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A particle performing linear S.H.M. of amplitude $0.1 \mathrm{~m}$ has displacement $0.02 \mathrm{~m}$ and acceleration $0.5 \mathrm{~m} / \mathrm{s}^2$. The maximum velocity of the particle in $\mathrm{m} / \mathrm{s}$ is
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The correct answer is:
0.50
Acceleration $\mathrm{a}=\omega^2 \mathrm{x}$
$$
\begin{aligned}
& \therefore \omega^2=\frac{\mathrm{a}}{\mathrm{x}}=\frac{0.5}{0.02}=25 \\
& \therefore \omega=5 \mathrm{rad} / \mathrm{s} \\
& \mathrm{V}_{\max }=\mathrm{A} \omega=0.1 \times 5=0.5 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \omega^2=\frac{\mathrm{a}}{\mathrm{x}}=\frac{0.5}{0.02}=25 \\
& \therefore \omega=5 \mathrm{rad} / \mathrm{s} \\
& \mathrm{V}_{\max }=\mathrm{A} \omega=0.1 \times 5=0.5 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
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