The Doppler effect: why a passing siren drops in pitch

Doppler effect

Why does an ambulance sound higher coming toward you and lower going away?

Set a sound source moving and watch its wavefronts crowd together ahead and spread out behind. The listener in front hears f₀·c/(c − v), the one behind f₀·c/(c + v); past the speed of sound the fronts pile into a shock cone.

High school
Moving source
Observed frequency
Wavelength compression
Shock waves and Mach cone

v_s\,(\text{m/s})

supersonic

200

f_0\,(\text{Hz})

500

t

8.00

f = f_0\,\dfrac{c}{c \mp v_s}

$M = v_s / c$: 0.59
$f_{\text{ahead}} = f_0\dfrac{c}{c - v_s}$: 1214.3
$f_{\text{behind}} = f_0\dfrac{c}{c + v_s}$: 314.8
$\lambda_{\text{ahead}}$: 0.280
$\lambda_{\text{behind}}$: 1.080
$\alpha\,(^\circ)$: —

Sourceslower than sound

Guided tour

1A source at rest
A $500\,\text{Hz}$ source standing still sends out evenly spaced circles. Both listeners hear $500\,\text{Hz}$ , and the wavelength is $c/f_0=340/500=0.68\,\text{m}$ in every direction.
2Moving at a tenth of the speed of sound
At $v_s=34\,\text{m/s}$ the source chases its own waves. Ahead the fronts are only $0.612\,\text{m}$ apart and the listener hears $500\times\dfrac{340}{306}\approx555.6\,\text{Hz}$ ; behind they are $0.748\,\text{m}$ apart, $\approx454.5\,\text{Hz}$ .
3Half the speed of sound
At $v_s=170\,\text{m/s}$ the effect is no longer subtle: an octave up ahead, $1000\,\text{Hz}$ , and down to $333.3\,\text{Hz}$ behind. Note it is not symmetric: the rise ahead is bigger than the drop behind.
4Faster than sound
At $400\,\text{m/s}$ the source outruns every front it emits. No sound reaches the listener ahead before the source does; instead the fronts pile up on a cone with half-angle $\sin\alpha=c/v_s$ , about $58.2^\circ$ . Crossing that cone is the sonic boom.

Explanation

1
Same waves, squeezed
The source still emits $f_0$ fronts per second, but it moves $v_s/f_0$ forward between two of them. Ahead each front is that much closer to the previous one, behind that much further:
$\lambda_{\text{ahead}}=\frac{c-v_s}{f_0},\qquad\lambda_{\text{behind}}=\frac{c+v_s}{f_0}$
Sound still travels at $c$ , so a shorter wavelength means more fronts per second: a higher pitch.
2
The formula
Dividing $c$ by those wavelengths gives what a listener at rest hears:
$f=f_0\,\frac{c}{c\mp v_s}$
minus when the source approaches, plus when it recedes. If instead the listener moves at $v_o$ toward a still source, $f=f_0\dfrac{c+v_o}{c}$ : the wavelength is unchanged but fronts are met more often. In both cases only motion along the line between them counts.
3
Where it is used
Police radar and weather radar measure speed from the frequency shift of a reflected wave. Doppler ultrasound measures blood flow. Light shows the same effect: galaxies moving away from us are redshifted, the first evidence that the universe is expanding.

Check yourself

Q01

A $680\,\text{Hz}$ siren approaches at $40\,\text{m/s}$ ( $c=340\,\text{m/s}$ ). You hear about

Q02

As a source moves away from you, compared with when it is still,

Q03

A jet flies at twice the speed of sound. The half-angle of its shock cone is

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