SATHEE: Wave Motion 4 Question 26

Wave Motion 4 Question 26

24. 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 through a rigid support. The wavelength of the sound emitted from the transmitter inside the water is $14.45 m m$ . Assume that attenuation of sound in water and air is negligible.

(2001, 10M)

(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 $5 m / s$ in the direction opposite to the river stream. Determine the frequency of the sound detected by the receiver.

(Temperature of the air and water $= 20^{\circ} C$ ; Density of river water $= 10^{3} k g / m^{3}$ ;

Bulk modulus of the water $= 2.088 \times 10^{9} P a$ ;

Gas constant $R = 8.31 J / m o l - K$ ;

Mean molecular mass of air $= 28.8 \times 10^{- 3} k g / m o l$ ; $C_{p} / C_{V}$ for air $= 1.4)$

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Answer:

Correct Answer: 24. (d)

Solution:

Velocity of sound in water is

$v_{w} = \sqrt{\frac{β}{ρ}} = \sqrt{\frac{2.088 \times 10^{9}}{10^{3}}} = 1445 m / s$

Frequency of sound in water will be

$\begin{aligned} f_{0} = \frac{v_{w}}{λ_{w}} = \frac{1445}{14.45 \times 10^{- 3}} H z \\ f_{0} = 10^{5} H z \end{aligned}$

(a) Frequency of sound detected by receiver (observer) at rest would be

$\begin{aligned} \to_{f_{0} v_{s}}^{\overset{Source}{⟶}} = 10 m / s \begin{array}{c} Observer \\ (At rest) \end{array} \\ ⟶ v_{r} = 2 m / s \\ \begin{matrix} f_{1} = f_{0} (\frac{v_{w} + v_{r}}{v_{w} + v_{r} - v_{s}}) = (10^{5}) (\frac{1445 + 2}{1445 + 2 - 10}) H z \\ f_{1} = 1.0069 \times 10^{5} H z \end{matrix} \end{aligned}$