Q. The bob of simple pendulum is hanging vertically down from a fixed identical bob by means of string of length $l$. If both bobs are charged with a charge with a charge $q$ each, time period of the pendulum is (ignore the radii of the bobs)
A) $2\pi\sqrt{\frac{l}{g + \left( \frac{q^{2}}{l^{2}m\ } \right)}}$
B) $2\pi\sqrt{\frac{l}{g - \left( \frac{q^{2}}{l^{2}m} \right)}}$
C) $2\pi\sqrt{\frac{l}{g}}$
D) $2\pi\sqrt{\frac{l}{g - \left( \frac{q2}{l} \right)}}$
Explanation
(c) There will be a electrostatic repulsion between two charged bobs, but it does not affect the motion of pendulum. Thus, time period of pendulum remains same $ie,T = 2\pi\sqrt{\frac{l}{g}}$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.