Saved Bookmarks
| 1. |
The minimum distance between two particle in same phase is 18 cm . If the velocity of the wave is 36m/s then find the time interval after which the given particle undergoes a phase change of pie |
| Answer» ELOCITY 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 ,thenϕ=2πxλ x=ϕλ2π=π3×0.72π=0.76m≈0.116m 2) Now in t=10−3s the wave movesv×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 becomesϕ=2π×0.350.7=π ,I hope this will help youIf not then COMMENT me | |