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RLC damped oscillator

  1. Feb 6, 2006 #1
    Hello,

    On this page:
    http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/Oscillations.htm
    It says (and shows) that in the case of a critically damped oscillation, it moves more quickly to zero than in the overdamped case.

    I don't understand why. The solution to this circuit is:

    [tex]Q(t) = A e^{(i\omega - \alpha )t}[/tex]

    Where [tex]\omega[/tex] is the square root of some expression that depends on R and L. The critically damped case corresponds to [tex]\omega = 0[/tex], while the overdamped case corresponds to the case in which [tex]\omega[/tex] is imaginary.

    So in the critically damped case the solution is:

    [tex]Q(t) = A e^{-\alpha t}[/tex]

    And in the overdamped case it is:

    [tex]Q(t) = A e^{(-P - \alpha )t}[/tex]

    Where P is some positive number, assuming [tex]\omega = \sqrt{-P^2}[/tex]. To my best understanding, the solution should move to zero quicker in the overdamped case. However, that's not the case, as I've seen on my webpages and read in many books.

    Can someone please explain this?

    Thanks,
    Chen
     
  2. jcsd
  3. Feb 6, 2006 #2

    vanesch

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  4. Feb 6, 2006 #3
    Yes, indeed. Thank you!

    Chen
     
  5. Feb 8, 2006 #4

    SGT

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