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In a simple pendulum of length I the bob is pulled aside from its equilibrium position through an angle $\theta$ and then released. The bob passes through the equilibrium position with speed
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The correct answer is:
$\sqrt{2 \mathrm{~g} l(1-\cos \theta)}$
If $l$ is length of pendulum and $\theta$ be angular amplitude then from principle of conservation of energy
(Initial total energy) $0+\mathrm{mgh}=\frac{1}{2} m v^{2}+0$ (Final total energy) $\Rightarrow v=\sqrt{2 g h}=\sqrt{2 g l(1-\cos \theta)}$
(Initial total energy) $0+\mathrm{mgh}=\frac{1}{2} m v^{2}+0$ (Final total energy) $\Rightarrow v=\sqrt{2 g h}=\sqrt{2 g l(1-\cos \theta)}$
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