﻿
Waves
.
Thegeneralsubjectofwavesissolargethatitwouldtakeseveralbooksjusttogetsomeideaof
ofthesimplestapplications,chosenbecausetheyappearinsomanycontexts.
7.1AString
Theprototypewaveisawaveonastring.Itcanbeaguitarstringorabullwhip,butthebasicanalysis
isthesameanditstartsfrom
~
F
=
m~a
.
Assumethatthestringisstretchedstraightalongthe
x
-axis,andI’llassumethatitsoscillations
occurinasingleplanesothatthe
y
-coordinatesayshowfarthestringisfromthisequilibriumposition.
Theshapeofthestringisthendenedbytheequation
y
=
f
(
x;t
),representingthedisplacementof
thestringfromtheaxisatpoint
x
andtime
t
.
The keytrickin setting up the equations ofmotion is toimagineslicingthe stringapartand
examiningthesegmentbetween
x
and
x
+
x
. Thissegmentwillthenhaveforcesactingonitcaused
bythepartofthestring(
<x
)pullinginonedirectionandpartofthestring(
>x
+
x
)pullingin
another. Therewillalsobe e theforcesbygravity and bythesurroundingair. . Ifthissortofdivision,
usingan imaginary cut in thestring,looks fairly plausibleand maybeeven commonplace,itwasn’t
whenthesubjectwasinvented. Thatyoucanimagineslicingthestring(withoutreallydoingit)was
thesubjectofgreatdebateandworrybythosewhodiditthersttime.
y
x
x
+
x
Aswithmostotherproblemsinvolvingoscillations,thisbecomessimpleonlywhenthedisplace-
mentsaresmall. Without t that assumptiontheresultingdierentialequationswill be extraordinarily
dicult. Ifyou’veeverlookedataguitaroranyotherstringedinstrumentbeingplayed,youknowthat
themotionofthestringisverytinyandsothissmallmotionapproximationwillbeaverygoodone.
Thelinearmassdensityis
=
dm=dx
andthetensioninthestringis
T
(
x
). Thistension,the
forcethatonepartofthestringexertsonthepartitisconnectedto,neednotbeaconstant,andthe
sameistrueofthemassdensity.Theinterval
x
issmall(eventually!0),sothemassintheinterval
is
m
=
(
x
)
x
.
x
x
+
x
(
x
)
(
x
+
x
)
Fig.7.1
Theforcesonthis
m
thesurroundingair.
F
y
T
(
x
)sin
(
x
)+
T
(
x
+
x
)sin
(
x
+
x
)
mg
b
xv
y
(7
:
1)
HereI’mtakingthesimplestmodelfortheairresistance,linearviscosity,withaproportionalityfactor
b
.
Thisstringishorizontal,sothattellsyouthegravitationalterm.Theanglethatthestringmakeswith
227
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7|Waves
228
the
x
-axisdependson
x
,sotogetthe
y
-componentoftheforceIneededtoindicatethatexplicitly.
Uptoherethepresentationisstraight-forward,I’vejustbeenverycarefulinexpressingtheforces.
Nowforatrick. Thedisplacementofthestringissmallandtheanglethatitmakeswiththeaxisis
small. ThatmeansthatIcanuseasmallangleapproximationforthesine. Nottheusualonethough.
smallangle =) ) sin
tan
Insteadofapproximatingthe sine by theta, approximate it by the tangentoftheta. . The e reason for
wantingtodothisisthatthetangentissimplythederivativeof
y
withrespectto
x
. Equation(7.1)
nowbecomes
F
y
T
(
x
)
@f
@x
(
x;t
)+
T
(
x
+
x
)
@f
@x
(
x
+
x;t
(
x
)
xg
b
x
@f
@t
(
x
+
x=
2
;t
)
(7
:
2)
Itookthelasttermatthecenterpointoftheinterval,butitwon’tmatterintheend. Iswitchedto
partialderivativenotationbecausetherearetwoindependentvariables,andforthe
x
-derivativetimeis
xed. Forthe
t
-derivative,positionisxed.Forthe
ma
sideoftheequation,youhave
F
y
=
ma
y
=
(
x
)
x
@
2
f
@t
2
(
x;t
)
Ifyouthinkthatthisoughttobeevaluatedat
x
+
x=
won’tmatter. IfInowtakethelimitas
x
!0,Igetthemostuninformativeequation,0=0. Not
x
andthentakethelimit.
1
x
T
(
x
)
@f
@x
(
x;t
)+
T
(
x
+
x
)
@f
@x
(
x
+
x;t
)
(
x
)
g
b
@f
@t
(
x
+
x=
2
;t
)
=
(
x
)
@
2
f
@t
2
(
x;t
)
Now as s 
x
! 0, , most t of the terms the equation simply y become e the function at
x
. The e rst
termis dierenthowever,anditgoes to0
=
[something(
x
+
x
) something(
x
)alldividedby
x
]Thelimitis
@
@x
T
@f
@x
g
b
@f
@t
=
@
2
f
@t
2
(7
:
3)
Have I left out any physics? ? Yes. . A A thin string g is easy y to bend, , but t a thicker one resists
deformation. It’sstiandI’veignoredthisproperty. . Isthatimportant? Itisifyouareaprofessional
thenextone.
F
x
onthemass
m
? Gobacktotheequation(7.1)andthepicturethataccom-
paniesit.
F
x
T
(
x
)cos
(
x
)+
T
(
x
+
x
)cos
(
x
+
x
)
(7
:
4)
Inthis example,because the string ishorizontalthere isno
x
-componentfromgravity. Thisimplies
thatwhenthestringishorizontal,thenforsmallanglesandformotioninthe
y
-direction,
F
x
=
ma
x
is zeroandthetensionisconstant. . Iwillleave e the
T
factorasisanyway,allowingthatitmayvary
with
x
,notbecauseit’sneededforthisproblemwiththestringbutbecauseit’sharmless,notcausing
anyextrawork. Alsothereareplentyofwaveproblemswheresomethinganalogoustothisfactorwill
showupandyoumayaswellgetusedtoithere. InmostcaseswhereI’mreallysolvingsomething,it
mg
inthis
equationbutnotinthe
F
y
equation. Inthatcase,puttheextraterm()
mg
intoEq.(7.4)andit
willtellyouhowthetensionvarieswith
x
.
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7|Waves
229
7.2Staticcase
Startwiththesimplestcase. Nothingismoving. . Astringisstretchedtautbetweentwopostsandit
sags. Letthetensionandthemassdensitybeconstanthere.
Theequation(7.1)fortheshapeofthestringisnow
T
@
2
f
@x
2
g
=0
(7
:
5)
Thisiseasytointegrate. (Dropthe
t
from
f
(
x;t
).)
f
(
x
)=
gx
2
=
2
T
+
Ax
+
B
Applytheboundaryconditionsthatthestringistiedatthetwoends
f
(0)=
f
(
L
)=0 =)
B
=0 and
gL
2
=
2
T
+
AL
=0
Thecurveisthen
f
(
x
)=
g
2
T
x
(
L
x
)
(7
:
6)
Theshapeofthewireshungbetweentelephonepolesinthissmallangleapproximation*isaparabola.
Thebiggerthetension,thelessitsags. Theheavieritis,themoreitsags. Alsolookatproblem7.33
7.3TheWaveEquation
InEq.(7.3),thesimplestcasethatexhibitswavescomesbyassumingthatthemassdensity andthe
tensionareconstantandthatgravityandairresistancearenegligible.
T
@f
@x
2
=
@
2
f
@t
2
or
@
2
f
@x
2
1
v
2
@
2
f
@t
2
=0
where
v
2
=
T
(7
:
7)
Thisiscalled\the"waveequation,asiftherewerenoothers.Itisthemostimportantwaveequation,
anditfullydeservesthedenitearticle. Theparameter
v
hasdimensionsofspeed,andthat’seasyto
seebecause
f
isthesameinthenumeratorinbothterms oftheequation,sothedenominators,
x
2
and
v
2
t
2
musthavethesamedimensions.
v
isthepropagationspeedofthewave.
Howdoyousolvethewaveequation? Therearemany y ways,startingwithinspiredguesswork
andcarryingonthroughseparationofvariablesandonintotheworldofcharacteristics.I’llmostlystay
withtherstmethod,thoughonedirectapproachtothesolutionappearsinproblem7.21. Fornow,
tryacosine,
f
(
x;t
)=
A
cos(
kx
!t
+
)andseewhathappens.
@
2
f
@x
2
1
v
2
@
2
f
@t
2
Ak
2
cos(
kx
!t
+
)+
A!
2
1
v
2
cos(
kx
!t
+
)
A
cos(
kx
!t
+
)
k
2
!
2
=v
2
=0
(7
:
8)
Thecosineisn’tidenticallyzeroand
A
betternotbeorthere’snothingthere. What’sleftdetermines
therelationbetween
k
and
!
.
k
2
!
2
=v
2
=0
;
or
v
2
=
!
2
k
2
=
T
(7
:
9)
* Thestaticcasecanbesolvedwithoutthesmallangleapproximation,andtheresultisacatenary
|ahyperboliccosine
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7|Waves
230
!
anythingthatIwant,
andaslongas
k
hasthecorrespondingvalue
!=v
thisworks.Comparetheharmonicoscillator,where
thereareexactlytwoindependentsolutionsandonefrequencyforthesystem,notaninnitenumber.
Thisrichnessofsolutionshereistypicalofpartialdierentialequationssuchas(7.7).
vt
kx
!t
+
=0
Fig.7.2
Is
v
reallyaspeed? Drawapictureofthefunction
y
versus
x
,andwheredoyounditspeak?
Apeakwilloccurwhereverthe argumentofthecosineiszerooramultipleof 2
. Thatdenethe
equationforthepositionofapeak. Pickone:
kx
!t
+
=0
implies
x
=
!
k
t
+
k
=
vt
+aconstant
(7
:
10)
Thepositionofthemaximummovesatpreciselythevalueof
v
,eithertotherightorleftdepending
onthesignof
!=k
.*Mostcommonlythefrequency,omega,istakenaspositiveand
k
canhaveeither
sign,therebydeterminingthedirectionofthewave’stravelasrightorleftas
k
ispositiveornegative.
k
iscalledthewavenumber,andhasdimensionsofangleperdistance,justas
!
’sisanglepertime.
Thedistanceinwhichthewaverepeatsisthewavelength,
. Thisimpliesthatj
k
j
=2
.
Theequation(7.7)islinear. Thatmeansthatthesumoftwosolutionsisasolution,andthat
youneedisthat
!
2
1
=k
2
1
=
!
2
2
=k
2
2
=
:::
=
v
2
.
A
1
cos(
k
1
x
!
1
t
+
1
)+
A
2
cos(
k
2
x
!
2
t
+
1
)+
A
3
cos(
k
3
x
!
3
t
+
3
)+
Thisisapictureofsuchacombinationformedbypickingasetofrandomvaluesofthe
A
’s,
k
’s,
and
’s(at
t
=0).SoundobeysthesamewaveequationastheoneI’mdescribing,andyouknowthat
soundwavescancarryenormouslycomplexinformation,whetherinmusicorinwords.
Are there any othersolutions besides thecosines (orsinesforthat matter)? ? Yes,justas s the
ordinarydierentialequation
md
2
x=dt
2
kx
hastwoarbitraryconstantsinitssolution,
A
cos
!t
+
B
sin
!t
,thewaveequationhastwoarbitraryfunctions!
y
=
f
(
x;t
)=
f
1
(
x
vt
)+
f
2
(
x
+
vt
)
(7
:
11)
Thedemonstrationofthisfactisnomorethanpluggingintotheequationandusingthechainrule.
@f
1
@x
=
f
0
1
(
x
vt
)
;
@
2
f
1
@x
2
=
f
00
1
(
x
vt
)
;
@f
1
@t
vf
0
1
(
x
vt
)
;
@
2
f
1
@t
2
=(
v
)
2
f
00
1
(
x
vt
)
Substitute these intoEq.(7.7) anditworks. . Note: : the\prime"notationisthederivativeof
f
with
respecttoitsargument,notthederivativewithrespectto
x
,sothatsin
0
=cos. Thesamecalculation
* Thisiscalledthephasevelocity todistinguishitfromotherdenitionsofvelocitythatwillshow
up later,in section 7.13. Thename e comes fromthefactthatyouarefollowingapointofconstant
phase,theargumentofthecosine,(
kx
!t
+
)inthewave.
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7|Waves
231
worksfor
f
2
(
x
+
vt
).Thismeansthatanywaveformwillmovealongthestring.Allthatyouhaveto
doistogetitstarted. ThecosinesolutionthatIstartedwithissimplyaspecialcaseofthisgeneral
solution. Andyes,Eq.(7.11)is themostgeneralsolutionforthewaveequation. Theproofistedious
butnotdicult,involvingonlyachangeofvariables.Seeproblem7.21.
Anyfunction (well,anyfunction withtwoderivatives)willprovide asolution,andthatiswhy
wavescancarrysomuchinformation. Tondthebehaviorofasinglemass,youneedtospecifyits
initialpositionandvelocity,andessentially thesamethinghappenshere. . Specify y theinitialposition
andinitialvelocityateachpointonthestring. Assume
f
(
x;
0)=
F
(
x
)
;
@f
@t
(
x;
0)=
G
(
x
)
(7
:
12)
ApplyEq.(7.11)forthegeneralsolution,thenthesebecome
f
1
(
x
)+
f
2
(
x
)=
F
(
x
)
and
vf
0
1
(
x
)+
vf
0
2
(
x
)=
G
(
x
)
Dierentiatetherstofthesetoget
f
0
1
(
x
)+
f
0
2
(
x
)=
F
0
(
x
)andyounowhavetwoalgebraicequations
forthetwounknowns
f
0
1
and
f
0
2
.
f
0
1
(
x
)=
1
2
F
0
(
x
1
2
v
G
(
x
)
;
f
0
2
(
x
)=
1
2
F
0
(
x
)+
1
2
v
G
(
x
)
Andnowdotwointegrals. Theseareallfunctionsofasinglevariable,so o nothingunusualhappens.
Youhaveconstantsofintegration,butthey’reeasytohandle.
f
1
(
x
)=
1
2
F
(
x
1
2
v
Z
x
0
G
(
x
0
)
dx
0
f
2
(
x
)=
1
2
F
(
x
)+
1
2
v
Z
x
0
G
(
x
0
)
dx
0
(7
:
13)
y
=
f
(
x;t
)=
1
2
F
(
x
vt
1
2
v
Z
x
vt
0
G
(
x
0
)
dx
0
+
1
2
F
(
x
+
vt
)+
1
2
v
Z
x
+
vt
0
G
(
x
0
)
dx
0
=
1
2
F
(
x
vt
)+
F
(
x
+
vt
)
+
1
2
v
Z
x
+
vt
x
vt
G
(
x
0
)
dx
0
(7
:
14)
Youshouldofcoursedirectlyverifythatthisexpressionsatisestheinitialconditions.Seeproblem7.8.
Example
Take
F
(
x
)=
A
(
L
2
x
2
)for
L<x<
+
L
and zeroelsewhere. . Taketheinitialvelocitytobe
zero:
G
0,then
f
(
x;t
)=
1
2
F
(
x
vt
)+
F
(
x
+
vt
)
(7
:
15)
Thisisapluckedstring,andthedisturbancegoesequallyinbothdirections. Forastruckstring,see
problem7.10.
7.4EnergyandPower
Waves can transportenergy. . Certainly y oceanwaves do, , and soundwaves s that are loud enoughcan
totheideasofkineticenergydensityandpotentialenergydensity.
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7|Waves
232
Startwiththekineticenergy;it’seasier. DothesamethingthatIdidtogetthewaveequation
intherstplace: lookatasmallpieceofthestringintheintervalfrom
x
to
x
+
x
. Ithasamass
m
(
x
)
x
.Itskineticenergyisthengivenbytheusual
mv
2
=
2formula.
Watchout!
v
isnot
v
.
The
v
thatshowsupinthewaveequation isthespeedwithwhichthedisturbancemovesalongthe
string,
p
T=
. The
v
thatappearsintheexpressionforkinetic energy is thespeed ofthemass
m
.
Theyarenotatallthesamething. Theformeristhespeedwithwhichashapemoves,notthespeed
withwhichamass moves. . Thatlatterone e occurs ataxedvalue of
x
soitis apartialderivative,
@f=@t
.
K
1
2
x
@f
@t
2
so
K
x
!
dK
dx
=
1
2
@f
@t
2
(7
:
16)
Thisisthekineticenergydensity(Joulespermeter).There’snothinginthisderivationthatrequires
tobeaconstant. Itappliesjustaswelltothegeneralequation,Eq.(7.3).
Fortherstsolution,thecosine,whatisthis?
f
(
x;t
)=
A
cos(
kx
!t
),so
dK
dx
=
1
2
A!
sin(
kx
!t
)
2
=
1
2
A
2
sin
2
(
kx
!t
)
f
dK=dx
Fig.7.3
Thisshowsthetwofunctions
f
and
dK=dx
onthesamegraph,andyoucanseethatthekineticenergy
density is zero whenthemagnitudeofthewave isatits maximumandviceversa. . Whenthestring
isatitspeak,themoleculesatthatpeakhavestoppedandhavenotyetturnedaround|that’szero
speed.
Thepotentialenergyis alittletrickier. . When n youusetheequation 
m
=
x
,themass is
whatoccurredalongtheaxis intheinterval
x
. Thelengthofthestringwhen n itis deformedcan’t
stayexactlythesamethough,becausethe
y
-coordinatesatitsendsaren’tthesameaswhenitwasat
rest. Itslengthis
=
p
(
x
)2+(
y
)2
x
y
Ittakessomeworktostretchthisstring. AllthatIhavetodoistocomputethisworkandthatwill
givemethepotentialenergyinthispieceofstretchedstring. Thetensioninthestringis
T
,or
T
(
x
),
andtheworkthatIdoinchangingthelengthofthisstringsegmentisthisforcetimesthestretch.
W
T
x
Toevaluatethis,usethebinomialseries.
=
x
s
1+
(
y
)2
(
x
)
2
=
x
1+
1
2
(
y
)
2
(
x
)
2
+
(Theslopeissmall.) Subtracttheoriginallength
x
and
W
T
x
1
2
(
y
)
2
(
x
)2
so
W
x
!
dW
dx
=
1
2
T
@y
@x
2
(7
:
17)
7|Waves
233
Thisisthepotentialenergydensity. Asbefore,thisderivationappliestothegeneralequation,forwhich
T
canbeafunctionofposition. Thetotalenergydensityisthesumofthese.
dE
dx
=
1
2
@f
@t
2
+
1
2
T
@f
@x
2
(7
:
18)
Howdoes energy  owdownthestring? ? Onepartofthestringpullsonitsneighboranddoes
workonit. Whatisthepowertransmitted?ThereareacoupleofwaystodothisandI’lldoonehere,
leavingthe other(much easier)methodtoyou inproblem 7.2. Thetotalenergywithinthe e interval
a<x<b
istheintegralovertheenergydensity.
E
(
t
)=
Z
b
a
dx
dE
dx
=
Z
b
a
dx
"
1
2
@f
@t
2
+
1
2
T
@f
@x
2
#
(7
:
19)
Notethatthisisafunction of timeas thewavemoves past. . Thetime e variabledoes not appearin
thelimits of the integral; ; they’re xed. . It t remains in the wave function
f
itself, because it is s just
the
x
-variablethatyouareintegrating. The
t
-variableisjustsittinginsidetheintegralwaitingtobe
dierentiated. (Lookahead d toEq.(7.25) to seeaconcreteexampleofthis.) ) Fromconservation n of
energy,thetotalpower owingintothisregionisthetime-derivativeof
E
.
dE
dt
=
d
dt
Z
b
a
dx
dE
dx
Thelimits
a
and
b
arenotfunctionsoftime,soIcansimplymovethederivativeundertheintegral,
makeitpartial,anddothedierentiationonthetwoterms.
dE
dt
=
Z
b
a
dx
@f
@t
@
2
f
@t
2
+
T
@f
@x
@
2
f
@x@t
(7
:
20)
Nowwhat? Ihaven’tusedthatfactthat
y
=
f
(
x;t
)satisesthewaveequation,andI’llhavetodo
thatorIcan’tgetanywhere.TheonlyobviousplacethatIcanapplytheequationisontherstterm.
Itbecomes
Z
b
a
dx
@f
@t
v
2
@
2
f
@x
2
+
T
@f
@x
@
2
f
@x@t
(7
:
21)
Now integrate e by y parts. . You u can n do it on n eitherterm; ; I’ll pickthe rst. . I’m m assumingthat
T
is
independentof
x
,thoughthemoregeneralcaseisnodierent: problem7.15
Z
b
a
dx
@f
@t
v
2
@
2
f
@x
2
=
@f
@t
v
2
@f
@x
b
a
Z
b
a
dx
@
2
f
@t@x
v
2
@f
@x
(7
:
22)
Putthisbackintotheequation(7.21)andthetwointegraltermscancel.Dothey? You’dbettercheck.
Allthat’sleftis
dE
dt
=
@f
@t
v
2
@f
@x
b
a
=
T
@f
@t
@f
@x
b
a
=
T
(
b
)
@f
@t
(
b;t
)
@f
@x
(
b;t
T
(
a
)
@f
@t
(
a;t
)
@f
@x
(
a;t
)
(7
:
23)
Thissaysthattherateofchangeofthetotalenergyinthisintervalcomesfromtwoterms:one
attherightendandoneattheleftend. ThisiswhatIoughttohaveexpected.Theenergywithinthis
7|Waves
234
intervalcancomeonlyfromsomethinghappeningattheends;itcan’tjumpoverandappearatsome
distantpoint.Icantheninterprettheexpression
P
T
@f
@t
@f
@x
(7
:
24)
as the powermoving (to the right) ) past t the point
x
. Why y the minus s sign? ? Look k at the term m in
Eq.(7.23) atthe point
a
. Ifthatterm(withitsminus s sign)is positive,itrepresentsenergy coming
intotheintervalfromtheleft. Havinggonethrough h this,youwillbetterappreciatethesimplicityof
thealternatederivationinproblem7.2.
Example
TakethecosinewavesolutionfromEq.(7.8),setting
=0forconvenience(
f
=
A
cos(
kx
!t
)),
andseewhatthetotalenergyinanintervalis.
dE
dx
=
1
2
@f
@t
2
+
1
2
T
@f
@x
2
;
and
f
(
x;t
)=
A
cos(
kx
!t
)
E
=
Z
b
a
dx
1
2
A
2
!
2
sin
2
()+
Tk
2
sin
2
()
=
A
2
Tk
2
Z
b
a
dx
sin
2
(
kx
!t
)
=
A
2
Tk
2
Z
b
a
dx
1 cos
2(
kx
!t
)

=
A
2
Tk
2
h
(
b
a
1
2
k
sin(2(
kb
!t
)) sin(2(
ka
!t
))
i
(7
:
25)
(RememberEq.(7.9).) Thersttermintheresultrepresentstheaverageenergyintheinterval,and
theothertermsrepresentthe uctuationsintotalenergyasthewaveentersat
a
andleavesat
b
.
Thepoweritselfinthisexampleis
P
T
@f
@t
@f
@x
T
+
A!
sin(
kx
!t
)

Ak
sin(
kx
!t
)
=+
A
2
Tk!
sin
2
(
kx
!t
)=
A
2
!
2
v
sin
2
(
kx
!t
)
(7
:
26)
Noticethatthepoweriszerowherethewaveisamaximumandviceversa. Orisit? Andwhatisthe
timederivativeoftheenergycomputedinEq.(7.25)anddoyoubelieveit?
7.5ComplexForm
Theequation(7.11)showed
f
(
x
vt
)asthemostgeneralsolutions ofthewaveequation. . Nothing
restricts
f
tobeingreal,andusingcomplexnotationprovidesagreatsimplicationinmanycalculations
e
i!t
oscillator.Intheeasycasesitdoesn’tmatter,butwhenthegoinggetstough
:::
.
kx
!t
)
;
use
e
ikx
i!t
(realpartunderstood)
Itwilltakesomepracticetogetusedtousingthiscomplexform,buteventuallyyouwillwonderwhy
someonewoulddoanythingelse. Ifyoudoubtthis,trysolvingproblems7.45or7.46usingsinesand
cosines,thensolvethemwithcomplexexponentials. Thelengthofonemethodismeasuredinpages;
theotherwayismeasuredinlines. Itisuptoyoutodecidewhichyouprefer. Note:Whenyoucompute
energyorpoweryoucan’tusethismethod. They y involvesquaresoftheamplitude,andtherealpart
of(
e
ix
)
2
isnotthesameas(cos
x
)
2
7|Waves
235
7.6ThreeDimensionalWaves
Waves on astringare deceptively simple becausethey canmoveonly inasingle dimension. . Sound
waves,lightwaves,andseismicwavesareexamplesofmorecomplexwavesthatarenotconnedtoa
pressurevariationfromthemeanairpressure,anditsatisesthesameequationasawaveonastring
butwithmorevariables.
@
2
f
@x
2
T
@
2
f
@t
2
=0
becomes
@
2
p
@x
2
+
@
2
p
@y
2
+
@
2
p
@z
2
1
v
2
@
2
p
@t
2
=0
(7
:
27)
Here
v
2
=
p
0
=
0
forordinaryfrequencies.
p
0
and
0
aretheaveragebackgroundairpressureand
density,while
p
(
x;y;z;t
)isthedeviationfromthisbackground. Thefactor
istheratioofspecic
heats that appears soofteninthesubjectof thermodynamics. . Forfrequencies s ofabout 10
8
Hz and
more,thefactordeviatesfrom
tosomethingthateventuallyapproachesoneas
!
increases.
isnotmuchdierentfromthewaveonastring,onlywithmorepossibilities. Atrigonometricfunction
oracomplexexponentialwillstillrepresentawaveofaparticularfrequency. Youcanwrite
p
=
A
cos
k
x
x
+
k
y
y
+
k
z
z
!t
=
A
cos
~
k
.
~r
!t
or
Ae
i
(
~
k
.
~r
!t
)
(7
:
28)
Putthisintotheequationtoseeifitworks.
@
2
p
@x
2
k
2
x
A
cos
k
x
x
+
k
y
y
+
k
z
z
!t
withsimilarresultsforthederivativeswithrespectto
y
andto
z
. Assemblethesetoget
A
k
2
x
+
k
2
y
+
k
2
z
cos
k
x
x
+
k
y
y
+
k
z
z
!t
1
v
2
A!
2
cos
k
x
x
+
k
y
y
+
k
z
z
!t
?
=
0
Therequirementforthistoworkisthatthecoecientofthecosinebezero.
k
2
x
+
k
2
y
+
k
2
z
+
!
2
v
2
=0  !
!
2
k
2
=
v
2
;
where
k
2
=
k
2
x
+
k
2
y
+
k
2
z
Itdoesn’tmatterifyouuseacosineorasineoracomplexexponential:
e
i
(
~
k
.
~r
!t
)
isjustasmucha
solutionasacosine.
~
k
analogoffrequency. Dothisyourselfforthecomplexexponential.
One aspect of this is actually easierif you use thethreedimensionalvectornotation: : Which
kx
!t
)or
sin(
kx
!t
)or
e
i
(
kx
!t
)
.Ifthewaveismovingleftyoucanchangethesigninfrontofthe
!
oryou
canchangethesigninfrontofthe
k
.Forthecosineitdoesn’tmatter,butforthesineitdoes,turning
thefunctionupsidedown. Inthreedimensions,whenyouchangedirectionsthere’sreallyoneobvious
thingtodo. Youchangethedirectionofthevector
~
k
. Thereisthennoambiguityinwhattodo.
7.7Re ections
Waves often meet barriers. . Lightwaveshit t glass. . Oceanwaveshit t areef. . Soundwaves s hitawall.
Theanalysisofalltheseissimilar,sospendingsometimelookingatre ectionsonastringwillnotbe
wasted.
Constructan example thattypies theprocessby tyingtwostrings together, onethickerand
heavierthan the other. . You u know w that lightis s partially re ected when it hits glass(just look in a
7|Waves
236
window),soitisnotsurprisingwhenthesamethinghappenshere. Letthemassdensitybe
1
tothe
leftoftheoriginand
2
totheright. Awavecomesinfromtheleftandsomeofitgetsthroughthe
boundary,butnotall.Someisre ected.Whatequationdoyouhavetosolve?Thesameequationthat
startedthischapter,Eq.(7.3).Afterdroppingthe
g
term,Iassumedthat
T
and
wereconstantin
ordertoproduce\the"waveequation. Now
isn’tconstant,but
T
willbefortherstexample.
T
@
2
f
@x
2
=
(
x
)
@
2
f
@t
2
!
@
2
f
@x
2
=
1
v
2(
x
)
@
2
f
@t
2
(7
:
29)
Themassdensityisconstantintwodomains,
=
1
for
x<
0and
2
for
x>
0. Ineachregionthere
arethecorresponding(constant)wavespeeds.
v
(
x
)=
v
1
=
q
T=
1
;
(
x<
0)
and
=
v
2
=
q
T=
2
;
(
x>
0)
This equation is exactly the same equation thatyou have to solve when light hits asheet of glass
comingalongthenormal. Thespeed
!=k
ofthelightwaveinairdecreasesinglassbyafactorof
n
,
theindexofrefraction.Thefunction
f
Themathematicsisthesame.
Ifyoulookonly at the left partof thestring oronly attherightpart, , you alreadyknowthe
solutions. Theyarewavessuchas
A
cos(
kx
!t
)or
B
cos(
kx
!t
),andallyouneedisthatthe
functions satisfythewaveequation,whichmeans
!=k
=
v
1
or=
v
2
inthetwohalves. Iftheyhave
thesamefrequencythen*
k
1
=
!=v
1
k
2
=
!=v
2
f
(
x;t
)=
A
cos(
k
1
x
!t
)+
B
cos(
k
1
x
!t
) (
x<
0)
C
cos(
k
2
x
!t
)+
D
cos(
k
2
x
!t
) (
x>
0)
orit =
Ae
i
(
k
1
x
!t
)
+
Be
i
k
1
x
!t
)
(
x<
0)
Ce
i
(
k
2
x
!t
)
+
De
i
k
2
x
!t
)
(
x>
0)
(7
:
30)
There’ssomethingmissing. Thisworksfor
x<
0andfor
x>
x
=0? Itdoesnot
haveasolutiontothedierentialequationifyoudon’thaveiteverywhere. Ifitmissesapoint,then
it’snotasolution.
BoundaryConditions
Fortheparticularproblemofastringwithtwomassdensities,theboundaryconditionsare
f
iscontinuous,
@f=@x
iscontinuous
(7
:
31)
and wheredothesecomefrom? ? Iwill l showtwodierent waysto get the answer. . One e byphysical
Thestringhasn’tbroken.Thatistherstcondition,thecontinuityof
f
.
Next,lookatthepointofattachment;didyoutieaknottoattachthestringstogether? Ifso
thatknothasamassandyoumustapply
~
F
=
m~a
toit. Inspecifyingthemassdensity
(
x
)Itacitly
assumedthatthereisnoknotat
x
=0,justsomesortofmasslessattachment.Theforce
F
y
onthat
zero-massconnectionmustbezero:
a
y
=
F
y
=m
=
F
y
=
0.Letthepoint
bealittleleftofzeroand
+
alittletotherightofzero. Theequation(7.2)thensays
F
y
(atzero)=
T
)
@f
@x
;t
)+
T
(+
)
@f
@x
(+
;t
) (inthelimit)
T
@f
@x
(0
;t
)+
T
@f
@x
(0+
;t
)=0
(7
:
32)