Waves and Sound

Waves and sound carry about 2 questions (our estimate), from a short list: v=f\lambda, the speed of sound and what changes it, resonance in a closed or open pipe, the fundamental of a stretched string, beats, decibels, and the Doppler effect.

Each has a one-line formula; the skill is choosing the right one and keeping track of which end of a pipe is closed.

4.1 Wave motion: travelling and stationary waves

A wave transfers energy without transferring matter. Transverse (displacement perpendicular to propagation: string, light) and longitudinal (compressions and rarefactions: sound). v=f\lambda.

A progressive wave y=A\sin(\omega t-kx) with \omega=2\pi f, k=\frac{2\pi}{\lambda}, v=\frac{\omega}{k}; phase difference =\frac{2\pi}{\lambda}\times path difference. Intensity \propto A^2f^2.

Stationary waves form by superposing two identical waves moving oppositely: y=2A\cos kx\sin\omega t.

Nodes (no motion) are \frac{\lambda}{2} apart, as are the antinodes; every point between two nodes is in phase; no energy is transmitted; amplitude varies with position.

A fixed end is a node (phase change \pi on reflection); a free end is an antinode.

4.2 Mechanical waves: speed of sound

Newton–Laplace: v=\sqrt{\dfrac{E}{\rho}}: solid \sqrt{Y/\rho}, liquid \sqrt{B/\rho}, gas \sqrt{\gamma P/\rho}=\sqrt{\gamma RT/M} (Laplace, adiabatic; Newton's isothermal assumption gave 280 m s⁻¹).

Typical values: air 332 m s⁻¹ at 0 °C (\approx331+0.61t), water about 1480, steel about 5000 m s⁻¹.

• Temperature: v\propto\sqrt{T} (kelvin).
• Pressure: no effect at constant temperature (P/\rho is constant).
• Humidity: moist air is less dense, so v rises.
• Gas type: lighter gas, faster sound (v_{H_2}\approx4v_{O_2} at the same T).
• Wind adds its component along the direction.

4.3 Waves in pipes and strings

Closed pipe (one end closed): node at the closed end, antinode at the open end; f_n=\dfrac{(2n-1)v}{4L} — only odd harmonics (f_1:f_2:f_3=1:3:5). Open pipe: antinodes at both ends; f_n=\dfrac{nv}{2L} — all harmonics (1:2:3).

A closed pipe's fundamental is half that of an open pipe of the same length. End correction e=0.6r, so L+0.6r is used.

Resonance tube: first resonance at l_1+e=\frac{\lambda}{4}, second at l_2+e=\frac{3\lambda}{4}; hence \lambda=2(l_2-l_1) and v=2f(l_2-l_1) (end correction cancels).

Stretched string (length L, tension T, mass per length \mu): v=\sqrt{\dfrac{T}{\mu}}; f_n=\dfrac{n}{2L}\sqrt{\dfrac{T}{\mu}}.

Laws of vibration: f\propto\frac{1}{L}, f\propto\sqrt{T}, f\propto\frac{1}{\sqrt{\mu}} (or \frac{1}{r} for a given material). The sonometer verifies them; a tuning fork of known frequency resonates with the wire.

4.4 Acoustic phenomena

• Intensity I=\frac{P}{A} (W m⁻²); threshold of hearing I_0=10^{-12} W m⁻²; level \beta=10\log_{10}\frac{I}{I_0} dB (a hundred-fold intensity is +20 dB; doubling adds 3 dB). Pressure amplitude p_0=\sqrt{2\rho vI}. Loudness is subjective (phon, sone).
• Pitch is set by frequency, loudness by amplitude (intensity), quality (timbre) by the number and strength of overtones.
• Beats: two close frequencies give |f_1-f_2| beats per second; adding wax to a fork lowers its frequency; filing raises it.
• Ultrasonic >20 kHz (sonar, imaging, cleaning, bats); infrasonic <20 Hz (earthquakes, elephants). Audible range 20 Hz – 20 kHz.
• Doppler effect: f'=f\dfrac{v+v_o}{v-v_s} (v_o positive toward the source, v_s positive toward the observer). Approach raises the pitch; recession lowers it. For a 500 Hz source approaching at 34 m s⁻¹: 500\times\frac{340}{306}\approx556 Hz. If the medium moves, include its speed.

(a) Organ pipes of different lengths

(b) 0.85 m pipe, v = 340 m/s: allowed frequencies

Chapter summary

• v=f\lambda; stationary-wave nodes are \frac{\lambda}{2} apart.
• Sound speed depends on temperature, medium and humidity; not on pressure alone.
• Closed pipes give odd harmonics; the resonance-tube method gives v=2f(l_2-l_1).
• A string's frequency rises with tension and falls with length and mass density.
• Beats count the frequency difference; a ten-fold intensity change is 10 dB.

Practice questions

Set A — Recall (10 questions)

1. A source of 500 Hz produces waves of wavelength 0.68 m. The speed of the wave isa170 m s⁻¹b680 m s⁻¹c340 m s⁻¹d1000 m s⁻¹
2. If the absolute temperature of a gas is quadrupled, the speed of sound in itahalvesbdoublescquadruplesdis unchanged
3. The fundamental frequency of an open pipe 0.5 m long is (v=340 m s⁻¹)a340 Hzb170 Hzc680 Hzd85 Hz
4. A pipe closed at one end can produceano resonanceball harmonicsconly even harmonicsdonly odd harmonics
5. A sound of intensity 10^{-6} W m⁻² has an intensity level ofa6 dBb120 dBc60 dBd600 dB
6. When a source of sound approaches a stationary observer, the observed frequencyaincreasesbdecreasescis unchangeddis zero
7. Ultrasonic waves have frequenciesabelow 20 Hzbabove 20 kHzcfrom 20 Hz to 20 kHzdof exactly zero
8. In a stationary wave of wavelength 1 m, the distance between consecutive nodes isa1 mb0.25 mc2 md0.5 m
9. If the tension in a stretched string is made four times, its fundamental frequencyabecomes four timesbhalvescdoublesdis unchanged
10. Two forks of 256 Hz and 260 Hz sounded together produce beats per second equal toa4b258c516d2

Answers and explanations — Set A

1. (c) — 500\times0.68=340.

2. (b) — v\propto\sqrt{T}, \sqrt{4}=2.

3. (a) — \frac{v}{2L}=\frac{340}{1}.

4. (d) — closed pipe: f=\frac{(2n-1)v}{4L}.

5. (c) — 10\log\frac{10^{-6}}{10^{-12}}=60.

6. (a) — approach compresses the waves.

7. (b) — above the audible limit.

8. (d) — \frac{\lambda}{2}.

9. (c) — f\propto\sqrt{T}.

10. (a) — 260-256.

Set B — Understanding (6 questions)

1. The speed of sound in air at 27 °C, taking v_0=331 m s⁻¹ at 0 °C and the rule v=331+0.61\,t, isa347 m s⁻¹b331 m s⁻¹c360 m s⁻¹d300 m s⁻¹
2. A pipe closed at one end is 0.85 m long (v=340 m s⁻¹). Its first overtone isa100 Hzb200 Hzc300 Hzd400 Hz
3. In a resonance-tube experiment with a 500 Hz fork, resonance occurs at air-column lengths 17 cm and 51 cm. The speed of sound isa340 m s⁻¹b170 m s⁻¹c680 m s⁻¹d510 m s⁻¹
4. A 1 m string under 100 N tension has linear density 0.01 kg m⁻¹. Its fundamental frequency isa25 Hzb50 Hzc100 Hzd200 Hz
5. A 500 Hz source approaches a stationary observer at 34 m s⁻¹ (v=340). The observed frequency is nearlya450 Hzb500 Hzc556 Hzd600 Hz
6. If the intensity of a sound is increased 100-fold, the level rises bya2 dBb10 dBc100 dBd20 dB

Answers and explanations — Set B

1. (a) — 331+0.61\times27=347.5.

2. (c) — f_1=\frac{340}{4\times0.85}=100 Hz; the first overtone is the third harmonic, 300 Hz.

3. (a) — v=2f(l_2-l_1)=2\times500\times0.34.

4. (b) — v=\sqrt{100/0.01}=100; f_1=\frac{100}{2\times1}.

5. (c) — 500\times\frac{340}{340-34}=555.6.

6. (d) — 10\log100=20.

Set C — Application (4 questions)

1. A fork of 256 Hz and an unknown fork give 4 beats per second. Loading the unknown fork with wax reduces the beats to 2 per second. The unknown frequency isa252 Hzb260 Hzc258 Hzd264 Hz
2. An open pipe has fundamental frequency 300 Hz. If one end is closed, its fundamental becomesa150 Hzb300 Hzc600 Hzd75 Hz
3. A whistle of 400 Hz on a train moving at 40 m s⁻¹ (v=340) is heard by a stationary person ahead of the train at nearlya400 Hzb360 Hzc467 Hzd453 Hz
4. At a point each of 10 identical sources produces a 50 dB level. The combined level isa500 dBb51 dBc60 dBd55 dB

Answers and explanations — Set C

1. (b) — loading lowers the unknown frequency, and the beats fell from 4 to 2, so the unknown was above 256: it was 260 Hz, loaded down to 258 Hz.

2. (a) — open: \frac{v}{2L}; closed: \frac{v}{4L}, half.

3. (d) — 400\times\frac{340}{300}=453 Hz.

4. (c) — intensities add: 10 times, +10 dB.

Group D — Model answers for long questions (10 marks each, for deeper revision)

1. Explain stationary waves in a stretched string, an open pipe and a closed pipe, with the harmonics and numericals. [10]
2. Derive the speed of sound in a gas, and explain beats, the Doppler effect and the intensity level of sound. [10]

(a) Organ pipe

(b) Chladni figure

(c) Guitar

(d) Oscilloscope

Model answer D1: Stationary waves in strings and pipes (10 marks)

Introduction. A stationary (standing) wave forms when two identical waves travel in opposite directions. It has fixed nodes (zero displacement) and antinodes (maximum displacement) a half wavelength \frac{\lambda}{2} apart, and it carries no net energy.

Stretched string (fixed at both ends). The speed is v = \sqrt{T/\mu} and the allowed wavelengths are \lambda_n = \frac{2L}{n}, so

All harmonics are present. Numerical: L = 0.6\ \text{m}, T = 100\ \text{N}, \mu = 0.01\ \text{kg m}^{-1}: v = 100\ \text{m s}^{-1}, f_1 = 100/1.2 = 83.3\ \text{Hz}, f_2 = 166.7\ \text{Hz}, f_3 = 250\ \text{Hz}.

Laws: f \propto 1/L, f \propto \sqrt{T}, f \propto 1/\sqrt{\mu}.

Open pipe (both ends open, antinodes at both ends). f_n = \frac{nv}{2L}. All harmonics are present. For L = 0.5\ \text{m} and v = 340\ \text{m s}^{-1}: f_1 = 340\ \text{Hz}, then 680 and 1020\ \text{Hz}.

Closed pipe (one end closed: node at the closed end, antinode at the open end). f_n = \frac{(2n-1)v}{4L}. Only odd harmonics are present. For L = 0.5\ \text{m}: f_1 = 170\ \text{Hz}, and the next are 510\ \text{Hz} and 850\ \text{Hz} (ratio 1:3:5).

End correction. The antinode is slightly outside the open end, so the effective length is L + e, with e \approx 0.6r. In the resonance-tube experiment, v = 2(l_2 - l_1)f eliminates e.

Source	Harmonics	Fundamental
String, both ends fixed	All	v/2L
Open pipe	All	v/2L
Closed pipe	Odd only	v/4L

Conclusion. The boundary conditions fix the wavelengths, so the tone of an instrument depends on its length and on which harmonics are present.

Model answer D2: Speed of sound, beats, Doppler effect and intensity (10 marks)

Introduction. Sound is a longitudinal mechanical wave that needs a medium.

Speed of sound. Newton assumed isothermal changes and got v = \sqrt{P/\rho} (about 280\ \text{m s}^{-1}), which is wrong. Laplace said that compressions are so fast that they are adiabatic, so

For air, \gamma = 1.4, P = 1.013\times10^5\ \text{Pa}, \rho = 1.29\ \text{kg m}^{-3}: v = 331.6\ \text{m s}^{-1} at 0^\circ\text{C}.

The speed increases about 0.61\ \text{m s}^{-1} per ^\circ\text{C}, so it is about 343\ \text{m s}^{-1} at 20^\circ\text{C}. It does not depend on pressure at constant temperature, and it increases with humidity.

Beats. Two sounds of nearby frequencies f_1 and f_2 give periodic loud-soft variations. The beat frequency is |f_1 - f_2|; for 256\ \text{Hz} and 260\ \text{Hz} it is 4 beats per second. Musicians use beats to tune instruments.

Doppler effect. The apparent frequency changes when the source or observer moves: f' = f\,\frac{v \pm v_o}{v \mp v_s}.

Numerical: a 340\ \text{Hz} siren moving at 20\ \text{m s}^{-1} towards a still observer (v = 340): f' = 340\times\frac{340}{320} = 361\ \text{Hz}, and when moving away f' = 340\times\frac{340}{360} = 321\ \text{Hz}.

It is used in radar, ultrasound blood-flow scans and speed guns.

Intensity level. The loudness is measured by \beta = 10\log_{10}\frac{I}{I_0} decibels, with I_0 = 10^{-12}\ \text{W m}^{-2}. For I = 10^{-6} the level is 60\ \text{dB}, and two such sources together (2\times10^{-6}) give 63\ \text{dB}, not 120\ \text{dB}.

Effect	Cause	Formula
Beats	Two close frequencies	\lvert f_1 - f_2\rvert
Doppler	Relative motion	f' = f\,\frac{v \pm v_o}{v \mp v_s}
Sound level	Logarithmic scale	10\log(I/I_0)

Conclusion. The adiabatic correction of Laplace gives the correct speed, and beats, the Doppler effect and the decibel scale describe how we hear sound.