Wavelength and wavefronts
To truly grasp the Doppler effect, we must separate two things. One is the physical pattern of the wave in the medium. The other is how an observer perceives that pattern.
What is wavelength?
Wavelength, \(\lambda\), is a purely spatial property of a wave. It is the smallest distance in space after which the pattern of the wave repeats itself. To measure it, you would take an instantaneous snapshot of the whole medium. Then you would measure the distance between two consecutive crests.
This is a measurement of a pattern in space at one fixed moment. So the motion of the observer has no role in deciding the wavelength. In simple words, the wavelength is like the length of the wave. A rod has a definite length, whoever looks at it. In the same way, the wave has a definite wavelength in the medium, however fast an observer moves.
The wavelength of a wave is the distance between two consecutive crests in the medium at one instant. It is fixed by the source and the medium. The motion of an observer does not change it.
Wavefronts from a source at rest
A wavefront is a surface of constant phase, such as a compression maximum, spreading outward from the source. Fig 5.1 shows a source at rest emitting wavefronts at a constant frequency.
Observe the key features of Fig 5.1:
- The source, the black dot, does not move.
- Each circle is a wavefront expanding outwards.
- The source is at rest, so every wavefront is emitted from the same central point. The result is a pattern of perfect concentric circles.
- The constant radial distance between any two adjacent circles is the wavelength, \(\lambda\). This wavelength is the same in all directions.
Wavefronts from a moving source
Now let the source move to the right at a constant speed below the speed of sound, a subsonic speed. This is a fundamentally different physical situation. Before you play it, predict what you will see.
A source moves to the right at half the speed of sound and emits a wavefront once every period. About which point does each wavefront expand?
Pay close attention to the process in the figure above:
- Each new wavefront is emitted from a new position, further to the right, where the source is at that moment.
- Once emitted, each wavefront expands as a perfect circle centred on the point from where it was emitted. It is not centred on the present, moving position of the source.
- This behaviour creates the pattern of the Doppler effect. Ahead of the source, to the right, the wavefronts are bunched up. This is a shorter, compressed wavelength \(\lambda'\). Behind the source, to the left, the fronts are spread apart. This is a longer, stretched wavelength.
The readout of the figure gives the spacings for \(v_S = 0.5c\). Ahead of the source, the fronts are half a wavelength apart. Behind it, they are one and a half wavelengths apart. The animation clearly shows that the source's motion physically changes the spatial pattern, the wavelength, of the wave in the medium.
Compare the two lessons. A moving observer meets the same wave more often: the wavelength in the air is unchanged. A moving source makes a different wave: the wavelength in the air itself is shorter ahead and longer behind. This is why the two formulas differ.
Deriving the changed wavelength
Let us put into equations what the animation shows. Let the source emit waves of frequency \(f\) and period \(T = 1/f\). Let the speed of the waves in the medium be \(v\), the same speed that the earlier lessons called \(c\).
Source moving towards the observer. A crest is emitted. In the time \(T\) before the next crest is emitted, this first crest travels a distance \(vT = \lambda\). In that same time, the source moves a distance \(v_S T\) in the same direction. The second crest is then emitted from this new position. The new, compressed wavelength \(\lambda'\) is the distance between the two crests.
\[ \lambda' = (\text{distance the crest travelled}) - (\text{distance the source travelled}) \]
Why. both start from the point of the first emission and move the same way for one period.
\[ \lambda' = vT - v_S T = (v - v_S)T \]
Why. in the time \(T\) the crest covers \(vT\) and the source covers \(v_S T\).
\[ \lambda' = \dfrac{v - v_S}{f} \]
Read it. substitute \(T = 1/f\).
Source moving away from the observer. In the same way, when the source moves away, it emits the second crest further back. The distance between the crests is increased:
\[ \lambda' = vT + v_S T = (v + v_S)T, \qquad \lambda' = \dfrac{v + v_S}{f} \]
An observer at rest measures these modified wavelengths. The observer finds the waves travelling at the normal speed \(v\). So the apparent frequency heard is \(f' = v/\lambda'\). With \(\lambda' = (v - v_S)/f\), this gives \(f' = fv/(v - v_S)\). That is exactly the formula derived from arrival times in the last lesson.
A source of period \(T\) moves at speed \(v_S\) towards an observer. The waves travel at \(v\). What is the distance between consecutive crests between the source and the observer?
A source of \(680\,\text{Hz}\) moves at \(34\,\text{m/s}\) towards an observer at rest. Take \(v = 340\,\text{m/s}\). What is the wavelength of the sound between the source and the observer, in metres?
A solved example
Let us apply the general formula to a classic problem. This example uses the speed of sound \(c = 340\,\text{m/s}\).
A train approaches a station at \(30\,\text{m/s}\), sounding a whistle with a true frequency of \(500\,\text{Hz}\). After passing the station, it continues to move away at the same speed. What frequencies are heard by a person standing on the platform (a) during the approach and (b) during the recession? Use \(c = 340\,\text{m/s}\) for the speed of sound.
Given: \(f = 500\,\text{Hz}\), \(c = 340\,\text{m/s}\), \(v_S = 30\,\text{m/s}\), and \(v_O = 0\), since the observer is at rest.
(a) The source moves towards the observer, so the frequency should increase. We use \((c - v_S)\) in the denominator: \(f'_{\text{approach}} = f\,c/(c - v_S)\). (b) The source moves away, so the frequency should decrease. We use \((c + v_S)\): \(f'_{\text{recede}} = f\,c/(c + v_S)\).
(a) \(f'_{\text{approach}} = 500 \times \dfrac{340}{340 - 30} = 500 \times \dfrac{340}{310} \approx 548.4\,\text{Hz}\). (b) \(f'_{\text{recede}} = 500 \times \dfrac{340}{340 + 30} = 500 \times \dfrac{340}{370} \approx 459.5\,\text{Hz}\).
The listener hears a sudden drop in pitch, from about \(548\,\text{Hz}\) to about \(460\,\text{Hz}\), as the train passes.
Play the three chips of Fig 5.3 in turn. The drop from \(548.4\,\text{Hz}\) to \(459.5\,\text{Hz}\) is about \(89\,\text{Hz}\). That is close to a musical interval of three semitones, which is easy to hear. On a real platform the drop happens in the second or two when the engine passes you.
Practice with the sign rule
Test your understanding with these three questions. Each one turns on the sign rule and on what moves relative to the medium.
True or false? If the source and the observer both move in the same direction at the same speed, there is no Doppler shift.
A police car with a \(1200\,\text{Hz}\) siren drives east, away from you. You drive west, away from it. The frequency you hear will be:
An ambulance at rest sounds a siren of frequency \(1000\,\text{Hz}\). You drive towards it at \(20\,\text{m/s}\). What frequency do you hear? Take \(c = 340\,\text{m/s}\).
In a hospital, a Doppler ultrasound probe sends sound of a few megahertz into the body. Red blood cells moving in an artery reflect it back at a slightly changed frequency. The machine compares the two frequencies and computes the speed of the blood. It often plays the difference as an audible whoosh, so a doctor can hear a narrowed artery. The same idea, with radio waves, lets the traffic police measure the speed of a car.
- Thinking that a moving observer changes the wavelength.The wavelength is a snapshot property of the medium. Only a moving source changes it; a moving observer changes only the rate of meeting crests.
- Centring every wavefront on the present position of a moving source.Each front grows about the point where it was emitted. The centres lag behind the source.
- Rounding the train answers before subtracting.Keep \(548.39\,\text{Hz}\) and \(459.46\,\text{Hz}\) until the end. The drop is about \(89\,\text{Hz}\).