Q.
A simple pendulum is released from $A$ as shown. If $m$ and $l$ represent the mass of the bob and length of the pendulum, the gain in kinetic energy at $B$ is
A) $\frac{mgl}{2}$
B) $\frac{mgl}{\sqrt{2}}$
C) $\frac{\sqrt{3}}{2}mgl$
D) $\frac{\sqrt{2}}{3}mgl$
Explanation
(c) Vertical height $= h = l\cos{30{^\circ}}$
Loss of potential energy $= mgh$
[image:6_image112.emf]
$= mgl\cos{30{^\circ}} = \frac{\sqrt{3}}{2}mgl$
$\therefore$ Kinetic energy gained $= \frac{\sqrt{3}}{2}mgl$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.