Doppler Effect (Acoustics)
The Doppler effect describes the frequency shift when source or observer move relative to each other: approach raises, receding lowers the perceived frequency.
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Formula
f_B = f_Q \cdot \frac{c \pm v_B}{c \mp v_Q}Variables & units – Doppler Effect (Acoustics)
| Symbol | Meaning | Unit |
|---|---|---|
| f_B | Frequency perceived by the observer | Hz |
| f_Q | Frequency emitted by the source | Hz |
| c | Speed of sound (air: 343 m/s) | m/s |
| v_B | Speed of the observer | m/s |
| v_Q | Speed of the source | m/s |
Derivation & background – Doppler Effect (Acoustics)
Christian Doppler described the effect in 1842. If the source moves towards the observer, the wavefronts are compressed (shorter wavelength, higher frequency); when receding they are stretched. Sign rule: upper sign for approach, lower for receding; a moving source acts in the denominator, a moving observer in the numerator, and the two cases are not symmetric. For light the relativistic Doppler effect applies, the basis of redshift in astronomy.
Exam blueprint
Validity range
Applies to sound in a stationary medium as long as source and observer move slower than the speed of sound. For light, the relativistic Doppler formula without a medium applies instead.
Derivation steps
A moving source compresses or stretches the spacing of the wavefronts.
- 1Between two crests the source travels v_Q·T_Q, so the wavelength becomes λ = (c − v_Q)·T_Q.
- 2A stationary observer hears f_B = c/λ = f_Q·c/(c − v_Q); a moving observer additionally changes the numerator.
Rearrangements
Source speed
For an approaching source and stationary observer, the working principle of speed radar.
Emitted frequency
Back-calculation from the heard to the actual frequency.
Task variant
A siren (800 Hz) recedes at 25 m/s. What frequency do you hear? (c = 343 m/s)
Receding source: f_B = f_Q·c/(c + v_Q) = 800 × 343/368 ≈ 746 Hz, the frequency drops.
You move at 20 m/s towards a stationary source (500 Hz). What do you hear?
Moving observer: f_B = f_Q·(c + v_B)/c = 500 × 363/343 ≈ 529 Hz.
Common mistakes
Treating moving source and moving observer as interchangeable.
The source acts in the denominator, the observer in the numerator; only for v << c do both approximately agree.
Choosing signs so the frequency drops during approach.
Sanity check: approach always raises the frequency, receding lowers it.
Substituting speeds in km/h.
Convert all speeds to m/s (divide by 3.6), c = 343 m/s.
Exam context
- Exams ask the four basic cases (source/observer, approach/recede), back-calculating the speed and qualitative explanations with wavefront sketches.
These mistakes cost points in real exams. The set drills them until they stick.
Formula cluster
Wave phenomena
Belongs with the wave equation and interference in acoustics and optics.
Worked example
An ambulance (f_Q = 700 Hz) approaches at v_Q = 30 m/s (c = 343 m/s): f_B = 700 × 343/(343 − 30) ≈ 767 Hz. When receding: f_B = 700 × 343/373 ≈ 644 Hz.
Applications
Radar speed measurement, Doppler sonography (blood flow), astronomy (red and blue shift), sirens, bat echolocation
Quanta exam set
Curated exam set for "Doppler Effect (Acoustics)":
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Which formula describes Doppler Effect (Acoustics)?
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How do you rearrange f_B = f_Q·(c ± v_B)/(c ∓ v_Q) for Source speed?
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Which common mistake happens with Doppler Effect (Acoustics)?
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Scientific sources
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Frequently asked questions about Doppler Effect (Acoustics)
How do you calculate the frequency in the Doppler effect?+
First pick the correct case. If the source moves towards a stationary observer, f_B = f_Q·c/(c − v_Q); if it recedes, the denominator is c + v_Q. If the observer moves towards a stationary source, f_B = f_Q·(c + v_B)/c. Example: an ambulance at 700 Hz approaching at 30 m/s gives f_B = 700 × 343/313 ≈ 767 Hz. All speeds must be in m/s; the speed of sound is 343 m/s at 20 °C. Always check the result against the basic rule: approach raises the pitch, receding lowers it.
Why does a passing siren sound high first and then low?+
While the vehicle approaches, it compresses the sound waves ahead of it: the crests arrive at shorter intervals and the frequency is raised. At the moment of passing you briefly hear the true emitted frequency. Afterwards the receding source stretches the waves behind it and the pitch audibly drops. At 700 Hz and 30 m/s the tone jumps from about 767 Hz to 644 Hz, a clearly noticeable interval of almost three semitones. Note: during the approach the pitch stays constantly high, it does not glide; the characteristic jump happens only at the moment of passing, when the radial velocity reverses.
Does it matter whether the source or the observer moves?+
No, for sound the two cases are physically different because the medium, the air, forms a preferred reference frame. A moving source changes the wavelength in the medium and appears in the denominator: f_B = f_Q·c/(c ∓ v_Q). A moving observer merely sweeps the unchanged waves faster or slower and appears in the numerator: f_B = f_Q·(c ± v_B)/c. Numerical example with v = 34.3 m/s (10 % of c) and approach: moving source gives the factor 1/0.9 ≈ 1.111, moving observer 1.1. Only for speeds small compared with c do both approximations merge; for light the distinction does not exist at all.
How do the police measure speed with the Doppler effect?+
A radar gun emits an electromagnetic wave of known frequency. The moving car reflects it and acts twice: once as a moving receiver, once as a moving transmitter. The returning wave is therefore shifted by Δf ≈ 2·f·v/c, twice as much as in the single Doppler effect. The measured frequency shift directly yields the speed: v = Δf·c/(2f). Since light rather than sound is used, the relativistic formula applies, but for v << c it reduces exactly to this simple approximation. Doppler sonography of blood flow and weather radar use the same principle.
What happens when the source becomes faster than sound?+
For v_Q → c the formula grows towards infinity and for v_Q > c it loses validity: ahead of the source the wave crests can no longer separate. The source overtakes its own wavefronts, which pile up into a cone, the Mach cone. Its half-angle follows from sin θ = c/v. For a supersonic aircraft you hear this cone as a sonic boom when it sweeps over you, continuously along the flight path, not just when breaking the sound barrier. The ratio v/c is the Mach number: Mach 2 means twice the speed of sound and a cone angle of 30°.
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How do you calculate with Doppler Effect (Acoustics)?
Here is how to work through a typical Doppler Effect (Acoustics) (f_B = f_Q·(c ± v_B)/(c ∓ v_Q)) task step by step:
- 1
Task
A siren (800 Hz) recedes at 25 m/s. What frequency do you hear? (c = 343 m/s)
Solution path
Receding source: f_B = f_Q·c/(c + v_Q) = 800 × 343/368 ≈ 746 Hz, the frequency drops.
- 2
Task
You move at 20 m/s towards a stationary source (500 Hz). What do you hear?
Solution path
Moving observer: f_B = f_Q·(c + v_B)/c = 500 × 363/343 ≈ 529 Hz.