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## Homework Statement

If the amplitude of a weakly damped oscillator decreased to 1/e of its initial value after n periods, show that the frequency of the oscillator must be approximately [1 − (8π

^{2}n

^{2})

^{-1}] times the frequency of an undamped oscillator with the same natural frequency.

## Homework Equations

The equation for a weakly damped harmonic oscillator is x(t) =Ae

^{-βt}cos(ω

_{1}t-δ) where ω

_{1}= sqrt(ω

_{o}

^{2}-β

^{2})

A = amplitude

β = decay constant

ω

_{1}= period for damped oscillator

ω

_{o}= natural frequency

T = period

t = time

δ = phase angle

## The Attempt at a Solution

Since the amplitude was initially Ae

^{-βt}and finally 1/e, I solved for t=1/β or β=1/t. I then took the equation ω

_{1}= sqrt(ω

_{o}

^{2}-β

^{2}) and did an expansion to the second term which gave me that ω

_{1}=ω

_{o}(1-β

^{2}/2ω

_{o}

^{2}). Ridding of β

^{2}for 1/t gives me ω

_{1}=ω

_{o}(1-1/2ω

_{o}

^{2}t

^{2}). I can then replace ω

_{o}with 2πn/T and then I would almost have the correct answer, but for that t= T which doesn't make sense. Also as an aside do I not have to include the phase angle (so I can make it zero) since I'm not measuring the oscillator in reference to another one? Or is it there due to impedance or something?

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