Q104 / 125 · BITSAT 2021 · Physics

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

  • A
    $\sqrt{2 g l(1+\cos \theta)}$
  • B
    $\sqrt{2 g l \sin \theta}$
  • C
    $\sqrt{2 \mathrm{~g} l}$
  • Correct answer
    D
    $\sqrt{2 \mathrm{~g} l(1-\cos \theta)}$

Solution

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) $\Rightarrow v=\sqrt{2 g h}=\sqrt{2 g l(1-\cos \theta)}$

Hence, the correct answer is option: 'D'.

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