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A receiver and a source of sonic oscillations of frequency v_(0)=2000Hz are located on the x axid. The source swings harmonically along that axis with a circulat frequency bandwidth registered by the stationary receiver be equal to Delta v=200 Hz ? The velocity of sound is equal to v=340 m//s. |
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Answer» Solution :Obviosuly the maximum FREQUENCY will be HEARD when the source is moving with maximum velocity towards the receiver and minimum frequency will be ehard when the source recedes with maximum velocity. As the source swing harmonically its maximum velocity equals `a OMEGA`. Hence `v_(max)=v_(0)(v)/( v- a omega)` and `v_(min )=v_(0)(v)/(v+a omega)` So the frequency band width `DELTA v = v_(max) - v _(min)= v_(0) v ((2 a omega)/( v^(2)-a^(2) omega^(2)))` or, `(Deltava^(2))omega^(2)+(2 v_(0)v a ) omega - Delta v v ^(2) = 0 ` so, `omega=(-2 v_(0)v a +- sqrt(4 v _(0)^(2) v ^(2) a ^(2) + Delta v^(2) a ^(2) v^(2)))/( 2 Delta v a ^(2))` On simplifying ( and taking `+` SIGN as `omega rarr0` if `Delta v rarr0)` `omega = ( v v_(0))/( Deltava)(sqrt(1+ ((Deltav)/( v_(0)))^(2))-1)` |
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