Q. The time period of a simple pendulum of length $L$ as measured in an elevator descending with acceleration $\frac{g}{3}$ is
A) $2\pi\sqrt{\frac{3L}{g}}$
B) $\pi\sqrt{\left( \frac{3L}{g} \right)}$
C) $2\pi\sqrt{\left( \frac{3L}{2g} \right)}$
D) $2\pi\sqrt{\left( \frac{2L}{3g} \right)}$
Explanation
(c) The effective acceleration in a lift descending with acceleration $\frac{g}{3}$ is $g_{eff} = g - \frac{g}{3} = \frac{2g}{3}$
$\therefore T = 2\pi\sqrt{\left( \frac{L}{g_{eff}} \right)} = 2\pi\sqrt{\left( \frac{L}{2g/3} \right)} = 2\pi\sqrt{\left( \frac{3L}{2g} \right)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.