Waves and Sound

Chapter 4 · Physics · 2 questions in the paper · 7 min read

Waves and Sound

What the syllabus asks for 4
  • Wave motion: travelling and stationary waves
  • Mechanical waves: velocity of sound in solid, liquid and gas; effect of temperature, pressure, humidity
  • Waves in pipes and strings: closed and open pipes, resonance tube, laws of vibration of a string
  • Acoustic phenomena: pressure amplitude, intensity level, quality and pitch, ultrasonic and infrasonic, Doppler effect

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.

What the paper asks from this chapter
  • v=f\lambda; wave equation y=A\sin(\omega t-kx), k=\frac{2\pi}{\lambda}; nodes and antinodes.
  • Speed of sound: v=\sqrt{\gamma P/\rho}, \propto\sqrt{T}, independent of pressure at fixed T, increases with humidity.
  • Pipes: closed f=\frac{(2n-1)v}{4L} (odd harmonics), open f=\frac{nv}{2L} (all harmonics); end correction 0.6r.
  • String: v=\sqrt{T/\mu}, f_1=\frac{1}{2L}\sqrt{T/\mu} and the laws of vibration.
  • Intensity level \beta=10\log\frac{I}{I_0} dB; beats |f_1-f_2|; Doppler f'=f\frac{v\pm v_o}{v\mp v_s}.

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.
Wave traps

The wavelength of the closed-pipe fundamental is 4L (not 2L). A change in pressure at constant temperature does not change the speed of sound. Light has no Doppler dependence on a medium but sound does.

Formula and fact sheet

Topic Result
wave speed v=f\lambda=\omega/k
gas v=\sqrt{\gamma P/\rho}\propto\sqrt{T}
closed / open pipe \frac{(2n-1)v}{4L} / \frac{nv}{2L}
string \frac{1}{2L}\sqrt{T/\mu}
level 10\log(I/I_0) dB
beats |f_1-f_2|
Doppler f\frac{v+v_o}{v-v_s}

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.