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Question

Two particles are executing S.H.M of the same amplitude of 20 cm with the same period along the same line about the same equilibrium position. The maximum distance between the two is 20 cm. Their phase difference (in radians) is equal to

A
π3
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B
π2
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C
2π3
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D
4π5
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Solution

The correct option is A π3
Given that
Amplitude of both the particles, A=20 cm
Let the equation of motion of the two particles be
x1=Asin(ωt+ϕ1)
x2=Asin(ωt+ϕ2)
The distance between the two particles is
x1x2=Asin(ωt+ϕ1)Asin(ωt+ϕ2)
Maximum distance between particles =20 cm [given]
20=2×20sin(ϕ1ϕ22).cos[ωt+(ϕ1+ϕ22)]
12=sin(ϕ1ϕ22).cos(ωt+(ϕ1+ϕ22)) ...(i)
Equation (i) will attain maximum value when the term cos(ωt+(ϕ1+ϕ22)) becomes 1
Then, phase difference sin1(12)=(ϕ1ϕ22)
ϕ1ϕ22=π6ϕ1ϕ2=π3

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