Q 11-13-025JEE MainMedium
A simple pendulum of length $l$ hangs from the roof of a car moving on a straight level road with acceleration $a$. Its period of small oscillation is
Answer: (C) $2\pi\sqrt{\dfrac{l}{\sqrt{g^2 + a^2}}}$
Concept. In the car's frame the bob feels $g$ downward and a pseudo acceleration $a$ backward; $g_{\text{eff}}$ is their vector sum.
Step 1. Effective gravity. $g_{\text{eff}} = \sqrt{g^2 + a^2}$.
Step 2. Period. $T = 2\pi\sqrt{\dfrac{l}{\sqrt{g^2 + a^2}}}$.
Answer. $2\pi\sqrt{\dfrac{l}{\sqrt{g^2 + a^2}}}$ (Option C).
Solution by Sreeraj P, M.Sc Physics