This document summarizes key concepts about physical properties of waves. It discusses types of waves including transverse and longitudinal waves. It defines important quantities used to describe waves such as wavelength, frequency, period, velocity and angular wave number. It also examines waves on a string, sound waves, the Doppler effect, and human perception of sound. Specific topics covered include how wave speed depends on the medium, mechanical models of sound waves, measuring sound intensity, the audible range of human hearing, and how the Doppler effect results from either a moving observer or moving source of sound.

1. Unit 12 Physical Properties of Waves I
12.1 Types of waves
12.2 Useful quantities in describing waves
12.3 Waves on a string
12.4 Sound waves
12.5 The frequency of a sound wave
12.6 Sound intensity
12.7 Human perception of sound
12.8 The Doppler effect
12.9 Superposition and interference
12.10 Standing waves
12.11 Beats
12.1 Types of waves
A disturbance that propagates from one place to another is referred to as a wave.
Waves propagate with well-defined speeds determined by the properties of the
material through which they travel. For example, sound waves have different speeds
in different materials. The following table lists a sampling of sound speed in various
materials.
Material Speed (m/s)
Aluminum 6420
Steel 5960
Copper 5010
Plastic 2680
Fresh water (20 oC) 1480
Air (20 oC) 343
Waves carry energy and propagate it when the waves travel. There are two typical
waves, namely, the transverse waves and the longitudinal waves.
1

2. Transverse waves:
The displacement of individual particles is perpendicular to the direction of
propagation of the wave, e.g. holding one end of a string with another end fixed on
the wall. When you swing your hands vertically, the waves propagate horizontally
along the string and the particles of string moves up and down.
Longitudinal waves:
The displacement of individual particles is in the same direction as the direction of
propagation of the waves, e.g. sound waves. The particles of air move back and forth
such that a series of compression and rarefaction are observed. Note that the particle
does not travel with the wave, but vibrating about its equilibrium position.
2

3. 12.2 Useful quantities in describing waves
Wavelength: The distance over which a wave repeats,
e.g. the distance between successive crests and the
distance between successive troughs. Wavelength is
quite often labeled as λ. The SI unit is, of course,
meter, m.
Angular wave number: The angular wave number is
2π
defined as k = . The SI unit is radian per meter.
λ
Angular frequency: It is the measure of how many
radians the waves change in one second. It is labeled as
ω.
Frequency: The number of oscillation per unit time, f,
where ω = 2πf .
Period: The time for one oscillation, it is labeled as T,
1 2π
where T = = .
f ω
Velocity: The distance that the wave travels per unit time is referred to as the velocity.
3

4. The wave equation is: y ( x, t ) = y m sin(kx − ωt ) , in general we have
y ( x, t ) = y m sin( kx − ωt + φ ) ,
where φ is the phase angle.
ω
A useful relation in waves: v = fλ = .
k
12.3 Waves on a string
The speed of a wave is determined by the properties of the medium through which it
propagates. For a string of length L, there are two factors that vary the speed of a
wave: (i) the tension in the string F, and (ii) the mass of the string. For the second
factor, we should say it more precisely that the speed of a wave varies with the
density of the string (mass per length) µ. The definition of µ is m/L. The unit is kg/m.
We can obtain the velocity v by dimensional analysis. Let the velocity relates the
tension of string F and the mass per unit length µ by
F
v=
µ
The proof is simple. Let v = F a µ b and consider the dimensions of the following
quantities.
[v] = [L][T]−1 (Unit: ms−1)
[F] = [M][L][T]−2 (Unit: kg ms−2)
[µ] = [M][L]−1 (Unit: kg m−1)
Comparing the dimension on both sides of v = F a µ b , we have three equations
[L]: 1=a–b
[T]: −1 = − 2a
[M]: 0=a+b
1 F
After solving, we find that a = −b = . Hence, we have v = .
2 µ
Example
A rope of length L and mass M hangs from a ceiling. If the bottom of the
rope is given a gentle wiggle, a wave will travel to the top of the rope. As
4

5. the wave travels upward does its speed (a) increase, (b) decrease, or (c) stay the same?
Answer:
Since the tension increases with the height, the speed of the wave increases when it
climbs up the rope. Note also that the tension of the rope increases from almost zero
at the bottom to Mg at the top of rope.
12.4 Sound waves
A mechanical model of a sound wave is provided by a slinky. Consider if a slinky is
oscillated at one end back and forth horizontally. Longitudinal wave travels in
horizontal direction with some regions are compressed and some regions are more
widely spaced, but these regions are distributed alternatively. If we plot the density
variation against the displacement x, we observe classical wave shape in the graph.
The rarefactions and compressions oscillate in a wave-like fashion. In the
compressions regions, the pressure is high, and in the rarefaction regions, the pressure
is low. The speed of sound is determined by the properties of the medium through
which it propagates. In air, under normal atmospheric pressure and temperature, the
speed of sound is approximately 343 m/s ≈ 770 mi/h. As the air is heated up to a
higher temperature, the air molecules moves faster and the speed of sound increases
as expected.
In a solid, the speed of sound is determined in part by the stiffness of the material.
The stiffer the material, the faster the sound wave, just as having more tension in a
string causes a faster wave. The speed of sound in steel is greater than that in plastic.
And both speeds are much higher than that in air.
Example
You drop a stone into a well that is 7.35 m deep. How long does it take
before you hear the splash?
5

6. Answer:
The time until the splash is heard is the sum of two time intervals.
t1: the time for the stone to drop a distance d and
t2: the time for the sound to travel a distance d.
1 2 2d 2(7.35)
Since d = gt1 , we obtain t1 = = = 1.22 s .
2 g 9.81
d 7.35
To calculate t2, we have d = vt 2 , and t 2 = = = 0.0214 s .
v 343
Hence, the sum of the two time intervals is (1.22 +0.0214) s = 1.24 s.
12.5 The frequency of a sound wave
Human can hear sounds between 20 Hz on the low frequency and 20,000 Hz on the
high frequency end. Sounds with frequencies above this range are referred as
ultrasonic, while those with frequencies lower than 20 Hz are classified as infrasonic.
12.6 Sound intensity
Intensity is a quantitative scale by which loudness may be measured. The intensity is
defined as the amount of energy that passes through a given area in a given time. This
is illustrated in the figure. If the energy E passes through the area A in the time t the
E
intensity, I, of the wave carrying the energy is I = , where E/t is the power.
At
P
Rewrite the expression again, we have I = .
A
The SI unit is W/m2. An example of intensity of light on the Earth’s upper atmosphere
coming from the Sun is about 1380 W/m2. A rock concert has an intensity of 0.1
W/m2, while the intensity of a classroom is 0.0000001 W/m2. The threshold of
hearing is 10−12 W/m2.
When we listen to a source of sound, such as a person speaking or a radio playing a
song, the loudness of the sound decreases as we move away from the source. The
surface area of a sphere from a distance r is 4π r 2 . The intensity of such sound is
P
I= .
4πr 2
6

7. 12.7 Human perception of sound
We can detect sounds that are about a million times fainter than a typical
conversation, and listen to sounds that are a million times louder before experiencing
pain. We are able to hear sounds over a wide range of frequencies, from 20 Hz to
20,000 Hz. Our perception of sound, for example the loudness seems to be “twice as
loud” if the intensity of the sound is about 10 times the original one. In the study of
sound, the loudness is measured by a convenient scale, which depends on the
logarithm of intensity.
I
Mathematically, the intensity level β is expressed in the form β = 10 log( ) . The
I0
intensity level β is dimensionless and the unit is given as decibel (dB), where I0 is the
intensity of the faintest sounds that can be heard. Experiments show that the lowest
detectable intensity is I 0 = 10 −12 W / m 2 . The smallest increase in intensity level that
can be detected by the human ear is about 1 dB. And, the loudness of a sound doubles
with each increase in intensity level of 10 dB.
Sound Decibels
Ear drum ruptures 160
Jet taking off 140
Loud rock band 120
Heavy traffic 90
Classroom 50
Whisper 20
Threshold of hearing 0
Example
If a sound has an intensity I = I0, the corresponding intensity level is
I0
β = 10 log( ) = 10 log 1 = 0 dB .
I0
Increasing the intensity by a factor of 10 makes the sound seem twice as loud. In
terms of decibels, we have
7

8. 10 I 0
β = 10 log( ) = 10 log 10 = 10 dB .
I0
A further increase in intensity by a factor of 10 double the loudness again.
100 I 0
β = 10 log( ) = 10 log 100 = 20 dB .
I0
Thus, the loudness of a sound doubles with each increase in intensity level of 10 dB.
The smallest increase in intensity level that can be detected by the human ear is about
1 dB.
Example
A crying child emits sound with an intensity of 8.0 × 10 −6 W / m 2 . Find
(a) the intensity level in decibels for
the child’s sounds, and
(b) the intensity level for this child and
its twin, both crying with identical
intensities.
Answer:
I
(a) As the intensity level is given by β = 10 log( ) , we substitute
I0
I = 8.0 × 10 −6 W / m 2 and the lowest detectable intensity I 0 = 10 −12 W / m 2 ,
hence β = 10 log(
8.0 × 10 −6
10 −12
[ ]
) = 10 log(8.0 × 10 −6 ) − log(10 −12 ) = 69 dB .
(b) When the twins cry, the intensity will be doubled,
I = 2 × (8.0 × 10 −6 W / m 2 ) = 1.6 × 10 −5 W / m 2 .
1.6 × 10 −5
The intensity level is β = 10 log( ) = 72 dB .
10 −12
Or, we can write
β = 10 log(
2 × 8.0 × 10 −6
10 −12
[ ]
) = 10 log (2) + log(8.0 × 10 −6 ) − log (10 −12 ) = 72 dB
8

9. N.B. We should note that double the intensity increases the intensity level by 3 dB,
since 10 log 2 ≈ 3 . Halved the intensity leads to a decrease of intensity level by 3 dB.
Obviously, ten times the intensity of sound gives an increase of 10 dB.
Example
Many animal species use sound waves that are too high or too low for human ears to
detect, e.g. bats and blue whales.
12.8 The Doppler effect
The relative motion between a source of sound and the receiver gives a change in
pitch. This is the Doppler effect. There are two cases for Doppler effect: Moving
observer and moving the source. For example, there is a change in pitch of a train
whistle or a car horn as the vehicle moves past us. Doppler effect applies to all wave
phenomena, not just to sound.
Example
For light, we observe a change in color, e.g. red-shifted in the color of their light when
the galaxies are moving away from the Earth. However, some galaxies are moving
toward us, and their light shows a blue shift.
12.81 Moving observer
A sound wave is emitted from a stationary source. The wave travels in the air with
velocity v, having frequency f and wavelength λ, where v = fλ. For an observer
9

10. moving toward the source with a speed u, the sound seems to have a higher speed, e.g.
v + u. As a result, more wavefronts move past the observer in a given time than if the
observer had been at rest. To the observer, the sound has a frequency, f’, that is higher
than the frequency of the source.
u u
1+1+
v' v + u v = v = (1 + u ) f > f
f '= = =
λ λ λ 1 v
v f
If the observer moves away from the source, the sound seems to have a lower speed,
e.g. v − u. As a result, less wavefronts move past the observer in a given time than if
the observer had been at rest. To the observer, the sound has a frequency, f’, that is
lower than the frequency of the source.
u u
1− 1−
v' v−u v = v = (1 − u ) f < f
f '= = =
λ λ λ 1 v
v f
Combing the two results, we have
f ' = (1 ± u / v) f ,
where plus sign is used when the observer moves toward the source and minus sign is
used when the observer moves away from the source.
12.82 Moving source
When the source moves, the Doppler effect is not due to the sound wave appearing to
have a higher or lower speed, but a variation in the magnitude of wavelength.
Consider, then, a source moving toward the observer with speed u, the sound waves
have one compression and then another compression in time T, where T = 1/f. The
wave travels a distance vT, and the source travels a distance uT. As a result, the new
wavelength of the sound waves is vT − uT = (v − u )T , which is shorter than that when
10

11. the source is at rest. The new frequency of the sound waves is obtained by v = f’λ’,
that is
v 1 ⎛ 1 ⎞
f '= = =⎜ ⎟f > f .
(v − u )T u ⎝1− u / v ⎠
(1 − )T
v
When the source reverses its direction, the new wavelength of the sound waves,
vT + uT = (v + u )T , is longer than that when the source is at rest.
v 1 ⎛ 1 ⎞
f '= = =⎜ ⎟f < f .
(v + u )T u ⎝1+ u / v ⎠
(1 + )T
v
Combine the two results, we have
⎛ 1 ⎞
f '= ⎜ ⎟f ,
⎝1∓ u / v ⎠
where minus sign is used when the source moves toward the observer, and the plus
sign when the source moves away from the observer.
12.83 General case
Doppler effect for both moving source and observer is concluded in a simple formula:
⎛ 1 ± uo / v ⎞
f '= ⎜
⎜1∓ u / v ⎟ f .
⎟
⎝ s ⎠
11

12. Example
A car moving at 18 m/s sounds its 550 Hz horn. A bicyclist on the sidewalk, moving
with a speed of 7.2 m/s, approaches the car. What frequency is heard by the bicyclist?
Answer:
As the car (source) and the bicyclist (observer) approach each other, we apply the
⎛ 1 + uo / v ⎞ 1 + 7.2 / 343 ⎞
⎜
formula f ' = ⎜ ⎟f =⎛
⎟ ⎜ ⎟ (550 Hz ) = 592.7 Hz .
⎝ 1 − us / v ⎠ ⎝ 1 − 18 / 343 ⎠
The right figure shows the Doppler shifted
frequency versus speed for a 400-Hz sound
source. The upper curve corresponds to a
moving source, the lower curve to a moving
observer. Notice that while the two cases
give similar results for low speed, the high-
speed behavior is quite different. In fact, the
Doppler frequency for the moving source
grows without limit for speeds near the speed
of sound, while the Doppler frequency for
the moving observer is relatively small.
12.84 Supersonic speed and shock waves
What happen when the speed of the source exceeds the speed of sound? The equations
derived above are no longer valid. For supersonic speeds, a V-shaped envelope is
observed, all wavefronts bunch are along this
envelop, which is in three dimensions. This cone is
called the Mach cone. A shock wave is said to exist
along the surface of this cone, because the bunching
of wavefronts causes an abrupt rose and fall of air
pressure as the surface passes through any point. The
Mach cone angle is given by
vt v
sin θ = = .
vs t vs
12

13. The ratio vs/v is the Mach number. The shock wave generated by a supersonic aircraft
or projectile produces a burst of sound, called a sonic boom.
12.9 Superposition and interference
The combination of two or more waves to form a resultant
wave is referred to as superposition. When waves are of
small amplitude, they superpose in the simplest of ways –
they just add.
For example, consider two waves on a string, as shown in
figure.
Example
Since two waves add, does the resultant wave y always have a greater amplitude than
the individual waves y1 and y2?
Answer:
The wave y is the sum of y1 and y2, but remember that y1 and y2 are sometimes
positive and sometimes negative. Thus, if y1 is positive at a given time, for example,
and y2 is negative, the sum y1 + y2 can be zero or even negative.
As simple as the principle of superposition is, it still leads to interesting
consequences. For example, consider the wave pulse on a string shown in the above
figure (a). When they combine, the resulting pulse has an amplitude equal to the sum
of the amplitudes of the individual pulses. This is referred to as constructive
interference. When two pulses like those in figure (b) may combine and gives a net
displacement of zero. That is the pulses momentarily cancel one another. This is
destructive interference.
13

14. It should also be noted that interference is not limited to waves on a string; all waves
exhibit interference effects. In fact, interference is one of the key characteristics that
define waves.
Suppose the two sources emit waves in phase. At point
A the distance to each source is the same, hence crest
meets crest and constructive interference results. At B
the distance from source 1 is greater than that from
source 2 by half a wavelength. The result is crest
meeting trough and destructive interference. Finally, at
C the distance from source 1 is one wavelength greater
than the distance from source 2. Hence, we find constructive interference at C, and the
waves are in phase again at C. If the sources had been opposite
in phase, then A and C would be points of destructive
interference, and B would be a point of constructive
interference.
Remarks:
• The system that when one source emits a crest, the other
emits a crest as well is referred to as synchronized
system. The sources are said to be in phase.
• In general, we can say that constructive and destructive interference occur
under the following conditions for two sources that are in phase:
i) Constructive interference occurs when the path length from
the two sources differs by 0, λ, 2λ, 3λ, ….
ii) Destructive interference occurs when the path length from
the two sources differs by λ/2, 3λ/2, 5λ/2, ….
Example
Two speakers separated by a distance of 4.30 m emit sound
of frequency 221 Hz. The speakers are in phase with one
another. A person listens from a location 2.8 m directly in
front of one of the speakers. Does the person hear
constructive or destructive interference?
14

15. Answer:
The wavelength of sound: λ = v / f = 343 m / s / 221 Hz = 1.55 m .
To determine the path difference, d = d2 – d1, we need to find d2 first, and
d 2 = D 2 + d12 = (4.30 m) 2 + (2.80 m) 2 = 5.13 m .
Now d = 5.13 m – 2.80 m = 2.33 m. The number of wavelength that fit into the path
d 2.33 m
difference: = = 1.50 . Since the path difference is 3λ/2 we expect destructive
λ 1.55 m
interference. In the ideal case, the person would hear no sound. As a practical matter,
some sound will be reflected from objects in the vicinity, resulting in a finite sound
intensity.
12.10 Standing waves
If you plucked a guitar string, or blown across the
mouth of a pop bottle to create a tone, you have
generated standing waves. In general, a standing
wave is one that oscillates with time, but remains
in its location. It is in this sense that the wave is
said to be “standing”. In some respects, a standing
wave can be considered as resulting from
constructive interference of a wave with itself.
15

16. 12.10.1 Waves on a string
A string is tied down at both ends. If the string is plucked
in the middle a standing wave results. This is the
fundamental mode of oscillation of the string. The
fundamental consists of one-half a wavelength between
the two ends of the string. Hence, its wavelength is 2L,
or we write λ = 2 L .
If the speed of waves on the string is v, it follows that the
frequency of the fundamental, f1, is determined by
v = λf 1 = (2 L) f 1 . Therefore,
v v
f1 = = .
λ 2L
Note that the fundamental frequency increases with the speed of the waves, and
decreases as the string is lengthened. Other than the fundamental frequency, there are
an infinite number of standing wave modes – or harmonics – for any given string.
Remarks:
• Points on a standing wave that stay fixed are
referred to as nodes, N.
• Halfway between any two nodes is a point on the
wave that has a maximum displacement, is called
an anti-node, A.
The second harmonics can be constructed by including
one more half wavelength in the standing wave. This
mode has one complete wavelength between the walls.
v v
f2 = = = 2 f1 .
λ L
Similarly, the third harmonic has one-and-a-half
wavelength in the length L, therefore, its frequency
16

17. v v
f3 = = = 3 f1 .
λ 2
L
3
Remark:
v
In general, we have f 1 = , f n = nf1 , and λ n = 2 L / n where n = 1, 2, 3, …. That is,
2L
all harmonics are present.
12.10.2 Vibrating columns of air
If you blow across the open end of a pop bottle, you hear a
tone of a certain frequency. If you pour some water into the
bottle and repeat the experiment, the sound you hear has a
higher frequency. The standing wave will have an antinode,
A, at the top (where the air is moving) and a node, N, at the
bottom (where the air cannot move.) The lowest frequency
standing wave has one-quarter of a wavelength fits into the
column of air in the bottom. Thus, we have the wave form N-
A in the pipe
1
λ=L
4
λ = 4L
The fundamental frequency, f1, is given by v = λf 1 = (4 L) f 1 . Or we can write
v
f1 = .
4L
The second harmonic is produced by adding
half a wavelength, i.e. N-A-N-A, therefore,
4
3λ / 4 = L , and hence λ = L . The frequency
3
v v v
is = = 3( ) = 3 f 1 .
λ 4
L
4L
3
Similarly, the next-higher harmonic is
represented by N-A-N-A-N-A. Inside the pipe,
v v v
we have standing waves 5λ / 4 = L , the frequency is = = 5( ) = 5 f1 .
λ 4
L
4L
5
17

18. Remark:
v
In general, we have f1 = , f n = nf1 and λ n = 4 L / n , where n = 1, 3, 5, …. That is
4L
odd harmonics are present.
Standing waves in a pipe that is open at both ends
have the following modes, as shown in the figure.
Remark:
v
In general, we have f1 = , f n = nf1 and
2L
λ n = 2 L / n , where n = 1, 2, 3, …. That is all
harmonics are present.
Example
An empty pop bottle is to be used as a musical instrument in a
band. In order to be tuned properly the fundamental frequency
of the bottle must be 440.0 Hz. If the bottle is 26.0 cm tall,
how high should it be filled with water to produce the desired
frequency?
Answer:
Since f1 = v / 4 L , we have
343 m / s
L = v / 4 f1 = = 0.195 m .
4(440.0 Hz )
The depth of water to be filled: h = H − L = 0.260 m − 0.195 m = 6.5 cm .
Example
If you fill your lungs with helium and speak you sound something like Donald Duck.
From this observation, we can conclude that the speed of sound in helium must be (a)
less than (b) the same as, or (c) greater than the speed of sound in air.
18

19. Answer:
When we speak with helium our words are higher pitched. Looking at the relation,
v
e.g. f1 = , the velocity of sound is increased if the length of vocal chords is fixed
2L
while the frequency is increased.
12.11 Beats
Beats can be considered as the interference pattern in time. To be specific, imagine
plucking two guitar strings that have slightly different frequencies. If you listen
carefully, you notice that the sound produced by the strings is not constant is time. In
fact, the intensity increases and decreases with a definite period. These fluctuations in
intensity are the beats, and the frequency of successive maximum intensities is the
beat frequency.
Consider two waves, with frequencies f1 = 1/T1 and f2 = 1/T2, interfere at a given,
fixed location. At this location, each wave moves up and down to the vertical
position, y, of each wave yields the following:
⎛ 2π ⎞
y1 = A cos⎜
⎜T t ⎟ = A cos(2πf1t )
⎟
⎝ 1 ⎠
⎛ 2π ⎞
y 2 = A cos⎜
⎜T t ⎟ = A cos(2πf 2 t )
⎟
⎝ 2 ⎠
If A = 1,we have the following plots, where y total = y1 + y 2 . Mathematically, we have
ytotal = y1 + y2 = A cos(2π f1t ) + A cos(2π f 2 t )
⎡ ⎛ f − f2 ⎞ ⎤ ⎡ ⎛ f1 + f 2 ⎞ ⎤
ytotal = 2 A cos ⎢ 2π ⎜ 1 ⎟ t ⎥ cos ⎢ 2π ⎜ 2 ⎟ t ⎥
⎣ ⎝ 2 ⎠ ⎦ ⎣ ⎝ ⎠ ⎦
19

20. ⎡ ⎛ f − f2 ⎞ ⎤
The first part of the ytotal is 2 A cos ⎢2π ⎜ 1 ⎟t ⎥ which gives the slowly-varying
⎣ ⎝ 2 ⎠⎦
amplitude of the beats. Since a loud sound is heard whenever this term is 2A or –2A,
the beat frequency is f beat = f 1 − f 2 . The rapid oscillations within each beat are due
⎡ ⎛ f + f2 ⎞ ⎤
to the second part of ytotal, cos ⎢2π ⎜ 1 ⎟t ⎥ . Now, beats can be understood as
⎣ ⎝ 2 ⎠⎦
oscillations at the average frequency, modulated by a slowly varying amplitude.
Example
Suppose two guitar strings have frequencies 438 Hz and 442 Hz. If you sound them
simultaneously you will hear the average frequency, 440 Hz, increasing and
decreasing in loudness with a beat frequency of 4 Hz. Beats can be used to tune a
musical instrument to a desired frequency. To tune a guitar string to 440 Hz, for
example, the string can be played simultaneously with a 440 –Hz fork. Listening to
the beats, the tension in the string can be increased or decreased until the beat
frequency becomes vanishingly small.
Example (Challenging)
An experimental way to tune the pop bottle is to
compare its frequency with that of a 440-Hz tuning fork.
Initially, a beat frequency of 4 Hz is heard. As a small
amount of water is added to that already present, the
beat frequency increases steadily to 5 Hz. What were the
initial and final frequencies of the bottle?
Answer:
Before extra water is added, possible frequency of the bottle is either 436 Hz or 444
Hz. After water is added, possible frequency of the bottle is either 435 Hz or 445 Hz.
But the frequency of the bottle should be increased as water is added. Hence, the
frequency of the bottle before adding extra water should be 444 Hz.
20