Question

A spring with spring constant 11.0 N/m hangs from the ceiling. A 540 g ball is attached to the spring and allowed to come to rest. It is then pulled down 7.80 cm and released.

Part A) What is the time constant if the ball's amplitude has decreased to 2.30 cm after 47.0 oscillations?

Answer 1

The natural oscillation frequency of the system is

f = 1/ (2pi) * sqrt (11/0.54) = 0.718 Hz

47 oscillations take

t = 47/0.718 = 65.45 sec

A model to be used for the oscillations is given by

y(t) = A exp (-t/tau) * cos (wt)

from the initial conditions, the equation above is re-written as

y(t) = 6.7*exp(-t/tau)*cos(wt)

The initial amplitude is 7.8 cm. After 65.45 secs, the amplitude becomes 2.3, thus the time constant tau can be calculated as

tau = 65.45 / ln(2.3/7.8) = 53.59 secs