Q. For a particle executing SHM, the kinetic energy k is given by $k = k_{0}\cos^{2}{\omega t.}$ The equation of its displacement can be
A) $\left( \frac{k_{0}}{m\omega^{2}} \right)^{1/2}\sin{\ \omega t}$
B) $\left( \frac{2k_{0}}{m\omega^{2}} \right)^{1/2}\sin{\ \omega t}$
C) $\left( \frac{2\omega^{2}}{m\ k_{0}} \right)^{1/2}\sin{\ \omega t}$
D) $\left( \frac{2k_{0}}{m\omega} \right)^{1/2}\sin{\ \omega t}$
Explanation
(d) If $m$ is the mass, $r$ is the amplitude of oscillation, then maximum kinetic energy,
$K_{0} = \frac{1}{2}m\omega^{2}r^{2}\ \ or\ \ \ \ r = \left( \frac{2K_{0}}{m\omega^{2}} \right)^{\frac{1}{2}}$
The displacement equation can be
$y = r\sin{\omega t} = \left( \frac{2K_{0}}{m\omega} \right)^{\frac{1}{2}}\sin{\omega t}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.