Question

A particle is executing simple harmonic motion. Its acceleration is given by $\overrightarrow{a}=-4{\pi }^2\overrightarrow{x}m/s^2$, where $\overrightarrow{x}$ is the displacement from the mean position. Find the time period of oscillation.

Answer 1

Acceleration in SHM is given by -${\omega }^{2}x$.
Comparing we get, $-{\omega }^2=-4{\pi }^2\Rightarrow {\omega }^2=4{\pi }^2=(2\pi {)}^2\therefore \omega =2\pi$
So, time period, $T=\frac{2\pi }{\omega }=\frac{2\pi }{2\pi }=1s$