Question

A particle performs $\text{SHM}$ along a straight line of length $4\times {10}^{-2}\text{m}$. When the velocity and acceleration are numerically equal, the corresonding displacement of the particle if the time period is $\frac{2\pi }{3}\text{sec}$ is:

• A. $2\pi \times {10}^{-2}\text{m}$
• B. $2\times {10}^{-5/2}\text{m}$
• C. $0.5\times {10}^{-2}\text{m}$
• D. $\frac{2\pi }{3}\times {10}^{-2}\text{m}$

Answer 1

The correct option is B $2\times {10}^{-5/2}\text{m}$

Given:
$2A=4\times {10}^{-2}\text{m}\Rightarrow A=2\times {10}^{-2}\text{m}$
$T=\frac{2\pi }{3}\text{sec}$

For $\text{SHM},$

$v=w\sqrt{A^2-y^2};a=-{\omega }^{2}y$

From question,

$a=v$ (numerically)

$\Rightarrow {\omega }^{2}y=w\sqrt{A^2-y^2}$

$\Rightarrow \frac{2\pi }{T}y=(A^2-y^2{)}^{1/2}$

$\Rightarrow \frac{2\pi }{\left(\frac{2\pi }{3}\right)}y=(A^2-y^2{)}^{1/2}$

$\Rightarrow 9y^2=A^2-y^2$

$\Rightarrow 10y^2=A^2=4\times {10}^{-4}$

$\Rightarrow y=\frac{2\times {10}^{-2}}{\sqrt{10}}=2\times {10}^{-5/2}\text{m}$

Hence, option $\left(b\right)$ is the correct answer.