A particle execute $S.H.M.$ along a straight line. The amplitude of oscillation is $2 \,cm$. When displacement of particle from the mean position is $1 \,cm$, the magnitude of its acceleration is equal to magnitude of its velocity. The time period of oscillation is ........
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(c)
$A=2 \,cm =2 \times 10^{-2}$
$a=-\omega^2 x$
and $v=\omega \sqrt{A^2-x^2}$
$\omega^2 \times 1 \times 10^{-2}=\omega \sqrt{(4-1) 10^{-4}}$
$\omega \times 1 \times 10^{-2}=\sqrt{3} \times 10^{-2}$
$\omega=\sqrt{3}$
$T=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{3}}$
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