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

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Aicharya · Accepted Answer

$\begin{aligned} v=\frac{d y}{d t}=A \omega \cos \omega t & =A \omega \sqrt{1-\sin ^2 \omega t} \ & =\omega \sqrt{A^2-y^2} \end{aligned}$ Here, $\mathrm{y}=\frac{\mathrm{a}}{2}$ $	herefore \mathrm{v}=\omega \sqrt{\mathrm{a}^2-\frac{\mathrm{a}^2}{4}}=\omega \sqrt{\frac{3 \mathrm{a}^2}{4}}=\frac{2 \pi}{\mathrm{T}} \frac{\mathrm{a} \sqrt{3}}{2}=\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}$ /

A simple pendulum performs simple harmonic motion about $\mathrm{x}=0$ with an amplitude a and time period…

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