Transcript Ch17 - Georgia State University

Ch17. The Principle of Linear
Superposition and Interference Phenomena
The Principle of Linear Superposition
1
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.
2
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
3
Constructive and Destructive
Interference of Sound WavesReading
content
When two waves always meet condensation-to-condensation
and rarefaction-to-rarefaction (or crest-to-crest and trough-totrough), they are said to be exactly in phase and to exhibit
4
constructive interference.
When two waves always meet condensation-to-rarefaction
(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.
5
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.
6
Interference effects can also be detected if the two
speakers are fixed in position and the listener moves
about the room.
7
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.
8
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.
9
Diffraction
10
The bending of a wave around an obstacle or the edges of
an opening is called diffraction. All kinds of waves exhibit
diffraction.
11
12
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.
13
14
Beats
15
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.
16
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.
17
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
18
Transverse Standing Waves
19
20
The antinodes are places where maximum vibration occurs.
21
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.
22
f1 = v/(2L)
or
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.
fn l n = v
=v
fn(2L/n) = v
23
Example 4. Playing a Guitar
The heaviest string on an electric guitar has a
linear density of m/L = 5.28 × 10–3 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.
24
(a)
(b)
This length is exactly half that determined in part (a)
because the frequencies have a ratio of 2:1.
25
Conceptual Example 5.
The Frets on a Guitar
26
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.
27
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
28
Longitudinal Standing Waves
Standing wave patterns can also be formed
from longitudinal waves.
29
fn
=
v/ln
30
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
31
adjusted to play a middle C?
(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.
32
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.
33
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?
1
l , (d) lowered
(a) antinode, (b) node, (c)
4
34
Complex Sound Waves
35
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.
36
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.
37
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?
38
39
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.
40
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)
443 Hz
440.0 Hz
437 Hz
} 3-Hz beat frequency
} 3-Hz beat frequency
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
41
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 .
443 Hz
440.0 Hz
437 Hz
Tuning
fork
Original
string
} beat frequency increases
} beat frequency decreases
Tightened
string
42
Problem 32
REASONING AND
SOLUTION We are
given
fj 
12
2 f j –1
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
 v 
f 0   v 
v
v
1 





1

1 
L0  L1 

 





12
f1   2 f 0 
2 f 0 2 f1  2 f 0 
2
1 

 L0 1  12   (0.628m)(0.0561)  0.0352m
2

43
b. The frequencies corresponding to the sixth and seventh
6
7
frets are f  12 2 f and
. The distance
12
f7  2 f0
6
0
between fret 6 and fret 7 is
 
 
v
v
v
v
L6  L7 



6
7
12
12
2 f6 2 f7 2 2 f 2 2 f
0
0
 
 v  1
 
 

6
 2 f 0   12 2
 
1 
7
12
2 
   
 1
 L0 

6
 12 2
1 
 (0.628m)(0.0397)  0.0249m
7
12
2 
   
44
Problem 50
t=1s
t=2s
t=3s
t=4s
0
2
4
6
8
10
45