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 $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$

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Aicharya · Accepted Answer

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

Frequency of a particle performing S.H.M. is 10 Hz . The particle is suspended from a vertical spring. At…

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