Two pendulums have time period T and 5T/4. They start SHM at the same time from the mean position. What will be the phase difference between them at the moment, when bigger pendulum completes one oscillation?
A
45∘
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B
90∘
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C
60∘
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D
30∘
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Solution
The correct option is B90∘ Bigger pendulum( greater (l)) will have greater time period. Time period of smaller pendulum T1=T Time period of bigger pendulum T2=5T4
They both start at the same time from mean position, after time t, phase difference between them is: Δϕ=ω1t−ω2t=(2πT−2π5T/4)×t when bigger pendulum completes oscillation, t=5T4 ⇒Δϕ=2π5T×5T4 ∴Δϕ=π2 Hence, option (b) is correct