Question

A particle performing $SHM$ has time period $\frac{2\pi }{\sqrt{3}}$ and path length $4cm$ . The displacement from mean position at which acceleration is equal to velocity is

• A. $0cm$
• B. $0.5cm$
• C. $1cm$
• D. $1.5cm$

Answer 1

Answer: (C)

Velocity $v=\omega \sqrt{A^2-x^2}$
and acceleration $={\omega }^{2}x$
Given, $\omega \sqrt{A^2-x^2}={\omega }^{2}x$
or $\sqrt{A^2-x^2}=\omega x$...(i)
Given, $T=\frac{2\pi }{\sqrt{3}}$
and $\omega =\frac{2\pi }{T}=\sqrt{3}$
Substituting the value of $\omega$ in $Eq$ (i), we get
$\sqrt{A^2-x^2}=\sqrt{3}x$
$\Rightarrow A=2x$
As amplitude $=\frac{\text{ path length }}{2}=2cm$
$\Rightarrow x=1cm$