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10|CoupledOscillators
337
Ofcourse,whenyoureducethistoasinglemassitistheusual
1
2
mv
2
+
1
2
kx
2
.
Whatisthisenergyforthemass-springsysteminEq.(10.1)?
M=
m
1
0
0
m
2
;
K=
k
1
+
k
2
k
2
k
2
k
2
+
k
3
E
=
1
2
(_
x
1
_
x
2
)
m
1
0
0
m
2

_x
1
_
x
2
+
1
2
(
x
1
x
2
)
k
1
+
k
2
k
2
k
2
k
2
+
k
3

x
1
x
2
=
1
2
m
1
_x
2
1
+
m
2
_x
2
2
+
1
2
(
k
1
+
k
2
)
x
2
1
+(
k
2
+
k
3
)
x
2
2
2
k
2
x
1
x
2
=
1
2
m
1
_
x
2
1
+
m
2
_
x
2
2
+
1
2
k
1
x
2
1
+
k
3
x
2
2
+
k
2
(
x
1
x
2
)
2
(10
:
15)
andthisisjustwhatyouwouldwritedownbylookingatthephysicalsystemdirectly|threesprings
compressing,twomassesmoving,andanenergyinapeartree.
Thekineticenergyandthepotentialenergyforaspringarebothsupposedtobepositivenumbers.
ThatisthereasonforthepositivedenitenessassumptionsinEq.(10.8).
10.2NormalModes
Thevariousfrequenciesofoscillationcorrespondtodierentshapesforthewaythatthemassesmove.
normalmeansperpendicular. Thedierent t oscillationsarenormal toeach other, and normalmeans
thatthescalarproductofonemodewithanothermodewillbezero.Whichscalarproduct?Probably
nottheoneyouarethinkingof.
Notation
Therewillbealotofsubscriptshere,andsomeofthemwillrefertodierentvectorswhileotherswill
refertodierentcomponents ofvectors.It’seasytogetthemconfused.There’snoacceptedstandard
waytohandlethis,soI’llmakeupmyown.
Subscriptsa,b,c,orthelikewillrepresentdierentvectors,andthe
subscripts
i
,
j
,
k
,etc.willrepresentcomponentsofaparticularvector.
BasicTheorem
This important result comes s without t solving a single e dierential l or algebraic equation. . It t involves
onlysomematrix manipulationtoderive thenecessary resultaboutnormalmodes. . Fortheequation
2
M+K
x=0,letx
a
andx
b
betwonon-zerosolutionsfortwovalues of
foundby settingthe
determinanttozero. FornowIdon’tneed d thevalues ofthe roots
,justthatthey exist. . Because
thesearelineardierentialequations,complexnotationisallowed.
2
a
M+K
x
a
=0
2
b
M+K
x
b
=0
Multiplytherstofthesebyx
y
b
andthesecondbyx
y
a
.
x
y
b
2
a
M+K
x
a
=0
x
y
a
2
b
M+K
x
b
=0
(10
:
16)
Bothofthesearenumbers. TakethecomplexconjugateofthesecondoneanduseEq.(10.12).
x
y
a
2
b
M+K
x
b
=0
!
x
y
b
2
b
M
y
+K
y
x
a
=0
(10
:
17)
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10|CoupledOscillators
338
RecalltheassumptionthatbothMand Karerealandsymmetric,soM
y
=Mand K
y
=K. That
meansthatIcancombinetheequations(10.16)and(10.17)as
x
y
b
2
a
M+K
x
a
=0
and
x
y
b
2
b
M+K
x
a
=0
(10
:
18)
NowsubtracttheseequationsandtheKtermscancel.
2
a
2
b
x
y
b
Mx
a
=0
(10
:
19)
ThisistheresultIwasaimingfor.FromthisequationIcandeducetwousefulconsequences,all
withoutsolvinganyequations.
Fortherstresulttakethespecialcaseinwhichthetwosolutionsarethesamesolution. x
a
=x
b
and
a
=
b
.
2
a
2
a
x
y
a
Mx
a
=0
Oneofmyassumptions(andhere’swhy)wasthattheproductx
y
a
Mx
a
>
0.Mispositivedenite.This
numbercan’tvanishunlessthexitselfiszero.Theonlywaythattheproductoftwonumberscanbe
zeroisifoneofthetwonumbersisitselfzero,sothisimpliesthattheotherfactor
2
a
2
a
=0
Thatis,
2
isreal. Itdoesn’tsay y ifit’spositiveornegative,butatleastit’snotcomplex. . Infactit
willbenegativefortheseequations,andthatiseasytoshowbyonefurthersimplemanipulation.Start
fromthesameequationandthistimetaketheproductwithx
a
itself:
2
a
M+K
x
a
=0  ! x
y
a
2
a
M+K
x
a
=0  !
2
a
x
y
a
Kx
a
x
y
a
Mx
a
(10
:
20)
Both the numeratorandthedenominatorare positive,andthat’sallittakes tomakethis negative.
Youcannowwriteitas
2
a
!
2
a
.
Nowgobacktothegeneralcase
2
a
2
b
x
y
b
Mx
a
=0
(10
:
21)
The
2
’sarerealnowsoIcandropacomplexconjugation.Ifthetwo
2
’saredierentthenthistime
therstfactorcannotbezeroandthatmeansthatthesecondfactormustbezero.
If
2
a
6=
2
b
;
then
x
y
b
Mx
a
=0
(10
:
22)
This is a scalar product. This is the sense in which the modes are \normal". . Their r scalar product
vanishes. Themassmatrixdenes s away tomultiplythetwovectors,x
a
and x
b
, and itisbyusing
thisparticulardenitionofscalarproductthatyougetthenicetheorem. Gobacktothematerialon
wavesand lookcloselyattheequations(7.48){(7.50). . Inaseeminglydistantcontextyou u aredoing
essentiallythesamemanipulationsashere. Inthoseequationsyouusedpartialintegration. Hereyou
usedmatrixtransposition,buttherearecloseparallels.* Alsolookbacktosection8.5whereyousaw
theorthogonalityoftheeigenvectorsofthetensorofinertia. Theproofthereisessentiallythesameas
fornormalmodeshere,justwithdierentnotation.
* This s parallel is much closer r than you u may y think. . The e bookLinear r Dierential Operators s by
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10|CoupledOscillators
339
Whathappens if
2
a
=
2
b
? Can n thateverhappen? ? In n one-dimensionalproblemsofthe sort
you encounter here, , it’s s not common, , but in n realistic c three-dimensional problems it is unavoidable.
Fortunatelythere’sacure.Youcanalwayschooselinearcombinationsofthese(\degenerate")vectors
sothattheyareorthogonal,thentheorthogonalitypropertywillholdforallyoureigenvectors.Seefor
exampleproblem10.40.
Theparticularvaluesoftheparameters
2
thatyougetfromequationssuchasEq.(10.6)are
called\characteristicvalues"or\eigenvalues"
y
of the system. . Thecorrespondingsolutionsforxare
then called characteristicvectorsorormore commonly, eigenvectors. . Inthislanguage,theequation
(10.21)statesthateigenvectorscorrespondingtodistincteigenvaluesareorthogonal.
10.3ScalarProducts
Whatmakesthisascalarproduct?Isitlegitimatetocallx
y
2
Mx
1
ascalarproduct?Yes.Toseewhy,it
isnecessarytounderstandwhatascalarproductis.
Thefundamentaldenitionofascalarproductisthatitisascalar-valuedfunctionoftwovector
variablesandthatitmustsatisfyacertainsetofrequirements. Callthefunction
f
fornow,sothat
f
(
~v
1
;~v
2
)isascalarand
f
hasthesefourproperties:
1
:f
(
~v
1
;~v
2
+
~v
3
)=
f
(
~v
1
;~v
2
)+
f
(
~v
1
;~v
3
)
2
:f
(
~v
1
;~v
2
)=
f
(
~v
1
;~v
2
)
3
:f
(
~v
1
;~v
2
)=
f
(
~v
2
;~v
1
)
4
:f
(
~v;~v
)0
;
and
f
(
~v;~v
)=0ifandonlyifthevector
~v
isitselfthezerovector.
(10
:
23)
1
;
2:\$\linearin2
nd
arg"
3:\$\conjugatesymmetric"
4:\$\positivedenite"
Any functionthatsatises these requirementsiscalledascalarproduct. . Ifyouareusingonly
realnumbersthenthecomplexconjugationinthethirdrequirementisunnecessary. Doesthefamiliar
dotproductsatisfythem?
f
(
~
A;
~
B
)=
~
A
.
~
B
=
AB
cos
iscertainlyascalar.
4.
~
A
.
~
A
0anditisneverzeroexceptwhen
~
A
isitselfzero.
3.
~
A
.
~
B
=
~
B
.
~
A
2.
~
A
.
(
~
B
)=
(
~
A
.
~
B
)
If
<
0rememberthatcos(
+
)= cos
.
1.Thisistheonlypartthattakesanyeort.Drawthepictureof
~
A
.
(
~
B
+
~
C
)assumingthateverything
is inasingleplane. . Thatoneis s notsohardtoprove|simplywriteoutthe meaningofeachterm.
Withalittlemorethoughtthesamepictureworksforthethreedimensionalcasetoo.
~
B
~
C
~
A
B
cos
1
C
cos
2
Fig.10.4
~
A
.
~
B
+
~
C
?
=
~
A
.
B
+
~
A
.
~
C
~
B
+
~
C
cos
3
=
B
cos
1
+
C
cos
2
Theprojectionsof
~
B
andof
~
C
on
~
A
sumtotheprojectionof
~
B
+
~
C
on
~
A
. Ididn’tdrawtheangles
explicitlybecauseitwouldmakethesketchtoocluttered;youcaneasilygureoutwhichangleiswhich.
y
Thisis anodd,sesquitranslatedword. . Itstarted d lifeinEnglishas characteristic value,thenit
wastranslatedintoGermanasEigenwerte,thenhalf-translatedbackaseigenvalue.Thereitstuck.
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10|CoupledOscillators
340
If you express displacement vectors in the plane in terms of rectangular coordinates, so that
~r
1
=
x
1
^
x
+
y
1
^
y
andsimilarlyfor
~r
2
,the standardscalarproductis
~r
1
.
~r
2
=
x
1
x
2
+
y
1
y
2
. Suppose
howeverthatyoudecidetomeasure
x
inmetersand
y
incentimeters. Thisproductwouldhavetobe
modiedtoaccountforthis,lookingsomethinglike
~r
1
.
~r
2
=
x
1
x
2
+
ky
1
y
2
,where
k
=10
4
m
2
=
cm
2
.
sixofMathematicalToolsforPhysicsformuchmoreonthesubject,withmanymoreexamples.
TheexampleEq.(10.22)islikethis,althoughthemassmatrixcamenaturallyoutofthecom-
scalarproduct(Eq.(10.23))iseasytocheck,andisessentiallyatranslationoftheassumptionsthatI
\real,symmetric"(10.8 !property3(10.23)
\positivedenite"(10.8 !property4(10.23)
10.4InitialValues
Justasinsection3.9,Eq.(3.68),youcanapplyinitialconditionstotheseresults,resultinginsometimes
unexpectedresults. Thematrixmethodsusedherewillorganizethecalculationssothattheybecome
simpletosetup. (ThoughImakenopromisethattheywillneverbetedious.) Theinitialvalueshere
areconditionsonxand_x.
Thesedierentialequationsarelinearandhomogeneous,meaningthatthesumofsolutionsis
asolution and aconstant times a a solutionis s one too. . Thekeyresultaboutthese e special forms of
matricesMandK,theonessatisfyingtherequirementsinEq.(10.8),isthatsucheigenvectors\span
thespace". Everyvectorcanbewrittenasthesumoftheseeigenvectors. . Theproof thatthisistrue
giventhestatedpropertiesofMandKissomethingthatIwillleavetoappropriatemathbooks.
Let x
a
(a = 1
;
2
;:::N
) be e eigenvectors, so o that
!
2
a
Mx
a
+Kx
a
= 0. . This s second order
dierential equation will require two initial l conditions, position n and velocity. . Call l them x
initial
and
_x
initial
.Thegeneralsolutionisalinearcombinationofallthex
a
,andthisis
x(
t
)=
N
X
a=1
a
x
a
cos
!
a
t
+
N
X
a=1
a
x
a
sin
!
a
t
(10
:
24)
where
2
a
!
2
a
. Attime
t
=0theinitialpositionandinitialvelocitiesare
N
X
a=1
a
x
a
=x
initial
and
N
X
a=1
a
!
a
x
a
=_x
initial
Theseare2
N
equationsin2
N
onality oftheeigenvectors. . Takethescalarproductofbothsidesoftherstoftheseequationswith
x
b
.Remember,thescalarproductyouneedistheonethatmakesthevectorsorthogonal,Eq.(10.22).
Also,rememberthenotation:x
b
istheb
th
eigenvector,but(x)
i
or
x
i
or(x
a
)
i
isthe
i
th
componentof
asinglevectorxorx
a
.
x
y
b
M
N
X
a=1
a
x
a
=
N
X
a=1
a
x
y
b
Mx
a
=
b
x
y
b
Mx
b
=x
y
b
Mx
initial
b
=
x
y
b
Mx
initial
x
y
b
Mx
b
(10
:
25)
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10|CoupledOscillators
341
Allbutoneterminthatsumonafromoneto
N
iszero.Thatonetermisa=b.
Fortheinitialvelocity,repeatthisforthesecondoftheinitialvalueequations.
N
X
a=1
a
!
a
x
i
=_x
initial
!
b
=
x
y
b
M_x
initial
!
b
x
y
b
Mx
b
(10
:
26)
Nowyouhaveallthefactors,
a
and
a
intheequation(10.24).
Youshouldgobacktotheprobleminsection3.9andrepeatthatcalculationusingthisformalism,
going all the way through h to o Eq. . (3.68), , but t in the process guring out what MK, and all the
orthogonalitypropertiesareforthis22case.
Other standard methods s of f solving g dierential equations s apply too: : The e method of f Green’s
functionsasinsection3.11willworkhere.Seeproblems10.22{10.24and10.34{10.36.
10.5TunedMassDampers
Thisisaverypracticalapplicationofnormalmodesusedforprovidingprotectionagainstearthquakes,
forminimizingoscillationsinthecrankshaftofanautomobileengine,forstabilizingtallsmokestacks
againstbuetingcausedbyvortexsheddingofthewind,
:::
more.
Averytallbuildingwillrespondtoanearthquakebyoscillating,aseverybuildingisnecessarily
Atallbuildingcanswayadisconcertingly large distancewhenan earthquake hits,and itbecomesa
verylargeoscillator. Ifanypartoftheearthquake’svibrationisneartothenaturalfrequencyofsuch
anoscillationyoucanhavearesonanteectlargeenoughtodestroytheentirestructure.
damage. Thisrstiswastefulofland,andthesecondiswastefulofmoney. Thentoothere’salways
onemore,stilllargerquakeyettocome.
Isthereabetterway?Somecleverengineerscameupwiththeideaofplacingalargemassatthe
topofthebuildingandconguringitsothatitcanoscillatehorizontally.Ifthebuildingstartstosway,
thenthismasscanbeconstructedsothatitwillmovetominimizethebuilding’smotion. Thedetails
willinvolvesomecarefulcalculationstooptimizetheeect,butthat’swhatmechanicalengineersare
paidtodo.
Toseetheideaofwhathappens,takeasimplermodel,onethatyounowhavesomeexperience
with:twomassescoupledbysprings.Applyanoscillatingforce,theearthquake,to
m
1
whichrepresents
thebuilding.Thetunedmassdamperis
m
2
,anditisattachedonlyto
m
1
,sothatitisnotincontact
withtheappliedforcefromtheground.Bothmasseshavedampinggivenbythetermsin _
x
.
F
0
e
i
t
k
1
k
2
m
1
m
2
Fig.10.5
m
1
x
1
k
1
x
1
k
2
(
x
1
x
2
b
1
x_
1
+
F
0
e
i
t
m
2
x
2
=
k
2
(
x
2
x
1
b
2
_
x
2
Translatethisintomatrixnotation.
Mx+B_x+Kx=f
(10
:
27)
M=
m
1
0
0
m
2
;
K=
k
1
+
k
2
k
2
k
2
k
2
;
B=
b
1
0
0
b
2
Theinhomogeneoussolutionhas
x
1
/
e
i
t
and
x
2
/
e
i
t
.Thenthisequation,(10.27),becomes
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10|CoupledOscillators
342
m
1
2
+
k
1
+
k
2
+
b
1
i
k
2
k
2
m
2
2
+
k
2
+
b
2
i

x
1
x
2
=
F
0
0
e
i
t
Thesolutioninvolvesnothingmorethaninvertinga22matrix:
a b
c d

x
y
=
e
f
!
x
y
=
1
det
d
b
c a

e
f
;
det=
bc
(10
:
28)
x
1
(
t
)=
F
0
det
m
2
2
+
k
2
+
b
2
i
e
i
t
;
x
2
(
t
)=
F
0
det
k
2
e
i
t
det=
m
1
2
+
k
1
+
k
2
+
b
1
i

m
2
2
+
k
2
+
b
2
i
k
2
2
(10
:
29)
Theamplitudesoftheseoscillationsaresimplythemagnitudesofthecoecientsof
e
i
t
ineach
x
(
t
m
1
isgreatly reduced,andgraphthemagnitude
ofitsmotion,j
x
1
j,versus  . . The
k
2
=0graph,with
x
2
(
t
) 0,isequivalenttohavingnosecond
masspresent,andthe
k
2
=0
:
1graphshowshoweectivelythismuchsmallertunedmassreducesthe
motionof
m
1
.
k
2
=0
k
2
=0
:
1
j
x
1
j
Fig.10.6
m
1
=1
k
1
=1
b
1
=0
:
04
m
2
=0
:
1
k
2
=0
:
1
b
2
=0
:
032
Theparameters forthegraphare picked simply toillustratetheeect,and inareal situation
youwillhavetochoosethemmuchmorecarefullyinordertoachieveyourgoal.Thentoo,whatisthat
dottedgraphinthepicture?Itisthemagnitudeofthemotionforthesecondmass,j
x
2
j,plottedtothe
samescale,andthisgraphimpliesthatthoughImayhaveminimizedthemotionofthebuildingitself,
thecostis tohavethemass damperswingbackandforthbyalmostasmuchasthebuildingwould
haveswayedanyway. Maybethisisacceptable,butclearlythisdesignrequiressomemoreanalysis.
10.6ChainofMasses_
ofthemotionoftheseconstituents? Picture e alineofmasses
m
connectedbyvery lightcords. . The
cordsareelasticandholdthemassesinline,butthemassescanmoveupanddownbyasmallamount.
Ignoregravity.
equilibrium
y
n
Fig.10.7
Lettheequilibriumdistancebetweenthemassesbe
,sothatthe
x
-coordinateofthe
n
th
mass
is
x
n
=
n‘
.Theverticalcoordinateofthe
n
th
massis
y
n
,andasusualthedeparturefromequilibrium
isassumedtobesmall,withj
y
n
y
n
1
j
. Letthetensionintheconnectingcordsbe
T
,thenthe
y
-componentofforceonthe
n
th
massis
F
y;n
T
(
y
n
y
n
1
)
=‘
T
(
y
n
y
n
+1
)
=‘
10|CoupledOscillators
343
Thisisjustliketheforcecalculationonacontinuousstring,asdoneinsection7.1.
T
isthemagnitude
oftheforcethecordexerts,andthefactormultiplyingitisthetrigonometricsin
tan
factorgiving
theverticalcomponentoftheforce.
Thisgivesthedierentialequations
m
d
2
y
n
dt
2
+
T
y
n
1
+2
y
n
y
n
+1
=0
(10
:
30)
Translatethisintothematrixnotation,MandK.
M=
m
0
B
B
B
B
B
B
B
@
.
.
.
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
.
.
.
1
C
C
C
C
C
C
C
A
K=
T
0
B
B
B
B
B
B
B
@
.
.
.
1
0
0
1
1
0
1
1
0
1
2
.
.
.
1
C
C
C
C
C
C
C
A
(10
:
31)
Howbigarethesematrices? Asbigasthechain. . There’sanother,equallyimportantquestion: : What
happensattheendsofthechain?Howisthechaintiedo?Thereareseveralpossiblewaystodoit.
Therstandseeminglythesimplestistotietheendsdownsothattheydon’tmove. Asecondistotie
theendstomasslessrings,asinproblem7.28. Thereisstillathirdway,andthatistowrapthechain
aroundinacircleandattachtheleftmostmasstotherightmostmassasifit’sjustanothermass. You
haveachoiceamongtheseboundaryconditions.
Fig.10.8
Thinkofthisthirdwayaswrappingthecordaroundasmoothcylindersothatitispulledtaut
and the masses are free to moveup and down along the surface of the cylinder. . If f this third way
lookspeculiarandperhapsarticial,maybeitis,butitisalsothesimplestmathematically,andinreal
problemsinvolvingcrystalstructure,it’sthemethodthatisbyfartheonemostcommonlyused.Inthe
studyofsolidstatephysicsitistheonlychoiceyouwilleversee. Ofcourseinthatcase,thewrapping
aroundisdoneinallthreedimensions,soit’salittlehardertopicture,thoughthemathematicsisno
dierent.
ThereisanotherquestionthatIskippedoverinsettingupthematrices(10.31).Whathappensin
thecornersofthematriceswheretherearethediagonaldots? Howdothematricesend?Thisquestion
andthechoicesofphysicalboundaryconditionsarereallythesamequestionindierentguises.
Case1: Thereare
N
masseswithindex
n
=1
;:::;N
andthe
y
-coordinates
y
0
and
y
N
+1
are
tieddowntozero.
!
Explicitlywriteoutallthreeequations(10.30)forthecase
N
=3.
10|CoupledOscillators
344
Yousee,becauseyoujustwroteoutthecase
N
=3explicitly,thattheKmatrixis
K=
T
0
B
B
B
B
B
B
@
1
1
1
zeros
1
1
1
.
.
.
zeros
1
1
2
1
C
C
C
C
C
C
A
(10
:
32)
Thesecondcaseisjustaseasyif youwriteoutthe
N
=3dierentialequationsagain. Remember
nowthattheringsontheend,
y
0
and
y
N
+1
havezeromass,sotheycannothaveanyverticalcomponent
offorce.Thinkoftheseasextracoordinates0and4thatobeydierentequations:
y
0
=
y
1
and
y
4
=
y
3
K=
T
0
B
B
B
B
B
B
@
1
1
1
1
1
1
.
.
.
1
1
1
1
C
C
C
C
C
C
A
(10
:
33)
Togureoutthethirdcase,youshouldwriteoutthedierentialequationsfor
N
=4. Now,
withthecordgoingaroundinacircle,the
N
th
(4
th
)massisdirectlyattachedtothe1
st
mass. Again,
that’slikehavingextracoordinates0and5with
y
5
y
1
and
y
0
y
4
.
K=
T
0
B
B
B
B
B
B
@
1
1
1
1
1
1
1
.
.
.
1
1
1
2
1
C
C
C
C
C
C
A
(10
:
34)
AllthreeoftheseKmatricessatisfytherequirementsofEq.(10.8),thoughprovingthepositivedenite
requirement,x
y
Kx
>
0,takessomeslightlymoreseriousknowledgeoflinearalgebra.
Nowto ndthe solutions to the equations of motion,and togetthe normalmodes: : With x
proportionalto
e
i!t
,thealgebraicequationstosolveare(
!
2
M+K)x=0. (Whyminus
i!t
?This
signisarbitrary,butitsconveniencewillbeclearwithinafewpages.) Tryingtosetthedeterminantof
this
N
N
matrixtozeroandsolvingan
N
th
-degreeequationismore-or-lesshopeless,butthere’san
easierway.
The easiest t of the three possibilities s that t I’ve written is the third one. . This s uses \periodic
boundary conditions",anditissimplerbecauseinthiscase you can easilyuse complexexponentials
in the calculations, whileintheothercases you willneedtousetrigonometricfunctions. . Thisform
ofboundaryconditionmaybelessfamiliartoyounow,butitsuseissocommonthatyoushouldget
usedtoit. Thewaytosolvethissystemistouseamethodverymuchliketheoneyouusetosolve
homogeneous,constant-coecientdierentialequations|assumeanexponentialform.Herehowever
theindependentvariableisnotpositionortime(
x
or
t
),buttheindex
n
e
x
youusesomethingmorelike
e
n
.Lookatthe
n
th
lineintheequation(
!
2
M+K)x=0.
y
n
(
t
)=
q
n
e
i!t

m!
2
q
n
+(
T=‘
)
q
n
1
+2
q
n
q
n
+1
=0
(10
:
35)
10|CoupledOscillators
345
This is of course nothingmore thatEq. . (10.30) for r asolution n proportional to
e
i!t
. Whatmakes
theperiodicboundary conditionseasiertousethantheothersisthatthis equationappliestoall the
masses.Youdon’thavetomakeexceptionsattheends.Whatyoudohaveisthatthe(
N
+1)
th
mass
andthe1
st
massarethesamemass. (Or0
th
and
N
th
|thesamethingandusuallyeasiertouse.)
Thisisalinear,constant-coecientdierenceequation,andyousolveitusingthemethodsso
familiarwithlinear,constant-coecient,dierential equations,onlyheretheindependentvariableis
n
t
. Assumean n exponential solution. . Doyoupicktheexponential l intheform
p
n
or
e
in
?
Bothwork,andif
p
=
e
i
theyarethesameanyway. Anticipatingthattheremaybewaveshere,I’ll
choosethesecondnotation:
q
n
=
e
in
m!
2
e
in
+(
T=‘
)
e
i
(
n
1)
+2
e
in
e
i
(
n
+1)
=0
m!
2
+(
T=‘
)
e
i
+2
e
i
m!
2
+(2
T=‘
)
1 cos
=0
(10
:
36)
Withasimpletrigonometricidentity,thelastoftheseequationsis
!
2
=
4
T
m‘
sin
2
(
=
2)  !
!
=2
r
T
m‘
sin(
=
2)
(10
:
37)
Take
!
tobepositivealways,allowing
tobepositiveornegative.Thisisachoice,butitisconvenient.
Howdoyouhandletheconditionthat
q
0
and
q
N
describethesamemass? Itisverystraight-
forward,anditdeterminestheallowedvaluesfor
.
q
N
=
q
0
!
e
iN
=
e
i
0
=1  !
N
= amultipleof2
!
=0
;
2
N
;
2
.
2
N
;
3
.
2
N
;
 (butseeEq.(10.39))
(10
:
38)
Howfardoes thissetof numbersgo? ? Ithastobenitebecause e there are only anitenumberof
modes.Exploreaspecialcaserst,andlookattheshapethatthefunction
q
n
takes.Itis
q
n
=
e
in
=
1
; e
i
;e
2
i
;e
3
i
;:::

The
=0caseistrivial|nothingmoves,asEq.(10.37)saysthat
!
=0inthiscase.* Nowcomes
therstnon-trivialcase,where
=1
.
2
=N
,andhereItaketherealparttointerpretit.
y
n
(
t
)=
q
n
e
i!
1
t
=
e
i
(2
n=N
i!
1
t
! cos
(2
n=N
!
1
t
At
t
=0,as
n
goesfromzeroto
N
1andthento
N
(whichisthesameaszero),thiscosinegoes
throughonefullcycle. Atlatertimesthiscosinewillmove,andthepeakwillbedenedbythepoint
(2
n=N
!
1
t
=0. Thisisawave,andithasthemaximumwavelengthpossiblealongtheloop.
Inthispicture,
N
=10and
t
=0.
n
=0
n
=10 !
n
=0
5
=1
.
2
=
10,
t
=0
Fig.10.9
Drawapictureatashorttimelater,andthecosinewillhavemovedtotherightandaroundthe
loop.Takethetime
!
1
t
=2
=
5asanexample
* Notreally. . LookbacktoEq.(10.30)andseewhathappensifallthe
y
n
havethesamevalue.
10|CoupledOscillators
346
n
=0
n
=10 !
n
=0
5
=1
.
2
=
10,
!
1
t
=2
=
5
Fig.10.10
Whatifyouchoosethenegativesignfor
? Simple:Thewaveismovingintheoppositedirection.
InEq.(10.38)howlargecan
be? Sofaryou’velookedatonly
=0
;
2
=N
. Aneasycase
totestis
=
N
.
2
=N
=2
. Thisisagaintrivial,justas
=0was,because
e
in
=
e
i
2
n
1.
Eq.(10.37)tellsyouthat
!
=0heretoo. Nothingmoves.
Whatif
comesfrom anindexone less than
N
?
=(
N
1)2
=N
=2
2
=N
. Then
e
in
=
e
i
(2
2
=N
)
n
=
e
2
in=N
. This s is identical tothe case that
2
=N
, and this s wave
is movingleft,having thesame wavelengthas inthese pictures,exceptforthedirection of motion.
It’s notanewwave, , only y thesameshape movingin the opposite direction. . You u cansee fromthis
reasoningthatgoingdownfromzeroordownfrom
N
givesthesamesetofsolutions,andthatmeans
thatthenumberofdierentsolutionscanbetakenfromtheset(0
;
1
;
2
;:::;N
1) orfromtheset
(0
;
1
;
2
;:::;
N=
2)asyouchoose. Notquite.Doesitmatterif
N
isevenorodd? Youshouldtry
thecases
N
=3and
N
=4,andsee. Youshouldinfactwriteoutallthesolutionsforallthemodes
for
N
=2
;
3
;
and4inordertogetafeelingforwhattheydo.
Cuttingthroughalltheverbiage,
0
<
2
instepsof 2
=N
(Seethegraph,Figure10.12.)
i.e.
=2
s=N; s
=0
;
1
;
2
:::;
(
N
1)
(10
:
39)
Howshortcanthewavelengthbe? Thatiswhen
isatthehalfwaypoint:
N=
2for
N
evenor
(
N
1)
=
2for
N
odd.Forthepictureshere,with
N
=10,thatwillbe
=5
.
2
=
10=
.
q
n
=
e
in
!
y
n
(
t
)=
e
in
i!
5
t
! cos(
n
!
5
t
)
(10
:
40)
0
10 !0
5
=5
.
2
=
10=
,
t
=0
Fig.10.11
Youcangetanideainthispicturewhythisrepresentstheshortestwavelengthavailableinthis
system. Ikepttheverticalscalethesameinallthreedrawings,eventhoughitmakesitappearvery
exaggeratedinthethirdone. Whichdirectionisitmoving,leftorright?Rememberthatsmallpositive
gavemotiontotheright. If
isjustlessthan
N
itmovedleft. Thisoneisinthecenter,sowhat
doesitdo? [WhatisthecosineofthedierenceoftwoanglesinEq.(10.40)?]
Lookatthe caseinwhichthere are many masses. . This s modelcanapproximateacontinuous
string,but itwill demonstrate some phenomenathatdon’t appearinthe simplermodelsof chapter
seven.Forexample,intheequation(10.37),
!
/sin(
=
2)impliesthatthereisamaximumfrequency
thata wave e can n have. . That t doesn’t happen with light orsound. . Or r does s it? ? With h sound, what
ifthefrequencyissohighthatthecorrespondingwavelengthis comparabletothedistancebetween
molecules? Sounddoesn’tmakemuchsenseanymore,sotherereallyis anupperlimitforsoundtoo.
Theletter\
"isthedistancebetweenthemasses,so
m=‘
=
istheaveragelinearmassdensity.
from
n
=0tothevalueof
n
suchthat
n
=2
. Thedistancecorrespondingtothis
n
is
n‘
. Eliminate
n
andthissays
=
n‘
=2
‘=
,whichinturnmeansthatthewavenumberis
k
=2
=
=
=‘
.