Two simple pendulum of lengths in the ratio 9: 1 are started at time t = 0. They will again be in phase. After shorter pendulum has made n oscillation, where n is:

Question

Two simple pendulum of lengths in the ratio 9: 1 are started at time t = 0. They will again be in phase. After shorter pendulum has made n oscillation, where n is: (A) 2 (B) 1 (C) 1/2 (D) 3/2

Tardigrade Admin · Accepted Answer

Given: The periods of pendulum are in the ratio of 3: 1. From the relation of time period T propto √l Suppose the pendulum be in phase after n. .oscillation of smaller pendulum. Then larger pendulum has made ( n-1 ) oscillation in the same duration it means ( n-1 )3=n× 1 Hence, n=(3/2)

Q. Two simple pendulum of lengths in the ratio 9 : 1 are started at time t = 0. They will again be in phase. After shorter pendulum has made n oscillation, where n is:

2

1

1/2

3/2

Given: The periods of pendulum are in the ratio of 3 : 1. From the relation of time period T∝l​ Suppose the pendulum be in phase after n. .oscillation of smaller pendulum. Then larger pendulum has made (n−1) oscillation in the same duration it means (n−1)3=n×1 Hence, n=23​