Frequency of a particle performing S.H.M. is 10 Hz . The particle is suspended from a vertical spring. At the highest point of its oscillation the spring is unstretched. Maximum speed of the particle is \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)\)

Time period of oscillation.
\(\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{~m}}{\mathrm{k}}} \\
& \mathrm{n}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{\mathrm{~m}}} \\
& \mathrm{n}^2=\frac{1}{4 \pi^2} \times \frac{\mathrm{k}}{\mathrm{~m}} \\
& \mathrm{k}=100 \times 4 \pi^2 \times \mathrm{m}=400 \pi^2 \mathrm{~m}...(i)
\end{aligned}\)
Amplitude at highest point of oscillation \(=\mathrm{A}\)
\(\begin{aligned}
\mathrm{F} & =\mathrm{kA}=\mathrm{mg} \\
\therefore \quad \mathrm{~A} & =\frac{\mathrm{mg}}{\mathrm{k}}...(ii) \\
\mathrm{v}_{\max } & =\omega \times \mathrm{A} \\
& =\omega \times \frac{\mathrm{mg}}{\mathrm{k}} ...[From(ii)]\\
& =2 \pi \mathrm{n} \times \frac{\mathrm{mg}}{\mathrm{k}} \\
& =\frac{(2 \times \pi \times 10) \times(\mathrm{m} \times 10)}{400 \pi^2 \mathrm{~m}} \\
& =\frac{1}{2 \pi} \mathrm{~m} / \mathrm{s}
\end{aligned}\)
...[From(i)]