Q. The KE and PE of a particle executing SHM of amplitude $a$ will be equal when displacement is
A) $\frac{a}{2}$
B) $a\sqrt{2}$
C) $2a$
D) $a/\sqrt{2}$
Explanation
(d) KE=PE
$\Rightarrow$ $\frac{1}{2}\ m\omega^{2}\left( a^{2} - y^{2} \right) = \frac{1}{2}\ m\omega^{2}y^{2}$
$\Rightarrow$ $y = \frac{a}{\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.