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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`
one year ago

```							period = 1/f = 1000/500 = 2.0 msec 1 ms = half period  phase shift = half period = 180° = PI radians RegardsArun (askIITians forum expert)
```
one year ago
```							Given velocity of the wave v=350m/sfrequency of the wave n=500HzSo wave length of the wave λ=vn=350500m=0.7m1) We are to find the distance between the two points which has 60∘ out of phase i.e the phase difference is ϕ=60∘=π3radAs 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,thenϕ=2πxλx=ϕλ2π=π3×0.72π=0.76m≈0.116m2) Now in t=10−3s the wave movesv×t=350×10−3=0.35mSo here path difference x=0.35mSo by the relation ϕ=2πxλ, the phase difference for x=0.35m becomesϕ=2π×0.350.7=π,
```
one year ago
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