JEE Main 2022 · 26 June · morning · Single correct · +4 / −1

Time period of a simple pendulum in a stationary lift is 'T'. If the lift accelerates with g/6 vertically upwards then the time period will be :
(Where g = acceleration due to gravity)

  1. (a) √(6/5) T
  2. (b) √(5/6) T
  3. (c) √(6/7) T
  4. (d) √(7/6) T
Answer and solution

Answer (c) √(6/7) T

Solution

Physics sir · 3 steps

An upward-accelerating lift makes gravity feel stronger: use g + g/6 in the period formula.

  1. Effective gravity. In a lift accelerating upwards with a, the bob feels a pseudo force ma downwards, on top of its weight. So the effective gravity is g_(eff) = g + a = g + g/6 = 7g/6.

    T = 2π√(l/g), T' = 2π√(l/g_(eff))

  2. Take the ratio. The length l and 2π cancel, so T'/T = √(g/g_(eff)) = √(g/(7g/6)) = √(6/7).
  3. So. T' = √(6/7) T, slightly shorter than T, which is option (c).

The idea to keep. In any accelerating frame, replace g by the effective gravity. Lift going up with acceleration a: g + a. Going down: g - a. Stronger gravity means a shorter period.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Simple Harmonic Motion

Asked about 130 times in JEE Main since 2020. This question is from 2022.