Q. The time period of a simple pendulum measured inside a stationary lift is found to be $T$. If the lift starts accelerating upwards with an acceleration $g/3$, the time period is
A) $T\sqrt{3}$
B) $T\sqrt{3}/2$
C) $T/\sqrt{3}$
D) $T/3$
Explanation
(b) $T = 2\pi\sqrt{\frac{l}{g}}\ $ and ${\ T}^{'} = 2\pi\sqrt{\frac{l}{4g/3}}$
[As $g^{'} = g + a = g + \frac{g}{3} = \frac{4g}{3}\rbrack$
$\therefore T^{'} = \frac{\sqrt{3}}{2}T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.