- #1
Allan McPherson
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Homework Statement
An oscillator when undamped has a time period T0, while its time period when damped. Suppose after n oscillations the amplitude of the damped oscillator drops to 1/e of its original value (value at t = 0).
(a) Assuming that n is a large number, show that $$\frac{T}{T_0}=\left (1+\frac{1}{4n^2\pi^2}\right) \approx 1 + \frac{1}{8n^2\pi^2}$$
(b) Assuming that n is a large number, show that $$\frac{\omega}{\omega_0} \approx 1 - \frac{1}{8n^2\pi^2}$$
Homework Equations
$$x(t) = A e^{-\gamma t}cos(\omega t + \phi)$$
$$T = \frac{2 \pi}{\omega}$$
$$\omega = \sqrt{\omega_0^2 - \gamma^2}$$
The Attempt at a Solution
[/B]
I've managed to get the equality for both parts. We want $$\frac{A}{e} = A e^{-\gamma t}cos(\omega t + \phi)$$
which is only true if
##\gamma n T = 1,## ##\omega n T = 2n\pi## and ##\phi = 2m\pi##
So $$\omega_0 = \gamma\sqrt{1 + 4n^2\pi^2}$$
And
$$\frac{T}{T_0} = \frac{\omega_0}{\omega}=\frac{\gamma\sqrt{1+4n^2\pi^2}}{2n\pi\gamma}$$ $$=\frac{\sqrt{1+4n^2\pi^2}}{4n^2\pi^2} = \left ( 1 + \frac{1}{4n^2\pi^2} \right )^{1/2}$$
The answer to part (b) will clearly involve taking the inverse of the right-hand side of the equality and applying the same or a similar approximation.
I know I've seen this approximation before, but I can't for the life of me remember how it works. l would appreciate any help that can be offered.