Approximating Damped Oscillator Time Period and Frequency with Large n

[Allan McPherson]

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 $\phi = 2m\pi$​

which implies $$T = \frac{1}{\gamma n} = \frac{2\pi}{\omega}$$
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.

 

[ehild]

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 $$\frac{T}{T_0}=\left (1+\frac{1}{4n^2\pi^2}\right) \approx 1 + \frac{1}{8n^2\pi^2}$$

You miss a square root:
$$\frac{T}{T_0}=\sqrt{\left (1+\frac{1}{4n^2\pi^2}\right)} \approx 1 + \frac{1}{8n^2\pi^2}$$

If x << 1,
$\sqrt{1+x}=\sqrt{(1+x/2)^2-(x/2)^2 }\approx \sqrt{(1+x/2)^2}=1+x/2$

Also $\frac{1}{1+x}$ is the sum of the geometric series 1-x+x^2-x^3+..., approximately 1-x if x << 1.

 

[haruspex]

$$=\frac{\sqrt{1+4n^2\pi^2}}{4n^2\pi^2} = \left ( 1 + \frac{1}{4n^2\pi^2} \right )^{1/2}$$

You meant
$$=\sqrt{\frac{1+4n^2\pi^2}{4n^2\pi^2}} = \left ( 1 + \frac{1}{4n^2\pi^2} \right )^{1/2}$$

 

[Allan McPherson]

Thank you. That clarifies the approximation for part (a). I'm not sure how that helps with part (b), though.

For part (b) I have $$\frac{\omega}{\omega_0} = \frac{2n\pi}{\sqrt{1+4n^2\pi^2}} = \sqrt{\frac{4n^2\pi^2}{1+4n^2\pi^2}} = \sqrt{1 - \frac{1}{4n^2\pi^2}}$$

 

[ehild]

Thank you. That clarifies the approximation for part (a). I'm not sure how that helps with part (b), though.

For part (b) I have $$\frac{\omega}{\omega_0} = \frac{2n\pi}{\sqrt{1+4n^2\pi^2}} = \sqrt{\frac{4n^2\pi^2}{1+4n^2\pi^2}} = \sqrt{1 - \frac{1}{4n^2\pi^2}}$$

Let be $x=-\frac{1}{4n^2\pi^2}$ . |x|<<1, apply the approxiamation for the square root as in Post #2.

 

[Allan McPherson]

Right. Looking at that again, I'm not sure how I thought that could be right. Thank you for your help.