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Damped Harmonic Oscillator - Help Please

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data

    A damped harmonic oscillator is displaced a distance xo from equilibrium and released with zero initial velocity. Find the motion in the underdamped, critically damped, and overdamped case.

    2. Relevant equations

    d2x/dt2 + 2K dx/dt + ω2x = 0

    Underdamped: x = C*e-Ktcos(ωt-[itex]\gamma[/itex])

    Overdamped: x = A*e-K-t+B*e-K+t

    Critically Damped: a*e-Kt*(1+bKt)


    3. The attempt at a solution

    I haven't attempted the solution because I'm not sure how to incorporate xo into the equations. I understand that at time t=0, x=xo, but how do I use this fact? Any help would be greatly appreciated. Thank you in advance.
     
  2. jcsd
  3. Oct 11, 2011 #2

    vela

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    Each of the solutions has two arbitrary constants. You need to determine those using the initial conditions. In the critically damped case, for example, when t=0, you have x(0)=a; therefore, you know that a=x0. By using the fact that the oscillator starts at rest, you can similarly solve for b.
     
  4. Oct 12, 2011 #3
    Hmm, is the equation I have for under damped correct then, since there is only one constant?
     
  5. Oct 12, 2011 #4
    I solved the equations of motion for the Under-Damped and Critical-Damped conditions, but I'm having trouble finding the Over-Damped constants.. Any advice?
     
    Last edited: Oct 12, 2011
  6. Oct 12, 2011 #5
    If I write the equation for over-damping using the initial conditions for t=0, I get: Xo=Ae^(-K) + Be^(-K)... which just gives me e^(-K)*(A+B). I'm not sure how to solve for A and B given this.. Any help before class would be greatly appreciated.. again thanks in advance.
     
  7. Oct 12, 2011 #6

    vela

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    Your solution for the overdamped case is incorrect.
     
  8. Oct 12, 2011 #7
    I believe that is just the solution for the real part? I'll have to look for the general solution..
     
  9. Oct 13, 2011 #8

    HallsofIvy

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    Your "underdamped" solution does have two constants, C and [itex]\gamma[/itex]

    Your "overdamped" solution, [itex]x= Ae^{-K-t}+Be^{-K+t}[/itex] should be [itex]x= Ae^{Kt}+ Be^{-Kt}[/itex].
     
  10. Oct 13, 2011 #9

    vela

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    HallsofIvy has uncharacteristically made a mistake regarding the overdamped solution. In the overdamped case, the characteristic equation, [itex]r^2+2Kr+\omega^2=0[/itex], has two real roots
    \begin{align*}
    r_+ &= {-K}+\sqrt{K^2-\omega^2} \\
    r_- &= {-K}-\sqrt{K^2-\omega^2}
    \end{align*}so the solution is
    \begin{align*}
    x(t) &= Ae^{r_+ t}+Be^{r_- t} \\
    &= Ae^{(-K+\sqrt{K^2-\omega^2})t}+Be^{(-K-\sqrt{K^2-\omega^2})t}
    \end{align*}So when t=0, you get x(0) = A+B.
     
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