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E2.5 · Analyse the Doppler effect for a moving sound source

Learn to analyse the doppler effect for a moving sound source through clear examples and targeted practice.

Ontario Grade 11 Physics

Waves and Sound

How a source’s motion changes the sound frequency heard by a stationary listener

A siren can sound higher in pitch as it approaches and lower after it passes. The siren’s own frequency does not need to change. Instead, its motion changes how closely its sound waves reach a listener. This change in received frequency is called the Doppler effect. In this lesson, the listener is stationary in the air, and only the sound source moves.

What you will learn

1. Prerequisite bridge: sound waves and frequency

Sound is a wave that travels through a material such as air. A sound source, such as a speaker or siren, vibrates and sends out wave fronts. A wave front is a surface where the sound wave has the same stage of vibration. In a simple drawing, each wave front can be shown as a line.
Frequency is the number of complete vibrations or wave fronts passing a point each second. It is a scalar: it has a size but no direction. Its SI unit is the hertz, written Hz\mathrm{Hz}, which means one cycle per second. The source frequency is the frequency produced by the source. The observed frequency is the frequency received by the listener.
Wave speed is the distance a sound wave travels each second. It is also a scalar, measured in metres per second (m/s\mathrm{m/s}). In still air, use the sound speed given in a problem. The source speed is the speed of the source relative to the air. Source velocity includes both speed and direction, so its direction matters in the calculation.
v=fλv=f\lambda

2. Physical situation, directions, and wave pattern

Set the system as the moving sound source, the sound travelling through still air, and a stationary listener. For this lesson, the listener is on one side of the source. Choose the positive direction along the line from the source toward the listener. A source moving in that direction is approaching, so its velocity is positive. A source moving away from the listener is receding, so its velocity is negative.
The source motion changes the spacing of the wave fronts. As the source moves toward the listener, it emits each new wave front from a position closer to the listener than the previous one. The wave fronts in front of the source are closer together. The listener receives more wave fronts each second, so the observed frequency is higher than the source frequency.
As the source moves away, each new wave front is emitted farther from the listener than the previous one. The wave fronts behind the source are farther apart. The listener receives fewer wave fronts each second, so the observed frequency is lower.
A simple wave sketch shows the pattern along the direction of travel. The listener is to the right. The source moves right, so the wave fronts in front of it are crowded together. The source’s motion does not make the sound travel faster through the air. The sound speed relative to still air stays the same.
source motion toward listener⇒fobserved>fsource\text{source motion toward listener}\Rightarrow f_{\text{observed}}>f_{\text{source}}

3. The model for a moving source

For a stationary listener and a moving source, use the Doppler relationship below. The symbol vv is the sound speed in air, fsourcef_{\text{source}} is the frequency produced by the source, and vsourcev_{\text{source}} is the signed source velocity along the line toward the listener. The observed frequency is fobservedf_{\text{observed}}.
The denominator uses a minus sign because the source’s signed velocity already records its direction. For an approaching source, vsourcev_{\text{source}} is positive, making the denominator smaller and the observed frequency larger. For a receding source, vsourcev_{\text{source}} is negative, making the denominator larger and the observed frequency smaller.
Use consistent SI units for both speeds. The speed ratio has no units, so the result has the same unit as the source frequency: hertz. The source speed must be less than the sound speed for this course-level model. Before calculating, identify the source direction and assign the correct sign.
fobserved=fsourcevv−vsourcef_{\text{observed}}=f_{\text{source}}\frac{v}{v-v_{\text{source}}}

4. Read the answer as a physical prediction

The equation predicts a frequency, but the direction and meaning of that frequency matter too. A higher observed frequency corresponds to a higher pitch; a lower observed frequency corresponds to a lower pitch. The equation describes the sound heard while the source is moving toward or away from a stationary listener.
A useful check is to compare the observed frequency with the source frequency before accepting the result. Also check that the calculated value has units of hertz, that the sign convention matches the motion, and that the size of the change is reasonable for the source speed. A source moving much slower than sound should produce a change, but not an enormous one.
[fobserved]=Hz[f_{\text{observed}}]=\mathrm{Hz}

Worked example

A source approaches a listener

A siren produces a frequency of 500 Hz500\ \mathrm{Hz}. It moves toward a stationary listener at 25.0 m/s25.0\ \mathrm{m/s}. Take the sound speed in air as 343 m/s343\ \mathrm{m/s}. Find the frequency heard.
  1. Set the direction and known values
    The system is the siren, still air, and a stationary listener. Positive is from the siren toward the listener. The siren approaches, so its source velocity is positive. The unknown is the observed frequency.
    fsource=500 Hz,v=343 m/s,vsource=+25.0 m/sf_{\text{source}}=500\ \mathrm{Hz},\quad v=343\ \mathrm{m/s},\quad v_{\text{source}}=+25.0\ \mathrm{m/s}
  2. Choose the relationship
    Use the moving-source model for a stationary listener. The positive source velocity makes the denominator smaller, which matches the expected increase in frequency.
    fobserved=fsourcevv−vsourcef_{\text{observed}}=f_{\text{source}}\frac{v}{v-v_{\text{source}}}
  3. Substitute and calculate
    Insert the values with their units. The speed units cancel in the ratio, leaving hertz.
    fobserved=(500 Hz)343 m/s343 m/s−25.0 m/s=539 Hzf_{\text{observed}}=(500\ \mathrm{Hz})\frac{343\ \mathrm{m/s}}{343\ \mathrm{m/s}-25.0\ \mathrm{m/s}}=539\ \mathrm{Hz}
Answer: The listener hears approximately 539 Hz539\ \mathrm{Hz}. The source is approaching, so the observed frequency is higher than 500 Hz500\ \mathrm{Hz}.
Check: The result has units of hertz and is above the source frequency, as expected for an approaching source. Three significant figures are appropriate.

Worked example

A source recedes from a listener

A speaker produces a tone of 720 Hz720\ \mathrm{Hz} while moving away from a stationary listener at 18.0 m/s18.0\ \mathrm{m/s}. Use 343 m/s343\ \mathrm{m/s} for the sound speed. Find the observed frequency.
  1. Set the direction and known values
    The system is the moving speaker, still air, and stationary listener. Positive points from the speaker toward the listener. The speaker moves away, so its velocity is negative. The unknown is the frequency received by the listener.
    fsource=720 Hz,v=343 m/s,vsource=−18.0 m/sf_{\text{source}}=720\ \mathrm{Hz},\quad v=343\ \mathrm{m/s},\quad v_{\text{source}}=-18.0\ \mathrm{m/s}
  2. Apply the moving-source relationship
    Use the signed velocity in the denominator. A negative source velocity increases the denominator, so the result should be lower than the source frequency.
    fobserved=fsourcevv−vsourcef_{\text{observed}}=f_{\text{source}}\frac{v}{v-v_{\text{source}}}
  3. Substitute and calculate
    Subtracting the negative source velocity adds its speed to the sound speed. The ratio is unitless, so the final unit remains hertz.
    fobserved=(720 Hz)343 m/s343 m/s−(−18.0 m/s)=684 Hzf_{\text{observed}}=(720\ \mathrm{Hz})\frac{343\ \mathrm{m/s}}{343\ \mathrm{m/s}-(-18.0\ \mathrm{m/s})}=684\ \mathrm{Hz}
Answer: The listener hears approximately 684 Hz684\ \mathrm{Hz}, which is lower than the speaker’s 720 Hz720\ \mathrm{Hz} tone.
Check: The units are hertz, and the frequency decreases for a receding source. The decrease is modest compared with the original frequency, which is reasonable for a source speed much lower than the sound speed.

Worked example

Find the source speed from the observed frequency

A source produces a tone of 600 Hz600\ \mathrm{Hz}. A stationary listener hears 660 Hz660\ \mathrm{Hz} while the source approaches. If sound travels at 343 m/s343\ \mathrm{m/s}, find the source speed.
  1. Set the direction and unknown
    The system is the source, still air, and stationary listener. Positive is toward the listener. Since the observed frequency is higher, the source is approaching and its velocity is positive. The unknown is its speed.
    fsource=600 Hz,fobserved=660 Hz,v=343 m/sf_{\text{source}}=600\ \mathrm{Hz},\quad f_{\text{observed}}=660\ \mathrm{Hz},\quad v=343\ \mathrm{m/s}
  2. Rearrange the model
    Start with the moving-source relationship and solve for the signed source velocity. The resulting positive value will confirm motion toward the listener.
    vsource=v(1−fsourcefobserved)v_{\text{source}}=v\left(1-\frac{f_{\text{source}}}{f_{\text{observed}}}\right)
  3. Substitute and calculate
    The frequency ratio has no units, so multiplying by sound speed gives a velocity in metres per second.
    vsource=(343 m/s)(1−600 Hz660 Hz)=31.2 m/sv_{\text{source}}=(343\ \mathrm{m/s})\left(1-\frac{600\ \mathrm{Hz}}{660\ \mathrm{Hz}}\right)=31.2\ \mathrm{m/s}
Answer: The source speed is 31.2 m/s31.2\ \mathrm{m/s} toward the listener.
Check: The answer has units of velocity and is positive under the stated convention. It is less than the sound speed. The observed frequency is higher than the source frequency, consistent with approach.

Common mistakes and how to avoid them

Using a positive source velocity for a source moving away.
Correction: With this lesson’s convention, motion toward the listener is positive and motion away is negative. Assign the sign before substituting.
Using the approaching-source sign for every situation.
Correction: Use the signed velocity in the same relationship. A negative value for a receding source makes the denominator larger.
Saying the source’s frequency changes just because it moves.
Correction: The source frequency can stay the same. Its motion changes wave-front spacing at the listener, changing the observed frequency.
Reporting a calculated frequency without checking whether it is higher or lower than the source frequency.
Correction: An approaching source should give a higher observed frequency; a receding source should give a lower one. Check this along with units and significant figures.

Lesson summary

Check your understanding

Question 1

A source emits 400 Hz400\ \mathrm{Hz} and moves away from a stationary listener. Which statement is correct?
  1. The listener hears a frequency above 400 Hz400\ \mathrm{Hz} because the wave fronts spread out.
  2. The listener hears a frequency below 400 Hz400\ \mathrm{Hz} because the wave fronts spread out.
  3. The listener hears exactly 400 Hz400\ \mathrm{Hz} because the source frequency cannot change.
  4. The sound speed increases because the source moves away.
Show answer and explanation
The listener hears a frequency below 400 Hz400\ \mathrm{Hz} because the wave fronts spread out.
A receding source spreads the wave fronts behind it. The stationary listener receives fewer wave fronts each second, so the observed frequency is below the source frequency.

Question 2

With positive defined as toward the listener, what signed source velocity should be used for a source moving away at 12.0 m/s12.0\ \mathrm{m/s}?
  1. +12.0 m/s+12.0\ \mathrm{m/s}
  2. −12.0 m/s-12.0\ \mathrm{m/s}
  3. 0 m/s0\ \mathrm{m/s}
  4. +343 m/s+343\ \mathrm{m/s}
Show answer and explanation
−12.0 m/s-12.0\ \mathrm{m/s}
Motion away is opposite the chosen positive direction, so the source velocity is negative.

Question 3

A 500 Hz500\ \mathrm{Hz} source approaches at 20.0 m/s20.0\ \mathrm{m/s}. Take the sound speed as 340 m/s340\ \mathrm{m/s}. What frequency is heard?
  1. 472 Hz472\ \mathrm{Hz}
  2. 500 Hz500\ \mathrm{Hz}
  3. 531 Hz531\ \mathrm{Hz}
  4. 560 Hz560\ \mathrm{Hz}
Show answer and explanation
531 Hz531\ \mathrm{Hz}
For approach, use positive source velocity. The calculation is fobserved=(500 Hz)(340 m/s)/(340 m/s−20.0 m/s)=531.25 Hzf_{\text{observed}}=(500\ \mathrm{Hz})(340\ \mathrm{m/s})/(340\ \mathrm{m/s}-20.0\ \mathrm{m/s})=531.25\ \mathrm{Hz}. To three significant figures, this is 531 Hz531\ \mathrm{Hz}, above the source frequency as expected.

Key terms

Doppler effect
For a moving sound source and a stationary listener, a change in the frequency received by the listener caused by the source’s motion.
Frequency
The number of complete vibrations or wave fronts passing a point each second, measured in hertz.
Wave front
A surface or line marking the same stage of a travelling wave.
Observed frequency
The frequency received by the stationary listener.
Source frequency
The frequency produced by the sound source.
Source velocity
The source’s speed and direction along the line toward or away from the stationary listener.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation E2.5. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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