Question

A boat is travelling in a river with a speed of $10\text{m/s}$ along the stream flowing with a speed of $2\text{m/s}$. 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 $14.45\text{mm}$. Assume that attenuation of sound in water and air is negligible. What will be the frequency detected by a receiver kept underwater downstream? [Given: Bulk modulus of water $=2.088\times {10}^9\text{Pa}$]

• A. $100696\text{Hz}$
• B. $236006\text{Hz}$
• C. $200696\text{Hz}$
• D. $54000\text{Hz}$

Answer 1

The correct option is A $100696\text{Hz}$


Given, speed of boat, $u=10\text{m/s}$
Speed of river stream, $w=2\text{m/s}$

Wavelength of sound wave inside water is $14.45\text{mm}$. Speed of sound inside water is given by

$v=\sqrt{\frac{K}{\rho }}$

Substituting the values of $K$ we get,

$v=\sqrt{\frac{2.088\times {10}^9}{1000}}=1445\text{m/s}$

Frequency of wave inside water is $n=\frac{v}{\lambda }$

Substituting the values of $v$ and $\lambda$

$n=\frac{1445}{14.45\times {10}^{-3}}={10}^5\text{Hz}$

In this case, source and medium both are in motion
Apparent frequency is given by

$n^{\prime }=\left[\frac{(v+w)}{(v+w)-u}\right]$

[since the receiver is kept downstream, source is approaching it]

Substituting the values in the above equation,

$n^{\prime }=\frac{(1445+2)}{(1445+2)-10}\times {10}^5$

$\Rightarrow n^{\prime }=\frac{1447}{1437}\times {10}^5\text{Hz}=1.00696\times {10}^5\text{Hz}$

$\therefore n^{\prime }=100696\text{Hz}$

Hence, the option (a) is correct answer.