A boat is travelling in a river with a speed `10 m//s` along the stream flowing with a speed `2 m//s` . From this boat , a sound transmitter is lowered into the river throught a rigid support. The wavelength of the sound emitted from the transmitter inside the water is `14.45 mm` . Aassume that attenuation of sound in water and air is neglisible.
(a) What will be the frequency delected by a receiver kept inside the river downstream ?
(b) The transmitter and the reciver are now pulled up into air. The air is blowing with a speed `5 m//s` in the direction opposite the river stream. Determine the frequency of the sound delected by the reciver.
(Temperature of the air and water = `20^(@) C` , Density of river water = `10^(3) kg//m^(3)` , Bulk modulus of the water = `2.088 xx10 ^(9) Pa` , Gas constant, `R= 8.31 J//mol - K` , Mean molecular mass of air = `28.8xx 10 ^(-3) kg//mol` , `C_(p)//C_(V)` for air = `1.4`

Answer

A, B, C, D

Explanation

Velocity of sound in water is

`upsilon_(omega)= sqrt((B)/(rho))= sqrt((2.088xx10^(9))/(10^(3))`
= `1445 m//s`
Frequency of sound in water will be
`f_(o) = (upsilon omega)/(lambda omega) = (1445)/(14.45xx10^(-3))H_(Z)`
`f_(o) = 10^(5) H_(Z)`
(a) Frequency of sound detected by (observer) at rest would be
`f_(1)= f_(o)((upsilon omega + upsilon_(r))/(upsilon omega + upsilon_(r) - upsilon_(s)))`
= `(10^(5)) ((1445 + 2)/(1445 + 2- 10)) H_(Z)`
`f_(1) = 1.0069 xx 10^(5) H_(Z)`
(b) Velocity of sound in air is
`upsilon _(a)` = `sqrt((gamma RT)/(M))`
Wind speed
`overset(V_(m)=5 m//s)(larr)`
source
`underset(V_(s)= 10 m//s)(rarr)` Observer(At rest)
= `sqrt(((1.4) (8.31) (20 + 273))/ (28.8xx10^(-3))`
= `344 m//s`
Frequency does not depend on the medium. Therefore, frequency in air is also `f_(o) = 10^(5) H_(Z)` .
`:.` Frequency of sound delected by receiver (observer ) in air would be
` f_(2) = f_(o) ((upsilon_(a) - upsilon_(omega))/(upsilon_(a) - upsilon _(omega - upsilon_(s)))) = 10 ^(5) [(344 - 5)/(344 - 5-10)] H_(Z)`
`f_(2) = 1.0304xx 10 ^(5) H_(Z)`