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MHT CET 2025 (26 Apr, Shift 2) · Simple Pendulum

Simple Pendulum question from MHT CET 2025 (26 Apr, Shift 2)

Q28MHT CET 2025 · 26 APR · SHIFT 2Medium

Simple Pendulum

A simple pendulum is suspended from ceiling of a lift when lift is at rest its period is ' T '. With what acceleration ' $a$ ' should lift be accelerated upward in order to reduce the period to ' $T/2$ '? (take ' g ' as acceleration due to gravity)

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Answer: (b)

Lift accelerating up: $T' = 2\pi\sqrt{\frac{l}{g+a}}$ .
$\frac{T'}{T} = \sqrt{\frac{g}{g+a}} = \frac{1}{2} \Rightarrow g + a = 4g$ .
So the answer is (b) $a = 3g$ .

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