The Doppler effect
Introduced| English | Meaning |
|---|---|
| pitch/pɪtʃ/ | The perceived highness or lowness of a sound, mainly related to its frequency. |
| frequency/ˈfriːkwənsi/ | The number of complete cycles per unit time. |
| Doppler effect/ˈdɒplə ɪˈfekt/ | A change in observed wave frequency caused by relative motion between a source and an observer. |
| source/sɔːs/ | The object or region that produces a wave. |
| wavefronts/ˈweɪvfrʌnts/ | wavefronts |
| observer/ɒbˈzɜːvə/ | A person or measuring system relative to which an event or motion is described. |
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:
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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.
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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.
A source of sound moves towards a stationary listener. Which are unchanged? Select all that apply.
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The medium fixes the speed and the source fixes what it emits. What the motion changes is the WAVELENGTH ahead of the source, and hence the frequency heard.
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:
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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.
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Approaching → minus in the formula → higher; receding → plus → lower; the switch between the two is the drop you hear as it passes.
Put the Doppler calculation in order.
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The last step catches every sign error for free: an approaching source must give a HIGHER observed frequency.
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?
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$f_{\text{o}} = \dfrac{340 \times 400}{340 - 20} = \dfrac{136000}{320} = 425\ \text{Hz}$.
A source moving away from you gives a ____ frequency than the source.
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Moving away spreads the wavefronts out, lengthening the wavelength and lowering the frequency.
A siren emitting 800 Hz approaches at 30 m/s. What frequency is heard, in Hz? (speed of sound 340 m/s)
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f = 800 x 340/(340 - 30) = 877 Hz. Approaching means minus on the bottom and an answer ABOVE 800 Hz; getting a lower number is the sign error announcing itself.
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?
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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.
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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 $v_{\text{s}}$