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MHT CET 2019 (3 May, Shift 1) · Simple Pendulum

Simple Pendulum question from MHT CET 2019 (3 May, Shift 1)

Q72MHT CET 2019 · 3 MAY · SHIFT 1Easy

Simple Pendulum

A person measures a time period of a simple pendulum inside a stationary lift and finds it to be $T$ . If the lift starts accelerating upwards with an acceleration $\left(\frac{g}{3}\right)$ , the time period of the pendulum will be

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Answer: (b)

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

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