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Un-damped Driven Harmonic Oscillator Question

  1. Oct 26, 2008 #1
    1. The problem statement, all variables and given/known data
    An un-damped driven harmonic oscillator satisfies the equation of motion: ma+kx=F(t) where we may write the un-damped angular frequency w-naught^2=k/m. The driving force F(t)=F-naught*sin(wt) is switched on at t=0. Find x(t) for t>0 for initial conditions x=0, v=0,at t=0.


    2. Relevant equations
    I know that this can be written in terms of a complimentary and a particular solution and that the complimentary solution will be in the form x(t)=Asin(w-naught*t-delta) and that I need to consider a particular solution in the form x(t)=Asin(wt) and determine A by plugging x(t) into the differential equation.


    3. The attempt at a solution
    The final answer is given as x(t)= -((F-naught/m)(w/w-naught)/(w-naught^2-w^2)) sin(w-naught*t) + ((F-naught/m)/(w-naught^2-w^2)) sin(wt)
    Ive done similar problems that have worked out but for some reason I can't get this to come out right. It's driving me nuts I've been working on it all weekend and have to turn this work in tomorrow morning.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 26, 2008 #2

    gabbagabbahey

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    What do you get for A when you plug your particular solution into the DE....what does that give you for your general solution?
     
  4. Oct 26, 2008 #3
    ok, i got that part. A will equal (F-naught/m)/(w-naught^2-w^2). I think I was just writing it wrong when I plugged into the DE. Now I just have to work through the complimentary part.
     
  5. Oct 27, 2008 #4
    hmm, I know this should be the easy part but I'm stuck again! I can't seem to figure out how to solve for A in the complimentary part.
     
  6. Oct 27, 2008 #5

    gabbagabbahey

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    You'll have to use the initial condition that you were given: x(0)=x'(0)=0....remember that these condition apply to the total solution, not just the complimentary part.
     
  7. Oct 27, 2008 #6
    thanks, thats whats probably getting me here. appreciate all the help
     
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