Damped Harmonic Oscillator/ME loss/freq question

[jssamp]

Homework Statement

A damped harmonic oscillator loses 6.0% of it's mechanical energy per cycle. (a) By what percentage does it's frequency differ from the natural frequency f$_{0} = (\frac{1}{2\pi})\sqrt{\frac{k}{m}}$? (b) After how many periods will the amplitude have decreased to $\frac{1}{e}$ of it's original value?

Homework Equations

natural frequency
$f_{0} = (\frac{1}{2\pi})\sqrt{\frac{k}{m}}$

damped frequency
$f' = \frac{1}{2\pi}\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}}}$

displacement for lightly damped harmonic oscillator
$x = Ae^{(\frac{-b}{2m})t}cos\omega't$

Total mechanical energy
$E = \frac{1}{2}kA^{2} = \frac{1}{2}mv^{2}_{max}$

And I know the mean half life, $\frac{2m}{b}$ is the time until oscillations reach 1/e of original.

The Attempt at a Solution

I used the A^2 expression for E and the A decay term, $Ae^{(\frac{-b}{2m})t}$ ,said it loses 6% of E when A^2 = .94A^2 (original) or in other words when $Ae^{(\frac{-b}{2m})t}$ = $\sqrt{0.94}$A
so, $e^{(\frac{-b}{2m})t}$ = $\sqrt{.94}$
$\frac{-b}{2m}t$ = $\frac{1}{2}$ln(.94)
t = $\frac{-m}{b}$ln(.94)
But this is time and I need it to be one cycle so do I plug the period in for t?
T = 1/f or 2∏ ω?

This is where I'm stuck. The answer is (a) -1.21x10^-3 % and (b) 32.3 periods but I don't see how to clear the unknowns with what is given. I could just copy the answer down but I want to know how to solve it. If anybody can give me hint it would be a great result for my first post here.

 

[vela]

Homework Statement

A damped harmonic oscillator loses 6.0% of it's mechanical energy per cycle. (a) By what percentage does it's frequency differ from the natural frequency f$_{0} = (\frac{1}{2\pi})\sqrt{\frac{k}{m}}$? (b) After how many periods will the amplitude have decreased to $\frac{1}{e}$ of it's original value?

Homework Equations

natural frequency
$f_{0} = (\frac{1}{2\pi})\sqrt{\frac{k}{m}}$

damped frequency
$f' = \frac{1}{2\pi}\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}}}$

displacement for lightly damped harmonic oscillator
$x = Ae^{(\frac{-b}{2m})t}cos\omega't$

Total mechanical energy
$E = \frac{1}{2}kA^{2} = \frac{1}{2}mv^{2}_{max}$

And I know the mean half life, $\frac{2m}{b}$ is the time until oscillations reach 1/e of original.

The Attempt at a Solution

I used the A^2 expression for E and the A decay term, $Ae^{(\frac{-b}{2m})t}$ ,said it loses 6% of E when A^2 = .94A^2 (original) or in other words when $Ae^{(\frac{-b}{2m})t}$ = $\sqrt{0.94}$A
so, $e^{(\frac{-b}{2m})t}$ = $\sqrt{.94}$
$\frac{-b}{2m}t$ = $\frac{1}{2}$ln(.94)
t = $\frac{-m}{b}$ln(.94)
But this is time and I need it to be one cycle so do I plug the period in for t?
T = 1/f or 2∏ ω?

Yes, that's essentially what you want to do.

In terms of angular frequencies, you have
$$\omega' = 2\pi f' = \sqrt{\omega_0^2-\frac{b^{2}}{4m^{2}}}$$ You also have
$$-\left(\frac{b}{2m}\right)T' = \log \sqrt{0.94}$$ where T'=1/f' is the period of (damped) oscillation. Use it to eliminate (b/2m) from the first equation. Then you'll be able to solve for ω' in terms of ω₀.

 

[jssamp]

Thanks. I beat my head against it a while longer and suddenly, there it was! This is one problem I won't forget after the test.