Q. A simple pendulum with a bob of mass $'m'$ oscillates from $A$ to $C$ and back to $A$ such that $PB$ is $H.$ If the acceleration due to gravity is $'g^{'},$ then the velocity of the bob as it passes through $B$ is
A) $mgH$
B) $\sqrt{2gH}$
C) $2gH$
D) Zero
Explanation
(b) At $B,$ the velocity is maximum. Using conservation of mechanical energy
$\mathrm{\Delta}PE = \mathrm{\Delta}KE \Rightarrow mgH = \frac{1}{2}mv^{2} \Rightarrow v = \sqrt{2gH}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.