Q. The kinetic energy of a particle executing S.H.M. is $16\ J$ when it is in its mean position. If the amplitude of oscillations is $25\ cm$ and the mass of the particle is $5.12\ kg,$ the time period of its oscillation is
A) $\frac{\pi}{5}s$
B) $2\pi\ s$
C) $20\pi\ s$
D) $5\pi\ s$
Explanation
(a) At mean position, the kinetic energy is maximum
Hence $\frac{1}{2}ma^{2}\omega^{2} = 16$
On putting the values we get
$\omega = 10 \Rightarrow T = \frac{2\pi}{\omega} = \frac{\pi}{5}s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.