Question

A particle performing SHM with amplitude $A$ and time period $T$ starts from mean position towards the positive extreme point as shown in the figure. Find the time taken by the particle (in seconds) to reach $x=\frac{A}{2}.$

• A. $t=\frac{T}{6}$
• B. $t=\frac{T}{8}$
• C. $t=\frac{T}{4}$
• D. $t=\frac{T}{12}$

Answer 1

The correct option is D $t=\frac{T}{12}$

For the particle starting from mean position $x=0$ at $t=0$, we can say that, $\varphi =0$
So, equation of motion can be written as
$x=A\sin (\omega t+0)\Rightarrow x=A\sin \omega t.......\left(1\right)$
Given, $x=+\frac{A}{2}$
$\frac{A}{2}=A\sin \omega t\Rightarrow \sin \omega t=+\frac{1}{2}$
We know, $\omega =\frac{2\pi }{T}$

Substituting this in the above equation, we get
$\sin \left(\frac{2\pi }{T}t\right)=+\frac{1}{2}$
$\Rightarrow \frac{2\pi }{T}t=\frac{\pi }{6}\text{or}\frac{5\pi }{6}......\left(2\right)$
Differentiating equation $\left(1\right)$ with respect to time we get,
$v=A\omega \cos \omega t$
From the data given in the figure, we can deduce that the particle is moving towards the positive extreme position. Hence, the velocity is positive.
$A\omega \cos \omega t>0\Rightarrow \cos \omega t>0\Rightarrow \omega t=\frac{\pi }{6}$
Using this, we can say from $\left(2\right)$ that,
$\frac{2\pi }{T}t=\frac{\pi }{6}\Rightarrow t=\frac{T}{12}\text{s}$
Thus, option (d) is the correct answer.