Q. A lift is ascending with an acceleration equal to g/3. Its time period of oscillation is$T$. What will be the time period of a simple pendulum suspended from its ceiling in stationary lift?
A) $2\ T$
B) $3\ T$
C) $(\sqrt{3/4)T}$
D) $2T/\sqrt{3}$
Explanation
(d) $T = 2\pi\sqrt{\frac{l}{g}};$
$T = 2\pi\sqrt{\frac{1}{g + g/3}} = 2\pi\sqrt{\frac{3l}{4g}} = \left( \sqrt{\frac{3}{4}} \right)T^{'}$
or $t^{'} = \frac{2T}{\sqrt{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.