A simple pendulum performs simple harmonic mo | vedclass

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(A) The velocity $v$ of a particle executing simple harmonic motion $(S.H.M.)$ at a displacement $x$ is given by the formula:$v = \omega \sqrt{A^2 - x

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

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(A) The velocity $v$ of a particle executing simple harmonic motion $(S.H.M.)$ at a displacement $x$ is given by the formula:$v = \omega \sqrt{A^2 - x^2}$Given that the angular frequency $\omega = \frac{2\pi}{T}$ and the displacement $x = \frac{A}{2}$,we substitute these values into the equation:$v = \frac{2\pi}{T} \sqrt{A^2 - (\frac{A}{2})^2}$Simplify the expression inside the square root:$v = \frac{2\pi}{T} \sqrt{A^2 - \frac{A^2}{4}}$$v = \frac{2\pi}{T} \sqrt{\frac{3A^2}{4}}$Extracting the terms from the square root:$v = \frac{2\pi}{T} \cdot \frac{A\sqrt{3}}{2}$$v = \frac{\pi A\sqrt{3}}{T}$Thus,the speed of the pendulum at $X = \frac{A}{2}$ is $\frac{\pi A\sqrt{3}}{T}$.

$A$ simple pendulum performs simple harmonic motion about $X = 0$ with an amplitude $A$ and time period $T$. The speed of the pendulum at $X = \frac{A}{2}$ will be

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