Question

When a mass of
m = 304g
is attached to a spring and the mass-spring system is set into oscillatory motion, the period of the motion is T = 0.467
Determine the amplitude (in meters) of the oscillation, if the total energy of the oscillating system is 0.163 J

Answer 1

The amplitude is 0.077 m

Explanation:

The period of a spring-mass system is given by

$T=2\pi \sqrt{\frac{m}{k}}$

where

m is the mass

k is the spring constant

For the system in this problem, we have

m = 304 g = 0.304 kg is the mass

T = 0.467 s is the period

Solving for k, we find the spring constant:

$k=(\frac{2\pi}{T})^2 m=(\frac{2\pi}{0.467})^2(0.304)=55.0 N/m$

The total energy of an oscillating system is given by

$E=\frac{1}{2}kA^2$

where A is the amplitude of the oscillation. In this problem, we know that

k = 55.0 N/m

E = 0.163 J is the total energy

Solving for A, we find the amplitude:

$A=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(0.163)}{55.0}}=0.077 m$

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