The time period of a simple pendulum inside a stationary lift is 3 second. When the lift moves upwards with an acceleration g / 3 , the time period will be ( g = acceleration due to gravity)

Step-by-step solution

The time period of a simple pendulum is given by T = 2 L g_ eff , where L denotes the pendulum length and g_ eff the effective gravitational acceleration. For a stationary lift, g_ eff = g and T₁ = 2 L g = 3 , s . When the lift accelerates upward with a = g/3 , the effective acceleration becomes g_ eff ' = g + a = 4g 3 . The new period is therefore T₂ = 2 L 4g 3 = 2 3L 4g = 3 2 2 L g . Substituting 3 for the stationary case yields T₂ = 3 2 3 = 3 2 = 1.5 , s . Final answer: 1.5 , s