The figure shows a graph between velocity and displacement (from mean position) of a particle performing $SHM\,:$
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by graph we can see that $V_{\max }=10 \mathrm{cm} / \mathrm{s}$ and $A=2.5 \mathrm{cm}$ and we know that $V_{\max }=A \omega$
$\Rightarrow 10=2.5 \omega$
$\Rightarrow \omega=4 s^{-1}$
$T=\frac{2 \pi}{4}=\frac{2 \pi}{4}=\frac{3.14}{2}=1.57 s $
$a_{\max }=A \omega^{2}=2.5 \times 4^{2}=40 \mathrm{cms}^{-2}$
$v(x=1 \mathrm{cm})=w \sqrt{A^{2}-x^{2}}=4 \sqrt{2.5^{2}-1^{2}}=4 \sqrt{5.25}=4 \sqrt{\frac{525}{100}}=4 \sqrt{\frac{21 \times 25}{100}}=$
$2 \sqrt{21} \mathrm{cm} / \mathrm{s} $
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