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Damped HO and Laplace transform method

  1. Oct 22, 2008 #1

    I am trying to solve the damped harmonic oscillator:

    [itex] \frac{d^2y}{dt^2}+\frac{b}{m} \frac{dy}{dt}+\frac{k}{m}y=0 [/itex]

    and I thought using the Laplace transform might do the trick. Anyway so I did the LT (and inserted the initial conditions that at t=0 y=A, and dy/dt=0) and obtained:

    [itex] Y(p)=\frac{A(p+\frac{b}{m})}{p^2+\frac{b}{m}p+\frac{k}{m} } [/itex]

    I can't seem to get this into a simple form so I can use the lookup tables to do the inverse transform. So I guess I have to somehow use the Bromwhich integral. Obviously the two poles are [itex]p=-\frac{b}{2m} \pm sqrt{ \frac{b^2}{4m^2}-\frac{k}{m} [/itex] and I recognise this from the actual solution so I think I'm on the right track.

    I just don't know how to get the residues and complete the inverse
  2. jcsd
  3. Oct 22, 2008 #2


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    Why go through all that?

    This is a linear, homogeneous d.e. with constant coefficients. It's characteristic equation is r2+ (b/m)r+ k/m= 0. Solve for r using the quadratic formula.

    If the roots are a[itex]\pm[/itex] bi, then the general solution is y(t)= eat(C cos(bt)+ D sin(bt))
    Last edited by a moderator: Oct 23, 2008
  4. Oct 22, 2008 #3
    I know it can be done by simpler means, it's just that I'm currently revising Laplace transforms and thought this would be a good exercise for me....until I got stuck :(
  5. Oct 22, 2008 #4

    Ben Niehoff

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    Using the two roots for p, you can factor the denominator and then decompose into partial fractions.

    It's a lot of algebra, but it should work.

    (Aside: Your tables don't have examples with a quadratic in the denominator?)
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