Damped Harmonic Motion Time Constant?

[m84uily]

Homework Statement

A spring with spring constant 17.0 N/m hangs from the ceiling. A 530 g ball is attached to the spring and allowed to come to rest. It is then pulled down 7.00 cm and released.

What is the time constant if the ball's amplitude has decreased to 3.50 cm after 41.0 oscillations?

k=17.0 N/m
m = 0.53kg
Ai = 0.07m
Af = 0.035
oscillations = 41

Homework Equations

T = period = s / oscillation
T = 2pi (m/k)^(1/2)
Us = (1/2)k(A)^2
Us(damped) = (1/2)k ((A)e^(-bt / 2m))^2

The Attempt at a Solution

Ei = (1/2)k(Ai^2) = (1/2)(17)(0.07^2) = 0.04165
Ef = (1/2)k(Af^2) = (1/2)(17)(0.035^2) = 0.0104125

Ef / Ei = 1/4


(1/2)k(A^2) = (1/2)k (A e^(-bt / 2m))^2 = (1/8)k(A^2)

(1/2)k (A e^(-bt / 2m))^2 = (1/8)k(A^2)

(1/2)e^(-bt / m) = 1/8

e^(-bt / m) = 1/4

-bt / m = ln(1/4)

b = - m ln(1/4) / t

2pi (m/k)^(1/2) = t / oscillation

t = 2pi (m/k)^(1/2 (oscillation)

t = 7.239

b = - (.53) ln(1/4) / (7.239) = 0.101

 

[anooop]

calculate time equivalent to 41 oscillation t=41*(2*pi)/(w1)
where w1 is damped oscillation freq w1^2=w0^2-(b/2m)^2

then use amplitude equation A=a*exp(-b/2m*t)

 

[m84uily]

Thanks!