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# A sinusoidal wave of frequency 500 Hz has a speed of 350 m/s. The phase difference between two displacements at a certain point at times 1 m apart is Ans: pi

## 3 Answers

2 years ago
period = 1/f = 1000/500 = 2.0 msec
1 ms = half period
phase shift = half period = 180° = PI radians

Regards
Arun (askIITians forum expert)
2 years ago

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=0.76m0.116m

2) Now in t=103s the wave moves

v×t=350×103=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,

10 months ago

iven 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=0.76m0.116m

2) Now in t=103s the wave moves

v×t=350×103=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,

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