Question

A suspended simple pendulum of length $l$ is making an angle $\theta$ with the vertical. On releasing, its velocity at lowers 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 $\sqrt{2gl(1-\cos \theta )}$

• From the diagram we see that loss in height is $l(1-cos\theta )$
• conservation of energy $\frac{m{v}^2}{2}=mgl(1-cos\theta )$
• so velocity $v=\sqrt[2]{22gl(1-cos\theta )}$ Option C is correct.