A simple pendulum performs S.H.M about x = 0 | vedclass

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(B) The velocity $v$ of a particle performing $S.H.M$ at a displacement $x$ is given by the formula: $v = \omega \sqrt{A^2 - x^2}$.Given the angular f

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$\frac{\sqrt{3} \pi A}{T}$

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(B) The velocity $v$ of a particle performing $S.H.M$ at a displacement $x$ is given by the formula: $v = \omega \sqrt{A^2 - x^2}$.Given the angular frequency $\omega = \frac{2\pi}{T}$ and the displacement $x = \frac{A}{2}$.Substituting these values into the velocity formula:$v = \frac{2\pi}{T} \sqrt{A^2 - (\frac{A}{2})^2}$$v = \frac{2\pi}{T} \sqrt{A^2 - \frac{A^2}{4}}$$v = \frac{2\pi}{T} \sqrt{\frac{3A^2}{4}}$$v = \frac{2\pi}{T} \cdot \frac{\sqrt{3}A}{2}$$v = \frac{\sqrt{3} \pi A}{T}$.

$A$ simple pendulum performs $S.H.M$ 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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