Ch17. The Principle of Linear Superposition and Interference Phenomena

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Ch17. The Principle of Linear Superposition and
Interference Phenomena
The Principle of Linear Superposition
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THE PRINCIPLE OF LINEAR SUPERPOSITION When two or
more waves are present simultaneously at the same
place, the resultant disturbance is the sum of
the disturbances from the individual waves.
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Check Your Understanding 1
The drawing shows two pulses traveling toward
each other at t 0 s. Each pulse has a constant
speed of 1 cm/s. When t 2 s, what is the height
of the resultant pulse at (a) x 2 cm, (b) x 4
cm, and (c) x 6 cm?
(a) 0 cm, (b) -2 cm, (c) 2 cm
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Constructive and Destructive Interference of
Sound WavesReading content
When two waves always meet condensation-to-condens
ation and rarefaction-to-rarefaction (or
crest-to-crest and trough-to-trough), they are
said to be exactly in phase and to exhibit
constructive interference.
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When two waves always meet condensation-to-rarefac
tion (or crest-to-trough), they are said to be
exactly out of phase and to exhibit destructive
interference.
In either case, this means that the wave patterns
do not shift relative to one another as time
passes. Sources that produce waves in this
fashion are called coherent sources.
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Destructive interference is the basis of a useful
technique for reducing the loudness of
undesirable sounds.
 ,   ,   
For two wave sources vibrating in phase, a
difference in path lengths that is zero or an
integer number (1, 2, 3,...) of wavelengths leads
to constructive interference a difference in
path lengths that is a half-integer number (
, , , ) of wavelengths leads to
destructive interference.
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Interference effects can also be detected if the
two speakers are fixed in position and the
listener moves about the room.
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Example 1.  What Does a Listener Hear?
Two in-phase loudspeakers, A and B, are separated
by 3.20 m. A listener is stationed at point C,
which is 2.40 m in front of speaker B. The
triangle ABC is a right triangle. Both speakers
are playing identical 214-Hz tones, and the speed
of sound is 343 m/s. Does the listener hear a
loud sound or no sound?
The listener hears a loud sound.
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Conceptual Example 2.  Out-of-Phase Speakers
To make a speaker operate, two wires must be
connected between the speaker and the receiver
(amplifier. To ensure that the diaphragms of two
speakers vibrate in phase, it is necessary to
make these connections in exactly the same way.
If the wires for one speaker are not connected
just as they are for the other speaker, the two
diaphragms will vibrate out of phase. Whenever
one diaphragm moves outward, the other will move
inward, and vice versa. Suppose that in Figures
17.3 and 17.4 the connections are made so that
the speaker diaphragms vibrate out of phase,
everything else remaining the same. In each case,
what kind of interference would result at the
overlap point?
In Figure 17.3 a rarefaction from the left now
meets a condensation from the right at the
overlap point, and destructive interference
results. In Figure 17.4, a condensation from the
left now meets a condensation from the right at
the overlap point, leading to constructive
interference.
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Diffraction
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The bending of a wave around an obstacle or the
edges of an opening is called diffraction. All
kinds of waves exhibit diffraction.
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Example 3.   Designing a Loudspeaker for Wide
Dispersion
A 1500-Hz sound and a 8500-Hz sound each emerges
from a loudspeaker through a circular opening
whose diameter is 0.30 m . Assuming that the
speed of sound in air is 343 m/s, find the
diffraction angle q  for each sound.
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Beats
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The number of times per second that the loudness
rises and falls is the beat frequency and is the
difference between the two sound frequencies.
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A 10-Hz sound wave and a 12-Hz sound wave, when
added together, produce a wave with a beat
frequency of 2 Hz. The drawings show the pressure
patterns (in blue) of the individual waves and
the pressure pattern (in red) that results when
the two overlap. The time interval shown is one
second.
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Check Your Understanding 2
A tuning fork of unknown frequency and a tuning
fork of frequency of 384 Hz produce 6 beats in 2
seconds. When a small piece of putty is attached
to the tuning fork of unknown frequency, the beat
frequency decreases. What is the frequency of
that tuning fork?
387 HZ
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Transverse Standing Waves
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The antinodes are places where maximum vibration
occurs.
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The nodes are places that do not vibrate at all.
The frequencies in this series (f1, 2f1, 3f1,
etc.) are called harmonics.
Frequencies above the fundamental are overtones.
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Standing waves arise because identical waves
travel on the string in opposite directions and
combine in accord with the principle of linear
superposition. A standing wave is said to be
standing because it does not travel in one
direction or the other, as do the individual
waves that produce it.
f1 v/(2L)
fnln v
v
or
fn(2L/n) v
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Example 4.   Playing a Guitar
The heaviest string on an electric guitar has a
linear density of m/L 5.28 103 kg/m and is
stretched with a tension of F 226 N. This
string produces the musical note E when vibrating
along its entire length in a standing wave at the
fundamental frequency of 164.8 Hz. (a) Find the
length L of the string between its two fixed ends
. (b) A guitar player wants the string to vibrate
at a fundamental frequency of 2 164.8 Hz
329.6 Hz, as it must if the musical note E is to
be sounded one octave higher in pitch. To
accomplish this, he presses the string against
the proper fret and then plucks the string . Find
the distance L between the fret and the bridge of
the guitar.
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(a)
(b)
This length is exactly half that determined in
part (a) because the frequencies have a ratio of
21.
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Conceptual Example 5.  The Frets on a Guitar
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The frets on the neck of a guitar. They allow the
player to produce a complete sequence of musical
notes using a single string. Starting with the
fret at the top of the neck, each successive fret
indicates where the player should press to get
the next note in the sequence. Musicians call the
sequence the chromatic scale, and every
thirteenth note in it corresponds to one octave,
or a doubling of the sound frequency. The spacing
between the frets is greatest at the top of the
neck and decreases with each additional fret
further on down. The spacing eventually becomes
smaller than the width of a finger, limiting the
number of frets that can be used. Why does the
spacing between the frets decrease going down the
neck?
D1 is greater than D2 , and the frets near the
top of the neck have more space between them than
those further down.
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Check Your Understanding 3
A standing wave that corresponds to the fourth
harmonic is set up on a string that is fixed at
both ends. (a) How many loops are in this
standing wave? (b) How many nodes (excluding the
nodes at the ends of the string) does this
standing wave have? (c) Is there a node or an
antinode at the midpoint of the string? (d) If
the frequency of this standing wave is 440 Hz,
what is the frequency of the lowest-frequency
standing wave that could be set up on this string?
(a) 4, (b) 3, (c) node, (d) 110 HZ
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Longitudinal Standing Waves
Standing wave patterns can also be formed from
longitudinal waves.
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f n v/ln
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Example 6.   Playing a Flute
When all the holes are closed on one type of
flute, the lowest note it can sound is a middle
C, whose fundamental frequency is 261.6 Hz. (a)
The air temperature is 293 K, and the speed of
sound is 343 m/s. Assuming the flute is a
cylindrical tube open at both ends, determine the
distance L , the distance from the mouthpiece to
the end of the tube. (This distance is only
approximate, since the antinode does not occur
exactly at the mouthpiece.)
(b) A flautist can alter the length of the flute
by adjusting the extent to which the head joint
is inserted into the main stem of the instrument.
If the air temperature rises to 305 K, to what
length must the flute be adjusted to play a
middle C?
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(a)
(b)
v305 K 1.02(v293 K) 1.02(343 m/s) 3.50
102 m/s
Thus, to play in tune at the higher temperature,
a flautist must lengthen the flute by 0.013 m.
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Standing waves can also exist in a tube with only
one end open. Here the standing waves have a
displacement antinode at the open end and a
displacement node at the closed end, where the
air molecules are not free to move.
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Check Your Understanding 4
A cylindrical bottle, partially filled with
water, is open at the top. When you blow across
the top of the bottle a standing wave is set up
inside it. Is there a node or an antinode (a) at
the top of the bottle and (b) at the surface of
the water? (c) If the standing wave is vibrating
at its fundamental frequency, what is the
distance between the top of the bottle and the
surface of the water? Express your answer in
terms of the wavelength l of the standing wave.
(d) If you take a sip, is the fundamental
frequency of the standing wave raised, lowered,
or does it remain the same?
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Complex Sound Waves
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The sound wave corresponding to a note produced
by a musical instrument or a singer is called a
complex sound wave because it consists of a
mixture of the fundamental and harmonic
frequencies.
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Concepts Calculations Example 7.  Diffraction
in Two Different Media
A sound wave with a frequency of 15 kHz emerges
through a circular opening that has a diameter of
0.20 m. Find the diffraction angle   when
the sound travels in air at a speed of 343 m/s
and in water at a speed of 1482 m/s.
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Concepts Calculations Example 8.  Standing
Waves of Sound
Two tubes of gas are identical and are open at
only one end. One tube contains neon (Ne) and the
other krypton (Kr). Both are monatomic gases,
have the same temperature, and may be assumed to
be ideal gases. The fundamental frequency of the
tube containing neon is 481Hz. What is the
fundamental frequency of the tube containing
krypton?
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Problem 1
REASONING AND SOLUTION In a time of t 1 s,
the pulse on the left has moved to the right a
distance of 1 cm, while the pulse on the right
has moved to the left a distance of 1 cm. Adding
the shapes of these two pulses when t 1 s
reveals that the height of the resultant pulse is
a. 2 cm at x 3 cm
b. 1 cm at x 4 cm.
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Problem 18
REASONING The beat frequency is the difference
between two sound frequencies. Therefore, the
original frequency of the guitar string (before
it was tightened) was either 3 Hz lower than that
of the tuning fork (440.0 Hz ? 3 Hz 337 Hz) or
3 Hz higher (440.0 Hz 3 Hz 443 Hz)
To determine which of these frequencies is the
correct one (437 or 443 Hz), we will use the
information that the beat frequency decreases
when the guitar string is tightened
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SOLUTION When the guitar string is tightened,
its frequency of vibration (either 437 or 443 Hz)
increases. As the drawing below shows, when the
437-Hz frequency increases, it becomes closer to
440.0 Hz, so the beat frequency decreases. When
the 443-Hz frequency increases, it becomes
farther from 440.0 Hz, so the beat frequency
increases. Since the problem states that the beat
frequency decreases, the original frequency of
the guitar string was 437 HZ .
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Problem 32
REASONING AND SOLUTION We are given
a. The length of the unfretted string is L0
v/(2f0) and the length of the string when it is
pushed against fret 1 is L1 v/(2f1). The
distance between the frets is
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b. The frequencies corresponding to the sixth and
seventh frets are and
. The distance
between fret 6 and fret 7 is
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Problem 50