Note 20
Superposition
Sections Covered in the Text: Chapter 21, except 21.8
In Note 18 we saw that a wave has the attributes of
displacement, amplitude, frequency, wavelength and
speed. We can now consider the effects that occur
when waves come together at the same point in space,
that is, when waves undergo the phenomena of superposition and interference. We shall see that such studies
shed light on, among other things, the production of
sound by musical instruments.
The “interfering” pulses move independently, and do
not affect one another, or physically change one
another, in any way.
The Principle of Superposition
It is possible for two or more waves to coexist at the
same point in space. When this happens it is important to know what the total wave amplitude is at that
point. This question is answered by the Principle of
Superposition. The principle may be stated in these
words:
When two or more waves are simultaneously present
at a single point in space, the displacement of the medium at that point is the sum of the displacements due
to each individual wave.
This simple, almost intuitive, principle can be shown
to apply to all common waves, be they mechanical,
sound or electromagnetic. A consequence is that the
various waves excited on a stretched string, for
example, are independent of one another. Two pulses
traveling in opposite directions on the string can pass
through one another and reappear on the other side
without being altered in any way.
This idea is illustrated in Figure 20-1 with graphically-constructed “wave pulses” of triangular and
rectangular shape. A triangular wave pulse is shown
moving to the right, a rectangular wave pulse moving
to the left, both at the speed of 1 m.s–1. Their positions
are shown in 5 frames—at the clocktimes of 0s, 1s, 2s,
3s and 4s. You can see that frames 3 and 4 show the
pulses in the process of passing through one another.
Frame 5 shows the pulses of frame 1 having passed
through one another without changing one another.
Let us focus on frames 3 and 4. When the amplitudes
of the two pulses have the same sign, the resultant
amplitude or displacement of the medium increases
as the two pulses pass through one another. This is
called constructive interference. What the figure doesn’t
show is that when the amplitudes have opposite signs
the resultant amplitude decreases as the two pulses
pass through one another. This is called destructive
interference. Notice that the meaning of the word
interference used here differs from its everyday usage.
Figure 20-1. Illustration of the principle of superposition using mathematical constructions of wave pulses on a line.
Wave pulses are complicated entities to describe
mathematically (and to produce experimentally). It is
simpler to deal with sinusoidal waves, which we do in
the next section.
Standing Waves
One could, at least in principle, induce two sinusoidal
waves to move in opposite directions and to pass
through one another by moving both ends of a very
long string up and down continuously. We begin by
considering the wave in qualitative terms. Then we
shall apply mathematics to the wave.
The transverse displacement of the string at any
point along the string would appear as shown in
Figure 20-2. This wave is called a standing wave
20-1
Note 20
because the crests and troughs “stand in place” as the
string oscillates. There are points on the string that do
not move. These points, spaced λ/2 apart, are called
nodes. Halfway between the nodes are points where
the string particles oscillate with maximum displacement. These points are called antinodes. The nodes of a
standing wave are points of destructive interference
where the waves are out of phase (phase difference is
π); antinodes are points of constructive interference
where the waves are in phase (phase difference is 0).
Now that we understand some qualitative aspects of a
standing wave, let us apply mathematics to the wave.
The Mathematics of a Standing Wave
Consider two waves that are identical in all respects
(same amplitude a and angular frequency ω) traveling
in opposite directions (right and left) on a line. From
what we know about the form of a traveling wave
(Note 18) we can write the wave disturbances as 1
DR = asin( kx − ωt ) and
DL = asin( kx + ωt ) .
…[20-1]
Applying the principle of superposition the resultant
of the two waves is€
€
D(x,t) = DR + DL = 2asin kx cos ωt . …[20-2]
You should be able to show that eq[20-2] follows
using the trigonometric identity
€
Figure 20-2. A graphical representation of a standing wave
on a string being vibrated at both ends. Shown are two
snapshots of the wave at two instants of clocktime.
Recall from Note 18 that the intensity I of a wave is
proportional to the square of the wave amplitude (I =
CA2). A graph of I(x) for the wave in Figure 20-2 is
shown in Figure 20-3. I(x) is a maximum at positions
of antinodes, a minimum at positions of nodes.
sin(α ± β ) = sin α cos β ± cosα sin β .
It is instructive to write eq[20-2] in the form
D(x,t) = A(x)cosωt
€
where
A(x) = 2asin kx .
…[20-3]
…[20-4]
A(x)€is the amplitude of the sinusoidal function.
Notice that eq[20-2] does not contain either of the
factors€
(kx–ωt) or (kx+ωt) that characterize the traveling waves of eq[20-1]. Eq[20-2] does not therefore
represent a traveling wave; it represents a standing
wave. The wave is “standing” in the sense of moving
neither to the right nor left. The “cosωt” factor shows
that every point x of the medium vibrates in simple
harmonic motion with frequency f = ω /2 π and with
the amplitude A(x). Several snapshots of the wave are
shown in Figure 20-4.
The nodes of the standing wave are the points at
which the amplitude is zero. They are located at
positions xm for which
A(x m ) = 2asin kx m = 0 ,
…[20-5]
or for which
1
Figure 20-3. A graphical representation of a standing wave
showing also the intensity I as a function of x.
20-2
€
We consider here a general, hypothetical line and use the letter
D to stand for a wave displacement that includes both transverse
and longitudinal displacements in any medium. Later, we shall
restrict our attention to transverse waves on a string or longitudinal
sound waves.
Note 20
Transverse Standing Waves
Figure 20-4. The net displacement resulting from two sinusoidal waves moving in opposite directions.
kx m =
2πx m
= mπ
λ
m = 0,1,2
…[20-6]
Thus the position x m of the mth node is
€
xm = m
λ
m = 0,1,2…
2
The procedure of setting up a standing wave on a
long string by moving both ends of it up and down
continuously is not a practical one. At least one end of
the string must be held fixed. To understand how a
standing wave might arise in this case it is necessary
to understand what happens when a traveling wave
encounters a boundary or a discontinuity between
two media.
The media in question might be the two strings of
different linear density (mass per unit length) joined
together as shown in Figure 20-5a. On the left of the
figure is shown a string with a larger linear density,
while on the right is one with a smaller linear density.
The junction between the strings is the discontinuity.
The tension in both strings is the same, so according
to eq[18-25] the wave speed is slower in the left
medium, faster in the right.
…[20-7]
where m is an integer.
By the same token, the antinodes of the standing
wave€are the points at which the amplitude is a
maximum. A(x) has the maximum value of Amax = 2a
at points xm where sinkx m = 1, or points xm such that
kx m =
2πx m 
1
=  m + π

λ
2
m = 0,1,2 …[20-8]
Figure 20-5a. A wave pulse is reflected and transmitted at a
discontinuity where the wave speed increases.
Thus the position x m of the mth antinode is
€

1 λ
xm = m + 
m = 0,1,2… …[20-9]

2 2
It is clear that antinodes lay halfway between nodes.
The intensity of the wave is proportional to the
square
€ of the wave amplitude, that is,
2
I(x) = [ A(x)] = 4a 2 sin 2 kx
= Imax sin 2 kx .
…[20-10]
€ function I(x) is consistent with the graph of
The
intensity vs x in Figure 20-3. The intensity is a maximum at €
values of x for which eq[20-9] holds, that is, at
the positions of the antinodes. As we have seen, the
intensity is a minimum at the positions of the nodes.
When a traveling wave (pulse) encounters the
junction from the left, some of the wave’s energy is
transmitted into the right medium, and some is reflected back into the left medium. The larger fraction of
the energy is transmitted (indicated by the pulse of
greater amplitude). However, neither the transmitted
nor the reflected pulse has the same amplitude as the
incident pulse.
In Figure 20-5b, an incident wave pulse encounters a
discontinuity at which the wave speed decreases. As
before, some of the wave’s energy is transmitted and
some is reflected. But this time the reflected pulse is
inverted. By this is meant a displacement D on the
incident wave becomes displacement –D on the reflected wave. Because sin(φ + π) = – sinφ, the reflected
wave has a phase change of π upon reflection.
The wave in Figure 20-5c reflects from a boundary,
not a discontinuity. The medium to the right of the
boundary can be thought of as a string medium of
infinite linear density. The reflected wave is again
20-3
Note 20
inverted, but there is no transmitted wave at all. All
the wave’s energy is reflected. And the wave speed
does not change.
Figure 20-5b. A wave pulse is reflected at a discontinuity
where the wave speed decreases.
and wavelength. Once reflected at the boundaries, the
traveling waves pass through one another and
produce a standing wave, as we have already seen.
Figure 20-6. A way of setting up a standing wave on a
stretched string tied at both ends.
The wave on the string is subject to boundary
conditions. Because the string is tied at both ends, the
displacements at x = 0 and x = L must be zero at all
times. In other words nodes must exist at these points.
Thus the boundary conditions are
D(x = 0,t) 
 = 0.
D(x = L,t)
Figure 20-5c. A wave pulse is totally reflected at a boundary. There is no transmitted wave. The wave speed does not
change.
We are now ready to apply these observations to
understand the production of standing waves on a
stretched string.
Recall that the displacement of a standing wave is
D(x,t) = (2asinkx)cosωt. This equation already satisfies
the first
€ boundary condition, eq[20-11a]. The second
boundary condition, eq[20-11b] is satisfied at all times
if
2asin kL = 0 .
…[20-12]
This is true if sinkL = 0, which in turn requires
€
Standing Waves on a Stretched String
The difference between a hypothetical line and a
stretched string is that a string must be supported.
Consider a string of length L rigidly tied at x = 0 and x
= L (Figure 20-6). In principle, we can set up a
standing wave on the string by “wiggling” or
continuously plucking the string. Traveling sinusoidal
waves are created which then travel in both
directions. They encounter the boundary at either end
and reflect. The speed of the reflected waves does not
change and therefore the wavelength and frequency
of the reflected sinusoidal waves does not change. In
addition, the reflected waves are of equal amplitude
20-4
…[20-11a, b]
kL =
2πL
= mπ m=1,2,3…
λ
…[20-13]
m = 0 is excluded because L cannot be zero. Eq[20-13]
can be satisfied only by values λm such that
€
λm =
2L
m
m=1,2,3…
…[20-14]
The natural frequencies of vibration f m of the string, in
Hz, are
€
€
fm =
v  vm 
=   m = 1, 2, 3… …[20-15]
λ  2L 
Note 20
Using the result for the speed of waves on a stretched
string under tension T derived in Note 18: v = (T/µ)1/2 ,
where T is the tension and µ is the mass per unit
length, eq[20-15] becomes
Let us consider a simple example.
Example Problem 20-1
A Standing Wave on a Stretched String
m T
fm =  
m = 1, 2, 3, … …[20-16]
 2L  µ
A string of length 2.50 m vibrates with a 100 Hz
standing wave. There are nodes 1.00 m and 1.50 m
from one end of the string, with no nodes in between.
Which harmonic is this, and what is the string’s
fundamental frequency?
The fact that the values fm depend on integers m is the
same as saying that the frequencies are quantized. That
is to
€ say, the string can support only certain discrete
frequencies
f m = mf1 m = 1,2,3…
Solution:
The standing wave looks like Figure 20-2. If there are
no nodes between 1.00 m and 1.50 m, then the node
spacing is λ/2 = 0.50 m. The number of half wavelengths that can fit into the string’s length L = 2.50 m
is m = 2.50/(1/2) = 5. This is the mode number, which
means that 100 Hz is the fifth harmonic. The fundamental frequency is therefore f1 = 100/5 = 20 Hz.
…[20-17]
f1 is called the fundamental frequency. The frequencies
f2, f 3, etc are called overtones or harmonics. Figure 20-7
shows €
the fundamental and three harmonics. Standing waves on a stretched string are a model of the
source of sound in all stringed musical instruments.
These include the piano, violin and cello amongst
others. 2
Standing Electromagnetic Waves
Standing electromagnetic waves also exist, as for
example in the cavity of a laser (Figure 20-8). Mirrors
at either end of the cavity play the role of boundary
reflectors, analogous to the boundaries at the ends of a
string tied at both ends.
Figure 20-8. A laser cavity contains a standing electromagnetic wave.
The boundary conditions on the electromagnetic wave
are the same as on a wave on a string, eqs[20-11], and
so eqs[20-14] and [20-16] also apply.
A typical laser cavity has a length L ≈ 30 cm and
operates in the red area of the electromagnetic
spectrum with λ ≈ 600 nm. The mode number of the
standing wave is, from eq[20-14]:
Figure 20-7. Standing mechanical waves on a stretched
string. Only the first 4 modes of vibration are shown here.
m=
2
This is the first example of the idea of quantization we have
encountered in this course. The concept is at the heart of quantum
physics. It will be discussed in some detail in PHYA21H3S.
2L 2 x 0.30 m
=
= 1,000,000 .
λ 6.00 x 10−7 m
The standing wave has about one million nodes! This
€
20-5
Note 20
is made possible by the very short wavelength of
light.
Various wind instruments (organ, flute, clarinet, etc)
produce musical sounds by virtue of the resonances of
air vibrating in a pipe or tube structure made of wood
or metal. The physics of tube structures we consider
next.
Conversely, where the tube is open to the atmosphere
an antinode must exist. The first three modes for the
three cases—a standing wave in a tube closed at both
ends (closed-closed), open at both ends (open-open),
and closed at one end (closed-open)—are illustrated
in Figures 20-10.
Standing Sound Waves
and Musical Acoustics
Air confined in a column or tube can support a
standing sound wave (Figure 20-9). We have seen in
Note 18 that a sound wave is a longitudinal wave; the
direction of alternate compressions and rarefactions of
air is in the direction of the tube axis. The point where
the tube is closed or capped must be the position of a
node (where the vibration of the air molecules is
minimum). The figure shows that the graphical representation of the m = 2 mode is the same as the m = 2
mode of a standing wave on a string. Keep in mind
that the representation is of a longitudinal displacement NOT a transverse displacement.
Figure 20-9. The m = 2 longitudinal standing wave inside a
closed column of air. The graphical representation is of a
longitudinal NOT a transverse displacement.
Sound produced by the setup in Figure 20-9 is of only
academic interest because it lacks the means to allow
the sound wave to leave the tube and reach the
listener (presumably outside the tube!). Thus practical
tubes are left open at one or both ends.
As we have stated earlier, at the point where the
tube is closed a node must exist in the standing wave.
20-6
Figures 20-10a and b. Standing sound waves in two types of
tube: (a) closed at both ends and (b) open at both ends.
Note that the standing wave in the open-open and
closed-closed tubes is the same except for a phase
inversion. In both cases there are m half-wavelengths
between the ends. Thus the wavelengths and frequencies of an open-open tube and a closed-closed tube are
Note 20
the same as those of a string tied at both ends:
Example Problem 20-2
The Length of an Organ Pipe
(open-open, closed-closed tube)

2L
 λm = m

m = 1,2,3…
 f m = m v = mf1

2L
…[20-18]
A tube closed at one end and open at the other end is
quite a different matter (Figure 20-10c). The funda€mental mode has only one-quarter of a wavelength in
a tube of length L, hence the m = 1 wavelength is λ1 =
4L. This is twice the λ1 wavelength of an open-open or
a closed-closed tube. Consequently, the fundamental
frequency of an open-closed tube is half that of an
open-open or a closed-closed tube of the same length.
An organ pipe open at both ends sounds its second
harmonic at a frequency of 523 Hz. What is the length
of the pipe?
Solution:
The second harmonic is the m = 2 mode, which for an
open-open tube has the frequency (eq[20-18]):
f2 = 2
Thus the length of the organ pipe is
L=
€
v
.
2L
v€ 343 m.s−1
=
= 0.656 m = 65.6 cm .
f2
523 Hz
Musical Instruments
We have seen then that the equation that governs the
sound produced by stringed instruments is
f1 =
Figure 20-10c. Three modes in an open-closed tube.
The possible wavelengths and frequencies of the
sound produced by an open-closed tube of length L
are:
(open-closed tube)

4L
 λm = m

m = 1,3,5… …[20-19]
 f m = m v = mf1

4L
Let us consider an example of an open-open organ
pipe.
€
v
1 T
=
.
2L 2L µ
…[20-20]
where T is the tension, L is the length and µ is the
mass per unit length of the string. The sound produced€ by a string is primarily at the fundamental
frequency f1 though some sound is also produced at
harmonic frequencies.3 This means that the note produced can be varied by varying L, T or µ.
Some instruments such as the piano are equipped
with strings of various L and µ. Clearly, the longer
and the fatter the string, the lower is the note
produced. Similar arguments can be made for
explaining the operation of many other stringed
instruments. We leave this subject to the interested
reader.
We have therefore seen some of the effects that occur
when two traveling waves of the same frequency
3
Many instruments can produce a middle “C”. However, it is
immediately evident to the average listener what type of
instrument produced the note. This is because different types of
instruments produce a different mix of harmonic frequencies in
addition to the fundamental frequency.
20-7
Note 20
move in opposite directions and interfere with one
another to produce a standing wave. We now move
on to the case of the interference of two traveling
waves moving in the same direction.
Interference in One Dimension
Two hypothetical schemes for obtaining two traveling
waves of the same frequency moving in the same
direction along a line are illustrated in Figures 20-11.
Suppose the waves have the same amplitude a, the
same wave number k and the same angular frequency
ω. We wish to establish what happens when the two
waves overlap.
where φ1 and φ2 are the phases of the waves.
The phase constants φ10 and φ20 are characteristics of
the sources of the waves, not the medium. For further
reinforcement of this idea, snapshot graphs at t = 0 of
waves emitted by three sources with phase constants
φ0 = 0 rad, φ 0 = π/2 rad and φ0 = π rad are drawn in
Figures 20-12. The phase constant determines what
the source is doing at t = 0. For example, Figure 20-12a
shows the meaning of a phase constant of 0 rad. This
would apply to a loudspeaker at its center position
while moving backward at t = 0. Similar arguments
can be applied to explain the meaning of the other
phase constants in the figure.
Figure 20-11a. A scheme for obtaining two traveling
electromagnetic waves of the same frequency moving in the
same direction.
Figure 20-11b. A scheme for obtaining two traveling sound
waves of the same frequency moving in the same direction.
We begin by considering the resultant wave at the
point of detection indicated in Figure 20-11b. The two
superposed waves, at distances x1 and x 2 from their
sources, can be written
D1(x1,t) = asin( kx1 − ωt + φ10 ) = asin φ1
…[20-21]
D2 (x 2 ,t) = asin( kx 2 − ωt + φ 20 ) = asin φ 2
€
€
20-8
Figure 20-12. Waves from three sources having phase constants φ0 = 0 rad, φ0 = π/2 rad and φ0 = π rad.
Two important special cases of the overlapped waves
are shown in Figures 20-13. Figure 20-13a shows the
crests and the troughs of the two waves aligned as
they travel along the x-axis. Figure 20-13b shows the
crests of one wave aligned with the troughs of the
other wave as they travel. Since the waves have the
same speed the relative position of the crests and
troughs remain fixed at all times. The position of the
Note 20
wave fronts are indicated in the middle panel of the
figures.
The two waves of Figure 20-13a have the same
displacement at every point and therefore the same
phase. That is, φ2 = φ1, or more precisely φ2 = φ1 ± 2πm
where m is an integer. These waves are in phase. The
resultant displacement A = 2a. This is maximum
constructive interference.
ence. The displacement of the waves is such that D1 =
–D2. Thus the net displacement of the resultant traveling wave is zero at every point along the axis. This
special case of interference (where the resultant amplitude is zero) is called perfect destructive interference.
The Phase Difference
The phases of the two waves are
φ1 = kx1 − ωt + φ10
φ 2 = kx 2 − ωt + φ 20 ,
…[20-22]
and €
the phase difference ∆φ is
Δφ €
= φ 2 − φ1 = ( kx 2 − ωt + φ 20 ) − ( kx1 − ωt + φ10 )
= k ( x 2 − x1 ) + (φ 20 − φ10 )
€
=
€
2π
Δx + Δφ 0 .
λ
…[20-23]
You can see that the phase difference consists of two
contributions: a phase difference due to
€
1 a path-length difference, and
2 a difference between the inherent phases of the
sources.
To see this note that ∆x = x 2 – x 1 is the difference
between the distances traveled by the waves from the
sources to the point of detection. This is called the
path-length difference. Obviously, if the sources are at
the same position in space then ∆x = 0 and ∆φ = ∆φ0.
The second contribution is due to the difference in
the inherent phase constants of the sources: ∆φ0 = φ20 –
φ10. If the sources are in phase (i.e., are identical) then
∆φ0 = 0. If the sources are at the same position in space
and are also in phase, then ∆φ = 0 and the waves
reaching the point of detection are also in phase.
The most general condition for maximum constructive interference is
Δφ = 2π
Figure 20-13. Constructive and destructive interference of
two waves traveling along the x-axis.
€
In Figure 20-13b the waves are 180˚ out of phase. This
alignment of the waves produces destructive interfer-
Δx
+ Δφ 0 = 2mπ rad m = 0,1… …[20-24]
λ
For identical sources that have ∆φ0 = 0 rad, maximum
constructive interference occurs when ∆x = m λ. That
is, two identical sources produce maximum constructive interference when the path-length difference is an
integer number of wavelengths (Figure 20-14).
20-9
Note 20
Figure 20-14. Two identical sources one wavelength apart
produce waves that are in phase.
The most general condition for perfect destructive
interference is
Δφ = 2π
€
Δx
+ Δφ 0
λ

1
= 2 m + π rad

2
m = 0,1…
…[20-25]
Two identical sources (∆φ0 = 0) produce perfect destructive interference when the path-length difference
is
€ a half-integer number of wavelengths. Figures 20-14
show three ways of producing perfect destructive
interference.
Let us consider an example of interference in onedimension when the sources are out of phase.
Example Problem 20-3
Interference Between Two Sound Waves
You are standing in front of two side-by-side loudspeakers playing sound of the same frequency (Figure
20-15). Initially there is almost no sound at all. Then
one of the speakers is moved slowly away from you.
The sound intensity increases as the separation
between the speakers increases, reaching a maximum
when the speakers are 0.75 m apart. Then, as the
speaker continues to move, the sound starts to
decrease. What is the distance between the speakers
when the sound intensity is again a minimum?
20-10
€
Figure 20-14. Three ways of producing perfect destructive
interference.
Solution:
Since initially there is almost no sound at all, the
sound waves are interfering destructively. The
sources are side-by-side (at roughly the same position
in space) so ∆x = 0. Since ∆φ = π initially, it follows
that ∆φ0 = π, that is, the speakers themselves are out of
phase. If the speakers are out of phase initially, then
they are always out of phase. Moving one of the
speakers does not change ∆φ0 but it does change ∆x.
Constructive interference is reached when
Δφ = 2π
Δx
Δx
+ Δφ 0 = 2π
+ π = 2π rad.
λ
λ
Note 20

 Δφ 
D = 2acos  sin( kx avg − ωt + (φ 0 ) avg ) . …[20-27]
 2 

€
Figure 20-15. The out-of-phase sources generate waves that
are in phase if the sources are one-half wavelength apart.
Thus it follows that ∆x = λ/2 when the intensity is a
maximum. Since ∆x = 0.75 m, λ = 2 x 0.75 m = 1.50 m.
The sound intensity will again be a minimum when
the speaker is moved one wavelength beyond its
starting position, or a distance of 1.50 m.
Let us now apply mathematics to the cases we have
just described qualitatively. Once again we use D to
indicate a wave of general (transverse or longitudinal)
sinusoidal type.
where ∆φ = φ 2 – φ1 is the phase difference between the
waves. xavg = (x 1 + x2)/2 is the average distance from the
point of detection to the two sources and (φ0)avg = (φ10
+ φ20)/2 is the average phase constant of the sources.
The sine term shows that the superposition of the
two traveling waves is itself a traveling wave.
The amplitude of the original traveling waves is
multiplied by the factor 2cos(∆φ/2). This is the major
result of interference.
The amplitude of the resultant wave depends on the
phase difference ∆φ between the wave components. If
∆ φ is a multiple of 2π , then the amplitude is
multiplied by 2. This is maximum constructive
interference. If the phase difference is an odd multiple
of π, then the amplitude is zero. This is perfectly
destructive interference. If the phase difference has any
other value then, in general, the interference can be
non-maximum constructive or imperfectly
destructive. For example, the interference patterns of
two component waves for three different arbitrary
values of ∆ φ are illustrated in Figure 20-16. Notice that
the resultant amplitude A is greater than 0 and less
than 2a. 5
The Mathematics of Interference
Consider two sinusoidal waves of equal amplitude a,
but different phase constant, moving in the same
direction along the x axis. According to the principle
of superposition, the net displacement D of the
medium is
D = D1 + D2 ,
= a[sin(kx1 − ωt + φ10 ) + sin( kx 2 − ωt + φ 20 )]
€
= asin φ1 + asin φ 2
…[20-26]
€ where the phases φ1 and φ2 have been defined. Using
the trigonometric identity
€
€
α − β  α + β 
sinα + sin β = 2cos
sin
,
 2   2 
we can write eq[20-26] as 4
Figure 20-16. The interference of two waves for three different values of the phase difference showing non-maximum
constructive and imperfectly destructive interference.
4
5
This requires a little algebra, which we leave as an exercise for
the interested reader.
You will not be held responsible for problems of non-maximum
constructive or imperfectly destructive interference in this course.
20-11
Note 20
Interference in Two and Three Dimensions
We have seen in Note 18 that traveling waves can
move in two-and three-dimensions. Ripples on the
surface of a pond and light and sound waves are good
examples. Figure 20-17 might represent a 2D or 3D
wave. Wave fronts (maximum displacements or
peaks) are separated by λ. Halfway between the wave
peaks are wave troughs. The waves move away from
the source with speed v.
constructive and destructive interference just
described.
Figure 20-17. Representation of a circular or spherical wave.
This kind of wave can be written
D(r,t) = asin( kr − ωt + φ 0 )
…[20-28]
where r is the distance measured outwards from the
source. Strictly speaking, for spherical and circular
waves a decreases as r increases. For simplicity, how€
ever, we shall assume a is constant over the region of
the wave we study.
Constructive and destructive interference of such
waves from two sources (Figure 20-18) depends on
how the peaks and troughs come together at a particular point. The orange dots indicate points at which
(at the particular instant represented) peaks from both
sources come together, or troughs from both sources
come together. These are points of maximum
constructive interference with A = 2a.
The blue dots indicate points at which peaks from
the one source and troughs from the other source
come together. These are points of perfectly destructive interference with A = 0.
Keep in mind that as time goes on the circles in the
figure move steadily outwards. The motion of the
wave, however, does not affect the points of
20-12
Figure 20-18. The overlapping ripple patterns of two sources.
Several points of maximum constructive interference
(orange dots) and perfectly destructive interference (blue
dots) are shown.
If the sources are identical (intrinsic phase difference
is zero) then the interference at any point can be
determined if the path-length difference at that point
is known. Let us suppose for argument that the ripple
patterns of two identical sources are as shown in
Figure 20-19. It follows that since the sources are
identical (resulting, say, from pebbles being dropped
simultaneously into a pond) ∆φ0 = 0 and the phase
difference between the two waves at any point is
determined by the path-length difference ∆r.
Consider point A. The path-length difference there is
∆ r = 3λ – 2λ = λ. Hence at that point maximum
constructive interference occurs. Consider point B. At
that point the path-length difference is ∆r = 3λ – 2.5λ =
0.5λ. Hence at B perfectly destructive interference
occurs.
Note 20
(b) How will the situation differ if the loudspeakers
are out of phase?
Solution:
Since the sources are in phase then ∆φ 0 = 0 and the
interference produced at the point of detection is due
to the path-length difference.
(a) Let r 1 and r2 be the distances from the sources to
the point of observation. These are:
r1 = (5.0 m) 2 + (1.0 m) 2 = 5.10 m
r2 = (5.0 m) 2 + (3.0 m) 2 = 5.83 m
€ Thus the path-length difference is ∆r = r – r = 0.73 m.
2
1
The wavelength of the sound wave is
Figure 20-19. When the sources are identical the path-length
difference ∆r determines whether the interference at a
particular point is constructive or destructive.
€
λ=
v 341 m.s−1
=
= 0.487 m .
f
700 Hz
In terms of wavelengths, the path-length difference is
∆r/λ = 1.50, or
Example Problem 20-4
Two Dimensional Interference Between the Sound from
Two Loudspeakers
Two loudspeakers are 2.0 m apart and in phase with
each other. Both emit sound waves at a frequency of
700 Hz into a room where the speed of sound is 341
m.s –1. A listener stands 5.0 m in front of the loudspeakers and 2.0 m to one side of the center (Figure
20-20).
3
Δr = λ .
2
€
The path-length difference is an integer multiple of
half-wavelengths, so the waves interfere
destructively.
€
(b) If the sources were out of phase (∆φ0 = π rad), then
the phase difference at the listener would be
Δφ = 2π
 3
Δr
+ Δφ 0 = 2π   + π rad = 4π rad .
2
λ
This is an integer multiple of 2 π rad so in this case the
interference would be constructive.
€
This concludes our study of waves.
Figure 20-20. Pictorial representation of the interference
between two loudspeakers.
(a) Is the interference at this point constructive,
destructive, or something in between?
20-13
Note 20
To Be Mastered
•
•
•
•
•
•
•
Definitions: Principle of Superposition, constructive interference, destructive interference
General expression for: wave intensity
Physics of: traveling wave pulses encountering a discontinuity
Physics of: standing waves on a stretched string
Physics of: standing waves in a pipe: closed-closed, open-open, open-closed
Physics of: interference in one-dimension, due to path-length difference, intrinsic phase difference of sources
Physics of: interference in two dimensions
Typical Quiz/Test/Exam Questions
1.
20-14