Two pendulums have time periods T and frac{5T | vedclass

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(B) Let the time periods be $T_1 = T$ and $T_2 = \frac{5T}{4}$.Since $T_2 > T_1$,the pendulum with time period $T_2$ is the bigger pendulum.After the

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(B) Let the time periods be $T_1 = T$ and $T_2 = \frac{5T}{4}$.Since $T_2 > T_1$,the pendulum with time period $T_2$ is the bigger pendulum.After the bigger pendulum completes one oscillation,the time elapsed is $t = T_2 = \frac{5T}{4}$.In this time $t$,the number of oscillations completed by the smaller pendulum is $n = \frac{t}{T_1} = \frac{5T/4}{T} = \frac{5}{4} = 1.25$ oscillations.This means the smaller pendulum completes $1$ full oscillation and an additional $0.25$ (or $\frac{1}{4}$) of an oscillation.At $t = T_2$,the bigger pendulum is at the mean position (phase $\phi_2 = 2\pi$ or $0$).The smaller pendulum has completed $1.25$ oscillations,meaning it is at the positive extreme position (phase $\phi_1 = 1.25 	imes 2\pi = 2.5\pi = 2\pi + \frac{\pi}{2}$).The phase difference is $\Delta\phi = |\phi_1 - \phi_2| = \frac{\pi}{2} = 90^o$.

Two pendulums have time periods $T$ and $\frac{5T}{4}$. 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 completed one oscillation (in $^o$)?

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