Two oscillating simple pendulums with time periods T and 4T 3 are in phase at a given time. They will be again in phase after an elapse of time

Step-by-step solution

Let the time elapsed be t . For the two pendulums to be in phase again, the time t must be an integer multiple of both their time periods. t = n₁ T₁ = n₂ T₂ Given T₁ = T and T₂ = 4T 3 . n₁ T = n₂ ( 4T 3 ) n₁ n₂ = 4 3 For the first time they are in phase again, the difference in the number of oscillations is 1 . Thus, n₁ = 4 and n₂ = 3 . The time elapsed is t = 4T . Alternatively, the time elapsed is the least common multiple of the two time periods. t = LCM (T, 4T 3 ) = 4T