Two simple pendulums of lengths 100 cm and 121 cm start oscillating simultaneously from their mean position in the same phase. How many oscillations will the shorter pendulum have completed when the two pendulums are next in phase together?

Step-by-step solution

Let T₁ and T₂ be the time periods of the shorter and longer pendulums, respectively. The time period of a simple pendulum is given by T = 2 L g . Thus, T L . For the two pendulums: T₁ T₂ = L₁ L₂ = 100 121 = 10 11 Let the two pendulums be in phase again after time t . The faster (shorter) pendulum will have completed one more oscillation than the slower (longer) pendulum. If the shorter pendulum completes n oscillations, the longer one completes (n - 1) oscillations in the same time t . t = n T₁ = (n - 1) T₂ Substit