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Damped Driven Harmonic Oscillator

  1. Jun 27, 2012 #1
    Just have a few questions regarding the method of solving the damped-driven harmonic oscillator.

    Once we have rewritten the differential equation in terms of z and it's derivatives, we try a solution [itex]z(t) = Ce^{i \omega t}[/itex]. When we sub in z and it's derivatives we then rewrite the complex constant C as [itex]Ae^{-i \delta}[/itex]. My book says that we can do this for any complex number. Why is this?

    When we solve for A and sub that into [itex]C=Ae^{-i \delta}[/itex] we find
    [itex]f_{0}e^{i \delta}=A(\omega_{0}^{2} - \omega^{2} +2i \beta \omega)[/itex].
    My book says we can then rewrite [itex]\delta[/itex] as [itex]\delta=tan^{-1}\left(\frac{2\beta\omega}{\omega_{0}^{2}-\omega^{2}}\right)[/itex]. Could someone please explain this? The right triangle explanation is confusing me.

    Thank you all.
     
  2. jcsd
  3. Jun 27, 2012 #2
    I asked a question about this on another forum. I never got an answer, but I know the answer involves taking the laplace transform....
     
  4. Jun 27, 2012 #3
    we can write any complex number in the form z = re^ig
    thus they have just written that complex number in a convenient form for them.
    that g (which i am using as my argument for the exponent) is just the angle in the polar representation of a complex number. thus it can be calculated from trigonometry if we know the adjacent and opposite sides of the triangle. this is where the arctan thing comes in.

    Do you know how to draw Argand Diagrams? if so, do that for every complex number you see in the derivation and hopefully your trig training will kick in and help you.

    hope that helps mate.
     
  5. Jun 27, 2012 #4
    Do you know about the trigonometric representation of complex numbers and Euler's formula?
    For you case, it will be:
    [itex]e^{-i \delta}=cos\delta - i sin\delta[/itex]
    You can use this to solve your equation and find δ.
     
  6. Jun 27, 2012 #5
    Awesome that makes sense, I think I'm clear on now. Thank you all for the help!
     
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