A boat is travelling in a river with a speed `10m//s` along with stream flowing 2m/s. From this boat, a sound transmitter is lowered into the river through a rigid support. The wavelength of the sound emitted from the transmitter inside the water is `14.45mm`. Assume that attenuation of sound in water and air is negligible.
(a) What will be the frequency detected by a receiver kept inside the river downstream?
(b) The transmitter and the receiver are now pulled up into air. the air is blowing with a speed `5m//s` in the direction opposite the river stream. Determine the frequency of the sound detected by the receiver.
(Temperature of the air and water = `20^(@)C`, Density of river water `= 10^(3) kg//m^(3)`,
Bulk modulus of the water `= 2.088 xx 10^(9)`Pa, gas constant `R = 8.31 J//mol-K`,
Mean molecular mass of air `= 28.8 xx 10^(-3) kg//mol`, `C_(P)//C_(V)` for air `= 1.4`)

Answer

(i) `100696 Hz
(ii) `1030338 Hz`

Explanation

Velocity of sound in water is
`V_(w)=sqrt((beta)/(rho))=sqrt((2.088xx10^(9))/(10^(3)))=1445m//s`
Frequency of sound in water will be
`f_(0)=(V_(w))/(lamda_(w))=(1445)/(14.45xx10^(-3))HzrArrf_(0)=10^(5)Hz`
(i) Frequency of sound detected by receiver (observer at rest would be).

`f_(1)=f_(0)((V_(w)+V_(tau))/(V_(w)+V_(tau)-V_(s)))=(10^(5))((1445+2)/(1445+2-10))Hz`
`f_(1)=1.0069xx10^(5)Hz`
(ii) Velocity of sound in air is
`V_(a)=sqrt((gammaRT)/(M))=sqrt(((1.4)(8.31)(20+273))/(28.8xx10^(-3)))=344m//s`
`:.` Frequency does not depend on the medium
Therefore, frequency in air is also `f_(0) = 10^(5) Hz`
`f_(0) = 10^(5) Hz`
`:.` frequency of sound detected by receiver (observer) in air would be
`f_(2)=f_(0)((V_(a)-w)/(V_(a)-w-V_(s)))=10^(6)[(344-5)/(344-5-10)]Hz`
`f_(2)=1.0304 xx10^(5)Hz`.