Question

A pendulum executes SHM. The acceleration of the bob of pendulum is ${\pi }^2{\text{m/s}}^2$ at a distance of $2\text{m}$ from the mean position. Find the time period of oscillation.

• A. $2\text{seconds}$
• B. $\sqrt{2}\text{seconds}$
• C. $2\sqrt{2}\text{seconds}$
• D. $\sqrt{2}{\pi }^2\text{seconds}$

Answer 1

The correct option is C $2\sqrt{2}\text{seconds}$

Given,
Acceleration of the bob of pendulum $\left(a\right)={\pi }^2{\text{m/s}}^2$
Position of pendulum from mean position $\left(x\right)=2\text{m}$
Since, the particle executes SHM,
$\left|a\right|={\omega }^{2}x$
${\omega }^2=\left|\frac{a}{x}\right|$
$\Rightarrow \omega =\sqrt{\left|\frac{a}{x}\right|}=\sqrt{\frac{{\pi }^2}{2}}=\frac{\pi }{\sqrt{2}}\text{rad/s}$
$\therefore$ Time period $\left(T\right)=\frac{2\pi }{\omega }=2\sqrt{2}\text{seconds}$.