04Sound Waves
Analyze sound as a longitudinal wave: calculate intensity and decibels, apply the Doppler effect, and analyze resonance in strings and air columns.
Sound is a mechanical wave that shapes communication, music, and medical imaging. The Doppler effect and resonance appear in everything from sirens to MRI machines.
How does sound travel through matter, and what determines whether a sound is loud, high-pitched, or capable of shattering glass?
Lesson Overview
Sound is a longitudinal mechanical wave — particles of the medium oscillate parallel to the direction of wave travel, creating alternating regions of compression and rarefaction. Because sound requires a medium, it cannot travel through a vacuum. In air at 20 °C the speed of sound is approximately v ≈ 343 m/s; more generally, v = √(B/ρ), where B is the bulk modulus and ρ is the density of the medium. The intensity of a sound wave (I = P/A, measured in W/m²) describes the power delivered per unit area, while the decibel scale (β = 10 log(I/I₀), with I₀ = 10⁻¹² W/m²) compresses this enormous range into a convenient logarithmic measure. Humans hear frequencies from roughly 20 Hz to 20 kHz; pitch corresponds to frequency and loudness to intensity. When a sound wave reflects inside a pipe or string, resonance produces standing waves at specific harmonic frequencies. Two slightly different frequencies produce beats at a rate equal to their frequency difference. The Doppler effect shifts the perceived frequency whenever the source or observer is in motion. At speeds exceeding the speed of sound, a shock wave (sonic boom) is produced.
Pipe Resonance
Open Pipe (both ends open)
Both ends are antinodes (pressure nodes). All harmonics are present.
fn = n·v / 2L n = 1, 2, 3, …
Fundamental (n=1): f₁ = v/2L
2nd harmonic (n=2): f₂ = v/L
3rd harmonic (n=3): f₃ = 3v/2L
Diagram: ∿∿∿ — half-wavelength fits in L for each harmonic.
Closed Pipe (one end closed)
Closed end is a node; open end is an antinode. Only odd harmonics are present.
fn = (2n−1)·v / 4L n = 1, 2, 3, …
Fundamental (n=1): f₁ = v/4L
3rd harmonic (n=2): f₃ = 3v/4L
5th harmonic (n=3): f₅ = 5v/4L
Diagram: ∿ — quarter-wavelength fits in L for the fundamental.
Decibel Reference Levels
| Level (dB) | Source | Intensity (W/m²) |
|---|---|---|
| 0 dB | Threshold of hearing | 10⁻¹² W/m² |
| 30 dB | Quiet whisper | 10⁻⁹ W/m² |
| 60 dB | Normal conversation | 10⁻⁶ W/m² |
| 90 dB | Lawnmower | 10⁻³ W/m² |
| 120 dB | Rock concert | 1 W/m² |
| 140 dB | Jet engine (nearby) | 100 W/m² |
Key Equations
Speed of sound (air, 20 °C)
v = 343 m/s
Decibel level
β = 10 log(I / I₀)
Intensity
I = P / A
Open pipe harmonics
fₙ = n·v / 2L
Closed pipe harmonics
fₙ = (2n−1)·v / 4L
Beat frequency
f_beat = |f₁ − f₂|
Worked Examples
A sound wave has an intensity of I = 10⁻⁶ W/m². Calculate the sound level in decibels. (I₀ = 10⁻¹² W/m²)
Use the decibel formula: β = 10 log(I / I₀)
Substitute: β = 10 log(10⁻⁶ / 10⁻¹²)
Simplify the ratio: 10⁻⁶ / 10⁻¹² = 10⁶
β = 10 log(10⁶) = 10 × 6 = 60 dB
An open pipe (both ends open) has length L = 0.85 m. With v = 340 m/s, find the fundamental frequency and the next two harmonics.
Formula for open pipe: fₙ = n·v / 2L
Fundamental (n=1): f₁ = (1 × 340) / (2 × 0.85) = 340 / 1.70 = 200 Hz
2nd harmonic (n=2): f₂ = (2 × 340) / 1.70 = 680 / 1.70 = 400 Hz
3rd harmonic (n=3): f₃ = (3 × 340) / 1.70 = 1020 / 1.70 = 600 Hz
A closed pipe (one end closed) has length L = 0.425 m. With v = 340 m/s, find the fundamental frequency and the next two resonant frequencies.
Formula for closed pipe: fₙ = (2n−1)·v / 4L
Fundamental (n=1): f₁ = (1 × 340) / (4 × 0.425) = 340 / 1.70 = 200 Hz
Next resonance (n=2): f₃ = (3 × 340) / 1.70 = 1020 / 1.70 = 600 Hz
Next resonance (n=3): f₅ = (5 × 340) / 1.70 = 1700 / 1.70 = 1000 Hz
Note: only odd harmonics (1st, 3rd, 5th, …) are present in a closed pipe.
Two tuning forks vibrate at f₁ = 440 Hz and f₂ = 444 Hz simultaneously. How many beats per second are heard?
Beat frequency formula: f_beat = |f₁ − f₂|
f_beat = |440 − 444| = |−4| = 4 Hz
This means the listener hears 4 amplitude pulses (beats) every second.
A car horn emits f = 500 Hz. The car moves at v_s = 25 m/s toward a wall. The speed of sound is v = 343 m/s. What reflected frequency does the driver hear?
Step 1 — Frequency hitting the wall (wall is stationary observer, source approaches):
f_wall = f × v / (v − v_s) = 500 × 343 / (343 − 25) = 500 × 343 / 318
f_wall ≈ 500 × 1.0786 ≈ 539.3 Hz
Step 2 — The wall reflects this frequency back. Now the wall acts as a stationary source and the driver (observer) moves toward it at v_o = 25 m/s:
f' = f_wall × (v + v_o) / v = 539.3 × (343 + 25) / 343
f' = 539.3 × 368 / 343 ≈ 539.3 × 1.0729 ≈ 578.6 Hz
Alternatively, the combined Doppler formula gives: f′ = 500 × (343 + 25)/(343 − 25) = 500 × 368/318 ≈ 578.6 Hz
Guided Problems
A loudspeaker delivers 0.02 W of acoustic power uniformly in all directions. What is the intensity at a distance of 2.0 m from the speaker?
Hint: Sound spreads over a sphere: A = 4πr². Substitute into I = P/A.
The sound level at a rock concert is 120 dB. What is the intensity in W/m²? (I₀ = 10⁻¹² W/m²)
Hint: Rearrange the decibel formula: I = I₀ × 10^(β/10). Substitute β = 120.
An open organ pipe resonates at a fundamental frequency of 256 Hz. If the speed of sound is 340 m/s, what is the length of the pipe?
Hint: Use f₁ = v/2L and solve for L: L = v/(2f₁).
A police siren emits 800 Hz. A stationary observer hears 850 Hz as the police car approaches. What is the speed of the police car? (v = 343 m/s)
Hint: Use the Doppler formula for a moving source and stationary observer: f′ = f·v/(v − v_s). Solve for v_s.
A guitar string and a tuning fork (440 Hz) are played together. The musician hears 3 beats per second. After tightening the string slightly, the beat rate increases to 5 beats per second. What was the original string frequency?
Hint: f_beat = |f_string − f_fork|. Tightening raises frequency — if beats increase, the string was below 440 Hz. So f_string = 440 − 3 = 437 Hz.
Vocabulary
Longitudinal wave
A wave in which particles oscillate parallel to the direction of wave propagation, creating compressions and rarefactions.
Example: Sound waves in air; a slinky pushed end-to-end.
Intensity
The power of a wave transmitted per unit area perpendicular to the direction of propagation. I = P/A, measured in W/m².
Example: A 1 W speaker at 1 m distance: I = 1/(4π·1²) ≈ 0.08 W/m².
Decibel (dB)
A logarithmic unit for sound level: β = 10 log(I/I₀), where I₀ = 10⁻¹² W/m² is the threshold of human hearing.
Example: 60 dB conversation; 120 dB concert.
Resonance
The reinforcement of vibration when a system is driven at one of its natural frequencies, producing standing waves of large amplitude.
Example: Blowing across a bottle top; organ pipes.
Harmonic
A frequency that is an integer multiple of the fundamental frequency. Open pipes support all harmonics; closed pipes support only odd harmonics.
Example: Fundamental 200 Hz → harmonics at 400, 600, 800 Hz (open pipe).
Beats
Periodic variations in amplitude caused by the superposition of two waves with slightly different frequencies. f_beat = |f₁ − f₂|.
Example: 440 Hz and 444 Hz together produce 4 beats per second.
Doppler effect
The change in observed frequency of a wave due to relative motion between the source and the observer.
Example: A siren sounds higher as an ambulance approaches and lower as it recedes.
Sonic boom
The explosive sound produced when an object travels faster than the speed of sound, creating a cone-shaped shock wave of constructively interfering wavefronts.
Example: A supersonic jet aircraft breaking the sound barrier.
Workbook Check
Interactive Practice — 5 Questions
Sound is classified as which type of wave?
A sound has intensity I = 10⁻⁸ W/m². What is its decibel level? (I₀ = 10⁻¹² W/m²)
A closed pipe (one end closed) of length L = 0.25 m resonates with v = 340 m/s. What is its fundamental frequency?
Two tuning forks produce 6 beats per second. One fork vibrates at 512 Hz. Which could be the frequency of the other fork?
As a sound source moves toward a stationary observer, the observer perceives the frequency as:
Independent Practice
A sound source emits 0.50 W of power uniformly in all directions. Calculate the intensity at a distance of 5.0 m.
The intensity of a sound is 1.0 × 10⁻⁵ W/m². Calculate the sound level in decibels. (I₀ = 1.0 × 10⁻¹² W/m²)
An ambulance siren emits a frequency of 800 Hz and moves toward a stationary observer at 30 m/s. Find the observed frequency. (v_sound = 343 m/s)
An open pipe resonates at its fundamental frequency of 170 Hz. Find the length of the pipe. (v_sound = 340 m/s)
★ A train moving at 40 m/s blows a horn at 500 Hz. (a) Find the frequency heard by a stationary observer as the train approaches. (b) Find the frequency as the train recedes. (c) Find the beat frequency heard by the observer as the train passes. (d) If the observer is also moving toward the train at 10 m/s (before it passes), find the observed frequency. (v_sound = 340 m/s)
ChallengeCommon Mistakes
Using the Doppler formula with the wrong sign — getting a frequency decrease when the source approaches
Doppler: f' = f(v ± v_o)/(v ∓ v_s). Use + in numerator when observer moves toward source; use − in denominator when source moves toward observer. Approaching → higher frequency
Confusing sound intensity (W/m²) with sound intensity level (dB)
Intensity I is power per area (W/m²). Intensity level β = 10 log(I/I₀) in decibels. A 10× increase in I adds only 10 dB
Thinking beats occur only when two sounds have very different frequencies
Beats occur when two sounds have SLIGHTLY different frequencies. Beat frequency f_beat = |f₁ − f₂|. Large differences produce a harsh dissonance, not audible beats
Forgetting that the speed of sound changes with temperature and medium
In air: v ≈ 331 + 0.6T m/s (T in °C). Sound travels faster in liquids and solids than in gases
Math Tips
Decibel addition: doubling the number of identical sources adds 3 dB (not double the dB). 10 identical sources add 10 dB. The dB scale is logarithmic
Open pipe resonance: f_n = nv/2L (all harmonics). Closed pipe (one end closed): f_n = nv/4L for odd n only (1st, 3rd, 5th harmonics)
Doppler memory trick: when source and observer approach each other, frequency goes UP (shorter wavelength). When they recede, frequency goes DOWN. Think of a passing ambulance
Intensity falls off as 1/r²: I₂/I₁ = r₁²/r₂². Doubling the distance reduces intensity by a factor of 4 (−6 dB)