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MHT CET 2023 (12 May, Shift 2) · Simple Pendulum

Simple Pendulum question from MHT CET 2023 (12 May, Shift 2)

Q44MHT CET 2023 · 12 MAY · SHIFT 2Easy

Simple Pendulum

In a stationary lift, time period of a simple pendulum is ' $T$ '. The lift starts accelerating downwards with acceleration $\left(\frac{g}{4}\right)$ , then the time period of the pendulum will be

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Answer: (b)

Lift accelerating down: $g_{eff} = g - \frac{g}{4} = \frac{3g}{4}$ .
$T' = T\sqrt{\frac{g}{g_{eff}}} = T\sqrt{\frac{4}{3}}$
So the answer is (b) $\frac{2}{\sqrt{3}}T$ .

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