Two simple pendulums of lengths 1 m and 16 m respectively are both given small displacements in the same direction at the same instant. They will again be in phase after the shorter pendulum has completed n oscillations where n is
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a
1/ 4
b
4 /3
c
4
d
5
answer is B.
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Detailed Solution
T1=2π1g and T2=2π16g ∴ T2T1=4 or v2v2=4 Now v1v2−1=4−1 or v1−v2v2=3 The two pendulums will swings together again if v1−v2=1 or 1v2=3 or v2=13 ∴ v1=1+v2=1=13+43