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Difference between Doppler Effect and Doppler Shift

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  • Last Updated : 05 Jan, 2023
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Doppler effect or Doppler shift phenomenon was described in 1842 by an Austrian physicist, Christian Doppler, and it is named after him. The Doppler effect or Doppler shift is a change (increase or decrease) in the frequency of a wave as the source and the observer move (towards or away from) each other relative to the medium. Based on the direction of the source and the observer and the magnitudes of their velocities, the observed frequency can be less or more than the source frequency. For example, the pitch of the sound of an ambulance siren changes as it passes us. It happens because of the relative velocity between the source and the observer. When the ambulance is approaching us, the relative velocity is negative, and the relative velocity is positive when it is moving away. So, that is the reason behind the difference in the pitch of the sound of the ambulance siren while it is approaching and while it is moving away. The Doppler effect applies to all types of waves, including sound and light. The Doppler effect is responsible for the Blue Shift or Red Shift phenomenon observed in light waves. The Doppler effect is used in various fields such as radar, astronomy, satellite communication and navigation, medical imaging, etc.

Doppler Shift in Sound

Doppler Shift in Sound

 

When a man is standing at the railway crossing, the frequency of the whistle emitted by the train while it is approaching and crossing changes. The reason behind this phenomenon is the Doppler shift. As the source moves toward the observer, the frequency of the sound wave will appear to increase, and as the source moves away from the observer, the frequency will appear to decrease. So, that is the reason behind the difference in the pitch of the sound of the whistle emitted by the train as it is approaching and moving away. If the source is moving toward the observer, then the velocity of the source is positive, and it is negative as it moves away from the observer. If the observer is moving toward the source, then the velocity of the observer is positive, and it is negative as it moves away from the source. 

Various cases in Doppler Shift in sound

 

Doppler Shift Formula

When the speed of the source and receiver is slower than the velocity of the waves in the medium, the relationship between the observed frequency (f) and the emitted frequency (f0) is described by the Doppler shift or Doppler effect formula.

f = [(c ± vr)/ (c ± vs)] f0

where,
c” is the velocity of waves in the medium,
vr” is the velocity of the receiver relative to the medium (if the receiver is moving towards the source, then “vr” is added to c, or if the receiver is moving away from the source, then “vr” is subtracted from c.)
vs” is the velocity of the source relative to the medium (if the source is moving away from the receiver, then “vs” is added to c, or if the source is moving towards the receiver, then “vs” is subtracted from c.) 
f” is the observed frequency, and
f0” is the emitted frequency.

Case 1: When the Source is Moving Toward the Observer at Rest 

If the source is moving toward the observer at rest (vr = 0) relative to the medium and the source is emitting waves with an actual frequency of f0, then the frequency at which the observer detects waves is given as follows:

f = f0

where,
c” is the velocity of waves in the medium,
vr” is the velocity of the receiver,
f” is the observed frequency, and
f0” is the emitted frequency.

Case 2: When the Source is Moving Away From the Observer at Rest 

If the source is moving away from the observer at rest (vr = 0) relative to the medium and the source is emitting waves with an actual frequency of f0, then the frequency at which the observer detects waves is given as follows:

f = f0

where,
c” is the velocity of waves in the medium,
vr” is the velocity of the receiver,
f” is the observed frequency, and
f0” is the emitted frequency.

Case 3: When the Observer is Moving Toward the Source at Rest

If the observer is moving toward the source at rest (vs = 0), then the equation for observed frequency is given as follows:

f = [(c + vr)/c] f0

where,
c” is the velocity of waves in the medium,
vr” is the velocity of the receiver,
f” is the observed frequency, and
f0” is the emitted frequency.

Case 4: When the Observer is Moving Away From the Source at Rest

If the observer is moving away from the source at rest (vs = 0), then the equation for observed frequency is given as follows:

f = [(c − vr)/c] f0

where,
c” is the velocity of waves in the medium,
vr” is the velocity of the receiver,
f” is the observed frequency, and
f0” is the emitted frequency.

Derivation of the Doppler Effect

There are two circumstances that must be taken into account to derive the Doppler effect:

Case 1: Moving Source and Stationary Observer

c = λs/T    ————— (1)
 
Where “c” is the wave velocity,

“λs” is the wavelength of the source, and

“T” is the time taken by the wave.

From equation (1),

T = λs/c   ————— (2)

Let the distance between the source and the stationary observer be “d”, and the velocity with which the source is moving towards a stationary observer be “vs“.

d = vsT    ————— (3)

By substituting the value of T in equation (3), we get,

d = vsλs/c ————— (4)

Now, the observed wavelength (λ0) will be equal to the difference between the wavelength of the source and the distance covered by the source, i.e.,

λ0 = λs – d ————— (5)

From equations (4) and (5) we get,

λ0 = λs – vsλs/c

λ0 = λs(1 – vs/c)

λ0 = λs[(c – vs)/c]

Make a note that the sign of the source velocity (vs) changes as the source moves away from the observer, but everything else in this formula remains the same.

Δλ (change in wavelength) = λs − λ0

We know that,

λ0 = λs – d

So, Δλ = λs − (λs − d)

Δλ = d = vsλs/c

∴ λ0 = λs(c−vs)/c

Δλ =  vsλs/c

Case 2: Moving Observer and Stationary Source

f0 = (c − v0)/λs   ————— (a)

Where “f0” is the observed frequency, and

“v0” is the velocity of the observer.

f0 = c/λ0    ————— (b)

From equations (a) and (b)

(c − v0)/λs = c/λ0

λs/ (c − v0) = λ0/c

λ0 = λsc/(c−v0)

λ0 = λs/ [(c−v0)/c]

λ0s/ [1− (v0/c)]

Δλ (change in wavelength) = λs − λ0

Now, by substituting the value of λ0, we get

Δλ = λs − [λsc/(c−v0)]

Δλ = (λs(c − v0) − λsc)/ (c − v0)

Δλ = −λsv0/ (c − v0)

∴ λ0 = λsc/(c−v0)

Δλ = −λsv0/ (c − v0)

Difference between the Doppler Shift and Doppler Effect

The term Doppler effect is defined as the change (increase or decrease) in the frequency of a wave as the source and the observer move (towards or away from) each other relative to the medium, whereas the term Doppler shift refers to the movement of the source or observer concerning the medium.

Applications of the Doppler Effect

Doppler Effect has various applications some of which are discussed below:

  • The Doppler effect helps in measuring the speed of oncoming vehicles.
  • It is also often used in robotics for dynamic real-time path planning.
  • It is used in astronomy to estimate the speed at which a star or a galaxy is approaching or receding from us.
  • It is also used in radar to calculate the speed of detected objects.
  • It is also used in satellite communication and satellite navigation.
  • In the medical field, it is used to measure the velocity of the blood.

Solved Examples on Doppler Effect and Doppler Shift

Example 1: Two vehicles A and B are moving toward each other at a speed of 80 m/s. If one of the vehicles honks with a frequency of 250 Hz, then what will be the frequency of the honk observed by the passenger sitting inside vehicle A? (The velocity of sound in air is 343 m/s).

Solution: 

Given data:

The velocity of vehicles A and B are 80m/s, i.e.,

vr = vs = 80 m/s

Emitted frequency (f0) = 250 Hz

As the observer and the source are moving toward each other, the apparent frequency (f) is given as follows:

f = [(c + vr)/ (c – vs)] f0

f = [(343 + 80)/ (343 – 80)] × 250

= (423/263) × 250

f = 402.0912 Hz ≈ 402 Hz

Hence, the apparent frequency (f) is approximately equal to 402 Hz.

Example 2: A man is standing at the railway crossing, and the frequency of the whistle emitted by the train is 1.05 kHz. What will be the apparent frequency heard by the man as the train approaches him at a speed of 360 km/h? (The velocity of sound in air is 343 m/s).

Solution: 

Given data:

Speed of the train (vs) = 360 km/h = 100 m/s

vr = 0

The velocity of sound in air (c) = 343 m/s

Emitted frequency (f0) = 1.05 kHz

Here, the source is moving toward the observer at rest. So, the apparent frequency (f) is given as follows:

f = f0

= [343/ (343 − 100)] × 1.05

= (343/243) × 1.05

= 1.48 kHz

Hence, the apparent frequency (f) is 1.48 kHz.

Example 3: A traffic policeman who was standing on a road whistled at a frequency of 480 Hz. What will be the apparent frequency heard by the car driver who is approaching the policeman at a speed of 90 km/h? (The velocity of sound in air is 343 m/s).

Solution: 

Given data:

The velocity of sound in air (c) = 343 m/s

Speed of the car (vr) = 90 km/h = 25 m/s

vs = 0

Emitted frequency (f0) = 480 Hz

Here, the observer is moving toward the source at rest. So, the apparent frequency (f) is given as follows:

f = [(c + vr)/c] f0

= [(343 + 25)/ 343] × 480

= (368/343) × 480

f = 514.985 Hz ≈ 515 Hz

Hence, the apparent frequency (f) is approximately equal to 515 Hz.

Example 4: Two trains A and B are moving away from other at a speed of 540 km/h. If the frequency of the whistle emitted by train B is 1.65 kHz, then what will be the apparent frequency of the whistle heard by the passenger sitting inside train A? (The velocity of sound in air is 343 m/s).

Solution:

Given data:

The velocity of sound in air (c) = 343 m/s

The velocity of trains A and B is 540 km/h, i.e., 

vs = vr = 540 km/h = 150 m/s

Emitted frequency (f0) = 1.65 kHz = 1650 Hz

As the observer and the source are moving away from each other, the apparent frequency (f) is given as follows:

f = [(c – vr)/ (c + vs)] f0

= [(343 – 150)/ (343 + 150)] ×1650

= (193/493) ×1650

= 645.943 ≈ 646 Hz

Hence, the apparent frequency (f) is approximately equal to 646 Hz.

Example 5: A traffic policeman who was standing on a road whistled at a frequency of 550 Hz. What will be the apparent frequency heard by the person driving a motorbike who is moving away from the policeman at a speed of 72 km/h? (The velocity of sound in air is 343 m/s).

Solution:

Given data:

The velocity of sound in air (c) = 343 m/s

Speed of the motorbike (vr) =72 km/h = 20 m/s

vs = 0

Emitted frequency (f0) = 550 Hz

Here, the observer is moving away from the source at rest. So, the apparent frequency (f) is given as follows:

f = [(c − vr)/c] f0

= [(343 − 20)/ 343] × 550

= (323/343) × 550

f = 517.930 Hz ≈ 518 Hz

Hence, the apparent frequency (f) is approximately equal to 518 Hz.

FAQs on Doppler Effect and Doppler Shift

Question 1: Does the Doppler effect apply to all waves?

Answer: 

Yes, the Doppler effect applies to all types of waves, like a sound wave that is a longitudinal wave or a light wave that is a transverse wave.

Question 2: What is meant by the Doppler effect?

Answer: 

The Doppler effect or Doppler shift is a change (increase or decrease) in the frequency of a wave as the source and the observer move (towards or away from) each other relative to the medium.

Question 3: What is the Doppler shift or Doppler effect formula?

Answer: 

When the speed of the source and receiver is slower than the velocity of the waves in the medium, the relationship between the observed frequency (f) and the emitted frequency (f0) is described by the Doppler shift or Doppler effect formula.

f = [(c ± vr)/ (c ± vs)] f0

where,
c” is the velocity of waves in the medium,
vr” is the velocity of the receiver relative to the medium (if the receiver is moving towards the source, then “vr” is added to c, or if the receiver is moving away from the source, then “vr” is subtracted from c.)
vs” is the velocity of the source relative to the medium (if the source is moving away from the receiver, then “vs” is added to c, or if the source is moving towards the receiver, then “vs” is subtracted from c.) 
f” is the observed frequency, and
f0” is the emitted frequency.

Question 4: Mention a few applications of the Doppler effect.

Answer: 

The Doppler effect helps in measuring the speed of the motor car. It is also often used in robotics for dynamic real-time path planning. It is used in astronomy to estimate the speed at which a star or a galaxy is approaching or receding from us. In the medical field, it is used to measure the velocity of the blood.

Question 5: What is the formula to find the apparent frequency when the observer is moving away from the source at rest?

Answer:

If the observer is moving away from the source at rest (vs = 0), then the equation for apparent frequency is given as follows:

f = [(c − vr)/c] f0

where,
c” is the velocity of waves in the medium,
vr” is the velocity of the receiver,
f” is the observed frequency, and
f0” is the emitted frequency.

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