# Physics Tutorial: The Doppler Effect

In this Physics tutorial, you will learn:

• What happens to the wavelength and frequency of a wave when the source and the receiver are moving in respect to each other?
• What is the Doppler Effect in waves?
• How the Doppler Effect is applied in sound waves?
• How the Doppler Effect is applied in light waves?
• What is the general equation of the Doppler Effect in waves?

## Introduction to The Doppler Effect

Most people are able to understand whether an ambulance car is approaching or moving away from them. Are you one of those? How do you understand this?

How the sound is emitted by a moving source compared to cases in which the source is stationary?

Do you think this phenomenon exists in other kind of waves? What can you say about light?

There is a common explanation for all the above questions. It is related to the relative movement of the wave source and the receiver, which will be discussed and explained in the following paragraphs.

## The Doppler Effect in Sound Waves

All sound waves have their specific wavelength and frequency depending on the pitch, as discussed in the Physics tutorial covering Sound Waves. Intensity and Sound Level. A sound wave produced by a stationary source propagates in spherical pattern as shown in the figure below.

The equidistant circles show that wavelength is the same in all directions. If we assume the medium as uniform, then the frequency of sound waves is the same in all directions as well.

When the source is moving in a certain direction, the wavelength decreases in the moving direction and increases in the opposite side as shown below.

The equidistant circles show that wavelength is the same in all directions. If we assume the medium as uniform, then the frequency of sound waves is the same in all directions as well.

When the source is moving in a certain direction, the wavelength decreases in the moving direction and increases in the opposite side as shown below.

Thus, on the left side we have f < f0 because λ > λ0 (wavelength and frequency are inversely proportional if the speed v is constant), while on the right side we have f > f0 because λ < λ0. You can compare the wavelengths by considering the distance between two adjacent circles, which represent the waveform. As a result, if the waves considered are sound waves, a listener (detector) on the left side will hear a lower-pitch sound than usual while a listener on the right side of the wavefront, will hear a higher-pitch sound than usual.

This effect of frequency change when the source and the detector are in relative motion to each other is known as the Doppler Effect.

Now, let's consider the Doppler Effect from the mathematical viewpoint. There are three possible cases of source and detector movement in respect to each other. They are:

1. Detector moving, source stationary
2. Source moving, detector stationary
3. Both detector and source are moving

In the following paragraphs, we will explain how the quantities involved in the Doppler Effect are related to each other in each case.

### a) Detector moving, source stationary

Not necessarily, a detector must be a complicated electronic apparatus; it can be simply a human ear. In this paragraph, we consider the situations in which the person who hears the sound is moving towards or away from a stationary sound source.

Let v be the speed of sound wave and vd the speed of human (the moving detector). Thus, the distance between two fixed points A and B will be the sum of the distances travelled by the two moving systems - the wave and the detector. Hence, since there is a single event to be considered (time is the same for both moving systems) the total distance AB travelled by the two approaching systems is

dAB = v × t + vd × t

The number of wavelengths N intercepted by the detector during this process is

N = dAB/λ
= v × t + vd × t/λ
= (v + vd) × t/λ

Hence, we can write

N/t = v + vd/λ

The left side of the last equation gives the new frequency f detected by the listener in motion. Hence, we obtain

f = v + vd/λ

Since the wavelength λ of the original wave (from the equation of waves v = λ × f0) is

λ = v/f0

we obtain after substituting it to the previous equation

f = v + vd/v/f0

or

f = v+vd/v × f0

The last equation represents the Doppler Effect equation for a detector moving towards a stationary source.

When the detector is moving away from a stationary source, we have for the distance AB:

dAB = v × t - vd × t

and after following the same procedure as before, we obtain for the new frequency detected by the listener:

f = v - vd/v × f0

#### Example 1

A stationary source emits sound waves of frequency 200 Hz. A listener is approaching the source at 20 m/s. What is the frequency detected by the listener if the sound speed is 343 m/s?

#### Solution 1

There are the following clues in this problem:

f0 = 200 Hz
v = 343 m/s
vd = 20 m/s
f = ?

Since the receiver is approaching the stationary source, we must apply the equation

f = v + vd/v × f0

to calculate the detected frequency of sound waves. Thus, after substitutions we obtain

f = 343 m/s+20 m/s/343 m/s × 200 Hz
= 211.7 Hz

As you see, the frequency detected by the listener is higher than the original frequency emitted by the source. As a result, the listener will hear a higher-pitch sound than it would hear if he was not moving.

### b) Source moving, detector stationary

In this case, there is no vd in the formula as the detector is not moving. Now, there is the source speed vs the new quantity to be considered.

During a given time interval t, the wavefront moves at the distance d0 = v × t and the source at the distance ds = vs × t. Therefore, the relative distance of the two moving systems is

d = d0 - ds
= v × t - vs × t

Since the distance d is a multiple of wavelength (d = N × λ) and the time t is a multiple of period (t = N × T), we can focus on one wavelength only. Therefore, the equation above becomes

λ = v × T - vs × T

The new frequency detected (based on the waves' equation) is

f = v/λ

Substituting the expression for the wavelength obtained earlier in the above equation, we obtain

f = v/v × T - vs × T

Given that T = 1/f0 where f0 is the original frequency emitted by the source, we obtain

f = v/v/f0 - vs/f0
= v/v - vs/f0

Thus, the frequency detected by a stationary listener when the sound source is approaching him, is

f = v/v-vs × f0

Using the same approach for a sound source moving away from a stationary detector, we obtain

f = v/v + vs × f0

#### Example 2

An ambulance car emits sound waves of frequency equal to 1500 Hz. If the car is moving away from a stationary man at 37 m/s, what is the frequency detected by the man's ear? Take the speed of sound in air equal to 343 m/s.

#### Solution 2

We have the following clues in this problem:

f0 = 1500 Hz
vs = 37 m/s
v = 343 m/s
f = ?

Since the source is moving away from the stationary receiver, we must use the equation

f = v/(v+vs ) × f0

to calculate the frequency of sound detected by him. Thus, substituting the above values, we obtain

f = 343 m/s/343 m/s + 37 m/s × 1500 Hz
= 343 m/s/380 m/s × 1500 Hz
= 1354 Hz

Therefore, the man in this will hear a lower-pitched sound than usual. This will decrease his state of alert, as he understands that the ambulance car is moving away from his location even though he does not see the car.

### c) Source and detector are both moving

In this case, we must consider both the aforementioned factors, i.e. we must include three speeds in the formula - the speed of wave v, that of the source vs and the speed of detector vd. The signs in the formula depend on the relative direction of motion. The general formula for the Doppler Effect for the new frequency in this case, is:

f = v ± vd/v ± vs × f0

#### Example 3

A source moving due right at 30 m/s emits an acoustic (sound) signal. A receiver moves towards the source (due left) at 20 m/s. What is the sound frequency detected by the receiver if the original sound emitted by the source has a frequency equal to 400 Hz? Take the speed of sound in air equal to 343 m/s.

#### Solution 3

There are two things to consider in this problem:

1. The source is moving towards the receiver (detector). This means we have to write v - vs in the denominator.
2. The receiver (detector) is moving towards the source. This means we have to write v + vd in the numerator.

Therefore, the Doppler Effect formula for this scenario becomes:

f = v + vd/v - vs × f0

where, v = 343 m/s, vd = 20 m/s, vs = 30 m/s and f0 = 400 Hz.

Substituting the above values, we obtain for the new frequency:

f = 343 m/s + 20 m/s/343 m/s - 30 m/s × 400 Hz
= 363 m/s/313 m/s × 400 Hz
= 463.9 Hz

This means the receiver will hear a higher-pitch sound than normal.

## The Doppler Effect in Light Waves

The Doppler Effect in light waves occurs in a similar way as in sound waves. The only difference is the introduction of a new factor, β = v/c in the formula of the new light frequency, where v is the relative speed of the moving object to the source and c is the speed of light in vacuum (c = 3 × 108 m/s). There are two possible cases in this regard:

### 1. When source and detector are separating from each other

The formula of frequency for the Doppler Effect in this case is:

f = f0 × √1 - β/1 + β

The new frequency will be lower than the original one. Therefore, a phenomenon known as "red-shift" does occur. It means the detected light frequency will shift towards the red part of visible light spectrum, as red light represents the part of the visible light spectrum with the lowest frequency.

### 2. When source and detector are approaching each other

In this case, we have:

f = f0 × √1 + β/1 - β

Obviously, the new frequency is higher than the original one. Therefore, a phenomenon known as "blue-shift" does occur. It means the detected light frequency will shift towards the blue part of visible light spectrum as blue light represents a part of the visible light spectrum with high frequency (it should have been violet, but violet belongs to the family of blue light).

### Example 4

A star emits an apparently blue light of frequency 7.5 × 1014 Hz but when measured by using special techniques, it results that the original frequency of the light emitted by this star is 5.45 × 1014 Hz (green light). What is happening to the star if we take the Earth as a frame of reference?

### Solution 4

Since the original light is green but it appears blue, there is a blue shift in the light frequency. This means the star is approaching the Earth.

To calculate the approaching speed v, we must therefore use the formula

f = f0 × √1 + β/1 - β

Thus, we have

7.5 × 1014 = 5.45 × 1014 × √1 + β/1 - β
1.376 = √1 + β/1 - β
1.894 = 1 + β/1 - β
1.894-1.894 × β = 1 + β
0.894 = 2.894 × β
β = 0.3089

Since

β = v/c

we obtain for the approaching speed of the star:

0.3089 = v/3×108
v = 0.3089 × 3 × 108
= 9.267 × 107 m/s
= 92 670 km/s

Remark! We can also use the wavelength instead of frequencies in all the above formulae. We must only transform them in the proper way in order to fit the theory.

## Physics Revision: The Doppler Effect Summary

When a wave source is moving in a certain direction, the wavelength decreases in the moving direction and increases in the opposite side. This means the sound frequency also changes compared to the original frequency emitted by the source.

By definition, the effect of frequency change when the source and the detector are in relative motion to each other is known as the Doppler Effect.

There are three possible cases of source and detector movement in respect to each other. They are:

1. Detector moving, source stationary
2. Source moving, detector stationary
3. Both detector and source are moving

When the detector is moving and the source is stationary, we have

f = v + vd/v × f0

for the new frequency when the detector is moving towards the source and

f = v - vd/v × f0

for the new frequency when the detector is moving away from the source.

If the source is moving and the detector is stationary, we have

f = v/v - vs × f0

when the source is moving towards the stationary detector and

f = v/v + vs × f0

when the source is moving away from the stationary detector.

When combining all the above equations, we obtain the general Doppler Effect formula

f = v ± vd/v ± vs × f0

The signs are determined by considering the relative movement of source and receiver in respect to each other.

The Doppler Effect in light waves occurs in a similar way as in sound waves. The only difference is the introduction of a new factor, β = v/c in the formula of the new light frequency, where v is the relative speed of the moving object to the source and c is the speed of light in vacuum (c = 3 × 108 m/s). There are two possible cases in this regard:

1. When source and detector are separating from each other. The formula of frequency for the Doppler Effect in this case is:

f = f0 × √1 - β/1 + β

The new frequency will be lower than the original one. Therefore, a phenomenon known as "red-shift" does occur. It means the detected light frequency will shift towards the red part of visible light spectrum, as red light represents the part of the visible light spectrum with the lowest frequency.

2. When source and detector are approaching each other. In this case, we have:

f = f0 × √1 + β/1 - β

Obviously, the new frequency is higher than the original one. Therefore, a phenomenon known as "blue-shift" does occur. It means the detected light frequency will shift towards the blue part of visible light spectrum as blue light represents a part of the visible light spectrum with high frequency (it should have been violet, but violet belongs to the family of blue light).

## Physics Revision Questions for The Doppler Effect

1. An ambulance car normally emits a sound wave of frequency 1500 Hz. A girl hears a 1560 Hz siren from such a car. How is the ambulance car moving? Take the speed of light in air equal to 343 m/s.

1. At 13.2 m/s away from the girl
2. At 13.2 m/s towards the girl
3. At 13.7 m/s away from the girl
4. At 13.7 m/s towards the girl

2. Two identical airplanes are approaching each other at equal speeds. Each of them emits a sound of frequency equal to 2000 Hz when at rest. What is the flying speed of each airplane if pilots hear a 4000 Hz sound wave emitted by the other airplane? Take the speed of sound in air equal to 343 m/s.

1. 57.7 m/s
2. 343 m/s
3. 228.7 m/s
4. 114.3 m/s

3. A star is moving away from our galaxy at 0.2 × c where c = 3 × 108 m/s is the speed of light in vacuum. As a result, the frequency of light it emits appears to be 4.3 × 1014 Hz (red light). What is the frequency of the original light emitted by the star?

1. 3.51 × 1014 Hz
2. 2.90 × 1014 Hz
3. 6.525 × 1014 Hz
4. 5.27 × 1014 Hz