An oscillating simple pendulum of time period T₀ is placed in a lift which is accelerating upwards with 2.5 m/s². The time period of pendulum in the lift will be (if g = 10 m/s²) (a) T₀ (b) 2/√(5)T₀ (c) 2/√(3)T₀ (d) T₀/4

Answer: (b). Lift accelerating up: geff = 10 + 2.5 = 12.5 m/s². T = T₀√((10)/(12.5)) = T₀√(4/5) So the answer is (b) 2/√(5)T₀.

An oscillating simple pendulum of time period T₀ is placed in a lift… | JEE Main 2022 (30 Jul, Shift 2)

Q53JEE Main 2022 · 30 JUL · SHIFT 2Easy

Simple Pendulum

An oscillating simple pendulum of time period $T_0$ is placed in a lift which is accelerating upwards with $2.5\text{ m/s}^2$ . The time period of pendulum in the lift will be (if $g = 10\text{ m/s}^2$ )

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Answer: (b)

Lift accelerating up: $g_{eff} = 10 + 2.5 = 12.5\text{ m/s}^2$ .
$T = T_0\sqrt{\frac{10}{12.5}} = T_0\sqrt{\frac{4}{5}}$
So the answer is (b) $\frac{2}{\sqrt{5}}T_0$ .

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