Unit 11
Waves and Sound
11.1
Types of waves
11.2
Useful quantities in describing waves
11.3
Waves on a string
11.4
Sound waves
11.5
The frequency of a sound wave
11.6
Sound intensity
11.7
Human perception of sound
11.8
The Doppler effect
11.9
Superposition and interference
11.10 Standing waves
11.11 Beats
11.1 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
o
Fresh water (20 C)
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
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
11.2 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 defined as
k=
2π
λ
. The SI unit is radian per meter.
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, where
T=
1 2π
.
=
f
ω
Velocity: The distance that the wave travels per unit time is
referred to as the velocity.
The wave equation is: y ( x, t ) = y m sin( kx − ωt ) , in general we have
y=
( x, t ) ym sin(kx − ωt + φ ) ,
where φ is the phase angle.
A useful relation in waves: v = fλ =
ω
k
.
Example
A sinusoidal wave with speed u = 0.08 m/s has a waveform at time t = 0 sec as shown in the
figure. (a) Write down the wave equation. (b) Sketch the waveform at t = T/8, where T is the
period.
y/m
u
0.04
0
0.2
0.4
0.6
x/m
3
Answer
(a) The sinusoidal wave has amplitude 0.04 m and a wavelength λ = 0.4 m. So, the wave
number k = 2π/ λ = 2π/(0.4 m) = 5π m–1. The angular speed ω = vk = (0.08 m s–1)(5π m–1) =
2π/5 s–1. The wave moves toward the right ensures the argument of the trigonometric
function has a form (kx − ωt ) . Hence, the wave equation can be written as
y=
( x, t ) ym sin(kx − ω t + φ ) and it becomes
=
y 0.04sin(5π x −
2π
t + φ ) , where φ is the phase
5
angle in the expression. Plugging in the condition when t = 0 sec, x = 0.1 m and y = 0.04 m,
2π
π



we have
=
0.04 0.04sin  5π (0.1) −
(0)t + φ  , that is sin  + φ  =
1 and φ =0. The wave
5


2

equation
is y 0.04sin(5π x −
=
2π
t) .
5
(b) The period T = 2π/ω = 5 sec. At time t = T/8
(i.e.
t
=
5/8
sec),
the
wave
has
y-
2π 5 

displacement
y 0.04sin  5π x −
(=
)
=
5 8 

y/m
u
t = T/8
t=0
0.04
0
0.05 0.2
0.4
0.6
π

0.04sin  5π x −  . The graph of it is a right
4

shift of the initial waveform by λ/8 = 0.05 m.
11.3 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
v=
F
µ
.
4
x/m
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
After solving, we find that a = −b =
1
. Hence, we have v =
2
F
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 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.
11.4 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.
5
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?
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.
Since d =
1 2
=
t1
gt1 , we obtain
2
2d
=
g
To calculate t2, we have d = vt2 , and t=
2
2(7.35)
= 1.22 s .
9.81
d 7.35
=
= 0.0214 s .
v 343
Hence, the sum of the two time intervals is (1.22 +0.0214) s = 1.24 s.
11.5 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.
Example
Many animal species use sound waves with frequencies that are too high or too low for
human ears to detect, e.g. bats and blue whales.
6
11.6 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 intensity, I, of the wave
carrying the energy is I =
E
, where E/t is the power. Rewrite the expression again, we have
At
I=
P
.
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
I=
P
.
4πr 2
11.7 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.
Mathematically, the intensity level β is expressed in the form β = 10 log(
I
) . The intensity
I0
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.
7
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
β = 10 log(
I0
) = 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
10 I 0
) = 10 log 10 = 10 dB .
I0
β = 10 log(
A further increase in intensity by a factor of 10 double the loudness again.
100 I 0
) = 10 log 100 = 20 dB .
I0
β = 10 log(
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.
8
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:
(a)
As the intensity level is given by β = 10 log(
and
the
β = 10 log(
(b)
When
lowest
detectable
I
) , we substitute I = 8.0 × 10 −6 W / m 2
I0
intensity
[
I 0 = 10 −12 W / m 2
,
hence
]
8.0 × 10 −6
) = 10 log(8.0 × 10 −6 ) − log(10 −12 ) = 69 dB .
10 −12
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 log (2) + log(8.0 × 10 −6 ) − log (10 −12 ) = 72 dB
−12
10
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.
11.8 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.
9
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.
11.8.1 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 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.
f '=
v'
λ
=
v+u
λ
u
u
1+
v = (1 + u ) f > f
v =
=
λ
1
v
f
v
1+
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−
v' v − u
v =
v = (1 − u ) f < f
f '= =
=
λ
1
λ
λ
v
v
f
1−
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.
10
11.8.2 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 the source is at rest. The new frequency of the sound waves is obtained by v =
f’λ’, that is
f '=
v
=
(v − u )T
1
 1 
=
f > f .
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.
f '=
v
=
(v + u )T
1
 1 
=
f < f .
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.
11
11.8.3 General case
Doppler effect for both moving source and observer is concluded in a simple formula:
 1 ± uo / v 
 f .
f ' = 
1

u
/
v
s


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 − 18 / 343 
 1 − us / v 
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.
11.8.4 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
12
any point. The Mach cone angle is given by
sin θ =
vt
v
= .
vs t vs
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.
11.9 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
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
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
d2 =
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.
11.10 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.
11.10.1
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 = λf1 = (2 L) f1 . Therefore,
15
f1 =
v
λ
=
v
.
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.
f2 =
v
v
= 2 f1 .
L
=
λ
Similarly, the third harmonic has one-and-a-half wavelength in the
length L, therefore, its frequency
f3 =
v
λ
=
v
= 3 f1 .
2
L
3
Remark:
In general, we have f1 =
v
, f n = nf1 , and λ n = 2 L / n where n = 1, 2, 3, ….
2L
That is, all harmonics are present.
11.10.2 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
16
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 = λf1 = (4 L) f1 .
Or we can write
f1 =
v
.
4L
The next harmonic is produced by adding half a wavelength, i.e. N-A-N-A, therefore,
3λ / 4 = L , and hence λ =
v
v
v
4
= 3( ) = 3 f1 (the third harmonic).
L . The frequency is =
4L
λ 4L
3
3
Similarly, the next-higher harmonic is represented by N-A-N-A-N-A. Inside the pipe, we
have standing waves 5λ / 4 = L , the frequency is
v
λ
=
v
v
= 5( ) = 5 f1 (the fifth harmonic).
4
4L
L
5
Remark:
In general, we have f1 =
v
, f n = nf1 and λ n = 4 L / n , where
4L
n = 1, 3, 5, …. That is 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:
In general, we have f1 =
v
, f n = nf1 and λ n = 2 L / n , where n = 1, 2, 3, …. That is all
2L
harmonics are present.
17
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
=
L v=
/ 4 f1
343 m / s
= 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 .
11.11 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  = A cos(2πf1t )
 T1 
 2π 
y 2 = A cos
t  = A cos(2πf 2 t )
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  
  f + f2  
ytotal = 2 A cos  2π  1
t  cos  2π  1

t
  2  
  2  
18
  f − f2  
The first part of the ytotal is 2 A cos 2π  1
t  which gives the slowly-varying amplitude
  2 
of the beats. Since a loud sound is heard whenever this term is 2A or –2A, the beat frequency
is f beat = f1 − f 2 . The rapid oscillations within each beat are due to the second part of ytotal,
  f + f2  
cos 2π  1
t  . Now, beats can be understood as oscillations at the average frequency,
  2 
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.
19
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