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Q.
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
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 $=\frac{\text { path length }}{2}=2 cm$
$\Rightarrow x=1 cm$