Two simple pendulums of length 5 m and 20 m are given small displacements in the same direction at the same time. The minimum number of oscillations, the shorter pendulum has completed, when the phase difference between them becomes zero again, is

Step-by-step solution

N S = Number of oscillations made by shorter length pendulum with time period T S . N L = Number of oscillations made by longer length pendulum with time period T L . If t is the time after which the pendulums are back in sync, then Then t = N S T S = N L T L ⇒ N S 2 π 5 g = N L × 2 π 20 g ∵     T = 2 π l g ⇒ N S = 2 N L i.e. if ⇒ N L = 1 N S = 2