Q. A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration $a,$ then the time period is given by $T = 2\pi\sqrt{\frac{1}{g'}},$ where $g'$ is equal to
A) $g$
B) $g - a$
C) $g + a$
D) $\sqrt{g^{2} + a^{2}}$
Explanation
(d) $g^{'} = \sqrt{g^{2} + a^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.