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A source of soud is moving along a circular orhit of dins 3 m with an angular velocity of 10 rad/s. sound detector located far away from the source is executing linear simple harmonic motion along the line BD with amplitude BC = CD = 6m. The frequency oscillation of the detector is (5//pi) rev/sec. The source is at the point A when the detector is at the point B. If the source emits a continuous sound wave of frequency 340 H, find the maximum and the minimum frequencies recorded by the detector velocity of sound :: 330 m/s) |
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Answer» Solution :Time period of circular motion `T=(2pi// OMEGA)=(2pi//10)` is same as that of SHM. i.e., `T=(1//f)=(pi//5)`, so both will COMPLETE one periodic motion in same time. Furthermore source is moving on a circle, its speed `V_(s)=r omega=3xx10=30m//s` and as detector is executing SHM. `V_(D)=omegasqrt(A^(2)-y^(2))=10 sqrt(6^(2)-y^(2))` i.e, `(V_(D))_(max)=60m//s` when y=0 So `f_(AP)` WIL be maximum when both move towards each other. `f_(max)=f|(V+V_(D))/(V-V_(S))|` with `V_(D)`= max i.e., the source is at M and detector at C and moving towards B, so `f_(max)=340[(330+60)/(330-30)]=442Hz` Similarly `f_(AP)` will be minimum when both are moving away from each other, i.e `f_(max)=f[(V-V_(D))/(V+V_(s))]` with `V_(D)` = max i.,e the source is at Nadn detector at C but moving towards D, `so f_(MIN) =340[(330+60)/(330+30)]=225Hz` |
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