Q. A particle is executing SHM with amplitude $a$. When the PE of a particle is one-fourth of its maximum value during the oscillation, its displacement from the equilibrium position will be
A) $\frac{a}{4}$
B) $\frac{a}{3}$
C) $\frac{a}{2}$
D) $\frac{2a}{3}$
Explanation
(c) Potential energy=$\frac{1}{2}m\omega^{2}y^{2} = \frac{1}{4} \times \frac{1}{2}m\omega^{2}a^{2}$
$y^{2} = \frac{a^{2}}{4} \Rightarrow y = \pm \frac{a}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.