# Time Constant problem

1. Jun 16, 2009

### Liketothink

1. The problem statement, all variables and given/known data
A spring with spring constant 18.39 N/m hangs from the ceiling. A 452 g ball is attached to the spring and allowed to come to rest. It is then pulled down 4.62 cm and released. What is the time constant if the ball’s amplitude has decreased to 2.85 cm after 32 oscillations?

2. Relevant equations
A^2=Ao^2e^-t/2T

3. The attempt at a solution
I solved for time which is t=32*period. I solved for period from spring constant and mass. Now I have -t/(ln(2.85^2/4.62^2)*2)= T but I got the wrong answer. Can someone help me please? Thank you.

2. Jun 16, 2009

### ideasrule

It looks like your math is right. What is the right answer?

3. Jun 16, 2009

### turin

Why is your exponent t/2T? I would expect either t/T or 2t/T, depending on the definition of time constant being used. In particular, I would expect:
$$A(t)=A_0e^{-t/T}$$
to be the relevant equation, where A is amplitude, t is elapsed time, and T is the time constant. Again, this is a matter of definition, but I would be surprised if your relevant equation is correct.

4. Jun 17, 2009

### ideasrule

I suspect that the "original" equation is:
$$E(t)=E_0e^{-t/T}$$
where E is the energy of the system. Since energy is proportional to the square of the maximum displacement, $$A(t)=A_0e^{-t/T}$$

5. Jun 17, 2009

### RTW69

I believe the original equation is A^2=Ao^2e^(-t/T) where t is time and T is the time contant. It looks like your t is correct. You have A and Ao.