
A whistle revolves in a circle with angular speed $\omega = 20\,rad/s$ using a string of length $50\,cm$. If the frequency of sound from the vessel is $385\,Hz$, then what is the minimum frequency heard by an observer which is far away from the centre? ( ${V_{sound}} = 340\,m/s$ )
A) $385\,Hz$
B) $374\,Hz$
C) $394\,Hz$
D) $333\,Hz$
Answer
216k+ views
Hint: We can use the Doppler effect of sound to solve this question. Whenever there is a relative motion between the observer and the source, we need to use the Doppler effect. Frequency heard will be minimum when the source is moving away from the observer.
Complete step by step solution:
It is given that a whistle is revolving in a circle with the help of a string.
The angular speed with which the whistle revolves is given as
$\omega = 20\,rad/s$
The length of the string is given as
$l = 50\,cm$
$ \Rightarrow l = 0.5\,m$
The original frequency of the whistle is
$f = 385\,Hz$
We need to find the minimum frequency heard by an observer.
We know that the Frequency heard will be minimum when the source is moving away from the observer.
Whenever there is a relative motion between the observer and the source of sound, we need to use the Doppler effect.
According to Doppler effect the frequency that is heard is given as
$f' = f\left( {\dfrac{{{v_{sound}} + {v_o}}}{{{v_{sound}} - {v_s}}}} \right)$
$f$ is the original frequency or the source frequency, ${v_{sound}}$ is the velocity of sound in the medium, ${v_o}$ is the velocity of the observer, ${v_s}$ is the velocity of the source.
Now let us find the velocity of the source.
The whistle is revolving in a circle with the length of a string as the radius.
We know that the angular velocity is the ratio of linear velocity to radius of the circular path.
$\omega = \dfrac{v}{R}$
Where, $\omega $ is angular velocity, $v$ is the linear velocity and R is the radius of the circular path.
From this we can find the linear velocity as
$v = \omega R$
Thus, the velocity of source can be written as
$ \Rightarrow {v_s} = \omega l$
Since l is the radius.
On substituting the given values, we get
$ \Rightarrow {v_s} = 20 \times 0.5\,\,m/s$
$ \Rightarrow {v_s} = 10\,\,m/s$
The observer is standing still so velocity of observer is
${v_o} = 0$
Velocity of sound is given as
${V_{sound}} = 340\,m/s$
Now let us substitute all the values in the equation for finding the frequency that is heard.
$f' = f\left( {\dfrac{{{v_{sound}} + {v_o}}}{{{v_{sound}} - {v_s}}}} \right)$
We need to take minus sign for the velocity of source because minimum frequency is heard when the source is moving away from the observer
$ \Rightarrow f' = 385\left( {\dfrac{{340 + 0}}{{340 - \left( { - 10} \right)}}} \right)$
$ \Rightarrow f' = 385\left( {\dfrac{{340}}{{340 + 10}}} \right)$
$ \Rightarrow f' = 374\,Hz$
This is the value of minimum frequency which is heard by an observer which is far from the centre.
So, the correct answer is option (B).
Note: Take care of the sign used while substituting the values for the velocity of observer and source in the equation of Doppler effect. We are asked to find the minimum frequency that is heard. Frequency will be minimum if the source is moving away from the observer. When the source is moving away from the observer, we need to take a negative sign and when the source is moving towards the observer, we need to take a positive sign.
Complete step by step solution:
It is given that a whistle is revolving in a circle with the help of a string.
The angular speed with which the whistle revolves is given as
$\omega = 20\,rad/s$
The length of the string is given as
$l = 50\,cm$
$ \Rightarrow l = 0.5\,m$
The original frequency of the whistle is
$f = 385\,Hz$
We need to find the minimum frequency heard by an observer.
We know that the Frequency heard will be minimum when the source is moving away from the observer.
Whenever there is a relative motion between the observer and the source of sound, we need to use the Doppler effect.
According to Doppler effect the frequency that is heard is given as
$f' = f\left( {\dfrac{{{v_{sound}} + {v_o}}}{{{v_{sound}} - {v_s}}}} \right)$
$f$ is the original frequency or the source frequency, ${v_{sound}}$ is the velocity of sound in the medium, ${v_o}$ is the velocity of the observer, ${v_s}$ is the velocity of the source.
Now let us find the velocity of the source.
The whistle is revolving in a circle with the length of a string as the radius.
We know that the angular velocity is the ratio of linear velocity to radius of the circular path.
$\omega = \dfrac{v}{R}$
Where, $\omega $ is angular velocity, $v$ is the linear velocity and R is the radius of the circular path.
From this we can find the linear velocity as
$v = \omega R$
Thus, the velocity of source can be written as
$ \Rightarrow {v_s} = \omega l$
Since l is the radius.
On substituting the given values, we get
$ \Rightarrow {v_s} = 20 \times 0.5\,\,m/s$
$ \Rightarrow {v_s} = 10\,\,m/s$
The observer is standing still so velocity of observer is
${v_o} = 0$
Velocity of sound is given as
${V_{sound}} = 340\,m/s$
Now let us substitute all the values in the equation for finding the frequency that is heard.
$f' = f\left( {\dfrac{{{v_{sound}} + {v_o}}}{{{v_{sound}} - {v_s}}}} \right)$
We need to take minus sign for the velocity of source because minimum frequency is heard when the source is moving away from the observer
$ \Rightarrow f' = 385\left( {\dfrac{{340 + 0}}{{340 - \left( { - 10} \right)}}} \right)$
$ \Rightarrow f' = 385\left( {\dfrac{{340}}{{340 + 10}}} \right)$
$ \Rightarrow f' = 374\,Hz$
This is the value of minimum frequency which is heard by an observer which is far from the centre.
So, the correct answer is option (B).
Note: Take care of the sign used while substituting the values for the velocity of observer and source in the equation of Doppler effect. We are asked to find the minimum frequency that is heard. Frequency will be minimum if the source is moving away from the observer. When the source is moving away from the observer, we need to take a negative sign and when the source is moving towards the observer, we need to take a positive sign.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

