Question

A suspended simple pendulum of length $l$ is making an angle $\theta$ with the vertical. On releasing, its velocity at lowest point will be:-

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

Answer 1

The correct option is C $\sqrt{2gL(1-\cos \theta )}$

$OC=OB\cos \theta =L\cos \theta$

$h=OC-OB$

$h=L(1-\cos \theta )$

KE at $C=$ PE at $B$

$\Rightarrow \frac{1}{2}m{v}^2=mgh$

$\Rightarrow v^2=2gh$

$\Rightarrow v^2=2gL(1-\cos \theta )$

$\Rightarrow v=\sqrt{2gL(1-\cos \theta )}$