Work Energy and Power - Result Question 23

26. In a simple pendulum of length $l$ 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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(a) $\sqrt{2 g \ell(1+\cos \theta)}$

(b) $\sqrt{2 g \ell \sin \theta}$

(c) $\sqrt{2 g \ell}$

(d) $\sqrt{2 g \ell(1-\cos \theta)}$

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Answer:

Correct Answer: 26. (d)

Solution:

  1. (d) If1 is length of pendulum and $\theta$ be angular amplitude then height

$h=AB-AC=1-1 \cos \theta=1(1-\cos \theta)$

At extreme position, potential energy is maximum and kinetic energy is zero; At mean (equilibrium) position potential energy is zero and kinetic energy is maximum, so from principle of conservation of energy.

$(KE+PE)$ at $P=(KE+PE)$ at $B$

$0+m g h=\frac{1}{2} m v^{2}+0$

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