Sound Waves

Lessons

Senior 6 · Physics

Sound Waves

5 lessons

Lesson 2 of 5

The Speed of Sound and the Behaviour of Sound Waves

25 min

Counting the seconds after lightning

During a thunderstorm the flash and the thunder are made at the same moment, but the thunder arrives later. Light takes a few microseconds to cross a kilometre; sound takes about three seconds. Count the seconds from flash to rumble and divide by three, and you have the distance to the lightning in kilometres. That rule of thumb is a measurement of the speed of sound, and this lesson asks what sets that speed and how sound behaves when it meets walls, warm air, doorways and other sound waves.

What sets the speed of sound?

A sound wave travels by one layer of the medium pushing the next. Two properties decide how fast: how strongly the medium resists being squeezed (its elasticity, measured by a modulus) and how much mass must be set moving (its density $\rho$ ). A stiffer medium makes sound faster; a denser one makes it slower.

Medium	Speed of longitudinal waves	Modulus used
thin solid rod	$v = \sqrt{\frac{Y}{\rho}}$	Young modulus $Y$
liquid or gas	$v = \sqrt{\frac{B}{\rho}}$	bulk modulus $B = -\frac{\Delta p}{\Delta V / V}$
ideal gas	$v = \sqrt{\frac{\gamma p}{\rho}} = \sqrt{\frac{\gamma R T}{M}}$	adiabatic bulk modulus $B = \gamma p$

Newton's value and Laplace's correction

Newton assumed that the compressions and rarefactions of a sound wave happen at constant temperature. For an isothermal change $pV = \text{constant}$ , which gives $B = p$ and $v = \sqrt{\frac{p}{\rho}}$ . His answer was about 15% too low. Laplace pointed out that the oscillations are far too fast for heat to flow between a compression and the neighbouring rarefaction: the changes are adiabatic. For an adiabatic change $pV^{\gamma} = \text{constant}$ . Differentiating, $V^{\gamma}\,dp + \gamma p V^{\gamma - 1}\,dV = 0$ , so

B = -V\frac{dp}{dV} = \gamma p \qquad \Rightarrow \qquad v = \sqrt{\frac{\gamma p}{\rho}}

For an ideal gas $p V = n R T$ and $\rho = \frac{nM}{V}$ , so $\frac{p}{\rho} = \frac{RT}{M}$ and

v = \sqrt{\frac{\gamma R T}{M}}

where $T$ is the absolute temperature in kelvin, $M$ the molar mass in $\mathrm{kg\,mol^{-1}}$ , $R = 8.31\,\mathrm{J\,mol^{-1}\,K^{-1}}$ and $\gamma$ the ratio of principal heat capacities (1.40 for air).

Definition

The bulk modulus $B$ of a fluid is the ratio of the pressure change to the fractional change in volume it causes: $B = -\frac{\Delta p}{\Delta V / V}$ , unit $\mathrm{Pa}$ . The minus sign makes $B$ positive, since a rise in pressure reduces the volume. For air, $\gamma = \frac{C_p}{C_V} = 1.40$ .

Important

What the gas formula tells you. (1) At a fixed temperature the speed does NOT depend on pressure: $\frac{p}{\rho}$ stays constant when you compress the gas. (2) $v \propto \sqrt{T}$ , so $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$ . (3) A lighter gas carries sound faster: in helium ( $M = 0.004\,\mathrm{kg\,mol^{-1}}$ ) sound is almost three times faster than in air. (4) Moist air is slightly less dense than dry air at the same temperature and pressure, so sound is slightly faster on a humid day.

Medium	Speed of sound ( $\mathrm{m\,s^{-1}}$ )
air at 0 °C	331
air at 20 °C	343
carbon dioxide at 0 °C	259
helium at 0 °C	965
fresh water at 20 °C	1480
sea water at 20 °C	1540
human soft tissue	1540
lead	1960
glass (Pyrex)	5640
steel	5960

Speeds from OpenStax College Physics, Table 17.2.1 (liquids at 20 °C; solids for bulk longitudinal waves).

Worked example 1: Newton against Laplace

At 0 °C and $p = 1.013 \times 10^{5}\,\mathrm{Pa}$ the density of air is $1.293\,\mathrm{kg\,m^{-3}}$ . Find the speed of sound predicted by each model.

v_{\text{Newton}} = \sqrt{\frac{p}{\rho}} = \sqrt{\frac{1.013 \times 10^{5}}{1.293}} = 280\,\mathrm{m\,s^{-1}}

v_{\text{Laplace}} = \sqrt{\frac{\gamma p}{\rho}} = \sqrt{\frac{1.40 \times 1.013 \times 10^{5}}{1.293}} = 331\,\mathrm{m\,s^{-1}}

Only Laplace's value agrees with the measured 331 m s⁻¹.

Worked example 2: a warm afternoon

The speed of sound in air is $331\,\mathrm{m\,s^{-1}}$ at 0 °C. Find it on an afternoon at 25 °C, and find the temperature at which it would be $350\,\mathrm{m\,s^{-1}}$ .

v_{25} = 331\sqrt{\frac{298}{273}} = 346\,\mathrm{m\,s^{-1}}

T = 273 \left(\frac{350}{331}\right)^{2} = 305\,\mathrm{K} \;(32\,^{\circ}\mathrm{C})

Always convert to kelvin: using $\sqrt{\frac{25}{0}}$ is meaningless.

Experiment 2.1: the echo method

Apparatus: two wooden blocks, a stopwatch, a measuring tape, and a tall, flat wall with open ground in front of it (a school building facing an empty playing field).

Figure 2.1: Measuring the speed of sound from echoes.

Measure a distance $d$ of about 80 m from the wall.
Student A beats the blocks together and adjusts the rhythm until each beat coincides with the echo of the one before.
Student B starts the stopwatch on a beat counted "zero" and stops it on beat 20.
Between two beats the sound travels to the wall and back, $2 d$ , so $v = \frac{2d}{t}$ , where $t$ is the time for ONE interval. Repeat, and average.

Warning

Safety. Choose a field away from roads and moving vehicles, and keep everyone well clear of the person clapping so the blocks are never close to anyone's ears or fingers. Do the experiment in dry, calm weather; stop at once if a thunderstorm approaches.

Worked example 3: analysing the echo results

With $d = 80.0\,\mathrm{m}$ , 20 intervals took 9.4 s. Find the speed of sound.

t = \frac{9.4}{20} = 0.47\,\mathrm{s}, \qquad v = \frac{2d}{t} = \frac{2 \times 80.0}{0.47} = 340\,\mathrm{m\,s^{-1}}

Timing 20 intervals instead of one divides the reaction-time error by 20.

Reflection: echoes and echo sounding

Sound reflects from any surface large compared with its wavelength, obeying the law of reflection. The ear separates two sounds only if they arrive at least about 0.1 s apart, so a distinct echo needs the reflector to be at least $\frac{343 \times 0.1}{2} \approx 17\,\mathrm{m}$ away. Nearer reflections merge with the original sound and make it linger: reverberation. Halls reduce it with curtains, soft seats and rough wall panels that absorb sound. Ships measure water depth by echo sounding: depth $= \frac{vt}{2}$ , where $t$ is the time for the pulse to go down and back.

Refraction: why sound carries at night

Sound changes speed when the temperature of the air changes, so its wavefronts bend, always towards the region where the speed is lower.

Figure 2.2: By day, warm air near the ground bends sound upwards. On a clear night the ground cools the air next to it, the rays bend back down, and a radio or a drum on the next hill sounds surprisingly close.

Diffraction: hearing round corners

Waves spread out after passing an obstacle or a gap, most strongly when the gap is similar in size to the wavelength. Speech (wavelengths of roughly 0.3 m to 3 m) spreads through a doorway about 1 m wide, so you hear a person in the next room without seeing them. Light, with wavelengths below a micrometre, does not spread noticeably, so the doorway casts a sharp shadow. Low notes (long wavelengths) diffract more than high notes, which is why, from music playing round a corner, the bass reaches you more clearly than the treble (both travel at the same speed).

Interference: loud and quiet spots

Two loudspeakers driven in phase by the same signal are coherent sources. At a point P the two waves add. If the path difference is a whole number of wavelengths they arrive in phase (constructive interference, loud); if it is an odd number of half-wavelengths they arrive in antiphase (destructive interference, quiet):

\text{loud: } \Delta = n\lambda, \qquad \text{quiet: } \Delta = \left(n + \tfrac{1}{2}\right)\lambda, \qquad n = 0, 1, 2, \ldots

Figure 2.3: Two coherent loudspeakers. Walking along a line in front of them, you hear the sound rise and fall.

Worked example 4: a quiet spot

In Figure 2.3 the generator is set to 686 Hz and $v = 343\,\mathrm{m\,s^{-1}}$ . Is P a loud or a quiet point? What is the lowest frequency that would make P loud?

\lambda = \frac{343}{686} = 0.500\,\mathrm{m}, \qquad \Delta = 4.25 - 4.00 = 0.25\,\mathrm{m} = 0.50\lambda

The path difference is half a wavelength, so the waves arrive in antiphase: P is a quiet point. P is loud when $\Delta = n\lambda$ ; the lowest non-zero frequency has $\lambda = 0.25\,\mathrm{m}$ , so $f = \frac{343}{0.25} = 1372\,\mathrm{Hz} \approx 1.37\,\mathrm{kHz}$ .

Warning

Common mistakes. (1) Using degrees Celsius in $v \propto \sqrt{T}$ . (2) Forgetting the factor 2 in echo problems: the sound goes there AND back. (3) Thinking a higher pressure makes sound faster in a gas: at constant temperature it does not. (4) Saying sound bends towards the faster region: it bends towards the SLOWER one.

Tip

In a gas, speed depends on temperature and molar mass only: $v = \sqrt{\frac{\gamma R T}{M}}$ . "Hot and light is fast."

Interference demo with two speakers (Physics Demos). Two speakers play the same 2000 Hz tone; watch the sound level rise and fall as the microphone moves through the sound field, and relate each loud and quiet spot to the path difference.

✦

Practice

Explain why sound travels faster in steel than in air even though steel is about 6000 times denser.
Thunder is heard 4.5 s after a flash of lightning. The air is at 20 °C. How far away was the lightning?
The speed of sound is 331 m s⁻¹ at 0 °C. Find it at 35 °C.
Calculate the speed of sound in helium at 0 °C ( $\gamma = 1.67$ , $M = 4.0 \times 10^{-3}\,\mathrm{kg\,mol^{-1}}$ ).
An echo sounder on a boat on Lake Kivu receives the echo from the lake bed 0.25 s after sending a pulse. Taking $v = 1480\,\mathrm{m\,s^{-1}}$ in fresh water, find the depth. (Illustrative data.)
Two loudspeakers in phase emit 850 Hz sound ( $v = 340\,\mathrm{m\,s^{-1}}$ ). A listener is 3.0 m from one speaker and 3.6 m from the other. Is the sound loud or quiet there? Show your reasoning.
A student says: "Sound carries further at night because the air is quieter." Evaluate this explanation and give a better one.

Show answersHide answers

Speed depends on $\sqrt{\frac{\text{modulus}}{\rho}}$ . Steel's elastic modulus is enormously greater than that of air (by far more than its density), so the ratio, and the speed, is much larger.
$d = vt = 343 \times 4.5 = 1.5 \times 10^{3}\,\mathrm{m}$ , about 1.5 km.
$v = 331\sqrt{\frac{308}{273}} = 352\,\mathrm{m\,s^{-1}}$
$v = \sqrt{\frac{1.67 \times 8.31 \times 273}{4.0 \times 10^{-3}}} = 9.7 \times 10^{2}\,\mathrm{m\,s^{-1}}$ , close to the 965 m s⁻¹ in the table.
depth $= \frac{vt}{2} = \frac{1480 \times 0.25}{2} = 185\,\mathrm{m}$
$\lambda = \frac{340}{850} = 0.40\,\mathrm{m}$ ; $\Delta = 0.6\,\mathrm{m} = 1.5\lambda$ , an odd number of half-wavelengths, so the sound is quiet (destructive interference).
Less background noise helps a little, but the main cause is refraction. On a clear night the ground cools the air next to it; sound is slower in the cooler air, so rays that head upwards are bent back down towards the ground and reach distant listeners.

Key points

$v = \sqrt{\frac{\text{modulus}}{\rho}}$ ; for a gas $v = \sqrt{\frac{\gamma p}{\rho}} = \sqrt{\frac{\gamma R T}{M}}$ (Laplace's adiabatic correction).
In a gas, $v \propto \sqrt{T}$ and does not depend on pressure.
Sound reflects (echoes, echo sounding), refracts towards slower air, diffracts through gaps near its wavelength, and interferes: loud for $\Delta = n\lambda$ , quiet for $\Delta = (n + \frac{1}{2})\lambda$ .

Sources and credits

17.2: Speed of Sound, Frequency, and Wavelength (including the table of speeds), from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
17.3: Speed of Sound, from University Physics Volume 1 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Video: Interference Demo: Speakers, Physics Demos, embedded from YouTube.

The echo experiment, the three diagrams, the worked examples and the practice exercises are written by Ganzaa.

1Counting the seconds after lightning

2What sets the speed of sound?

3Newton's value and Laplace's correction

4Experiment 2.1: the echo method

5Reflection: echoes and echo sounding

6Refraction: why sound carries at night

7Diffraction: hearing round corners

8Interference: loud and quiet spots