Question

A SONAR system fixed in a submarine operates at a frequency $40.0 \mathrm{kHz}$. An enemy submarine moves towards the SONAR with a speed of $360 \mathrm{~km} \mathrm{~h}^{-1}$. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be $1450 \mathrm{~m} \mathrm{~s}^{-1}$

Answer 1

Sonar travels at the same rate as sound. This varies with salinity, temperature, and pressure underwater, while it varies with temperature and humidity when traveling through the air. Write down the given data in the question Calculate the apparent frequency of the sonar system similarly calculate the apparent frequency of the enemy sonar system and equate both the equations.

Complete step-by-step solution:
Operating frequency of the SONAR system, $\mathbf{v}=40 \mathrm{kHz}$
Speed of the enemy submarine,
$v_{\mathrm{e}}=360 \mathrm{~km} / \mathrm{h}$
$=100 \mathrm{~m} / \mathrm{s}$
Speed of sound in water,
$v=1450 \mathrm{~m} / \mathrm{s}$
The source is at rest and the observer (enemy submarine) is moving toward it. Hence, the apparent frequency $(v)$ received and reflected by the submarine is given by the relation,
Frequency of the SONAR system,
$v=40 \mathrm{kHz}=40 \times 10^{3} \mathrm{~Hz}$
The energy submarine first acts as an observer which is approaching the source (SONAR) with a speed of:
$v_{0}=360 \mathrm{~km} / \mathrm{hr}=360 \times \dfrac{5}{18} \mathrm{~m} / \mathrm{s}$
$=100 \mathrm{~m} / \mathrm{s}$
Apparent frequency of sound waves is,
$v^{\prime}=\left(\dfrac{v+v_{0}}{v}\right) v$
$=\dfrac{1450+100}{1450} \times 4 \times 10^{4}$
This frequency is the frequency of sound received by submarines.
Now, the enemy submarine will reflect the waves of frequency $v$ ' and thus acts as a source which is moving towards the observer (SONAR) with a speed given by,
$v_{s}=100 \mathrm{~m} / \mathrm{s}$
Therefore,
Apparent frequency is given by,
$v_{a}^{\prime}=\left(\dfrac{v}{v-v_{s}}\right) \times v^{\prime}$
$=\dfrac{1450}{1450-100} \times \dfrac{1450+100}{1450} \times 4 \times 10^{4}$
$=\dfrac{1550}{1350} \times 4 \times 10^{4}$
$=45.92 \times 10^{3} \mathrm{~Hz}$
$=45.92 \mathrm{KHz}$

Note: Typically, sonar arrays are designed to record sounds in specific frequency ranges. Sounds that are higher in frequency than the array's intended range may cause the system to become confused; it may be able to detect the presence of an important contact but unable to locate it. The information contained in sound waves emitted by or reflected from the object is detected by sonar apparatus and analyzed.