Two particles P and Q describe SHM of same amplitude A, same frequency f along the same straight line. The maximum distance between the two particles is A√2. The phase difference between the particle is
A
Zero
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B
π2
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C
π6
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D
π3
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Solution
The correct option is Bπ2 Equation of motion of particle 1, x1=Asin(ωt+ϕ1) Equation of motion of particle 2, x2=Asin(ωt+ϕ2) At a particular instant, separation between two particles is given by, |x1−x2|=A[sin(ωt+ϕ1)−sin(ωt+ϕ2)] ∵[sinC−sinD=2cos(C+D2)sin(C−D2)] |x1−x2|=2Acos(ωt+ϕ1+ϕ22)sin(ϕ1−ϕ22) Given maximum distance between two particle is A√2 ∴ To maximize |x1−x2|, we choose cos(ωt+ϕ1+ϕ22)=1 ⇒A√2=2Asin(ϕ1−ϕ22) ⇒sin(ϕ1−ϕ22)=1√2 ⇒sin(ϕ1−ϕ22)=sin(π4) ⇒ϕ1−ϕ22=π4 ⇒ϕ1−ϕ2=π2