First slide

Applications of SHM

Question

Two pendulums have time periods T and 5T/4. They start S.H.M. at the same time from the mean position. What will be the phase difference between them after the bigger
pendulum has complete one oscillation?

Moderate

A

45^o
B

90^o
C

60^o
D

30^o

Solution

$As \frac{5 T}{4} = T + \frac{T}{4}$
By the time, the bigger pendulum makes one full oscillation, the smaller pendulum will make $(1 + \frac{1}{4})$ oscillation. The bigger pendulum will be in the mean position and the smaller one will be in the positive extreme position. Thus, phase difference = 90^o

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