In the figure shown, a lift goes downward with a constant retardation. An observer in the lift observes a simple pendulum with time period 2s. The length of the string which is used to suspend the bal

Question

In the figure shown, a lift goes downward with a constant retardation. An observer in the lift observes a simple pendulum with time period 2s. The length of the string which is used to suspend the ball from the ceiling is 2.0 m. If the magnitude of retardation of the lift is $α m / s^{2}$ then find the value of $\frac{α}{2}$ .

TestHub · Accepted Answer

Lift retards while moving down, so acceleration is upward: g eff ​ = g + α . T = 2 π g + α L ​ ​ with T = 2 s , L = 2 m , g = 10 . 10 + α = 2 π 2 = 20 so α = 10 and α /2 = 5 . Answer: 5

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