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A simple pendulum is suspended from the roof of a lift that is initially at rest, and its period is T. The lift then moves upward such that the distance it covers is given by y = t², where y is in metres and t is in seconds. Taking g = 10 m/s², what will be the new time period of the pendulum?
- √(4/5) T
- √(5/6) T
- √(5/4) T
- √(6/5) T
Correct answer: √(5/6) T
Solution
From y = t^2, acceleration = d^2y/dt^2 = 2 m/s^2 upward, so effective gravity = g + a = 12 m/s^2. New period T' = T*sqrt(g/g_eff) = T*sqrt(10/12) = sqrt(5/6) T.
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