Question

A boat is travelling in a river with a speed `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 1

Correct Answer - A::B::C::D
Velocity of sound in water is
`nu_(w) = sqrt((B)/(rho)) = sqrt((2.088 xx 10^(9))/(10^(3))) = 1445 m//s`
Frequency of sound in water will be
`f_(0) = nu_(w)/(lambda_(w)) = (1445)/(14.45 xx 10^(-3))H_(Z)`
`f_(0) = 10^(5) H_(Z)`
(a) Frequency of sound detected by reciver (observer) at rest would be

`f_(1) = f_(0) ((nu_(w) + nu _(r))/(nu_(w) + nu_(r) - nu_(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
Air speed
`overset(V_(a) = s m//s)(larr)`
`overset (Source) underset(V_(s) = 10 m//s)(rarr) Observer at (rest)`
`nu_(a) = sqrt((gamma RT)/(M)) = sqrt(((1.4)(8.31)(20 + 273))/(28.8 xx 10^(-3))`
`= 344 m//s`
Frequency does not depend on the medium. Therefore, frequency in air is also `f_(0) = 10^(5) H_(Z)`.
`:.` Frequency of sound detected by reciver (observer) in air would be
`f_(2) = f_(0) ((nu_(a) - nu _(w))/(nu_(a) - nu_(w) - nu_(s)))`
`= 10^(5) [(344 - 5)/(344 - 5 - 10)] H_(Z)`
`f_(2) = 1.0304 xx 10^(5) H_(Z)`