Question

A particle executes $\text{SHM}$ of amplitude $25\text{cm}$ and time period $3\text{s}$. What is the minimum time required for the particle to move between two points located at $12.5\text{cm}$ on either side of the mean position.

• A. $0.8\text{s}$
• B. $0.5\text{s}$
• C. $0.3\text{s}$
• D. $1.0\text{s}$

Answer 1

The correct option is B $0.5\text{s}$


Let $C$ and $D$ be the two extreme position of the particle, with $O$ as mean position.

Displacements to the right of $O$ are taken as positive while those to the left of $O$ are taken as negative.

The displacement at any instant of a particle executing $\text{SHM}$ is given by

$x\left(t\right)=A\sin (\omega t+\varphi )$

Here, $x\left(t\right)=25\sin (\frac{2\pi }{3}t+\varphi )$

Let's suppose that at $t=0$, the particle is at point $\text{D}$.

So, $x\left(0\right)=25=25\sin \varphi$

$\Rightarrow \varphi =\frac{\pi }{2}$

$\therefore x\left(t\right)=25\sin (\frac{2\pi }{3}t+\frac{\pi }{2})=25\cos (\frac{2\pi }{3}t)$

Now, assume that particle reaches at $\text{A}$ at $t=t_1$ and at $\text{B}$ at $t=t_2$.

$\Rightarrow x\left(t_1\right)=-12.5\text{cm}$

And, $x\left(t_2\right)=+12.5\text{cm}$

$\therefore -12.5=25\cos \frac{2\pi }{3}{t}_1$

$\Rightarrow \cos \frac{2\pi }{3}{t}_1=-0.5$

And, $+12.5=25\cos \frac{2\pi }{3}{t}_2$

And, $\cos \frac{2\pi }{3}{t}_2=0.5$

$\therefore \frac{2\pi }{3}{t}_1=\frac{2\pi }{3}$ and $\frac{2\pi }{3}{t}_2=\frac{\pi }{3}$

$\Rightarrow t_1=1\text{s}$ and $t_2=0.5\text{s}$

$\therefore (t_1-t_2{)}_{min}=(1-0.5)=0.5\text{s}$

Hence, option $\left(B\right)$ is correct.

Alternate solution: