Discussion
Time periods of pendulums A and B are T and \(\frac{5T}{2}\) . If they start executing S.H.M. at the same time from the mean position, the phase difference between them after the bigger pendulum has completed one oscillation is
Answer: E
The phase difference is \(\pi\).
The bigger (longer) pendulum B has period \(\frac{5T}{2}\), so it completes one oscillation at \(t = \frac{5T}{2}\). In that time A completes
\[ \frac{5T/2}{T} = 2.5 \text{ oscillations} \]
B is back at its starting phase (a whole cycle), while A is half a cycle beyond a whole number of cycles. The phase difference is \(0.5 \times 2\pi = \pi\): A is at the mean position moving the opposite way to B. The smaller fractions of \(\pi\) do not match half a cycle.
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