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 $14.45 mm$. 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 $=10^{3} kg / m^{3}$;
Bulk modulus of the water $=2.088 \times 10^{9} Pa$;
Gas constant $R=8.31 J / mol-K$;
Mean molecular mass of air $=28.8 \times 10^{-3} kg / mol$; $C _p / C _V$ for air $\left.=1.4\right)$
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Answer:
Correct Answer: 24. (a) $1.0069 \times 10^5 \mathrm{~Hz}$ (b) $1.0304 \times 10^5 \mathrm{~Hz}$
Solution:
- Velocity of sound in water is
$$ v _w=\sqrt{\frac{\beta}{\rho}}=\sqrt{\frac{2.088 \times 10^{9}}{10^{3}}}=1445 m / s $$
Frequency of sound in water will be
$$ \begin{aligned} & f _0=\frac{v _w}{\lambda _w}=\frac{1445}{14.45 \times 10^{-3}} Hz \\ & f _0=10^{5} Hz \end{aligned} $$
(a) Frequency of sound detected by receiver (observer) at rest would be
$$ \begin{aligned} & \xrightarrow[f _0 \quad v _s]{\stackrel{\text { Source }}{\longrightarrow}}=10 m / s \quad \begin{array}{c} \text { Observer } \\ \text { (At rest) } \end{array} \\ & \longrightarrow v _r=2 m / s \\ & \begin{array}{l} f _1=f _0\left(\frac{v _w+v _r}{v _w+v _r-v _s}\right)=\left(10^{5}\right)\left(\frac{1445+2}{1445+2-10}\right) Hz \\ f _1=1.0069 \times 10^{5} Hz \end{array} \end{aligned} $$
(b) Velocity of sound in air is
$$ \begin{aligned} v _a & =\sqrt{\frac{\gamma R T}{M}}=\sqrt{\frac{(1.4)(8.31)(20+273)}{28.8 \times 10^{-3}}} \\ & =344 m / s \end{aligned} $$
$\therefore$ Frequency does not depend on the medium.
Therefore, frequency in air is also $f _0=10^{5} Hz$.
$\therefore$ Frequency of sound detected by receiver (observer) in air would be
$$ f _2=f _0\left(\frac{v _a-w}{v _a-w-v _s}\right) $$
$$ \begin{aligned} & =10^{5}\left[\frac{344-5}{344-5-10}\right] Hz \\ f _2 & =1.0304 \times 10^{5} Hz \end{aligned} $$
