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 $x=\frac{2 \sqrt{2}}{3} A$