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 (in degree) between them after the bigger pendulum has completed one oscillation?

Answer 1

Given, time period for the bigger pendulum,

$T_1=\frac{5T}{4}=(T+\frac{T}{4})$

And time period for smaller pendulum,

$T_2=T$

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.

Hence, phase difference = ${90}^{\circ }$
Final answer: ${90}^{\circ }$