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A simple pendulum performs SHM about x = 0 with an amplitude A and time period T. The speed of the pendulum at x = A/2 will be:

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$\frac{A}{T}$

$\frac{(\sqrt{3}) πA}{T}$

$\frac{πA (\sqrt{3})}{2 T}$

$\frac{3 π^{2} A}{T}$

answer is B.

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Detailed Solution

$v = ω \sqrt{A^{2} - y^{2}} at y = \frac{A}{2}, v = \frac{2 π}{T} \sqrt{A^{2} - \frac{A^{2}}{4}} = \frac{\sqrt{3} πA}{T}$

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