Calculating Time Constant for Oscillating Spring: Step-by-Step Guide

[Liketothink]

Homework Statement

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?

Homework Equations

A^2=Ao^2e^-t/2T

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.

 

[ideasrule]

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

 

[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.

 

[ideasrule]

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.

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, $$A(t)=A_0e^{-t/T}$$

 

[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.