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.
The equation for a weakly damped harmonic oscillator is x(t) =Ae-βtcos(ω1t-δ) where ω1 = sqrt(ωo2-β2)
A = amplitude
β = decay constant
ω1 = period for damped oscillator
ωo = 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(ωo2-β2) and did an expansion to the second term which gave me that ω1=ωo(1-β2/2ωo2). Ridding of β2 for 1/t gives me ω1=ωo(1-1/2ωo2t2). I can then replace ωo 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?