You wait at a railway level crossing while a train approaches, sounding its horn. The horn sounds high and sharp as the engine comes near. The moment the engine passes, the note drops, and it stays low as the train moves away. The driver hears one steady note all the time. The horn itself never changed.
The Doppler effect
When a vehicle passes you sounding its horn, you notice something interesting. The pitch is higher as the vehicle approaches and lower as it moves away. This apparent change in frequency, caused by relative motion between a source and an observer, is called the Doppler effect.
Let a source of waves and an observer move relative to each other along the line joining them. Then the frequency received by the observer differs from the frequency emitted by the source. This apparent change of frequency is the Doppler effect.
The Doppler effect occurs with all types of waves: sound, light and water waves. In this chapter we focus on sound waves. For sound, every speed is measured in the air, or in whatever medium carries the sound.
Which kinds of waves show the Doppler effect?
Building intuition: the conveyor belt
Before the mathematics, let us build an intuition with an analogy. Imagine a source placing chocolates on a conveyor belt at a constant rate. That rate is a frequency. An observer at the other end of the belt receives the chocolates. Now, what happens if the source or the observer starts moving?
Play each chip of Fig 3.1 for its \(10\,\text{s}\). S places \(26\) dots in every case. With both at rest, O receives one dot every \(0.4\,\text{s}\), the same rate at which S places them. When S moves towards O, the dots it places lie closer together on the belt. They then reach O every \(0.28\,\text{s}\), and O's counter ends at \(30\). When O walks towards S, O meets the dots every \(0.31\,\text{s}\) or so, and the counter ends at \(32\).
As you can see from the analogy:
- When the source moves towards the observer, the chocolates are bunched up on the belt and arrive more often.
- When the observer moves towards the source, the observer intercepts the chocolates at a faster rate.
- The opposite effects occur when either one moves away from the other.
For sound waves, the chocolates stand for compression pulses, the maxima of the sound wave. The rate at which the source emits them is the true frequency \(f\). The rate at which the observer receives them is the apparent frequency \(f'\). Let us now put this idea into mathematics.
In the conveyor-belt analogy for sound, what do the chocolates stand for, and what does the rate of receiving them stand for?
The mathematics of the Doppler effect
We will derive the Doppler formula from the time interval between the arrivals of consecutive compression maxima at the observer. Let \(c\) denote the speed of sound in the medium. The examples of this chapter take \(c = 340\,\text{m/s}\).
Case 1: source at rest, observer moving
Consider a source at rest in the medium, emitting sound of frequency \(f\). The time between the emissions of consecutive maxima is the period \(T = 1/f\). The maxima travel through the medium at speed \(c\). Since the source is at rest, consecutive maxima are separated by a distance \(cT\) in the medium.
Suppose an observer moves towards the source at speed \(v_O\). Consider the moment the observer receives one compression maximum. The next maximum is then at a distance \(cT\) away. It travels towards the observer at speed \(c\), while the observer moves towards it at speed \(v_O\). So the relative speed of approach is
\[ v_{\text{rel}} = c + v_O \]
This is the relative velocity on a line, from the chapter Relative Velocity in One Dimension.
\[ v_{\text{rel}} = c + v_O \]
Why. the maximum and the observer move towards each other, so their speeds add.
\[ T' = \dfrac{cT}{c + v_O} \]
Why. the gap \(cT\) to the next maximum closes at the relative speed; time equals distance divided by speed.
\[ f' = \dfrac{1}{T'} = \dfrac{c + v_O}{cT} \]
Why. the apparent frequency is the reciprocal of the observed period.
\[ f' = \dfrac{c + v_O}{c} \cdot f \]
Read it. \(1/T = f\).
When the observer moves towards a source at rest, the frequency increases. Fig 3.2 shows the count. In \(6T\) the observer at rest receives \(6\) maxima. Moving towards S at \(0.5c\), the observer receives \(9\), because arrivals come every \(0.67T\).
Observer moving away
Now let the observer move away from the source at speed \(v_O\). The next maximum still moves towards the observer at \(c\), but the observer now runs from it. The relative speed of approach is \(c - v_O\). Following the same logic,
\[ T' = \dfrac{cT}{c - v_O}, \qquad f' = f\left(\dfrac{c - v_O}{c}\right) \]
When the observer moves away from a source at rest, the frequency decreases. In Fig 3.2, the receding observer receives only \(3\) maxima in \(6T\).
The moving observer does not change the sound in the air at all. The maxima are still \(cT\) apart in the medium. The observer only meets them more often, or less often, than a listener at rest would.
A fire station siren at rest sounds at \(680\,\text{Hz}\). A student rides a scooter at \(20\,\text{m/s}\) along the road through the station. Take \(c = 340\,\text{m/s}\). What frequency does the student hear (a) while riding towards the siren, and (b) after passing it, while riding away?
(a) Towards the source: \(f' = f(c + v_O)/c = 680 \times 360/340 = 720\,\text{Hz}\).
(b) Away from the source: \(f' = f(c - v_O)/c = 680 \times 320/340 = 640\,\text{Hz}\).
The shift is \(v_O/c = 20/340\) of \(680\,\text{Hz}\), that is, \(40\,\text{Hz}\) each way. The pitch drops by \(80\,\text{Hz}\) as the student passes.
A source at rest emits sound of \(680\,\text{Hz}\). An observer moves straight towards it at \(17\,\text{m/s}\). Take \(c = 340\,\text{m/s}\). What frequency does the observer hear, in Hz?
An observer moves directly away from a source at rest at exactly the speed of sound, \(c\). What frequency does the observer hear?
When you walk towards a source at rest, the wavelength of the sound in the air does not change. Why, then, does the frequency you hear rise?
The frequency heard is the number of maxima reaching the ear per second. The maxima are still \(cT\) apart, but you move to meet them, so the gap closes at \(c + v_O\) instead of \(c\). You meet each next maximum after \(cT/(c + v_O)\), which is shorter than \(T\). So more maxima arrive per second, and \(f' = f(c + v_O)/c\).
In 1842 the Austrian physicist Christian Doppler read a paper in Prague on the colours of double stars. He argued that a star moving towards the Earth should look bluer and one moving away should look redder. The waves from the star would reach us more often, or less often, than they left it. His idea about star colours was mostly wrong in detail. The principle was right, and it applies to every kind of wave.
Now return to the level crossing. As the engine approaches, the horn's maxima reach you more often than the horn emits them, so you hear a higher note. After the engine passes, they reach you less often, and the note is lower. The driver moves with the horn, so the driver hears the true note throughout. In the next lesson the moving source gets its own derivation.
- Adding the observer's speed to the source frequency, for example \(1000 + 20 = 1020\,\text{Hz}\).The speed changes the rate of meeting maxima. The factor is \((c + v_O)/c\), so the shift is \(f\,v_O/c\).
- Thinking that a moving observer changes the wavelength in the air.Only the source is in the air making the wave. A moving observer meets the same maxima, \(cT\) apart, at a different rate.
- Using \(c + v_O\) for an observer moving away.Motion that brings the two closer raises the frequency. Away from the source the numerator is \(c - v_O\).