Question

A simple pendulum is swinging in vertical plane. The ratio of its potential energy when it is making ${45}^0$ and ${90}^0$ with the vertical is

• A. $1:1$
• B. $1:(\sqrt{2}+1)$
• C. $4:(\sqrt{5}+2)$
• D. $\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$

Answer 1

The correct option is D

$\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$

Consider the problem

Potential energy of pendulum swinging in vertical plane can be given as

$\begin{array}{c}U=mgL\left(1-\cos \theta \right)\\ At\theta ={45}^o\\ U_{45}=mgL\left(1-\frac{1}{\sqrt{2}}\right)\\ U_{45}=mgL\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)\\ At\theta ={90}^o\\ U_{90}=mgL\left(1-0\right)\\ U_{90}=mgL\\ \frac{U_{45}}{U_{90}}=\frac{mgL}{mgL}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)\\ U_{45}:U_{90}=\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)\end{array}$

Hence, Option $D$ is the correct answer.