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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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$\frac{πA}{T}$

$\frac{πA \sqrt{3}}{T}$

$\frac{πA \sqrt{3}}{2 T}$

$\frac{3 π^{2} A}{T}$

answer is A.

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Detailed Solution

Velocity of a particle executing S.H.M. is given by

$v = ω \sqrt{a^{2} - x^{2}} = \frac{2 π}{T} \sqrt{A^{2} - \frac{A^{2}}{4}} = \frac{2 π}{T} \sqrt{\frac{3 A^{2}}{4}} = \frac{πA \sqrt{3}}{T}$

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