1. Nov 26, 2004

### physicsss

A physical pendulum consists of an L = 70 cm long, 100 g mass, uniform wooden rod hung from a nail near one end (Fig. 14-38). The motion is damped because of friction in the pivot. The damping force is approximately proportional to d(theta)/dt. The rod is set in oscillation by displacing it 15° from its equilibrium position and releasing it. After 10 seconds, the amplitude of the oscillation has been reduced to 4° . Assume that the angular displacement can be written as

theta= A*e^(-alpha*t)*cos(w'*t).

http://www.webassign.net/gianpse3/14-38alt.gif

(a) Find alpha

(b) Find the approximate period of the motion.

(c) Find how long it takes for the amplitude to be reduced to 1/2 of its original value.

Don't I need b to find alpha, since damping force is proportional to dtheta/dt, or F=b(dtheta/dt)?

2. Nov 26, 2004

### ceptimus

The formula shows that the amplitude changes with time

e^(-alpha*t)

The other bits of the formula are the initial amplitude A, and the oscillating part (the cos term)

You know that the amplitude decreases from 15° to 4° in ten seconds, so you have:

e^(-10 alpha) = 4 / 15

3. Nov 26, 2004

### physicsss

Can someone verify if I have found the period corectly?

formulas:
alpha=b/(2m), b=2*m*a
w'=sqrt( k/m - b^2/(4m^2) )
t=(2*pi)/w'

I found the k/m to be 3/2(g/L), where L is the length of the rod. Then I plugged everything in.

4. Nov 27, 2004