The Doppler effect
| English | Chinese | Pinyin |
|---|---|---|
| pitch | 音调 | yīn diào |
| frequency | 频率 | pín lǜ |
| Doppler effect | 多普勒效应 | duō pǔ lè xiào yìng |
| source | 波源 | bō yuán |
| wavefronts | 波前 | bō qián |
| observer | 观察者 | guān chá zhě |
The passing siren
- An ambulance races past and its siren suddenly drops in pitch 音调.
- The siren itself never changed — your ear heard a different frequency 频率.
- This is the Doppler effect 多普勒效应.
Why the pitch changes
- A source moving toward you bunches the wavefronts 波前 ahead → shorter $\lambda$ → higher pitch.
- Moving away, the wavefronts spread out → longer $\lambda$ → lower pitch.
Doppler effect
Send the source moving and watch the wavefronts bunch up ahead (higher pitch) and stretch out behind — the siren effect, controlled by the source's speed.
As a sound source moves toward you, the pitch you hear is:
Moving toward you bunches the wavefronts, shortening the wavelength and raising the frequency you hear.
A source moving toward an observer bunches the wavefronts ahead of it.
Yes — each new wavefront is sent from a point a little closer, so they crowd together in front.
What does not change
- The source 波源 still emits $f_{\text{s}}$ waves every second — its frequency is fixed.
- The waves still travel through the air at the same speed $v$; the moving source cannot push them faster.
- Only the spacing of the wavefronts changes, so only the observed wavelength and frequency change.
Where the formula comes from
- In one period $T = \dfrac{1}{f_{\text{s}}}$ the source moves $v_{\text{s}}T$, so the wavelength ahead of it is squeezed to $\lambda' = \lambda - v_{\text{s}}T = \dfrac{v - v_{\text{s}}}{f_{\text{s}}}$.
- The observer 观察者 hears $f_{\text{o}} = \dfrac{v}{\lambda'}$.
- Behind the source the wavelength is stretched instead: $\lambda' = \dfrac{v + v_{\text{s}}}{f_{\text{s}}}$.
The formula
- For a moving source and a still observer: $f_{\text{o}} = \dfrac{v\,f_{\text{s}}}{v \pm v_{\text{s}}}$.
- Use minus when approaching (higher pitch), plus when receding (lower pitch).
A moving source squashes the wavefronts ahead of it, raising the observed frequency
In $f_{\text{o}} = \dfrac{v\,f_{\text{s}}}{v \pm v_{\text{s}}}$, for a source moving toward the observer you use:
A smaller denominator gives a larger $f_{\text{o}}$ — the higher pitch you expect when approaching.
Match each situation to what a stationary listener hears.
Approaching → minus in the formula → higher; receding → plus → lower; the switch between the two is the drop you hear as it passes.
Worked example: approaching
- A horn at $f_{\text{s}} = 800\ \text{Hz}$ moves at $30\ \dfrac{\text{m}}{\text{s}}$ toward you; $v = 340\ \dfrac{\text{m}}{\text{s}}$.
- Approaching → minus sign: $f_{\text{o}} = \dfrac{340 \times 800}{340 - 30}$.
- $f_{\text{o}} = \dfrac{272000}{310} \approx 877\ \text{Hz}$ — higher, as expected.
A horn ($f_{\text{s}} = 400\ \text{Hz}$) moves at $20\ \dfrac{\text{m}}{\text{s}}$ toward you; $v = 340\ \dfrac{\text{m}}{\text{s}}$. What frequency do you hear?
$f_{\text{o}} = \dfrac{340 \times 400}{340 - 20} = \dfrac{136000}{320} = 425\ \text{Hz}$.
A source moving away from you gives a ____ frequency than the source.
Moving away spreads the wavefronts out, lengthening the wavelength and lowering the frequency.
Worked example: a source going round in a circle
A siren on a car driving round a circular track is heard by someone standing outside the track. The frequency they hear varies between a maximum of $1000\ \text{Hz}$ and a minimum of $818\ \text{Hz}$. Take $v = 340\ \dfrac{\text{m}}{\text{s}}$. Find the speed of the car and the siren's own frequency.
- Maximum is heard when the car moves straight toward the listener: $1000 = \dfrac{340\,f_{\text{s}}}{340 - v_{\text{s}}}$.
- Minimum when it moves straight away: $818 = \dfrac{340\,f_{\text{s}}}{340 + v_{\text{s}}}$.
- Divide the two equations so $f_{\text{s}}$ cancels: $\dfrac{1000}{818} = \dfrac{340 + v_{\text{s}}}{340 - v_{\text{s}}}$, giving $v_{\text{s}} = 34\ \dfrac{\text{m}}{\text{s}}$.
- Source frequency: $f_{\text{s}} = \dfrac{1000 \times (340 - 34)}{340} = 900\ \text{Hz}$.
- Check: the true frequency lies between the two heard values, and the car's speed is about a tenth of the speed of sound — sensible for a racing car.
A siren of frequency $900\ \text{Hz}$ moves directly away from you at $34\ \dfrac{\text{m}}{\text{s}}$. The speed of sound is $340\ \dfrac{\text{m}}{\text{s}}$. What frequency do you hear, in Hz?
Receding → plus sign: $f_{\text{o}} = \dfrac{340 \times 900}{340 + 34} = \dfrac{306000}{374} = 818\ \text{Hz}$.
Three traps. The $v$ on top is the speed of the wave ($340\ \dfrac{\text{m}}{\text{s}}$ for sound), never the speed of the source. The $\pm$ sign belongs to the source speed. And while a source drives straight at you at constant speed, the pitch you hear is constant — it does not climb as the car gets nearer; it changes only as the car passes and its direction relative to you changes. Louder is not higher.
A police car drives straight toward you at a constant speed with its siren on. The pitch you hear rises steadily as it gets closer.
While it approaches head-on at constant speed the observed frequency is constant — it depends on the speed, not the distance. The sound gets louder, not higher. The pitch only drops as the car passes.
You've got it
- a moving source changes the frequency you hear — the Doppler effect; the source frequency and wave speed do not change
- toward → bunched wavefronts → higher; away → spread out → lower
- $f_{\text{o}} = \dfrac{v\,f_{\text{s}}}{v \pm v_{\text{s}}}$ (minus for approaching); divide the max and min equations to find $v_{\text{s}}$