Posted by **Michael** on Friday, November 19, 2010 at 10:02pm.

A spring with spring constant 14.0 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to rest. It is then pulled down 8.00 cm and released.

What is the time constant if the ball's amplitude has decreased to 2.90 cm after 31.0 oscillations?

- Physics -
**drwls**, Saturday, November 20, 2010 at 1:17am
"Time constant" usually refers to nonoscillatory motion, with an exponential decay. Do you want the "damping constant" that relates the friction force to velocity? Or the constant term that describes the exponential decay of the amplitude of the oscillations?

The solution to the equation of motion is of the form

X = 8 exp(-t/T) cos wt

w = sqrt(k/m) = 5.34 rad/s

(Actually, the damping reduces the frequency of oscillation slightly)

Period = 2 pi/w = 1.18 s

31 oscillations require 36 seconds.

exp(-36/T) = 2.9/8.0 = 0.363

T = 35 seconds is an effective time constant for decay.

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