Question

Two sources A and B are sounding notes of frequency 680Hz. A listener moves from A to B with a constant velocity $u$ . If the speed of the sound is $340{\text{m}}{{\text{s}}^{ - 1}}$ , then find the value of $u$ so that he hears 10 beats per second.
A) $2{\text{m}}{{\text{s}}^{ - 1}}$
B) $2 \cdot 5{\text{m}}{{\text{s}}^{ - 1}}$
C) ${\text{3m}}{{\text{s}}^{ - 1}}$
D) ${\text{3}} \cdot {\text{5m}}{{\text{s}}^{ - 1}}$

Answer 1

As the listener moves from source A to source B, the frequency heard by the listener from both the sources will be different. This change in frequency due to the motion of the listener is referred to as the doppler effect. Generally, the frequency change can also be due to the motion of the source or due to the motion of both the observer and the source. The given beat frequency refers to the difference between the two apparent frequencies heard by the listener.

Formula used:
The apparent frequency heard by the moving observer is given by, ${f_{apparent}} = f\left( {\dfrac{{v \pm {v_o}}}{v}} \right)$ where $f$ is the frequency of the sound wave produced by the stationary source, $v$ is the velocity of sound and ${v_o}$ is the velocity of the observer.

Complete step by step solution:
Step 1: List the known parameters involved in the given problem.
The frequency of the sound wave produced by both sources A and B is given to be ${f_A} = {f_B} = f = 680{\text{Hz}}$ .
The velocity of the sound wave in air is given to be $v = 340{\text{m}}{{\text{s}}^{ - 1}}$ .
The velocity of the observer is given to be $u$ .
The beat frequency is given to be $\Delta f = 10{\text{bps}}$ .
Let ${f_1}$ and ${f_2}$ be the apparent frequencies heard by the listener from sources A and B respectively.

Step 2: Express the relation for the apparent frequencies heard by the moving observer.
The apparent frequency heard by the observer moving from source A can be expressed as
${f_1} = f\left( {\dfrac{{v - u}}{v}} \right)$ ------- (1)
Similarly, the apparent frequency heard by the observer moving to source B can be expressed as ${f_2} = f\left( {\dfrac{{v + u}}{v}} \right)$ ------- (2)

Step 3: Express the beat frequency heard by the observer.
The beat frequency heard by the observer is the difference between the apparent frequencies ${f_1}$ and ${f_2}$ .
i.e., $\Delta f = {f_2} - {f_1}$ ------- (3)
Substituting equations (1) and (2) in (3) we get, $\Delta f = f\left( {\dfrac{{v + u}}{v}} \right) - f\left( {\dfrac{{v - u}}{v}} \right)$
$\Rightarrow \Delta f = \dfrac{f}{v}\left( {v + u - v\left( { - u} \right)} \right) = \dfrac{{2uf}}{v}$
$\Rightarrow u = \dfrac{{v\Delta f}}{{2f}}$ -------- (4)
Equation (4) gives the constant velocity of the listener.
Substituting for $\Delta f = 10{\text{bps}}$ , $f = 680{\text{Hz}}$ and $v = 340{\text{m}}{{\text{s}}^{ - 1}}$ in equation (4) we get, $u = \dfrac{{340 \times 10}}{{2 \times 680}} = 2 \cdot 5{\text{m}}{{\text{s}}^{ - 1}}$
$\therefore$ the velocity of the listener is obtained to be $u = 2 \cdot 5{\text{m}}{{\text{s}}^{ - 1}}$ .

So the correct option is B.

Note: The listener is mentioned to be moving from source A to source B. This means that it moves away from source A and moves towards source B. The velocity of the listener will be positive if he moves towards the source and his velocity will be taken to be negative if he moves away from the source. So the velocity of the observer in equation (1) is expressed as $- u$ while that in equation (2) is expressed as $+ u$. Conversion of the unit of frequency from beats per second to Hertz is not needed as $1{\text{bps}} = 1{\text{Hz}}$ .