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NEET 1991 (AIPMT) · Simple Pendulum

Simple Pendulum question from NEET 1991 (AIPMT)

Q92NEET 1991 · AIPMTMedium

Simple Pendulum

A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration $a$ , then the time period is given by $T = 2\pi\sqrt{(l/g)}$ , where $g$ is equal to

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Answer: (d)

In the trolley frame the bob feels gravity $g$ down and pseudo acceleration $a$ horizontally.
These are perpendicular, so $g_{eff} = \sqrt{g^2 + a^2}$ .
So the answer is (d) $\sqrt{g^2+a^2}$ .

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