Sound as a wave: pitch, timbre and the Doppler effect
Why does the siren of a passing ambulance sound as if it drops in pitch just as it goes by, even though the siren itself keeps sounding exactly the same?
Prior knowledge
- Sound is a longitudinal wave, and the speed of a wave is $v = f\lambda$chapter How waves form and travel
- When waves meet, their displacements add upchapter When waves combine: superposition and interference
- The natural frequencies of a string and of a tube open at both ends are multiples of the fundamental frequencychapter Standing waves: nodes, antinodes and natural frequencies
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 5.
Think about it
You are standing by the road when an ambulance drives past with its siren sounding. As the car approaches, the siren sounds bright and high. Just as the car passes you, the sound drops clearly lower and stays low as the car moves away. To the driver of the ambulance, the siren sounds the same all the time.
How can the motion of the car change the pitch you hear, even though the siren itself does not change?
Optional Review: natural frequencies of a string
Predict
Make your predictions before investigating 2 questions
Prediction 1: an approaching siren
Prediction 2: which one moves?
Investigate
A loudspeaker emits sound, and two listeners listen to it on different sides. Move the source or the listeners and follow the wavefronts and the frequencies heard.
Simulation
How to use
Choose the Situation. When the source moves, the loudspeaker travels to the right towards listener A and away from listener B. When the listeners move, the loudspeaker is at rest and both listeners travel to the left: A towards the loudspeaker, B away from it. Adjust the frequency $f_0$ and the speed of the source. The grey circles are wavefronts, that is, compressions, and the orange ring at a listener's ear glows whenever a wavefront arrives. The readings show the frequency and the wavelength each listener hears in their direction, and the frequency bar below shows $f_\text{B}$, $f_0$ and $f_\text{A}$. The picture starts from the beginning, before the source and the listener meet. Speed changes only the pace of the animation.
Assumptions of the model
The speed of sound is $340\,\text{m/s}$ relative to still air, and there is no wind. The source and the listeners move along the same straight line at a steady speed that is less than the speed of sound. The picture is slowed down by a factor of about $400$, and in it one metre is $100$ pixels. Sound does not weaken on the way, and the model has no reflections. The frequencies are values calculated by the model, not measurements.
How the calculation works
Wavefront $k$ leaves at the time $t_k = \tfrac{k}{f_0}$ from the point where the source is at that moment and travels at speed $v$. For two successive wavefronts the simulation solves the moment when the wavefront reaches a moving or a stationary listener. The frequency heard is one divided by the difference of these arrival times. The Doppler formula is not used; the result comes from the geometry of the wavefronts. The wavelength is the distance of two successive wavefronts at the same moment in the direction of the listener.
Experiment 1: the source moves
Test prediction 1. The frequency of the loudspeaker is $400\,\text{Hz}$, and the loudspeaker moves towards listener A and away from listener B. Change only the speed of the source and record the speeds $0\,\text{m/s}$, $34\,\text{m/s}$ and $102\,\text{m/s}$ (each is a required row, because the feedback of the predictions uses them). Also watch whether $f_\text{A}$ changes as the loudspeaker gets closer to listener A in the picture.
Initial state: situation: source moves, source frequency $f_0$ = $400\,\text{Hz}$, speed of the source = $0\,\text{m/s}$. Kept constant: situation, source frequency $f_0$. Varied: speed of the source.
| speed of the source | $f_\text{A}$ $(\text{Hz})$ | $f_\text{B}$ $(\text{Hz})$ | $\lambda _\text{A}$ $(\text{m})$ |
|---|
Experiment 2: the listeners move
Test prediction 2. The loudspeaker is now at rest and again plays a sound of $400\,\text{Hz}$. Listener A moves towards the loudspeaker and listener B away from it. Change only the speed of the listeners and record the same speeds $0\,\text{m/s}$, $34\,\text{m/s}$ and $102\,\text{m/s}$ as in experiment 1 (each is a required row). Compare the table with the table of experiment 1: which changes the frequency more, the motion of the source or of the listener? Does the wavelength change?
Initial state: situation: listeners move, source frequency $f_0$ = $400\,\text{Hz}$, speed of the listeners = $0\,\text{m/s}$. Kept constant: situation, source frequency $f_0$. Varied: speed of the listeners.
| speed of the listeners | $f_\text{A}$ $(\text{Hz})$ | $f_\text{B}$ $(\text{Hz})$ | $\lambda _\text{A}$ $(\text{m})$ |
|---|
Conclusion: why does the frequency change?
On the basis of your tables, write: how do approaching and moving away change the frequency heard? Does the frequency change when the source comes closer at a steady speed? Why do a moving source and a moving listener not give exactly the same frequency at the same speed? Use the words wavelength and wavefront.
Model answer
When the source and the listener approach each other, the frequency heard is higher than $400\,\text{Hz}$, and when they move apart, lower. At a steady speed the frequency stays the same all the way, even though the distance changes: at $34\,\text{m/s}$ listener A heard $444\,\text{Hz}$ all the time from the moving source. In front of a moving source the wavefronts are closer together, because the source follows right behind the previous wavefront: the wavelength shortened from $0.85\,\text{m}$ to $0.765\,\text{m}$. When the listener moves, the wavelength stays $0.85\,\text{m}$, but the listener meets the wavefronts more often, and heard $440\,\text{Hz}$. The difference grows with the speed: at $102\,\text{m/s}$ the frequencies were $571\,\text{Hz}$ and $520\,\text{Hz}$. The values are calculated by the model, and in the model the air is at rest.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Read an FFT spectrum
Made-up example data
A student sang one long vowel and made an FFT graph of the sound (made-up sample data). The graph has five narrow peaks.
Make an FFT graph yourself
Make an FFT graph of a sound with the school's measuring program (for example Graphical Analysis) and a microphone or a sound sensor. The steps are the same whatever the program:
- Connect the microphone and check that the program shows the pressure or the sound signal as a function of time.
- Choose a high sampling rate, for example $20\,000$ samples per second, and a short measuring time, for example $0.1\,\text{s}$. The sampling rate must be at least twice the greatest frequency that you want to see.
- Sound the note close to the microphone: strike a tuning fork, play a long note on an instrument or sing a vowel. Start the measurement when the sound is steady.
- Check the time graph: it must show a periodic curve that is not clipped at the top or the bottom.
- Choose FFT as the type of the graph. Read the frequencies of the peaks, for example with the examine tool.
If you do not have a microphone, use this made-up sample data: a $440\,\text{Hz}$ tuning fork, one peak at $440\,\text{Hz}$; the same note sung, peaks at $440$, $880$, $1320$ and $1760\,\text{Hz}$ with strengths in the ratio $1.0 : 0.6 : 0.4 : 0.2$. Otherwise measure at least two sounds, for example a tuning fork and your own voice on the same note. Record for each the fundamental frequency and the frequencies of the next three peaks. Are they multiples of the fundamental frequency? How do the spectra differ, and what does the difference tell about timbre?
Model answer
For example, the spectrum of a $440\,\text{Hz}$ tuning fork has one clear peak at about $440\,\text{Hz}$. When a singer sings the vowel a on the same note, the spectrum has peaks at about $440\,\text{Hz}$, $880\,\text{Hz}$, $1320\,\text{Hz}$ and $1760\,\text{Hz}$, that is, at whole-number multiples of the fundamental frequency. The pitch is the same, because the fundamental frequency is the same. The timbre is different, because the singing has strong overtones and the tuning fork has almost none. The measured frequencies differ by a few hertz: with a measuring time of $0.1\,\text{s}$ the spectrum resolves frequencies only at intervals of about $\tfrac{1}{0.1\,\text{s}} = 10\,\text{Hz}$.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Read beats from a graph
Made-up example data
Two tuning forks sound at the same time, and a microphone records their combined sound for one second. The beat frequency is how many times per second the sound grows louder. The vibration is so dense that in the graph it appears as a filled area; the width of the area tells how strong the sound is at each moment.
How large is the beat frequency $f_\text{h}$? Read the time you need from the graph and give the answer in hertz.
Hint 1: a guiding question
A loud moment is where the filled area is at its widest. At which moments are there such points?
Hint 2: what is the connection
The time between two successive loud moments is the period of the beats $T_\text{h}$, and $f_\text{h} = \tfrac{1}{T_\text{h}}$.
Hint 3: an intermediate step
The loud moments are $0\,\text{s}$, $0.25\,\text{s}$, $0.50\,\text{s}$ … so $T_\text{h} = 0.25\,\text{s}$.
Show the solution
Given
from the graph, the loud moments (filled area at its widest) are $t = 0$, $0.25$, $0.50$, $0.75$ and $1.00\,\text{s}$
Find
the beat frequency $f_\text{h}$ in hertz
Principle
Two sounds of almost the same frequency reinforce each other when they are in phase and weaken each other when they are in opposite phase. The time between the loud moments is the period of the beats $T_\text{h}$, and the number of loud moments per second is $f_\text{h}$.
Reading the graph
The filled area is at its widest at the moments $0\,\text{s}$, $0.25\,\text{s}$, $0.50\,\text{s}$, $0.75\,\text{s}$ and $1.00\,\text{s}$. So there are four intervals in a second, and the time between two successive loud moments is $T_\text{h} = 0.25\,\text{s}$.
Formula
$f_\text{h} = \dfrac{1}{T_\text{h}}$
Conversions
The time is already in seconds, so the frequency comes out in hertz ($\tfrac{1}{\text{s}} = \text{Hz}$).
Substitution
Answer
$4.0\,\text{Hz}$
Two significant figures like the time $0.25\,\text{s}$ read from the graph.
Reasonableness
The graph has four narrowing points per second and four intervals between loud moments, which corresponds to four beats per second. If one of the tuning forks is $440\,\text{Hz}$, the other is, by $f_\text{h} = |f_1 - f_2|$, either $436\,\text{Hz}$ or $444\,\text{Hz}$; the difference is less than $10\,\text{Hz}$, so the beats are clearly audible.
Theory summary
Key idea
Sound is a longitudinal pressure wave
The cone of a loudspeaker or the string of a guitar vibrates and pushes the air particles alternately forward and backward. The particles do not travel along with the sound. They vibrate in place along the direction in which the wave travels, so sound in air is a longitudinal wave (chapter How waves form and travel). Where the particles bunch together, a compression forms, where the pressure is slightly higher than the normal air pressure $p_0$. Between the compressions lie rarefactions, where the pressure is lower. The pressure variation is small: in ordinary speech it is about $0.02\,\text{Pa}$, while the air pressure is about $101\,\text{kPa}$. Sound needs a medium such as a gas, a liquid or a solid. In a vacuum, sound does not travel.
The distance between two successive compressions is the wavelength $\lambda$. As for all waves, the speed of sound satisfies $v = f\lambda$, where $f$ is the frequency (chapter How waves form and travel). When sound passes from one substance to another, its frequency stays the same, because the particles at the boundary vibrate in step with the incoming wave. The speed and the wavelength, however, change.
The speed of sound
The speed of sound depends on the medium: the less a substance can be compressed and the lower its density, the faster sound travels in it. Water is much denser than air, but it is so hard to compress that sound travels in water more than four times as fast as in air. In a gas the speed increases as the temperature rises, because the particles move faster and pass the pressure variation on faster. In air the speed increases by about $0.6\,\text{m/s}$ for every degree Celsius. In this chapter we use the value $v = 340\,\text{m/s}$ for air, which corresponds to a temperature of about $15\,{}^\circ\text{C}$, as in the earlier chapters.
| Medium | $v$ (m/s) |
|---|---|
| air, $0\,{}^\circ\text{C}$ | 331 |
| air, $15\,{}^\circ\text{C}$ | 340 |
| air, $20\,{}^\circ\text{C}$ | 343 |
| water, $20\,{}^\circ\text{C}$ | 1480 |
| steel | about 5900 |
Determining the speed of sound
The speed of sound is determined by measuring how long sound takes to travel a known distance. In the echo method, two wooden blocks are clapped together at a distance $d$ from a wall, and the time $t$ after which the echo returns is measured. The sound travels to the wall and back, a distance $2d$, so $v = \tfrac{2d}{t}$. A more accurate value comes from two microphones a distance $d$ apart on the same straight line as the sound source. The measuring program records the signals of both microphones, and the difference $\Delta t$ in the arrival times of the sound is read from the graph, giving $v = \tfrac{d}{\Delta t}$. A third way is a resonance tube, where the wavelength is obtained from the length of a standing wave (chapter Standing waves: nodes, antinodes and natural frequencies).
Pitch and the range of hearing
Pitch is determined by frequency: the higher the frequency, the higher the sound. When the frequency doubles, the sound rises by an octave. For example, the tuning note A is $440\,\text{Hz}$ and the A an octave higher is $880\,\text{Hz}$. A young person hears frequencies from about $20\,\text{Hz}$ to about $20\,\text{kHz}$, and the highest audible frequency falls with age. The wavelengths of audible sounds in air range from about $17\,\text{m}$ to $1.7\,\text{cm}$. Sound lower than the range of hearing is infrasound and higher is ultrasound. They are longitudinal waves just like audible sound, but the human ear does not sense them.
Infrasound is produced, for example, by earthquakes, thunder and volcanic eruptions, and elephants communicate partly with infrasound. Infrasound is weakly absorbed in air, so it can travel hundreds of kilometres. Bats and dolphins use ultrasound for echolocation: the short wavelength is reflected even by small targets such as insects. In technology, ultrasound is used, for example, in cleaning equipment, in looking for flaws in materials and in the echo sounders of boats. Medical ultrasound imaging comes later in this course, and the loudness of sound and decibels are covered in the chapter Intensity and the decibel scale.
Timbre and the harmonic series
A flute, a violin and a singer can produce the same note, and yet they sound different. The difference is in the timbre of the sound. The string or the air column of an instrument vibrates at several of its natural frequencies at once (chapter Standing waves: nodes, antinodes and natural frequencies). The lowest of them is the fundamental frequency $f_1$, and the higher ones are overtones. In a string and in a tube open at both ends the overtones are whole-number multiples of the fundamental frequency, and these frequencies form the harmonic series
$$f_n = n f_1, \quad n = 1, 2, 3, \ldots$$
where $f_n$ is the $n$th frequency of the series (Hz) and $f_1$ is the fundamental frequency (Hz). We hear the pitch from the fundamental frequency. The timbre depends on how strong the overtones are compared with the fundamental frequency. A tuning fork produces an almost pure sine wave that contains only the fundamental frequency.
The FFT spectrum
A microphone measures the variation of air pressure as a function of time. From this series of measurements the measuring program calculates the spectrum of the sound, that is, the FFT graph (Fast Fourier Transform). The spectrum shows which frequencies are in the sound and how strong they are. Each sinusoidal component of the sound appears as a peak at its own frequency, in the same way as in the chapter When waves combine: superposition and interference a complicated wave was built as a sum of sine waves. In the spectrum of an instrument the peaks are at the frequencies of the harmonic series, so the spacing of neighbouring peaks is the fundamental frequency. The fundamental frequency is usually also the frequency of the lowest peak, but the highest peak is not always at the fundamental frequency.
Beats and the Doppler effect
Beats
When two sounds whose frequencies $f_1$ and $f_2$ differ only slightly sound at the same time, their pressure variations add up. At times the waves are in phase and reinforce each other, at times in opposite phase and weaken each other. We hear one sound whose pitch corresponds to the average of the frequencies and whose loudness alternately grows and fades. This is called beats. In one second one wave completes $|f_1 - f_2|$ more vibrations than the other, and each time the waves return to the same phase. So the number of loud moments per second is
$$f_\text{h} = |f_1 - f_2|$$
where $f_\text{h}$ is the beat frequency (Hz) and $f_1$ and $f_2$ are the frequencies of the sounds (Hz). The time between two loud moments, the period of the beats, is $T_\text{h} = \tfrac{1}{f_\text{h}}$. The subscript h, used in the figures as well, comes from the Finnish edition of this chapter: for the beat frequency and the period of the beats it stands for huojunta (beats), and for the speed of the listener below it stands for havaitsija (observer).
Beats are used for tuning instruments. When a string is tightened towards the frequency of the reference note, the beats slow down and stop when the frequencies are equal. Beats stand out clearly when the frequency difference is at most about $10\,\text{Hz}$. With a larger difference the ear hears a rough sound or two separate sounds.
The Doppler effect when the source moves
When a sound source moves, each wavefront starts from the point where the source is at that moment and spreads through the air at speed $v$ in every direction. The period of the source is $T_0 = \tfrac{1}{f_0}$. During that time the previous wavefront travels a distance $v T_0$, but the source travels after it a distance $v_\text{s} T_0$. In front of the source the spacing of the wavefronts, that is, the wavelength, is therefore $(v - v_\text{s}) T_0$, and behind the source it is $(v + v_\text{s}) T_0$. A listener at rest hears the frequency $f = \tfrac{v}{\lambda}$, so
$$f = f_0 \cdot \frac{v}{v \mp v_\text{s}}$$
where $f$ is the frequency heard (Hz), $f_0$ is the frequency of the source (Hz), $v$ is the speed of sound (m/s) and $v_\text{s}$ is the speed of the source (m/s). The minus sign is chosen when the source approaches the listener and the plus sign when the source moves away.
The Doppler effect when the listener moves
When the source is at rest and the listener moves, the wavelength does not change. The listener, however, meets wavefronts more often when moving towards them, because the wavefronts approach at speed $v + v_\text{h}$. When moving away, the wavefronts catch up with the listener at speed $v - v_\text{h}$. The frequency heard is the rate at which the wavefronts are met:
$$f = f_0 \cdot \frac{v \pm v_\text{h}}{v}$$
where $v_\text{h}$ is the speed of the listener, or observer (m/s), and the other quantities are as above. The plus sign is chosen when the listener approaches the source and the minus sign when the listener moves away.
Signs and assumptions
The sign can be checked with a rule: approaching raises the frequency heard and moving away lowers it. So choose the sign that gives $f > f_0$ when approaching and $f < f_0$ when moving away. The speeds are measured relative to the air, which is at rest, that is, there is no wind. The formulas apply to motion along the straight line between the source and the listener, and they hold when only one of them moves and the speed of the source is less than the speed of sound. When the speed is much smaller than the speed of sound, both formulas give almost the same change, about $f_0 \cdot \tfrac{v_\text{s}}{v}$ or $f_0 \cdot \tfrac{v_\text{h}}{v}$. The Doppler effect is used to measure speeds: an ultrasound device measures the flow speed of blood.
Common misconception
Many people think that the sound of an approaching siren keeps rising as the car gets closer. The frequency heard, however, depends on the speed, not on the distance. When an ambulance drives at a steady speed straight towards the listener, the frequency heard is the same throughout the approach and higher than $f_0$, and at the moment of passing it drops. As the car approaches, the loudness grows, not the pitch. If the listener stands at the side of the road, the change happens gradually during the passing, because only the part of the velocity that points towards the listener or away from them affects the frequency heard.
Another mistake is to think that the motion of the source changes the speed of sound or the frequency of the source itself. Sound travels relative to the air at speed $v$ whether the source moves or not, and the siren vibrates at the frequency $f_0$ all the time. A moving source changes the wavelength, and a moving listener changes how often they meet wavefronts. A third mistake is to read the pitch from a spectrum as the frequency of the highest peak: the fundamental frequency is the lowest frequency of the harmonic series and the spacing of the peaks, even if some overtone is stronger than it.
Summary
- Sound is a longitudinal pressure wave in which compressions and rarefactions travel through a medium. The speed of sound depends on the medium and, in a gas, on the temperature; in air it is about $340\,\text{m/s}$. The speed is determined by measuring the travel time of sound over a known distance.
- Pitch is determined by frequency. The human range of hearing is about $20\,\text{Hz}$–$20\,\text{kHz}$; lower sound is infrasound and higher is ultrasound.
- Timbre depends on the strengths of the overtones. The partials of an instrument form the harmonic series $f_n = n f_1$, and the FFT spectrum shows them as peaks.
- When two sounds of almost the same frequency sound together, beats are heard with the frequency $f_\text{h} = |f_1 - f_2|$.
- The Doppler effect: approaching raises and moving away lowers the frequency heard. For a moving source $f = f_0 \cdot \tfrac{v}{v \mp v_\text{s}}$ and for a moving listener $f = f_0 \cdot \tfrac{v \pm v_\text{h}}{v}$, the upper sign for approaching.
Optional Return to your predictions
Compare your predictions with the values of the model in the experiments and with the Doppler formulas. How does the situation of prediction 1 (you stand on the line of the car) differ from the roadside at the start, where the frequency slides? Then return to the question at the start: why does the siren sound as if it drops in pitch just at the moment of passing? Why does the siren sound the same to the driver all the time?
Would you change your prediction?
Apply
Worked example
Problem
A train runs along a straight track at a steady speed of $108\,\text{km/h}$ and sounds a horn whose frequency is $311\,\text{Hz}$. A passenger stands on the platform right next to the track, so the train comes straight towards them and then moves straight away from them. What frequency does the passenger hear as the train approaches and what frequency as it moves away, and by how much does the frequency drop at the moment the train passes? The air is calm, and the speed of sound is $340\,\text{m/s}$.
Step 1/7
Given and asked. The frequency of the horn is $f_0 = 311\,\text{Hz}$, the speed of the train is $v_\text{s} = 108\,\text{km/h}$ and the speed of sound is $v = 340\,\text{m/s}$. We are asked for the frequency heard as the train approaches ($f_1$) and as it moves away ($f_2$), and for the drop in frequency $\Delta f = f_1 - f_2$.
Step 2/7
Principle. The sound source moves and the listener is at rest, so we use the Doppler formula for a moving source. In front of the approaching train the wavefronts bunch together, so the sign is chosen so that $f_1 > f_0$. Behind the receding train the wavefronts spread out, and $f_2 < f_0$.
Step 3/7
Formula solved.
$$\begin{aligned} f_1 &= f_0 \cdot \frac{v}{v - v_\text{s}} \\ f_2 &= f_0 \cdot \frac{v}{v + v_\text{s}} \\ \Delta f &= f_1 - f_2 \end{aligned}$$
Step 4/7
Unit conversions. The speed of the train is converted to m/s so that it can be subtracted from the speed of sound:
Step 5/7
Substitution. $$\begin{aligned} f_1 &= f_0 \cdot \frac{v}{v - v_\text{s}} \\ \quad &= 311\,\text{Hz} \cdot \frac{340\,\text{m/s}}{(340 - 30.0)\,\text{m/s}} \\ \quad &= 341.10\,\text{Hz} \\[6pt] f_2 &= f_0 \cdot \frac{v}{v + v_\text{s}} \\ \quad &= 311\,\text{Hz} \cdot \frac{340\,\text{m/s}}{(340 + 30.0)\,\text{m/s}} \\ \quad &= 285.78\,\text{Hz} \\[6pt] \Delta f &= f_1 - f_2 \\ \quad &= 341.10\,\text{Hz} - 285.78\,\text{Hz} \\ \quad &= 55.32\,\text{Hz} \end{aligned}$$
Step 6/7
Answer. The passenger hears the approaching train at $341\,\text{Hz}$ and the receding train at $286\,\text{Hz}$, to three significant figures like the given data. At the moment of passing, the frequency drops by about $55\,\text{Hz}$: the difference is accurate to about a hertz, because each frequency is known to about half a hertz.
Step 7/7
Reasonableness. At a low speed the change on each side is about
so the whole drop is about $55\,\text{Hz}$, as calculated. The rise on approach, $30\,\text{Hz}$, is slightly larger than the fall on moving away, $25\,\text{Hz}$, because in front we divide by the smaller number $v - v_\text{s}$. The speed of the train is less than a tenth of the speed of sound, so the change on each side is only about $9\,\%$ of the frequency $f_0$.
Optional Find the error
Optional Apply to a new situation
A student whirls a buzzer tied to the end of a string in a horizontal circle of radius $1.00\,\text{m}$. One revolution takes $0.500\,\text{s}$, and the frequency of the buzzer is $2.00\,\text{kHz}$. A listener standing far away in the same plane hears the pitch vary.
What is the greatest frequency the listener hears? The air is calm, and the speed of sound is $340\,\text{m/s}$. Answer in hertz.
Hint 1: a guiding question
At which point of the circle does the buzzer move straight towards the listener? How large is its speed then?
Hint 2: what is the connection
In uniform circular motion the speed is $v_\text{s} = \tfrac{2\pi r}{T}$ (chapter Moving in a circle: period and speed). The greatest frequency is heard when the buzzer approaches the listener at this speed.
Hint 3: an intermediate step
$f = f_0 \cdot \tfrac{v}{v - v_\text{s}}$.
Show the solution
Given
radius of the circle $r = 1.00\,\text{m}$, period of revolution $T = 0.500\,\text{s}$, frequency of the buzzer $f_0 = 2.00\,\text{kHz}$, speed of sound $v = 340\,\text{m/s}$
Find
the greatest frequency heard $f$ in hertz
Principle
The buzzer moves in a circle at a steady speed, and the distant listener is in the plane of the circle. The frequency heard is affected only by the part of the velocity that points towards the listener or away from them. The greatest frequency is heard when the buzzer moves straight towards the listener with its whole speed. The listener is at rest, so we use the moving-source formula with the sign for approaching.
Formula
so
Conversions
the other quantities are in base units.
Substitution
Answer
$2.08\,\text{kHz}$
That is, about $2080\,\text{Hz}$, three significant figures like the radius, the period and the frequency.
Reasonableness
The speed of the buzzer is about $\tfrac{12.6}{340} \approx 3.7\,\%$ of the speed of sound, so the frequency rises by about $3.7\,\%$, that is, about $75\,\text{Hz}$. When the buzzer moves away, the frequency is correspondingly about $1929\,\text{Hz}$, so the listener hears the sound waver between about $1930\,\text{Hz}$ and $2080\,\text{Hz}$ twice a second.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- Two flutes play at the same time notes with the frequencies $440\,\text{Hz}$ and $437\,\text{Hz}$. What does a listener hear?
Answer
One sound whose pitch corresponds to a frequency of about $438.5\,\text{Hz}$ and whose loudness waxes and wanes:
$f_\text{h}$$= |440\,\text{Hz} - 437\,\text{Hz}|$$= 3\,\text{Hz}$that is, three loud moments per second.
- The same echo measurement is made on a hot summer day and in a winter frost. On which occasion does the echo return sooner, and why?
Answer
On the hot summer day. In a gas the speed of sound increases as the temperature rises, in air by about $0.6\,\text{m/s}$ for every degree Celsius, so in warm air sound covers the same distance in a shorter time.
1. A moving listener
2. Measuring the speed of sound
Two microphones are placed $1.20\,\text{m}$ apart on the same straight line as a loudspeaker. The loudspeaker emits a short click, and from the signals recorded by the measuring program it is read that the click arrives at the farther microphone $3.50\,\text{ms}$ later than at the nearer one.
What is the speed of sound according to the measurement? Answer in m/s.
Hint 1: a guiding question
What distance does the sound travel in the time in which the click goes from the nearer microphone to the farther one?
Hint 2: what is the connection
Sound travels the distance $d$ between the microphones in the time $\Delta t$, so $v = \tfrac{d}{\Delta t}$.
Hint 3: an intermediate step
Convert the time to seconds: $3.50\,\text{ms} = 3.50 \cdot 10^{-3}\,\text{s}$.
Show the solution
Given
distance of the microphones $d = 1.20\,\text{m}$, difference in arrival times $\Delta t = 3.50\,\text{ms}$
Find
the speed of sound $v$ in m/s
Principle
The microphones are on the same straight line as the loudspeaker, so the sound travels to the farther microphone the same distance as to the nearer one and in addition the distance $d$ between the microphones. The time difference $\Delta t$ is therefore the time in which sound travels the distance $d$ at a steady speed.
Formula
$d = v \Delta t$, so $v = \dfrac{d}{\Delta t}$
Conversions
$3.50\,\text{ms} = 3.50 \cdot 10^{-3}\,\text{s}$
Substitution
Answer
$343\,\text{m/s}$
Three significant figures like the distance and the time difference.
Reasonableness
The result matches the table value $343\,\text{m/s}$ for air at $20\,{}^\circ\text{C}$, that is, room temperature. If the time difference were misread by $0.05\,\text{ms}$, the result would change by about $5\,\text{m/s}$, so reading the time difference accurately decides the accuracy of the measurement.
Suggestion: one answer does not show everything you can do, but review the theory summary and do the core exercises.
Optional 3. Explain timbre
A violin and a flute play the same note with the fundamental frequency $440\,\text{Hz}$, and yet they sound different. Explain what is the same and what is different in the sounds, and how the difference would show in their FFT graphs. Use the concepts fundamental frequency, overtone and harmonic series.
Model answer
Both sounds have the fundamental frequency $440\,\text{Hz}$, so the ear hears the same pitch. However, both instruments also vibrate at their overtones at the same time. These are the multiples of the fundamental frequency, $880\,\text{Hz}$, $1320\,\text{Hz}$ and so on, and they form the harmonic series $f_n = n f_1$. Timbre depends on how strong the overtones are compared with the fundamental frequency, and it is different for different instruments. In the FFT graphs both have peaks at the same frequencies, but the heights of the peaks are different: in the sound of the violin there are strong overtones far up the series, in the sound of the flute the fundamental frequency dominates and the overtones are weaker.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Review the chapter's concepts
Cards you mark Review later collect in your review deck.
- Resultant wave
- The wave formed by the combined effect of two or more waves that meet at the same place.
- Beats
- A periodic variation in loudness caused by the interference of two waves of nearly equal frequency.
- Partial
- A single oscillation component of a sound with a given frequency.
- Doppler effect
- The change in the observed frequency when the source and the observer move relative to each other.
- Range of hearing
- The range of frequencies a person can hear; about $16\,\text{Hz}$–$20\,\text{kHz}$.
- Infrasound
- Sound whose frequency is below the lower limit of human hearing.
- Ultrasound
- Sound whose frequency is above the upper limit of human hearing.
Exercises
Level 1 · basic practice 5 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 14-1Core
A student stands $86.0\,\text{m}$ from the wall of a school and claps two wooden blocks together. The timing app of a phone measures that the echo returns $0.505\,\text{s}$ after the clap.
What is the speed of sound according to the measurement?
Answer
$341\,\text{m/s}$
Show the solution
Given
distance to the wall $d = 86.0\,\text{m}$, return time of the echo $t = 0.505\,\text{s}$
Find
the speed of sound $v$ in $\text{m/s}$
Principle
The sound travels to the wall and is reflected back, so during the return time it covers a distance $2d$. In air sound travels at a steady speed, so the speed is the distance travelled divided by the time.
Formula
$2d = v t$, so $v = \dfrac{2d}{t}$
Conversions
None needed: the distance is in metres and the time in seconds.
Substitution
$v$$= \dfrac{2 \cdot 86.0\,\text{m}}{0.505\,\text{s}}$$= \dfrac{172.0\,\text{m}}{0.505\,\text{s}}$$= 340.59\,\text{m/s}$Answer
$341\,\text{m/s}$
Three significant figures like the distance and the time.
Reasonableness
The result is close to the table value $340\,\text{m/s}$ for air at about $15\,{}^\circ\text{C}$. If the distance had not been multiplied by two, the speed would have come out as about $170\,\text{m/s}$, which is clearly too small. With timing started by hand, the reaction time can still change the result by several per cent.
- Exercise 14-2Core
A bat emits ultrasound for echolocation with a frequency of $45.0\,\text{kHz}$.
What is the wavelength of the sound in air when the speed of sound is $340\,\text{m/s}$? Give the answer in millimetres.
Answer
$7.56\,\text{mm}$
Show the solution
Given
frequency of the ultrasound $f = 45.0\,\text{kHz}$, speed of sound in air $v = 340\,\text{m/s}$
Find
the wavelength $\lambda$ in millimetres
Principle
Ultrasound is a longitudinal wave just like audible sound, so the basic wave equation $v = f\lambda$ holds for it.
Formula
$v = f\lambda$, so $\lambda = \dfrac{v}{f}$
Conversions
$f$$= 45.0\,\text{kHz}$$= 45.0 \cdot 1000\,\text{Hz}$$= 45\,000\,\text{Hz}$finally $1\,\text{m} = 1000\,\text{mm}$.
Substitution
$\lambda$$= \dfrac{340\,\text{m/s}}{45\,000\,\text{Hz}}$$= 0.0075556\,\text{m}$$= 7.5556\,\text{mm}$Answer
$7.56\,\text{mm}$
Three significant figures like the frequency $45.0\,\text{kHz}$ and the speed $340\,\text{m/s}$.
Reasonableness
At the upper limit of hearing, $20\,\text{kHz}$, the wavelength is $1.7\,\text{cm}$, and at more than twice the frequency the wavelength is less than half of that. A wavelength of a few millimetres is the size of an insect, so a bat can detect insects by echolocation.
- Exercise 14-3
The fundamental frequency of the D string of a cello is $147\,\text{Hz}$. The string vibrates at the frequencies of the harmonic series at the same time.
What is the frequency of the fourth member of the series, that is, the third overtone?
Answer
$588\,\text{Hz}$
Show the solution
Given
fundamental frequency $f_1 = 147\,\text{Hz}$
Find
the frequency $f_4$ of the fourth member of the harmonic series
Principle
A string fixed at both ends vibrates at all its natural frequencies at once, which are whole-number multiples of the fundamental frequency and form the harmonic series. The first member of the series is the fundamental frequency, so the fourth member is the third overtone.
Formula
$f_n = n f_1$, so $f_4 = 4 f_1$
Conversions
None needed: the frequency is in hertz.
Substitution
$f_4$$= 4 \cdot 147\,\text{Hz}$$= 588\,\text{Hz}$Answer
$588\,\text{Hz}$
Three significant figures like the fundamental frequency.
Reasonableness
The members of the series are $147\,\text{Hz}$, $294\,\text{Hz}$, $441\,\text{Hz}$ and $588\,\text{Hz}$, so the fourth member is two octaves above the fundamental: two doublings give four times the frequency.
- Exercise 14-4
An ice-cream van drives straight away from a listener at $15.0\,\text{m/s}$ and plays a melody, one note of which is $415\,\text{Hz}$.
What frequency of this note does a listener at rest hear? The air is calm, and the speed of sound is $340\,\text{m/s}$.
Answer
$397\,\text{Hz}$
Show the solution
Given
frequency of the note $f_0 = 415\,\text{Hz}$, speed of the car $v_\text{s} = 15.0\,\text{m/s}$ away from the listener, speed of sound $v = 340\,\text{m/s}$
Find
the frequency heard $f$
Principle
The sound source moves and the listener is at rest, so we use the Doppler formula for a moving source. The source moves away, so the wavefronts spread out behind it and the frequency heard is lower than $f_0$: the denominator gets a plus sign.
Formula
$f$$= f_0 \cdot \dfrac{v}{v + v_\text{s}}$Conversions
None needed: the speeds are in $\text{m/s}$ and the frequency in hertz.
Substitution
$f$$= 415\,\text{Hz} \cdot \dfrac{340\,\text{m/s}}{340\,\text{m/s} + 15.0\,\text{m/s}}$$= 415\,\text{Hz} \cdot \dfrac{340}{355}$$= 397.46\,\text{Hz}$Answer
$397\,\text{Hz}$
Three significant figures like the given data.
Reasonableness
The frequency drops because the source moves away. The speed of the car is about $4.4\,\%$ of the speed of sound, and the frequency drops by about $\tfrac{415 - 397}{415} \approx 4.2\,\%$, so the change is of the same order as the ratio of the speeds.
- Exercise 14-5
Classify the sounds as infrasound, audible sound or ultrasound: a) the lowest part of a thunder rumble, $12\,\text{Hz}$, b) a reference tuning note, $440\,\text{Hz}$, c) a dog whistle, $35\,\text{kHz}$, d) the sound of ultrasound imaging, $2.5\,\text{MHz}$. Justify using the limits.
Answer
a) infrasound; b) audible sound; c) ultrasound; d) ultrasound
Show the solution
Find
the class of each sound: infrasound, audible sound or ultrasound
Principle
A human hears frequencies from about $20\,\text{Hz}$ to about $20\,\text{kHz}$. Sound below $20\,\text{Hz}$ is infrasound and sound above $20\,\text{kHz}$ is ultrasound. All of them are longitudinal pressure waves of the same kind; the class depends only on the frequency.
Model answer
a) $12\,\text{Hz} < 20\,\text{Hz}$, so the lowest part of the thunder is infrasound. b) $440\,\text{Hz}$ lies between the limits $20\,\text{Hz}$ and $20\,\text{kHz}$, so the tuning note is audible sound. c) $35\,\text{kHz} > 20\,\text{kHz}$, so the sound of the dog whistle is ultrasound: a dog hears it, but a human does not. d) $2.5\,\text{MHz} = 2\,500\,000\,\text{Hz}$ is many orders of magnitude above the range of hearing, so the sound of ultrasound imaging is ultrasound.
Level 2 · applied practice 4 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 14-6Core
A piano tuner strikes a tuning fork with a frequency of $262\,\text{Hz}$ and at the same time plays a piano string that sounds slightly lower than the tuning fork. He counts $15$ loud moments, that is, beats, in $5.0\,\text{s}$.
What is the frequency of the string?
Answer
$259\,\text{Hz}$
Show the solution
Given
frequency of the tuning fork $f_1 = 262\,\text{Hz}$, $N = 15$ loud moments in a time $t = 5.0\,\text{s}$, the string is lower than the tuning fork
Find
the frequency of the string $f_2$
Principle
Two sounds of almost the same frequency produce beats, and the number of loud moments per second equals the difference of the frequencies. The beat frequency is obtained by counting the loud moments per unit time. Because the string is lower, its frequency is smaller than that of the tuning fork by the beat frequency.
Formula
$f_\text{h} = \dfrac{N}{t}$ and $f_\text{h} = |f_1 - f_2| = f_1 - f_2$, so $f_2 = f_1 - \dfrac{N}{t}$
Conversions
None needed: the time is in seconds, so $\tfrac{N}{t}$ comes out in hertz.
Substitution
$f_\text{h}$$= \dfrac{15}{5.0\,\text{s}}$$= 3.0\,\text{Hz}$$f_2$$= 262\,\text{Hz} - 3.0\,\text{Hz}$$= 259\,\text{Hz}$Answer
$259\,\text{Hz}$
; the difference $3.0\,\text{Hz}$ is known to a tenth of a hertz, so the result is given in whole hertz like the frequency of the tuning fork.
Reasonableness
Three loud moments per second are easy to count by ear, and the frequency difference is less than $10\,\text{Hz}$, so the beats stand out clearly. When the tuner tightens the string, the frequency rises towards $262\,\text{Hz}$ and the beats slow down.
- Exercise 14-7Core
A loudspeaker emits sound in air with a wavelength of $0.850\,\text{m}$. The sound passes from the surface of a lake into the water.
What is the wavelength of the sound in water? The speed of sound in air is $340\,\text{m/s}$ and in water $1480\,\text{m/s}$.
Answer
$3.70\,\text{m}$
Show the solution
Given
wavelength in air $\lambda_\text{a} = 0.850\,\text{m}$, speed of sound in air $v_\text{a} = 340\,\text{m/s}$ and in water $v_\text{w} = 1480\,\text{m/s}$
Find
the wavelength in water $\lambda_\text{w}$
Principle
When sound passes from air into water, the frequency stays the same, because the particles at the boundary vibrate in step with the incoming wave. The speed changes, so the wavelength changes. In both substances $v = f\lambda$ holds.
Formula
$f$$= \dfrac{v_\text{a}}{\lambda_\text{a}}$$\lambda_\text{w}$$= \dfrac{v_\text{w}}{f}$Conversions
None needed: the speeds are in $\text{m/s}$ and the wavelength in metres.
Substitution
$f$$= \dfrac{340\,\text{m/s}}{0.850\,\text{m}}$$= 400\,\text{Hz}$$\lambda_\text{w}$$= \dfrac{1480\,\text{m/s}}{400\,\text{Hz}}$$= 3.70\,\text{m}$Answer
$3.70\,\text{m}$
Three significant figures like the given data.
Reasonableness
Sound travels $\tfrac{1480}{340} \approx 4.4$ times as fast in water as in air, and at the same frequency the wavelength grows in the same ratio: $0.850\,\text{m} \cdot 4.35 \approx 3.7\,\text{m}$.
- Exercise 14-8
A driver drives at $90.0\,\text{km/h}$ straight towards a junction where a stationary warning siren sounds at $650\,\text{Hz}$.
What frequency does the driver hear? The air is calm, and the speed of sound is $340\,\text{m/s}$.
Answer
$698\,\text{Hz}$
Show the solution
Given
frequency of the siren $f_0 = 650\,\text{Hz}$, speed of the driver $v_\text{h} = 90.0\,\text{km/h}$ towards the siren, speed of sound $v = 340\,\text{m/s}$
Find
the frequency heard by the driver $f$
Principle
The source is at rest and the listener moves, so the wavelength does not change, but the driver meets the wavefronts more often, because they approach at speed $v + v_\text{h}$. We use the Doppler formula for a moving listener with the plus sign for approaching.
Formula
$f$$= f_0 \cdot \dfrac{v + v_\text{h}}{v}$Conversions
$v_\text{h}$$= 90.0\,\tfrac{\text{km}}{\text{h}}$$= 90.0 \cdot \tfrac{1000\,\text{m}}{3600\,\text{s}}$$= 25.0\,\tfrac{\text{m}}{\text{s}}$Substitution
$f$$= 650\,\text{Hz} \cdot \dfrac{340\,\text{m/s} + 25.0\,\text{m/s}}{340\,\text{m/s}}$$= 650\,\text{Hz} \cdot \dfrac{365}{340}$$= 697.79\,\text{Hz}$Answer
$698\,\text{Hz}$
Three significant figures like the given data.
Reasonableness
The frequency rises because the driver approaches. The rise is
$f_0 \cdot \tfrac{v_\text{h}}{v}$$= 650\,\text{Hz} \cdot \tfrac{25.0}{340}$$\approx 48\,\text{Hz}$which is about $7\,\%$, like the ratio of the speeds $\tfrac{25}{340}$.
- Exercise 14-9
The engine of a racing car sounds at $425\,\text{Hz}$ in the pit at rest. When the car drives at the same engine speed straight away from a spectator standing on the extension of the straight, the spectator hears the frequency $398\,\text{Hz}$.
How fast is the car moving? The air is calm, and the speed of sound is $340\,\text{m/s}$.
Answer
$23\,\text{m/s}$
Show the solution
Given
frequency of the engine at rest $f_0 = 425\,\text{Hz}$, frequency heard as the car moves away $f = 398\,\text{Hz}$, speed of sound $v = 340\,\text{m/s}$
Find
the speed of the car $v_\text{s}$
Principle
The car, that is, the sound source, moves straight away from the spectator, and the spectator is at rest, so the Doppler formula for a moving source applies with the plus sign for moving away. The engine speed is the same, so the engine's own frequency is $f_0$. The speed of the source is solved from the formula.
Formula
$f$$= f_0 \cdot \dfrac{v}{v + v_\text{s}}$so
$v + v_\text{s}$$= \dfrac{f_0 v}{f}$$v_\text{s}$$= v \left(\dfrac{f_0}{f} - 1\right)$Conversions
None needed: the frequencies are in hertz, and the unit cancels in their ratio.
Substitution
$v_\text{s}$$= 340\,\text{m/s} \cdot \left(\dfrac{425\,\text{Hz}}{398\,\text{Hz}} - 1\right)$$= 340\,\text{m/s} \cdot 0.067839$$= 23.065\,\text{m/s}$Answer
$23\,\text{m/s}$
Two significant figures, because the result depends on the frequency difference $27\,\text{Hz}$, which is known to two figures.
Reasonableness
$23\,\text{m/s}$ is about $83\,\text{km/h}$, which is a believable speed on a racetrack. Check: $425\,\text{Hz} \cdot \tfrac{340}{340 + 23.07} \approx 398\,\text{Hz}$, as measured.
Level 3 · challenging 3 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 14-10
A listener stands right next to a straight road and measures with a phone app the frequency of the siren of a passing ambulance: $476\,\text{Hz}$ as the car approaches and $404\,\text{Hz}$ as it moves away. The speed of the ambulance is constant, but the siren's own frequency is not known.
How fast is the ambulance moving? The air is calm, and the speed of sound is $340\,\text{m/s}$.
Answer
$28\,\text{m/s}$
Show the solution
Given
frequency heard when approaching $f_1 = 476\,\text{Hz}$ and when moving away $f_2 = 404\,\text{Hz}$, speed of sound $v = 340\,\text{m/s}$; the frequency of the siren $f_0$ is unknown
Find
the speed of the ambulance $v_\text{s}$
Principle
The listener is at rest and the source moves straight towards them and straight away from them at the same speed, so the moving-source formula holds in both phases, with the minus sign when approaching and the plus sign when moving away. The unknown $f_0$ cancels when the frequencies are divided by each other.
Formula
$f_1$$= f_0 \cdot \dfrac{v}{v - v_\text{s}}$$f_2$$= f_0 \cdot \dfrac{v}{v + v_\text{s}}$so
$\dfrac{f_1}{f_2}$$= \dfrac{v + v_\text{s}}{v - v_\text{s}}$Cross-multiplying, $f_1 v - f_1 v_\text{s} = f_2 v + f_2 v_\text{s}$, so
$v_\text{s} (f_1 + f_2)$$= v (f_1 - f_2)$$v_\text{s}$$= v \cdot \dfrac{f_1 - f_2}{f_1 + f_2}$Conversions
None needed: the unit of the frequencies cancels, and the speed comes out in $\text{m/s}$.
Substitution
$v_\text{s}$$= 340\,\text{m/s} \cdot \dfrac{476\,\text{Hz} - 404\,\text{Hz}}{476\,\text{Hz} + 404\,\text{Hz}}$$= 340\,\text{m/s} \cdot \dfrac{72}{880}$$= 27.818\,\text{m/s}$Answer
$28\,\text{m/s}$
Two significant figures, because the frequency difference $72\,\text{Hz}$ is known to two figures.
Reasonableness
$28\,\text{m/s}$ is about $100\,\text{km/h}$, which suits an emergency drive. The siren's own frequency is
$f_0$$= f_1 \cdot \tfrac{v - v_\text{s}}{v}$$\approx 437\,\text{Hz}$and it lies between the frequencies heard, as it should. Also
$f_2$$= 437\,\text{Hz} \cdot \tfrac{340}{367.8}$$\approx 404\,\text{Hz}$agrees.
- Exercise 14-11
A guitarist plays the A string at the same time as a $440\,\text{Hz}$ reference note and hears a beat frequency of $4\,\text{Hz}$, that is, four loud moments every second. When she tightens the string slightly, the beat frequency grows to $6\,\text{Hz}$.
What was the frequency of the string before tightening? Justify why the answer is unique.
Answer
$444\,\text{Hz}$
Show the solution
Given
reference note $f_0 = 440\,\text{Hz}$, beat frequency before tightening $f_\text{h} = 4\,\text{Hz}$ and after tightening $6\,\text{Hz}$
Find
the frequency of the string $f$ before tightening
Principle
The beat frequency is the absolute value of the frequency difference, so from the beats alone the frequency of the string could be either $440\,\text{Hz} - 4\,\text{Hz}$ or $440\,\text{Hz} + 4\,\text{Hz}$. Tightening increases the speed of the wave in the string and therefore raises the frequency of the string. Because the beats sped up, that is, the difference grew, the rising frequency moved away from the reference note: the string was already higher than the reference note. If the string had been $436\,\text{Hz}$, tightening would have taken it towards $440\,\text{Hz}$ and the beats would have slowed down.
Formula
$f_\text{h} = |f - f_0|$, and when $f > f_0$, $f = f_0 + f_\text{h}$
Conversions
None needed: all the frequencies are in hertz.
Substitution
$f$$= 440\,\text{Hz} + 4\,\text{Hz}$$= 444\,\text{Hz}$Answer
$444\,\text{Hz}$
In whole hertz like the reference note and the beat frequency.
Reasonableness
After tightening, the frequency of the string is $440\,\text{Hz} + 6\,\text{Hz} = 446\,\text{Hz}$, so it rose by $2\,\text{Hz}$, which suits a small tightening. To tune the string, it must therefore be loosened until the beats stop.
- Exercise 14-12
A student measures the FFT spectrum of a low note of a clarinet.
The spectrum has clear peaks at the frequencies $147\,\text{Hz}$, $441\,\text{Hz}$, $735\,\text{Hz}$ and $1029\,\text{Hz}$, and almost nothing in between. a) What is the fundamental frequency of the note? b) Which members of the harmonic series are missing from the spectrum? c) What does this tell about the standing wave of the air column of the clarinet?
Answer
Fundamental frequency $147\,\text{Hz}$; the even multiples ($294\,\text{Hz}$, $588\,\text{Hz}$, $882\,\text{Hz}$) are missing, so the air column vibrates like a tube closed at one end.
Show the solution
Find
the fundamental frequency of the note, the members of the harmonic series missing from the spectrum, and what they mean for the air column of the clarinet
Principle
The peaks in the spectrum of an instrument are at the natural frequencies of the air column, and the fundamental frequency is the lowest natural frequency. In a tube open at both ends the natural frequencies are all multiples $n f_1$ of the fundamental frequency, but in a tube closed at one end only the odd multiples, because the closed end is always a node and the open end an antinode (chapter Standing waves: nodes, antinodes and natural frequencies).
Model answer
The lowest peak is $147\,\text{Hz}$, and the others are its multiples: $441\,\text{Hz} = 3 \cdot 147\,\text{Hz}$, $735\,\text{Hz} = 5 \cdot 147\,\text{Hz}$ and $1029\,\text{Hz} = 7 \cdot 147\,\text{Hz}$. So the fundamental frequency is $147\,\text{Hz}$. The even members of the harmonic series, $294\,\text{Hz}$, $588\,\text{Hz}$ and $882\,\text{Hz}$, are almost entirely missing. Only the odd multiples are natural frequencies of a tube closed at one end, so the air column of the clarinet vibrates like a tube with one end closed: the mouthpiece end, where the reed is, acts as the closed end with a node of the air motion, and the bell end as the open end with an antinode. The missing overtones give the clarinet its dark timbre.
Photographs
- Kennedy Road ambulance panning. Jason Zhang. source, CC BY-SA 3.0
