20-1

Note 20 SuperpositionSections Covered in the Text: Chapter 21, except 21.8

In Note 18 we saw that a wave has the attributes ofdisplacement, amplitude, frequency, wavelength andspeed. We can now consider the effects that occurwhen waves come together at the same point in space,that is, when waves undergo the phenomena of super-position and interference. We shall see that such studiesshed light on, among other things, the production ofsound by musical instruments.

The Principle of SuperpositionIt is possible for two or more waves to coexist at thesame point in space. When this happens it is import-ant to know what the total wave amplitude is at thatpoint. This question is answered by the Principle ofSuperposition. The principle may be stated in thesewords:

When two or more waves are simultaneously presentat a single point in space, the displacement of the med-ium at that point is the sum of the displacements dueto each individual wave.

This simple, almost intuitive, principle can be shownto apply to all common waves, be they mechanical,sound or electromagnetic. A consequence is that thevarious waves excited on a stretched string, forexample, are independent of one another. Two pulsestraveling in opposite directions on the string can passthrough one another and reappear on the other sidewithout being altered in any way.

This idea is illustrated in Figure 20-1 with graph-ically-constructed “wave pulses” of triangular andrectangular shape. A triangular wave pulse is shownmoving to the right, a rectangular wave pulse movingto the left, both at the speed of 1 m.s–1. Their positionsare shown in 5 frames—at the clocktimes of 0s, 1s, 2s,3s and 4s. You can see that frames 3 and 4 show thepulses in the process of passing through one another.Frame 5 shows the pulses of frame 1 having passedthrough one another without changing one another.

Let us focus on frames 3 and 4. When the amplitudesof the two pulses have the same sign, the resultantamplitude or displacement of the medium increasesas the two pulses pass through one another. This iscalled constructive interference. What the figure doesn’tshow is that when the amplitudes have opposite signsthe resultant amplitude decreases as the two pulsespass through one another. This is called destructiveinterference. Notice that the meaning of the wordinterference used here differs from its everyday usage.

The “interfering” pulses move independently, and donot affect one another, or physically change oneanother, in any way.

Figure 20-1. Illustration of the principle of superposition us-ing mathematical constructions of wave pulses on a line.

Wave pulses are complicated entities to describemathematically (and to produce experimentally). It issimpler to deal with sinusoidal waves, which we do inthe next section.

Standing WavesOne could, at least in principle, induce two sinusoidalwaves to move in opposite directions and to passthrough one another by moving both ends of a verylong string up and down continuously. We begin byconsidering the wave in qualitative terms. Then weshall apply mathematics to the wave.

The transverse displacement of the string at anypoint along the string would appear as shown inFigure 20-2. This wave is called a standing wave

Note 20

20-2

because the crests and troughs “stand in place” as thestring oscillates. There are points on the string that donot move. These points, spaced λ/2 apart, are callednodes. Halfway between the nodes are points wherethe string particles oscillate with maximum displace-ment. These points are called antinodes. The nodes of astanding wave are points of destructive interferencewhere the waves are out of phase (phase difference isπ); antinodes are points of constructive interferencewhere the waves are in phase (phase difference is 0).

Figure 20-2. A graphical representation of a standing waveon a string being vibrated at both ends. Shown are twosnapshots of the wave at two instants of clocktime.

Recall from Note 18 that the intensity I of a wave isproportional to the square of the wave amplitude (I =CA2). A graph of I(x) for the wave in Figure 20-2 isshown in Figure 20-3. I(x) is a maximum at positionsof antinodes, a minimum at positions of nodes.

Figure 20-3. A graphical representation of a standing waveshowing also the intensity I as a function of x.

Now that we understand some qualitative aspects of astanding wave, let us apply mathematics to the wave.

The Mathematics of a Standing WaveConsider two waves that are identical in all respects(same amplitude a and angular frequency ω) travelingin opposite directions (right and left) on a line. Fromwhat 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 resultantof the two waves is

€

D(x, t) = DR + DL = 2asinkxcosωt . …[20-2]

You should be able to show that eq[20-2] followsusing the trigonometric identity

€

sin α ± β( ) = sinα cosβ ± cosα sinβ .

It is instructive to write eq[20-2] in the form

€

D(x, t) = A(x)cosωt …[20-3]

where

€

A(x) = 2asinkx . …[20-4]

A(x) is the amplitude of the sinusoidal function.Notice that eq[20-2] does not contain either of thefactors (kx–ωt) or (kx+ωt) that characterize the travel-ing waves of eq[20-1]. Eq[20-2] does not thereforerepresent a traveling wave; it represents a standingwave. The wave is “standing” in the sense of movingneither to the right nor left. The “cosωt” factor showsthat every point x of the medium vibrates in simpleharmonic motion with frequency f = ω/2π and withthe amplitude A(x). Several snapshots of the wave areshown in Figure 20-4.

The nodes of the standing wave are the points atwhich the amplitude is zero. They are located atpositions xm for which

€

A(xm ) = 2asinkxm = 0 , …[20-5]

or for which

1 We consider here a general, hypothetical line and use the letterD to stand for a wave displacement that includes both transverseand longitudinal displacements in any medium. Later, we shallrestrict our attention to transverse waves on a string or longitudinalsound waves.

Note 20

20-3

Figure 20-4. The net displacement resulting from two sinu-soidal waves moving in opposite directions.

€

kxm =2πxmλ

= mπ m = 0,1,2 …[20-6]

Thus the position xm of the mth node is

€

xm = m λ2

m = 0,1,2… …[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 amaximum. A(x) has the maximum value of Amax = 2aat points xm where sinkxm = 1, or points xm such that

€

kxm =2πxmλ

= m +12

π m = 0,1,2 …[20-8]

Thus the position xm of the mth antinode is

€

xm = m +12

λ2

m = 0,1,2… …[20-9]

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,

€

I(x) = A(x)[ ]2 = 4a2 sin2 kx

€

= Imax sin2 kx . …[20-10]

The function I(x) is consistent with the graph ofintensity vs x in Figure 20-3. The intensity is a maxi-mum at values of x for which eq[20-9] holds, that is, atthe positions of the antinodes. As we have seen, theintensity is a minimum at the positions of the nodes.

Transverse Standing WavesThe procedure of setting up a standing wave on along string by moving both ends of it up and downcontinuously is not a practical one. At least one end ofthe string must be held fixed. To understand how astanding wave might arise in this case it is necessaryto understand what happens when a traveling waveencounters a boundary or a discontinuity betweentwo media.

The media in question might be the two strings ofdifferent linear density (mass per unit length) joinedtogether as shown in Figure 20-5a. On the left of thefigure 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 accordingto eq[18-25] the wave speed is slower in the leftmedium, faster in the right.

Figure 20-5a. A wave pulse is reflected and transmitted at adiscontinuity where the wave speed increases.

When a traveling wave (pulse) encounters thejunction from the left, some of the wave’s energy istransmitted into the right medium, and some is ref-lected back into the left medium. The larger fraction ofthe energy is transmitted (indicated by the pulse ofgreater amplitude). However, neither the transmittednor the reflected pulse has the same amplitude as theincident pulse.

In Figure 20-5b, an incident wave pulse encounters adiscontinuity at which the wave speed decreases. Asbefore, some of the wave’s energy is transmitted andsome is reflected. But this time the reflected pulse isinverted. By this is meant a displacement D on theincident wave becomes displacement –D on the reflec-ted wave. Because sin(φ + π) = – sinφ, the reflectedwave 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 theboundary can be thought of as a string medium ofinfinite linear density. The reflected wave is again

Note 20

20-4

inverted, but there is no transmitted wave at all. Allthe wave’s energy is reflected. And the wave speeddoes not change.

Figure 20-5b. A wave pulse is reflected at a discontinuitywhere the wave speed decreases.

Figure 20-5c. A wave pulse is totally reflected at a bound-ary. There is no transmitted wave. The wave speed does notchange.

We are now ready to apply these observations tounderstand the production of standing waves on astretched string.

Standing Waves on a Stretched StringThe difference between a hypothetical line and astretched 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 astanding wave on the string by “wiggling” orcontinuously plucking the string. Traveling sinusoidalwaves are created which then travel in bothdirections. They encounter the boundary at either endand reflect. The speed of the reflected waves does notchange and therefore the wavelength and frequencyof the reflected sinusoidal waves does not change. Inaddition, the reflected waves are of equal amplitude

and wavelength. Once reflected at the boundaries, thetraveling waves pass through one another andproduce a standing wave, as we have already seen.

Figure 20-6. A way of setting up a standing wave on astretched string tied at both ends.

The wave on the string is subject to boundaryconditions. Because the string is tied at both ends, thedisplacements at x = 0 and x = L must be zero at alltimes. In other words nodes must exist at these points.Thus the boundary conditions are

€

D(x = 0,t)D(x = L,t)

= 0 . …[20-11a, b]

Recall that the displacement of a standing wave isD(x,t) = (2asinkx)cosωt. This equation already satisfiesthe first boundary condition, eq[20-11a]. The secondboundary condition, eq[20-11b] is satisfied at all timesif

€

2asinkL = 0 . …[20-12]

This is true if sinkL = 0, which in turn requires

€

kL =2πLλ

= mπ m=1,2,3… …[20-13]

m = 0 is excluded because L cannot be zero. Eq[20-13]can be satisfied only by values λm such that

€

λm =2Lm

m=1,2,3… …[20-14]

The natural frequencies of vibration fm of the string, inHz, are

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fm =vλ

=vm2L

m = 1, 2, 3… …[20-15]

Note 20

20-5

Using the result for the speed of waves on a stretchedstring under tension T derived in Note 18: v = (T/µ)1/2,where T is the tension and µ is the mass per unitlength, eq[20-15] becomes

€

fm =m2L

Tµ

m = 1, 2, 3, … …[20-16]

The fact that the values fm depend on integers m is thesame as saying that the frequencies are quantized. Thatis to say, the string can support only certain discretefrequencies

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fm = mf1 m = 1,2,3… …[20-17]

f1 is called the fundamental frequency. The frequenciesf2, f3, etc are called overtones or harmonics. Figure 20-7shows the fundamental and three harmonics. Stand-ing waves on a stretched string are a model of thesource of sound in all stringed musical instruments.These include the piano, violin and cello amongstothers. 2

Figure 20-7. Standing mechanical waves on a stretchedstring. Only the first 4 modes of vibration are shown here.

2 This is the first example of the idea of quantization we haveencountered in this course. The concept is at the heart of quantumphysics. It will be discussed in some detail in PHYA21H3S.

Let us consider a simple example.

Example Problem 20-1A Standing Wave on a Stretched String

A string of length 2.50 m vibrates with a 100 Hzstanding wave. There are nodes 1.00 m and 1.50 mfrom one end of the string, with no nodes in between.Which harmonic is this, and what is the string’sfundamental frequency?

Solution:The standing wave looks like Figure 20-2. If there areno nodes between 1.00 m and 1.50 m, then the nodespacing is λ/2 = 0.50 m. The number of half wave-lengths that can fit into the string’s length L = 2.50 mis m = 2.50/(1/2) = 5. This is the mode number, whichmeans that 100 Hz is the fifth harmonic. The funda-mental frequency is therefore f1 = 100/5 = 20 Hz.

Standing Electromagnetic WavesStanding electromagnetic waves also exist, as forexample in the cavity of a laser (Figure 20-8). Mirrorsat either end of the cavity play the role of boundaryreflectors, analogous to the boundaries at the ends of astring tied at both ends.

Figure 20-8. A laser cavity contains a standing electromag-netic wave.

The boundary conditions on the electromagnetic waveare the same as on a wave on a string, eqs[20-11], andso eqs[20-14] and [20-16] also apply.

A typical laser cavity has a length L ≈ 30 cm andoperates in the red area of the electromagneticspectrum with λ ≈ 600 nm. The mode number of thestanding wave is, from eq[20-14]:

€

m =2Lλ

=2 x 0.30 m

6.00 x 10−7m=1,000,000 .

The standing wave has about one million nodes! This

Note 20

20-6

is made possible by the very short wavelength oflight.

Various wind instruments (organ, flute, clarinet, etc)produce musical sounds by virtue of the resonances ofair vibrating in a pipe or tube structure made of woodor metal. The physics of tube structures we considernext.

Standing Sound Wavesand Musical Acoustics

Air confined in a column or tube can support astanding sound wave (Figure 20-9). We have seen inNote 18 that a sound wave is a longitudinal wave; thedirection of alternate compressions and rarefactions ofair is in the direction of the tube axis. The point wherethe tube is closed or capped must be the position of anode (where the vibration of the air molecules isminimum). The figure shows that the graphical rep-resentation of the m = 2 mode is the same as the m = 2mode of a standing wave on a string. Keep in mindthat the representation is of a longitudinal displace-ment NOT a transverse displacement.

Figure 20-9. The m = 2 longitudinal standing wave inside aclosed column of air. The graphical representation is of alongitudinal NOT a transverse displacement.

Sound produced by the setup in Figure 20-9 is of onlyacademic interest because it lacks the means to allowthe sound wave to leave the tube and reach thelistener (presumably outside the tube!). Thus practicaltubes are left open at one or both ends.

As we have stated earlier, at the point where thetube is closed a node must exist in the standing wave.

Conversely, where the tube is open to the atmospherean antinode must exist. The first three modes for thethree cases—a standing wave in a tube closed at bothends (closed-closed), open at both ends (open-open),and closed at one end (closed-open)—are illustratedin Figures 20-10.

Figures 20-10a and b. Standing sound waves in two types oftube: (a) closed at both ends and (b) open at both ends.

Note that the standing wave in the open-open andclosed-closed tubes is the same except for a phaseinversion. In both cases there are m half-wavelengthsbetween the ends. Thus the wavelengths and frequen-cies of an open-open tube and a closed-closed tube are

Note 20

20-7

the same as those of a string tied at both ends:

(open-open, closed-closed tube)

€

λm =2Lm

fm = m v2L

= mf1

m = 1,2,3… …[20-18]

A tube closed at one end and open at the other end isquite a different matter (Figure 20-10c). The funda-mental mode has only one-quarter of a wavelength ina tube of length L, hence the m = 1 wavelength is λ1 =4L. This is twice the λ1 wavelength of an open-open ora closed-closed tube. Consequently, the fundamentalfrequency of an open-closed tube is half that of anopen-open or a closed-closed tube of the same length.

Figure 20-10c. Three modes in an open-closed tube.

The possible wavelengths and frequencies of thesound produced by an open-closed tube of length Lare:

(open-closed tube)

€

λm =4Lm

fm = m v4L

= mf1

m = 1,3,5… …[20-19]

Let us consider an example of an open-open organpipe.

Example Problem 20-2The Length of an Organ Pipe

An organ pipe open at both ends sounds its secondharmonic at a frequency of 523 Hz. What is the lengthof the pipe?

Solution:The second harmonic is the m = 2 mode, which for anopen-open tube has the frequency (eq[20-18]):

€

f2 = 2 v2L

.

Thus the length of the organ pipe is

€

L =vf2

=343 m.s−1

523 Hz= 0.656 m = 65.6 cm .

Musical InstrumentsWe have seen then that the equation that governs thesound produced by stringed instruments is

€

f1 =v2L

=12L

Tµ

. …[20-20]

where T is the tension, L is the length and µ is themass per unit length of the string. The sound prod-uced by a string is primarily at the fundamentalfrequency f1 though some sound is also produced atharmonic frequencies.3 This means that the note pro-duced can be varied by varying L, T or µ.

Some instruments such as the piano are equippedwith strings of various L and µ. Clearly, the longerand the fatter the string, the lower is the noteproduced. Similar arguments can be made forexplaining the operation of many other stringedinstruments. We leave this subject to the interestedreader.

We have therefore seen some of the effects that occurwhen two traveling waves of the same frequency

3 Many instruments can produce a middle “C”. However, it isimmediately evident to the average listener what type ofinstrument produced the note. This is because different types ofinstruments produce a different mix of harmonic frequencies inaddition to the fundamental frequency.

Note 20

20-8

move in opposite directions and interfere with oneanother to produce a standing wave. We now moveon to the case of the interference of two travelingwaves moving in the same direction.

Interference in One DimensionTwo hypothetical schemes for obtaining two travelingwaves of the same frequency moving in the samedirection along a line are illustrated in Figures 20-11.Suppose the waves have the same amplitude a, thesame wave number k and the same angular frequencyω. We wish to establish what happens when the twowaves overlap.

Figure 20-11a. A scheme for obtaining two travelingelectromagnetic waves of the same frequency moving in thesame direction.

Figure 20-11b. A scheme for obtaining two traveling soundwaves of the same frequency moving in the same direction.

We begin by considering the resultant wave at thepoint of detection indicated in Figure 20-11b. The twosuperposed waves, at distances x1 and x2 from theirsources, can be written

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D1(x1,t) = asin kx1 −ωt + φ10( ) = asinφ1…[20-21]

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D2(x2,t) = asin kx2 −ωt + φ20( ) = asinφ2

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 furtherreinforcement of this idea, snapshot graphs at t = 0 ofwaves emitted by three sources with phase constantsφ0 = 0 rad, φ0 = π/2 rad and φ0 = π rad are drawn inFigures 20-12. The phase constant determines whatthe source is doing at t = 0. For example, Figure 20-12ashows the meaning of a phase constant of 0 rad. Thiswould apply to a loudspeaker at its center positionwhile moving backward at t = 0. Similar argumentscan be applied to explain the meaning of the otherphase constants in the figure.

Figure 20-12. Waves from three sources having phase con-stants φ0 = 0 rad, φ0 = π/2 rad and φ0 = π rad.

Two important special cases of the overlapped wavesare shown in Figures 20-13. Figure 20-13a shows thecrests and the troughs of the two waves aligned asthey travel along the x-axis. Figure 20-13b shows thecrests of one wave aligned with the troughs of theother wave as they travel. Since the waves have thesame speed the relative position of the crests andtroughs remain fixed at all times. The position of the

Note 20

20-9

wave fronts are indicated in the middle panel of thefigures.

The two waves of Figure 20-13a have the samedisplacement at every point and therefore the samephase. That is, φ2 = φ1, or more precisely φ2 = φ1 ± 2πmwhere m is an integer. These waves are in phase. Theresultant displacement A = 2a. This is maximumconstructive interference.

Figure 20-13. Constructive and destructive interference oftwo waves traveling along the x-axis.

In Figure 20-13b the waves are 180˚ out of phase. Thisalignment of the waves produces destructive interfer-

ence. The displacement of the waves is such that D1 =–D2. Thus the net displacement of the resultant travel-ing wave is zero at every point along the axis. Thisspecial case of interference (where the resultant amp-litude is zero) is called perfect destructive interference.

The Phase DifferenceThe phases of the two waves are

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φ1 = kx1 −ωt + φ10

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φ2 = kx2 −ωt + φ20 , …[20-22]

and the phase difference ∆φ is

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Δφ = φ2 −φ1 = kx2 −ωt + φ20( ) − kx1 −ωt + φ10( )

€

= k x2 − x1( ) + φ20 −φ10( )

€

=2πλΔx + Δφ0 . …[20-23]

You can see that the phase difference consists of twocontributions: a phase difference due to

1 a path-length difference, and

2 a difference between the inherent phases of thesources.

To see this note that ∆x = x 2 – x 1 is the differencebetween the distances traveled by the waves from thesources to the point of detection. This is called thepath-length difference. Obviously, if the sources are atthe same position in space then ∆x = 0 and ∆φ = ∆φ0.

The second contribution is due to the difference inthe 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 spaceand are also in phase, then ∆φ = 0 and the wavesreaching the point of detection are also in phase.

The most general condition for maximum construc-tive interference is

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Δφ = 2π Δxλ

+ Δφ0 = 2mπ rad m = 0,1… …[20-24]

For identical sources that have ∆φ0 = 0 rad, maximumconstructive interference occurs when ∆x = mλ. Thatis, two identical sources produce maximum construc-tive interference when the path-length difference is aninteger number of wavelengths (Figure 20-14).

Note 20

20-10

Figure 20-14. Two identical sources one wavelength apartproduce waves that are in phase.

The most general condition for perfect destructiveinterference is

€

Δφ = 2π Δxλ

+ Δφ0

€

= 2 m +12

π rad m = 0,1… …[20-25]

Two identical sources (∆φ0 = 0) produce perfect des-tructive interference when the path-length differenceis a half-integer number of wavelengths. Figures 20-14show three ways of producing perfect destructiveinterference.

Let us consider an example of interference in one-dimension when the sources are out of phase.

Example Problem 20-3Interference Between Two Sound Waves

You are standing in front of two side-by-side loud-speakers playing sound of the same frequency (Figure20-15). Initially there is almost no sound at all. Thenone of the speakers is moved slowly away from you.The sound intensity increases as the separationbetween the speakers increases, reaching a maximumwhen the speakers are 0.75 m apart. Then, as thespeaker continues to move, the sound starts todecrease. What is the distance between the speakerswhen the sound intensity is again a minimum?

Figure 20-14. Three ways of producing perfect destructiveinterference.

Solution:Since initially there is almost no sound at all, thesound waves are interfering destructively. Thesources are side-by-side (at roughly the same positionin space) so ∆x = 0. Since ∆φ = π initially, it followsthat ∆φ0 = π, that is, the speakers themselves are out ofphase. If the speakers are out of phase initially, thenthey are always out of phase. Moving one of thespeakers does not change ∆φ0 but it does change ∆x.

Constructive interference is reached when

€

Δφ = 2π Δxλ

+ Δφ0 = 2π Δxλ

+ π = 2π rad.

Note 20

20-11

Figure 20-15. The out-of-phase sources generate waves thatare in phase if the sources are one-half wavelength apart.

Thus it follows that ∆x = λ/2 when the intensity is amaximum. Since ∆x = 0.75 m, λ = 2 x 0.75 m = 1.50 m.The sound intensity will again be a minimum whenthe speaker is moved one wavelength beyond itsstarting position, or a distance of 1.50 m.

Let us now apply mathematics to the cases we havejust described qualitatively. Once again we use D toindicate a wave of general (transverse or longitudinal)sinusoidal type.

The Mathematics of InterferenceConsider two sinusoidal waves of equal amplitude a,but different phase constant, moving in the samedirection along the x axis. According to the principleof superposition, the net displacement D of themedium is

€

D = D1 + D2 ,

€

= a sin kx1 −ωt + φ10( ) + sin kx2 −ωt + φ20( )[ ]

€

= asinφ1 + asinφ2 …[20-26]

where the phases φ1 and φ2 have been defined. Usingthe trigonometric identity

€

sinα + sinβ = 2cos α −β2

sin α + β

2

,

we can write eq[20-26] as 4

4 This requires a little algebra, which we leave as an exercise forthe interested reader.

€

D = 2acos Δφ2

sin kxavg −ωt + (φ0)avg( ) . …[20-27]

where ∆φ = φ2 – φ1 is the phase difference between thewaves. xavg = (x1 + x2)/2 is the average distance from thepoint 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 thetwo traveling waves is itself a traveling wave.

The amplitude of the original traveling waves ismultiplied by the factor 2cos(∆φ/2). This is the majorresult of interference.

The amplitude of the resultant wave depends on thephase difference ∆φ between the wave components. If∆ φ is a multiple of 2π , then the amplitude ismultiplied by 2. This is maximum constructiveinterference. If the phase difference is an odd multipleof π , then the amplitude is zero. This is perfectlydestructive interference. If the phase difference has anyother value then, in general, the interference can benon-maximum constructive or imperfectlydestructive. For example, the interference patterns oftwo component waves for three different arbitraryvalues of ∆φ are illustrated in Figure 20-16. Notice thatthe resultant amplitude A is greater than 0 and lessthan 2a. 5

Figure 20-16. The interference of two waves for three differ-ent values of the phase difference showing non-maximumconstructive and imperfectly destructive interference. 5 You will not be held responsible for problems of non-maximumconstructive or imperfectly destructive interference in this course.

Note 20

20-12

Interference in Two and Three DimensionsWe have seen in Note 18 that traveling waves canmove in two-and three-dimensions. Ripples on thesurface of a pond and light and sound waves are goodexamples. Figure 20-17 might represent a 2D or 3Dwave. Wave fronts (maximum displacements orpeaks) are separated by λ. Halfway between the wavepeaks are wave troughs. The waves move away fromthe source with speed v.

Figure 20-17. Representation of a circular or spherical wave.

This kind of wave can be written

€

D(r,t) = asin kr −ωt + φ0( ) …[20-28]

where r is the distance measured outwards from thesource. Strictly speaking, for spherical and circularwaves a decreases as r increases. For simplicity, how-ever, we shall assume a is constant over the region ofthe wave we study.

Constructive and destructive interference of suchwaves from two sources (Figure 20-18) depends onhow the peaks and troughs come together at a partic-ular point. The orange dots indicate points at which(at the particular instant represented) peaks from bothsources come together, or troughs from both sourcescome together. These are points of maximumconstructive interference with A = 2a.

The blue dots indicate points at which peaks fromthe one source and troughs from the other sourcecome together. These are points of perfectly destruc-tive interference with A = 0.

Keep in mind that as time goes on the circles in thefigure move steadily outwards. The motion of thewave, however, does not affect the points of

constructive and destructive interference justdescribed.

Figure 20-18. The overlapping ripple patterns of two sources.Several points of maximum constructive interference(orange dots) and perfectly destructive interference (bluedots) are shown.

If the sources are identical (intrinsic phase differenceis zero) then the interference at any point can bedetermined if the path-length difference at that pointis known. Let us suppose for argument that the ripplepatterns of two identical sources are as shown inFigure 20-19. It follows that since the sources areidentical (resulting, say, from pebbles being droppedsimultaneously into a pond) ∆φ0 = 0 and the phasedifference between the two waves at any point isdetermined by the path-length difference ∆r.

Consider point A. The path-length difference there is∆ r = 3λ – 2λ = λ . Hence at that point maximumconstructive interference occurs. Consider point B. Atthat point the path-length difference is ∆r = 3λ – 2.5λ =0.5λ. Hence at B perfectly destructive interferenceoccurs.

Note 20

20-13

Figure 20-19. When the sources are identical the path-lengthdifference ∆r determines whether the interference at aparticular point is constructive or destructive.

Example Problem 20-4Two Dimensional Interference Between the Sound fromTwo Loudspeakers

Two loudspeakers are 2.0 m apart and in phase witheach other. Both emit sound waves at a frequency of700 Hz into a room where the speed of sound is 341m.s–1. A listener stands 5.0 m in front of the loud-speakers and 2.0 m to one side of the center (Figure20-20).

Figure 20-20. Pictorial representation of the interferencebetween two loudspeakers.

(a) Is the interference at this point constructive,destructive, or something in between?

(b) How will the situation differ if the loudspeakersare out of phase?

Solution:Since the sources are in phase then ∆φ0 = 0 and theinterference produced at the point of detection is dueto the path-length difference.(a) Let r1 and r2 be the distances from the sources tothe point of observation. These are:

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r1 = (5.0 m)2 + (1.0 m)2 = 5.10 m

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r2 = (5.0 m)2 + (3.0 m)2 = 5.83 m

Thus the path-length difference is ∆r = r2 – r1 = 0.73 m.The wavelength of the sound wave is

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λ =vf

=341 m.s−1

700 Hz= 0.487 m .

In terms of wavelengths, the path-length difference is∆r/λ = 1.50, or

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Δr =32λ .

The path-length difference is an integer multiple ofhalf-wavelengths, so the waves interferedestructively.(b) If the sources were out of phase (∆φ0 = π rad), thenthe phase difference at the listener would be

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Δφ = 2π Δrλ

+ Δφ0 = 2π 32

+ π rad = 4π rad .

This is an integer multiple of 2π rad so in this case theinterference would be constructive.

This concludes our study of waves.

Note 20

20-14

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

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