Question

Two submarines are approaching each other in a calm sea. The first submarine travels at a speed of 36 km h⁻¹ and the other at 54 km h⁻¹ relative to the water. The first submarine sends a sound signal (sound waves in water are also called sonar) at a frequency of 2000 Hz. (a) At what frequency is this signal received from the second submarine. At what frequency is this signal received by the first submarine. Take the speed of of the sound wave in water to be 1500 m s⁻¹.

Answer 1

Given:
Velocity of water v = 1500 m/s
Frequency of sound signal $f_0$ = 2000 Hz
Velocity of first submarine v_s = 36 kmh⁻¹ = $36\times \frac{5}{18}m/s$ = 10 m/s
Velocity of second submarine $v_0$ = 54 km h⁻¹ = $54\times \frac{5}{18}$ m/s = 15 m/s
Frequency received by the first submarine $\left(f_1\right)$ is given by:

$f_1=\left(\frac{v+v_0}{v-v_{\mathrm{s}}}\right)f_0$

On substituting the values, we get:

$$\begin{gathered} f_1=\left(\frac{1500+15}{1500-10}\right)\times \left(2000\right) \\ =2034\mathrm{Hz} \end{gathered}$$

(b) Here, $f_0=2034\mathrm{Hz}$.

Apparent frequency received by second submarine $\left(f_2\right)$ is given by:

$$\begin{gathered} f_2=\left(\frac{1500+10}{1500-15}\right)\times \left(2034\right) \\ =2068\mathrm{Hz} \end{gathered}$$