Question Detail
Question
A simple pendulum attached to the ceiling of a stationary lift has a time period of 1 s. The distance y covered by the lift moving downward varies with time as y = 3.75t², where y is in meters and t is in seconds. If g = 10 m/s², what is the period T of the pendulum when the motion is downward?
Correct Answer
Option C is the correct answer.
Detailed Explanation
y = 3.75t² => acceleration a = d²y/dt² = 7.5 m/s² (downward). Effective g' = g - a = 10 - 7.5 = 2.5 m/s². T = 2π√(L/g'). T₀ = 2π√(L/g) = 1 s. T/T₀ = √(g/g') = √(10/2.5) = √4 = 2. T = 2 × 1 = 2 s.
Hint
Effective gravity in a downward accelerating lift is g' = g - a.