Question

A particle executes SHM with a time period of $8\text{s}$. Find the time taken by the particle to go directly from its mean position to half of its amplitude.

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

Answer 1

The correct option is B $\frac{2}{3}\text{s}$

If the particle starts from the mean position in positive direction, then equation of motion can be given by
$x=A\sin \left(\omega t\right)$
By given condition,
$\frac{A}{2}=A\sin \left(\omega t\right)$
$\Rightarrow \omega t={\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{6}$
$\Rightarrow t=\frac{\pi }{6\omega }$
Given,
$T=\frac{2\pi }{\omega }=8\text{s}$
$\Rightarrow \omega =\frac{\pi }{4}\text{rad/s}$
Putting the value of $\omega$
$\therefore t=\frac{\pi }{6}\times \frac{4}{\pi }=\frac{2}{3}\text{s}$

Why this question? Tips: A particle executing SHM covers first $\frac{A}{2}$ distance in $t\text{s}$ and second $\frac{A}{2}$ distance in $2t\text{s}$, when particle start from mean position.