Two simple pendulums of lengths 1.44 m and 1.00 m start oscillating in phase from their mean positions. Taking the acceleration due to gravity g = ^2 m/s ^2 , what is the minimum time elapsed before they are again in phase at the mean position?

Step-by-step solution

Time period of a simple pendulum is given by T = 2 l g . For the first pendulum: T₁ = 2 1.44 ^2 = 2 1.2 = 2.4 s . For the second pendulum: T₂ = 2 1.00 ^2 = 2 1.0 = 2.0 s . Let the pendulums be in phase again after n oscillations of the shorter pendulum (which is faster) and (n-1) oscillations of the longer pendulum. Time elapsed, t = nT₂ = (n-1)T₁ n(2.0) = (n-1)(2.4) 2.0n = 2.4n - 2.4 0.4n = 2.4 n = 6 The total time elapsed is t = 6 2.0 = 12 s . Answer: 12 s