Physics · Book 2 · Grades 10–12

High School Physics

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 λ=v/f\lambda = v/f apart.

Proposition 21.2 (Speed of sound)

Sound travels at about 340 m/s340\,\mathrm{m}/\mathrm{s} in air at 20∘C20{}^{\circ}\mathrm{C}, 1500 m/s1500\,\mathrm{m}/\mathrm{s} in water and 5000 m/s5000\,\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 m/s331\,\mathrm{m}/\mathrm{s} at 0∘C0{}^{\circ}\mathrm{C}) — why winds drift out of tune as the hall warms.

Example 21.4 (The sizes of sounds)

The relation λ=v/f\lambda = v/f sizes any sound: in air a 20 Hz20\,\mathrm{Hz} rumble spans 340/20=17 m340/20 = 17\,\mathrm{m} — a building — and a 20 kHz20\,\mathrm{kHz} hiss fits in 1.7 cm1.7\,\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 ff, the higher the note; concert A is 440 Hz440\,\mathrm{Hz}, and doubling ff 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 Hz20\,\mathrm{Hz} to 20 kHz20\,\mathrm{kHz} (Chapter 8); below lies infrasound, above lies ultrasound — bats at 50 kHz50\,\mathrm{kHz}, medical probes at megahertz. A piano only samples the middle: 27.5 to 4186 Hz27.5\text{ to }4186\,\mathrm{Hz}.

21.3 Sound intensity and the decibel scale

Definition 21.7 (Sound intensity)

The sound intensity II is the acoustic power crossing one square metre facing the wave: I=P/SI = P/S, in W/m2\mathrm{W}/\mathrm{m}^{2}. The quietest audible sound is about I0=1.0×10−12 W/m2I_0 = 1.0 \times 10^{-12}\,\mathrm{W}/\mathrm{m}^{2}; pain begins near 1 W/m21\,\mathrm{W}/\mathrm{m}^{2}.

Proposition 21.8 (Inverse-square attenuation)

A small source radiating a power PP equally in all directions gives, at distance rr, I=P/(4πr2)∝1/r2I = P/(4\pi r^2) \propto 1/r^2: doubling the distance divides the intensity by four.

Proof. The whole power PP crosses the sphere of area 4πr24\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 II is

L=10log⁡10(I/I0),I0=1.0×10−12 W/m2,L = 10 \log_{10}(I/I_0), \qquad I_0 = 1.0 \times 10^{-12}\,\mathrm{W}/\mathrm{m}^{2},

in decibels (dB\mathrm{dB}): hearing threshold 0 dB0\,\mathrm{dB}, pain near 120 dB120\,\mathrm{dB}.

Proposition 21.10 (Decibel arithmetic)

Multiplying II by kk adds 10log⁡10k10\log_{10}k decibels. In particular ×10\times 10 adds 10 dB10\,\mathrm{dB}; ×2\times 2 adds 10log⁡102≈3.0 dB10\log_{10}2 \approx 3.0\,\mathrm{dB}; NN incoherent identical sources add intensities, never levels, giving L1+10log⁡10NL_1 + 10\log_{10}N; doubling the distance to a small source (II divided by four) removes about 6.0 dB6.0\,\mathrm{dB}.

Proof. 10log⁡10(kI/I0)=10log⁡10(I/I0)+10log⁡10k10\log_{10}(kI/I_0) = 10\log_{10}(I/I_0) + 10\log_{10}k; distance: Proposition 21.8, k=1/4k = 1/4. ∎

Method 21.11 (Working in decibels)

  1. L=10log⁡10(I/I0)L = 10\log_{10}(I/I_0); back: I=I0×10L/10I = I_0 \times 10^{L/10}.
  2. Never add levels: convert to intensities, add, convert back — or +3 dB+3\,\mathrm{dB} per doubling, +10 dB+10\,\mathrm{dB} per tenfold.
  3. Each doubling of distance to a small source costs 6 dB6\,\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 dB20\,\mathrm{dB} step is a hundredfold intensity jump; beyond 85 dB85\,\mathrm{dB} (red zone) exposure damages hearing.

Remark 21.12 (Hearing damage)

Hearing loss is a dose: 85 dB85\,\mathrm{dB} is safe for eight hours, and every 3 dB3\,\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 f1f_1 — the fundamental — is in general a sum of sines at the frequencies fn=nf1f_n = n f_1, n=1,2,3,…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 fnf_n, the height that harmonic’s strength. The ear hears f1f_1 as the pitch and the mix of strengths as the timbre — the sound’s color: instruments playing the same note share f1f_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 Hz220\,\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 Hz220\,\mathrm{Hz} adds strong harmonics at 440 Hz, 660 Hz and 880 Hz440\,\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 LL, fixed at both ends, carrying waves of speed vv, vibrates steadily only in patterns fitting a whole number nn of half-wavelengths between the imposed end nodes:

λn=2Ln,fn=n v2L=nf1,n=1,2,3,…\lambda_n = \frac{2L}{n}, \quad f_n = n\,\frac{v}{2L} = n f_1, \quad n = 1, 2, 3, \dots

The lowest mode f1=v/(2L)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: nn half-wavelengths fit in LL, so λn=2L/n\lambda_n = 2L/n and fn=nf1f_n = nf_1.

Example 21.19 (A violin’s A string)

A violin’s A string has L=32.8 cmL = 32.8\,\mathrm{cm} and sounds f1=440 Hzf_1 = 440\,\mathrm{Hz}: waves run along it at v=2Lf1=2×0.328×440≈289 m/sv = 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 LL and raises every fnf_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 f1f_1 leaves a guitar, a violin and a piano each sounding like itself.

21.6 Exercises

Exercise 21.1 ★

In air (v=340 m/sv = 340\,\mathrm{m}/\mathrm{s}), find the wavelength of (a) a 55 Hz55\,\mathrm{Hz} bass note; (b) a 1.0 kHz1.0\,\mathrm{kHz} beep; (c) a 15 kHz15\,\mathrm{kHz} whine. Compare the largest to the smallest.

Solution

Solution of Exercise 21.1.

λ=v/f\lambda = v/f: (a) 340/55≈6.2 m340/55 \approx 6.2\,\mathrm{m}; (b) 0.34 m0.34\,\mathrm{m}; (c) 2.3 cm2.3\,\mathrm{cm}. Ratio 15000/55≈27015000/55 \approx 270.

Exercise 21.2 ★

Classify as infrasound, audible sound or ultrasound: 8 Hz8\,\mathrm{Hz}; 100 Hz100\,\mathrm{Hz}; 15 kHz15\,\mathrm{kHz}; 40 kHz40\,\mathrm{kHz}; 3 MHz3\,\mathrm{MHz}. Of the audible ones, which sounds lower in pitch?

Solution

Solution of Exercise 21.2.

Infrasound: 8 Hz8\,\mathrm{Hz}. Audible: 100 Hz100\,\mathrm{Hz} and 15 kHz15\,\mathrm{kHz}. Ultrasound: 40 kHz40\,\mathrm{kHz} and 3 MHz3\,\mathrm{MHz}. Lower pitch: 100 Hz100\,\mathrm{Hz} (lower frequency).

Exercise 21.3 ★

A worker strikes a rail 1.0 km1.0\,\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 s1000/5000 = 0.20\,\mathrm{s}; air: 1000/340≈2.9 s1000/340 \approx 2.9\,\mathrm{s}. Two clicks, 2.7 s2.7\,\mathrm{s} apart — the rail first.

Exercise 21.4 ★

(a) The sound level for I=1.0×10−5 W/m2I = 1.0 \times 10^{-5}\,\mathrm{W}/\mathrm{m}^{2}? (b) The intensity at 60 dB60\,\mathrm{dB}? (c) Name an everyday sound near each level.

Solution

Solution of Exercise 21.4.

(a) L=10log⁡10107=70 dBL = 10\log_{10}10^{7} = 70\,\mathrm{dB}. (b) I=10−12×106=1.0×10−6 W/m2I = 10^{-12} \times 10^{6} = 1.0 \times 10^{-6}\,\mathrm{W}/\mathrm{m}^{2}. (c) 70 dB70\,\mathrm{dB}: a vacuum cleaner; 60 dB60\,\mathrm{dB}: a conversation.

Exercise 21.5 ★

A violin G string sounds f1=196 Hzf_1 = 196\,\mathrm{Hz}: give its next three harmonics. Why does a 12 kHz12\,\mathrm{kHz} whistle, however rich at the source, reach the ear as a pure tone?

Solution

Solution of Exercise 21.5.

392 Hz, 588 Hz and 784 Hz392\,\mathrm{Hz}\text{, }588\,\mathrm{Hz}\text{ and }784\,\mathrm{Hz}. Its higher harmonics (n≥2n \ge 2) start at 24 kHz24\,\mathrm{kHz}, above the audible ceiling: only the 12 kHz12\,\mathrm{kHz} fundamental is heard.

Exercise 21.6 ★★

A siren radiates P=0.50 WP = 0.50\,\mathrm{W} equally in all directions. Compute the intensity and the sound level 5.0 m5.0\,\mathrm{m} away.

Solution

Solution of Exercise 21.6.

I=0.50/(4π×25)≈1.6×10−3 W/m2I = 0.50/(4\pi \times 25) \approx 1.6 \times 10^{-3}\,\mathrm{W}/\mathrm{m}^{2}; L=10log⁡10(1.6×109)≈92 dBL = 10\log_{10}(1.6 \times 10^{9}) \approx 92\,\mathrm{dB}.

Exercise 21.7 ★★

One violin gives 68 dB68\,\mathrm{dB} at your seat. What level do two give? Four? How many violins would it take to reach 78 dB78\,\mathrm{dB}?

Solution

Solution of Exercise 21.7.

Two: +3 dB→71 dB+3\,\mathrm{dB} \to 71\,\mathrm{dB}; four: 74 dB74\,\mathrm{dB}. +10 dB+10\,\mathrm{dB} needs ×10\times 10 in intensity: 1010 violins.

Exercise 21.8 ★★

The level is 95 dB95\,\mathrm{dB} at 2.0 m2.0\,\mathrm{m} from a small loudspeaker. Predict the level at 8.0 m8.0\,\mathrm{m}: below the 85 dB85\,\mathrm{dB} mark?

Solution

Solution of Exercise 21.8.

r×4r \times 4: I÷16I \div 16, so −12 dB-12\,\mathrm{dB}: 83 dB83\,\mathrm{dB}. Yes — just below the 85 dB85\,\mathrm{dB} mark.

Exercise 21.9 ★★

A 440 Hz440\,\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≈0.77 m340/440 \approx 0.77\,\mathrm{m}; water: 1500/440≈3.4 m1500/440 \approx 3.4\,\mathrm{m}. The source fixes the frequency; the wavelength λ=v/f\lambda = v/f adjusts to the medium.

Exercise 21.10 ★★

A spectrum shows peaks at 130 Hz, 260 Hz, 390 Hz and 520 Hz130\,\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.

f1=130 Hzf_1 = 130\,\mathrm{Hz} (the peak spacing). Same: the bar positions nf1nf_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 cm65.0\,\mathrm{cm} sounds f1=110 Hzf_1 = 110\,\mathrm{Hz}. Find the wave speed, then the vibrating lengths giving the octave (220 Hz220\,\mathrm{Hz}) and the fifth (165 Hz165\,\mathrm{Hz}).

Solution

Solution of Exercise 21.11.

v=2Lf1=2×0.650×110=143 m/sv = 2Lf_1 = 2 \times 0.650 \times 110 = 143\,\mathrm{m}/\mathrm{s}. Octave: L′=v/(2×220)=L/2=32.5 cmL' = v/(2 \times 220) = L/2 = 32.5\,\mathrm{cm}; fifth: L′=143/330≈43.3 cm=2L/3L' = 143/330 \approx 43.3\,\mathrm{cm} = 2L/3.

Exercise 21.12 ★★★

A medical probe sends 5.0 MHz5.0\,\mathrm{MHz} ultrasound into soft tissue (v=1540 m/sv = 1540\,\mathrm{m}/\mathrm{s}): the wavelength? Echoes return after 26 µs26\,\text{µ}\mathrm{s} and 58 µs58\,\text{µ}\mathrm{s} from an organ’s two walls: find their depths and the organ’s thickness.

Solution

Solution of Exercise 21.12.

λ=1540/5.0×106≈0.31 mm\lambda = 1540/5.0 \times 10^{6} \approx 0.31\,\mathrm{mm}. Depth =vt/2= vt/2: 1540×26×10−6/2≈2.0 cm1540 \times 26 \times 10^{-6}/2 \approx 2.0\,\mathrm{cm} and 1540×58×10−6/2≈4.5 cm1540 \times 58 \times 10^{-6}/2 \approx 4.5\,\mathrm{cm}; thickness ≈2.5 cm\approx 2.5\,\mathrm{cm}.

Exercise 21.13 ★★★

An outdoor stage produces 110 dB110\,\mathrm{dB} at 3.0 m3.0\,\mathrm{m}. Treating it as a small source, at what distance does the level fall to 85 dB85\,\mathrm{dB}? Give two reasons the real distance differs.

Solution

Solution of Exercise 21.13.

−25 dB-25\,\mathrm{dB} means I÷102.5≈316I \div 10^{2.5} \approx 316, so r×316≈17.8r \times \sqrt{316} \approx 17.8: r≈3.0×17.8≈53 mr \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 W/m21\,\mathrm{W}/\mathrm{m}^{2}) to I0I_0. Taking the eardrum as 0.5 cm20.5\,\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=10121/10^{-12} = 10^{12}. Eardrum S=5×10−5 m2S = 5 \times 10^{-5}\,\mathrm{m}^{2}: received power 5×10−17 W5 \times 10^{-17}\,\mathrm{W} at threshold, 5×10−5 W5 \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 dB80\,\mathrm{dB} at the mixing desk. What level do 22 give? 1010? 100100? What does each extra +10 dB+10\,\mathrm{dB} cost — what does this say of loudness races?

Solution

Solution of Exercise 21.15.

22: 83 dB83\,\mathrm{dB}; 1010: 90 dB90\,\mathrm{dB}; 100100: 100 dB100\,\mathrm{dB}. Each extra +10 dB+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 cmL = 65.0\,\mathrm{cm} and must sound f1=110 Hzf_1 = 110\,\mathrm{Hz}. Take 340 m/s340\,\mathrm{m}/\mathrm{s} for sound in air, and admit f1=v/(2L)f_1 = v/(2L) (Proposition 21.17), vv being the wave speed on the string.

Part I — Laying out the fretboard.

  1. Compute the wave speed vv on the string.
  2. Compute the fundamental’s wavelength on the string, then the wavelength of the 110 Hz110\,\mathrm{Hz} sound in the air. Why do they differ at the same frequency?
  3. The octave (220 Hz220\,\mathrm{Hz}): what vibrating length gives it? Where must the twelfth fret sit, as a fraction of LL?
  4. The fifth (165 Hz165\,\mathrm{Hz}): what vibrating length gives it?
  5. Each fret raises the note one semitone, a factor 21/122^{1/12}. Check with logarithms how many semitones separate 110 Hz110\,\mathrm{Hz} from 220 Hz220\,\mathrm{Hz}; length at the first fret?

Part II — Harmonics and timbre.

  1. List the first five harmonics of the open A string.
  2. A violinist’s A sounds 440 Hz440\,\mathrm{Hz}. Which harmonic of the guitar string is that, and what musical interval separates it from the fundamental? And 330 Hz330\,\mathrm{Hz}?
  3. The luthier plucks near the bridge, then over the fingerboard: same pitch, different color. Explain with spectra.
  4. A light touch at the midpoint while plucking forces a node there. Which harmonics survive, and what pitch is heard?
  5. 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=6.0×10−3 WP = 6.0 \times 10^{-3}\,\mathrm{W} as a small source, equally in all directions.

  1. Compute the intensity at r=3.0 mr = 3.0\,\mathrm{m} from one instrument.
  2. Deduce the sound level there.
  3. A quartet plays: what level do four such instruments give at the same spot? Why is it not four times the level?
  4. Twelve instruments play: what is the level at 3.0 m3.0\,\mathrm{m}?
  5. How far back must a listener sit for the ensemble to fall to 85 dB85\,\mathrm{dB}?
  6. Safety rule: 85 dB85\,\mathrm{dB} is safe for eight hours, and every +3 dB+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 mm45\,\mathrm{mm} thick neck blank with a 2.0 MHz2.0\,\mathrm{MHz} tester; in this wood take v=5.0×103 m/sv = 5.0 \times 10^{3}\,\mathrm{m}/\mathrm{s}.

  1. Compute the ultrasound wavelength in the wood. Why must a flaw-hunting wave be ultrasonic rather than audible?
  2. Compute the echo delay from the back face of the blank.
  3. An echo arrives after 7.2 µs7.2\,\text{µ}\mathrm{s}: how deep is the flaw?
  4. 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=2Lf1=2×0.650×110=143 m/sv = 2Lf_1 = 2 \times 0.650 \times 110 = 143\,\mathrm{m}/\mathrm{s}.

2. On the string λ1=2L=1.30 m\lambda_1 = 2L = 1.30\,\mathrm{m}; in air λ=340/110≈3.1 m\lambda = 340/110 \approx 3.1\,\mathrm{m}. Same ff, but λ=v/f\lambda = v/f and the wave speeds differ.

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

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

5. log⁡(220/110)/log⁡21/12=12\log(220/110)/\log 2^{1/12} = 12 semitones — twelve frets to the octave. First fret: L1=L/21/12=65.0/1.059≈61.4 cmL_1 = L/2^{1/12} = 65.0/1.059 \approx 61.4\,\mathrm{cm}.

6. 110 Hz, 220 Hz, 330 Hz, 440 Hz and 550 Hz110\,\mathrm{Hz}\text{, }220\,\mathrm{Hz}\text{, }330\,\mathrm{Hz}\text{, }440\,\mathrm{Hz}\text{ and }550\,\mathrm{Hz}.

7. 440 Hz=4f1440\,\mathrm{Hz} = 4f_1: the fourth harmonic, two octaves up. 330 Hz=3f1330\,\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 f1f_1, so same pitch — different spectrum, different timbre.

9. Only harmonics with a node at the midpoint survive: the even ones, 220 Hz, 440 Hz and 660 Hz220\,\mathrm{Hz}\text{, }440\,\mathrm{Hz}\text{ and }660\,\mathrm{Hz}, …The pitch jumps an octave, to 220 Hz220\,\mathrm{Hz}.

10. nf1≤20 kHznf_1 \le 20\,\mathrm{kHz}: n≤20000/110=181.8n \le 20000/110 = 181.8, so 181181 harmonics.

11. I=P/(4πr2)=6.0×10−3/(4π×9.0)≈5.3×10−5 W/m2I = 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=10log⁡10(5.3×107)≈77 dBL = 10\log_{10}(5.3 \times 10^{7}) \approx 77\,\mathrm{dB}.

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

14. +10log⁡1012≈+10.8 dB→88 dB+10\log_{10}12 \approx +10.8\,\mathrm{dB} \to 88\,\mathrm{dB}.

15. Target I=I0×108.5≈3.2×10−4 W/m2I = I_0 \times 10^{8.5} \approx 3.2 \times 10^{-4}\,\mathrm{W}/\mathrm{m}^{2} with 12P=7.2×10−2 W12P = 7.2 \times 10^{-2}\,\mathrm{W}: r=12P/(4πI)≈18.1≈4.3 mr = \sqrt{12P/(4\pi I)} \approx \sqrt{18.1} \approx 4.3\,\mathrm{m}.

16. 88 dB88\,\mathrm{dB} is +3 dB+3\,\mathrm{dB} over the eight-hour limit: the safe time halves once — four hours. The recital fits.

17. λ=5000/2.0×106=2.5 mm\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 λ\lambda of tens of centimetres and diffracts around any small defect.

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

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

20. The card: v=143 m/sv = 143\,\mathrm{m}/\mathrm{s} on the string; frets at 32.5 cm32.5\,\mathrm{cm} (twelfth) and 61.4 cm61.4\,\mathrm{cm} (first); ensemble safe beyond ≈4.3 m\approx 4.3\,\mathrm{m}; flaw 18 mm18\,\mathrm{mm} deep — four numbers, one chapter.

Terms defined in this chapter

See all 431 terms in the glossary