Question

A particle executes linear $\text{SHM}$ with a time period of $10\text{s}$. Find the time taken by the particle to go directly from its mean position (at $t=0)$ to $\frac{1}{2}$ of its amplitude.

• A. $\frac{5}{6}\text{s}$
• B. $\frac{5\pi }{6}\text{s}$
• C. $1.2\text{s}$
• D. $\frac{5}{3}\text{s}$

Answer 1

The correct option is A $\frac{5}{6}\text{s}$

If the particle starts from the mean position in positive direction, $x=\text{A sin}\omega t$
from given data, $\frac{A}{2}=\text{A sin}\omega t$
or $\frac{1}{2}=\text{sin}\omega t$
$\Rightarrow \omega t=\frac{\pi }{6}\Rightarrow t=\frac{\pi }{6\omega }$
$\Rightarrow t=\frac{T}{12}(\because \omega =\frac{2\pi }{T})$
Given, $T=10\text{s}$
$\therefore t=\frac{10}{12}=\frac{5}{6}\text{s}$