A simple pendulum performs simple harmonic motion about $X = 0$ with an amplitude $A$ and time period $T$. The speed of the pendulum at $X = \frac{A}{2}$ will be
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(a)Velocity of a particle executing $S.H.M.$ is given by
$v = \omega \sqrt {{a^2} - {x^2}} $
$= \frac{{2\pi }}{T}\sqrt {{A^2} - \frac{{{A^2}}}{4}} $
$= \frac{{2\pi }}{T}\sqrt {\frac{{3{A^2}}}{4}}$
$= \frac{{\pi A\sqrt 3 }}{T}$.
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