A 2.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.The displacement is given by the expression x(t) = 12.0 cm sin($\omega$ t).What is the period of the simple harmonic motion?
A) 4.75 sec
B) 4.27 sec
C) 3.95 sec
D) 3.36 sec
E) 2.96 sec

Question

Quizplus · Accepted Answer

The angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m}}$. Here, $k = 9.00 \, \text{N/m}$ and $m = 2.00 \, \text{kg}$. So, $\omega = \sqrt{\frac{9.00}{2.00}} = \sqrt{4.5} \approx 2.12 \, \text{rad/s}$. The period $T$ is given by $T = \frac{2\pi}{\omega}$. Thus, $T \approx \frac{2\pi}{2.12} \approx 2.96 \, \text{sec}$.

A 200 Kg Mass Is Connected to a Spring with a a Spring