Question

A particle executes SHM with a time period of $4\text{s}$. Find the time taken by the particle to go directly from its mean position to half of its amplitude.

• A. $\frac{1}{6}\text{s}$
• B. $\frac{1}{3}\text{s}$
• C. $\frac{1}{2}\text{s}$
• D. $\frac{2}{5}\text{s}$

Answer 1

The correct option is B $\frac{1}{3}\text{s}$

$\because$ Particle is executing SHM, so we can write equation of motion as
$x=A\sin (\omega t+\varphi ).......\left(1\right)$
At $t=0,x=0$ [assuming the particle starts from mean position]
$\therefore \varphi =0$
From the data given in the question, let us suppose the particle is moving towards positive extreme position, So, from equation $\left(1\right)$,
$\frac{A}{2}=A\sin \omega t\Rightarrow \sin \omega t=\frac{1}{2}\Rightarrow \omega t=\frac{\pi }{6}\text{or}\frac{5\pi }{6}$
Differentiating $\left(1\right)$ with respect to time we get,
$v=A\omega \cos (\omega t+\varphi )$
Since the particle is moving towards the positive extreme position, velocity is positive.
$\Rightarrow A\omega \cos \omega t>0\Rightarrow \cos \omega t>0\Rightarrow \omega t=\frac{\pi }{6}$
We know that, $\omega =\frac{2\pi }{T}$
$\Rightarrow \frac{2\pi }{T}t=\frac{\pi }{6}\Rightarrow t=\frac{T}{12}\text{s}$
Given, Time period $\left(T\right)=4\text{s}$
$\therefore t=\frac{1}{3}\text{s}$
Thus, option (b) is the correct answer.