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Amplitude of harmonic oscillator

  1. Jan 30, 2010 #1
    1. The problem statement, all variables and given/known data
    I have a simple harmonic oscillator system with the driving force a sinusoidal term. The question is to find the general solution and the amplitude of the steady state solution

    2. Relevant equations

    I found the steady state part of the solution. It is of the form:

    3. The attempt at a solution

    I'm told that the amplitude of the system in steady state should be 1/sqrt(f(w)) but I can't derive it. Can anyone help ?

  2. jcsd
  3. Jan 30, 2010 #2
    use trig identity and reduce the equation to sin(A-B), amplitude is the factor by which the stated term is raised
  4. Jan 30, 2010 #3
    But the coefficients of each sinusoidal term are different........
  5. Feb 1, 2010 #4
    is your steady state solution correct? it gets crazy if i reduce it to some trig identity
    and what is f(w) equal to?
  6. Feb 1, 2010 #5
    Yes I'm sure it's correct. This is homework so I'm loathed to post the whole details. However, take a look here:
    Look at the section "Driven Harmonic Oscillator" on page 4. The solution given is exactly what I want (with lambda, w_0 and F_0/m all equal to 1). I just can't derive it from my own solution. By hand I just used a trial solution of p.cos(wt) + q.sin(wt) and equated coefficients, which lead to the form I posted above. I also used Maple to check this, and it agrees.
  7. Feb 1, 2010 #6


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    The (1-w^2) factor isn't under a square root?
  8. Feb 1, 2010 #7
    no, it's not......
  9. Feb 1, 2010 #8
    Another look at this in Maple:
    [Sorry, fiddling with latex.........]
    Last edited: Feb 1, 2010
  10. Feb 1, 2010 #9
    In Maple:
    [tex]{\it s2s}\, := \,{\frac { \left( -{\omega_{{0}}}^{2}+{\omega}^{2} \right) \sin \left( \omega\,t \right) +\cos \left( \omega\,t \right) \omega\,\lambda}{{\omega}^{4}+ \left( {\lambda}^{2}-2\,{\omega_{{0}}}^{2} \right) {\omega}^{2}\\

    which simplifies to:

    [tex]{\frac {\cos \left( \omega\,t-\arctan \left( -{\omega_{{0}}}^{2}+{\omega}^{2},\omega\,\lambda \right) \right) }{ \sqrt{{\lambda}^{2}{\omega}^{2}+{\omega_{{0}}}^{4}-2\,{\omega_{{0}}}^{2}{\omega}^{2}\\

    And that's what I'm trying to achieve by hand....

    Maple code is

  11. Feb 1, 2010 #10


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    Oh, that helps. Note that the denominator is [itex](\omega^2-\omega_0^2)^2+(\lambda\omega)^2[/itex]. Consider a right triangle with one leg of length [itex]\omega^2-\omega_0^2[/itex] and the other of length [itex]\lambda\omega[/itex]. Does that help?
  12. Feb 1, 2010 #11
    Indeed, yes. So easy now. Thanks !
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