Question

A particle executing SHM with an amplitude A. The displacement of the particle when its potential energy is half of its total energy is

• A. $\frac{A}{\sqrt{2}}$
• B. $\frac{A}{2}$
• C. $\frac{A}{4}$
• D. $\frac{A}{3}$

Answer 1

Answer: (D)

$\frac{A}{3}$

Answer 2

The period of harmonic oscillator does not depend on the amplitude of oscillation.

Energy of oscillator, $E = \frac{1}{2}m\omega^{2}A^{2}$

$E \propto A^{2}$

So, if amplitude (1) is doubled its energy becomes 4 times.