A 0.10-kg mass stretches a massless spring 0.20 m from its equilibrium position.If this same mass is set into vibration on this spring,the frequency is
A)0.023 Hz
B)1.1 Hz
C)2.0 Hz
D)7.0 Hz
E)13 Hz

Question

Quizplus · Accepted Answer

The frequency of a mass-spring system is given by $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$, where $k$ is the spring constant and $m$ is the mass. First, find $k$ using $F = kx$, where $F = mg$ and $x = 0.20$ m. With $g = 9.8 \, \text{m/s}^2$, $k = \frac{mg}{x} = \frac{0.10 \times 9.8}{0.20} = 4.9 \, \text{N/m}$. Then, $f = \frac{1}{2\pi}\sqrt{\frac{4.9}{0.10}} \approx 1.1 \, \text{Hz}$.

A 010-Kg Mass Stretches a Massless Spring 0