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JEE PHYSICS MCQ: OSCILLATIONS

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The maximum height of a simple pendulum with respect to the equilibrium position is h\mathrm{h} and the maximum velocity of the bob of pendulum is v\mathrm{v}. Then
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Correct option: C
v=2gh\mathrm{v}=\sqrt{2 \mathrm{gh}}
Option C (v=2gh\mathrm{v}=\sqrt{2 \mathrm{gh}}) is correct because it is derived from the conservation of energy, equating the change in kinetic energy to the change in potential energy.

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