A simple pendulum performs simple harmonic mo | vedclass

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(C) The velocity of a particle in simple harmonic motion is given by $v = \omega \sqrt{a^2 - x^2}$.Given the angular frequency $\omega = \frac{2\pi}{T

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

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(C) The velocity of a particle in simple harmonic motion is given by $v = \omega \sqrt{a^2 - x^2}$.Given the angular frequency $\omega = \frac{2\pi}{T}$.At position $x = \frac{a}{2}$,the speed is:$v = \omega \sqrt{a^2 - (\frac{a}{2})^2}$$v = \omega \sqrt{a^2 - \frac{a^2}{4}}$$v = \omega \sqrt{\frac{3a^2}{4}}$$v = \omega \cdot \frac{a\sqrt{3}}{2}$Substituting $\omega = \frac{2\pi}{T}$:$v = \frac{2\pi}{T} \cdot \frac{a\sqrt{3}}{2}$$v = \frac{\pi a \sqrt{3}}{T}$

$A$ simple pendulum performs simple harmonic motion about $x=0$ with an amplitude '$a$' and time period '$T$'. The speed of the pendulum at $x=\frac{a}{2}$ is

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