A small sphere oscillates simple harmonically in a watch glass whose radius of curvature is 1.6 m . The period of oscillation of the sphere in second is (acceleration due to gravity, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

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Aicharya · Accepted Answer

$\begin{aligned} \mathrm{T} & =2 \pi \sqrt{\frac{l}{\mathrm{~g}}} \ & =2 \pi \sqrt{\frac{1.6}{10}} \quad \ldots(\because \mathrm{R}=l) \ & =0.8 \pi\end{aligned}$

A small sphere oscillates simple harmonically in a watch glass whose radius of curvature is 1.6 m . The…

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