Q. A pendulum bob on a $2\ m$ string is displaced $60{^\circ}$ from the vertical and then released. What is the speed of the bob as it passes through the lowest point in its path
A) $\sqrt{2}\ m/s$
B) $\sqrt{9.8}\ m/s$
C) $4.43\ m/s$
D) $1/\sqrt{2}\ m/s$
Explanation
(c) $v = \sqrt{2gl(1 - \cos{\theta)}} = \ \sqrt{2 \times 9.8 \times 2\left( 1 - \cos{60{^\circ}} \right)} = 4.43\ m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.