The maximum velocity of a particle, executing S.H.M. with an amplitude 7 mm is $4.4 \mathrm{~ms}^{-1}$. The period of oscillation is $\left[\pi=\frac{22}{7}\right]$.

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Aicharya · Accepted Answer

$\begin{array}{ll} & \mathrm{V}_{\max }=\mathrm{A} \omega \ 	herefore \quad & 4.4=7 	imes 10^{-3}\left(\frac{2 \pi}{\mathrm{~T}}ight) \ 	herefore \quad & \mathrm{T}=\frac{7 	imes 10^{-3} 	imes 2 	imes 22}{4.4 	imes 7}=0.01 \mathrm{~s} \end{array}$ $\begin{aligned} & \frac{mv^2}{r}= \frac{G M m}{r^2} \ & v=\sqrt{\frac{G M}{r}} \ & L=m v r \ &=m \sqrt{\frac{G M}{r}} r \ &=m(G M r)^{1 / 2} \end{aligned}$

The maximum velocity of a particle, executing S.H.M. with an amplitude 7 mm is $4.4 \mathrm{~ms}^{-1}$. The…

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