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Q.

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
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