Q. The total energy of the body executing S.H.M. is $E.$ Then the kinetic energy when the displacement is half of the amplitude, is
A) $\frac{E}{2}$
B) $\frac{E}{4}$
C) $\frac{3E}{4}$
D) $\frac{\sqrt{3}}{4}E$
Explanation
(c) Total energy in SHM $E = \frac{1}{2}m\omega^{2}a^{2};$ (where $a =$ amplitude)
Kinetic energy $K = \frac{1}{2}m\omega^{2}\left( a^{2} - y^{2} \right) = E - \frac{1}{2}m\omega^{2}y^{2}$
When $y = \frac{a}{2} \Rightarrow K = E - \frac{1}{2}m\omega^{2}\left( \frac{a^{2}}{4} \right) = E - \frac{E}{4} = \frac{3E}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.