Q. A particle of mass $m$ executes simple harmonic motion with amplitude $a$ and frequency v. The average kinetic energy during its motion from the position of equilibrium to the end is
A) $\pi^{2}ma^{2}v^{2}$
B) $\frac{1}{4}ma^{2}v^{2}$
C) $4\pi^{2}ma^{2}v^{2}$
D) $2\pi^{2}ma^{2}\ v^{2}$
Explanation
(a) Average kinetic energy of particle
$= \ \frac{1}{4}\ ma^{2}\omega^{2}$
$= \ \frac{1}{4}\ ma^{2}(2\pi v)^{2} = \pi^{2}v^{2}ma^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.