Q. A particle is vibrating in a simple harmonic motion with an amplitude of $4\ cm.$ At what displacement from the equilibrium position, is its energy half potential and half kinetic
A) $1\ cm$
B) $\sqrt{2}\ cm$
C) $3\ cm$
D) $2\sqrt{2}\ cm$
Explanation
(d) Let $x$ be the point where K.E. = P.E.
Hence $\frac{1}{2}m\omega^{2}\left( a^{2} - x^{2} \right) = \frac{1}{2}m\omega^{2}x^{2}$
$\Rightarrow \ 2x^{2} = a^{2}\ \Rightarrow \frac{a}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.