PHY300

1. Giambattista Physics
Chapter 12
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Chapter 12: Sound
12.1 Sound Waves
12.2 The Speed of Sound Waves
12.3 Amplitude and Intensity of Sound Waves
12.4 Standing Sound Waves
12.5 Timbre
12.6 The Human Ear
12.7 Beats
12.8 The Doppler Effect
12.9 Echolocation and Medical Imaging

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12.1 Sound Waves
In a sound wave, the uniform distribution of molecules is
disturbed.
In some regions ( compressions ), the molecules are bunched
together and the pressure is higher than the average pressure.
In other regions ( rarefactions ), the molecules are spread out
and the pressure is lower than average.

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Sound Waves: Pressure Variation

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Sound Waves: Displacement of Air Elements

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Frequency Ranges of Animal Hearing
The human ear responds to sound waves within a limited range
of frequencies.
We generally consider the audible range to extend from 20 Hz to
20 kHz. (The terms infrasound and ultrasound refer to sound
waves with frequencies below 20 Hz and above 20 kHz,
respectively.)
Even for a person with excellent hearing, the sensitivity of the
human ear declines rapidly below 100 Hz and above 10 kHz.

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12.2 The Speed of Sound Waves
For string waves, the restoring force is characterized by the
tension in the string F , and the inertia is characterized by the
linear mass density  (mass per unit length).
The speed of transverse waves on a string is
For sound waves in a fluid, the restoring force is characterized
by the bulk modulus B.

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Temperature Dependence of the Speed of
Sound in a Gas
The bulk modulus B of an ideal gas is directly proportional to the
density ρ and to T, the absolute temperature ( B ∝ ρT ).

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Approximate Formula for Speed of Sound in Air
An approximate formula that can be used for the speed of
sound in air is
accurate to better than 1% all the way from −66°C to + 89°C.

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Speed of Sound in a Solid
The speed of sound in a solid depends on the Young’s modulus Y
and the shear modulus S.
For sound waves traveling along the length of a thin solid rod,
the speed is approximately

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Speed of Sound in Various Materials

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Example 12.1
The speed of sound in hydrogen gas at 0°C is almost as large as
the speed of sound in mercury, even though the density of
mercury is 150 000 times larger than the density of hydrogen.
How is that possible? Shouldn’t the speed in mercury be much
smaller, since it has so much more inertia?

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Example 12.1 Solution
The speed of sound depends on two characteristics of the
medium: the restoring force (measured by the bulk modulus)
and the inertia (measured by the density).
The bulk modulus of mercury is much larger than the bulk
modulus of hydrogen.

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12.3 Amplitude and Intensity of Sound Waves
Because we can describe a sound wave in two ways—pressure
or displacement—the amplitude of a sound wave can take one
of two forms: the pressure amplitude p0 or the displacement
amplitude s0.
For a harmonic sound wave at angular frequency ω, an
advanced analysis shows that
where v is the speed of sound and ρ is the mass density of the
medium.

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Perceived Loudness
Perceived loudness turns out to be roughly proportional to the
logarithm of the amplitude. If the amplitude of a sound wave
doubles repeatedly, the perceived loudness does not double; it
increases by a series of roughly equal steps.
Discussions of loudness are more often phrased in terms of
intensity rather than amplitude since we are interested in how
much energy the sound wave carries.

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Sound Intensity
Intensity is proportional to amplitude squared:
This is also true for waves other than sound.

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Example 12.2
The song of the Brown Creeper ( Certhia americana ) is high in
frequency—about 8 kHz.
Many people who have lost some of their high-frequency
hearing can’t hear it at all.
Suppose that you are out in the woods and hear the song. If the
intensity of the song at your position is 1.4 × 10−8 W/m2 and the
frequency is 6.0 kHz, what are the pressure and displacement
amplitudes? (Assume the temperature is 20°C.)

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Example 12.2 Solution

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Threshold of Hearing
The perception of loudness by the human ear is roughly
proportional to the logarithm of the intensity, making us capable
of hearing sound over a wide range of intensities.
By convention, we compare intensities to a reference level
I0 = 10−12 W/m2 known as the threshold of hearing because it is
roughly the lowest intensity sound wave that can be heard under
ideal conditions by a person with excellent hearing.

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Sound Intensity Level
A sound intensity I is compared with the reference level I0 by
taking the ratio of the two intensities.
Suppose a sound has an intensity of 10−5 W/m2 ; the ratio is
so the intensity is 107 times that of the hearing threshold.
The power to which 10 is raised is the sound intensity level b in
units of bels.

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Sound Intensity Level: Decibels
Since log10 (10x ) = x , the sound intensity level b in decibels is

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Example 12.3
The sound intensity 0.250 m from a roaring lion is 0.250 W/m2.
What is the sound intensity level in decibels? (Use the usual
reference level of I0 = 1.00 × 10−12 W/m2.)
Strategy
We are given the intensity in W/m2 and asked for the intensity
level in dB.
First we find the ratio of the given intensity to the reference
level. Then we take the logarithm of the result (to get the level in
bels) and multiply by 10 (to convert from bels to dB).

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Example 12.3 Solution

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Incoherent Waves
For two incoherent waves, the intensity due to the two waves
together at a point is the sum of the two intensities:
This is not true for two coherent waves, where the total
intensity depends on the phase difference between the waves.

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Example 12.4
A metal lathe in a workshop produces a 90.0-dB sound intensity
level at a distance of 1 m.
What is the intensity level when a second identical lathe starts
operating? Assume the listener is at the same distance from both
lathes.

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Example 12.4 Strategy
The noise is coming from two different machines and, thus, they
are incoherent sources.
We cannot add 90.0 dB to 90.0 dB to get 180.0 dB, which would
be a senseless result—two lathes are not going to drown out a
jet engine at close range.
Instead, what doubles is the intensity. We must work in terms of
intensity rather than intensity level.

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Example 12.4 Solution
First find the intensity due to one lathe:

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Intensity Level Rule of Thumb
A useful rule of thumb:
Every time the intensity increases by a factor of 10, the intensity
level adds 10 dB; since log10 2 = 0.30, adding 3.0 dB to the
intensity level doubles the intensity.

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Comparing Intensity Levels
Decibels can also be used in a relative sense; instead of
comparing an intensity to I0, we can compare two intensities
directly.
Suppose we have two intensities I1 and I2 and two corresponding
intensity levels β1 and β2. Then

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Example 12.5
At a distance of 30 m from a jet engine, the sound intensity level
is 130 dB.
Assume the engine is an isotropic source of sound and ignore
reflections and absorption. At what distance is the intensity level
110 dB—still quite loud but below the threshold of pain?

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Example 12.5 Strategy
The intensity level drops 20 dB. According to the rule of thumb,
each 10-dB change represents a factor of 10 in intensity.
Therefore, we must find the distance at which the intensity is 2
factors of 10 smaller—that is, 1 /100 the original intensity.
The intensity is proportional to 1/r2 since we assume an isotropic
source.

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Example 12.5 Solution

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12.4 Standing Sound Waves
Pipe Open at Both Ends

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Pipe Open at Both Ends Continued
Standing sound waves (thin pipe open at both ends):
where n = 1, 2, 3, . . .

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Pipe Closed at One End

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Pipe Closed at One End Continued
Standing sound waves (thin pipe closed at one end):
where n = 1, 3, 5, 7, . . .

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Example 12.6
A thin hollow tube of length 1.00 m is
inserted vertically into a tall container of
water. A tuning fork ( f = 520.0 Hz) is
struck and held near the top of the tube
as the tube is slowly pulled up and out
of the water.
At certain distances (L) between the top
of the tube and the water surface, the
otherwise faint sound of the tuning fork
is greatly amplified. At what values of L
does this occur? The temperature of the
air in the tube is 18°C.

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Example 12.6 Strategy
The sound is amplified due to resonance; when the frequency of
the tuning fork matches one of the natural frequencies of the air
column, a large-amplitude standing wave builds up in the
column.

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Example 12.6 Solution

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12.5 Timbre
The standing wave on a string or in a column of air is almost
always the superposition of many standing wave patterns at
different frequencies.
The lowest frequency in a complex sound wave is called the
fundamental; the rest of the frequencies are called overtones.
All the overtones of a periodic sound wave have frequencies that
are integral multiples of the fundamental; the fundamental and
the overtones are then called harmonics.

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Timbre Continued
Middle C played on an oboe
does not sound the same as
middle C played on a trumpet.
What is different about the two
sounds is the tone quality, or
timbre (pronounced “tamber”).
Any periodic wave, no matter
how complicated, can be
decomposed into a set of
harmonics, each of which is a
simple sinusoid.

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12.6 The Human Ear

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Loudness
Although loudness is most closely correlated to intensity level, it
also depends on frequency (as well as other factors).
Curves of equal
loudness.

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Pitch
Pitch is the perception of frequency.
Our sense of pitch is a logarithmic function of frequency, just as
loudness is approximately a logarithmic function of intensity.

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Localization
The ear has several different tools it uses to localize sounds:
• The principal method for high-frequency sounds (>4 kHz) is
the difference in intensity sensed by the two ears. The head
casts a “sound shadow,” so a sound coming from the right
has a larger intensity at the right ear than at the left ear.
• The shape of the pinna makes it slightly preferential to
sounds coming from the front. This helps with front-back
localization for high-frequency sounds.
• For lower-frequency sounds, both the difference in arrival
time and the phase difference between the waves arriving at
the two ears are used for localization.

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12.7 Beats
When two sound waves are close in frequency (within about 15
Hz of one another), the superposition of the two produces a
pulsation that we call beats.
Beats can be produced by any kind of wave; they are a general
result of the principle of superposition when applied to two
waves of nearly the same frequency.

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Beats Continued

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Example 12.7
A piano tuner strikes his tuning fork ( f = 523.3 Hz) and strikes a
key on the piano at the same time. The two have nearly the
same frequency; he hears 3.0 beats per second.
As he tightens the piano string, he hears the beat frequency
gradually decrease to 2.0 beats per second when the two sound
together.
(a) What was the frequency of the piano string before it was
tightened?
(b) By what percentage did the tension increase?

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Example 12.7 Strategy
The beat frequency is the difference between the two
frequencies; we only have to determine which is higher.
The wavelength of the string is determined by its length, which
does not change.
The increase in tension increases the speed of waves on the
string, which in turn increases the frequency.

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Example 12.7 Solution 1
(a)
As the tension increases gradually, the beat frequency
decreases, which means that the frequency of the piano string is
getting closer to the frequency of the tuning fork.
Therefore, the string frequency must be 3.0 Hz lower than the
tuning fork frequency:

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Example 12.7 Solution 2
(b)

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12.8 The Doppler Effect
A police car races by, its sirens screaming. As it passes, we hear
the pitch change from higher to lower.
The frequency change is called the Doppler effect.

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Moving Observer

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Doppler shift (moving source and/or observer)
o
o s
s
v v
f
v
f
v



vo and vs are positive in the direction of propagation of the wave
(from source to observer) and negative in the opposite direction.

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Problem-Solving Strategy: Doppler Effect
• Determine vs and vo; they are positive for motion in the
direction the wave travels (from source to observer) and
negative in the opposite direction (from observer to source).
• Use the frequency-wavelength relationship (wave speed =
frequency times wavelength):
• Some Doppler effect problems involve reflected waves. One
way to handle a reflected wave is to think of the reflecting
surface as first observing the wave and then re-emitting it at
the same frequency.

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Example 12.8
A monorail train approaches a platform at a speed of 10.0 m/s
while it blows its whistle.
A musician with perfect pitch standing on the platform hears the
whistle as “middle C,” a frequency of 261 Hz. There is no wind
and the temperature is a chilly 0°C.
What is the observed frequency of the whistle when the train is
at rest?

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Example 12.8 Strategy
In this case, the source—the whistle—is moving and the
observer is stationary.
The source is moving toward the observer, so vs is positive.
With the source approaching the observer, the observed
frequency is higher than the source frequency.
When the train is at rest, there is no Doppler shift; the observed
frequency then is equal to the source frequency.

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Example 12.8 Solution

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Example 12.9
Two cars, with equal ground speeds, are moving in opposite
directions away from each other on a straight highway.
One driver blows a horn with a frequency of 111 Hz; the other
measures the frequency as 105 Hz.
If the speed of sound is 338 m/s and there is no wind, what is
the ground speed of each car?

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Example 12.9 Strategy
The sound wave travels from source to observer.
The source moves opposite the direction of the wave, so vs is
negative.
The observer moves in the direction of the wave, so vo is
positive. The speeds are the same, so vs = −vo.

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Example 12.9 Solution

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Shock Waves
When a plane moves at the speed of sound, the wave crests pile
up on one another since the plane moves to the right as fast as
the wave crests. When the plane is supersonic, the wave crests
pile up along the cone indicated by the black lines.

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12.9 Echolocation and Medical Imaging
• Animal Echolocation (e.g., bats, dolphins)
• Sonar and Radar
• Medical ultrasound imaging

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Sound Waves: Pressure Variation Appendix
A sound wave generated by a loudspeaker, shown as a snapshot
at one time (t = 0). Numerous small dots illustrate the locations
of air particles in the wave. The elements of the air are clustered
closer together in regions marked “compressions” and farther
apart in regions marked “rarefactions”. Vector arrows indicate
the force F on air particles midway between the rarefactions and
compressions. The force vectors are directed away from the
compressions (higher pressure), toward the rarefactions (lower
pressure).
Accompanying the image of the air particles is a graph of
pressure variation versus distance from the speaker, which
appears as a sinusoid. The pressure is high in compressions and
low in rarefactions.

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Sound Waves: Displacement of Air Elements
Appendix
A sound wave generated by a loudspeaker, shown as a snapshot
at one time (t = 0). Numerous small dots illustrate the locations
of air particles in the wave. The elements of the air are clustered
closer together in regions marked “compressions” and farther
apart in regions marked “rarefactions”.
Accompanying the image of the air particles is a graph of the
displacement s of air elements versus distance from the speaker,
which appears as a sinusoid. A positive s is a displacement to the
right, while negative s represents a displacement to the left. The
displacement is always toward compressions, so just to the left
of a compression the displacement is positive, while to the right
of a compression the displacement is negative. The displacement
is zero at the center of compressions and rarefactions.

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12.4 Standing Sound Waves Appendix
The first three standing sound wave patterns for a pipe open at
both ends. The open ends are pressure nodes and displacement
antinodes. The compressions (maximum pressure) and
rarefactions (minimum pressure) are sketched along with arrows
showing the air displacement at a single instant in time
(displacement is always toward compressions).
Accompanying each standing wave pattern are graphs of
displacement and pressure variation in the pipe. Displacement
nodes correspond to pressure antinodes, and vice versa.
The first standing wave pattern has one-half wavelength in the
pipe length L. The second has one complete wavelength, while
the third has 1.5 wavelengths within the length L.

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Pipe Closed at One End Appendix
The first three standing sound wave patterns for a pipe closed at
one end. The open end is a pressure node and displacement
antinode, while the closed end is a pressure antinode and
displacement node. The compressions (maximum pressure) and
rarefactions (minimum pressure) are sketched along with arrows
showing the air displacement at a single instant in time
(displacement is always toward compressions).
Accompanying each standing wave pattern are graphs of
displacement and pressure variation in the pipe. Displacement
nodes correspond to pressure antinodes, and vice versa.
The first standing wave pattern has one-quarter wavelength in
the pipe length L. The second has 3/4 wavelength, while the
third has 5/4 wavelengths within the length L.

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Loudness Appendix
Graphs of intensity level in dB versus frequency showing lines of
equal perceived loudness. Two points are marked on a particular
curve as an example: a 40 dB sound at 1000 Hz is as loud as a 62
dB sound at 100 Hz. The curves are a minimum between 3 kHz
and 4 kHz and increase steeply below 800 Hz and above 10 kHz.
At any given frequency between 800 Hz and 10 kHz, the curves
are approximately evenly spaced. The threshold of hearing is
shown by the lowest curve in the set; The threshold of hearing is
at an intensity level of 0 dB or lower only in the frequency range
of about 1–6 kHz.

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Beats Continued Appendix
An example of the superposition of two sound waves with
different frequencies, resulting in beats. The two sound waves
are sinusoidal functions with slightly different frequencies. The
superposition of the two waves is a sinusoidal function within an
envelope of a lower frequency sinusoid. That is, the amplitude of
the superposition varies gradually between maximum and
minimum values.

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Moving Observer Appendix
An observer moves at speed vo away from a stationary sound
source. The positive direction is from the source to observer. The
velocity of the observer is away from the source, in the positive
direction. The distance between wave crests is λ = vTs. During
the observed period To, the observer moves a distance voTo and
the wave moves a distance vTo, so λ = vTo − voTo.