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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 $\mathrm{x}=\mathrm{a} / 2$ will be :
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Verified Answer
The correct answer is:
$\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}$
$\quad \mathrm{v}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2}$
$$
=\frac{2 \pi}{\mathrm{T}} \sqrt{\mathrm{a}^2-\frac{\mathrm{a}^2}{4}}=\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}
$$
$$
=\frac{2 \pi}{\mathrm{T}} \sqrt{\mathrm{a}^2-\frac{\mathrm{a}^2}{4}}=\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}
$$
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