Time period of a simple pendulum is T inside a lift when the lift is stationary. If the lift moves upwards with an acceleration g/2, the time period of pendulum will be :

Solution

Physics sir · 3 steps

In an accelerating lift, swap g for the effective g the bob feels; that is the whole trick.

1. Effective gravity. In the lift's frame, a pseudo force ma acts downward when the lift accelerates upward. It adds to the weight, so the bob feels g_(eff) = g + a = g + g/2 = 3g/2.
2. Period formula. The length stays the same, and only g changes. Since T = 2π√(L/g), the period varies as 1/√g.

   T'/T = √(g/g_(eff)) = √(g/(3g/2)) = √(2/3)

3. So. T' = √(2/3) T. Stronger effective gravity makes the pendulum swing faster, so the period gets shorter. The answer is (b).

The idea to keep. For a pendulum in a lift, use g + a when the lift accelerates up and g - a when it accelerates down. Then T ∝ 1/√g_(eff).

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