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The quality factor

  1. Oct 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that the fractional change in the resonant frequency ([itex]\Delta \omega/ \omega_{0}[/itex]) of a lightly damped simple harmonic oscillator is ≈ [itex](8Q^{2})^{-1}[/itex].



    2. Relevant equations
    Is this a formula for the resonant frequency?
    [tex]
    \omega_{m} = \omega_{0}\left(1 - \frac{1}{2Q^{2}}\right)^{1/2}
    [/tex]
    How would I use this formula?



    3. The attempt at a solution
    Right now I'm in a place where I'm not sure what this question is asking. I'm trying to move to a situation where I understand what is being asked. What does fractional change in the resonant frequency mean? What parameter is changing that causes the resonant frequency to change?
     
  2. jcsd
  3. Oct 11, 2013 #2

    NascentOxygen

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    When Q is infinite, the resonant frequency =ω0

    At any practical value of Q, the resonant frequency changes from this a little, and becomes
    resonant frequency =ωm as given in that equation.

    I'm speculating that they want you to use the first term of a Taylor Series expansion to approx a power of ½ by something to a power of 1.
     
  4. Oct 11, 2013 #3

    haruspex

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    That formula will not lead to the desired answer. Are you sure it isn't [itex]\omega_{m} = \omega_{0}\left(1 - \frac{1}{(2Q)^{2}}\right)^{1/2}[/itex]? That's what seems to be implied by http://en.wikipedia.org/wiki/Harmonic_oscillator.
     
  5. Oct 11, 2013 #4
    so the fractional change means: how far away from [itex]\omega_{0}[/itex] is [itex]\omega_{m}[/itex]? How do I apply the taylor series expansion to this question?
     
  6. Oct 11, 2013 #5

    NascentOxygen

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    Fractional change would be ( ωm - ω0 ) / ω0

    See how much you can simplify that expression.
     
  7. Oct 11, 2013 #6
    I'm getting
    [tex]
    \left(1-\frac{1}{2Q^{2}}\right)^{1/2}-1
    [/tex]
     
  8. Oct 11, 2013 #7

    haruspex

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    ... where Q is large, right? So expand the square root expression according to binomial / Taylor and take the first few terms as an approximation. You'll have to figure out how many terms to take.
     
  9. Oct 11, 2013 #8
    thank you very kindly for your help. I see what you mean about how my formula needs parenthesis.
     
  10. Oct 11, 2013 #9

    haruspex

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    ... where Q is large, right? So expand the square root expression according to binomial / Taylor and take the first few terms as an approximation. You'll have to figure out how many terms to take.
     
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