Doppler Effect: Definition, Formula, Uses and Solved Questions

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We often come across an instance where the police use radar to detect a speeding vehicle or a siren on an ambulance or a fire engine. As the fast moving siren passes by us, the pitch of the siren abruptly drops. It is the doppler effect. Doppler effect is referred to as the change in wave frequency during the relative motion between a wave source and its observer.

The topic Doppler Effect is covered in the Unit 10 Chapter 15 i.e, Waves of CBSE Class 11 Physics. The whole unit 10 (Oscillations and Waves) will carry a weightage of 23 periods and students can expect around 5 marks from this topic in the CBSE class 11 examination.

Table of Content

Define Doppler Effect
Use of the Doppler Effect
The Disadvantage of Doppler Effect
Doppler Effect Formula
Sample Questions

Define Doppler Effect

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The fluctuation in the intensity of sound because of distance is known as the doppler effect. The decrease in the sound of the loudspeaker when a person moves away and the gradual increase in the sound when he comes near is an example of the doppler effect. The concept is important for scientists and astronomers to conclude several scientific experiments. Doppler effect is recognized as the motion-related frequency change. It is applicable both for sound waves as well as electromagnetic waves.

Doppler Effect

Two different people at either side of an ambulance can hear the sound of an ambulance differently. A person can hear the distinct sound of an ambulance when he is near it but it fades away as the vehicle starts to speed up. The doppler effect is also applicable to the intensity of lights. The intensity of light increases and decreases as one moves towards or further off from the light source.

Wave Speed	Wave Power
Wave Function	LC Oscillation

Use of the Doppler Effect

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The use of the doppler effect in real life is applicable in the following situations:

In the field of medicine: Doctor use it for Blood flow measurement and heat beat

Doppler Effect - Blood flow measurement

Vibration measurement

Doppler Effect - Vibration measurement

Used by astrophysicists to measure the velocity of stars
Airports to guide aircrafts
Detect enemy aircraft (Radars and Sirens)

Read More: Laplace Correlation

Doppler Effect - Downlink Jamming

Measurement in velocity in the field of medical science, military, and astrophysics.
Ultrasonic waves or sonography
Echocardiogram

Doppler Effect - Echocardiogram

Development biology
Communication with satellites

AC Voltage Applied to a Resistor	AC Voltage Applied to an Inductor
AC Voltage Applied to a Capacitor	AC Voltage Applied to a Series LCR Circuit

The Disadvantage of the Doppler Effect

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The receiver and the source need to be in a straight line for the effect to take place. It should be noted that the velocity of sound is more than that of the receiver and the source.

Situations under which the frequency changes:

The source is moving but the observer is at rest
The source is at rest but the observer is in motion
The source and the observer are moving

There is a presence or absence of relative pressure between the medium and the observer which is why situations (a) and (b) have their difference.

Doppler Effect Formula

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Due to the relative motion between the source of the sound and the observer, the doppler effect implies the apparent change in the frequency of waves. By using the following equation, we can deduce the apparent frequency in the Doppler effect.

f' = (v ± v 0 / v \(\mp\) vs ) fo

Where f = frequency received by the observer

F0 = original frequency of the source

When the observer moves away from the source, the numerator is subtracted whereas the denominator is subtracted when two bodies are moving towards each other.

Wavelength (λ) = vf

Read Also: Displacement Current Important Notes

Doppler Effect: Sample Questions

Ques: The speed of a rocket towards a stationary object is 200ms-1. A wave of frequency of 1000Hz is emitted by the moving rocket. An echo is created when some of the sound reaching the target gets reflected. Calculate (a) the frequency of sound which is detected by the target (b) the frequency of echo which is detected by the rocket. (5 marks)

Ans: The rocket is moving with a speed of 200ms-1 towards a stationary object. As the source is reaching a stationary object, v0 = 0 and vs will be replaced by –vs. since the speed can be compared to the velocity of sound, thus,

v = v₀ (1 + vs / v )-1

v = 1000Hz [ 1 – 200ms-1 / 330 ms -1 ]-1

= 2540 Hz

The target is the source of echo which makes it the actual source now. The rocket's detector being able to detect echo makes it the detector or the observer. Thus vs = 0 and a positive value is carried by v0 .

V= the frequency of the sound emitted by the source.

The frequency registered by the rocket is

v' = v [(v + v 0)/v]

= 2540Hz * [(200ms-1 + 330 ms-1)/ 330ms-1

= 4080Hz

Ques: A sound source of 790 Hz frequency moves away from an observer who Is at rest at the speed of 15m/s. What frequency does the observer hear? 340m/s is the speed of sound. (2 marks)

Ans: fo = f s [(v+ v0)/(v +vs)]

= 790Hz [(340 m/s + 0m/s)/ (340m/s + 15m/s) = 757Hz

Ques: A person runs away from the speaker at 3m/s creating a sound wave of 200Hz. What frequency does he/she hear? The speed of sound is 340m/s. (3 marks)

Ans: f= [(c + vr) * f0]/ (c + vs)

where, f = frequency heard by the recipient

vr = velocity of the receiver

vs = velocity of the source

f0 = original frequency

here,

vs = 0 because the speaker is stationary

as the person is walking towards the speaker, vr is negative. Thus the original frequency will be higher than the frequency heard.

f = [(c + vr)* fo] / (c + vs)

= (340 m/s- 3 m/s) * 200 Hz/ 340 m/s

= 198.2 Hz

Ques: Two cars approaching each other at 50m/s. one car starts beeping its horn at 475 Hz frequency. What is the wavelength of the horn as heard by the other driver? Velocity of sound is 343 m/s. (3 marks)

Ans: f0 = fs * (v ± v0 )/ (v \(\mp\) vs )

The frequency heard will be higher because the cars are approaching others. This information helps you to determine the signs of the equation. The signs of the equation will help to make the coefficient greater than one.

f0 = fs * (v + v0) / (v – vs)

= 475 * (343 + 50)/ (343 – 50) = 637 Hz

Wavelength (λ) = vf

λ = (343 m/s)/ 637 Hz = 0.54 m

Ques: You are jogging on the pavement at a rate of 3m/s. a police car is patrolling behind you at the rate of 4m/s when it turns on its siren. 10,000 Hz is the frequency of the siren. Is the perceived frequency higher or lower than the original frequency? (3 marks)

Ans: As the overall distance between the siren and you are decreasing, the perceived frequency will be higher.

According to the doppler effect, the perceived frequency for the emitted waves will be higher if two objects are moving closer towards each other. The overall distance between you and the siren is decreasing as you are jogging away from the car at 3 m/s but the car is traveling at 4m/s. the frequency heard will be higher than the original frequency.

Also read:

Cells in Series and in Parallel	Mechanical Properties of Fluids	Moving Charges and Magnetism
Surface Tension	Magnetic Force and Magnetic Field	Unit of Viscosity
Buoyant Force	Mechanical Properties of Solid	Poisson’s Ratio
Shearing Stress	Pascal’s Law	Hydraulic Machines
Surface Energy	Bernoulli’s principle	Venturi-meter

Doppler Effect: Definition, Formula, Uses and Solved Questions

Define Doppler Effect

Use of the Doppler Effect

The Disadvantage of the Doppler Effect

Doppler Effect Formula

Doppler Effect: Sample Questions

CBSE CLASS XII Related Questions

1.
Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

2.
Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

3.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

4.
If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

5.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

6.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

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Doppler Effect: Definition, Formula, Uses and Solved Questions

Define Doppler Effect

Use of the Doppler Effect

The Disadvantage of the Doppler Effect

Doppler Effect Formula

Doppler Effect: Sample Questions

CBSE CLASS XII Related Questions

1.Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

2. Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

3.Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

4. If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

5.Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

6.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

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SUBSCRIBE TO OUR NEWS LETTER

1.
Two thin lenses of focal length \( f_1 \) and \( f_2 \) are placed in contact with each other coaxially. Prove that the focal length \( f \) of the combination is given by \[ f = \frac{f_1 f_2}{f_1 + f_2}. \]

2.
Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

3.
Write the expression for the magnetic field due to a current element in vector form. Consider a 1 cm segment of a wire, centered at the origin, carrying a current of 10 A in positive x-direction. Calculate the magnetic field \( \mathbf{B} \) at a point \( (1 \, \text{m}, 1 \, \text{m}, 0) \).

4.
If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

5.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

6.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be: