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) $2\pi\sqrt{\dfrac{l}{\sqrt{g^2 + a^2}}}$
In the car's frame the bob feels gravity $g$ downward and a pseudo force $ma$ backward. The effective gravity is their vector sum, $g_{eff} = \sqrt{g^2 + a^2}$:
$$T = 2\pi\sqrt{\frac{l}{\sqrt{g^2 + a^2}}}$$
Solution by Sreeraj P, M.Sc Physics