Standing Waves

Reflected
pulse
A wave pulse reflected
at a free end is not inverted
Free
end
g
pplyin
the
ncep
Co
1. a) Look up the speed of sound in oak (Table 13.1). Calculate the change
in wavelength for a sound generated in air by a 250 Hz tuning
fork as it enters the oak, given that the air temperature is 20°C.
b) Repeat the exercise in part a) for sound travelling from air to water.
ts
Free
end
Fig.14.8
a
Incident
pulse
An acoustical engineer is someone who deals with the impact of noise
on people. Noise affects all of us and can induce fatigue, interrupt communication, and possibly affect safety. Noise from highways, airports,
and industry is usually contained by various barriers. The design and
production of shapes and materials for barriers, along with their positioning relative to noise generators and receivers, are aspects of this job.
g
uttin
er
it all
g eth
To
Absorption, transmission
(no phase change)
ds up
Spee
reflection
v2 v1
(no phase change)
p
Fig.14.9 When Sound Waves Meet a Boundary
How does
speed change
at boundary? S
lows
Absorption, transmission
d
v2 own
(no phase change)
v1
reflection
(with phase change)
14.4 Standing Waves — A Special Case
of Interference
One of the most important instances of wave interference occurs when a
particular wave reflects back on itself (with or without phase inversion)
with the same frequency, wavelength, and speed. Figure 14.10 demonstrates
how two wave trains with the same characteristics interfere with each other
as they meet head on.
The red and blue waves in Fig. 14.10 are approaching component waves,
and the black wave shows the resultant wave pulse, calculated using the principle of superposition. The resulting wave form, called a standing wave, has
a particular pattern, which is illustrated in Fig. 14.11.
chapt e r 14 : More than Meets the Ear
487
Fig.14.10
Interference of two
similar wave forms produces a
standing wave
1
Fig.14.11
Antinode
Wave
motion
Nodes
1/ 2
Supercrests
Supertroughs
Wave
motion
1
A standing wave
For image
see student
text.
Fig.14.12
Nodes are 2 apart
1
1/ 2
1/ 2
dn
dn
A standing wave is characterized by points in which the medium does not
vibrate, called nodes. These nodes are interspersed between sections of
medium that alternate between constructive “supercrests” and “supertroughs,”
called antinodes. In longitudinal sound waves, they are called supercompressions and superrarefactions.
Figure 14.12 illustrates that the distance between successive nodes, the
inter-nodal distance (dn), is equivalent to one-half the wavelength of the
wave source.
dn 12
The number of inter-nodal distances is always one less than the number of
nodes, just as your hand has five fingers, but only four spaces between them.
In a medium such as a guitar string, the boundaries between the string
and the next medium can be considered “fixed” because a guitar string is
attached at both ends. Therefore, the standing wave is terminated at each
end with a node. Other mediums, such as a car antenna (Fig. 14.13), have
“free ends” (attached or closed at one end only). Their standing waves terminate with an antinode.
Fig.14.13 The wind can produce
a free-end standing wave in a car
radio antenna
1
3
2
3
x
(a)
488
(b)
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example 4
Standing waves in a duck pondt
A standing wave occurs in a duck pond when one duck repeatedly tries
to “jump” for food, as shown in Fig. 14.14.
(a) The nodes are at every 38 cm. What wavelength of wave is our hungry
duck producing?
Fig.14.14
1
2
38 cm
(b) If the wave speed in the pond is 0.95 m/s, how often does the duck
“jump”?
Solution and Connection to Theory
Given
dn 38 cm
v 0.95 m/s
(a) dn 12
38 cm 12
fduck ?
?
2(38 cm) 76 cm
Therefore, the wavelength of the duck’s wave is 76 cm or 0.76 m.
(b) v f
Rearranging the equation and solving for f,
0.95 m/s
v
f 1.25 Hz
0.76 m
Therefore, our duck jumps for food 1.25 times every second.
example 5
Wavelength and speed of a standing wave
A standing wave has a distance of 45 cm between four consecutive nodes.
What is the wavelength of the wave? What is the speed of the wave in the
medium if the frequency of the source is 30 Hz?
chapt e r 14 : More than Meets the Ear
489
Fig.14.15
1/ 2
1
1/ 2
2
1/ 2
3
dn
dn
4
dn
L 45 cm
Solution and Connection to Theory
Given
L 45 cm
f 30 Hz
?
v?
There are four nodes, so there must be three inter-nodal distances. The
length of the wave is therefore
L 3dn 3冢12 冣
(L)23 (45 cm)23 30 cm
Therefore, the wavelength of the standing wave is 30 cm.
For the speed,
v f 30 Hz(30 cm) 900 cm/s
ts
Co
pplyin
the
ncep
g
a
From the measured wavelength and frequency, the speed of the wave is
900 cm/s or 9.0 m/s.
1. State the conditions needed to produce standing waves.
2. a) A standing wave is produced in a vibrating car antenna as the
car moves along a slightly rough highway. The wave has three
nodes in a distance of 30 cm. Calculate the wavelength of the
standing wave.
b) Assume that the wave’s frequency is 20 Hz. Calculate the velocity of the wave.
14.5 Resonance
In the last section, we found that under the right conditions, a standing
wave can be formed in certain mediums such as a guitar string. Strings with
fixed ends may vibrate with one large loop (antinode) between them. This
implies that the string has a specific frequency with which it can vibrate.
The lowest frequency of a string is called its fundamental frequency
or first harmonic, f0. Higher frequencies of the string are integer multiples
of the fundamental frequency, 2 f0, 3 f0, etc. They are called harmonics (second, third, etc.) or overtones (first, second, third, etc.). (See Fig. 14.16.)
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Frequency f1
1st harmonic
(fundamental)
Fig.14.16
1/ 2
Frequency 2f1
2nd harmonic (1st overtone)
Frequency 3f1
3rd harmonic 3/ 2
(2nd overtone)
The first three harmonics of a vibrating string
Mechanical Resonance
On a swing, if you want to swing with greater and greater amplitude, you
have to push yourself at the right time in order to match your push frequency with the natural frequency of the swing.
Mechanical resonance is the vibrating response of an object to a periodic force from a source that has the same frequency as the natural
frequency of the object.
Figure 14.17 uses a speaker and a tuning fork to illustrate the basic
concepts of mechanical resonance. The tuning fork has a natural resonant
frequency with which it vibrates. The speaker, if tuned to emit the same
frequency, will cause the tines of the tuning fork to oscillate. If the speaker
is turned off after a few moments, we would hear the tuning fork continue
to sound as if it had been struck. One of the most famous examples of
mechanical resonance was the collapse of the Tacoma Narrows Bridge
(Washington State, 1940), shown in Fig. 14.18.
Fig.14.17
The speaker causes the
tuning fork to vibrate
ftuning fork
f
ning fork
Fig.14.18
Mechanical resonance in the Tacoma
Narrows Bridge (Washington State, 1940)
For image
see student
text.
For image
see student
text.
For image
see student
text.
chapt e r 14 : More than Meets the Ear
491
The bridge had a design flaw that gave it a natural frequency that was
coincidentally similar to the frequency of the wind gusts at that particular
spot. Large amplitude oscillations were caused in the bridge structure by
mechanical resonance and the structure crumbled under the stress. Figure
14.19 summarizes other examples of mechanical resonance.
Fig.14.19 Further examples
of mechanical resonance
Example of Mechanical Resonance
Photo
(a) Rocking a car out of an ice rut
If the people pushing the car and the driver pushing down on the accelerator do so at the natural
rocking frequency of the car, it rocks back and
forth with ever-increasing amplitude until it can
finally be pushed out of the rut.
For image
see student
text.
(b) Pushing a child on a swing
Push a child on the swing at the correct frequency,
always at the same point in the cycle, and the
child’s amplitude of swinging increases dramatically.
For image
see student
text.
(c) Pendulum clocks
ts
Co
pplyin
the
ncep
g
a
For hundreds of years, people depended on the
constant natural frequency of a swinging pendulum to control the speed of a clock. Today, the
source of vibration is often a quartz crystal.
An Example of Resonance
The frequency at which a person walks can coincide with the resonance frequency of water swishing back and forth in a bucket. If you
carry two buckets of water while walking, the amplitude of the wave
action builds up and your leg will get wet. However, if your walking frequency doesn’t match the resonant frequency of the water in the buckets, the water forms little surface ripples and becomes more stable.
For the following resonance situations, explain what is happening
and suggest a solution for getting rid of the resonance.
1. a) A truck drives down your street and the windows of your house
rattle.
b) Soldiers marching in step across a long bridge cause it to vibrate
up and down with an ever-increasing amplitude.
c) The low hum of large motors in a factory cause workers’ internal
organs, like the heart, to vibrate, making the workers feel sick.
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u n i t d: Waves and Sound
d) A tuning fork vibrating at 250 Hz causes another tuning fork
with the same frequency to also vibrate.
e) A sharp note maintained by a singer shatters a thin-walled glass.
2. Find other examples of resonance and explain how the resonance
can be eliminated.
Fig.14.20A
Joe Carter after hitting
a home run in the final game of the
1993 World Series
The Sweet Spot
Anyone who has ever hit a ball using a baseball bat knows the effect of
a stinger: the feeling in your hands caused by the vibrating bat after
impact with the ball. In some cases, the whole arm is jarred, and in
other instances, the bat may break. We have also experienced the
smooth, effortless hit. The point of contact where this type of hit
occurs is called the “sweet spot.”
If the bat hits the ball at a node (point of destructive interference), the
bat doesn’t excite any resonance modes and hence doesn’t vibrate. This hit
is smooth and effortless. If the bat hits the ball at a maximum (antinode),
then a corresponding resonance mode is excited and the bat vibrates, causing the bat to “sting” the hands. In many hits, the duration of bat-ball contact is enough to excite the fundamental and second resonance modes,
both with roughly equal amplitudes (illustrated in Fig. 14.20B). The ideal
spot to hit the ball is about halfway between the nodes of each vibration
(the sweet spot). At this spot, each resonance mode has a tiny amplitude,
thus creating an effect similar to hitting the ball at a pure node.
3. Speculate on the difference in energy transfer to the ball when the
bat hits the ball at the sweet spot as opposed to away from the sweet
spot, especially at an antinode.
4. Research the different materials that baseball bats are made of. How
would the material affect the resonance of the bat?
5. Research the various ways in which a bat is “doctored.” Do these
techniques change the location of the sweet spot on the bat?
For image
see student
text.
Fig.14.20B
d
The sweet spot
Sweet spot
Node 1
Node 2
Ball
Fundamental
mode
Second mode
14.6 Acoustic Resonance and Musical
Instruments
Acoustic resonance is the process responsible for the sound waves that come
from various musical instruments. Like mechanical resonance, the instrument is
tuned to a particular natural frequency. When stimulated by a source of vibration that has the same frequency, a standing resonant wave is set up in the
instrument and large amplitude oscillations result. We hear these oscillations as
the characteristic “note” that is being played. The best way to examine acoustic
resonance is to look at two different types of musical instruments that support
standing waves in different ways. We will look at wind instruments and stringed
instruments that support standing waves in air columns and strings respectively.
chapt e r 14 : More than Meets the Ear
493
Wind Instruments — Standing Waves in Air Columns
Fig.14.21
A bottle acts like a closed
air column
If you have ever taken a breath and blown air over the opening of a pop bottle to make a sound like a fog horn then you have experienced resonance in
air columns (see Fig. 14.21).
Wind instruments include traditional brass or woodwind instruments,
such as a trumpet or clarinet. Figure 14.22 illustrates how the vibration source
for a brass instrument is the musician blowing air through vibrating compressed lips. A thin sliver of wood, called a reed, does the same for a woodwind.
Fig. 14.22
Vibrations in a brass instrument
Brass instrument
Mouthpiece
No matter what the source of vibration, these instruments are nothing
more than elaborate air columns in which a standing wave is formed. It is
important to note that it is virtually impossible to model the behaviour of
longitudinal sound waves with pictures. For all intents and purposes, longitudinal wave compressions and rarefactions will be modelled with crests
and troughs respectively, using transverse waves. Figure 14.23 uses a tuning fork to show how standing waves can be set up in open and closed air
columns. Note that sound waves reflect at closed and open ends in the same
way that waves in springs reflect at fixed and free ends.
Fig.14.23
The relationship between
transverse and longitudinal waves
Open
end
(a)
Before reflection
Compression
(longitudinal wave)
During reflection
Reflected pulse
(same phase)
After reflection
494
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Closed
end
(b)
Before reflection
Wave pulse
(transverse wave)
During reflection
Reflected pulse
(opposite phase)
After reflection
The waves that are produced migrate from the source to the open end of the
air column, where some of their energy is transmitted. Much of the wave
energy is reflected back inside the air column, without any phase change. This
type of reflection is the same as that witnessed in solid materials, such as the
car antenna (see Fig. 14.13). In closed air columns where the sound wave meets
a “fixed” end, there is reflection in the opposite phase. Either type of reflected
wave meets more waves from the source. Acoustic resonance is achieved when
a standing wave is formed inside the air column and the entire instrument
begins to vibrate. Figure 14.24 illustrates how standing waves look in open and
closed air columns.
f 1536 Hz
11/2 dn
First resonant
length
f 1536 Hz
3/ 4
3/ 4
dn
f 512 Hz
Third resonant length
dn
1/ 4
f 512 Hz
Second resonant length
dn
Second resonant
length
1
1
f 512 Hz
Same length, different wavelength and frequency
Second resonant length
First resonant
length
Second resonant
length
f 1024 Hz
Third resonant
length
Same length, different wavelength and frequency
1/ 2
f 512 Hz
Third resonant length
f 512 Hz
f 512 Hz
Resonant lengths and
frequencies for open and closed
air columns
Closed Same frequency and wavelength, different resonant length
dn
11/4 dn
f 2560 Hz
Third resonant
length
Open Same frequency and wavelength, different resonant length
Fig.14.24
11/4 11/2 From the know your “knodes” rule, you can easily see that when
standing waves are created, only a few wavelengths are possible. This means
that the air column is also tuned to resonate with only a few different frequencies. If we recall that the internodal distance for any standing wave, dn,
is equal to 12, then the specific resonant lengths of the air column can be
compared to the wavelength, as shown in Table 14.1.
chapt e r 14 : More than Meets the Ear
KNOW YOUR “KNODES”
Standing wave patterns have a node
at any closed end and an antinode
at any open end of the medium in
which they travel.
495
Table 14.1
Lengths of Air Columns
Closed Air Column
Number of
nodes (n)
Number of
internodal
distances
Length of
column in
wavelengths ()
Number of
internodal
distances
Length of
column in
wavelengths ()
First
1
1
2
1
4
1
1
2
Second
2
3
2
3
4
2
1
Third
3
5
2
5
4
3
3
2
Fourth
4
7
2
7
4
4
2
Based on n
2n 1
2
(2n – 1)
4
n
n
2
Resonant
length
General
statement
example 6
Recall that the distance between
1
consecutive nodes, dn, is 2.
Open Air Column
Finding wavelength
An air column that is open at both ends is 1.50 m long. A specific frequency is heard resonating from the column. What is the longest wavelength and its associated frequency that could be responsible for this
resonance? The speed of sound is 345 m/s.
Solution and Connection to Theory
Given
The longest wave occurs at the first resonant length, where n 1.
Therefore, L 12
Fig. 14.25
1/ 4
1/ 4
L 1.50 m
1
2
1.50 m 2(1.50 m) 3.00 m
Therefore, the maximum wavelength is 3.00 m.
v f
345 m/s
v
f 115 Hz
3.00 m
Therefore, the frequency of this wave is 115 Hz.
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v 345 m/s
example 7
Finding frequency
A closed air column resonates at two consective lengths of 94.0 cm and
156 cm. If the speed of sound is 350 m/s, what is the resonant frequency
of the air column?
Solution and Connection to Theory
Given
L1 94.0 cm
L2 156 cm
v 350 m/s
The difference in length between any two resonant lengths is always 12.
1
2
L2 L1 156 cm 94.0 cm 62 cm or 0.62 m
2(0.62 m) 1.24 m
350 m/s
v
Finally, f 282 Hz
1.24 m
Therefore, the resonant frequency of the column is 282 Hz.
d
m
etho
pr
o ces
Fig.14.26 Solving Problems With Air Columns
Set
dn (1/2) L
How many internodal
distances, dn?
YES
Givens
L or v or T, f
v 332 0.6T
and f
Draw standing
wave diagram
Are you
comparing
more than one
resonant
length?
s
of
NO
Which resonant
length (n)?
n
Ope
mn?
colu
Clos
ed
colu
mn?
Set
n
L
2
Set
(2n 1)
L
4
Universal wave
v
equation f
Wind instruments change their frequency in two ways. They can have the
lengths of their air columns adjusted so they are tuned to different fundamental frequencies, or they can have the source vibrate at another resonant
frequency for a particular air column length. In brass instruments, such as
a trumpet (Fig. 14.27A) or tuba, the musician pushes valves that open passages to more tubing, which effectively lengthens the instrument. A trombone (Fig. 14.27B) has a free-moving slide that allows the musician to
continually adjust the length of the air column and frequency of the sound.
chapt e r 14 : More than Meets the Ear
497