High School Physics · Grades 10–12

21Sound and Acoustics

Pluck a guitar’s low string: the room fills with a low A, yet nothing traveled to your ear but squeezed and stretched air. This chapter follows that pattern — its speed, how frequency becomes pitch and power loudness, and why one note differs on every instrument.

21.1 Sound is a pressure wave

Definition 21.1 (Sound wave)

A sound wave is a mechanical pressure wave (Chapter 20): a vibrating surface creates traveling compressions (slight overpressure) and rarefactions (underpressure). Air parcels oscillate along the travel direction (sound is longitudinal), and a medium is needed: a bell under a vacuum jar falls silent (Chapter 8).

A loudspeaker drives a longitudinal wave: parcels bunch into compressions one wavelength = v/f apart. — A loudspeaker drives a longitudinal wave: parcels bunch into compressions one wavelength $\lambda = v/f$ apart.

Proposition 21.2 (Speed of sound)

Sound travels at about $340\,\mathrm{m}/\mathrm{s}$ in air at $20{}^{\circ}\mathrm{C}$ , $1500\,\mathrm{m}/\mathrm{s}$ in water and $5000\,\mathrm{m}/\mathrm{s}$ in steel: the stiffer the medium, the faster.

Proof. Admitted at this level. ∎

Remark 21.3 (Warm air is faster)

These are measured values; predicting them is done in the Year 1 volume. Sound speeds up in warmer air ( $331\,\mathrm{m}/\mathrm{s}$ at $0{}^{\circ}\mathrm{C}$ ) — why winds drift out of tune as the hall warms.

Example 21.4 (The sizes of sounds)

The relation $\lambda = v/f$ sizes any sound: in air a $20\,\mathrm{Hz}$ rumble spans $340/20 = 17\,\mathrm{m}$ — a building — and a $20\,\mathrm{kHz}$ hiss fits in $1.7\,\mathrm{cm}$ ; the metre-sized ones bend easily around doors and corners.

21.2 Pitch, loudness and the ear

Definition 21.5 (Pitch and loudness)

Pitch is the perception of frequency: the higher $f$ , the higher the note; concert A is $440\,\mathrm{Hz}$ , and doubling $f$ raises the note one octave. Loudness is the perception of amplitude: the stronger the pressure oscillation, the louder; amplification changes loudness, not pitch.

Remark 21.6 (The ear’s window)

Hearing spans about $20\,\mathrm{Hz}$ to $20\,\mathrm{kHz}$ (Chapter 8); below lies infrasound, above lies ultrasound — bats at $50\,\mathrm{kHz}$ , medical probes at megahertz. A piano only samples the middle: $27.5\text{ to }4186\,\mathrm{Hz}$ .

21.3 Sound intensity and the decibel scale

Definition 21.7 (Sound intensity)

The sound intensity $I$ is the acoustic power crossing one square metre facing the wave: $I = P/S$ , in $\mathrm{W}/\mathrm{m}^{2}$ . The quietest audible sound is about $I_0 = 1.0 \times 10^{-12}\,\mathrm{W}/\mathrm{m}^{2}$ ; pain begins near $1\,\mathrm{W}/\mathrm{m}^{2}$ .

Proposition 21.8 (Inverse-square attenuation)

A small source radiating a power $P$ equally in all directions gives, at distance $r$ , $I = P/(4\pi r^2) \propto 1/r^2$ : doubling the distance divides the intensity by four.

Proof. The whole power $P$ crosses the sphere of area $4\pi r^2$ ; divide. ∎

Definition 21.9 (Sound level)

The ear spans twelve powers of ten, so intensities are quoted logarithmically. The sound level of $I$ is

L = 10 \log_{10}(I/I_0), \qquad I_0 = 1.0 \times 10^{-12}\,\mathrm{W}/\mathrm{m}^{2},

in decibels ( $\mathrm{dB}$ ): hearing threshold $0\,\mathrm{dB}$ , pain near $120\,\mathrm{dB}$ .

Proposition 21.10 (Decibel arithmetic)

Multiplying $I$ by $k$ adds $10\log_{10}k$ decibels. In particular $\times 10$ adds $10\,\mathrm{dB}$ ; $\times 2$ adds $10\log_{10}2 \approx 3.0\,\mathrm{dB}$ ; $N$ incoherent identical sources add intensities, never levels, giving $L_1 + 10\log_{10}N$ ; doubling the distance to a small source ( $I$ divided by four) removes about $6.0\,\mathrm{dB}$ .

Proof. $10\log_{10}(kI/I_0) = 10\log_{10}(I/I_0) + 10\log_{10}k$ ; distance: Proposition 21.8, $k = 1/4$ . ∎

Method 21.11 (Working in decibels)

$L = 10\log_{10}(I/I_0)$ ; back: $I = I_0 \times 10^{L/10}$ .
Never add levels: convert to intensities, add, convert back — or $+3\,\mathrm{dB}$ per doubling, $+10\,\mathrm{dB}$ per tenfold.
Each doubling of distance to a small source costs $6\,\mathrm{dB}$ .

A ladder of everyday levels: each 20\, dB step is a hundredfold intensity jump; beyond 85\, dB (red zone) exposure damages hearing. — A ladder of everyday levels: each $20\,\mathrm{dB}$ step is a hundredfold intensity jump; beyond $85\,\mathrm{dB}$ (red zone) exposure damages hearing.

Remark 21.12 (Hearing damage)

Hearing loss is a dose: $85\,\mathrm{dB}$ is safe for eight hours, and every $3\,\mathrm{dB}$ more — a doubling of intensity — halves the safe time; lost hair cells do not grow back.

21.4 Timbre and harmonics

Definition 21.13 (Harmonics)

A periodic sound of frequency $f_1$ — the fundamental — is in general a sum of sines at the frequencies $f_n = n f_1$ , $n = 1, 2, 3, \dots$ , its harmonics. A sound reduced to its fundamental is a pure tone, like a tuning fork’s.

Definition 21.14 (Spectrum and timbre)

The spectrum of a sound is the bar chart of its harmonics: one bar per $f_n$ , the height that harmonic’s strength. The ear hears $f_1$ as the pitch and the mix of strengths as the timbre — the sound’s color: instruments playing the same note share $f_1$ , differing only in their spectra.

The same A (220\, Hz): waveforms (top), spectra (bottom) — same fundamental and pitch; the violin’s harmonic ladder is its timbre. — The same A ( $220\,\mathrm{Hz}$ ): waveforms (top), spectra (bottom) — same fundamental and pitch; the violin’s harmonic ladder is its timbre.

Example 21.15 (Why you recognize the caller)

A flute’s A is nearly a pure tone; a violin bowing the same $220\,\mathrm{Hz}$ adds strong harmonics at $440\,\mathrm{Hz}\text{, }660\,\mathrm{Hz}\text{ and }880\,\mathrm{Hz}$ . Same pitch and loudness, yet no one confuses them: a voice, too, is a spectrum you know by heart.

21.5 The vibrating string

Definition 21.16 (Standing wave)

A wave confined between the fixed ends of a string can settle into a standing wave: an oscillation in place, motionless points — nodes — alternating with points of maximal swing — antinodes. Nothing travels any more; the string vibrates in a fixed pattern.

Proposition 21.17 (Modes of a string)

A string of length $L$ , fixed at both ends, carrying waves of speed $v$ , vibrates steadily only in patterns fitting a whole number $n$ of half-wavelengths between the imposed end nodes:

\lambda_n = \frac{2L}{n}, \quad f_n = n\,\frac{v}{2L} = n f_1, \quad n = 1, 2, 3, \dots

The lowest mode $f_1 = v/(2L)$ is the fundamental; the others are exactly its harmonics.

Proof. Admitted at this level. ∎

Remark 21.18 (Where the honest proof lives)

That confined waves pick exactly these modes follows from superposing the wave with its reflections, a calculus carried out in the Year 1 volume. The geometry convinces: both ends must stay still, so a whole number of half-wavelengths must fit.

The first three modes of a string fixed at both ends: n half-wavelengths fit in L, so _n = 2L/n and f_n = nf_1. — The first three modes of a string fixed at both ends: $n$ half-wavelengths fit in $L$ , so $\lambda_n = 2L/n$ and $f_n = nf_1$ .

Example 21.19 (A violin’s A string)

A violin’s A string has $L = 32.8\,\mathrm{cm}$ and sounds $f_1 = 440\,\mathrm{Hz}$ : waves run along it at $v = 2Lf_1 = 2 \times 0.328 \times 440 \approx 289\,\mathrm{m}/\mathrm{s}$ — set by tension and mass, tuned by the peg. Pressing a finger shortens $L$ and raises every $f_n$ at once: the spectrum slides up.

Remark 21.20 (Why instruments differ)

How the string is excited — plucked, bowed, struck — sets how much of each mode is present, and the body amplifies some harmonics more than others; excitation plus body fix the spectrum, so the same $f_1$ leaves a guitar, a violin and a piano each sounding like itself.

21.6 Exercises

Exercise 21.1 ★

In air ( $v = 340\,\mathrm{m}/\mathrm{s}$ ), find the wavelength of (a) a $55\,\mathrm{Hz}$ bass note; (b) a $1.0\,\mathrm{kHz}$ beep; (c) a $15\,\mathrm{kHz}$ whine. Compare the largest to the smallest.

Solution

Solution of Exercise 21.1.

$\lambda = v/f$ : (a) $340/55 \approx 6.2\,\mathrm{m}$ ; (b) $0.34\,\mathrm{m}$ ; (c) $2.3\,\mathrm{cm}$ . Ratio $15000/55 \approx 270$ .

Exercise 21.2 ★

Classify as infrasound, audible sound or ultrasound: $8\,\mathrm{Hz}$ ; $100\,\mathrm{Hz}$ ; $15\,\mathrm{kHz}$ ; $40\,\mathrm{kHz}$ ; $3\,\mathrm{MHz}$ . Of the audible ones, which sounds lower in pitch?

Solution

Solution of Exercise 21.2.

Infrasound: $8\,\mathrm{Hz}$ . Audible: $100\,\mathrm{Hz}$ and $15\,\mathrm{kHz}$ . Ultrasound: $40\,\mathrm{kHz}$ and $3\,\mathrm{MHz}$ . Lower pitch: $100\,\mathrm{Hz}$ (lower frequency).

Exercise 21.3 ★

A worker strikes a rail $1.0\,\mathrm{km}$ away. Compute the travel times in the steel and in the air. What does a listener with one ear pressed to the rail hear?

Solution

Solution of Exercise 21.3.

Steel: $1000/5000 = 0.20\,\mathrm{s}$ ; air: $1000/340 \approx 2.9\,\mathrm{s}$ . Two clicks, $2.7\,\mathrm{s}$ apart — the rail first.

Exercise 21.4 ★

(a) The sound level for $I = 1.0 \times 10^{-5}\,\mathrm{W}/\mathrm{m}^{2}$ ? (b) The intensity at $60\,\mathrm{dB}$ ? (c) Name an everyday sound near each level.

Solution

Solution of Exercise 21.4.

(a) $L = 10\log_{10}10^{7} = 70\,\mathrm{dB}$ . (b) $I = 10^{-12} \times 10^{6} = 1.0 \times 10^{-6}\,\mathrm{W}/\mathrm{m}^{2}$ . (c) $70\,\mathrm{dB}$ : a vacuum cleaner; $60\,\mathrm{dB}$ : a conversation.

Exercise 21.5 ★

A violin G string sounds $f_1 = 196\,\mathrm{Hz}$ : give its next three harmonics. Why does a $12\,\mathrm{kHz}$ whistle, however rich at the source, reach the ear as a pure tone?

Solution

Solution of Exercise 21.5.

$392\,\mathrm{Hz}\text{, }588\,\mathrm{Hz}\text{ and }784\,\mathrm{Hz}$ . Its higher harmonics ( $n \ge 2$ ) start at $24\,\mathrm{kHz}$ , above the audible ceiling: only the $12\,\mathrm{kHz}$ fundamental is heard.

Exercise 21.6 ★★

A siren radiates $P = 0.50\,\mathrm{W}$ equally in all directions. Compute the intensity and the sound level $5.0\,\mathrm{m}$ away.

Solution

Solution of Exercise 21.6.

$I = 0.50/(4\pi \times 25) \approx 1.6 \times 10^{-3}\,\mathrm{W}/\mathrm{m}^{2}$ ; $L = 10\log_{10}(1.6 \times 10^{9}) \approx 92\,\mathrm{dB}$ .

Exercise 21.7 ★★

One violin gives $68\,\mathrm{dB}$ at your seat. What level do two give? Four? How many violins would it take to reach $78\,\mathrm{dB}$ ?

Solution

Solution of Exercise 21.7.

Two: $+3\,\mathrm{dB} \to 71\,\mathrm{dB}$ ; four: $74\,\mathrm{dB}$ . $+10\,\mathrm{dB}$ needs $\times 10$ in intensity: $10$ violins.

Exercise 21.8 ★★

The level is $95\,\mathrm{dB}$ at $2.0\,\mathrm{m}$ from a small loudspeaker. Predict the level at $8.0\,\mathrm{m}$ : below the $85\,\mathrm{dB}$ mark?

Solution

Solution of Exercise 21.8.

$r \times 4$ : $I \div 16$ , so $-12\,\mathrm{dB}$ : $83\,\mathrm{dB}$ . Yes — just below the $85\,\mathrm{dB}$ mark.

Exercise 21.9 ★★

A $440\,\mathrm{Hz}$ note passes from air into water. Compute its wavelength in each medium. The diver hears the same pitch: which quantity is fixed by the source, and which one adjusts?

Solution

Solution of Exercise 21.9.

Air: $340/440 \approx 0.77\,\mathrm{m}$ ; water: $1500/440 \approx 3.4\,\mathrm{m}$ . The source fixes the frequency; the wavelength $\lambda = v/f$ adjusts to the medium.

Exercise 21.10 ★★

A spectrum shows peaks at $130\,\mathrm{Hz}\text{, }260\,\mathrm{Hz}\text{, }390\,\mathrm{Hz}\text{ and }520\,\mathrm{Hz}$ of decreasing heights: what is the fundamental? Another instrument plays the same note at the same loudness: what is the same in its spectrum, what differs?

Solution

Solution of Exercise 21.10.

$f_1 = 130\,\mathrm{Hz}$ (the peak spacing). Same: the bar positions $nf_1$ (same pitch) and the total intensity; different: the bar heights — the mix that makes the timbre.

Exercise 21.11 ★★

A guitar string of length $65.0\,\mathrm{cm}$ sounds $f_1 = 110\,\mathrm{Hz}$ . Find the wave speed, then the vibrating lengths giving the octave ( $220\,\mathrm{Hz}$ ) and the fifth ( $165\,\mathrm{Hz}$ ).

Solution

Solution of Exercise 21.11.

$v = 2Lf_1 = 2 \times 0.650 \times 110 = 143\,\mathrm{m}/\mathrm{s}$ . Octave: $L' = v/(2 \times 220) = L/2 = 32.5\,\mathrm{cm}$ ; fifth: $L' = 143/330 \approx 43.3\,\mathrm{cm} = 2L/3$ .

Exercise 21.12 ★★★

A medical probe sends $5.0\,\mathrm{MHz}$ ultrasound into soft tissue ( $v = 1540\,\mathrm{m}/\mathrm{s}$ ): the wavelength? Echoes return after $26\,\text{µ}\mathrm{s}$ and $58\,\text{µ}\mathrm{s}$ from an organ’s two walls: find their depths and the organ’s thickness.

Solution

Solution of Exercise 21.12.

$\lambda = 1540/5.0 \times 10^{6} \approx 0.31\,\mathrm{mm}$ . Depth $= vt/2$ : $1540 \times 26 \times 10^{-6}/2 \approx 2.0\,\mathrm{cm}$ and $1540 \times 58 \times 10^{-6}/2 \approx 4.5\,\mathrm{cm}$ ; thickness $\approx 2.5\,\mathrm{cm}$ .

Exercise 21.13 ★★★

An outdoor stage produces $110\,\mathrm{dB}$ at $3.0\,\mathrm{m}$ . Treating it as a small source, at what distance does the level fall to $85\,\mathrm{dB}$ ? Give two reasons the real distance differs.

Solution

Solution of Exercise 21.13.

$-25\,\mathrm{dB}$ means $I \div 10^{2.5} \approx 316$ , so $r \times \sqrt{316} \approx 17.8$ : $r \approx 3.0 \times 17.8 \approx 53\,\mathrm{m}$ . Real halls differ: reflections from ground and walls add intensity, while air and the crowd absorb it (and the stage is not an isotropic point source).

Exercise 21.14 ★★★

Find the ratio of the pain threshold ( $1\,\mathrm{W}/\mathrm{m}^{2}$ ) to $I_0$ . Taking the eardrum as $0.5\,\mathrm{cm}^{2}$ , compute the power received at each threshold; comment on the ear as a detector.

Solution

Solution of Exercise 21.14.

$1/10^{-12} = 10^{12}$ . Eardrum $S = 5 \times 10^{-5}\,\mathrm{m}^{2}$ : received power $5 \times 10^{-17}\,\mathrm{W}$ at threshold, $5 \times 10^{-5}\,\mathrm{W}$ at pain — the ear detects tens of attowatts and still works at a trillion times more: a detector with a twelve-decade dynamic range.

Exercise 21.15 ★★★

Each identical festival loudspeaker alone gives $80\,\mathrm{dB}$ at the mixing desk. What level do $2$ give? $10$ ? $100$ ? What does each extra $10\,\mathrm{dB}$ cost — what does this say of loudness races?

Solution

Solution of Exercise 21.15.

$2$ : $83\,\mathrm{dB}$ ; $10$ : $90\,\mathrm{dB}$ ; $100$ : $100\,\mathrm{dB}$ . Each extra $10\,\mathrm{dB}$ costs a tenfold in loudspeakers: loudness races are exponentially expensive (and dangerous long before they are won).

21.7 Problem: The Luthier’s Workshop

Problem 21.1

Weekend problem — the luthier’s workshop: a guitar string laid out fret by fret, its harmonics tuned into a timbre, a recital kept below the danger line, and an ultrasound hunt for a flaw in the wood

A luthier is finishing a guitar. Its A string has vibrating length $L = 65.0\,\mathrm{cm}$ and must sound $f_1 = 110\,\mathrm{Hz}$ . Take $340\,\mathrm{m}/\mathrm{s}$ for sound in air, and admit $f_1 = v/(2L)$ (Proposition 21.17), $v$ being the wave speed on the string.

Part I — Laying out the fretboard.

Compute the wave speed $v$ on the string.
Compute the fundamental’s wavelength on the string, then the wavelength of the $110\,\mathrm{Hz}$ sound in the air. Why do they differ at the same frequency?
The octave ( $220\,\mathrm{Hz}$ ): what vibrating length gives it? Where must the twelfth fret sit, as a fraction of $L$ ?
The fifth ( $165\,\mathrm{Hz}$ ): what vibrating length gives it?
Each fret raises the note one semitone, a factor $2^{1/12}$ . Check with logarithms how many semitones separate $110\,\mathrm{Hz}$ from $220\,\mathrm{Hz}$ ; length at the first fret?

Part II — Harmonics and timbre.

List the first five harmonics of the open A string.
A violinist’s A sounds $440\,\mathrm{Hz}$ . Which harmonic of the guitar string is that, and what musical interval separates it from the fundamental? And $330\,\mathrm{Hz}$ ?
The luthier plucks near the bridge, then over the fingerboard: same pitch, different color. Explain with spectra.
A light touch at the midpoint while plucking forces a node there. Which harmonics survive, and what pitch is heard?
How many harmonics of the open string lie, in principle, within the audible range?

Part III — The recital. At the guitar’s first concert, each instrument radiates an acoustic power $P = 6.0 \times 10^{-3}\,\mathrm{W}$ as a small source, equally in all directions.

Compute the intensity at $r = 3.0\,\mathrm{m}$ from one instrument.
Deduce the sound level there.
A quartet plays: what level do four such instruments give at the same spot? Why is it not four times the level?
Twelve instruments play: what is the level at $3.0\,\mathrm{m}$ ?
How far back must a listener sit for the ensemble to fall to $85\,\mathrm{dB}$ ?
Safety rule: $85\,\mathrm{dB}$ is safe for eight hours, and every $3\,\mathrm{dB}$ halves the safe time. How long may the front-row listener of question 14 safely stay?

Part IV — The flaw in the wood. Before varnishing, the luthier probes a $45\,\mathrm{mm}$ thick neck blank with a $2.0\,\mathrm{MHz}$ tester; in this wood take $v = 5.0 \times 10^{3}\,\mathrm{m}/\mathrm{s}$ .

Compute the ultrasound wavelength in the wood. Why must a flaw-hunting wave be ultrasonic rather than audible?
Compute the echo delay from the back face of the blank.
An echo arrives after $7.2\,\text{µ}\mathrm{s}$ : how deep is the flaw?
Write the workshop card: wave speed on the string, twelfth- and first-fret lengths, safe listening distance, flaw depth.

Solution

Solution of Problem 21.1.

1. $v = 2Lf_1 = 2 \times 0.650 \times 110 = 143\,\mathrm{m}/\mathrm{s}$ .

2. On the string $\lambda_1 = 2L = 1.30\,\mathrm{m}$ ; in air $\lambda = 340/110 \approx 3.1\,\mathrm{m}$ . Same $f$ , but $\lambda = v/f$ and the wave speeds differ.

3. $L' = v/(2 \times 220) = L/2 = 32.5\,\mathrm{cm}$ : the twelfth fret sits at the midpoint, $L/2$ from the bridge.

4. $L' = 143/(2 \times 165) \approx 43.3\,\mathrm{cm} = 2L/3$ .

5. $\log(220/110)/\log 2^{1/12} = 12$ semitones — twelve frets to the octave. First fret: $L_1 = L/2^{1/12} = 65.0/1.059 \approx 61.4\,\mathrm{cm}$ .

6. $110\,\mathrm{Hz}\text{, }220\,\mathrm{Hz}\text{, }330\,\mathrm{Hz}\text{, }440\,\mathrm{Hz}\text{ and }550\,\mathrm{Hz}$ .

7. $440\,\mathrm{Hz} = 4f_1$ : the fourth harmonic, two octaves up. $330\,\mathrm{Hz} = 3f_1$ : the third harmonic, an octave plus a fifth.

8. The pluck position sets the mode mix: near the bridge the high harmonics are strong (bright spectrum), over the fingerboard the fundamental dominates (mellow). Same $f_1$ , so same pitch — different spectrum, different timbre.

9. Only harmonics with a node at the midpoint survive: the even ones, $220\,\mathrm{Hz}\text{, }440\,\mathrm{Hz}\text{ and }660\,\mathrm{Hz}$ , …The pitch jumps an octave, to $220\,\mathrm{Hz}$ .

10. $nf_1 \le 20\,\mathrm{kHz}$ : $n \le 20000/110 = 181.8$ , so $181$ harmonics.

11. $I = P/(4\pi r^2) = 6.0 \times 10^{-3}/(4\pi \times 9.0) \approx 5.3 \times 10^{-5}\,\mathrm{W}/\mathrm{m}^{2}$ .

12. $L = 10\log_{10}(5.3 \times 10^{7}) \approx 77\,\mathrm{dB}$ .

13. $I \times 4$ : $+6\,\mathrm{dB} \to 83\,\mathrm{dB}$ . Levels are logarithms: intensities add, levels do not.

14. $+10\log_{10}12 \approx +10.8\,\mathrm{dB} \to 88\,\mathrm{dB}$ .

15. Target $I = I_0 \times 10^{8.5} \approx 3.2 \times 10^{-4}\,\mathrm{W}/\mathrm{m}^{2}$ with $12P = 7.2 \times 10^{-2}\,\mathrm{W}$ : $r = \sqrt{12P/(4\pi I)} \approx \sqrt{18.1} \approx 4.3\,\mathrm{m}$ .

16. $88\,\mathrm{dB}$ is $+3\,\mathrm{dB}$ over the eight-hour limit: the safe time halves once — four hours. The recital fits.

17. $\lambda = 5000/2.0 \times 10^{6} = 2.5\,\mathrm{mm}$ . A wave only reflects off flaws at least about a wavelength across: audible sound in wood has $\lambda$ of tens of centimetres and diffracts around any small defect.

18. $t = 2d/v = 2 \times 0.045/5000 = 18\,\text{µ}\mathrm{s}$ .

19. $d = vt/2 = 5000 \times 7.2 \times 10^{-6}/2 = 18\,\mathrm{mm}$ .

20. The card: $v = 143\,\mathrm{m}/\mathrm{s}$ on the string; frets at $32.5\,\mathrm{cm}$ (twelfth) and $61.4\,\mathrm{cm}$ (first); ensemble safe beyond $\approx 4.3\,\mathrm{m}$ ; flaw $18\,\mathrm{mm}$ deep — four numbers, one chapter.