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# A SONAR inside the sea works at $40{\text{kHz}}$. A submarine is approaching it with a velocity of $360{\text{km}}{{\text{h}}^{ - 1}}$. If the speed of sound in water is $1450{\text{m}}{{\text{s}}^{ - 1}}$, then find the apparent frequency of waves after reflection from the submarine.A) $11 \cdot 5{\text{kHz}}$B) $36 \cdot 8{\text{kHz}}$C) $42 \cdot 9{\text{kHz}}$D) $98 \cdot 6{\text{kHz}}$

Last updated date: 21st Jun 2024
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Hint:The given situation can be examined in two cases. In the first case, the SONAR acts as the source of the sound wave and the approaching submarine is the observer. As the submarine is moving towards the SONAR, the frequency heard by the submarine will be different. This concept of change in frequency related to motion is referred to as the Doppler effect. Now when the sound waves get reflected from the submarine, the submarine becomes the source and the stationary SONAR becomes the observer of the reflected sound waves.

Formulas used:
-The frequency of the sound waves, produced by a stationary source, as heard by a moving observer is given by, $f = {f_s}\left( {1 + \dfrac{{{v_o}}}{v}} \right)$ where ${f_s}$ is the frequency of the sound wave produced by the source, $v$ is the velocity of sound in air and ${v_o}$ is the velocity of the observer.
-The frequency of the sound wave, produced by a moving source, as heard by a stationary observer is given by, $f = {f_s}\left( {1 + \dfrac{{{v_s}}}{v}} \right)$ where ${f_s}$ is the frequency of the sound wave produced by the source, $v$ is the velocity of sound in air and ${v_s}$ is the velocity of the source.

Complete step by step solution:
Step 1: List the given parameters.
The frequency of the sound produced by the SONAR inside the sea is given to be ${f_{sonar}} = 40{\text{kHz}}$ .
The SONAR is stationary and so ${v_{sonar}} = 0$ .
The velocity of the submarine is given to be ${v_{sub}} = 360{\text{km}}{{\text{h}}^{ - 1}}$ .
The velocity of sound in water is given to be $v = 1450{\text{m}}{{\text{s}}^{ - 1}}$.
We consider two cases. Case 1 is when the SONAR is the stationary source and the submarine is the observer of the sound wave. Case 2 is when the submarine is the moving source and SONAR is the stationary observer for the reflected sound waves.

Step 2: Express the relation for the apparent frequency in case 1.
Case 1: Source is SONAR and observer is submarine.
The velocity of the observer is ${v_{sub}} = {v_o} = 360{\text{km}}{{\text{h}}^{ - 1}}$ and the frequency of the sound wave produced by the source is ${f_{sonar}} = {f_s} = 40{\text{kHz}}$.
Then the apparent frequency of the wave heard by the observer (submarine) is given by, $f = {f_s}\left( {1 + \dfrac{{{v_o}}}{v}} \right)$ -------- (1)
Substituting for ${v_o} = 100{\text{m}}{{\text{s}}^{ - 1}}$, $v = 1450{\text{m}}{{\text{s}}^{ - 1}}$ and ${f_s} = 40{\text{kHz}}$ in equation (1) we get, $f = 40 \times {10^3}\left( {1 + \dfrac{{100}}{{1450}}} \right) = 42758 \cdot 62{\text{kHz}}$
Thus the apparent frequency heard by the submarine is $f = 42 \cdot 76{\text{kHz}}$ .

Step 3: Express the relation for the apparent frequency in case 2.
Case 2: Source is submarine and observer is SONAR.
The velocity of the source is ${v_{sub}} = {v_s} = 360{\text{km}}{{\text{h}}^{ - 1}}$ and the frequency of the sound wave produced by the source (reflected sound wave) is $f = {f_s}^\prime = 42 \cdot 76{\text{kHz}}$.
Then the apparent frequency of the wave heard by the observer (SONAR) is given by, $f' = {f_s}^\prime \left( {1 + \dfrac{{{v_s}}}{v}} \right)$ -------- (2)
Substituting for ${v_s} = 100{\text{m}}{{\text{s}}^{ - 1}}$, $v = 1450{\text{m}}{{\text{s}}^{ - 1}}$ and ${f_s}^\prime = 42 \cdot 76{\text{kHz}}$ in equation (2) we get, $f' = 42 \cdot 76 \times {10^3}\left( {1 + \dfrac{{100}}{{1450}}} \right) = 45708 \cdot 96{\text{kHz}}$
Thus the apparent frequency of the reflected sound wave is $f' = 45 \cdot 71{\text{kHz}} \approx {\text{42}} \cdot {\text{9kHz}}$ .

So the closest option is C.

Note:While substituting values of physical quantities in any equation, make sure that all the quantities are expressed in their respective S.I. units. Here, the velocity of the submarine is given as ${v_{sub}} = 360{\text{km}}{{\text{h}}^{ - 1}}$ so we convert it into its S.I. unit as ${v_{sub}} = 360 \times \dfrac{5}{{18}} = 100{\text{m}}{{\text{s}}^{ - 1}}$ before substituting in equations (1) and (2).