Question

The equation of motion of a particle executing simple harmonic motion is $a+16{\pi }^{2}x=0$. In this equation, $a$ is linear acceleration in ${\text{m/s}}^2$ of the particle at displacement $x$ in $\text{m}$. The time period of this simple harmonic motion is

• A. $\frac{1}{4}\text{s}$
• B. $\frac{1}{2}\text{s}$
• C. $\frac{1}{3}\text{s}$
• D. $\frac{1}{5}\text{s}$

Answer 1

The correct option is B $\frac{1}{2}\text{s}$

Given:
$a+16{\pi }^{2}x=0$
$\Rightarrow a=-16{\pi }^{2}x$
On comparing it with the acceleration of $\text{SHM}$,
$a=-{\omega }^{2}x$
We get, ${\omega }^2=16{\pi }^2$
$\Rightarrow \omega =4\pi$
We know, time period of $\text{SHM}$,
$T=\frac{2\pi }{\omega }$
$\Rightarrow T=\frac{2\pi }{4\pi }=\frac{1}{2}\text{s}$