7.3 - The Doppler effect for sound

7.3 - The Doppler effect for sound

Explain why a moving sound source is heard at a different frequency by a stationary observer. Choose the sign in the moving-source Doppler equation from the wavefront spacing, calculate observed frequencies, and work backwards to a source speed.

Moving sources change wavefront spacing

Why does a steady siren sound higher as its source approaches you and lower as it moves away? The source need not change its own oscillation frequency. Its motion changes the spacing of the sound waves reaching you.

Imagine a source emitting one compression every period TsT_s. Each compression travels through still air at sound speed vv. Before the next compression is emitted, the source has moved a distance vsTsv_sT_s. Ahead of the source, the next compression starts closer to the previous one; behind it, the next starts farther from the previous one.

[DIAGRAM: asset_name: Wavefronts from a moving source; asset_slug: moving-source-wavefronts; caption: Motion compresses wavefront spacing ahead and increases it behind; description: Circular wavefronts emitted at equal time intervals have centres along the past positions of a source moving to the right. All wavefronts expand at the same speed in the stationary medium. The current source is marked at the rightmost position. Smaller spacing is labelled ahead and larger spacing behind; a rightward arrow shows source motion.]
Motion compresses wavefront spacing ahead and increases it behind

The source frequency fsf_s counts oscillations per second at the source. The observed frequency fof_o counts compressions arriving per second at the observer. With a shorter wavelength ahead but the same sound speed through the medium, fo=v/λf_o=v/\lambda is higher. Behind the source the wavelength is longer, so the observed frequency is lower.

Doppler effect for a moving sound source

The observed frequency differs from the emitted frequency when a sound source moves relative to a stationary observer. Approaching source motion increases the observed frequency; receding source motion decreases it.

The air is not carried forward at the source speed. The wavefronts travel at the sound speed appropriate to the medium; source motion changes where successive wavefronts begin. This distinction is the physical reason for the frequency shift.

Model limit

Use this course's equation for a moving source and an observer stationary relative to the medium. The simple calculations treat motion along the source–observer line and a source slower than sound. A moving-observer formula is a different case and is not required here.

Choosing the Doppler sign

A period Ts=1/fsT_s=1/f_s separates successive emissions. The earlier wavefront travels vTsvT_s, while the source moves vsTsv_sT_s. Their difference gives the wavelength ahead; their sum gives the wavelength behind.

λahead=(v−vs)Tsλbehind=(v+vs)Ts\lambda_{\mathrm{ahead}}=(v-v_s)T_s \qquad \lambda_{\mathrm{behind}}=(v+v_s)T_s

Substituting each wavelength into fo=v/λf_o=v/\lambda gives the supplied equation.

Moving-source Doppler equation

fo=fsvv±vsf_o=\frac{f_sv}{v\pm v_s}

Here fof_o is observed frequency and fsf_s source frequency, both in Hz\mathrm{Hz}; vv is sound speed and vsv_s source speed relative to the medium, both in m s−1\mathrm{m\,s^{-1}}. For an approaching source use v−vsv-v_s; for a receding source use v+vsv+v_s. The observer is stationary relative to the medium and the source motion is along the observation line.

On the data sheet.

You can choose the sign without memorising a disconnected rule. A higher frequency needs a smaller denominator: subtract for approach. A lower frequency needs a larger denominator: add for recession.

Worked example 1: an approaching source

Question. A source emits 500 Hz500\,\mathrm{Hz} sound and moves directly towards a stationary observer at 20 m s−120\,\mathrm{m\,s^{-1}}. Sound speed is 340 m s−1340\,\mathrm{m\,s^{-1}}. Determine the observed frequency.

Plan. Approach means compressed wavefronts and the smaller denominator.

fo=500×340340−20 Hz=531.25 Hzf_o=\frac{500\times340}{340-20}\,\mathrm{Hz}=531.25\,\mathrm{Hz}

Answer. About 530 Hz530\,\mathrm{Hz}.

Check. The result exceeds the emitted 500 Hz500\,\mathrm{Hz}, as required for approach.

If this same source moves directly away at the same speed,

fo=500×340340+20 Hz=472.2 Hzf_o=\frac{500\times340}{340+20}\,\mathrm{Hz}=472.2\,\mathrm{Hz}

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