The Doppler effect for sound waves

Section: Waves | Syllabus: Cambridge AS Level Physics (9702)

The Doppler Effect

Think of a stationary car sounding its horn. When you stand beside it you hear the horn at a certain pitch. We detect the pitch of a sound by its frequency, which is the number of complete waves reaching your ear per second.

The apparent change in wave frequency with relative motion of a source and an observer is called the Doppler effect.

Doppler Effect The change in observed frequency (and wavelength) of a wave when there is relative motion between the source and the observer. Key Point

The Doppler effect is only noticeable if the speed of the source relative to the observer is a significant proportion of the wave speed.

Understanding the Effect

Stationary Source

For a stationary car, sound waves move towards an observer with compressions equally spaced. The wavelength doesn't change, so the number of compressions reaching the observer per second is constant and determined by the wave speed.

Source Moving Towards Observer

When the car moves quickly towards the observer:

The frequency of sound waves leaving the car is the same (the horn makes the same sound)
The speed of sound waves in air doesn't change (determined by air properties)
However, the wavelength between the car and observer becomes shorter
More compressions reach the observer per second
This is detected as an increase in pitch (higher frequency)

Source Moving Away From Observer

When the car moves quickly away from the observer:

The compressions arrive less frequently
The wavelength between source and observer is longer
The pitch heard by the observer is lower (lower frequency)

FIG 7.5: The Doppler Effect

Show a moving sound source (e.g., car with horn) emitting circular wavefronts. Demonstrate three scenarios: (a) Stationary source with evenly spaced wavefronts. (b) Source moving right: wavefronts compressed in front (shorter λ, higher f) and stretched behind (longer λ, lower f). Label observer positions and indicate that observer in front hears higher pitch, observer behind hears lower pitch.

Deriving the Doppler Equation

From the wave equation v = fλ, we know that frequency is inversely proportional to wavelength for constant wave speed v.

Consider the case where the source is moving towards the observer:

Step 1: A compression that leaves the source travels a distance v m in one second.

Step 2: The source travels towards the observer with speed v_s, so moves v_s m in one second.

Step 3: Another compression, leaving one second after the first, will be a distance (v - v_s) m behind the first compression.

Step 4: The number of compressions leaving per second is f_s.

Step 5: So the distance (v - v_s) m contains f_s wavelengths:

v - v_s = f_s × _o

Step 6: The observed wavelength is: _o = (v - v_s/f_s)

Step 7: The observed frequency f_o = v ÷ λ_o:

f_o = (f_s · v/v - v_s)

Similarly, for source moving away:

f_o = (f_s · v/v + v_s)

The General Doppler Equation

f_o = (f_s · v/v ± v_s)

where:

f_o = observed frequency (Hz)
f_s = source frequency (Hz)
v = speed of sound in the medium (m s⁻¹)
v_s = speed of the source (m s⁻¹)

Sign Convention

Use minus (−) when source moves towards observer → f_o > f_s
Use plus (+) when source moves away from observer → f_o < f_s
When the distance between source and observer is increasing, the sign is positive
When the distance is decreasing, the sign is negative

Worked Examples

Worked Example 1: Police Car Siren

Question: A police car is travelling on a straight road at a constant speed of 35 m s⁻¹. It has a siren that emits a sound of frequency 1200 Hz. The police car passes a person standing close to the road. The speed of sound in air is 340 m s⁻¹. Calculate the frequency that the person observes when (a) the car is approaching, (b) the car is moving away.

Solution

(a) Car approaching (use minus):

f_o = (f_s · v/v - v_s) = (1200 × 340/340 - 35) = (408000/305) = 1300 Hz (2 sf)

(b) Car moving away (use plus):

f_o = (f_s · v/v + v_s) = (1200 × 340/340 + 35) = (408000/375) = 1100 Hz (2 sf)

Worked Example 2: Finding Source Speed

Question: The driver of a train sounds a horn that operates at 1350 Hz. A person close to the track hears the horn at a frequency of 1450 Hz. The speed of sound in air is 343 m s⁻¹. Calculate the speed of the train.

Solution

Since f_o > f_s, the train is approaching. Use minus sign.

f_o = (f_s · v/v - v_s)

Rearranging: f_o(v - v_s) = f_s · v

f_o · v - f_o · v_s = f_s · v

f_o · v_s = f_o · v - f_s · v

v_s = (f_o · v - f_s · v/f_o) = (1450 × 343 - 1350 × 343/1450)

v_s = (497350 - 463050/1450) = (34300/1450) = 23.7 m s^-1

Worked Example 3: Motorcycle Horn

Question: A motorcycle horn has a frequency of 1.1 kHz. An observer sees the motorcycle approaching and hears the horn with an apparent frequency of 1.2 kHz. Calculate the speed of the motorcycle. (Speed of sound = 340 m s⁻¹)

Solution

f_s = 1100 Hz, f_o = 1200 Hz, v = 340 m s⁻¹

1200 = (1100 × 340/340 - v_s)

340 - v_s = (1100 × 340/1200) = 311.67

v_s = 340 - 311.67 = 28 m s^-1

Worked Example 4: Train Engine Sound

Question: The sound produced by the engine of a train has a wavelength of 17.4 cm when emitted in air. An observer sees the train approaching at a speed of 25.0 m s⁻¹. Take the speed of sound in air as 351 m s⁻¹. (a) Calculate the wavelength received by the observer. (b) Calculate the frequency received by the observer.

Solution

First find source frequency: f_s = v/λ = 351/0.174 = 2017 Hz

(a) Using λ_o = (v - v_s)/f_s:

_o = (351 - 25/2017) = (326/2017) = 0.162 m (16.2 cm)

(b) f_o = v/λ_o = 351/0.162 = 2170 Hz

Or directly: f_o = (2017 × 351/351 - 25) = 2170 Hz

Key Ideas Summary

If a source of sound waves moves with a speed that is a significant proportion of the wave speed, this causes changes in the observed wave properties
When the source moves towards a stationary observer, the observer detects an increase in frequency (higher pitch)
When the source moves away from a stationary observer, the observer detects a decrease in frequency (lower pitch)
An increase in frequency is heard as a higher pitch
A decrease in frequency is heard as a lower pitch
The wave speed in the medium remains constant-only the observed wavelength and frequency change

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The Doppler effect for sound waves