First slide

Simple hormonic motion

Question

Two pendulums have time periods T and $\frac{5 T}{4}$ . 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°
B

90°
C

60°
D

30°

Solution

$\frac{5 T}{4} = T + \frac{T}{4}$
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°

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