Question

A particle performing $SHM$ has time period $\frac{2\pi}{\sqrt3}$ and path length $4\, cm$ . The displacement from mean position at which acceleration is equal to velocity is

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

Answer 1

Answer: (C)

$1 \,cm$

Answer 2

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
$\sqrt{A^{2}-x^{2}}=\sqrt{3} x$
$\Rightarrow A=2 x$
As amplitude $=\frac{\text { path length }}{2}=2 cm$
$\Rightarrow x=1 cm$