Question

The bob of a pendulum of length $l$ is pulled a side from its equilibrium position through an angle $\theta$ and then released. The bob will, then pass through its equilibrium position with a speed $v$, where $v$ equals to

• A. $\sqrt{2gl\left(1-\sin\theta\right)}$
• B. $\sqrt{2gl\left(1+\cos\,\theta\right)}$
• C. $\sqrt{2gl\left(1-\cos\theta\right)}$
• D. $\sqrt{2gl\left(1+\sin\theta\right)}$

Answer 1

Answer: (C)

Suppose bob rises up a height $h$ as shown then after releasing, potential energy at extreme position becomes kinetic energy of mean position.

$mgh = \frac{1}{2} mv^{2}_{max}$
$v_{\max }=\sqrt{2 g h}$ ...(i)
From the figure,
$\cos \theta = \frac{l-h}{l}$
$\Rightarrow l\cos \theta=l-h$
$h=l(1-\cos \theta)$
From Eq. (i),
$v_{\max }=\sqrt{2 g l(1-\cos \theta)}$