Two pendulums have time periods T and 5T/4. T | vedclass

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(A) Let the time periods be $T_1 = T$ and $T_2 = 5T/4$. Since $T_1 < T_2$,the pendulum with period $T_1$ is the smaller pendulum.Let $n$ be the number

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$5$

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(A) Let the time periods be $T_1 = T$ and $T_2 = 5T/4$. Since $T_1 Let $n$ be the number of oscillations of the smaller pendulum and $m$ be the number of oscillations of the larger pendulum when they are again in the same phase.For them to be in the same phase,the total time elapsed must be equal: $n T_1 = m T_2$.Substituting the values: $n T = m (5T/4)$.This simplifies to $n = 5m/4$,or $4n = 5m$.For the pendulums to be in the same phase again,$n$ and $m$ must be the smallest integers satisfying this equation.The smallest integer solution is $n = 5$ and $m = 4$.Thus,after $5$ oscillations of the smaller pendulum,they will be in the same phase again.

Two pendulums have time periods $T$ and $5T/4$. They start $SHM$ at the same time from the mean position. After how many oscillations of the smaller pendulum will they be again in the same phase?

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