Superposition of Waves - hedberg.ccnysites.cuny.edu · These two wave pulses are moving towards...
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Superposition of Waves
Whathappenswhentwowavestouch.
Principle of Superposition
Whentwomarblescollide,theybounceback.That'showmaterialobjectsbehave.Twodifferentobjectscan'toccupythesamespaceatthesametime.
Waves,ontheotherhand,canoccupythesamespace.Thesetwowavepulsesaremovingtowardseachother,thenstarttooverlap,thencontinueon.
Principle of Superposition
Thisisoneofthemajorconceptsofphysics-superposition.
Theconceptofsuperpositionisaveryfundamentalpartofphysics.Itistheunderlyingcauseformanyphenomenainoptics,acoustics,quantummechanics,andothersubfields.Superpositionalsoplaysalargeroleinengineeringfields,especiallyelectrical.Ifyouwanttosendseveralchunksofinformationdownthesameelectricalwire,you'llrelyonaspectsofsuperposition.We'lllookathowitoccursinsimplesystemslikeawaveonastringaswellasmorecomplicatedsituationslike2dimensionalacousticinterference.Later,you'llusethesameconceptsandtoolstotreatmorecomplicatedsituations.
Whateverthetwowavesare,allweneedtodoisaddthemupinordertofindthenewwave.
Thisanimationshowstwodifferentwaveapproaching,interacting,thenreceding.
(x, t) = (x, t) + (x, t)y′ y1 y2
PHY 208 - superposition
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0 1 2 3 4 5 6 7 8 9 10
2 3 4 5 6 7 8
1 m/s 1 m/s
2 3 4 5 6 7 82 3 4 5 6 7 8
2 3 4 5 6 7 8
A B
DC
Thesetwowaves(theredandtheblue)areaddedtogethertogetthepurple
Redcurve:wave1:
Bluecurve:
Quick Question 1
Thesetwowavesareapproachingeachotheratt=0.Whatwillthesumlooklikeatt=2s?
Math of Superposition
Let'saddtwowavestravelinginthesamedirectiononthesamestring.( , ,and arethesame)
wave1:
wave2:
Since,
withatrigidentity(below)
Thisisaveryusefultrigidentity:
Theonlydifferencebetweenthesetwowavesisthephasefactorthatappearsinthesecondone.Thisjustindicatesthatthewavesmighthavedifferentamplitudesatt=0.
The sum of two waves
k ω A
(x, t) = A sin(kx − ωt)y1
(x, t) = A sin(kx − ωt+ ϕ)y2
(x, t)y′ ==
(x, t) + (x, t)y1 y2A sin(kx − ωt) + A sin(kx − ωt+ ϕ)
(1)(2)
(x, t) = [2Acos ] sin(kx − ωt+ )y′ϕ
2ϕ
2
sinα + sinβ = 2 sin (α + β) cos (α − β)12
12
ϕ
(x, t) = A sin(kx − ωt)y1
(x, t) = A sin(kx − ωt+ ϕ)2
PHY 208 - superposition
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wave. wave2:
PurpleCurve:
Summary of interference types
Interferenceisusedtodescribethisphenomenon.
PhasedifferenceinAmplitude InterferenceType
Degrees Radians wavelength0º 0 0 FullyConstructive120º .33 Intermediate
180º .5 0 Fullydestructive240º .67 Intermediate
360º 1 FullyConstructive
(Applicableforwaveswiththesameamplitudeandwavelengthtravelinginthesamedirection.)
Waves traveling in opposite directions.
Herearetwowaveswithequalamplitudesandfrequenciestravelinginoppositedirectionsonastring.(Thebluewaveisthesumofthetworedwaves)
Asyouwatchtheanimation,keepaneyeonthedarkblueline.Thisisthesumofthetworedlines.You'llnotethatitdoesn'tappeartomovinginthexdirection,onlytheydirection.Therearealsosomepointsthatnevermoveintheydirection(i.e.havenodisplacment,
(x, t) = A sin(kx − ωt+ ϕ)y2
(x, t) = [2Acos ] sin(kx − ωt+ )y′ϕ
2ϕ
2
2A2π3 A
π4π3 A
2π 2A
PHY 208 - superposition
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ever),andotherpointsthatoscillatebetween+Aand-A(maximumdisplacement).
Math of Superposition for opposite direction
Let'saddtwowavestravelinginoppositedirectiononthesamestring.( , ,and arethesame)
wave1:
wave2:
Since,
withatrigindentity(below)
Thisisnotofthestandardtravelingwaveformat!Indeed,thisequationdescribesastandingwave.
Standing waves
Fortravelingwaves,theamplitudeofdisplacementofeachelementwasthesame.Theywouldallgetdisplacedtoamaximum .Inastandingwave,theamplitudeswillbepositiondependent.
The argumentofthesinefunctionleadstothisphenomenon.
Nodes
Since ,wecandeterminewhereexactlytheamplitudeswillbezero.
Whenever , ,we'llobtainazeroforthedisplacement.
Rearranging:
.
Anti-Nodes
Likewise,when =1,theropewillundergoamaximumdisplacement.
k ω A
(x, t) = A sin(kx − ωt)y1
(x, t) = A sin(kx + ωt)y2
(x, t)y′ ==
(x, t) + (x, t)y1 y2A sin(kx − ωt) + A sin(kx + ωt)
(3)(4)
(x, t) = [2A sinkx] cos(ωt)y′
sinα + sinβ = 2 sin (α + β) cos (α − β)12
12
A
(x, t) = [2A sin ] cos(ωt)y′ (kx) positiondependent
kx
sin(nπ) = 0
kx = nπ n = 0,1,2,3...
x = nλ
2
sin(kx)
, , , . . .
PHY 208 - superposition
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Ifweoscillateafixedstringinsuchawaythatthereisanodeateachendpoint,wehaveeffectivelysetupastandingwave.Thiswillhappenwhenthefrequencyofoscillationisinresonancewiththestringcharacteristics.
Thus,onlywavelengthswhich'fit'inthestringwillcreateresonantoscillations.
Thesearecalled'harmonicmodes'(shownare1,2,and3).
L
Nodes and Antinodes
Hereisourstandingwave.
Onenodeandoneantinodearepointedout.(Althoughtherearemanymore)
Inabstractmath-land,allwehavetodotocreateastandingwaveisadduptwowavefunctions.Inthephysicaluniversehowever,wehavetodoalittlebitmore.Tounderstandhowastandingwaveiscreatedinaphysicalsystem,we'llneedtoseewhathappenswhenwavesbounce.
Boundary (hard)
Ifwesendawavepulsedownastring,wherethestringisfixedatthefarend,weseethatthewaveformflips.
Anotherwayofphrasingitistosay,thewaveformisinverteduponreflection,orundergoesaphaseflipof180°
Boundary (soft)
Ifwesendawavepulsedownastring,wherethestringislooseatthefarend,weseethatthewaveformdoesnotflip.
Creation of a standing wave
Standing waves and resonance
Theimagehereshowsthefirst3harmonicmodesofthestring,modes1,2and3.Thefirstmodeisoftencalledthefundamental,orlowestharmonic.
Resonance frequencies
Whatdeterminestheresonancefrequencies?
kx =
=
, , , . . .π
23π
25π
2
(n+ )π for n = 0,1,2...12
(5)
(6)
PHY 208 - superposition
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Rememberthat .Now,thevelocityofthewavewavegivenbythephysicalcharacteristicsofthestring:thelinearmassdensity, ,andthetension, .
Thereforewecanwrite:
Quick Question 2
Whenawireundertensionoscillatesinitsthirdharmonicmode,howmanywavelengthsareobserved?
1. 1/32. 2/33. 1/24. 3/25. 2
Quick Question 3
Whichofthesecouldbethefrequencyofastandingwavewithawavespeedof12m/sasitoscillatesona4.0-mstringfixedatbothends?(itmightnotbethelowestharmonic)
A. 2.5HzB. 5.0HzC. 10HzD. 15HzE. 20Hz
Example Problem: Piano String
ThelowestnoteonmostpianosisA0.Ithasafrequencyof27.5Hz.Thevibratingsectionofthiswireonagrandpianois1.9meterslong.[1meterofpianowirehasamassof200g]Whatisthetensioninthestring?
Interference
Consideraspeakerplayingaprettysinusoidalwavewithawavelength .
Then,anotherspeakerplayingtheexactsamepitchisplacedinfrontofSpeaker1,sothatthetwospeakersareexactlyonewavelengthapart.
Thetwosignals,orwaves,willconstructivelyinterferecreatingasignaloflargeramplitude.
v = λf
μ τ
f = = n = nv
λ
v
2L
τ
μ
−−√ 12L
Example Problem #1:
λ
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spkr 1 spkr 2
sum of 1+ 2
Now,let’simaginethesecondspeakerwasproducingawaveexactlyoppositetothespeaker1wave.
Thesumofthesetwowaveswillnowbeequaltozero,sincetheyarealwayscreatingdestructiveinterference.
Thesetwowavesarecalled“outofphase”.
spkr 1 spkr 2
sum of 1+ 2
Math of 1-D audio interference
IfwanttoknowwhatthesoundislikeatpointA,we’llneedtoknowhowfaritisfrombothsources:
spkr 1 spkr 2
A
If isequaltowavelengthtimesawholenumber,thentheamplitudeoftheoscillationsisincreased:
If isequaltowavelengthtimeshalfanintegernumber,thentheamplitudeoftheoscillationsisdecreased:
Two speakers
Δd
Δd = nλ
Δd
Δd = nλ
2
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Math of 1d interference
Thephasedifferencebetweentwowaveswilldeterminewhethertheinterferenceisconstructiveordestructive.Forconstructive:
andfordestructive:
Theequationsoftwotravelingwaves(travelinginthesamedirectionwiththesamefrequencyandwavelength)aregivenby:
and
Forthesewaves, meanstheintitialphaseconstantofwave1.
Thus,thephases,ortheargumentsofthesinetermsaregivenby:
and
Ifwesubtractthesetwophases,thatisfind (akathephasedifference):
Andso,forthecaseoftwosoundsourcesinonedimension,ifwewanttofigureoutwhenconstructiveinterferencewilloccur,basedoneithertheinitialphaseconstantsofthetwowaves,ortheseparationinspace:
Orfordestructiveinterference:
1d-interference
Hereisthephasedifferenceequation:
Wecanseethattherearetwocontributions:
1. Thepath-lengthdifference, ,inproportiontothewavelength.
Δϕ = 0, 2π, 4π…
Δϕ = π, 3π, 5π…
= A sin(k − ωt+ )y1 x1 ϕ10
= A sin(k − ωt+ )y2 x2 ϕ20
ϕ10
= k − ωt+ϕ1 x1 ϕ10
= k − ωt+ϕ2 x2 ϕ20
Δϕ
Δϕ = − = k( − ) + ( − ) = ( − ) + Δϕ2 ϕ1 x2 x1 ϕ20 ϕ102π
λx2 x1 ϕ0
Δϕ = ( − ) + Δϕ = 0, 2π, 4π…2π
λx2 x1
Δϕ = ( − ) + Δϕ = π, 3π, 5π…2π
λx2 x1
Δϕ = ( − ) + Δ2π
λx2 x1 ϕ0
−x2 x1
PHY 208 - superposition
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Consider2points.Weneedtofigureouthowfartheyarefromeachsoundsource.
point 1
point 2
crest
trough
L1L2
L2L1
2. Theinherentphasedifferencebetweenthetwooscillators.
Ifwehavetwoidenticalsourcesthatareinphase(i.e. ),thenonlythepathlengthdifferencewilldetermineifthewavesconstructivelyordestructivelyinterfere.
where isanintegerwillcreatemaximumconstructiveinterference.Thismakessense.Ifthewavesareseparatedbyawholenumberofwavelengths,thenit'sessentiallythesameasiftheyarenotseparatedatall.
Sounds waves in 2-dimensions (or 3)
Hereisasinglesource.Imaginejustonespeaker,creatingpressurewavesintheair.
Nowweaddasecondsource.
We'llneedtobeverycarefulwithinterpretingthecontrast(i.ecolorscheme)ofthisplot.Ifapointisredorblue,thatmeansthemedium'sdisplacementislarge.Ifapointiscoloredwithwhite,thatmeansthemedium'sdisplacementiszero.Mostpositionswillalternatebetweenred,white,blue,white,redastimeadvances,andthewavepropagates.However,someregions,asyoucanseeintheanimatedversion,remainwhiteatalltimes.Thesearetheregionsofdestructiveinterferences,wherethewavesfromthetwosorcesinterferedestructively.Attheselocations,thereisnodisplacementofthemedium,andthusnosoundisheard.
Constructive and Destructive Interference
Wecanfindsomegeneralguidelinestodetermineifwe'llhaveconstructiveordestructiveinterferencebasedonthepositionofthelistener.
1. Case1:Thepathlengthdifferenceisequaltoawholenumberofwavelengths:
Thisyieldsconstructiveinterference2. Case2:Thepathlengthdifferenceisequaltoahalfnumberofwavelengths:
Thisyieldsdestructiveinterference
Rephrased in terms of phase.
Δ = 0ϕ0
Δx = mλ
m
ΔL = nλ
ΔL = (n+ ) λ12
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Mathematicallyspeaking,interferenceisdeterminedbythephasedifferencebetweentwowaves.
If is0, ,oranymultipleof ,constructiveinterferenceoccurs
i.e. where
Thisiscontrarytodestructiveinterferencewhichoccursatoddmultiplesof
i.e. where
ϕ 2π 2π
ϕ = m× 2π m = 0,1,2,…
2π
ϕ = (2m+ 1)π m = 0,1,2,…
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Quick Question 4
crest trough trough crest
12
3
Herearetwosoundsourceemittingwoundwavesinphase.Thesolidlinesarethemaxiumumpressureregions,thedashedlinesshowthelocationoftheminimumpressureregions,at
Atpoint1,theinterferenceis
1. MaximumConstructive2. Constructive,butlessthanmaximum3. PerfectlyDestructive4. Destructive,butonlypartially5. Nointerference
Quick Question 5
Atpoint2,theinterferenceis
1. MaximumConstructive2. Constructive,butlessthanmaximum3. PerfectlyDestructive4. Destructive,butonlypartially5. Nointerference
Quick Question 6
Atpoint3,theinterferenceis
1. MaximumConstructive2. Constructive,butlessthanmaximum3. PerfectlyDestructive4. Destructive,butonlypartially5. Nointerference
t = t0
Example Problem #2:
PHY 208 - superposition
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Twospeakersinaplaneare2.0mapartandinphasewitheachother.Bothemit700HzSoundwavesintoaroomwherethespeedofsoundis341m/s.Alistenerstands5.0minfrontofthespeakersand2.0mtoonesideofthecenter.Describetheinterferenceatthispointinspace.
Reflecting sound waves
Justlikewithawaveonastring,iftheconditionsareright,we'llobtainastandingwaveinsidethetube.
And,justlikewithwavesonastring,onlycertainwavelengthswill'fit'.
Pressure and Displacement in tubes
closed-closed open-open∆(pressure)
displacement
∆(pressure)
displacement
m = 1
m = 2
m = 3
m = 1
m = 2
m = 3
Thisfigureshowsthefirstthreemodesofoscillationforstandingwavesintubes.Ontheleft,weseeatubethatisclosedonboth
Example Problem #2:PHY 208 - superposition
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closed-open∆(pressure)
displacement
m = 1
m = 3
m = 5
…arejusttubeswithairflowing.Thelengthofthetubeisoneimportantaspect.Differentnotesarecreatedbycoveringholes,whicheffectivelychangethelengthofthetube.
ends.Sincethetubeisclosed,particlesofaircannotmovepasttheboundary.Thus,thedisplacementgraphsshowanodeattheendsofthetube.However,inthepressuregraph,theendsareoccupiedbyanti-nodes.Ontheright,thesameisillustratedbutfortubesthatareopenonbothends.Now,theparticlesarefreetomoveinandoutofthetubeattheend.However,sincethepressureattheendofthetubeissetbyatmosphericpressure,thisvaluecannotchange.Therefore,weseenodesinthepressuregraphsattheends,andanti-nodesinthedisplacementgraphs.
Asyoucanseeinthetopleft,thefundamentalmodeforaclosed-closedtubehasafrequencywhosewavelengthisequaltotwicethelengthofthetube.Thesecondharmonicfrequencycontains1fullwavelength.
Pressure and Displacement in tubes
Theseplotsshowtubesthatareclosedononeend,andopenontheother.Thefundamentalmodeforaclosed-opentubewillhaveawavelengthequalto4timesthelengthofthetube.
Wind Instruments...
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Tuningforksareverycleanoscillators.
Theends(calledtines)movebackandforthinaverysinusoidalmotion.
Thismotioncreatesthepressurewavesthatwehearasapuresinewave.
Thewavelengthandthusthefrequencyaregivenbythegeometryandmaterialsofthetuningfork.
disp
lace
men
t
time
Tuning Forks
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Hereisatuningfork
Andhereisthefrequencyspectrumofthesoundthetuningforkmakes.Thespectrumisagraphofthesignalstrengthversesfrequency.Wecanseethatonefrequencyhasaveryhighvalue.
Sound, in general
Rarelydowehearperfectsinusoidaloscillations.Mostinstruments,noises,vocalizationsaremixturesofmanydifferentoscillations.
ThisismiddleConthepiano.Youcanseethefundamentalat261Hertz.Buttherearealotofhighertonesalsopresent.Thezoominshowsevenmorelittlepeaks.
Beats
Sofar,allthistalkofinterferencehasbeenabouttwosourceswiththesamefrequency.Inreallife,that’susuallynotthecase.
Herearetwoplotsofsinewavescomingfromtwospeakerslocatedatthesameplace.
Thetopsoundhasafrequencyof1Hertzwhilethebottomhasafrequencyof1.1Hertz.
It’shardtotelltheyaredifferent.
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5 10 15 20
2
1
1
2
region of constructive interference
region of destructive interference
5 10 15 20
2
1
1
2
overlap
out of phase
Leteachofthetwosourcewavesbegivenby:
Theresultantdisplacementisthen:
= cos t and = cos ts1 sm ω1 s2 sm ω2
s = + = (cos t+ cos t)s1 s2 sm ω1 ω2 (7)
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5 10 15 20
2
1
1
2
region of constructive interference
region of destructive interference
Since ,theterminbracketscanbeconsideredamodulatedenvelopeoscillatingatangularfrequency .
Itsminandmaxvaluesoccurtwiceinagivencycle.
Thisoscillationbetweenaminandmaxiswhatwehearasbeats.
or,intermsoffrequency, :
M
L
Mass(kg) (Hz)2.00 684.00 976.00 1178.00 13510.00 152
Usingatrigidentity,thisbecomes:
if and ,thentheaboveequationbecomesmoretidy:
Aheavymassissuspendedfroma1.65mlongsteelwire.(Thewirehasamassof5.85g)Thefrequenciesofthe3rdharmonicoscillationofthewireasafunctionofmassaregivenbelowinthetable.Usethisdatatodetermineavalueof.
Enter:Fit{{0,0},{2,68^2},{4,97^2},{6,117^2},{8,135^2},{10,152^2}}atwolframalpha.
Link
s = 2 cos[ ( − ) t] cos[ ( + ) t]sm
12
ω1 ω212
ω1 ω2 (8)
= ( − )ω′ 12
ω1 ω2 ω = ( + )12
ω1 ω2
s(t) = [2 cos t] cosωtsm ω′
s(t) = [2 cos t] cosωtsm ω′
ω≫ ω′
ω′
= 2 = 2( )( − ) = −ωbeat ω′ 12
ω1 ω2 ω1 ω2
f
= −fbeat f1 f2
f3
g
PHY 208 - superposition
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