Relationship between period and time in oscillators

[Vitani11]

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 − (8π²n²)^-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^-βtcos(ω₁t-δ) where ω₁ = sqrt(ω_o²-β²)
A = amplitude
β = decay constant
ω₁ = 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 ω₁ = sqrt(ω_o²-β²) and did an expansion to the second term which gave me that ω₁=ω_o(1-β²/2ω_o²). Ridding of β² for 1/t gives me ω₁=ω_o(1-1/2ω_o²t²). 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?

 

[vela]

You're introducing $n$ in the wrong place. If $1/\beta$ is the time for $n$ periods, you can say $1/\beta = n(2\pi/\omega_1)$.

 

[Vitani11]

Okay, can you explain why that is?

 

[vela]

How are the angular frequency of the damped oscillator $\omega_1$ and the period $T$ related?