Two pendulums X and Y of time periods 4s and 4.2s are made to vibrate simultaneously. They are initially in same phase. After how many vibrations of X, they will be in the same phase again?
A
30
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B
25
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C
21
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D
26
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Solution
The correct option is C21 ω1t=ω2t+2π(θ1=θ2+2π) ∴t=2πω1−ω2=2π(2π/T1)−(2π/T2) t=T1T2T2−T1 t=4×4.24.2−4=84 Number of vibration of X in this time are, N1=tT1=844=21