# Damped Oscillator and time

1. Aug 13, 2009

### iamtrojan3

1. The problem statement, all variables and given/known data
A mass M is suspended from a spring and oscillates with a period of 0.880 s. Each complete oscillation results in an amplitude reduction of a factor of 0.96 due to a small velocity dependent frictional effect. Calculate the time it takes for the total energy of the oscillator to decrease to 0.50 of its initial value.

2. Relevant equations

N oscillations=(initial amplitude)x(factor)^N
E=Eo*e^(-t/Tau)
Tau = m/b

3. The attempt at a solution
I really have no idea on how to approach this problem. I need to find tau, which is m/b, but idk what b is. if i have tau, the E on both sides cancel and i'm left with
1/2 = e^-t/tau. t = tau ln (2)
So basicly i need to find tau.

Thanks!

2. Aug 13, 2009

### kuruman

Ignore m/b. What fraction of the initial energy is left after one oscillation?
Ans. 0.962
After two oscillations?
Ans. 0.962*0.962
After n oscillations?
.....

3. Aug 13, 2009

### iamtrojan3

i got 16.97 as n. it works because .96^17 = ~ 0.499

So if thats true, 0.88 which is the period * 16.97 which gives me 14.94 seconds.

This makes sense except the answer's still wrong?

4. Aug 13, 2009

### iamtrojan3

OK i'm retarded. n = 7.47

My friend here said not to do .96squared and woulnd't tell me why. So i blame her.

Thanks again =D

5. Aug 13, 2009

### kuruman

Your friend is correct. I got 7.53 s (close enough). Initially, I assumed linearity where there was none.

I will get you started. Assume that the rate of change of the amplitude is proportional to the amplitude. Call the proportionality constant C. Then

$$\frac{dA}{dt} = - c A$$

Solve this equation for A(t), and use the fact that A(0.88) = 0.96 A0

Once you have A(t) you can find E(t), etc. etc.

This is a good problem. I learned something from it. Thanks.