Question

A boat is travelling in a river with a speed of 10m/s along the stream flowing 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 downstream?
b. The transmitter and the receiver are now pulled up into air. The air is blowing with a speed of 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 is ${20}^{\circ }C$; density of river water is ${10}^{3}kg/m^3$ ; bulk modulus of the water is $2.088\times {10}^9$ Pa, gas constant R = 8.31J/mol – K; mean molecular mass of air is for air is $28.8\times {10}^{-3}kg/mol:\frac{c_p}{c_v}$ for air is 1.4.)

• A. 100545 Hz, 103646 Hz
• B. 100100 Hz, 103103 Hz
• C. 100696 Hz, 103040 Hz
• D. 100425 Hz, 103954 Hz

Answer 1

The correct option is C 100696 Hz, 103040 Hz



a) Speed of boat $\mu =10m/s$
Speed of river stream w=2 m/s
Wavelength of sound wave inside water is 14.45 mm.
Speed of sound inside water is $v=\sqrt{\frac{B}{\rho }}=\sqrt{\left[\frac{2.088\times {10}^9}{{10}^3}\right]}$
$=\sqrt{2.088}\times {10}^{3}m/s=1445m/s$
Frequency of wave inside water is
$n=\frac{v}{\lambda }=\frac{1445}{14.45\times {10}^{-3}}={10}^{5}Hz$
In this case source and medium both are in motion.
Apparent frequency is given by
$n^{\prime }=\frac{(v+w)}{(v+w)-\mu }n=\frac{(1445+2)}{(1445+2)-10}\times {10}^5=\frac{1447}{1437}\times {10}^{5}Hz=1.00696\times {10}^{5}Hz=100696Hz$

b) Speed of sound air
$v=\sqrt{\frac{\gamma RT}{M}}=\sqrt{\frac{1.4\times 8.3\times 293}{28.8\times {10}^{-3}}}=344m/s$
Speed of air w=5m/s
$n^{\prime }=\frac{(v-w)}{(v-w)-\mu }n=\frac{344-5}{344-5-10}\times {10}^5=\frac{339}{329}\times {10}^{5}Hz=1.03040\times {10}^{5}Hz=103040Hz$