Two pendulums of length 121   cm and 100   cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is

Step-by-step solution

Time period of a pendulum, T ∝ l ⇒ T 1 T 2 = l 1 l 2 If n is the number of vibrations after which the pendulums are again in phase, the number of vibration of the longer pendulum will be n - 1 . Therefore, T 1 × n - 1 T 2 × n = 1 ⇒ 121 n - 1 100 n = 1 ⇒ 11 n - 11 = 10 n ⇒ n = 11