Q. A particle of mass 10 $g$ is describing S.H.M. along a straight line with period of $2\ s$ and amplitude of $10\ cm.$ Its kinetic energy when it is at $5\ cm$ from its equilibrium position is
A) $37.5\pi^{2}ergs$
B) $3.75\pi^{2}ergs$
C) $375\pi^{2}ergs$
D) $0.375\pi^{2}ergs$
Explanation
(c) Kinetic energy $K = \frac{1}{2}m\omega^{2}(a^{2} - y^{2})$
$= \frac{1}{2} \times 10 \times \left( \frac{2\pi}{2} \right)^{2}\left\lbrack 10^{2} - 5^{2} \right\rbrack = 375\pi^{2}ergs$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.