Two sources ${S_1}$ and ${S_2}$ of same frequency $f$ emit sound. The sources are moving as shown in figure 1 with speed $u$each. A stationary observer hears that sound. The beat frequency is? ( $v = 0$ velocity of sound)

A. $\dfrac{{2{v^2}f}}{{{v^2} - {u^2}}}$
B. $\dfrac{{2u}}{v}f$
C. $\dfrac{{2{u^2}f}}{{{v^2} - {u^2}}}$
D. $\dfrac{{2uvf}}{{{v^2} - {u^2}}}$
Answer
289.8k+ views
Hint: In the case of a problem based on wave phenomena, we know that all the parameters such as i.e., frequency, amplitude, time period, etc., vary with each other in some way hence, analyze every option given in this question and use mathematical calculation by finding the apparent frequency of both the sources with the information given.
Formula used:
The expression of apparent frequency (${f_1}$) of source ${S_1}$ is,
${f_1} = f\left( {\dfrac{{v - 0}}{{v - u}}} \right) = \dfrac{v}{{v - u}}f$
Here, $v$ is the velocity of sound and $u$ is the velocity of source.
Complete step by step solution:
Two Sources ${S_1}$ and ${S_2}$ of the same frequency $f$ emit sound (given) and these sources are moving in the same direction as shown in figure 2. The velocity of the two sources is again same i.e., $u$(given). And, velocity of sound $v = 0$ (given). As the observer is stationary, therefore, the direction of sound waves must be opposite in direction as shown in figure 2.

Now, Apparent frequency of source ${S_1}$ will be:
${f_1} = f\left( {\dfrac{{v - 0}}{{v - u}}} \right) = \dfrac{v}{{v - u}}f$ … (1)
And, Apparent frequency of source ${S_2}$ will be:
${f_2} = f\left( {\dfrac{{v - 0}}{{v + u}}} \right) = \dfrac{v}{{v + u}}f$ … (2)
We know that Beat Frequency $f'$ can be calculated as:
$f' = {f_1} - {f_2}$
Substitute the values of ${f_1}$ and ${f_2}$ in the above expression, we get
$f' = \dfrac{v}{{v - u}}f - \dfrac{v}{{v + u}}f$
Taking $vf$common, we get
$f' = vf\left( {\dfrac{{v + u - (v - u)}}{{(v - u)(v + u)}}} \right)$
$\Rightarrow f' = vf\left( {\dfrac{{2u}}{{{v^2} - {u^2}}}} \right)$........(Because, ${a^2} - {b^2} = (a - b)(a + b)$)
Thus, according to the given question, the beat frequency for the two sources is $\dfrac{{2uvf}}{{{v^2} - {u^2}}}$.
Hence, the correct option is D.
Note: Since this is a conceptual-based problem on wave and acoustics hence, it is essential that given sources and their conditions must be analyzed very carefully to give a precise explanation of the solution. While writing an answer always remember to use the mathematical proven relations to provide an accurate solution.
Formula used:
The expression of apparent frequency (${f_1}$) of source ${S_1}$ is,
${f_1} = f\left( {\dfrac{{v - 0}}{{v - u}}} \right) = \dfrac{v}{{v - u}}f$
Here, $v$ is the velocity of sound and $u$ is the velocity of source.
Complete step by step solution:
Two Sources ${S_1}$ and ${S_2}$ of the same frequency $f$ emit sound (given) and these sources are moving in the same direction as shown in figure 2. The velocity of the two sources is again same i.e., $u$(given). And, velocity of sound $v = 0$ (given). As the observer is stationary, therefore, the direction of sound waves must be opposite in direction as shown in figure 2.

Now, Apparent frequency of source ${S_1}$ will be:
${f_1} = f\left( {\dfrac{{v - 0}}{{v - u}}} \right) = \dfrac{v}{{v - u}}f$ … (1)
And, Apparent frequency of source ${S_2}$ will be:
${f_2} = f\left( {\dfrac{{v - 0}}{{v + u}}} \right) = \dfrac{v}{{v + u}}f$ … (2)
We know that Beat Frequency $f'$ can be calculated as:
$f' = {f_1} - {f_2}$
Substitute the values of ${f_1}$ and ${f_2}$ in the above expression, we get
$f' = \dfrac{v}{{v - u}}f - \dfrac{v}{{v + u}}f$
Taking $vf$common, we get
$f' = vf\left( {\dfrac{{v + u - (v - u)}}{{(v - u)(v + u)}}} \right)$
$\Rightarrow f' = vf\left( {\dfrac{{2u}}{{{v^2} - {u^2}}}} \right)$........(Because, ${a^2} - {b^2} = (a - b)(a + b)$)
Thus, according to the given question, the beat frequency for the two sources is $\dfrac{{2uvf}}{{{v^2} - {u^2}}}$.
Hence, the correct option is D.
Note: Since this is a conceptual-based problem on wave and acoustics hence, it is essential that given sources and their conditions must be analyzed very carefully to give a precise explanation of the solution. While writing an answer always remember to use the mathematical proven relations to provide an accurate solution.
Recently Updated Pages
Environmental Chemistry Chapter for JEE Main Chemistry

Chemical Bonding and Molecular Structure Chapter for JEE Main Chemistry

Chelating Ligand, Ambidentate Ligand, and Flexidentate Ligand for JEE Exam

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Maths Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 2) Chemistry Question Paper with Answer Key

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2026-27 Free PDF Download (Login Required)

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27 Free PDF Download (Sign-in Required)

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Notes Class 11 Physics Chapter 2 - Motion in a Straight Line - 2026-27 PDF Download (Login Required)

