Question

The kinetic energy of a particle executing SHM will be equal to $(1/8{)}^{\text{th }}$ of its potential energy when its displacement from the mean position is (where $A$ is the amplitude)

• A. $A\sqrt{2}$
• B. $\frac{A}{2}$
• C. $\frac{2\sqrt{2}}{3}A$
• D. $A\sqrt{\frac{2}{3}}$

Answer 1

Answer: (C)

Kinetic energy when the displacement of the particle is $x$
is given by $K=\frac{1}{2}m{\omega }^2\left(A^2-x^2\right)$
Potential energy at this instant is given by $U=\frac{1}{2}m{\omega }^2{x}^2$
As $K=\frac{1}{8}U$
$\therefore \frac{1}{2}m{\omega }^2\left(A^2-x^2\right)=\frac{1}{8}\left(\frac{1}{2}m{\omega }^2{x}^2\right)$
or $\left(A^2-x^2\right)=\frac{x^2}{8}$
or $x=\frac{2\sqrt{2}}{3}A$