Lightly Damped Simple Harmonic Oscillator Resonant Frequency Formula Explained

[mbigras]

Homework Statement

Show that the fractional change in the resonant frequency (\Delta \omega/ \omega_{0}) of a lightly damped simple harmonic oscillator is ≈ (8Q^{2})^{-1}.

Homework Equations

Is this a formula for the resonant frequency?
<br /> \omega_{m} = \omega_{0}\left(1 - \frac{1}{2Q^{2}}\right)^{1/2}<br />
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?

 

[NascentOxygen]

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 =ω_m 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.

 

[haruspex]

Is this a formula for the resonant frequency?
<br /> \omega_{m} = \omega_{0}\left(1 - \frac{1}{2Q^{2}}\right)^{1/2}<br />

That formula will not lead to the desired answer. Are you sure it isn't \omega_{m} = \omega_{0}\left(1 - \frac{1}{(2Q)^{2}}\right)^{1/2}? That's what seems to be implied by http://en.wikipedia.org/wiki/Harmonic_oscillator.

 

[mbigras]

so the fractional change means: how far away from \omega_{0} is \omega_{m}? How do I apply the taylor series expansion to this question?

 

[NascentOxygen]

Fractional change would be ( ω_m - ω₀ ) / ω₀

See how much you can simplify that expression.

 

[mbigras]

I'm getting
<br /> \left(1-\frac{1}{2Q^{2}}\right)^{1/2}-1<br />

 

[haruspex]

I'm getting
<br /> \left(1-\frac{1}{2Q^{2}}\right)^{1/2}-1<br />

... 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.

 

[mbigras]

thank you very kindly for your help. I see what you mean about how my formula needs parenthesis.

 

[haruspex]

I'm getting
<br /> \left(1-\frac{1}{2Q^{2}}\right)^{1/2}-1<br />

... 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.