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Chapter 8 • Theory & Derivations

Sound Waves

A comprehensive physical and psychoacoustic study of sound waves, decibel intensity levels, human pitch and loudness perception, 3D spherical wavefront propagation, inverse-square law, interference and diffraction, acoustic radiation efficiency, beats, Tartini combination tones, Doppler frequency shifts, and supersonic Mach shock cones.

§8.1 Intensity and Sound Intensity Levels: The Decibel Scale

Sound is a mechanical longitudinal compression wave propagating through an elastic medium. Human auditory perception spans an extraordinary dynamic range of twelve orders of magnitude in acoustic power.

1. Acoustic Intensity

The acoustic intensity ($I$) is defined as the time-averaged sound energy transmitted per unit time across a unit area perpendicular to the direction of wave propagation (W/m²): $$I = \langle p(t) v(t) \rangle = \frac{p_0^2}{2 \rho_0 c}$$ where:
  • $p_0$: Maximum acoustic pressure amplitude (Pa).
  • $\rho_0$: Equilibrium medium density ($1.204 \text{ kg/m}^3$ for dry air at 20°C).
  • $c$: Speed of sound ($343.2 \text{ m/s}$ in air at 20°C).
  • $\rho_0 c$: Specific acoustic impedance of air ($Z_0 \approx 413 \text{ Pa}\cdot\text{s/m} = 413 \text{ Rayls}$).
The threshold of human hearing at 1000 Hz is internationally defined as the reference intensity: $$I_0 = 1.00 \times 10^{-12} \text{ W/m}^2 \quad (\text{corresponding to } p_0 \approx 20 \ \mu\text{Pa})$$ The threshold of pain corresponds to $I \approx 1.0 \text{ W/m}^2$ ($p_0 \approx 29 \text{ Pa}$).

2. The Decibel (dB) Sound Level Scale

Because the human ear responds logarithmically (Weber-Fechner Law), acoustic levels are measured on the logarithmic Decibel Scale: $$\beta = 10 \log_{10}\left( \frac{I}{I_0} \right) \quad (\text{dB})$$ In terms of sound pressure level (SPL): $$\text{SPL} = 20 \log_{10}\left( \frac{p_{\text{rms}}}{p_{\text{ref}}} \right) \quad (\text{dB, with } p_{\text{ref}} = 20 \ \mu\text{Pa})$$ Key logarithmic benchmarks:
  • Threshold of hearing ($I = I_0$): $\beta = 10 \log_{10}(1) = 0\text{ dB}$.
  • Whisper: $\approx 20 - 30\text{ dB}$.
  • Normal conversation: $\approx 60\text{ dB}$.
  • Heavy city traffic: $\approx 80 - 85\text{ dB}$.
  • Rock concert / Jet engine takeoff at 50 m: $\approx 120 - 130\text{ dB}$ (threshold of pain).
Rule of Thumb:
  • Doubling sound intensity ($I \to 2I$) produces a $+3.01\text{ dB}$ increase ($\Delta\beta = 10\log_{10} 2 = 3.01\text{ dB}$).
  • A tenfold increase in intensity ($I \to 10I$) produces a $+10\text{ dB}$ increase.

§8.2 Loudness, Pitch, and Human Psychoacoustics

Human auditory perception differentiates between physical stimulus parameters (intensity, frequency, spectral content) and subjective psychoacoustic sensations (loudness, pitch, timbre).

1. Pitch and Fundamental Frequency

Pitch is the subjective sensation that orders sounds on a musical frequency scale from low/bass to high/treble. For pure sinusoidal tones, pitch is predominantly determined by frequency $f$. For complex multi-harmonic musical tones, the human brain perceives the pitch corresponding to the fundamental frequency $f_1$, even if the fundamental physical harmonic is filtered out or missing (the "missing fundamental" psychoacoustic illusion).

2. Loudness and Equal-Loudness Contours (Fletcher-Munson Curves)

Loudness is the subjective psychological magnitude of sound sensation. The human ear does not exhibit a flat frequency response; it is most sensitive in the range $2000 - 5000\text{ Hz}$ (due to acoustic resonance in the ear canal). Harvey Fletcher and Wilden A. Munson (1933) established the standard Equal-Loudness Contours:
  • Phon Scale: A sound has a loudness level of $L$ phons if it is judged to be equally loud as a $1000\text{ Hz}$ reference tone having a sound pressure level of $L$ dB.
  • Sone Scale: Developed by S. S. Stevens. 1 sone is defined as the loudness of a 1000 Hz tone at 40 dB SPL (40 phons). Subjective loudness doubles with every $+10\text{ phon}$ increase: $$S = 2^{(L_{\text{phon}} - 40)/10} \quad (\text{sones})$$

3. Timbre and Spectral Distribution

Timbre (tone quality) is the acoustic characteristic that enables a listener to distinguish between two musical instruments (e.g., a trumpet and an oboe) playing the exact same pitch at the exact same loudness. Timbre is determined by:
  • The relative amplitude and harmonic distribution of overtones (Fourier spectrum).
  • Temporal envelope attack, decay, sustain, and release (ADSR) transients.

§8.3 Spherical Waves in Three Dimensions and the Inverse-Square Law

In an isotropic 3D medium, a point acoustic source radiates sound uniformly in all spatial directions.

1. The 3D Wave Equation in Spherical Coordinates

The linear acoustic wave equation in three dimensions is: $$\nabla^2 p = \frac{1}{c^2} \frac{\partial^2 p}{\partial t^2}$$ For a spherically symmetric wave where pressure depends only on radial distance $r$ from the source: $$\nabla^2 p = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial p}{\partial r} \right) = \frac{1}{r} \frac{\partial^2 (r p)}{\partial r^2}$$ Substituting into the wave equation: $$\frac{\partial^2 (r p)}{\partial r^2} = \frac{1}{c^2} \frac{\partial^2 (r p)}{\partial t^2}$$ Defining the auxiliary variable $\psi(r, t) = r p(r, t)$, this reduces to the 1D classical wave equation! The general solution for outgoing expanding spherical waves is: $$p(r, t) = \frac{A}{r} \cos(k r - \omega t + \phi)$$ Critical Consequence: The pressure amplitude of a spherical wave decreases inversely with radial distance: $$p_0(r) \propto \frac{1}{r}$$

2. The Inverse-Square Law of Acoustic Intensity

Because acoustic intensity is proportional to pressure amplitude squared ($I \propto p_0^2$): $$I(r) = \frac{P_{\text{source}}}{4\pi r^2} \propto \frac{1}{r^2}$$ where $P_{\text{source}}$ is the total acoustic power output of an omnidirectional point source (Watts). Intensity obeys the Inverse-Square Law: $$\frac{I_2}{I_1} = \left(\frac{r_1}{r_2}\right)^2$$ On the decibel scale, doubling the distance from a point source reduces the sound level by exactly $6.02\text{ dB}$: $$\beta_2 - \beta_1 = 10 \log_{10}\left( \frac{I_2}{I_1} \right) = 10 \log_{10}\left( \frac{r_1}{r_2} \right)^2 = 20 \log_{10}\left( \frac{r_1}{r_2} \right) = 20 \log_{10}(0.5) = -6.02\text{ dB}$$

§8.4 Interference, Diffraction, and Radiation Efficiency of Sound Sources

Acoustic waves exhibit classical interference and diffraction phenomena governed by wave superposition and boundary conditions.

1. Interference of Coherent Sound Waves

When two coherent loudspeakers emit sound waves of wavelength $\lambda$, the resultant pressure amplitude at observation point $P$ separated from the sources by paths $r_1$ and $r_2$ is: $$\Delta r = |r_1 - r_2|$$
  • Constructive Interference (Maximum Loudness): $$\Delta r = n \lambda, \quad n = 0, 1, 2, \dots$$
  • Destructive Interference (Silence / Minimum): $$\Delta r = \left(n + \frac{1}{2}\right) \lambda$$
Quincke's Interference Tube: Sound enters a branch split into two paths of lengths $L_1$ and $L_2$. Sliding one tube by $\Delta L$ produces constructive or destructive interference at the listener's ear, allowing direct precision measurement of acoustic wavelength $\lambda = 2 \Delta L$.

2. Diffraction of Sound Waves

Sound waves diffract around obstacles and through doorways because their acoustic wavelengths ($\lambda \sim 0.1 - 3\text{ m}$) are comparable to everyday architectural dimensions ($D \sim 1\text{ m}$). By Airy's circular aperture diffraction formula: $$\sin\theta \approx 1.22 \frac{\lambda}{D}$$
  • Low-frequency bass notes (100 Hz, $\lambda = 3.4\text{ m}$): $\lambda \gg D$, sound bends around obstacles into acoustic shadow zones.
  • High-frequency treble notes (10 kHz, $\lambda = 3.4\text{ cm}$): $\lambda \ll D$, sound forms sharp directional beams and geometric acoustic shadows.

3. Radiation Efficiency of Acoustic Sources

The ability of a vibrating body to convert mechanical vibrational power into acoustic radiated power is quantified by its radiation efficiency: $$\eta_{rad} = \frac{R_{rad}}{\rho_0 c A}$$ where $R_{rad}$ is the real acoustic radiation resistance.
  • Monopole (Pulsating sphere): Net volume displacement changes. High acoustic radiation efficiency at low frequencies.
  • Dipole (Unbaffled vibrating loudspeaker cone): Two out-of-phase pulsating sources separated by small distance. Air simply sloshes back and forth between front and back without radiating efficiently. Installing a baffle board or enclosed cabinet prevents dipole cancellation, dramatically boosting bass output!

§8.5 Acoustic Beats and Combination Tones

When two sound waves of slightly different frequencies are sounded simultaneously, the human ear perceives acoustic beats and combination tones.

1. Mathematical Theory of Beats

Consider two acoustic pressure signals of equal amplitude $p_0$ and neighboring frequencies $f_1$ and $f_2$ ($f_1 \approx f_2$): $$p_1(t) = p_0 \cos(2\pi f_1 t), \quad p_2(t) = p_0 \cos(2\pi f_2 t)$$ By superposition: $$p(t) = p_1(t) + p_2(t) = 2 p_0 \cos\left[ 2\pi \left( \frac{f_1 - f_2}{2} \right) t \right] \cos\left[ 2\pi \left( \frac{f_1 + f_2}{2} \right) t \right]$$ The resultant wave represents a carrier vibration at the average frequency $\bar{f} = \frac{f_1 + f_2}{2}$ whose envelope amplitude is slowly modulated: $$A_{\text{mod}}(t) = 2 p_0 \left| \cos\left( \pi (f_1 - f_2) t \right) \right|$$ Since acoustic intensity is proportional to amplitude squared: $$I(t) \propto A_{\text{mod}}^2(t) = 4 p_0^2 \cos^2\left( \pi (f_1 - f_2) t \right) = 2 p_0^2 \left[ 1 + \cos\left( 2\pi (f_1 - f_2) t \right) \right]$$ The intensity surges from 0 to $4 p_0^2$ at a rate called the Beat Frequency ($f_{\text{beat}}$): $$f_{\text{beat}} = |f_1 - f_2|$$ Piano tuners adjust wire tension until beat frequency drops to zero, achieving unison tuning.

2. Tartini Combination Tones

In 1714, Italian violinist Giuseppe Tartini discovered that when two loud, pure tones of frequencies $f_1$ and $f_2$ ($f_2 > f_1$) are sounded together, the ear perceives additional tones that are not physically present in the acoustic sound field! These are subjective combination tones generated by the non-linear elasticity of the human eardrum and cochlea: $$x_{\text{cochlea}} = a_1 p + a_2 p^2 + a_3 p^3 + \dots$$ When $p = p_1 \cos(\omega_1 t) + p_2 \cos(\omega_2 t)$, the quadratic term $p^2$ yields: $$p^2 = \frac{1}{2}p_1^2 + \frac{1}{2}p_2^2 + \frac{1}{2}p_1^2 \cos(2\omega_1 t) + \frac{1}{2}p_2^2 \cos(2\omega_2 t) + p_1 p_2 [\cos((\omega_2 - \omega_1)t) + \cos((\omega_2 + \omega_1)t)]$$ This generates:
  • Difference Tone (Tartini Tone): $f_{\text{diff}} = f_2 - f_1$ (very prominent, used by organ builders to generate deep 32-foot bass notes using smaller 16-foot pipes).
  • Summation Tone: $f_{\text{sum}} = f_1 + f_2$ (fainter, higher in pitch).
  • Cubic Combination Tones: $2f_1 - f_2$ and $2f_2 - f_1$ arising from the cubic term $a_3 p^3$.

§8.6 The Doppler Effect and Supersonic Shock Waves

Christian Doppler (1842) demonstrated that the observed frequency of a wave depends on the relative motion between the wave source, the observer, and the propagating medium.

1. The Classical Acoustic Doppler Equation

Let:
  • $c$: Speed of sound in the still medium.
  • $v_s$: Velocity of the sound source along the line connecting source and observer (positive when moving toward observer).
  • $v_o$: Velocity of the observer along the connecting line (positive when moving toward source).
  • $v_w$: Velocity of wind/medium along the line of propagation.
  • $f_0$: Emitted source frequency.
The general observed frequency $f'$ is: $$f' = f_0 \left( \frac{c \pm v_o}{c \mp v_s} \right)$$ Sign conventions:
  • Observer moving toward stationary source ($v_o > 0, v_s = 0$): Observer intercepts more wavefronts per second: $$f' = f_0 \left( \frac{c + v_o}{c} \right) > f_0$$
  • Source moving toward stationary observer ($v_s > 0, v_o = 0$): Wavefronts are compressed ahead of the source ($\lambda' = \frac{c - v_s}{f_0}$): $$f' = f_0 \left( \frac{c}{c - v_s} \right) > f_0$$
  • Approaching systems: Frequency shifts higher (blueshift).
  • Receding systems: Frequency shifts lower (redshift).

2. Supersonic Motion and Mach Shock Waves

When the source speed $v_s$ equals the speed of sound $c$ ($M = v_s / c = 1$), wavefronts pile up ahead of the source into a singular pressure barrier. When the source travels faster than sound ($M > 1$, supersonic): The circular wavefronts emitted at successive positions lag behind the source, and their envelope forms a conical wavefront called the Mach Cone: The half-angle of the cone (Mach Angle $\theta$) is: $$\sin\theta = \frac{c t}{v_s t} = \frac{c}{v_s} = \frac{1}{M}$$ Across this conical discontinuity, pressure, density, and temperature jump discontinuously. When this conical shock front sweeps past an observer on the ground, the abrupt double pressure jump produces an explosive Sonic Boom.

3. Modern Technical Applications of the Doppler Effect

  • Medical Color Doppler Echocardiography: Measures blood flow velocity and detects heart valve regurgitation non-invasively via ultrasound reflected from erythrocytes.
  • Radar Speed Guns: Police microwave radar detects vehicle speed via $\Delta f = \frac{2 v}{c} f_0$.
  • Astronomical Redshifts: Hubble's discovery of cosmic expansion via Doppler redshift of spectral absorption lines in distant galaxies.
Standard University Exam Examination Problems

Rigorous Analytical & Numerical Solved Problems

Comprehensive step-by-step mathematical proofs, dimensional evaluations, and calculations matching B.Sc. Honors university examinations.

Undergraduate Classical Exam Standard Example 8.1: Decibel Sound Intensity Addition and Distance Attenuation

A small construction generator acts as an omnidirectional point acoustic source radiating sound power $P_{\text{sound}} = 0.500\text{ W}$ in an open field.\n(a) Determine the acoustic intensity $I$ and sound level $\beta$ at distance $r_1 = 5.00\text{ m}$,\n(b) Find the sound level $\beta_2$ at distance $r_2 = 25.0\text{ m}$, and\n(c) If four identical generators operate simultaneously at the original location, what is the combined sound level in decibels at $r_1 = 5.00\text{ m}$? Take $I_0 = 1.00 \times 10^{-12}\text{ W/m}^2$.

Step 1: Calculate intensity and sound level at 5.00 m
$$I_1 = \frac{P_{\text{sound}}}{4\pi r_1^2} = \frac{0.500}{4\pi \times (5.00)^2} = \frac{0.500}{100\pi} = \frac{0.500}{314.16} = 1.5915 \times 10^{-3} \text{ W/m}^2$$ $$\beta_1 = 10 \log_{10}\left( \frac{I_1}{I_0} \right) = 10 \log_{10}\left( \frac{1.5915 \times 10^{-3}}{1.00 \times 10^{-12}} \right) = 10 \log_{10}(1.5915 \times 10^9)$$ $$\beta_1 = 10 \times (9 + \log_{10} 1.5915) = 10 \times (9 + 0.2018) = 92.02 \text{ dB}$$

At 5.0 m, the generator produces a loud industrial level of 92.0 dB.

Step 2: Attenuation over distance to 25.0 m
$$\beta_2 = \beta_1 - 20 \log_{10}\left(\frac{r_2}{r_1}\right) = 92.02 - 20 \log_{10}\left(\frac{25.0}{5.00}\right) = 92.02 - 20 \log_{10}(5)$$ $$\beta_2 = 92.02 - 20 \times 0.69897 = 92.02 - 13.98 = 78.04 \text{ dB}$$

Increasing the distance fivefold reduces the sound level by 14.0 dB to 78.0 dB.

Step 3: Superposition of four identical incoherent sources
$$I_{\text{tot}} = 4 I_1$$ $$\beta_{\text{tot}} = \beta_1 + 10 \log_{10}(4) = 92.02 + 10 \times 0.60206 = 92.02 + 6.02 = 98.04 \text{ dB}$$

Quadrupling the acoustic power adds $+6.02$ dB, raising the level to 98.0 dB.

Precision Acoustic Calibration Problem Example 8.2: Acoustic Beats and Tuning Fork Calibration

A standard calibration tuning fork $A$ has a known frequency $f_A = 440.0\text{ Hz}$. When sounded simultaneously with an unknown fork $B$, $4.00\text{ beats per second}$ are heard. When a small piece of beeswax is attached to the prong of fork $B$ (which increases its effective inertia), the beat frequency decreases to $2.00\text{ beats per second}$.\n(a) Explain why attaching wax alters the frequency of fork $B$,\n(b) Determine the exact original frequency $f_B$ of fork $B$, and\n(c) What would happen to the beat frequency if even more wax were added until $f_B$ drops further by $4.00\text{ Hz}$?

Step 1: Frequency candidates from initial beat frequency
$$f_{\text{beat}} = |f_A - f_B| = 4.00 \text{ Hz}$$ $$f_B = f_A \pm 4.00 = 440.0 \pm 4.00 \implies f_B = 444.0 \text{ Hz} \quad \text{or} \quad 436.0 \text{ Hz}$$

Initial beat frequency yields two possible mathematical solutions: 444.0 Hz or 436.0 Hz.

Step 2: Effect of mass loading and unique identification
$$f_{\text{fork}} = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \implies \frac{df}{dm} < 0$$ $$\text{Adding wax increases mass, so } f_B' < f_B$$ $$\text{If } f_B = 436.0 \text{ Hz}: \text{ lowering it would yield } f_B' < 436.0 \implies |440.0 - f_B'| > 4.00 \text{ Hz (beat rate increases).}$$ $$\text{If } f_B = 444.0 \text{ Hz}: \text{ lowering it toward 440 Hz yields } |440.0 - f_B'| < 4.00 \text{ Hz (beat rate decreases).}$$ $$\text{Because the observed beat frequency dropped to } 2.00 \text{ Hz}, \text{ the true original frequency was:}$$ $$f_B = 444.0 \text{ Hz}$$

Because adding inertia lowered the beat count, $f_B$ must have been initially higher than 440 Hz.

Step 3: Further wax addition analysis
$$f_B'' = 442.0 - 4.00 = 438.0 \text{ Hz}$$ $$f_{\text{beat}}'' = |440.0 - 438.0| = 2.00 \text{ Hz}$$

Adding more wax lowers $f_B$ through unison (0 beats at 440 Hz) down to 438 Hz, where beats reappear at 2.0 Hz.

Honors Doppler Effect Exam Standard Example 8.3: Doppler Shift with Reflected Acoustic Echo and Beats

A train locomotive moves at constant speed $v_s = 20.0\text{ m/s}$ directly toward a sheer vertical rock cliff. The locomotive engineer sounds the train whistle at frequency $f_0 = 500.0\text{ Hz}$. The speed of sound in still air is $c = 340.0\text{ m/s}$.\n(a) What frequency $f_{\text{cliff}}$ is received by a stationary observer standing at the base of the cliff?\n(b) What frequency $f'_{\text{echo}}$ of the reflected echo is heard by the train engineer aboard the moving locomotive?\n(c) What beat frequency $f_{\text{beat}}$ does the engineer hear between the direct whistle and the reflected echo from the cliff?

Step 1: Frequency incident on the cliff
$$f_{\text{cliff}} = f_0 \left( \frac{c}{c - v_s} \right) = 500.0 \times \left( \frac{340.0}{340.0 - 20.0} \right) = 500.0 \times \frac{340.0}{320.0} = 500.0 \times 1.0625 = 531.25 \text{ Hz}$$

Wavefronts are compressed ahead of the moving locomotive, striking the cliff at 531.25 Hz.

Step 2: Frequency of reflected echo received by the engineer
$$\text{The cliff acts as a stationary source re-radiating sound at } f_{\text{cliff}} = 531.25 \text{ Hz}.$$ $$\text{The engineer is an observer moving toward this stationary source with speed } v_o = v_s = 20.0 \text{ m/s}:$$ $$f'_{\text{echo}} = f_{\text{cliff}} \left( \frac{c + v_s}{c} \right) = f_0 \left( \frac{c + v_s}{c - v_s} \right)$$ $$f'_{\text{echo}} = 500.0 \times \left( \frac{340.0 + 20.0}{340.0 - 20.0} \right) = 500.0 \times \frac{360.0}{320.0} = 500.0 \times 1.125 = 562.50 \text{ Hz}$$

Because the engineer is in motion both when emitting and when intercepting the echo, the Doppler factor applies twice.

Step 3: Beat frequency heard by the engineer
$$f_{\text{beat}} = f'_{\text{echo}} - f_0 = 562.50 - 500.0 = 62.50 \text{ Hz}$$

The direct whistle (500 Hz) and reflected echo (562.5 Hz) superpose to produce a distinct rapid beat frequency of 62.5 Hz.