Question

A body executes S.H.M. with an amplitude A At what displacement from the mean position is the potential energy of the body one-fourth of its total energy?

• A. $\frac{A}{4}$
• B. $\frac{A}{2}$
• C. $\frac{3A}{4}$
• D. Some other fraction of $A$

Answer 1

The correct option is B $\frac{A}{2}$

$\begin{array}{c}accordingtoquestion\\ potentialenergy=\frac{1}{4}\left(totalenergy\right)\\ \Rightarrow \frac{1}{2}m{\omega }^2{y}^2=\frac{1}{4}\left(\frac{1}{2}m{\omega }^2{A}^2\right)\\ \Rightarrow y=\frac{A}{2}\end{array}$

Hence, the option $B$ is the correct answer.