Q. The period of a simple pendulum inside a stationary lift is $T$. The lift accelerates upwards with an acceleration of $\frac{g}{3}$. The time period of pendulum will be
A) $\sqrt{2}\ T$
B) $\frac{T}{\sqrt{2}}$
C) $\frac{\sqrt{3}}{2}T$
D) $\frac{T}{3}$
Explanation
(c) When lift accelerates upwards, then effective acceleration on the pendulum
$g_{eff} = g + \frac{g}{3} = \frac{4g}{3}$
$\therefore$ Time period $T^{'} = 2\pi\sqrt{\frac{l}{g_{eff}}} = 2\pi\sqrt{\frac{l}{\frac{4g}{3}}}$
$= \frac{\sqrt{3}}{2}.2\pi\sqrt{\frac{l}{g}}$
$= \ \frac{\sqrt{3}}{2}\ T$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.