The maximum velocity of a body undergoing $S.H.M$. is $0.2\,m/s$ and its acceleration at $0.1\,m$ from the mean position is $0.4\,m/s^2$. The amplitude of the $S.H.M.$ is .... $m$
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$\mathrm{V}_{\max }=\omega \mathrm{A}=0.2 \mathrm{m} / \mathrm{s}$
$a=-\omega^{2} x=0.4 \mathrm{m} / \mathrm{s}^{2}$
$\omega^{2}=4 \quad \Rightarrow \quad \omega=2$
$\omega A=0.2$
$2 \times \mathrm{A}=0.2$
$\mathrm{A}=0.1 \mathrm{m}$
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