The quality factor

  • Thread starter mbigras
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Homework Statement


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].



Homework 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?



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?
 

Answers and Replies

  • #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.
 
  • #3
haruspex
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Is this a formula for the resonant frequency?
[tex]
\omega_{m} = \omega_{0}\left(1 - \frac{1}{2Q^{2}}\right)^{1/2}
[/tex]
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.
 
  • #4
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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?
 
  • #5
NascentOxygen
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Fractional change would be ( ωm - ω0 ) / ω0

See how much you can simplify that expression.
 
  • #6
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I'm getting
[tex]
\left(1-\frac{1}{2Q^{2}}\right)^{1/2}-1
[/tex]
 
  • #7
haruspex
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I'm getting
[tex]
\left(1-\frac{1}{2Q^{2}}\right)^{1/2}-1
[/tex]
... 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.
 
  • #8
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thank you very kindly for your help. I see what you mean about how my formula needs parenthesis.
 
  • #9
haruspex
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I'm getting
[tex]
\left(1-\frac{1}{2Q^{2}}\right)^{1/2}-1
[/tex]
... 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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