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Given velocity of the wave v=350m/s
frequency of the wave n=500Hz
So wave length of the wave λ=vn=350500m=0.7m
1) We are to find the distance between the two points which has 60∘ out of phase i.e the phase difference is ϕ=60∘=π3rad
As we know that for path difference λ there is phase difference 2π,
we can say , if ϕ is the phase difference for path difference x,then
ϕ=2πxλ
x=ϕλ2π=π3×0.72π=0.76m≈0.116m
2) Now in t=10−3s the wave moves
v×t=350×10−3=0.35m
So here path difference x=0.35m
So by the relation ϕ=2πxλ, the phase difference for x=0.35m becomes
ϕ=2π×0.350.7=π,
iven velocity of the wave v=350m/s
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