Two simple pendulums of length 1 m and 16 m are in phase at the mean position at a certain instant of time. If T is the time period of shorter pendulum then the minimum time after which they will again in phase is

Since the time period of shorter pendulum is T.Therefore time period of pendulum of length 16is 4T. let t be time when they are in phase.let n be no of oscillation of slow(length 16) and (n+1) that of fast(length=1) t=n×4T=(n+1)T4n=n+1n=1/3 for 16 cm lengthfor shorter pendulum n+1=4/3 so4/3 ×T is minimum time after which they will be in phase again