Question

An undamped 2.89-kg horizontal spring oscillator has a spring constant of 24.3 n/m. while oscillating, it is found to have a speed of 3.89 m/s as it passes through its equilibrium position. what is its amplitude of oscillation?

Answer 1

The speed of the system when it passes through the equilibrium position is $v=3.89 m/s$, and this corresponds to the maximum speed of the oscillator.

The angular frequency of the oscillator is given by
$\omega= \sqrt{ \frac{k}{m} }$
where k is the spring constant, and m the mass. Substituting the data, we find
$\omega= \sqrt{ \frac{24.3 N/m}{2.89 kg} }=2.9 s^{-1}$

and since the amplitude of the oscillation, A, is related to the maximum speed by
$v_{max}=\omega A$
we can re-arrange this equation to calculate A:
$A= \frac{v_{max}}{\omega}= \frac{3.89 m/s}{2.9 s^{-1}}=1.34 m$