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JEE PHYSICS MCQ: OSCILLATIONS

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A simple pendulum executing S.H.M. about x=0x=0 with period TT and amplitude AA. Its speed when at a distance A2\frac{A}{2} is
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Correct option: A
πA3 T\frac{\pi \mathrm{A} \sqrt{3}}{\mathrm{~T}}
Option A (πA3 T\frac{\pi \mathrm{A} \sqrt{3}}{\mathrm{~T}}) is correct because the speed of a particle in SHM at a distance y from the mean position is given by v=ωA2−y2v = \omega \sqrt{A^2 - y^2}, where ω=2πT\omega = \frac{2\pi}{T}.

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