. One of them should be

    -(gamma) - [(gamma)^2 - (omega0)^2 ]^1/2
  4. Apr 24, 2007 #3

  5. Apr 24, 2007 #4


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    Sorry, I thought the typo was the reason you couldn't do the question.

    sqrt(gamma^2 + omega0^2) = gamma sqrt(1 + omega0^2/gamma^2)

    gamma >> omega0 so omega0/ gamma is small.

    Use the biominal theorem to approximate the square root.
  6. Apr 24, 2007 #5
    yeh okay. that was obvious :redface:

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