
Education Journalist | Study Abroad Lead
Doppler Effect is a phenomenon that is observed when the source of the waves moves with respect to an observer. It is a characteristic of sound waves in general. The Doppler Effect describes the apparent shift in sound frequency when the observer and the medium are both moving in the same direction. It is determined by three factors: the source's velocity, the medium's velocity, and the observer's velocity. An example of the Doppler Effect is an ambulance crossing you with its siren blaring.
| Table of Content |
Key Takeaways: Doppler Effect, Doppler Shift, Observed Frequency, Doppler Effect Formula, Satellite communication
What is Doppler Effect?
[Click Here for Sample Questions]
The Doppler Effect, also known as the Doppler shift, is the result of a change in the frequency of sound waves caused by movement.
The Doppler effect is a phenomenon caused by a moving wave source that causes an apparent upward shift in frequency for observers who are approaching the source and a visible downward change in frequency for observers who are retreating from the source.
- When the source and the observer move relative to each other, the frequency that is observed by the observer (fa) is different from the actual frequency that is produced by the source (f0).
- When the source of the sound waves is moving towards the observer, an upward shift in frequency is observed.
- As for the observers from whom the source is receding, a downward shift in frequency will be observed.
- It is crucial to note that the impact isn't caused by a change in the source's frequency.
- The Doppler effect may be seen in any wave type, including water waves, sound waves, and light waves.
- Doppler effect is known to our encounters with sound waves.
Do check out:
Doppler Effect Formula
[Click Here for Previous Year Questions]
If the source of the sound and the listener move in relation to each other, the sound heard by the listener changes. When the listener and the source get near enough, the frequency heard by the listener is higher than the sound produced by the source. When the listener and the source moves away from each other, the frequency heard by the listener is lower than the frequency heard by the source. The frequency of the sound is measured in Hertz (Hz) where one Hertz is one cycle per second. (1 Hz = 1s-1 = 1 cycle/s).
As a result, the formula for the doppler effect is:
fL = \(\frac {v + vl} {v + vs}\)fs
In the Doppler Effect formula,
- fL is the frequency of sound that the listener perceives (Hz, or 1/s)
- v is the speed of sound in the medium (in m/s)
- vL is the listener's velocity (m/s)
- vs is the velocity of the sound source (in m/s)
- fs is the frequency of sound emitted by the source (Hz, or 1/s)
a) Source Moving Towards the Observer at Rest
In this case, the velocity of the observer velocity is zero, so vo is equal to zero. Substituting this into the Doppler effect formula, we get the equation of the Doppler effect when a source moves towards an observer at rest-
fL = \(\frac {v} {v - vs}\)fs
b) Source Moving Away from the Observer at Rest
Since the velocity of the observer, in this case, is zero, we can eliminate vo from the equation. However, the source moves away from the observer, so its velocity becomes negative to indicate the direction. Therefore, the equation now becomes-
fL = \(\frac {v} {v - (- vs)}\)fs
c) Observer Moving Towards a Stationary Source
In this case, vs will become zero, hence we get the following doppler effect formula:
fL = \(\frac {v\ +\ v_{0}} {v}\)fs
d)Observer Moving Away from a Stationary Source
Since the observer is moving away, the velocity of the observer in this situation becomes negative. So, instead of adding vo, we can now subtract it, since vo is negative.
fL = \(\frac {v\ -\ v_{0}} {v}\)fs
Doppler Effect Formula – Frequency
[Click Here for Sample Questions]
The Doppler shift, also known as the Doppler effect, describes the connection between the perceived frequency and the emitted frequency when the velocity of the source and receiver is slower than the waves’ velocity in the medium. When the receiver and source speeds are smaller than the wave velocity, the formula is as follows:
Observed Frequency: F = (\(1 + \frac {\bigtriangleup v} {c}\))f0
Change in Frequency: Δf = \(\frac {\bigtriangleup v} {cf_0}\)
Where,
- Δf = f – f0, Δv = vr – vs, the velocity of the receiver relative to the source;
- C is the wave's amplitude in the medium;
- vr is the receiver's speed in relation to the medium (positive if the receiver moves towards the source and negative if it moves in the opposite direction).
- vs denotes the source's velocity in relation to the medium (positive if the source moves away from the receiver and negative if it moves in the opposite direction).
- The frequency observed is denoted by the letter f.
- The frequency transmitted is f0, and the transmitter and receiver are traveling toward each other.
Doppler Effect Examples
[Click Here for Previous Year Questions]
Consider the following scenario: You are standing on the sidewalk when an ambulance whizzes by. Do you notice how the siren's tone alters as it travels a certain distance?
It gets louder as it gets closer to you, but there is another aspect of the sound that changes. The pitch of the vehicle increases as it approaches you and decreases as it pulls away. The frequency of the waves, or how many waves travel through an area per unit of time, causes the pitch to shift.

Doppler Effect observed in a moving ambulance
- In the case of an ambulance, you are standing motionless as the vehicle approaches.
- As sound waves approach you, they compress, causing the frequency to rise, resulting in a higher pitch.
- However, when the vehicle moves away from you, the sound waves spread more apart, lowering the frequency and hence the pitch.
When the wave source travels in relation to the observer, the Doppler effect is noticed.
Application of Doppler Effect
[Click Here for Sample Questions]
The applications of the doppler effect are –
-
Sirens
The principle behind a siren is that it begins at a pitch higher than its stationary pitch as it moves away from the observer and then returns to a lower pitch as it moves away from the observer. Siren's speed is calculated as follows:
Vradial = vs.cosθ
where θ is the angle formed between the line of sight of the object and the advancing velocity.

Application of Doppler Effect in Sirens
-
Astronomy
Astronomers utilize measurements of electromagnetic waves emitted by stars in distant galaxies to support the idea that the universe is expanding. Furthermore, the Doppler effect will be used to analyze the fine features of stars within galaxies.
-
Velocity Profile Measurement
The real-time performance velocity profile of any liquid containing suspended particles, such as dust, emulsions, and gas bubbles, is calculated using an ultrasonic Doppler Velocimeter.
Things to Remember
- The Doppler Effect is defined as a shift in wave frequency caused by relative motion between the wave source and the observer.
- Distinct stars have different absorption lines at different frequencies, however, the Doppler Effect can only be seen when the absorption lines are distant from the stated frequencies.
- The Doppler effect formula can be written as: fL=\(\frac {v + vl} {v + vs}\)fs
- The Doppler Effect may be used in a variety of ways. Its use can be found in:
- Sirens
- Astronomy
- Radars
- Medical imaging and blood flow management
- Flow management
- Velocity profile management
- Satellite communication
- Audio
- Vibration measurement
Also Read:
Sample Questions
Ques. A vehicle driver is driving on a road that runs beside railway rails. When the train gets close enough, it sounds its horn, which produces a single frequency of 420.0 Hz. The train is moving at 32.0 m/s while the driver is going at 18.0 m/s. Furthermore, the sound travels at a speed of 340.0 m/s. Determine the frequency of the sound that the car's driver will hear. (5 marks)
Ans. We must first create a coordinate system before we can discover the frequency. The listener to the source is considered to be the positive direction.
The train's horn is the source, therefore the train's velocity is negative while the driver's car's velocity is positive.
Given: v = 340.0 m/s,
vL = +18.0 m/s,
vs = -32.0 m/s, and
fs = 420.0 Hz
As a result, we'll use the Doppler Effect Formula to calculate the frequency, which is:
fL=\(\frac {v + vl} {v + vs}\)fs
fL=\(\frac {340.0 m/s+18.0 m/s} {340.0 m/s-32.0 m/s}\)(420.0 Hz)
fL=\(\frac {358.0 m/s} {308.0 m/s}\)(420.0 Hz)
fL≅(1.162)(420.0 Hz)
fL≅488.2 Hz
As a result, the train's horn will have a frequency of 488.2 Hz, which the driver will hear.
Ques. A submarine moves through the water at an average speed of 8.00 m/s, generating a 1400 Hz sonar sound. In water, sound travels at 530 meters per second. A second submarine is stationed such that both submarines are traveling in the same direction. The second submarine is traveling at a speed of 6.00 meters per second. (5 marks)
(A) As the first submarine approaches, what frequency is noticed by an observer on the second submarine?
(B) The submarines pass each other with only a fraction of a second between them. As the submarines withdraw from it, what frequency is observed by an observer aboard the second submarine?
(C) As the two submarines approach each other, some of the first submarine's noises reverberate off the second submarine and return to it. What would be the frequency of the sound if it were heard by a first submarine?
Ans. Given: Frequency f=1400 Hz,
The velocity of first submarine, Vs=8 m/s,
The velocity of the second submarine, V0=6 m/s.
Speed of sound in water V=530 m/s
(A) The Apparent frequency is given by,
f' = (\(\frac {v + v_0} {v - v_s}\)) f
f' = (\(\frac {530+6} {530-8}\)) 1400
f'=1437.54 Hz
(B) To locate the Doppler-shifted frequency heard by the observer in the second submarine, follow these steps:
f' = (\(\frac {v + v_0} {v - v_s}\)) f
f' = (\(\frac {530 + (-6)} {530 - (-8)}\)) 1400
ff=1363.56 Hz
(C) Part (A) shows how sound with an apparent frequency of 1436.5 Hz is reflected from a moving source (sub B) and perceived by a moving observer.
f'' = (\(\frac {v + v_0} {v - v_s}\)) f'
f''= (\(\frac {530+6} {530-8}\)) 1437.54
f'=1476 Hz
Ques. Assume a train with a 150-Hz horn is moving at 35.0 m/s through calm air at 340 m/s. (5 marks)
(a) As the train approaches and passes, what frequencies can a stationary spectator at the tracks' edge notice?
(b) When the railway engineer is on board the train, what frequency does he notice?
Ans. a. Enter known values into f0= fs( \(\frac {v} {v-v_s}\) ):
f0=fs(\(\frac {v} {v-v_s}\)) = (150 Hz) (\(\frac {340 m/s} {340 m/s-35.0 m/s}\))
We can calculate the frequency that will be observed by a person who is not moving as the train approaches:
f0=(150 Hz)(1.11)=167 Hz
To calculate the frequency heard by a stationary individual while the train recedes, use the same equation with the plus sign:
f0=fs \(\frac {v} { v+v_s}\) = (150 Hz) (\(\frac {340 m/s} {340 m/s-35.0 m/s}\))
Calculate the second frequency:
fo=(150 Hz)(0.907)=136 Hz
- Because the relative velocity between both is zero, it seems plausible that the engineer would get the same frequency as the horn.
- The speeds are vs = vo = 35.0 m/s, Relative to the medium (air).
- The moving observer experiences the first Doppler shift, whereas the moving source experiences the second. Apply the following formula:
f0 = [fs (\(\frac {v \pm v_0} {v} \))] (\(\frac {v} {v \mp v_s}\))
The Doppler-shifted frequency owing to a moving observer is the quantity in square brackets. The influence of the moving source is the factor on the right.
We must use the plus sign for vobs since the train engineer is traveling toward the horn; but, because the horn is simultaneously going away from the engineer, we must also use the plus sign for vs. However, because the engineer and the horn are traveling at the same speed, vs=v0. As a consequence, everything but fs cancels out, leaving us with
fo= fs
Ques. At a pace of 15m/s, a sound source with a frequency of 790Hz travels away from a stationary observer. What is the frequency of the observer's hearing?
Speed of sound: 340 meters per second (3 marks)
Ans. The Doppler effect is given by the equation below in this case.
fo = fs \(\frac {(v + v_0)} {(v - v_s)}\)
Using the problem's values, we can deduce that v0 is zero and vf is 15 m/s. v is 340 m/s and fs is 790 Hz.
fo= 790 Hz \(\frac {(340 m/s+0 m/s)} {(340 m/s+15 m/s)}\) = 757 Hz
Ques. An 880Hz siren is emitted by a fire vehicle. The pitch of the truck as it approaches an observer on the sidewalk is 950Hz. After the truck has passed and is driving away, what pitch does he hear? Assume that the truck's speed remains constant and that the sound velocity in air is 340m/s. (5 marks)
Ans. The equation for the Doppler effect is fobserved = fsource \(\frac {v} {v \pm v_{source}}\)
When the source and observer are moving farther away, the + sign is used, and when they are moving closer together, the - sign is used. V = 340 m/s is the speed of sound, fsource is the frequency emitted by the source,
f observed is the frequency perceived by the observer, and
vsource is the relative velocity between the source and the observer in these equations.
We can use this equation to solve for the truck's velocity during the first part of the motion, as the truck approaches the observer.
950 = 880 \(\frac {340} {340-v_source}\) vsource ≈ 25\(\frac {m} {s}\)
When the vehicle moves further away from the observer, we may enter this velocity back into the equation and solve for it.
fobserved . fobserved = 880 \(\frac {340} {340+25}\) ≈ 820 Hz
Ques. A speaker is put up at a nearby concert to create low-pitched base sounds in the 20Hz to 200Hz range, which may be represented as sine waves. The sound waves produced by the speaker may be represented as a cylindrical pipe with one end closed that travels through the air at v = 331\(\frac {m} {s}\)+0.6\(\frac {m} {s}\)(T), where T is the temperature in oC. As a person approaches the speaker, the frequency that he or she hears changes. Why? (3 marks)
Ans. This is a Doppler effect problem since it asks us how the frequency varies as one item travels directly towards another. Recalling our Doppler effect formula, we can see that
f = \(\frac {(c + v_r) * f_0} {c + v_s}\),
where f is the frequency received by the receiver (the concert goer),
v r is the receiver's velocity,
v s is the source's velocity, and
f0 is the original frequency.
Because the speaker is not moving in our situation, vs is zero. When a person walks towards the speaker, vr is positive, therefore the frequency heard is greater than the initial frequency.
Ques. A sound's source is moving away from the listener. The source appears to be __________, according to the listener. (3 marks)
Ans. The Doppler effect is calculated as follows:
f = f0 (\(\frac {c \pm v_obs} {c \pm v_s}\))
The Doppler effect affects just the frequency of sound; velocity and loudness are unaffected. The velocity of the source is added to the speed of light when it moves away from the observer.
f = f0 (\(\frac {c} {c + v_s}\))
This raises the denominator's value while lowering the observed frequency's value. Pitch or tone is related to frequency; a lower observed frequency correlates to a lower perceived pitch.
Ques. Two vehicles are approaching each other at 50 mph when one of them begins to blast its horn at 475 Hz. What is the horn's wavelength as seen by the other driver? vsound=343m/s (5 marks)
Ans. The Doppler equation is:
fo= fs*\(\frac {v \pm v_0} {v \mp v_s}\)
The frequency heard will rise as the automobiles get closer to each other. This essential understanding enables you to identify the equation's signs. To produce a coefficient higher than 1, the top of the fraction will be added and the bottom will be subtraction.
fo= fs * \(\frac {v + v_0} {v - v_s}\)
Using the given values to solve:
fo = 475 * \(\frac {343+50} {343-50}\) =637 Hz
We can solve for the wavelength now that we know the frequency:
λ = vf λ = \(\frac {343 (\frac {m} {s})} {637}\) Hz=0.54 m
Ques. A balloon ascending vertically upwards at a constant velocity of 5 m/s drops a body producing sound with a frequency of 350 Hz. (Velocity of sound in air is 335 m/s; acceleration due to gravity is 10 m/s2) The frequency of sound felt by the observer in the balloon 2s after the release is (3 marks)
Ans. The velocity of the observer. vo=5 ms-1
The velocity of sound v3=gt=10(2)=20 ms-1
According to the Doppler effect:
frequency f' = (\(\frac {v - v_3} {v + v_0}\)) f
=(\(\frac {335 - 20} {335 + 5}\)) 350
=324.26 Hz
Ques. With a radius of r and a time period of T, a sound source spins in a circular orbit. The circle is d meters away from the observer. If the sound velocity is represented by x, what will be the difference between the minimum and maximum frequencies that can be seen by the stationary observer. (the original frequency is n) (3 marks)
Ans. When the source is approaching, the frequency is highest, and when it is heading away, the frequency is lowest.
f1 =( \(\frac {v + v_0} {v}\) ) f
f2 = (\(\frac {v + v_0} {v}\)) f
f1 – f2 = \(\frac {2 v_0 f}{v}\) = \(\frac {2 \phi} {x}\) (\(\frac {2 \pi} {T}\))n = \(\frac {4 \pi \phi} {nx}\) [\(\because\)v=φω]
For Latest Updates on Upcoming Board Exams, Click Here:https://t.me/class_10_12_board_updates
Do Check Out:






Comments