1 Simple

HarmonicMotion

AtfirstsighttheeightphysicalsystemsinFigure1.1appeartohavelittleincommon.

1.1(a)isasimplependulum,amassmswingingattheendofalightrigidrodoflengthl.

1.1(b)isaflatdiscsupportedbyarigidwirethroughitscentreandoscillatingthrough

smallanglesintheplaneofitscircumference.

1.1(c)isamassfixedtoawallviaaspringofstiffnesssslidingtoandfrointhex

directiononafrictionlessplane.

1.1(d)isamassmatthecentreofalightstringoflength2lfixedatbothendsundera

constanttensionT.Themassvibratesintheplaneofthepaper.

1.1(e)isafrictionlessU-tubeofconstantcross-sectionalareacontainingalengthlof

liquid,density

,oscillatingaboutitsequilibriumpositionofequallevelsineach

limb.

1.1(f)isanopenflaskofvolumeVandaneckoflengthlandconstantcross-sectional

areaAinwhichtheairofdensity

vibratesassoundpassesacrosstheneck.

1.1(g)isahydrometer,abodyofmassmfloatinginaliquidofdensity

withaneckof

constantcross-sectionalareacuttingtheliquidsurface.Whendepressedslightly

fromitsequilibriumpositionitperformssmallverticaloscillations.

1.1(h)isanelectricalcircuit,aninductanceLconnectedacrossacapacitanceCcarrying

achargeq.

Allofthesesystemsaresimpleharmonicoscillatorswhich,whenslightlydisturbedfrom

theirequilibriumorrestpostion,willoscillatewithsimpleharmonicmotion.Thisisthe

mostfundamentalvibrationofasingleparticleorone-dimensionalsystem.Asmall

displacementxfromitsequilibriumpositionsetsuparestoringforcewhichisproportional

toxactinginadirectiontowardstheequilibriumposition.

Thus,thisrestoringforceFmaybewritten

F¼!sx

wheres,theconstantofproportionality,iscalledthestiffnessandthenegativesignshows

thattheforceisactingagainstthedirectionofincreasingdisplacementandbacktowards

ThePhysicsofVibrationsandWaves,6thEdition

H.J.Pain

#2005JohnWiley&Sons,Ltd

1

(a)

(b)

c

Ix

l

θ

m

mx +

mg = 0

x l

ml + m

g = 0

..

.. ω2 =

g/ l

mg s

in ∼ m

g

~ m

gx l

~~

mg

I + c

=

0..

ω2 =

c l

θ θ

θ θ

θ θ

θ

m

x

s

(c)

(d)

x

m

TT

2 l

mx +

sx =

0

ω2 =

s/m

mx +

2T =

0x l

ω2 =

2 T

l m

. .

. .

2x

x

x

x

pp

AV

l

(f)

(e)

p lx +

2 p

g x

= 0

..

ω2 =

2g/l

p A

lx +

γ pxA

2

v= 0

ω2 =

γ p

Al pV

..

2SimpleHarmonicMotion

theequilibriumposition.Aconstantvalueofthestiffnessrestrictsthedisplacementxto

smallvalues(thisisHooke’sLaw

ofElasticity).Thestiffnesssisobviouslytherestoring

forceperunitdistance(ordisplacement)andhasthedimensions

force

distance"MLT!2

L

Theequationofmotionofsuchadisturbedsystem

isgivenbythedynamicbalance

betweentheforcesactingonthesystem,whichbyNewton’sLaw

is

masstimesacceleration¼restoringforce

or

m€ xx¼!sx

wheretheacceleration

€ xx¼d2x

dt2

Thisgives

m€ xxþsx¼0

c

q

L

x

A

m

p

(h)

(g)

mx

+ A

pgx =

0..

ω2

= A

pg/m

Lq

+q c

= 0

..

ω2 =

1 Lc

Figure1.1

Simple

harmonic

oscillators

withtheireq

uationsofmotionan

dan

gularfreq

uen

cies

!of

oscillation.(a)A

simple

pen

dulum.(b)A

torsional

pen

dulum.(c)A

massonafrictionless

plane

connectedbyaspringto

awall.(d)Amassat

thecentreofastringunde

rco

nstan

ttensionT.

(e)A

fixedlength

ofnon-visco

usliquid

inaU-tubeofco

nstan

tcross-section.(f)Anacoustic

Helmholtz

resonator.(g)Ahyd

rometer

massm

inaliquid

ofden

sity

.(h)Anelectrical

LCresonan

tcircuit

SimpleHarmonicMotion

3

or

€ xxþs mx¼0

wherethedimensionsof

s mareMLT!2

ML

¼T!2¼

"2

HereTisatime,orperiodofoscillation,thereciprocalof"whichisthefrequencywith

whichthesystem

oscillates.

However,whenwesolvetheequationofmotionweshallfindthatthebehaviourofx

withtimehasasinusoidalorcosinusoidaldependence,anditwillprovemoreappropriate

toconsider,not",buttheangularfrequency

!¼2#"sothattheperiod

T¼1 "¼2#

ffiffiffiffi

m s

r

wheres=misnowwrittenas

!2.Thustheequationofsimpleharmonicmotion

€ xxþs mx¼0

becomes

€ xxþ!2x¼0

ð1:1Þ

(Problem

1.1)

Displacementin

Simple

HarmonicMotion

Thebehaviourofasimpleharmonicoscillatorisexpressedintermsofitsdisplacementx

fromequilibrium,itsvelocity

_ xx,anditsacceleration€ xxatanygiventime.Ifwetrythesolution

x¼Acos!t

whereAisaconstantwiththesamedimensionsasx,weshallfindthatitsatisfiesthe

equationofmotion

€ xxþ!2x¼0

for

_ xx¼!A!sin!t

and

€ xx¼!A!2cos!t¼!!2x

4SimpleHarmonicMotion

Anothersolution

x¼Bsin!t

isequallyvalid,whereBhasthesamedimensionsasA,forthen

_ xx¼B!cos!t

and

€ xx¼!B!2sin!t¼!!2x

Thecompleteorgeneralsolutionofequation(1.1)isgiven

bytheadditionor

superpositionofbothvaluesforxsowehave

x¼Acos!tþBsin!t

ð1:2Þ

with

€ xx¼!!2ðAcos!tþBsin!tÞ¼!!2x

whereAandBaredeterminedbythevaluesofxand_ xxataspecifiedtime.Ifwerewritethe

constantsas

A¼asin&

and

B¼acos&

where&isaconstantangle,then

A2þB2¼a2ðsin2&þcos2&Þ¼

a2

sothat

a¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

A2þB2

p

and

x¼asin&cos!tþacos&sin!t

¼asinð!tþ&Þ

Themaximumvalueofsin(!tþ&)isunitysotheconstantaisthemaximumvalueofx,

knownastheamplitudeofdisplacement.Thelimitingvaluesofsinð!tþ&Þare&1sothe

system

willoscillatebetweenthevaluesofx¼&aandweshallseethatthemagnitudeofa

isdeterminedbythetotalenergyoftheoscillator.

Theangle

&iscalledthe‘phaseconstant’forthefollowingreason.Simpleharmonic

motionisoftenintroducedbyreferenceto‘circularmotion’becauseeachpossiblevalueof

thedisplacementxcanberepresentedbytheprojectionofaradiusvectorofconstant

lengthaonthediameterofthecircletracedbythetipofthevectorasitrotatesinapositive

DisplacementinSimpleHarmonicMotion

5

anticlockwisedirectionwithaconstantangularvelocity

!.Eachrotation,astheradius

vectorsweepsthroughaphaseangleof2#rad,thereforecorrespondstoacomplete

vibrationoftheoscillator.Inthesolution

x¼asinð!tþ&Þ

thephaseconstant&,measuredinradians,definesthepositioninthecycleofoscillationat

thetimet¼0,sothatthepositioninthecyclefromwhichtheoscillatorstartedtomoveis

x¼asin&

Thesolution

x¼asin!t

definesthedisplacementonlyofthatsystem

whichstartsfromtheoriginx¼0attime

t¼0buttheinclusionof&inthesolution

x¼asinð!tþ&Þ

where&maytakeallvaluesbetweenzeroand2#allowsthemotiontobedefinedfromany

startingpointinthecycle.ThisisillustratedinFigure1.2forvariousvaluesof&.

(Problems1.2,1.3,1.4,1.5)

Velocity

andAccelerationin

Simple

HarmonicMotion

Thevaluesofthevelocityandaccelerationinsimpleharmonicmotionfor

x¼asinð!tþ&Þ

aregivenby

dx dt¼

_ xx¼a!cosð!tþ&Þ

φ4φ3

φ2

φ1

φ1

φ2

φ3

φ4

φ5

φ6

φ0 φ

6

φ5

= 2

70°

= 9

0° = 0 a

a

ωt

φx = a

Sin

(ωt + )

Figure

1.2

Sinusoidal

displacemen

tofsimple

harmonic

oscillatorwithtime,

showingvariationof

startingpointin

cyclein

term

sofphasean

gle

&

6SimpleHarmonicMotion

and

d2x

dt2¼

€ xx¼!a!2sinð!tþ&Þ

Themaximumvalueofthevelocitya!iscalled

thevelocityamplitudeandthe

accelerationamplitudeisgivenbya!2.

FromFigure1.2weseethatapositivephaseangleof#=2radconvertsasineintoa

cosinecurve.Thusthevelocity

_ xx¼a!cosð!tþ&Þ

leadsthedisplacement

x¼asinð!tþ&Þ

byaphaseangleof#=2radanditsmaximaandminimaarealwaysaquarterofacycle

aheadofthoseofthedisplacement;thevelocityisamaximumwhenthedisplacementis

zeroandiszeroatmaximumdisplacement.Theaccelerationis‘anti-phase’(#rad)with

respecttothedisplacement,beingmaximumpositivewhenthedisplacementismaximum

negativeandviceversa.ThesefeaturesareshowninFigure1.3.

Often,therelativedisplacementormotionbetweentwooscillatorshavingthesame

frequencyandamplitudemaybeconsideredintermsoftheirphasedifference

&1!&2

whichcanhaveanyvaluebecauseonesystem

mayhavestartedseveralcyclesbeforethe

otherandeachcompletecycleofvibrationrepresentsachangeinthephaseangleof

&¼2#.Whenthemotionsofthetwosystemsarediametricallyopposed;thatis,onehas

x =

a s

in(ω

t + )

x =

aω

cos(ω

t + )

ωt

ωt

ωt

x =

−aω

2 s

in(ω

t + )

aω

2

aωa

Acceleration xVelocity xDisplacement x

φ

φ

φ

Figure

1.3

Variation

with

time

ofdisplacemen

t,velocity

and

acceleration

insimple

harmonic

motion.Displacemen

tlagsvelocity

by#=2

radan

dis#radoutofphasewiththeacceleration.Th

e

initialphaseco

nstan

t&

istakenas

zero

VelocityandAccelerationinSimpleHarmonicMotion

7

x¼þawhilsttheotherisatx¼!a,thesystemsare‘anti-phase’andthetotalphase

difference

&1!&2¼n#rad

wherenisanoddinteger.Identicalsystems‘inphase’have

&1!&2¼2n#rad

wherenisanyinteger.Theyhaveexactlyequalvaluesofdisplacement,velocityand

accelerationatanyinstant.

(Problems1.6,1.7,1.8,1.9)

Non

-linea

rity

Ifthestiffnesssisconstant,thentherestoringforceF¼!sx,whenplottedversusx,will

produceastraightlineandthesystem

issaidtobelinear.Thedisplacementofalinear

simpleharmonicmotionsystem

followsasineorcosinebehaviour.Non-linearityresults

whenthestiffnesssisnotconstantbutvarieswithdisplacementx(seethebeginningof

Chapter14).

Energy

ofaSimple

HarmonicOscillator

Thefactthatthevelocityiszeroatmaximumdisplacementinsimpleharmonicmotionand

isamaximumatzerodisplacementillustratestheimportantconceptofan

exchange

betweenkineticandpotentialenergy.Inanidealcasethetotalenergyremainsconstantbut

thisisneverrealizedinpractice.Ifnoenergyisdissipatedthenallthepotentialenergy

becomeskineticenergyandviceversa,sothatthevaluesof(a)thetotalenergyatanytime,

(b)themaximumpotentialenergyand(c)themaximumkineticenergywillallbeequal;

thatis

Etotal¼KEþPE¼KEmax¼PEmax

Thesolutionx¼asin(!tþ&)impliesthatthetotalenergyremainsconstantbecausethe

amplitudeofdisplacementx¼&aisregainedeveryhalfcycleatthepositionofmaximum

potentialenergy;whenenergyislosttheamplitudegraduallydecaysasweshallseelaterin

Chapter2.Thepotentialenergyisfoundbysummingallthesmallelementsofworksx.dx

(forcesxtimesdistancedx)donebythesystemagainsttherestoringforceovertherange

zerotoxwherex¼0giveszeropotentialenergy.

Thusthepotentialenergy¼

ð

x 0

sx'dx¼

1 2sx2

Thekineticenergyisgivenby1 2m_ xx2sothatthetotalenergy

E¼

1 2m_ xx2þ1 2sx2

8SimpleHarmonicMotion

SinceEisconstantwehave

dE dt¼ðm

€ xxþsxÞ_ xx¼0

givingagaintheequationofmotion

m€ xxþsx¼0

Themaximumpotentialenergyoccursatx¼&aandistherefore

PEmax¼

1 2sa2

Themaximumkineticenergyis

KEmax¼ð1 2m_ xx2Þ m

ax¼

1 2ma2!2½cos2ð!tþ&Þ) m

ax

¼1 2ma2!2

whenthecosinefactorisunity.

Butm!2¼ssothemaximumvaluesofthepotentialandkineticenergiesareequal,

showingthattheenergyexchangeiscomplete.

Thetotalenergyatanyinstantoftimeorvalueofxis

E¼

1 2m_ xx2þ1 2sx2

¼1 2ma2!2½cos2ð!tþ&Þþ

sin2ð!tþ&Þ)

¼1 2ma2!2

¼1 2sa2

asweshouldexpect.

Figure1.4showsthedistributionofenergyversusdisplacementforsimpleharmonic

motion.Notethatthepotentialenergycurve

PE¼

1 2sx2¼

1 2ma2!2sin2ð!tþ&Þ

isparabolicwithrespecttoxandissymmetricaboutx¼0,sothatenergyisstoredinthe

oscillatorbothwhenxispositiveandwhenitisnegative,e.g.aspringstoresenergy

whethercompressedorextended,asdoesagasincompressionorrarefaction.Thekinetic

energycurve

KE¼

1 2m_ xx2¼

1 2ma2!2cos2ð!tþ&Þ

isparabolicwithrespecttobothxand_ xx.Theinversionofonecurvewithrespecttothe

otherdisplaysthe#=2phasedifferencebetweenthedisplacement(relatedtothepotential

energy)andthevelocity(relatedtothekineticenergy).

Foranyvalueofthedisplacementxthesumoftheordinatesofbothcurvesequalsthe

totalconstantenergyE.

EnergyofaSimpleHarmonicOscillator

9

(Problems1.10,1.11,1.12)

Simple

HarmonicOscillationsin

anElectricalSystem

Sofarwehavediscussedthesimpleharmonicmotionofthemechanicalandfluidsystems

ofFigure1.1,chieflyintermsoftheinertialmassstretchingtheweightlessspringof

stiffnesss.Thestiffnesssofaspringdefinesthedifficultyofstretching;thereciprocalof

thestiffness,thecomplianceC(wheres¼1=C)definestheeasewithwhichthespringis

stretched

andpotentialenergystored.ThisnotationofcomplianceCisusefulwhen

discussingthesimpleharmonicoscillationsoftheelectricalcircuitofFigure1.1(h)and

Figure1.5,whereaninductanceLisconnectedacrosstheplatesofacapacitanceC.The

forceequationofthemechanicalandfluidexamplesnowbecomesthevoltageequation

Energy

Tota

l energ

y E

= K

E +

PE

E

E 2

E 2

1 2K

E =

m

x 2

1 2= E

− sx

2

1 2P

E =

sx

2

−a

a 2−

a 2

+a

x

Dis

pla

cem

ent

Figure

1.4

Parabolicrepresentation

ofpotential

energyan

dkinetic

energyofsimple

harmonic

motionversusdisplacemen

t.Inversionofonecu

rvewithrespectto

theother

showsa90*phase

difference.Atan

ydisplacemen

tvalue

the

sum

ofthe

ordinates

ofthe

curves

equalsthe

total

constan

ten

ergyE

I+

+

−

−q c

Lq +

q c= 0

LdI

dt

Figure

1.5

Electrical

system

whichoscillatessimple

harmonically.Th

esum

ofthevo

ltag

esaround

thecircuit

isgiven

byKirch

hoff’slaw

asLdI=

dtþ

q=C¼

0

10

SimpleHarmonicMotion

(balanceofvoltages)oftheelectricalcircuit,buttheformandsolutionoftheequationsand

theoscillatorybehaviourofthesystemsareidentical.

Intheabsenceofresistancetheenergyoftheelectricalsystem

remainsconstantandis

exchangedbetweenthemagneticfieldenergystoredintheinductanceandtheelectricfield

energystoredbetweentheplatesofthecapacitance.Atanyinstant,thevoltageacrossthe

inductanceis

V¼!LdI

dt¼!Ld2q

dt2

whereIisthecurrentflowingandqisthechargeonthecapacitor,thenegativesign

showingthatthevoltageopposestheincreaseofcurrent.Thisequalsthevoltageq=C

acrossthecapacitancesothat

L€ qqþq=C¼0

ðKirchhoff’sLawÞ

or

€ qqþ!2q¼0

where

!2¼1 LC

Theenergystoredinthemagneticfieldorinductivepartofthecircuitthroughoutthe

cycle,asthecurrentincreasesfrom0toI,isformedbyintegratingthepoweratanyinstant

withrespecttotime;thatis

EL¼

ð

VI'dt

(whereVisthemagnitudeofthevoltageacrosstheinductance).

So

EL¼

ð

VIdt¼

ð

LdI

dtIdt¼

ð

I 0

LIdI

¼1 2LI2¼

1 2L_ qq2

Thepotentialenergystoredmechanicallybythespringisnowstoredelectrostaticallyby

thecapacitanceandequals

1 2CV2¼q2

2C

SimpleHarmonicOscillationsinanElectricalSystem

11

Comparisonbetweentheequationsforthemechanicalandelectricaloscillators

mechanical(force)!m€ xxþsx¼0

electrical(voltage)!L€ qqþq C¼0

mechanical(energy)!

1 2m_ xx2þ1 2sx2¼E

electrical(energy)!1 2L_ qq2þ1 2

q2

C¼E

showsthatmagneticfieldinertia(definedbytheinductanceL)controlstherateofchange

ofcurrentforagivenvoltageinacircuitinexactlythesamewayastheinertialmass

controlsthechangeofvelocityforagivenforce.Magneticinertialorinductivebehaviour

arisesfromthetendencyofthemagneticfluxthreadingacircuittoremainconstantand

reactiontoanychangeinitsvaluegeneratesavoltageandhenceacurrentwhichflowsto

opposethechangeofflux.ThisisthephysicalbasisofFleming’sright-handrule.

SuperpositionofTwoSimple

HarmonicVibrationsin

One

Dimension

(1)Vibration

sHav

ingEq

ual

Freq

uen

cies

Inthefollowingchaptersweshallmeetphysicalsituationswhichinvolvethesuperposition

oftwoormoresimpleharmonicvibrationsonthesamesystem.

Wehavealreadyseen

howthedisplacementinsimpleharmonicmotionmay

be

representedinmagnitudeandphasebyaconstantlengthvectorrotatinginthepositive

(anticlockwise)sensewithaconstantangularvelocity

!.Tofindtheresultingmotionofa

system

whichmovesinthexdirectionunderthesimultaneouseffectoftwosimple

harmonicoscillationsofequalangularfrequenciesbutofdifferentamplitudesandphases,

wecanrepresenteachsimpleharmonicmotionbyitsappropriatevectorandcarryouta

vectoraddition.

Ifthedisplacementofthefirstmotionisgivenby

x1¼a1cosð!tþ&1Þ

andthatofthesecondby

x2¼a2cosð!tþ&2Þ

thenFigure1.6showsthattheresultingdisplacementamplitudeRisgivenby

R2¼ða1þa2cos'Þ2þða2sin'Þ2

¼a2 1þa2 2þ2a1a2cos'

where'¼

&2!&1isconstant.

12

SimpleHarmonicMotion

Thephaseconstant(ofRisgivenby

tan(¼a1sin&1þa2sin&2

a1cos&1þa2cos&2

sotheresultingsimpleharmonicmotionhasadisplacement

x¼Rcosð!tþ(Þ

anoscillationofthesamefrequency

!buthavinganamplitudeRandaphaseconstant(.

(Problem

1.13)

(2)Vibration

sHav

ingDifferentFreq

uen

cies

Supposewenowconsiderwhathappenswhentwovibrationsofequalamplitudesbut

differentfrequenciesaresuperposed.Ifweexpressthem

as

x1¼asin!1t

and

x2¼asin!2t

where

!2>

!1

y

x

a2

a1

Ra

2

2

a2 s

in δ

a2 c

os δ

2 − 1 = δ

φθ

φφ

f1

Figure

1.6

Addition

ofvectors,ea

chrepresenting

simple

harmonic

motion

along

the

xaxis

at

angularfreq

uen

cy!to

givearesultingsimple

harmonic

motiondisplacemen

tx¼

Rco

sð!

tþ(Þ

---

hereshownfort¼

0

SuperpositionofTwoSimpleHarmonicVibrationsinOneDimension

13

thentheresultingdisplacementisgivenby

x¼x1þx2¼aðsin!1tþsin!2tÞ

¼2asinð!

1þ!2Þt

2cosð!

2!!1Þt

2

ThisexpressionisillustratedinFigure1.7.Itrepresentsasinusoidaloscillationatthe

averagefrequencyð!

1þ!2Þ=2havingadisplacementamplitudeof2awhichmodulates;

thatis,variesbetween2aandzeroundertheinfluenceofthecosinetermofamuchslower

frequencyequaltohalfthedifferenceð!

2!!1Þ=2betweentheoriginalfrequencies.

When

!1and!2arealmostequalthesinetermhasafrequencyveryclosetoboth

!1

and!2whilstthecosineenvelopemodulatestheamplitude2aatafrequency(!

2!!1)=2

whichisveryslow.

Acousticallythisgrowthanddecayoftheamplitudeisregisteredas‘beats’ofstrong

reinforcementwhentwosoundsofalmostequalfrequencyareheard.Thefrequencyofthe

‘beats’isð!

2!!1Þ,thedifference

betweentheseparatefrequencies(nothalfthe

difference)becausethemaximumamplitudeof2aoccurstwiceineveryperiodassociated

withthefrequency(!

2!!1Þ=2.Weshallmeetthissituationagainwhenweconsider

thecouplingoftwooscillatorsinChapter4andthewavegroupoftwocomponentsin

Chapter5.

2a

2a

x

ω2 − ω

1

2t

ωt

cos

ω2 +

ω1

2t

sin

Figure

1.7

Superpositionoftw

osimple

harmonic

displacemen

tsx1¼

asin!

1tan

dx2¼

asin!

2t

when

!2>

!1.Th

eslow

cos½ð!

2!!

1Þ=2)t

envelope

modulatesthe

sin½ð!

2þ!

1Þ=2)t

curve

betwee

nthevalues

x¼&2a

14

SimpleHarmonicMotion

SuperpositionofTwoPerpendicularSimple

Harmonic

Vibrations

(1)Vibration

sHav

ingEq

ual

Freq

uen

cies

Supposethataparticlemovesunderthesimultaneousinfluenceoftwosimpleharmonic

vibrationsofequalfrequency,onealongthexaxis,theotheralongtheperpendicularyaxis.

Whatisitssubsequentmotion?

Thisdisplacementsmaybewritten x¼a1sinð!tþ&1Þ

y¼a2sinð!tþ&2Þ

andthepathfollowedbytheparticleisformedbyeliminatingthetimetfromthese

equationstoleaveanexpressioninvolvingonlyxandyandtheconstants&1and&2.

Expandingtheargumentsofthesineswehave

x a1

¼sin!tcos&1þcos!tsin&1

and

y a2

¼sin!tcos&2þcos!tsin&2

Ifwecarryouttheprocess

x a1

sin&2!y a2

sin&1

$%

2

þy a2

cos&1!x a1

cos&2

$%

2

thiswillyield

x2

a2 1

þy2

a2 2

!2xy

a1a2

cosð&

2!&1Þ¼

sin2ð&

2!&1Þ

ð1:3Þ

whichisthegeneralequationforanellipse.

Inthemostgeneralcasetheaxesoftheellipseareinclinedtothexandyaxes,butthese

becometheprincipalaxeswhenthephasedifference

&2!&1¼

# 2

Equation(1.3)thentakesthefamiliarform

x2

a2 1

þy2

a2 2

¼1

thatis,anellipsewithsemi-axesa1anda2.

SuperpositionofTwoPerpendicularSimpleHarmonicVibrations

15

Ifa1¼a2¼athisbecomesthecircle

x2þy2¼a2

When

&2!&1¼0;2#;4#;etc:

theequationsimplifiesto

y¼a2

a1

x

whichisastraightlinethroughtheoriginofslopea2=a

1.

Againfor&2!&1¼

#,3#,5#,etc.,weobtain

y¼!a2

a1

x

astraightlinethroughtheoriginofequalbutoppositeslope.

Thepathstracedoutbytheparticleforvariousvaluesof'¼

&2!&1areshownin

Figure1.8andaremosteasilydemonstratedonacathoderayoscilloscope.

When

&2!&1¼0;#;2#;etc:

andtheellipsedegeneratesintoastraightline,theresultingvibrationlieswhollyinone

planeandtheoscillationsaresaidtobeplanepolarized.

δ = 0

δ = π 4

δ =

π 2δ =

δ =

π3 4

π

δ =

π5 4

δ =

π3 2

δ =

π7 4

δ = 2π

δ =

π 49

2 −

1

= δ

x =

a s

in (ω

t +

1)

y = a sin (ωt + 2)

φφ

φ

φ

Figure

1.8

Pathstraced

byasystem

vibratingsimultan

eouslyin

twoperpen

diculardirectionswith

simple

harmonic

motionsofeq

ual

freq

uen

cy.Th

ephasean

gle

'is

thean

gle

bywhichtheymotion

lead

sthexmotion

16

SimpleHarmonicMotion

Conventiondefinestheplaneofpolarizationasthatplaneperpendiculartotheplane

containingthevibrations.Similarlytheothervaluesof

&2!&1

yieldcircularorellipticpolarizationwherethetipofthevectorresultanttracesoutthe

appropriateconicsection.

(Problems1.14,1.15,1.16)

, Polarization

Polarizationisafundamentaltopicinopticsandarisesfromthesuperpositionoftwo

perpendicularsimpleharmonicopticalvibrations.WeshallseeinChapter8thatwhena

lightwaveisplanepolarizeditselectricalfieldoscillationlieswithinasingleplaneand

tracesasinusoidalcurvealongthedirectionofwavemotion.Substancessuchasquartzand

calcitearecapableofsplittinglightintotwowaveswhoseplanesofpolarizationare

perpendiculartoeachother.Exceptinaspecifieddirection,knownastheopticaxis,these

waveshavedifferentvelocities.Onewave,theordinaryorOwave,travelsatthesame

velocityinalldirectionsanditselectricfieldvibrationsarealwaysperpendiculartothe

opticaxis.TheextraordinaryorEwavehasavelocitywhichisdirection-dependent.Both

ordinaryandextraordinarylighthavetheirownrefractiveindices,andthusquartzand

calciteareknownasdoublyrefractingmaterials.Whentheordinarylightisfaster,asin

quartz,acrystalofthesubstanceisdefinedaspositive,butincalcitetheextraordinarylight

isfasteranditscrystalisnegative.Thesurfaces,spheresandellipsoids,whicharetheloci

ofthevaluesofthewavevelocitiesinanydirectionareshowninFigure1.9(a),andfora

Optic a

xis

O v

ibra

tion

E v

ibra

tion

x

y

xE

elli

psoid

O s

phere

z

y

O s

phere

E e

llipsoid

Optic a

xis

z

Quartz (+ve

)C

alc

ite (−ve

)

Figure1.9a

Ordinary(spherical)an

dextrao

rdinary(elliposoidal)wavesurfaces

indoubly

refracting

calcitean

dquartz.In

calcitetheEwaveis

faster

than

theOwave,

exceptalongtheopticaxis.In

quartz

theOwaveis

faster.Th

eOvibrationsarealwaysperpen

dicularto

theopticaxis,an

dtheOan

d

Evibrationsarealwaystangen

tial

totheirwavesurfaces

, Thissectionmaybeomittedatafirstreading.

Polarization

17

givendirectiontheelectricfieldvibrationsoftheseparatewavesaretangentialtothe

surfaceofthesphereorellipsoidasshown.Figure1.9(b)showsplanepolarizedlight

normallyincidentonacalcitecrystalcutparalleltoitsopticaxis.Withinthecrystalthe

fasterEwavehasvibrationsparalleltotheopticaxis,whiletheOwavevibrationsare

perpendiculartotheplaneofthepaper.Thevelocitydifferenceresultsinaphasegainof

theEvibrationovertheOvibrationwhichincreaseswiththethicknessofthecrystal.

Figure1.9(c)showsplanepolarizedlightnormallyincidentonthecrystalofFigure1.9(b)

withitsvibrationatanangleof45*oftheopticaxis.Thecrystalsplitsthevibrationinto

Pla

ne p

ola

rized

light norm

ally

incid

ent

O v

ibra

tion

to

pla

ne o

f paper

E v

ibra

tion

Optic

axis

Calc

ite

cry

sta

l

Figure1.9b

Planepolarizedlightnorm

ally

inciden

tonacalcitecrystalface

cutparallelto

itsoptic

axis.Th

ead

vance

oftheEwaveovertheOwaveis

equivalen

tto

againin

phase

E

O

45°

E v

ibra

tion 9

0°

ahead in p

hase

of O

vib

ration

O

E (O

ptic a

xis

)

Calc

ite

cry

sta

l

Optic a

xis

Phase d

iffe

rence

causes rota

tion o

fre

sultin

g e

lectric

field

vecto

r

Sin

usoid

al

vib

ration o

fele

ctric

fie

ld

Figure

1.9c

ThecrystalofFig.1.9cis

thicken

oughto

produceaphasegainof#=2

radin

the

EwaveovertheO

wave.

Wavereco

mbinationonleavingthecrystalproducescircularlypolarized

light

18

SimpleHarmonicMotion

equalEandOcomponents,andforagiventhicknesstheEwaveemergeswithaphasegain

of90*overtheOcomponent.Recombinationofthetwovibrationsproducescircularly

polarizedlight,ofwhichtheelectricfieldvectornowtracesahelixintheanticlockwise

directionasshown.

(2)VibrationsHav

ingDifferentFreq

uen

cies(Lissa

jousFigu

res)

Whenthefrequenciesofthetwoperpendicularsimpleharmonicvibrationsarenotequal

theresultingmotionbecomesmorecomplicated.Thepatternswhicharetracedarecalled

LissajousfiguresandexamplesoftheseareshowninFigure1.10wheretheaxial

frequenciesbearthesimpleratiosshownand

'¼

&2!&1¼0(ontheleft)

¼# 2(ontheright)

IftheamplitudesofthevibrationsarerespectivelyaandbtheresultingLissajousfigure

willalwaysbecontainedwithintherectangleofsides2aand2b.Thesidesoftherectangle

willbetangentialtothecurveatanumberofpointsandtheratioofthenumbersofthese

tangentialpointsalongthexaxistothosealongtheyaxisistheinverseoftheratioofthe

correspondingfrequencies(asindicatedinFigure1.10).

2a

2b

2b

2a

2b

2a

2a

2b

ωx

ωy

= 3

ωx

ωy

= 2

ωy

ωx

= 3

ωy

ωx

= 2

δ =

0

π 2δ =

Figure

1.10

Simple

Lissajous

figuresproduced

by

perpen

dicularsimple

harmonic

motionsof

differentan

gularfreq

uen

cies

Polarization

19

SuperpositionofaLargeNumbernofSimpleHarmonicVibrations

ofEqualAmplitudeaandEqualSuccessive

PhaseDifference

d

Figure1.11showstheadditionofnvectorsofequallengtha,eachrepresentingasimple

harmonicvibrationwithaconstantphasedifference

'fromitsneighbour.Twogeneral

physicalsituationsarecharacterizedbysuchasuperposition.ThefirstismetinChapter5

asawavegroupproblemwherethephasedifference

'arisesfromasmallfrequency

difference,'!,betweenconsecutivecomponents.ThesecondappearsinChapter12where

theintensityofopticalinterferenceanddiffractionpatternsareconsidered.There,the

superposedharmonicvibrationswillhavethesamefrequencybuteachcomponentwillhave

aconstantphasedifferencefromitsneighbourbecauseoftheextradistanceithastravelled.

Thefiguredisplaysthemathematicalexpression

Rcosð!tþ+Þ¼

acos!tþacosð!tþ'Þþacosð!tþ2'Þ

þ'''þ

acosð!tþ½n!1)'Þ

AB

a

a

a

a

a

a

a

C

r

O

rr α

δ

δ

δδ

δ

δ

δ

δ

90° −

290° −

2n δ

n δ

2n δ

R = 2r s

in

2 δa =

2r sin

Figure

1.11

Vectorsuperposition

ofa

largenumber

nofsimple

harmonic

vibrationsofeq

ual

amplitudeaan

deq

ual

successive

phasedifference

'.Th

eam

plitudeoftheresultan

t

R¼

2rsin

n' 2¼

asinn'=

2

sin'=

2

anditsphasewithrespectto

thefirstco

ntributionis

given

by

+¼ðn!

1Þ'=2

20

SimpleHarmonicMotion

whereRisthemagnitudeoftheresultantand+isitsphasedifferencewithrespecttothe

firstcomponentacos!t.

Geometricallyweseethateachlength a¼2rsin' 2

whereristheradiusofthecircleenclosingthe(incomplete)polygon.

FromtheisoscelestriangleOACthemagnitudeoftheresultant

R¼2rsinn' 2¼asinn'=2

sin'=2

anditsphaseangleisseentobe

+¼OAAB!OAAC

IntheisoscelestriangleOAC

OOAC¼90*!n' 2

andintheisoscelestriangleOAB

OAAB¼90*!' 2

so

+¼

90*!' 2

$%

!90*!n' 2

$%

¼ðn!1Þ' 2

thatis,halfthephasedifferencebetweenthefirstandthelastcontributions.Hencethe

resultant

Rcosð!tþ+Þ¼

asinn'=2

sin'=2cos

!tþðn!1Þ' 2

&'

Weshallobtainthesameresultlaterinthischapterasanexampleontheuseofexponential

notation.

Forthemomentletusexaminethebehaviourofthemagnitudeoftheresultant

R¼asinn'=2

sin'=2

whichisnotconstantbutdependsonthevalueof'.Whennisverylarge'isverysmall

andthepolygonbecomesanarcofthecirclecentreO,oflengthna¼A,withRasthe

chord.Then

+¼ðn!1Þ' 2-n' 2

SuperpositionofaLargeNumbernofSimpleHarmonicVibrations

21

and

sin' 2!

' 2-

+ n

Hence,inthislimit,

R¼asinn'=2

sin'=2¼asin+

+=n¼nasin+

+¼Asin+

+

ThebehaviourofAsin+=+versus+isshowninFigure1.12.Thepatternissymmetric

aboutthevalue+¼0andiszerowheneversin+¼0exceptat

+!0thatis,whensin

+=+!1.When

+¼0,'¼0andtheresultantofthenvectorsisthestraightlineoflength

A,Figure1.12(b).As'increasesAbecomesthearcofacircleuntilat+¼

#=2thefirstand

lastcontributionsareoutofphaseð2+¼

#ÞandthearcAhasbecomeasemicircleof

whichthediameteristheresultantRFigure1.12(c).Afurtherincreasein

'increases+and

curlstheconstantlengthAintothecircumferenceofacircle(+¼

#)withazeroresultant,

Figure

1.12(d).At+¼3#=2,Figure

1.12(e)thelength

Aisnow3/2

times

the

circumferenceofacirclewhosediameteristheamplitudeofthefirstminimum.

, SuperpositionofnEqual

SHMVectorsofLengthawith

Random

Phase

Whenthephasedifferencebetweenthesuccessivevectorsofthelastsectionmaytake

randomvalues

&betweenzeroand2#(measuredfromthexaxis)thevectorsuperposition

andresultantRmayberepresentedbyFigure1.13.

(b)

(c)

(e)

(d)

0

RA

2A

A

A=na

A =

R =

α

α

2π

ππ

π 2

2

π 23

3circum

fere

nce

A s

inα

Figure

1.12

(a)GraphofA

sin+=+

versus+,showingthemag

nitudeoftheresultan

tsfor(b)

+¼

0;(c)+¼

#/2

;(d)+¼

#an

d(e)+¼

3#/2

, Thissectionmaybeomittedatafirstreading.

22

SimpleHarmonicMotion

ThecomponentsofRonthexandyaxesaregivenby

Rx¼acos&1þacos&2þacos&3...acos&n

¼a

X

n

i¼1

cos&i

and

Ry¼a

X

n

i¼1

sin&i

where

R2¼R2 xþR2 y

Now

R2 x¼a2

X

n

i¼1

cos&i

!

2

¼a2

X

n

i¼1

cos2&iþ

X

n i¼1

i6¼j

cos&i

X

n

j¼1

cos&j

2 4

3 5

Inthetypicalterm2cos&icos&jofthedoublesummation,cos&iandcos&jhaverandom

valuesbetween&1andtheaveragedsumofsetsoftheseproductsiseffectivelyzero.

Thesummation

X

n

i¼1

cos2&i¼ncos2&

R

x

y

Figure1.13

Theresultan

tR¼

ffiffiffi n

paofnvectors,ea

choflength

a,havingrando

mphase.

Thisresult

isim

portan

tin

opticalinco

heren

cean

din

energyloss

from

waves

from

random

dissipationprocesses

SuperpositionofnEqualSHMVectorsofLengthawithRandomPhase

23

thatis,thenumberoftermsntimestheaveragevaluecos2&whichistheintegratedvalue

ofcos2&overtheintervalzeroto2#dividedbythetotalinterval2#,or

cos2&¼1 2#

ð

2#

0

cos2&d&¼1 2¼sin2&

So

R2 x¼a2

X

n

i¼1

cos2&i¼na2cos2&i¼na2

2

and

R2 y¼a2

X

n

i¼1

sin2&i¼na2sin2&i¼na2

2

giving

R2¼R2 xþR2 y¼na2

or

R¼

ffiffiffi n

pa

Thus,theamplitudeRofasystem

subjectedtonequalsimpleharmonicmotionsof

amplitudeawithrandomphasesinonly

ffiffiffi n

pawhereas,ifthemotionswereallinphaseR

wouldequalna.

Sucharesultillustratesaveryimportantprincipleofrandombehaviour.

(Problem

1.17)

App

lica

tion

s

IncoherentSources

inOptics

Theresultaboveisdirectlyapplicabletotheproblemof

coherenceinoptics.Lightsourceswhichareinphasearesaidtobecoherentandthis

conditionisessentialforproducingopticalinterferenceeffectsexperimentally.Ifthe

amplitudeofalightsourceisgivenbythequantityaitsintensityisproportionaltoa2,n

coherentsourceshavearesultingamplitudenaandatotalintensityn2a2.Incoherent

sourceshaverandomphases,nsuchsourceseachofamplitudeahavearesultingamplitude

ffiffiffi n

paandatotalintensityofna2.

Random

Processes

and

Energy

Absorption

Fromourpresentpointofviewthe

importanceofrandombehaviouristhecontributionitmakestoenergylossorabsorption

fromwavesmovingthroughamedium.Weshallmeetthisinallthewaveswediscuss.

24

SimpleHarmonicMotion

Randomprocesses,forexamplecollisionsbetweenparticles,inBrownianmotion,areof

greatsignificanceinphysics.Diffusion,viscosityorfrictionalresistanceandthermal

conductivityarealltheresultofrandomcollisionprocesses.Theseenergydissipating

phenomenarepresentthetransportofmass,momentumandenergy,andchangeonlyinthe

directionofincreasingdisorder.They

areknownas‘thermodynamicallyirreversible’

processesandareassociatedwiththeincreaseofentropy.Heat,forexample,canflowonly

fromabodyatahighertemperaturetooneatalowertemperature.Usingtheearlier

analysiswherethelengthaisnolongerasimpleharmonicamplitudebutisnowthe

averagedistanceaparticletravelsbetweenrandomcollisions(itsmeanfreepath),wesee

thatafternsuchcollisions(with,onaverage,equaltimeintervalsbetweencollisions)the

particlewill,onaverage,havetravelledonlyadistance

ffiffiffi n

pafromitspositionattimet¼0,

sothatthedistancetravelledvariesonlywiththesquarerootofthetimeelapsedinsteadof

beingdirectlyproportionaltoit.Thisisafeatureofallrandomprocesses.

Notalltheparticlesofthesystem

willhavetravelledadistance

ffiffiffi n

pabutthisdistanceis

themostprobableandrepresentsastatisticalaverage.

Randombehaviourisdescribed

bythediffusionequation(seethelastsectionof

Chapter7)andaconstantcoefficientcalledthediffusivityoftheprocesswillalways

arise.Thedimensionsofadiffusivityarealwayslength2/timeandmustbeinterpretedin

termsofacharacteristicdistanceoftheprocesswhichvariesonlywiththesquarerootof

time.

SomeUsefulMathem

atics

TheEx

ponen

tial

Series

Bya‘naturalprocess’ofgrowthordecaywemeanaprocessinwhichaquantitychanges

byaconstantfractionofitselfinagivenintervalofspaceortime.A5%perannum

compoundinterestrepresentsanaturalgrowthlaw;attenuationprocessesinphysicsusually

describenaturaldecay.

Thelawisexpresseddifferentiallyas

dN N¼&+dx

or

dN N¼&+dt

whereNisthechangingquantity,+isaconstantandthepositiveandnegativesigns

representgrowthanddecayrespectively.ThederivativesdN/dxordN/dtaretherefore

proportionaltothevalueofNatwhichthederivativeismeasured.

IntegrationyieldsN¼N0e&+xorN¼N0e&+twhereN0isthevalueatxort¼0ande

istheexponentialorthebaseofnaturallogarithms.Theexponentialseriesisdefinedas

ex¼1þxþx2

2!þx3

3!þ'''þ

xn

n!þ'''

andisshowngraphicallyforpositiveandnegativexinFigure1.14.Itisimportanttonote

thatwhatevertheformoftheindexofthelogarithmicbasee,itisthepowertowhichthe

SomeUsefulMathematics

25

baseisraised,andisthereforealwaysnon-dimensional.Thuse+xisnon-dimensionaland+

musthavethedimensionsofx!1.Writing

e+x¼1þ+xþð+xÞ22!þð+xÞ33!þ'''

itfollowsimmediatelythat

d dxðe

+xÞ¼

+þ2+2

2!xþ3+3

3!x2þ'''

¼+1þ+xþð+xÞ22!þð+xÞ33!

!

þ'''

"#

¼+e+x

Similarly

d2

dx2ðe

+xÞ¼

+2e+x

InChapter2weshallused(e

+t )=dt¼

+e+tandd2(e

+t )=dt2¼

+2e+tonanumberof

occasions.

Bytakinglogarithmsitiseasily

shownthatexey¼exþysince

logeðexeyÞ¼

logeexþlogeey¼xþy.

TheNotationi¼

ffiffiffiffiffiffiffi

!1

p

Thecombinationoftheexponentialserieswiththecomplexnumbernotationi¼

ffiffiffiffiffiffiffi

!1

pis

particularlyconvenientinphysics.Hereweshallshowthemathematicalconveniencein

expressingsineorcosine(oscillatory)behaviourintheform

eix¼cosxþisinx.

0x

y 1

y =

ex

y =

e−x

Figure

1.14

Thebeh

aviouroftheexponen

tial

series

y¼

exan

dy¼

e!x

26

SimpleHarmonicMotion

InChapter3weshallseetheadditionalmeritofiinitsroleofvectoroperator.

Theseriesrepresentationofsinxiswritten

sinx¼x!x3

3!þx5

5!!x7

7!'''

andthatofcosxis

cosx¼1!x2

2!þx4

4!!x6

6!'''

Since

i¼

ffiffiffiffiffiffiffi

!1

p;i2¼!1;i3¼!i

etc.wehave

eix¼1þixþðixÞ22!þðixÞ33!þðixÞ44!þ'''

¼1þix!x2

2!!ix3

3!þx4

4!þ'''

¼1!x2

2!þx4

4!þix!x3

3!þx5

5!þ'''

$%

¼cosxþisinx

Wealsoseethat

d dxðeixÞ¼

ieix¼icosx!sinx

Oftenweshallrepresentasineorcosineoscillationbytheformeixandrecovertheoriginal

formbytakingthatpartofthesolutionprecededbyiinthecaseofthesine,andthereal

partofthesolutioninthecaseofthecosine.

Exam

ples

(1)Insimpleharmonicmotion(€ xxþ!2x¼0)letustrythesolutionx¼aei!tei&,wherea

isaconstantlength,and&(andthereforeei&)isaconstant.

dx dt¼

_ xx¼i!aei!tei&¼i!x

d2x

dt2¼

€ xx¼i2!2aei!tei&¼!!2x

Therefore

x¼aei!tei&¼aeið!tþ

&Þ

¼acosð!tþ&Þþ

iasinð!tþ&Þ

isacompletesolutionof€ xxþ!2x¼0.

SomeUsefulMathematics

27

Onp.6weusedthesineformofthesolution;thecosineformisequallyvalidandmerely

involvesanadvanceof#=2inthephase&.

(2)

eixþe!ix¼21!x2

2!þx4

4!!'''

$%

¼2cosx

eix!e!ix¼2ix!x3

3!þx5

5!!'''

$%

¼2isinx

(3)Onp.21weusedageometricalmethodtoshowthattheresultantofthesuperposed

harmonicvibrations

acos!tþacosð!tþ'Þþacosð!tþ2'Þþ'''þ

acosð!tþ½n!1)'Þ

¼asinn'=2

sin'=2cos

!tþ

n!1

2

$%

'

12

Wecanderivethesameresultusingthecomplexexponentialnotationandtakingthereal

partoftheseriesexpressedasthegeometricalprogression

aei!tþaeið!tþ

'Þþaeið!tþ2'Þþ'''þ

aei½!tþðn!1Þ')

¼aei!t ð1þzþz2þ'''þ

zðn!1Þ Þ

wherez¼ei'.

Writing

SðzÞ¼

1þzþz2þ'''þ

zn!1

and

z½SðzÞ)¼zþz2þ'''þ

zn

wehave

SðzÞ¼

1!zn

1!z¼1!ein'

1!ei'

So

aei!t SðzÞ¼

aei!t1!ein'

1!ei'

¼aei!tein'=2ðe!in'=2!ein'=2Þ

ei'=2ðe!i'=2!ei'=2Þ

¼aei½!tþ

n!12ðÞ')sinn'=2

sin'=2

28

SimpleHarmonicMotion

withtherealpart

¼acos

!tþ

n!1

2

$%

'

&'

sinn'=2

sin'=2

whichrecoverstheoriginalcosineterm

fromthecomplexexponentialnotation.

(Problem

1.18)

(4)Supposewerepresentaharmonicoscillationbythecomplexexponentialform

z¼aei!t

whereaistheamplitude.Replacingiby!idefinesthecomplexconjugate

z,¼ae!i!t

TheuseofthisconjugateisdiscussedmorefullyinChapter3butherewecannotethatthe

productofacomplexquantityanditsconjugateisalwaysequaltothesquareofthe

amplitudefor

zz,¼a2ei!te!i!t¼a2eði!

iÞ!t¼a2e0

¼a2

(Problem

1.19)

Problem

1.1

Theequationofmotion

m€ xx¼!sx

with

!2¼s m

appliesdirectlytothesystem

inFigure1.1(c).

IfthependulumbobofFigure1.1(a)isdisplacedasmalldistancexshowthatthestiffness(restoring

forceperunitdistance)ismg=landthat!2¼g=lwheregistheaccelerationduetogravity.Nowuse

thesmallangulardisplacement(insteadofxandshowthat

!isthesame.

InFigure1.1(b)theangularoscillationsarerotationalsothemassisreplacedbythemomentof

inertiaIofthediscandthestiffnessbytherestoringcoupleofthewirewhichisCrad!1ofangular

displacement.Showthat

!2¼C=I.

InFigure1.1(d)showthatthestiffnessis2T=landthat

!2¼2T=lm.

InFigure1.1(e)showthatthestiffnessofthesystem

in2 Ag,whereAistheareaofcrosssection

andthat

!2¼2g=lwheregistheaccelerationduetogravity.

SomeUsefulMathematics

29

InFigure1.1(f)onlythegasintheflaskneckoscillates,behavingasapistonofmass Al.Ifthe

pressurechangesarecalculatedfromtheequationofstateusetheadiabaticrelationpV

,¼constant

andtakelogarithmstoshowthatthepressurechangeintheflaskis

dp¼!,pdV V¼!,pAx V;

wherexisthegasdisplacementintheneck.Henceshowthat

!2¼

,pA=l V.Notethat

,pisthe

stiffnessofagas(seeChapter6).

InFigure1.1(g),ifthecross-sectionalareaoftheneckisAandthehydrometerisadistancexabove

itsnormalfloatinglevel,therestoringforcedependsonthevolumeofliquiddisplaced(Archimedes’

principle).Showthat

!2¼g A=m.

Checkthedimensionsof!2foreachcase.

Problem

1.2

ShowbythechoiceofappropriatevaluesforAandBinequation(1.2)thatequallyvalidsolutions

forxare

x¼acosð!tþ&Þ

x¼asinð!t!&Þ

x¼acosð!t!&Þ

andcheckthatthesesolutionssatisfytheequation

€ xxþ!2x¼0

Problem

1.3

ThependuluminFigure1.1(a)swingswithadisplacementamplitudea.Ifitsstartingpointfromrest

is

ðaÞx¼a

ðbÞx¼!a

findthedifferentvaluesofthephaseconstant&forthesolutions

x¼asinð!tþ&Þ

x¼acosð!tþ&Þ

x¼asinð!t!&Þ

x¼acosð!t!&Þ

Foreachofthedifferentvaluesof&,findthevaluesof!tatwhichthependulumswingsthroughthe

positions

x¼þa=

ffiffiffi 2

p

x¼a=2

30

SimpleHarmonicMotion

and

x¼0

forthefirsttimeafterreleasefrom

x¼&a

Problem

1.4

Whentheelectroninahydrogenatomboundtothenucleusmovesasmalldistancefromits

equilibriumposition,arestoringforceperunitdistanceisgivenby

s¼e2=4#- 0r2

wherer¼0:05nmmaybetakenastheradiusoftheatom.Showthattheelectroncanoscillatewith

asimpleharmonicmotionwith

!0-4:5010!16rads!1

Iftheelectronisforcedtovibrateatthisfrequency,inwhichregionoftheelectromagneticspectrum

woulditsradiationbefound?

e¼1:6010!19C;electronmassme¼9:1010!31kg

-0¼8:85010!12N!1m!2C2

Problem

1.5

Showthatthevaluesof!2forthethreesimpleharmonicoscillations(a),(b),(c)inthediagramare

intheratio1:2:4.

m

mm

ss

ss s

(a)

(b)

(c)

Problem

1.6

Thedisplacementofasimpleharmonicoscillatorisgivenby

x¼asinð!tþ&Þ

Iftheoscillationstartedattimet¼0fromapositionx0withavelocity

_ xx¼

v0showthat

tan&¼

!x0=v0

and

a¼ðx2 0þv2 0=!2Þ1

=2

SomeUsefulMathematics

31

Problem

1.7

Aparticleoscillateswithsimpleharmonicmotionalongthexaxiswithadisplacementamplitudea

andspendsatimedtinmovingfromxtoxþdx.Showthattheprobabilityoffindingitbetweenx

andxþdxisgivenby

dx

#ða2!x2Þ1

=2

(inwavemechanicssuchaprobabilityisnotzeroforx>a).

Problem.1.8

Manyidenticalsimpleharmonicoscillatorsareequallyspacedalongthexaxisofamediumanda

photographshowsthatthelocusoftheirdisplacementsintheydirectionisasinecurve.Ifthe

distance

/separatesoscillatorswhichdifferinphaseby2#radians,whatisthephasedifference

betweentwooscillatorsadistancexapart?

Problem

1.9

Amassstandsonaplatform

whichvibratessimpleharmonicallyinaverticaldirectionata

frequencyof5Hz.Showthatthemasslosescontactwiththeplatform

whenthedisplacement

exceeds10!2m.

Problem

1.10

AmassMissuspendedattheendofaspringoflengthlandstiffnesss.Ifthemassofthespringism

andthevelocityofanelementdyofitslengthisproportionaltoitsdistanceyfromthefixedendof

thespring,showthatthekineticenergyofthiselementis

1 2

m ldy

34

y lv

34

2

wherevisthevelocityofthesuspendedmassM.Hence,byintegratingoverthelengthofthespring,

showthatitstotalkineticenergyis1 6mv2and,fromthetotalenergyoftheoscillatingsystem,show

thatthefrequencyofoscillationisgivenby !2¼

s

Mþm=3

Problem

1.11

Thegeneralform

fortheenergyofasimpleharmonicoscillatoris

E¼

1 2mass(velocity)2þ1 2stiffness(displacement)2

SetuptheenergyequationsfortheoscillatorsinFigure1.1(a),(b),(c),(d),(e),(f)and(g),anduse

theexpression

dE dt¼0

toderivetheequationofmotionineachcase.

32

SimpleHarmonicMotion

Problem

1.12

Thedisplacementofasimpleharmonicoscillatorisgivenbyx¼asin!t.Ifthevaluesofthe

displacementxandthevelocity

_ xxareplottedonperpendicularaxes,eliminatettoshowthatthelocus

ofthepoints(x;_ xx)isanellipse.Showthatthisellipserepresentsapathofconstantenergy.

Problem

1.13

InChapter12theintensityofthepatternwhenlightfromtwoslitsinterferes(Young’sexperiment)

willbeseentodependonthesuperpositionoftwosimpleharmonicoscillationsofequalamplitudea

andphasedifference

'.Showthattheintensity

I¼R2/4a2cos2'=2

Betweenwhatvaluesdoestheintensityvary?

Problem

1.14

Carryouttheprocessindicatedinthetexttoderiveequation(1.3)onp.15.

Problem

1.15

Theco-ordinatesofthedisplacementofaparticleofmassmaregivenby

x¼asin!t

y¼bcos!t

Eliminatettoshowthattheparticlefollowsanellipticalpathandshowbyaddingitskineticand

potentialenergyatanypositionx,ythattheellipseisapathofconstantenergyequaltothesumof

theseparateenergiesofthesimpleharmonicvibrations.

Provethatthequantitymðx_ yy!y_ xxÞisalsoconstant.Whatdoesthisquantityrepresent?

Problem

1.16

Twosimpleharmonicmotionsofthesamefrequencyvibrateindirectionsperpendiculartoeach

otheralongthexandyaxes.Aphasedifference

'¼

&2!&1

existsbetweenthem

suchthattheprincipalaxesoftheresultingellipticaltraceareinclinedatan

angletothexandyaxes.Showthatthemeasurementoftwoseparatevaluesofx(ory)issufficientto

determinethephasedifference.

(Hint:useequation(1.3)andmeasurey(max),andyfor(x¼0.)

Problem

1.17

Takearandomgroupofn>7valuesof&intherange02&2#andform

theproduct

X

n i¼1

i6¼j

cos&i

X

n

j¼1

cos&j

Showthattheaveragevalueobtainedforseveralsuchgroupsisnegligiblewithrespectton=2.

SomeUsefulMathematics

33

Problem

1.18

Usethemethodofexample(3)(p.28)toshowthat

asin!tþasinð!tþ'Þþasinð!tþ2'Þþ'''þ

asin½!tþðn!1Þ')

¼asin

!tþðn!1Þ

2'

&'

sinn'=2

sin'=2

Problem

1.19

Ifwerepresentthesumoftheseries

acos!tþacosð!tþ'Þþacosð!tþ2'Þþ'''þ

acos½!tþðn!1Þ')

bythecomplexexponentialform

z¼aei!t ð1þei'þei2'þ'''þ

eiðn!1Þ'Þ

showthat

zz,¼a2sin2n'=2

sin2'=2

Summary

ofIm

portantResults

SimpleHarmonicOscillator(massm,stiffnesss,amplitudea)

Equationofmotion€ xxþ!2x¼0where!2¼s=m

Displacementx¼asinð!tþ&Þ

Energy¼

1 2m_ xx2þ1 2sx2¼

1 2m!2a2¼

1 2sa2¼constant

Supe

rpos

ition(A

mplitude

andPhas

e)of

twoSH

Ms

One-dimen

sion

al

Equal

!,differentamplitudes,phasedifference

',resultantRwhereR2¼a2 1þa2 2þ

2a1a2cos'

Different!,equalamplitude,

x¼x1þx2¼aðsin!1tþsin!2tÞ

¼2asinð!

1þ!2Þt

2cosð!

2!!1Þt

2

Two-dimen

sion

al:pe

rpen

dicu

larax

esEqual

!,differentamplitude—

givinggeneralconicsection

x2

a2 1

þy2

a2 2

!2xy

a1a2

cosð&

2!&1Þ¼

sin2ð&

2!&1Þ

(basisofopticalpolarization)

34

SimpleHarmonicMotion

SuperpositionofnSHMVectors(equalamplitudea,constantsuccessivephasedifference

')

TheresultantisRcosð!tþ+Þ,where R¼asinn'=2

sin'=2

and

+¼ðn!1Þ'=2

Importantinopticaldiffractionandwavegroupsofmanycomponents.

SomeUsefulMathematics

35

2 DampedSimple

Harm

onicMotion

Initia

lly

we

dis

cuss

edth

eca

seof

idea

lsi

mple

har

monic

motion

wher

eth

eto

tal

ener

gy

rem

ained

const

antan

dth

edis

pla

cem

entfo

llow

eda

sine

curv

e,ap

par

ently

for

anin

finite

tim

e.In

pra

ctic

eso

me

ener

gy

isal

way

sdis

sipat

edby

are

sist

ive

or

vis

cous

pro

cess

;fo

r

exam

ple

,th

eam

plitu

de

of

afr

eely

swin

gin

gpen

dulu

mw

ill

alw

ays

dec

ayw

ith

tim

eas

ener

gy

islo

st.The

pre

sence

of

resi

stan

ceto

motion

mea

ns

that

anoth

erfo

rce

isac

tive,

whic

his

taken

asbei

ng

pro

portio

nal

toth

evel

oci

ty.

The

fric

tional

forc

eac

tsin

the

direc

tion

opposi

teto

that

of

the

vel

oci

ty(s

eeFig

ure

2.1

)an

dso

New

ton’s

Sec

ond

law

bec

om

es

m€ xx¼!sx!r_ xx

wher

er

isth

eco

nst

ant

of

pro

portio

nal

ity

and

has

the

dim

ensi

ons

of

forc

eper

unit

of

vel

oci

ty.The

pre

sence

of

such

ate

rmw

illal

way

sre

sult

inen

ergy

loss

.

The

pro

ble

mnow

isto

find

the

beh

avio

urofth

edis

pla

cem

entx

from

the

equat

ion

m€ xxþr_ xxþsx¼

0ð2

:1Þ

wher

eth

eco

effici

entsm

,r

ands

are

const

ant.

When

thes

eco

effici

ents

are

const

ant

aso

lution

of

the

form

x¼C

e!t

can

alw

ays

be

found.

Obvio

usl

y,si

nce

anex

ponen

tial

term

isal

way

snondim

ensi

onal

,C

has

the

dim

ensi

ons

ofx

(ale

ngth

,sa

y)

and

!has

the

dim

ensi

ons

ofin

ver

setim

e,T!

1.W

esh

all

see

that

ther

ear

eth

ree

poss

ible

form

sof

this

solu

tion,

each

des

crib

ing

adiffe

rent

beh

avio

urofth

edis

pla

cem

entx

with

tim

e.In

two

ofth

ese

solu

tionsC

appea

rsex

plici

tly

as

aco

nst

antle

ngth

,butin

the

third

case

itta

kes

the

form

C¼AþBt%

ThePhysicsofVibrationsandWaves,

6th

Editio

nH

.J.

Pai

n#

2005

John

Wiley

&Sons,

Ltd

%The

num

ber

ofco

nstan

tsal

low

edin

the

gen

eral

solu

tion

ofa

diffe

rential

equat

ion

isal

way

seq

ual

toth

eord

er(that

is,th

ehig

hes

tdiffe

rential

coef

fici

ent)

ofth

eeq

uat

ion.The

two

val

uesA

andB

are

allo

wed

bec

ause

equat

ion

(2.1

)is

seco

nd

ord

er.Theval

ues

ofth

eco

nstan

tsar

ead

justed

tosa

tisf

yth

e

initia

lco

nditio

ns.

37

wher

eA

isa

length

,B

isa

vel

oci

tyan

dtis

tim

e,giv

ingC

the

over

alldim

ensi

ons

of

a

length

,as

we

expec

t.Fro

mour

poin

tofvie

wth

isca

seis

notth

em

ost

importan

t.

Tak

ingC

asa

const

antle

ngth

giv

es_ xx¼

!C

e!tan

d€ xx¼

!2C

e!t ,

soth

ateq

uat

ion

(2.1

)

may

be

rew

ritten

Ce

!t ðm

!2þr!þsÞ¼

0

soth

atei

ther

x¼C

e!t¼

0(w

hic

his

triv

ial)

or

m!

2þr!þs¼

0

Solv

ing

the

quad

ratic

equat

ion

in!

giv

es

!¼!r

2m&

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

r2

4m

2!s m

r

Note

thatr=

2m

andðs

=mÞ1

=2,an

dth

eref

ore

,!,al

lhav

eth

edim

ensi

ons

of

inver

setim

e,

T!

1,w

hic

hw

eex

pec

tfr

om

the

form

ofe

!t .

The

dis

pla

cem

entca

nnow

be

expre

ssed

as

x1¼C

1e!rt

=2mþðr

2=4m

2!s=mÞ

1=2t ;

x2¼C

2e!rt

=2m!ðr

2=4m

2!s=mÞ

1=2t

orth

esu

mofboth

thes

ete

rms

x¼x

1þx

2¼C

1e!rt

=2mþðr

2=4m

2!s=mÞ

1=2tþC

2e!rt

=2m!ðr

2=4m

2!s=mÞ

1=2t

The

bra

cketðr

2=4m

2!s=mÞca

nbe

posi

tive,

zero

orneg

ativ

edep

endin

gon

the

rela

tive

mag

nitude

of

the

two

term

sin

side

it.

Eac

hof

thes

eco

nditio

ns

giv

esone

of

the

thre

e

poss

ible

solu

tions

refe

rred

toea

rlie

ran

dea

chso

lution

des

crib

esa

par

ticu

lar

kin

dof

m

Frictional

force

F = −rx

s

x

Figure

2.1

Simple

harmonic

motionsystem

withadam

pingorfrictional

force

r_ xxactingag

ainst

the

directionofmotion.Th

eeq

uationofmotionis

m€ xxþ

r_ xxþ

sx¼

0

38

DampedSimpleHarmonicMotion

behaviour.Weshalldiscuss

thesesolutionsin

order

ofincreasingsignificance

from

our

pointofview;thethirdsolutionistheoneweshallconcentrateuponthroughouttherestof

this

book.

Theconditionsare:

(1)Bracketpositive

ðr2=4m

2>s=mÞ.

Here

the

dam

ping

resistance

term

r2=4m

2

dominates

thestiffnessterm

s=m,andheavydam

pingresultsin

adeadbeatsystem

.

(2)Bracketzero

ðr2=4m

2¼s=mÞ.

The

balance

between

the

two

term

sresultsin

a

criticallydampedsystem

.

Neither

(1)nor(2)gives

oscillatory

behaviour.

(3)Bracketnegativeðr

2=4m

2<s=mÞ.Thesystem

islightlydam

ped

andgives

oscillatory

dampedsimpleharmonicmotion.

Case

1.Hea

vyDamping

Writingr=2m

¼pand

ðr2=4m

2#s=mÞ1

=2¼q,wecanreplace

x¼C

1e

#rt=2mþ

ðr2=4m

2#s=mÞ1

=2tþC

2e

#rt=2m#

ðr2=4m

2#s=mÞ1

=2t

by

x¼

e#pt ðC

1eqtþC

2e

#qtÞ;

wheretheC

1andC

2arearbitrary

invaluebuthavethesamedim

ensionsasC(note

that

twoseparatevalues

ofC

areallowed

because

thedifferential

equation(2.1)is

second

order).

IfnowF

¼C

1þC

2andG

¼C

1#C

2,thedisplacementisgiven

by

x¼

e#ptF 2

ðeqtþe

#qt Þ

þG 2

ðeqt#e

#qt Þ

!

or

x¼

e#pt ðF

coshqtþGsinhqtÞ

This

represents

non-oscillatory

behaviour,buttheactual

displacementwilldependupon

theinitial(orboundary)conditions;that

is,thevalueofxat

timet¼

0.Ifx¼

0att¼

0

thenF

¼0,and

x¼Ge

#rt=2msinh

r2

4m

2#s m

"#

1=2

t

Figure

2.2

illustratessuch

behaviourwhen

aheavilydam

ped

system

isdisturbed

from

equilibrium

byasudden

impulse(that

is,given

avelocity

att¼

0).Itwillreturn

tozero

DampedSimpleHarmonicMotion

39

displacementquite

slowly

withoutoscillating

aboutits

equilibrium

position.More

advancedmathem

aticsshowsthatthevalueofthevelocitydx=dtvanishes

only

once

sothat

thereis

only

onevalueofmaxim

um

displacement.

(Problem

2.1)

Case

2.CriticalDamping

ðr2=4

m2¼

s=mÞ

UsingthenotationofCase1,weseethatq

¼0andthatx¼

e#pt ðC

1þC

2Þ.This

is,in

fact,thelimitingcase

ofthebehaviourofCaseIasqchanges

from

positiveto

negative.In

this

case

thequadraticequationin

$has

equal

roots,which,in

adifferential

equation

solution,dem

andsthatCmustbewrittenC

¼A

þBt,whereAisaconstantlength

andBa

given

velocity

whichdependsontheboundaryconditions.Itiseasily

verified

thatthevalue

x¼

ðAþBtÞe

#rt=2m

¼ðA

þBtÞe

#pt

satisfiesm€ xxþr_ xxþsx

¼0whenr2=4m

2¼s=m.

(Problem

2.2)

Applicationto

aDamped

MechanicalOscillator

Criticaldam

pingis

ofpractical

importance

inmechanical

oscillators

whichexperience

sudden

impulses

andarerequired

toreturn

tozero

displacementin

theminim

um

time.

Suppose

such

asystem

has

zero

displacementatt¼

0andreceives

anim

pulsewhichgives

itan

initialvelocityV.

Time

r increasing

Displacement

Heavy damping

r 2

4m 2

s m

>

Figu

re2.2

Non-oscillatory

behaviourofdam

ped

simple

harmonic

system

withheavy

dam

ping

(wherer2=4

m2>

s=m)afterthesystem

has

beengiven

animpulsefrom

arest

positionx¼

0

40

DampedSimpleHarmonicMotion

Thenx¼

0(sothatA

¼0)and_ xx¼V

att¼

0.However,

_ xx¼B½ð#ptÞe

#ptþe

#pt &

¼Batt¼

0

sothatB

¼V

andthecomplete

solutionis

x¼Vte

#pt

Themaxim

um

displacementxoccurswhen

thesystem

comes

torestbefore

returningto

zero

displacement.Atmaxim

um

displacement

_ xx¼Ve

#pt ð1

#ptÞ

¼0

thusgiving

ð1#ptÞ

¼0,i.e.t¼

1=p.

Atthistimethedisplacementistherefore

x¼Vte

#pt¼V p

e#1

¼0:368V p

¼0:3682mV

r

Thecurveofdisplacementversustimeis

shownin

Figure

2.3;thereturn

tozero

ina

critically

dam

ped

system

isreached

inminimum

time.

Case

3.Damped

Simple

Harm

onic

Motion

Whenr2=4m

2<s=mthedam

pingislight,andthisgives

from

thepresentpointofviewthe

most

importantkindofbehaviour,oscillatorydampedsimpleharmonicmotion.

r 2

4m 2

s m

2m r

t =

m r=

Displacement

Time

0

Critical

damping

2Ve−1

x =

Figu

re2.3

Limitingcase

ofnon-oscillatory

behaviourofdam

ped

simple

harmonic

system

where

r2=4

m2¼

s=m

(criticaldam

ping)

DampedSimpleHarmonicMotion

41

Theexpression

ðr2=4m

2#s=mÞ1

=2is

anim

aginary

quantity,thesquarerootofa

negativenumber,whichcanberewritten

'r2

4m

2#s m

"#

1=2

¼'

ffiffiffiffiffiffiffi

#1

ps m

#r2

4m

2

"#

1=2

¼'is m

#r2

4m

2

"#

1=2

ðwherei¼

ffiffiffiffiffiffiffi

#1

pÞ

sothedisplacement

x¼C

1e

#rt=2me

þiðs

=m#r2=4m

2Þ1

=2tþC

2e

#rt=2me

#iðs

=m#r2=4m

2Þ1

=2t

Thebracket

has

thedim

ensionsofinverse

time;

that

is,offrequency,andcanbewritten

ðs=m

#r2=4m

2Þ1

=2¼

!0 ,

sothat

the

second

exponential

becomes

ei!

0 t¼

cos!

0 tþ

isin!

0 t:This

showsthat

thebehaviourofthedisplacementxis

oscillatory

withanew

frequency

!0 <

!¼

ðs=mÞ1

=2,thefrequency

ofidealsimpleharmonicmotion.Tocompare

thebehaviourofthedam

ped

oscillatorwiththeidealcase

weshould

liketo

express

the

solutionin

aform

similar

tox¼Asinð!

0 tþ'Þas

intheidealcase,where!

has

been

replacedby!

0 .Wecandothisbywriting

x¼

e#rt=2mðC

1ei!

0 tþC

2e

#i!

0 tÞ

Ifwenow

choose

C1¼A 2iei'

and

C2¼

#A 2ie

#i'

whereAand'(andthusei')areconstantswhichdependonthemotionatt¼

0,wefind

aftersubstitution

x¼Ae

#rt=2m½e

ið!0 tþ'Þ#e

#ið!

0 tþ'Þ&

2i

¼Ae

#rt=2msinð!

0 tþ'Þ

This

procedure

isequivalentto

imposingtheboundaryconditionx¼Asin'att¼

0

uponthesolutionforx.Thedisplacementtherefore

variessinusoidally

withtimeas

inthe

case

ofsimple

harmonic

motion,butnow

has

anew

frequency

!0 ¼

s m#r2

4m

2

"#

1=2

42

DampedSimpleHarmonicMotion

anditsam

plitudeAismodified

bytheexponential

term

e#rt=2m,aterm

whichdecayswith

time. Ifx¼

0att¼

0then

'¼

0;Figure

2.4

showsthebehaviourofxwith

time,

its

oscillationsgradually

decayingwiththeenvelopeofmaxim

um

amplitudes

followingthe

dotted

curvee

#rt=2m.TheconstantAisobviouslythevalueto

whichtheam

plitudewould

haverisenat

thefirstmaxim

um

ifnodam

pingwerepresent.

Thepresence

oftheforcetermr_ xxin

theequationofmotiontherefore

introducesaloss

of

energywhichcausestheam

plitudeofoscillationto

decay

withtimeas

e#rt=2m.

(Problem

2.3)

MethodsofDes

cribingth

eDam

pingofan

Oscillato

r

Earlier

inthischapterwesaw

that

theenergyofan

oscillatorisgiven

by

E¼

1 2ma2!

2¼

1 2sa

2

that

is,proportional

tothesquareofitsam

plitude.

Wehavejustseen

that

inthepresence

ofadam

pingforcer_ xxtheam

plitudedecayswith

timeas

e#rt=2m

sothat

theenergydecay

willbeproportional

to

ðe#rt=2mÞ2

that

is,e

#rt=m.Thelarger

thevalueofthedam

pingforcerthemore

rapid

thedecay

ofthe

amplitudeandenergy.Thuswecanuse

theexponentialfactorto

express

theratesatwhich

theam

plitudeandenergyarereduced.

τ′

τ′

2

t

r t

2m

r 2

4m

2

s m

e

<

Displacement

−

Figu

re2.4

Dam

ped

oscillatory

motionwheres=m

>r2=4m

2.Theam

plitudedecayswithe

#rt=2

m,

andthereducedangularfrequency

isgiven

by!

02¼

s=m

#r2=4

m2

MethodsofDescribingtheDampingofanOscillator

43

Loga

rithmic

Decremen

t

Thismeasurestherate

atwhichtheamplitudediesaw

ay.Suppose

intheexpression

x¼Ae

#rt=2msinð!

0 tþ'Þ

wechoose

'¼

(=2

andwewrite

x¼A

0e

#rt=2mcos!

0 t

withx¼A

0att¼

0.Itsbehaviourwillfollow

thecurvein

Figure

2.5.

Iftheperiodofoscillationis)

0 where!

0 ¼2(=)

0 ,then

oneperiodlatertheam

plitudeis

given

by

A1¼A

0e

ð#r=2mÞ)

0

sothat

A0

A1

¼er)

0 =2m

¼e*

A0

At

A2

t0

τ ′

τ ′

τ ′

τ ′

e

r2m

t

e

r2m

(2 )

e

r2m

−

−

−

Figu

re2.5

Thelogarithmic

ratio

ofanytwo

amplitudes

oneperiod

apartis

thelogarithmic

decrement,defined

as*¼

logeðA

n=A

nþ1Þ¼

r)0 =2m

44

DampedSimpleHarmonicMotion

where

*¼

r 2m

)0 ¼

loge

A0

A1

iscalled

thelogarithmicdecrement.(N

otethatthisuse

of*differsfrom

thatin

Figure

1.11).

Thelogarithmic

decrement*isthelogarithm

oftheratiooftwoam

plitudes

ofoscillation

whichareseparated

byoneperiod,thelarger

amplitudebeingthenumeratorsince

e*>

1.

Sim

ilarly

A0

A2

¼erð2)

0 Þ=2m

¼e2*

and

A0

An

¼en*

Experim

entally,

thevalueof*is

bestfoundbycomparingam

plitudes

ofoscillations

whichareseparated

bynperiods.Thegraphof

loge

A0

An

versusnfordifferentvalues

ofnhas

aslope*.

RelaxationTimeorModulusof

Decay

Another

way

ofexpressingthedam

pingeffect

isbymeansofthetimetaken

forthe

amplitudeto

decay

to

e#1¼

0:368

ofitsoriginalvalueA

0.Thistimeiscalled

therelaxationtimeormodulusofdecayandthe

amplitude

At¼A

0e

#rt=2m

¼A

0e

#1

atatimet¼

2m=r.

Measuringthenaturaldecay

interm

softhefractione

#1oftheoriginal

valueisavery

commonprocedure

inphysics.Thetimeforanaturaldecay

process

toreachzero

is,of

course,

theoreticallyinfinite.

(Problem

2.4)

TheQuality

FactororQ-valueof

aDamped

Simple

Harm

onic

Oscillator

This

measurestherate

atwhichtheenergydecays.

Since

thedecay

oftheam

plitudeis

representedby

A¼A

0e

#rt=2m

MethodsofDescribingtheDampingofanOscillator

45

thedecay

ofenergyis

proportional

to A2¼A

2 0e

ð#rt=2mÞ2

andmay

bewritten

E¼E

0e

ð#r=mÞt

whereE

0is

theenergyvalueatt¼

0.

ThetimefortheenergyEto

decay

toE

0e

#1isgiven

byt¼m=r

sduringwhichtimethe

oscillatorwillhavevibratedthrough!

0 m=rrad.

Wedefinethequalityfactor

Q¼

!0 m r

asthenumberofradiansthroughwhichthedampedsystemoscillatesasitsenergy

decaysto

E¼E

0e

#1

Ifrissm

all,thenQ

isverylargeand

s m*

r2

4m

2

sothat

!0 +

!0¼

s m

%&

1=2

Thus,wewrite,to

averyclose

approxim

ation,

Q¼

!0m r

whichisaconstantofthedam

ped

system

.

Sincer=m

now

equals!

0=Q

wecanwrite

E¼E

0e

ð#r=mÞt

¼E

0e

#!

0t=Q

Thefact

thatQ

isaconstantð¼

!0m=rÞ

implies

that

theratio

energystoredin

system

energylost

per

cycle

46

DampedSimpleHarmonicMotion

isalso

aconstant,for

Q 2(

¼!

0m

2(r

¼+0m r

isthenumber

ofcycles(orcomplete

oscillations)

throughwhichthesystem

moves

in

decayingto

E¼E

0e

#1

andif

E¼E

0e

ð#r=mÞt

theenergylostper

cycleis

#"E

¼dE dt"t¼

#r

mE1 +0

where"t¼

1=+

0 ¼)

0 ,theperiodofoscillation.

Thus,theratio

energystoredin

system

energylost

per

cycle

¼E

#"E

¼+

0 m r+

+0m r

¼Q 2(

InthenextchapterweshallmeetthesamequalityfactorQin

twoother

roles,thefirstas

ameasure

ofthepower

absorptionbandwidth

ofadam

ped

oscillatordriven

nearits

resonantfrequency

andagainas

thefactorbywhichthedisplacementoftheoscillator

isam

plified

atresonance.

Example

ontheQ-valueofaDamped

Simple

Harm

onic

Oscillator

Anelectronin

anatom

whichis

freely

radiatingpower

behaves

asadam

ped

simple

harmonic

oscillator.

Iftheradiatedpower

isgiven

byP

¼q2!

4x2 0=12("0c3W

atawavelength

of0.6mm

(6000A),showthattheQ-valueoftheatom

isabout108andthatitsfree

radiationlifetime

isabout10

#8s(thetimeforitsenergyto

decay

toe

#1ofitsoriginal

value).

q¼

1:6

,10

#19C

1=4("0¼

9,109mF

#1

me¼

9,10

#31kg

c¼

3,108ms#1

x0¼

maxim

um

amplitudeofoscillation

TheradiatedpowerPis

#+"E,where#"Eistheenergyloss

per

cycle,andtheenergyof

theoscillatorisgiven

byE

¼1 2me!

2x2 0.

MethodsofDescribingtheDampingofanOscillator

47

Thus,Q

¼2(E=#"E

¼+(me!

2x2 0=P

,andinsertingthevalues

abovewith!

¼2(+

¼2(c=-,wherethewavelength

-isgiven,yieldsaQ

valueof-

5,107.

TherelationQ

¼!tgivest,theradiationlifetime,

avalueof-

10

#8s.

Energy

Dissipation

Wehaveseen

that

thepresence

oftheresistiveforcereducestheam

plitudeofoscillation

withtimeas

energyis

dissipated.

Thetotalenergyremainsthesum

ofthekinetic

andpotential

energies

E¼

1 2m_ xx2þ

1 2sx

2

Now,however,dE=dtisnotzero

butnegativebecause

energyis

lost,so

that

dE dt¼

d dtð1 2m_ xx2þ

1 2sx

2Þ¼

_ xxðm€ xxþsx

Þ

¼_ xxð#r_ xxÞ

form_ xxþr_ xxþsx

¼0

i.e.dE=dt¼

#r_ xx2,whichistherateofdoingwork

againstthefrictionalforce(dim

ensions

offorce

,velocity

¼force

,distance/tim

e).

(Problems2.5,2.6)

Damped

SHM

inanElectricalCircuit

Theforceequationin

themechanical

oscillatorisreplacedbythevoltageequationin

the

electrical

circuitofinductance,resistance

andcapacitance

(Figure

2.6).

IR

IR

++

+

++

−−

dI

dt

L

dI

dt

L

q C

q C= 0

−

Figu

re2.6

Electrical

circuitofinductance,capacitance

andresistance

capable

ofdam

ped

simple

harmonic

oscillations.

Thesum

ofthevoltages

aroundthecircuit

isgiven

from

Kirchhoff’s

law

asLdI dtþ

RIþ

q C¼

0

48

DampedSimpleHarmonicMotion

Wehave,

therefore,

LdI dtþRIþq C

¼0

or

L€ qqþR_ qqþq C

¼0

andbycomparisonwiththesolutionsforxin

themechanical

case

weknow

immediately

that

thecharge

q¼q0e

#Rt=2L'

ðR2=4L2#1=LCÞ1

=2t

which,for1=LC>R

2=4L2,gives

oscillatory

behaviourat

afrequency

!2¼

1 LC

#R

2

4L2

From

theexponential

decay

term

weseethatR=L

has

thedim

ensionsofinverse

timeT

#1

or!,so

that

!Lhas

thedim

ensionsofR;that

is,!Lismeasuredin

ohms.

Sim

ilarly,since

!2¼

1=LC;!L

¼1=!C,so

that

1=!C

isalso

measuredin

ohms.

We

shalluse

theseresultsin

thenextchapter.

(Problems2.7,2.8,2.9)

Problem

2.1

Theheavilydam

ped

simple

harmonic

system

ofFigure

2.2

isdisplacedadistanceF

from

its

equilibrium

positionandreleased

from

rest.Show

that

intheexpressionforthedisplacement

x¼

e#pt ðF

coshqtþGsinhqtÞ

where

p¼

r 2m

and

q¼

r2

4m

2#s m

"#

1=2

that

theratio

G F¼

r

ðr2#4msÞ

1=2

Problem

2.2

Verifythat

thesolution

x¼

ðAþBtÞe

#rt=2m

MethodsofDescribingtheDampingofanOscillator

49

satisfies

theequation

m€ xxþr_ xxþsx

¼0

when

r2=4m

2¼s=m

Problem

2.3

Thesolutionfordam

ped

simple

harmonic

motionis

given

by

x¼

e#rt=2mðC

1ei!

0 tþC

2e

#i!

0 tÞ

Ifx¼Acos'att¼

0,findthevalues

ofC

1andC

2to

showthat

_ xx+

#!

0 Asin'att¼

0only

ifr=m

isverysm

allor'

+(=2.

Problem

2.4

AcapacitanceC

withachargeq0att¼

0discharges

througharesistanceR.Use

thevoltage

equationq=C

þIR

¼0to

show

that

therelaxationtimeofthisprocess

isRC

s;that

is,

q¼q0e

#t=RC

(Note

thatt=RC

isnon-dim

ensional.)

Problem

2.5

Thefrequency

ofadam

ped

simple

harmonic

oscillatorisgiven

by

!02

¼s m

#r2

4m

2¼

!2 0#r2

4m

2

(a)If!

2 0#!

02¼

10

#6!

2 0show

thatQ

¼500andthat

thelogarithmic

decrement*¼

(=500.

(b)If!

0¼

106andm

¼10

#10Kgshow

that

thestiffnessofthesystem

is100Nm

#1,andthat

the

resistiveconstantris

2,10

#7N

.sm

#1.

(c)Ifthemaxim

um

displacementatt¼

0is10

#2m,showthattheenergyofthesystem

is5,10

#3

Jandthedecay

toe

#1ofthisvaluetakes

0.5

ms.

(d)Show

that

theenergyloss

inthefirstcycleis

2(

,10

#5J.

Problem

2.6

Show

that

thefractional

changein

theresonantfrequency

!0ð!

2 0¼s=mÞofadam

ped

simple

harmonic

mechanical

oscillatoris

+ð8Q

2Þ#

1whereQ

isthequalityfactor.

Problem

2.7

Show

that

thequalityfactorofan

electrical

LCR

series

circuitisQ

¼!

0L=Rwhere!

2 0¼

1=LC

Problem

2.8

Aplasm

aconsistsofan

ionized

gas

ofionsandelectronsofequal

number

densities

ðni¼ne¼nÞ

havingcharges

ofopposite

sign

'e,

andmassesmiandme,respectively,wheremi>me.Relative

50

DampedSimpleHarmonicMotion

displacementbetweenthetwospeciessetsuparestoring

+ + + + + + + + +

− − − − − − − − −

E

x

l

electric

fieldwhichreturnstheelectronsto

equilibrium,theionsbeingconsidered

stationary.In

the

diagram,aplasm

aslab

ofthicknesslhas

allitselectronsdisplacedadistancexto

givearestoring

electric

fieldE

¼nex="0,where"0is

constant.Show

that

therestoringforceper

unitarea

onthe

electronsisxn2e2l="0andthat

they

oscillate

simple

harmonically

withangularfrequency

!2 e¼

ne2=m

e"0.This

frequency

iscalled

theelectronplasm

afrequency,andonly

those

radio

waves

of

frequency

!>

!ewillpropagatein

such

anionized

medium.Hence

thereflectionofsuch

waves

from

theionosphere.

Problem

2.9

Asimple

pendulum

consistsofamassm

attheendofastringoflengthlandperform

ssm

all

oscillations.

Thelength

isveryslowly

shortened

whilst

thependulum

oscillatesmanytimes

ata

constantam

plitudel.

where.is

verysm

all.Show

that

ifthelength

ischanged

by

#"lthework

doneis

#mg"l(owingto

theelevationofthepositionofequilibrium)together

withan

increase

in

thependulum

energy

"E

¼mg.2 2#ml_ ..2

!

"l

where.2istheaveragevalueof.2duringtheshortening.If.¼

.0cos!t,show

that

theenergyof

thependulum

atanyinstantmay

bewritten

E¼ml2!

2.2 0

2¼mgl.

2 0

2

andhence

show

that

"E E

¼#1 2

"l l¼

"+ +

that

is,E=+,theratiooftheenergyofthependulum

toitsfrequency

ofoscillationremainsconstant

during

the

slowly

changing

process.(This

constantratio

under

slowly

varying

conditionsis

importantin

quantum

theory

wheretheconstantiswritten

asamultiple

ofPlanck’sconstant,h.)

MethodsofDescribingtheDampingofanOscillator

51

Summary

ofIm

portantResults

Damped

Simple

Harm

onic

Motion

Equationofmotionm€ xxþr_ xxþsx

¼0

Oscillationswhen

s m>

r2

4m

2

Displacementx¼Ae

#rt=2mcosð!

0 tþ'Þw

here

!02

¼s m

#r2

4m

2

AmplitudeDecay

Logarithmicdecrement*—

thelogarithm

oftheratiooftwosuccessiveam

plitudes

one

period)

0apart

*¼

loge

An

Anþ1

¼r)

0

2m

RelaxationTime

Tim

eforam

plitudeto

decay

toA

¼A

0e

#rt=2m

¼A

0e

#1;that

is,t¼

2m=r

Energy

Decay

QualityfactorQ

isthenumber

ofradiansduringwhichenergydecreases

toE

¼E

0e

#1

Q¼

!0m r

¼2(energystoredin

system

energylostper

cycle

E¼E

0e

#rt=m

¼E

0e

#1

whenQ

¼!

0t

Indam

ped

SHM

dE dt¼

ðm€ xxþsx

Þ_ xx¼

#r_ xx2

(work

rate

ofresistiveforce)

Forequivalentexpressionsin

electrical

oscillators

replacem

byL,rbyRandsby1=C

.

Forceequationsbecomevoltageequations.

52

DampedSimpleHarmonicMotion

3 TheForced

Oscillator

TheOperationofiuponaVector

Wehavealreadyseenthataharmonicoscillationcanbeconvenientlyrepresentedbythe

formei!

t .Inadditiontoitsmathematicalconvenienceicanalsobeusedasavector

operatorofphysicalsignificance.Wesaythatwheniprecedesoroperatesonavectorthe

directionofthatvectoristurnedthroughapositiveangle(anticlockwise)of!=2,i.e.i

actingasanoperatoradvancesthephaseofavectorby90 .Theoperator!irotatesthe

vectorclockwiseby!=2andretardsitsphaseby90 .Themathematicsofiasanoperator

differsinnowayfromitsuseas

ffiffiffiffiffiffiffi

!1

pandfromnowonitwillplaybothroles.

Thevectorr¼

aþibisshowninFigure3.1,wherethedirectionofbisperpendicularto

thatofabecauseitisprecededbyi.Themagnitudeormodulusorriswritten

r¼

jrj¼

ða2þ

b2Þ1=2

and

r2¼

ða2þ

b2Þ¼

ðaþibÞða!ibÞ¼

rr( ;

whereða

!ibÞ¼

r(isdefinedasthecomplexconjugateofða

þibÞ;thatis,thesignofiis

changed.

Thevectorr(¼

a!ibisalsoshowninFigure3.1.

Thevectorrcanbewrittenasaproductofitsmagnitude

r(scalarquantity)anditsphase

ordirectionintheform(Figure3.1)

r¼

rei$¼

rðcos$þisin$Þ

¼aþib

showingthat

a¼

rcos$and

b¼

rsin$.

The

Physi

cs

of

Vib

rati

ons

and

Waves,6thEditionH.J.Pain

#2005JohnWiley&Sons,Ltd

53

Itfollowsthat

cos$¼

a r¼

a

ða2þ

b2Þ1=2

and

sin$¼

b r¼

b

ða2þ

b2Þ1=2

givingtan$¼

b=a.

Similarly

r(¼

re!i$¼

rðcos$!

isin$Þ

cos$¼

a r;sin$¼

!b randtan$¼

!b a

ðFigure3:1Þ

Thereadershouldconfirmthattheoperatorirotatesavectorby!=2inthepositive

direction(asstatedinthefirstparagraphofp.53)bytaking$¼!=2intheexpression

r¼

rei$¼

rðcos!=2þisin!=2Þ

Notethat$¼

!!=2in

r¼

re!i!=2rotatesthevectorinthenegativedirection.

Vectorform

ofOhm’sLaw

Ohm’sLawisfirstmetasthescalarrelation

V¼

IR,where

Visthevoltageacrossthe

resistance

Rand

Iisthecurrentthroughit.Itsscalarformstatesthatthevoltageandcurrent

arealwaysinphase.Bothwillfollowasinð!

tþ$Þoracosð!

tþ$Þcurve,andthevalue

of$willbethesameforbothvoltageandcurrent.

However,thepresenceofeitherorbothoftheothertwoelectricalcomponents,

inductance

Landcapacitance

C,willintroduceaphasedifferencebetweenvoltageand

r

r*

aa

ib −ib

φ φ

φφr = r

ei φr*

= r

e−i

r c

os

φir cos

φ−ir cos

Figure

3.1

Vectorrepresentationusingioperatorandexponentialindex.Star

superscriptindicates

complexconjugatewhere!ireplacesi

54

The

Forc

ed

Osc

illa

tor

current,andOhm’sLawtakesthevectorform

V¼

IZ

e;

where

Ze,calledthe

impedance,replacestheresistance,andisthevectorsumofthe

effectiveresistancesof

R,

L,and

Cinthecircuit.

Whenanalternatingvoltage

Vaoffrequency!isappliedacrossaresistance,inductance

andcondenserinseriesasinFigure3.2a,thebalanceofvoltagesisgivenby

Va¼

IRþ

LdI

dtþ

q=C

andthecurrentthroughthecircuitisgivenby

I¼

I 0ei!

t .Thevoltageacrosstheinductance

VL¼

LdI

dt¼

Ld dt

I 0ei!

t¼i!

LI 0ei!

t¼i!

LI

But!

L,aswesawattheendofthelastchapter,hasthedimensionsofohms,beingthe

valueoftheeffectiveresistancepresentedbyaninductance

Ltoacurrentoffrequency

!.

Theproduct!

LIwithdimensionsofohmstimescurrent,i.e.volts,isprecededbyi;this

tellsusthatthephaseofthevoltageacrosstheinductanceis90 aheadofthatofthecurrent

throughthecircuit.

Similarly,thevoltageacrossthecondenseris

q C¼1 C

ð

Idt¼1 C

I 0

ð

ei!

tdt¼1

i!C

I 0ei!

t¼

!iI !C

(since1=i¼

!i).

Again1=!

C,measuredinohms,isthevalueoftheeffectiveresistancepresentedbythe

condensertothecurrentoffrequency

!.Now,however,thevoltage

I=!

Cacrossthe

condenserisprecededby!iandthereforelagsthecurrentby90 .Thevoltageandcurrent

acrosstheresistanceareinphaseandFigure3.2bshowsthatthevectorformofOhm’s

Lawmaybewritten

V¼

IZ

e¼

I½Rþið!

L!1=!

CÞ*,wheretheimpedance

Ze¼

Rþið!

L!1=!

CÞ.Thequantities

!Land1=!

Carecalled

reacta

ncesbecausethey

++

+−

−−

IR

I = I

0e

iωt

q C

dI

dt

L

Va

Figure

3.2a

Anelectricalforced

oscillator.ThevoltageVaisapplied

totheseries

LCRcircuitgiving

Va¼

LdI=dtþ

IRþ

q=C

Vecto

rfo

rmof

Ohm

’sLaw

55

introduceaphaserelationshipaswellasaneffectiveresistance,andthebracket

ð!L!1=!

CÞisoftenwritten

Xe,thereactivecomponentofZ

e.

Themagnitude,inohms,i.e.thevalueoftheimpedance,is

Ze¼

R2þ

!L!1 !C

"#

2"

#

1=2

andthevectorZ

emayberepresentedbyitsmagnitudeandphaseas

Ze¼

Zeei$¼

Zeðcos$þisin$Þ

sothat

cos$¼

R Ze

;sin$¼

Xe

Ze

and

tan$¼

Xe=R;

where$isthephasedifferencebetweenthetotalvoltageacrossthecircuitandthecurrent

throughit.

Thevalueof$canbepositiveornegativedependingontherelativevalueof!

Land

1=!

C:when

!L>1=!

C;$ispositive,butthefrequencydependenceofthecomponents

showthat$canchangebothsignandsize.

ThemagnitudeofZ

eisalsofrequencydependentandhasitsminimumvalue

Ze¼

R

when

!L¼1=!

C.

InthevectorformofOhm’sLaw,V¼

IZ

e.IfV¼

V0ei!

tandZ

e¼

Zeei$,thenwehave

I¼

V0ei!

t

Zeei$

¼V0

Ze

eið!

t!$Þ

givingacurrentofamplitude

V0=Z

ewhichlagsthevoltagebyaphaseangle$.

TheIm

pedance

ofaMechanicalCircuit

Exactlysimilarargumentsholdwhenweconsidernotanelectricaloscillatorbuta

mechanicalcircuithavingmass,stiffnessandresistance.

R

iωL

iXe =

i ω

L −

ωC

−i

1

ωC1

i ω

L −

ωC1

φ RZe

Figure

3.2b

Vectoradditionofresistance

andreactancesto

givetheelectrical

impedance

Z e¼

Rþið!

L!1=!CÞ

56

The

Forc

ed

Osc

illa

tor

Themechanicalimpedanceisdefinedastheforcerequiredtoproduceunitvelocityin

theoscillator,i.e.

Zm¼

F=vorF¼

vZ

m.

Immediately,wecanwritethemechanicalimpedanceas

Zm¼

rþi!

m!

s !

&'

¼rþiX

m

where

Zm¼

Zmei$

and

tan$¼

Xm=r

$beingthephasedifferencebetweentheforceandthevelocity.Themagnitudeof

Zm¼

½r2þð!

m!

s=!Þ2*1=2.

Mass,likeinductance,producesapositivereactance,andthestiffnessbehavesinexactly

thesamewayasthecapacitance.

BehaviourofaForced

Oscillator

Wearenowinapositiontodiscussthephysicalbehaviourofamechanicaloscillatorof

mass

m,stiffness

sandresistance

rbeingdrivenbyanalternatingforce

F0cos!

t,where

F0

istheamplitudeoftheforce(Figure3.3).Theequivalentelectricaloscillatorwouldbean

alternatingvoltage

V0cos!

tappliedtothecircuitofinductance

L,capacitance

Cand

resistance

RinFigure3.2a.

Themechanicalequationofmotion,i.e.thedynamicbalanceofforces,isgivenby

m€ xxþ

r_ xxþ

sx¼

F0cos!

t

andthevoltageequationintheelectricalcaseis

L€ qqþ

R_ qqþ

q=C

¼V0cos!

t

Weshallanalysethebehaviourofthemechanicalsystembuttheanalysisfitstheelectrical

oscillatorequallywell.

mr

s

F0 c

os ω

t

Figure

3.3

Mechanical

forced

oscillatorwithforceF 0cos!tapplied

todam

ped

mechanical

circuit

ofFigure

2.1

Behavio

ur

of

aForc

ed

Osc

illa

tor

57

Thecompletesolutionfor

xintheequationofmotionconsistsoftwoterms:

(1)a‘transient’termwhichdiesawaywithtimeandis,infact,thesolutiontotheequation

m€ xxþ

r_ xxþ

sx¼0discussedinChapter2.Thiscontributestheterm

x¼

Ce!

rt=2meiðs=m

!r2=4

m2Þ1=2

t

whichdecayswithe!

rt=2m.Thesecondterm

(2)iscalledthe‘steadystate’term,anddescribesthebehaviouroftheoscillatorafterthe

transienttermhasdiedaway.

Bothtermscontributetothesolutioninitially,butforthemomentweshallconcentrate

onthe‘steadystate’termwhichdescribestheultimatebehaviouroftheoscillator.

Todothisweshallrewritetheforceequationinvectorformandrepresentcos!

tbyei!

t

asfollows:

m€ xxþ

r_ xxþ

sx¼

F0ei!

tð3:1Þ

Solvingforthevectorxwillgivebothitsmagnitudeandphasewithrespecttothedriving

force

F0ei!

t .Initially,letustrythesolutionx¼

Aei!

t ,whereAmaybecomplex,sothatit

mayhavecomponentsinandoutofphasewiththedrivingforce.

Thevelocity

_ xx¼i!

Aei!

t¼i!

x

sothat

€ xx¼i2!2x¼

!!2x

andequation(3.1)becomes ð!

A!2mþi!

Arþ

AsÞei!

t¼

F0ei!

t

whichistrueforall

twhen

A¼

F0

i!rþðs!!2mÞ

or,aftermultiplyingnumeratoranddenominatorby!i

A¼

!iF0

!½rþið!

m!

s=!Þ*¼

!iF0

!Z

m

Hence

x¼

Aei!

t¼

!iF0ei!

t

!Z

m

¼!iF0ei!

t

!Z

mei$

¼!iF0eið!

t!$Þ

!Z

m

58

The

Forc

ed

Osc

illa

tor

where

Zm¼

½r2þð!

m!

s=!Þ2*1=2

Thisvectorformofthe

steady

statebehaviourof

xgivesthreepiecesofinformationand

completelydefinesthemagnitudeofthedisplacement

xanditsphasewithrespecttothe

drivingforceafterthetransienttermdiesaway.Ittellsus

1.Thatthephasedifference$existsbetween

xandtheforcebecauseofthereactivepart

ð!m!

s=!Þofthemechanicalimpedance.

2.Thatanextradifferenceisintroducedbythefactor!iandevenif$werezerothe

displacement

xwouldlagtheforce

F0cos!

tby90 .

3.Thatthemaximumamplitudeofthedisplacement

xis

F0=!

Zm.Weseethatthisis

dimensionallycorrectbecausethevelocity

x=thasdimensions

F0=Z

m.

Havingused

F0ei!

ttorepresentitsrealpart

F0cos!

t,wenowtaketherealpartofthe

solution

x¼

!iF0eið!

t!$Þ

!Z

m

toobtaintheactualvalueofx.(Iftheforcehadbeen

F0sin!

t,wewouldnowtakethatpart

ofxprecededbyi.)

Now

x¼

!iF0

!Z

m

eið!

t!$Þ

¼!iF0

!Z

m

½cosð!

t!$Þþisinð!

t!$Þ*

¼!iF0

!Z

m

cosð!

t!$Þþ

F0

!Z

m

sinð!

t!$Þ

Thevalueof

xresultingfrom

F0cos!

tistherefore

x¼

F0

!Z

m

sinð!

t!$Þ

[thevalueof

xresultingfrom

F0sin!

twouldbe!

F0cosð!

t!$Þ=!

Zm*.

Notethatbothofthesesolutionssatisfytherequirementthatthetotalphasedifference

betweendisplacementandforceis$plusthe!!=2termintroducedbythe!ifactor.When

$¼0thedisplacement

x¼

F0sin!

t=!

Zm

lagstheforce

F0cos!

tbyexactly90 .

Behavio

ur

of

aForc

ed

Osc

illa

tor

59

Tofindthevelocityoftheforcedoscillationinthesteadystatewewrite

v¼

_ xx¼

ði!Þð!iF0Þ

!Z

m

eið!

t!$Þ

¼F0

Zm

eið!

t!$Þ

Weseeimmediatelythat

1.Thereisnoprecedingifactorsothatthevelocity

vandtheforcedifferinphaseonly

by$,andwhen

$¼0thevelocityandforceareinphase.

2.Theamplitudeofthevelocityis

F0=Z

m,whichweexpectfromthedefinitionof

mechanicalimpedance

Zm¼

F=v.

Againwetaketherealpartofthevectorexpressionforthevelocity,whichwill

correspondtotherealpartoftheforce

F0ei!

t .Thisis

v¼

F0

Zm

cosð!

t!$Þ

Thus,the

velo

cit

yis

alw

ays

exactl

y90

ahead

of

the

dis

pla

cem

ent

inphaseanddiffers

fromtheforceonlybyaphaseangle$,where

tan$¼!

m!

s=!

r¼

Xm r

sothataforce

F0cos!

tgivesadisplacement

x¼

F0

!Z

m

sinð!

t!$Þ

andavelocity

v¼

F0

Zm

cosð!

t!$Þ

(Problems3.1,3.2,3.3,3.4)

BehaviourofVelocity

vvin

MagnitudeandPhaseversusDriving

ForceFrequencyx

Thevelocityamplitudeis

F0

Zm

¼F0

½r2þð!

m!

s=!Þ2*1=2

sothatthemagnitudeofthevelocitywillvarywiththefrequency

!because

Zmis

frequencydependent.

60

The

Forc

ed

Osc

illa

tor

Atlowfrequencies,theterm

!s=!isthelargesttermin

Zmandtheimpedanceissaidto

be

stif

fnes

scontr

oll

ed.Athighfrequencies!

misthedominanttermandtheimpedanceis

mass

contr

oll

ed.Atafrequency

!0where!0m¼

s=!0,theimpedancehasitsminimum

value

Zm¼

randisarealquantitywithzeroreactance.

Thevelocity

F0=Z

mthenhasitsmaximumvaluev¼

F0=r,and!0issaidtobethe

frequencyof

velo

cit

yre

sonance.Notethattan$¼0at!0,thevelocityandforcebeingin

phase.

Thevariationofthemagnitudeofthevelocitywithdrivingfrequency,!,isshownin

Figure3.4,theheightandsharpnessofthepeakatresonancedependingon

r,whichisthe

onlyeffectivetermof

Zmat!0.

Theexpression

v¼

F0

Zm

cosð!

t!$Þ

where

tan$¼!

m!

s=!

r

showsthatforpositive$;thatis,!

m>

s=!,thevelocityvwilllagtheforcebecause

!$

appearsintheargumentofthecosine.Whenthedrivingforcefrequency!isveryhighand

!!

1,then

$!90 andthevelocitylagstheforcebythatamount.

When

!m<

s=!;$isnegative,thevelocityisaheadoftheforceinphase,andatlow

drivingfrequenciesas!!0theterm

s=!!

1and$!

!90 .

Thus,atlowfrequenciesthevelocityleadstheforce($negative)andathighfrequencies

thevelocitylagstheforce($positive).

Atthefrequency!0,however,!0m¼

s=!0and$¼0,sothatvelocityandforcearein

phase.Figure3.5showsthevariationof$with!forthevelocity,theactualshapeofthe

curvesdependinguponthevalueof

r.

Velocity

F0 r

ω0 =

(s/m

)

ω1 2

Figure

3.4

Velocity

offorced

oscillatorversusdrivingfrequency!.Maximum

velocityv m

ax¼

F 0=r

at!2 0¼

s=m

Behavio

ur

of

Velo

cit

yv

inM

agnit

ude

and

Phase

vers

us

Dri

vin

gForc

eF

requency!61

(Problem

3.5)

BehaviourofDisplacementversusDrivingForceFrequencyx

Thephaseofthedisplacement

x¼

F0

!Z

m

sinð!

t !Þ

isat

all

tim

esex

actl

y90"

beh

ind

that

of

the

vel

oci

ty.

Whil

stth

egra

ph

of!

ver

sus!

rem

ains

the

sam

e,th

eto

tal

phas

edif

fere

nce

bet

wee

nth

edis

pla

cem

ent

and

the

forc

e

involv

esth

eex

tra

90"

reta

rdat

ion

intr

oduce

dby

the

ioper

ator.

Thus,

atver

ylo

w

freq

uen

cies

,w

her

e!¼ "=2

rad

and

the

vel

oci

tyle

ads

the

forc

e,th

edis

pla

cem

ent

and

the

forc

ear

ein

phas

eas

we

should

expec

t.A

thig

hfr

equen

cies

the

dis

pla

cem

ent

lags

the

forc

eby"

rad

and

isex

actl

yout

of

phas

e,so

that

the

curv

esh

ow

ing

the

phas

ean

gle

bet

wee

nth

edis

pla

cem

entan

dth

efo

rce

iseq

uiv

alen

tto

the!

ver

sus!

curv

e,dis

pla

ced

by

anam

ount

equal

to"=2

rad.

This

issh

ow

nin

Fig

ure

3.6

.

The

ampli

tude

of

the

dis

pla

cem

ent

x¼

F0=!

Zm,

and

atlo

wfr

equen

cies

Zm¼

½r2þð!

m

s=!Þ2'1=2!

s=!

,so

that

x)

F0=ð!

s=!Þ¼

F0=s:

Total phase

angle (radians)

between

x and F

x and F in phase

x lags F by rad

x lags F

π 2

π 2

π 2−

π 2−

0

ω0

r increasing

Phase angle

(red)

φ0

ω

− π

Figu

re3.6

Variationoftotalphaseangle

betweendisplacementanddrivingforceversusdriving

frequency!.Thetotalphaseangle

is ! "=2

rad

Phase angle

(radians)

between

V and F

0 π 2π 2

v leads F

v lags F

r increasing

v and F

in phase

+

ω

φ

−

Figu

re3.5

Variationofphaseangle!versusdrivingfrequency,where!isthephaseanglebetween

thevelocity

oftheforced

oscillatorandthedrivingforce.!¼

0at

velocity

resonance.Each

curve

represents

afixedresistance

value

62

The

Forc

ed

Osc

illa

tor

Athig

hfr

equen

cies

Zm!

!m

,so

that

x)

F0=ð!

2mÞ,

whic

hte

nds

toze

roas!

bec

om

es

ver

yla

rge.

At

ver

yhig

hfr

equen

cies

,th

eref

ore

,th

edis

pla

cem

ent

ampli

tude

isal

most

zero

bec

ause

of

the

mas

s-co

ntr

oll

edor

iner

tial

effe

ct.

The

vel

oci

tyre

sonan

ceocc

urs

at!

2 0¼

s=m

,w

her

eth

eden

om

inat

or

Zm

of

the

vel

oci

ty

ampli

tude

isa

min

imum

,but

the

dis

pla

cem

ent

reso

nan

cew

ill

occ

ur,

since

x¼ðF

0=!

ZmÞ

sinð!

t !Þ,

when

the

den

om

inat

or!

Zm

isa

min

imum

.T

his

takes

pla

cew

hen

d d!ð!

ZmÞ¼

d d!!½r

2þð!

m

s=!Þ2'1=2¼

0

i.e.

when

2!

r2þ

4!

mð!

2m

sÞ¼

0

or

2!½r

2þ

2mð!

2m

sÞ'¼

0

soth

atei

ther

!¼

0

or

!2¼

s m

r2

2m

2¼!

2 0

r2

2m

2

Thus

the

dis

pla

cem

ent

reso

nance

occ

urs

ata

freq

uen

cysl

ightl

yle

ssth

an!

0,

the

freq

uen

cyof

vel

oci

tyre

sonan

ce.

For

asm

all

dam

pin

gco

nst

ant

ror

ala

rge

mas

sm

thes

e

two

reso

nan

ces,

for

all

pra

ctic

alpurp

ose

s,occ

ur

atth

efr

equen

cy!

0.

Den

oti

ng

the

dis

pla

cem

ent

reso

nan

cefr

equen

cyby

!r¼

s m

r2

2m

2

!

1=2

we

can

wri

teth

em

axim

um

dis

pla

cem

ent

as

xm

ax¼

F0

!rZ

m

The

val

ue

of!

rZ

mat!

ris

easi

lysh

ow

nto

be

equal

to!0 r

wher

e

!02¼

s m

r2

4m

2¼!

2 0

r2

4m

2

The

val

ue

of

xat

dis

pla

cem

ent

reso

nan

ceis

ther

efore

giv

enby

xm

ax¼

F0

!0 r

wher

e

!0 ¼

!2 0

r2

4m

2

!

1=2

Behavio

ur

of

Dis

pla

cem

ent

vers

us

Dri

vin

gForc

eF

requencyx

63

Sin

cex

max¼

F0=!0 r

atre

sonan

ce,th

eam

pli

tude

atre

sonan

ceis

kep

tlo

wby

incr

easi

ng

ran

dth

evar

iati

on

of

xw

ith!

for

dif

fere

ntval

ues

of

ris

show

nin

Fig

ure

3.7

.A

neg

ligib

le

val

ue

of

rpro

duce

sa

larg

eam

pli

fica

tion

atre

sonan

ce:th

isis

the

bas

isof

hig

hse

lect

ivit

yin

atu

ned

radio

circ

uit

(see

the

sect

ion

inth

isch

apte

ron

Qas

anam

pli

fica

tion

fact

or)

.

Kee

pin

gth

ere

sonan

ceam

pli

tude

low

isth

epri

nci

ple

of

vib

rati

on

insu

lati

on.

(Problems3.6,3.7)

Problem

onVibrationInsu

lation

Aty

pic

alvib

rati

on

insu

lato

ris

show

nin

Fig

ure

3.8

.A

hea

vy

bas

eis

support

edon

a

vib

rati

ng

floor

by

asp

ring

syst

emof

stif

fnes

ss

and

vis

cous

dam

per

r.T

he

insu

lato

rw

ill

gen

eral

lyoper

ate

atth

em

ass

contr

oll

eden

dof

the

freq

uen

cysp

ectr

um

and

the

reso

nan

t

freq

uen

cyis

des

igned

tobe

low

erth

anth

era

nge

of

freq

uen

cies

likel

yto

be

met

.Suppose

the

ver

tica

lvib

rati

on

of

the

floor

isgiv

enby

x¼

Aco

s!

tab

outit

seq

uil

ibri

um

posi

tion

and

yis

the

corr

espondin

gver

tica

ldis

pla

cem

entof

the

bas

eab

outit

sre

stposi

tion.T

he

funct

ion

of

the

insu

lato

ris

tokee

pth

era

tio

y=A

toa

min

imum

.

The

equat

ion

of

moti

on

isgiv

enby

m€ yy¼

rð_ yy

_ xxÞ

sðy

xÞ

ωω0

F0

S

Displacement x

r increasing

Figu

re3.7

Variationofthedisplacementofaforced

oscillatorversusdrivingforcefrequency!for

variousvalues

of

r

64

The

Forc

ed

Osc

illa

tor

whic

h,

ify

x¼

X,

bec

om

es

m€ XXþ

r_ XXþ

sX¼

m€ xx¼

mA!

2co

s!

t

¼F

0co

s!

t;

wher

e

F0¼

mA!

2

Use

the

stea

dy

stat

eso

luti

on

of

Xto

show

that

y¼

F0

!Z

m

sinð!

t !Þþ

Aco

s!

t

and

(noti

ng

that

yis

the

super

posi

tion

of

two

har

monic

com

ponen

tsw

ith

aco

nst

ant

phas

e

dif

fere

nce

)sh

ow

that

ym

ax

A¼ðr

2þ

s2=!

2Þ1=2

Zm

wher

e

Z2 m¼

r2þð!

m

s=!Þ2

Note

that

ym

ax

A>

1if

!2<

2s m

r

y

Vibrating floor

x = A cos ωt

Fixed reference level

Heavy base

Equilibrium

rest position

of base

Figu

re3.8

Vibrationinsulator.Aheavy

basesupported

byaspringandviscousdam

per

system

ona

vibratingfloor

Pro

ble

mon

Vib

rati

on

Insu

lati

on

65

soth

ats=

msh

ould

be

aslo

was

poss

ible

togiv

epro

tect

ion

agai

nst

agiv

enfr

equen

cy!.

(a)

Show

that

ym

ax

A¼

1fo

r!

2¼

2s m

(b)

Show

that

ym

ax

A<

1fo

r!

2>

2s m

(c)

Show

that

if!

2¼

s=m

,th

eny

max=A

>1

but

that

the

dam

pin

gte

rmr

ishel

pfu

lin

kee

pin

gth

em

oti

on

of

the

bas

eto

are

asonab

lylo

wle

vel

.

(d)

Show

that

if!

2>

2s=

m,

then

ym

ax=A

<1

but

dam

pin

gis

det

rim

enta

l.

Sign

ifican

ceoftheTw

oCo

mponen

tsoftheDisplace

men

tCu

rve

Any

single

curv

eof

Fig

ure

3.7

isth

esu

per

posi

tion

of

the

two

com

ponen

tcu

rves

(a)

and

(b)

inFig

ure

3.9

,fo

rth

edis

pla

cem

ent

xm

aybe

rew

ritt

en

x¼

F0

!Z

m

sinð!

t !Þ¼

F0

!Z

m

ðsin!

tco

s!

cos!

tsi

n!Þ

(b)F0ω

ω

r

r

r 2+Xm

2

2m

(a)

F0ω

ω0

ω0

ω0

r 2+

+

Xm

2

Xm

F0 l ω

′r

F0 l ω

0r

F0 l 2ω0r F0

S

r2m

−

−

Figu

re3.9

AtypicalcurveofFigure

3.7

resolved

into

its‘anti-phase’component(curve(a))andits

‘90"outofphase’

component(curve

(b)).Curve

(b)represents

the

resistive

fraction

ofthe

impedance

andcurve(a)thereactive

fraction.Curve(b)correspondsto

absorptionandcurve(a)to

anomalousdispersionofan

electromagneticwaveinamediumhavingan

atomicormolecularresonant

frequency

equal

tothefrequency

ofthewave

66

The

Forc

ed

Osc

illa

tor

or,

since

cos!¼

r Zm

and

sin!¼

Xm

Zm

as

x¼

F0

!Z

m

r Zm

sin!

t

F0

!Z

m

Xm

Zm

cos!

t

The

cos!

tco

mponen

t(w

ith

aneg

ativ

esi

gn)

isex

actl

yan

ti-p

has

ew

ith

resp

ect

toth

e

dri

vin

gfo

rce

F0

cos!

t.It

sam

pli

tude,

plo

tted

ascu

rve

(a)

may

be

expre

ssed

as

F

0 !

Xm

Z2 m

¼F

0mð!

2 0 !

2Þ

m2ð!

2 0 !

2Þ2þ!

2r

2ð3:2Þ

wher

e!

2 0¼

s=m

and!

0is

the

freq

uen

cyof

vel

oci

tyre

sonan

ce.

The

sin!

tco

mponen

tla

gs

the

dri

vin

gfo

rce

F0

cos!

tby

90" .

Its

ampli

tude

plo

tted

as

curv

e(b

)bec

om

es

F0 !

r

r2þ

X2 m

¼F

0!

r

m2ð!

2 0 !

2Þ2þ!

2r

2

We

see

imm

edia

tely

that

at!

0cu

rve

(a)

isze

roan

dcu

rve

(b)

isnea

rit

sm

axim

um

butth

ey

com

bin

eto

giv

ea

max

imum

at!

wher

e

!2¼!

2 0

r2

2m

2

the

reso

nan

tfr

equen

cyfo

ram

pli

tude

dis

pla

cem

ent.

Thes

ecu

rves

are

par

ticu

larl

yfa

mil

iar

inth

est

udy

of

opti

caldis

per

sion

wher

eth

efo

rced

osc

illa

tor

isan

elec

tron

inan

atom

and

the

dri

vin

gfo

rce

isth

eosc

illa

ting

fiel

dvec

tor

of

an

elec

trom

agnet

icw

ave

of

freq

uen

cy!.W

hen

!is

the

reso

nan

tfr

equen

cyof

the

elec

tron

in

the

atom

,th

eat

om

abso

rbs

ala

rge

amount

of

ener

gy

from

the

elec

trom

agnet

icw

ave

and

curv

e(b

)is

the

shap

eof

the

char

acte

rist

icab

sorp

tion

curv

e.N

ote

that

curv

e(b

)re

pre

sents

the

dis

sipat

ing

or

abso

rbin

gfr

acti

on

of

the

imped

ance

r

ðr2þ

X2 mÞ1=2

and

that

par

tof

the

dis

pla

cem

ent

whic

hla

gs

the

dri

vin

gfo

rce

by

90" .

The

vel

oci

ty

asso

ciat

edw

ith

this

com

ponen

tw

ill

ther

efore

be

inphas

ew

ith

the

dri

vin

gfo

rce

and

itis

this

par

tof

the

vel

oci

tyw

hic

hap

pea

rsin

the

ener

gy

loss

term

r_ xx

2due

toth

ere

sist

ance

of

the

osc

illa

tor

and

whic

hgiv

esri

seto

abso

rpti

on.

Sig

nifi

cance

of

the

Tw

oC

om

ponents

of

the

Dis

pla

cem

ent

Curv

e67

On

the

oth

erhan

d,

curv

e(a

)re

pre

sents

the

reac

tive

or

ener

gy

stori

ng

frac

tion

of

the

imped

ance

Xm

ðr2þ

X2 mÞ1=2

and

the

reac

tive

com

ponen

tsin

am

ediu

mdet

erm

ine

the

vel

oci

tyof

the

wav

esin

the

med

ium

whic

hin

turn

gover

ns

the

refr

active

index

n.In

fact

,cu

rve

(a)

isa

gra

ph

of

the

val

ue

of

n2

ina

regio

nof

anom

alous

dis

per

sion

wher

eth

e!

axis

repre

sents

the

val

ue

n¼

1.

Thes

ere

gio

ns

occ

ur

atev

ery

reso

nan

tfr

equen

cyof

the

const

ituen

tat

om

sof

the

med

ium

.W

esh

all

retu

rnto

this

topic

late

rin

the

book.

(Problems3.8,3.9,3.10)

Power

Supplied

toOscillatorbytheDrivingFo

rce

Inord

erto

mai

nta

inth

est

eady

stat

eosc

illa

tions

of

the

syst

emth

edri

vin

gfo

rce

must

repla

ceth

een

ergy

lost

inea

chcy

cle

bec

ause

of

the

pre

sence

of

the

resi

stan

ce.

We

shal

l

now

der

ive

the

most

import

ant

resu

ltth

at:

‘in

the

stea

dy

stat

eth

eam

pli

tude

and

phas

eof

adri

ven

osc

illa

tor

adju

stth

emse

lves

so

that

the

aver

age

pow

ersu

ppli

edby

the

dri

vin

gfo

rce

just

equal

sth

atbei

ng

dis

sipat

edby

the

fric

tional

forc

e’.

The

inst

anta

neous

pow

er

Psu

ppli

edis

equal

toth

epro

duct

of

the

inst

anta

neous

dri

vin

g

forc

ean

dth

ein

stanta

neous

velo

cit

y;

that

is,

P¼

F0

cos!

tF

0

Zm

cosð!

t !Þ

¼F

2 0

Zm

cos!

tco

sð!

t !Þ

The

avera

ge

pow

er

Pav¼

tota

lw

ork

per

osc

illa

tion

osc

illa

tion

per

iod

;P

av¼ð

T 0

Pdt

Tw

her

eT¼

osc

illa

tion

per

iod

¼F

2 0

ZmT

ð

T 0

cos!

tco

sð!

t !Þd

t

¼F

2 0

ZmT

ð

T 0

½cos

2!

tco

s!þ

cos!

tsi

n!

tsi

n!Þd

t

¼F

2 0

2Z

m

cos!

68

The

Forc

ed

Osc

illa

tor

bec

ause

ð

T 0

cos!

t+

sin!

tdt¼

0

and

1 T

ð

T 0

cos

2!

tdt¼

1 2

The

pow

ersu

ppli

edby

the

dri

vin

gfo

rce

isnot

store

din

the

syst

em,

but

dis

sipat

edas

work

expen

ded

inm

ovin

gth

esy

stem

agai

nst

the

fric

tional

forc

er_ xx.

The

rate

of

work

ing

(inst

anta

neo

us

pow

er)

by

the

fric

tional

forc

eis

ðr_ xxÞ_ xx¼

r_ xx

2¼

rF

2 0

Z2 m

cos

2ð!

t !Þ

and

the

aver

age

val

ue

of

this

over

one

per

iod

of

osc

illa

tion

1 2

rF2 0

Z2 m

¼1 2

F2 0

Zm

cos!

for

r Zm

¼co

s!

This

pro

ves

the

init

ial

stat

emen

tth

atth

epow

ersu

ppli

edeq

ual

sth

epow

erdis

sipat

ed.

Inan

elec

tric

alci

rcuit

the

pow

eris

giv

enby

VIco

s!,w

her

eV

and

Iar

eth

ein

stan

taneo

us

r.m

.s.

val

ues

of

volt

age

and

curr

ent

and

cos!

isknow

nas

the

pow

er

facto

r.

VIco

s!¼

V2

Ze

cos!¼

V2 0

2Z

e

cos!

since

V¼

V0 ffiffiffi 2

p

(Problem

3.11)

VariationofPav

withx.Abso

rptionRes

onan

ceCu

rve

Ret

urn

ing

toth

em

echan

ical

case

,w

ese

eth

atth

eav

erag

epow

ersu

ppli

ed

Pav¼ðF

2 0=2

ZmÞc

os!

isa

max

imum

when

cos!¼

1;

that

is,w

hen

!¼

0an

d!

m

s=!¼

0or!

2 0¼

s=m

.T

he

forc

ean

dth

evel

oci

tyar

eth

enin

phas

ean

dZ

mhas

its

min

imum

val

ue

of

r.T

hus

Pav

(max

imum

)¼

F2 0=2r

Vari

ati

on

of

Pav

wit

h!

.A

bso

rpti

on

Res

onan

ceC

urv

e69

Agra

ph

of

Pav

ver

sus!

,th

efr

equen

cyof

the

dri

vin

gfo

rce,

issh

ow

nin

Fig

ure

3.1

0.L

ike

the

curv

eof

dis

pla

cem

ent

ver

sus!

,th

isgra

ph

mea

sure

sth

ere

sponse

of

the

osc

illa

tor;

the

shar

pnes

sof

its

pea

kat

reso

nan

ceis

also

det

erm

ined

by

the

val

ue

of

the

dam

pin

gco

nst

ant

r,w

hic

his

the

only

term

rem

ainin

gin

Zm

atth

ere

sonan

cefr

equen

cy!

0.T

he

pea

kocc

urs

atth

efr

equen

cyof

vel

oci

tyre

sonan

cew

hen

the

pow

erab

sorb

edby

the

syst

emfr

om

the

dri

vin

gfo

rce

isa

max

imum

;th

iscu

rve

isknow

nas

the

abso

rpti

on

curv

eof

the

osc

illa

tor

(it

issi

mil

arto

curv

e(b

)of

Fig

ure

3.9

).

TheQ-Valuein

Term

softheRes

onan

ceAbso

rptionBan

dwidth

Inth

ela

stch

apte

rw

edis

cuss

edth

equal

ity

fact

or

of

anosc

illa

tor

syst

emin

term

sof

ener

gy

dec

ay.W

em

ayder

ive

the

sam

epar

amet

erin

term

sof

the

curv

eof

Fig

ure

3.1

0,w

her

eth

e

shar

pnes

sof

the

reso

nan

ceis

pre

cise

lydefi

ned

by

the

rati

o

Q¼

!0

!2 !

1

;

wher

e!

2an

d!

1ar

eth

ose

freq

uen

cies

atw

hic

hth

epow

ersu

ppli

ed

Pav¼

1 2P

av(m

axim

um

)

The

freq

uen

cydif

fere

nce

!2 !

1is

oft

enca

lled

the

ban

dw

idth

.

ω0

ω2

ωω1

F 02

Pav(max)

2r

4r

=

F 02

Figu

re3.10

Graph

ofaveragepower

versus!

supplied

toan

oscillatorbythedrivingforce.

Bandwidth

!2 !1ofresonance

curve

defines

response

interm

softhe

quality

factor,

Q¼

!0=ð!2 !1Þ,

where!

2 0¼

s=m

70

The

Forc

ed

Osc

illa

tor

Now

Pav¼

rF2 0=2

Z2 m¼

1 2P

av(m

axim

um

)¼

1 2F

2 0=2

r

when

Z2 m¼

2r

2

that

is,

when

r2þ

X2 m¼

2r

2or

Xm¼!

m

s=!¼-

r:

If!

2>!

1,

then

!2m

s=!

2¼þ

r

and

!1m

s=!

1¼

r

Eli

min

atin

gs

bet

wee

nth

ese

equat

ions

giv

es

!2 !

1¼

r=m

soth

at

Q¼!

0m=r

Note

that!

1¼!

0

r=2m

and!

2¼!

0þ

r=2m

are

the

two

signifi

cant

freq

uen

cies

in

Fig

ure

3.9

.T

he

qual

ity

fact

or

of

anel

ectr

ical

circ

uit

isgiv

enby

Q¼!

0L

R;

wher

e

!2 0¼ðL

CÞ

1

Note

that

for

hig

hval

ues

of

Q,

wher

eth

edam

pin

gco

nst

ant

ris

smal

l,th

efr

equen

cy!0

use

din

the

last

chap

ter

todefi

ne

Q¼!0 m=r

moves

ver

ycl

ose

toth

efr

equen

cy!

0,an

dth

e

two

defi

nit

ions

of

Qbec

om

eeq

uiv

alen

tto

each

oth

eran

dto

the

thir

ddefi

nit

ion

we

mee

tin

the

nex

tse

ctio

n.

TheQ-Valueas

anAmplificationFa

ctor

We

hav

ese

enth

atth

eval

ue

of

the

dis

pla

cem

ent

atre

sonan

ceis

giv

enby

Am

ax¼

F0

!0 r

wher

e!02¼

s m

r2

4m

2

The

Q-V

alu

eas

an

Am

pli

ficati

on

Facto

r71

At

low

freq

uen

ciesð!!

0Þt

he

dis

pla

cem

ent

has

aval

ue

A0¼

F0=s

,so

that

Am

ax

A0

!

2

¼F

2 0

!02

r2

s2

F2 0

¼m

2!

4 0

r2½!

2 0

r2=4

m2'

¼!

2 0m

2

r2½1

1=4

Q2'1=2'¼

Q2

½1

1=4

Q2'

Hen

ce:

Am

ax

A0

¼Q

½1

1=4

Q2'1=2)

Q1þ

1

8Q

2

$%

)Q

for

larg

eQ

.

Q = 5

Q = 4

Q = 3

Q = 2

ω0

F0

ω

SF0S

Displacement in units of

Figu

re3.11

Curves

ofFigure

3.7

nowgiven

interm

softhequalityfactor

Qofthesystem

,where

Q

isam

plificationat

resonance

oflowfrequency

response

x¼

F 0=s

72

The

Forc

ed

Osc

illa

tor

Thus,

the

dis

pla

cem

ent

atlo

wfr

equen

cies

isam

pli

fied

by

afa

ctor

of

Qat

dis

pla

cem

ent

reso

nan

ce.

Fig

ure

3.7

isnow

show

nas

Fig

ure

3.1

1w

her

eth

eQ

-val

ues

hav

ebee

nat

tach

edto

each

curv

e.In

tunin

gra

dio

circ

uit

s,th

eQ

-val

ue

isuse

das

am

easu

reof

sele

ctiv

ity,

wher

e

the

shar

pnes

sof

resp

onse

allo

ws

asi

gnal

tobe

obta

ined

free

from

inte

rfer

ence

from

signal

s

atnea

rby

freq

uen

cies

.In

conven

tional

radio

circ

uit

sat

freq

uen

cies

of

one

meg

acycl

e,

Transient

vecto

r

A2

A3

A4

A0A1

A2

A3

A4

A1

BOB

Steady state vector

0

(b)

(a)

At t = 0 , transient vector = BO = BA0

t = 0

t

Figu

re3.12

(a)Thesteadystateoscillation(heavy

curve)

ismodulatedbythetransientwhich

decaysexponentially

withtime.

(b)In

thevectordiagram

of(b)OBis

theconstantlength

steady

statevectorandBA1isthetransientvector.Each

vectorrotatesanti-clockwisewithitsownangular

velocity.At

t¼

0thevectors

OBandBA0areequal

andopposite

onthehorizontalaxis

andtheir

vectorsumiszero.Atsubsequenttimes

thetotalam

plitudeisthelength

ofOA1whichchangesas

A

traces

acontractingspiral

aroundB.Thepoints

A1,A2,A3andA4indicatehowtheam

plitudeis

modifiedin

(a)

The

Q-V

alu

eas

an

Am

pli

ficati

on

Facto

r73

Q-v

alues

are

of

the

ord

erof

afe

whundre

d;

athig

her

radio

freq

uen

cies

reso

nan

tco

pper

cavit

ies

hav

eQ

-val

ues

of

about30

000

and

pie

zo-e

lect

ric

cryst

als

can

pro

duce

Q-v

alues

of

500

000.

Opti

cal

abso

rpti

on

incr

yst

als

and

nucl

ear

mag

net

icre

sonan

ces

are

oft

en

des

crib

edin

term

sof

Q-v

alues

.T

he

Moss

bau

eref

fect

innucl

ear

physi

csin

volv

esQ

-val

ues

of

10

10.

TheEffect

oftheTran

sien

tTerm

Thro

ughout

this

chap

ter

we

hav

eco

nsi

der

edonly

the

stea

dy

stat

ebeh

avio

ur

wit

hout

acco

unti

ng

for

the

tran

sien

tte

rmm

enti

oned

on

p.

58.

This

term

mak

esan

init

ial

contr

ibuti

on

toth

eto

tal

dis

pla

cem

ent

but

dec

ays

wit

hti

me

ase

rt=2m.

Its

effe

ctis

bes

t

dis

pla

yed

by

consi

der

ing

the

vec

tor

sum

of

the

tran

sien

tan

dst

eady

stat

eco

mponen

ts.

The

stea

dy

stat

ete

rmm

aybe

repre

sente

dby

avec

tor

of

const

ant

length

rota

ting

anti

clock

wis

eat

the

angula

rvel

oci

ty!

of

the

dri

vin

gfo

rce.

The

vec

tor

tip

trac

esa

circ

le.

Upon

this

issu

per

pose

dth

etr

ansi

ent

term

vec

tor

of

dim

inis

hin

gle

ngth

whic

hro

tate

san

ti

clock

wis

ew

ith

angula

rvel

oci

ty!0 ¼

ðs=m

r2=4

m2Þ1=2.It

sti

ptr

aces

aco

ntr

acti

ng

spir

al.

The

locu

sof

the

mag

nit

ude

of

the

vec

tor

sum

of

thes

ete

rms

isth

een

vel

ope

of

the

var

yin

gam

pli

tudes

of

the

osc

illa

tor.

This

envel

ope

modula

tes

the

stea

dy

stat

eosc

illa

tions

of

freq

uen

cy!

ata

freq

uen

cyw

hic

hdep

ends

upon!0

and

the

rela

tive

phas

ebet

wee

n!

t

and!0 t.

Thus,

inFig

ure

3.1

2(a

)w

her

eth

eto

tal

osc

illa

tor

dis

pla

cem

ent

isze

roat

tim

et¼

0w

e

hav

eth

est

eady

stat

ean

dtr

ansi

entvec

tors

equal

and

opposi

tein

Fig

ure

3.1

2(b

)butbec

ause

!6¼!0

the

rela

tive

phas

ebet

wee

nth

evec

tors

wil

lch

ange

asth

etr

ansi

ent

term

dec

ays.

The

vec

tor

tip

of

the

tran

sien

tte

rmis

show

nas

the

dott

edsp

iral

and

the

tota

lam

pli

tude

assu

mes

the

var

yin

gle

ngth

sO

A1,

OA

2,

OA

3,

OA

4,

etc.

(Problems3.12,3.13,3.14,3.15,3.16,3.17,3.18)

Problem

3.1

Show

,if

F0

ei!

tre

pre

sents

F0

sin!

tin

the

vec

tor

form

of

the

equat

ion

of

motion

for

the

forc

ed

osc

illa

tor

that

x¼

F0

!Z

m

cosð!

t !Þ

and

the

vel

oci

ty

v¼

F0

Zm

sinð!

t !Þ

Problem

3.2

The

dis

pla

cem

entof

afo

rced

osc

illa

tor

isze

roat

tim

et¼

0an

dits

rate

of

gro

wth

isgover

ned

by

the

rate

of

dec

ayof

the

tran

sien

tte

rm.

Ifth

iste

rmdec

ays

toe

kof

its

ori

gin

alval

ue

ina

tim

et

show

that

,fo

rsm

alldam

pin

g,th

eav

erag

era

teof

gro

wth

of

the

osc

illa

tions

isgiv

enby

x0=t¼

F0=2k

m!

0

wher

ex

0is

the

max

imum

stea

dy

stat

edis

pla

cem

ent,

F0

isth

efo

rce

amplitu

de

and!

2 0¼

s=m

.

74

The

Forc

ed

Osc

illa

tor

Problem

3.3

The

equat

ion

m€ xxþ

sx¼

F0

sin!

tdes

crib

esth

em

otion

of

anundam

ped

sim

ple

har

monic

osc

illa

tor

dri

ven

by

afo

rce

of

freq

uen

cy!.Show

,by

solv

ing

the

equat

ion

invec

tor

form

,th

atth

est

eady

stat

e

solu

tion

isgiv

enby

x¼

F0

sin!

t

mð!

2 0 !

2Þ

wher

e!

2 0¼

s m

Sket

chth

ebeh

avio

ur

of

the

amplitu

de

of

xver

sus!

and

note

that

the

chan

ge

of

sign

as!

pas

ses

thro

ugh!

0defi

nes

aphas

ech

ange

of"

rad

inth

edis

pla

cem

ent.

Now

show

that

the

gen

eral

solu

tion

for

the

dis

pla

cem

ent

isgiv

enby

x¼

F0

sin!

t

mð!

2 0 !

2Þþ

Aco

s!

0tþ

Bsi

n!

0t

wher

eA

and

Bar

eco

nst

ant.

Problem

3.4

Inpro

ble

m3.3

,if

x¼

_ xx¼

0at

t¼

0sh

ow

that

x¼

F0

m

1

ð!2 0 !

2Þ

sin!

t

! !0

sin!

0t

!

and,

by

wri

ting!¼!

0þ"!

wher

e"!=!

0/

1an

d"!

t/

1,

show

that

nea

rre

sonan

ce,

x¼

F0

2m!

2 0

ðsin!

0t !

0tco

s!

0tÞ

Sket

chth

isbeh

avio

ur,

noting

that

the

seco

nd

term

incr

ease

sw

ith

tim

e,al

low

ing

the

osc

illa

tions

to

gro

w(r

esonan

cebet

wee

nfr

eean

dfo

rced

osc

illa

tions)

.N

ote

that

the

conditio

n"!

t/

1fo

cuse

s

atte

ntion

on

the

tran

sien

t.

Problem

3.5

What

isth

egen

eral

expre

ssio

nfo

rth

eac

cele

ration

_ vvof

asi

mple

dam

ped

mec

han

ical

osc

illa

tor

dri

ven

by

afo

rce

F0

cos!

t?D

eriv

ean

expre

ssio

nto

giv

eth

efr

equen

cyof

max

imum

acce

lera

tion

and

show

that

ifr¼

ffiffiffiffiffiffi

smp

,th

enth

eac

cele

ration

amplitu

de

atth

efr

equen

cyof

vel

oci

tyre

sonan

ce

equal

sth

elim

itof

the

acce

lera

tion

amplitu

de

athig

hfr

equen

cies

.

Problem

3.6

Pro

ve

that

theexact

amplitu

de

atth

edis

pla

cem

ent

reso

nan

ceof

adri

ven

mec

han

ical

osc

illa

tor

may

be

wri

tten

x¼

F0=!0 r

wher

eF

0is

the

dri

vin

gfo

rce

amplitu

de

and

!02¼

s m

r2

4m

2

Problem

3.7

Ina

forc

edm

echan

ical

osc

illa

tor

show

that

the

follow

ing

are

freq

uen

cyin

dep

enden

t(a

)th

e

dis

pla

cem

ent

amplitu

de

atlo

wfr

equen

cies

(b)

the

vel

oci

tyam

plitu

de

atvel

oci

tyre

sonan

cean

d(c

)

the

acce

lera

tion

amplitu

de

athig

hfr

equen

cies

,ð!!1Þ.

The

Effect

of

the

Tra

nsi

ent

Term

75

Problem

3.8

InFig

ure

3.9

show

that

for

smal

lr,

the

max

imum

val

ue

of

curv

e(a

)is)

F0=2!

0r

at

!1¼!

0

r=2m

and

its

min

imum

val

ue

is)

F0=2!

0r

at!

2¼!

0þ

r=2m

.

Problem

3.9

The

equat

ion€ xxþ!

2 0x¼ð

eE

0=mÞc

os!

tdes

crib

esth

em

otion

of

abound

undam

ped

elec

tric

char

ge

eof

mas

sm

under

the

influen

ceof

anal

tern

atin

gel

ectr

icfiel

dE¼

E0

cos!

t.For

an

elec

tron

num

ber

den

sity

nsh

ow

that

the

induce

dpola

riza

bility

per

unit

volu

me

(the

dynam

ic

susc

eptibility)

of

am

ediu

m

)e¼

nex

"0E¼

ne

2

"0mð!

2 0 !

2Þ

(The

per

mittivity

of

am

ediu

mis

defi

ned

as"¼"

0ð1þ)Þw

her

e"

0is

the

per

mittivity

of

free

spac

e.

The

rela

tive

per

mittivity"

r¼"="

0is

called

the

die

lect

ric

const

antan

dis

the

squar

eof

the

refr

active

index

when

Eis

the

elec

tric

fiel

dof

anel

ectr

om

agnet

icw

ave.

)

Problem

3.10

Rep

eatPro

ble

m3.9

for

the

case

of

adam

ped

osc

illa

tory

elec

tron,by

takin

gth

edis

pla

cem

entx

asth

e

com

ponen

tre

pre

sente

dby

curv

e(a

)in

Fig

ure

3.9

tosh

ow

that

"r¼

1þ)¼

1þ

ne

2mð!

2 0 !

2Þ

"0½m

2ð!

2 0 !

2Þ2þ!

2r

2'

Infa

ct,

Fig

ure

3.9

(a)

plo

ts"

r¼"="

0.

Note

that

for

!/

!0;

"r)

1þ

ne

2

"0m!

2 0

and

for

!1

!0;

"r)

1

ne

2

"0m!

2

Problem

3.11

Show

that

the

ener

gy

dis

sipat

edper

cycl

eby

the

fric

tional

forc

er_ xx

atan

angula

rfr

equen

cy!

isgiv

en

by"r!

x2 m

ax.

Problem

3.12

Show

that

the

ban

dw

idth

of

the

reso

nan

ceab

sorp

tion

curv

edefi

nes

the

phas

ean

gle

range

tan!¼-

1.

Problem

3.13

An

alte

rnat

ing

voltag

e,am

plitu

de

V0

isap

plied

acro

ssan

LC

Rse

ries

circ

uit.Show

that

the

voltag

eat

curr

ent

reso

nan

ceac

ross

eith

erth

ein

duct

ance

or

the

conden

ser

isQ

V0.

76

The

Forc

ed

Osc

illa

tor

Problem

3.14

Show

that

ina

reso

nan

tLC

Rse

ries

circ

uit

the

max

imum

pote

ntial

acro

ssth

eco

nden

ser

occ

urs

ata

freq

uen

cy!¼!

0ð1

1=2Q

2 0Þ1=2

wher

e!

2 0¼ðL

CÞ

1an

dQ

0¼!

0L=R

.

Problem

3.15

InPro

ble

m3.1

4sh

ow

that

the

max

imum

pote

ntial

acro

ssth

ein

duct

ance

occ

urs

ata

freq

uen

cy

!¼!

0ð1

1=2Q

2 0Þ

1=2.

Problem

3.16

Lig

ht

of

wav

elen

gth

0.6mm

(6000

A)

isem

itte

dby

anel

ectr

on

inan

atom

beh

avin

gas

alightly

dam

ped

sim

ple

har

monic

osc

illa

tor

with

aQ

-val

ue

of

5+

10

7.Show

from

the

reso

nan

ceban

dw

idth

that

the

wid

thof

the

spec

tral

line

from

such

anat

om

is1:2+

10

14

m.

Problem

3.17

Ifth

eQ

-val

ue

of

Pro

ble

m3.6

ishig

hsh

ow

that

the

wid

thof

the

dis

pla

cem

ent

reso

nan

cecu

rve

is

appro

xim

atel

yffiffiffi 3

pr=

mw

her

eth

ew

idth

ism

easu

red

bet

wee

nth

ose

freq

uen

cies

wher

ex¼

xm

ax=2.

Problem

3.18

Show

that

,in

Pro

ble

m3.1

0,th

em

ean

rate

of

ener

gy

abso

rption

per

unit

volu

me;

that

is,th

epow

er

supplied

is

P¼

ne

2E

2 0

2

!2r

m2ð!

2 0 !

2Þ2þ!

2r

2

Summary

ofIm

portantResults

Mec

han

ical

Imped

ance

Zm¼

F=v

(forc

eper

unit

vel

oci

ty)

Zm¼

Zm

ei!¼

rþ

ið!m

s=!Þ

wher

eZ

2 m¼

r2þð!

m

s=!Þ2

sin!¼!

m

s=!

Zm

;co

s!¼

r Zm

;ta

n!¼!

m

s=!

r

!is

the

phas

ean

gle

bet

wee

nth

efo

rce

and

vel

oci

ty.

Forc

edOsc

illa

tor

Equat

ion

of

moti

on

m€ xxþ

r_ xxþ

sx¼

F0

cos!

t

(Vec

tor

form

)m€ xxþ

r_ xxþ

sx¼

F0

ei!

t

Use

x¼

Ae

i!t

togiv

est

eady

stat

edis

pla

cem

ent

x¼

iF

0

!Z

m

eið!

t !Þ

The

Effect

of

the

Tra

nsi

ent

Term

77

and

vel

oci

ty

_ xx¼

v¼

F0

Zm

eið!

t !Þ

When

F0

ei!

tre

pre

sents

F0

cos!

t x¼

F0

!Z

m

sinð!

t !Þ

v¼

F0

Zm

cosð!

t !Þ

Maxim

um

velocity¼

F0 r

atvelocity

reso

nan

tfr

equen

cy!

0¼ðs=mÞ1=2

Maxim

um

displacement¼

F0

!0 r

wher

e!0 ¼

ðs=m

r2=4

m2Þ1=2

at

displacement

reso

nan

tfr

equen

cy!¼ðs=m

r2=2

m2Þ1=2

Pow

erAbso

rbed

by

Osc

illa

tor

from

Drivi

ng

Forc

e

Osc

illa

tor

adju

sts

ampli

tude

and

phas

eso

that

pow

ersu

ppli

edeq

ual

spow

erdis

sipat

ed.

Pow

erab

sorb

ed¼

1 2ðF

2 0=Z

mÞc

os!

(cosf

ispow

erfa

ctor)

Max

imum

pow

erab

sorb

ed¼

F2 0

2r

at!

0

Max

miu

mpow

er

2ab

sorb

ed¼

F2 0

4r

at!

1¼!

0

r 2m

and!

2¼!

0þ

r 2m

Qual

ity

fact

or

Q¼!

0m r¼

!0

!2 !

1

Q¼

max

imum

dis

pla

cem

ent

atdis

pla

cem

ent

reso

nan

ce

dis

pla

cem

ent

as!!

0

¼Aðm

axÞ

F0=s

For

equiv

alen

tex

pre

ssio

ns

for

elec

tric

alosc

illa

tors

repla

cem

by

L,r

by

R,s

by

1=C

and

F0

by

V0

(volt

age)

.

78

The

Forc

ed

Osc

illa

tor

4 CoupledOscillations

Theprecedingchaptershaveshowninsomedetailhowasinglevibratingsystem

will

behave.Oscillators,however,rarely

existin

completeisolation;wavemotionowes

its

existencetoneighbouringvibratingsystemswhichareabletotransmittheirenergytoeach

other.

Such

energytransfertakesplace,ingeneral,becausetwooscillatorsshareacommon

component,capacitanceorstiffness,inductanceormass,orresistance.Resistancecoupling

inevitablybringsenergylossandarapiddecay

inthevibration,butcouplingbyeitherof

theothertwoparametersconsumesnopower,andcontinuousenergytransferovermany

oscillatorsispossible.Thisisthebasisofwavemotion.

Weshallinvestigatefirstamechanical

example

ofstiffnesscouplingbetweentwo

pendulums.Twoatomssetinacrystallatticeexperience

amutualcouplingforceand

wouldbeamenabletoasimilartreatment.Then

weinvestigatean

exampleofmass,or

inductive,coupling,andfinallyweconsiderthecoupledmotionofan

extended

arrayof

oscillatorswhichleadsusnaturallyintoadiscussiononwavemotion.

Stiffness(orCapacitance)CoupledOscillators

Figure4.1showstwoidenticalpendulums,eachhavingamassmsuspendedonalightrigid

rodoflengthl.Themassesareconnectedbyalightspringofstiffnessswhosenatural

lengthequalsthedistancebetweenthemasseswhenneitherisdisplacedfromequilibrium.

Thesmalloscillationswediscussarerestricted

totheplaneofthepaper.

Ifxandyaretherespectivedisplacementsofthemasses,then

theequationsof

motionare

m€ xx¼!mgx l!sðx!yÞ

and

m€ yy¼!mgy lþsðx!yÞ

ThePhysicsofVibrationsandWaves,6thEdition

H.J.Pain

#2005JohnWiley

&Sons,Ltd

79

Theserepresentthenormalsimpleharmonicmotiontermsofeach

pendulumplusacoup-

lingtermsðx!yÞfromthespring.Weseethatifx>ythespringisextended

beyondits

normallengthandwillactagainsttheaccelerationofxbutinfavouroftheaccelerationofy.

Writing!2 0¼g=l,where!0isthenaturalvibrationfrequency

ofeach

pendulum,gives

€ xxþ!2 0x¼!s mðx!yÞ

ð4:1Þ

€ yyþ!2 0y¼!s mðy!xÞ

ð4:2Þ

Insteadofsolvingtheseequationsdirectlyforxandywearegoingtochoosetwonew

coordinates

X¼xþy

Y¼x!y

Theimportance

ofthisapproachwillemergeasthischapterproceeds.Addingequations

(4.1)and(4.2)gives

€ xxþ€ yyþ!2 0ðxþyÞ¼0

thatis

€ XXþ!2 0X¼0

andsubtracting(4.2)from(4.1)gives

€ YYþð!

2 0þ2s=mÞY¼0

Themotionofthecoupledsystem

isthusdescribedintermsofthetwocoordinatesXandY,

each

ofwhichhasan

equationofmotionwhichissimpleharmonic.

IfY¼0,x¼yatalltimes,sothatthemotioniscompletelydescribed

bytheequation

€ XXþ!2 0X¼0

thenthefrequencyofoscillationisthesameasthatofeitherpenduluminisolationandthe

stiffnessofthecouplinghasnoeffect.Thisisbecausebothpendulumsarealwaysswinging

inphase(Figure4.2a)andthelightspringisalwaysatitsnaturallength.

y

s

ll

x

Figure

4.1

Twoiden

ticalpen

dulums,

each

alightrigid

rodoflength

lsupportingamass

man

d

coupledbyaweightlessspringofstiffness

san

dofnaturallength

equal

totheseparationofthe

massesat

zero

displacemen

t

80

CoupledOscillations

IfX¼0,x¼!yatalltimes,sothatthemotioniscompletelydescribed

by

€ YYþð!

2 0þ2s=mÞY¼0

Thefrequency

ofoscillationisgreaterbecause

thependulumsarealwaysoutofphase

(Figure

4.2b)so

thatthespringiseither

extended

orcompressed

andthecouplingis

effective.

Norm

alCoordinates,Degrees

ofFreedom

andNorm

alModes

ofVibration

ThesignificanceofchoosingXandYtodescribethemotionisthattheseparametersgivea

verysimpleillustrationofnormalcoordinates.

%Normalcoordinatesarecoordinatesinwhichtheequationsofmotiontaketheform

ofa

setoflineardifferentialequationswithconstantcoefficientsinwhicheach

equation

containsonlyonedependentvariable(oursimpleharmonicequationsinXandY).

%AvibrationinvolvingonlyonedependentvariableX(orY)iscalled

anormalmodeof

vibrationandhasitsownnormalfrequency.Insuch

anormalmodeallcomponentsof

thesystem

oscillatewiththesamenormalfrequency.

%Thetotalenergyofan

undamped

system

may

beexpressed

asasumofthesquaresof

thenormalcoordinatesmultiplied

byconstantcoefficientsandasumofthesquaresof

thefirsttimederivatives

ofthecoordinates

multiplied

byconstantcoefficients.The

energyofacoupledsystem

when

theXandYmodesarebothvibratingwouldthen

be

expressed

intermsofthesquaresofthevelocitiesanddisplacementsofXandY.

%Theimportanceofthenormalmodesofvibrationisthatthey

areentirelyindependent

ofeach

other.Theenergyassociated

withanormalmodeisneverexchangedwith

anothermode;thisiswhywecanaddtheenergiesoftheseparatemodestogivethe

totalenergy.Ifonlyonemodevibratesthesecondmodeofoursystem

willalwaysbeat

rest,acquiringnoenergyfromthevibratingmode.

%Eachindependentway

bywhichasystem

may

acquireenergyiscalled

adegreeof

freedomtowhichisassigneditsownparticularnormalcoordinate.Thenumberofsuch

ll

ll

(a)

(b)

Figure

4.2

(a)Th

e‘in

phase’

mode

ofvibration

given

by€ XXþ!

2 0X¼

0,where

Xis

thenorm

al

coordinate

X¼

xþ

yan

d!

2 0¼

g=l.(b)‘Outofphase’

modeofvibrationgiven

by€ YYþð!

2 0þ

2s=

mÞ

where

Yis

thenorm

alco

ordinate

Y¼

x!

y

NormalCoordinates,DegreesofFreedomandNormalModesofVibration

81

differentwaysinwhichthesystem

cantakeupenergydefinesitsnumberofdegreesof

freedom

anditsnumber

ofnormal

coordinates.Eachharmonicoscillatorhas

two

degreesoffreedom,itmay

takeupbothpotentialenergy(normalcoordinateX)and

kineticenergy(normalcoordinate

_ XX).Inourtwonormalmodestheenergiesmay

be

written

EX¼a_ XX2þbX2

ð4:3aÞ

and

EY¼c_ YY2þdY2

ð4:3bÞ

wherea,b,canddareconstant.

Oursystem

oftwocoupledpendulumshas,then,fourdegrees

offreedomandfour

normalcoordinates.

Anyconfigurationofourcoupledsystem

mayberepresentedbythesuper-positionofthe

twonormalmodes

X¼xþy¼X0cosð!

1tþ$1Þ

and

Y¼x!y¼Y0cosð!

2tþ$2Þ

whereX0andY0arethenormalmodeamplitudes,whilst!2 1¼g=land!2 2¼ðg=lþ2s=mÞ

arethenormalmodefrequencies.Tosimplifythediscussionletuschoose

X0¼Y0¼2a

andput

$1¼$2¼0

Thependulumdisplacementsarethen

given

by

x¼

1 2ðXþYÞ¼acos!1tþacos!2t

and

y¼

1 2ðX!YÞ¼acos!1t!acos!2t

withvelocities

_ xx¼!a!1sin!1t!a!2sin!2t

and

_ yy¼!a!1sin!1tþa!2sin!2t

82

CoupledOscillations

Nowletussetthesystem

inmotionbydisplacingtherighthandmassadistancex¼2a

andreleasingbothmassesfromrestsothat

_ xx¼

_ yy¼0attimet¼0.

Figure4.3showsthatourinitialdisplacementx¼2a,y¼0att¼0may

beseen

asa

combinationofthe‘inphase’modeðx¼y¼asothatxþy¼X0¼2aÞandofthe‘outof

phase’modeðx¼!y¼aso

thatY0¼2aÞ.Afterrelease,themotionoftherighthand

pendulumisgiven

by

x¼acos!1tþacos!2t

¼2acosð!

2!!1Þt

2cosð!

1þ!2Þt

2

andthatofthelefthandpendulumisgiven

by

y¼acos!1t!acos!2t

¼!2asinð!

1!!2Þt

2sinð!

1þ!2Þt

2

¼2asinð!

2!!1Þt

2sinð!

1þ!2Þt

2

Ifweplotthebehaviouroftheindividualmassesbyshowinghowxandychangewithtime

(Figure4.4),weseethatafterdrawingthefirstmassasideadistance2aandreleasingitx

followsaconsinusoidalbehaviouratafrequency

whichistheaverageofthetwonormal

modefrequencies,butitsamplitudevariescosinusoidallywithalowfrequency

whichis

halfthedifference

betweenthenormalmodefrequencies.Ontheotherhand,y,which

startedatzero,vibratessinusoidallywiththeaveragefrequencybutitsamplitudebuildsup

to2aandthen

decayssinusoidallyatthelowfrequencyofhalfthedifferencebetweenthe

normalmodefrequencies.Inshort,theydisplacementmassacquiresalltheenergyofthex

displacementmasswhichisstationarywhenyisvibratingwithamplitude2a,butthe

energyisthenreturnedtothemassoriginallydisplaced.Thiscompleteenergyexchangeis

only

possiblewhen

themassesareidenticalandtheratioð!

1þ!2Þ=ð!

2!!1Þisan

integer,otherwiseneitherwilleverbequitestationary.Theslowvariationofamplitudeat

halfthenormalmodefrequency

difference

isthephenomenonof‘beats’whichoccurs

betweentwooscillationsofnearlyequalfrequencies.Weshalldiscussthisfurtherinthe

sectiononwavegroupsinChapter5.

y = 0

a2a

a

YX

− a

a

++

Figure4.3

Thedisplacemen

tofonepen

dulum

byan

amount2

ais

shownas

theco

mbinationofthe

twonorm

alco

ordinates

Xþ

Y

NormalCoordinates,DegreesofFreedomandNormalModesofVibration

83

Theimportantpointtorecognize,however,isthatalthoughtheindividualpendulums

may

exchangeenergy,thereisnoenergyexchangebetweenthenormalmodes.Figure4.3

showed

theinitialconfigurationx¼2a,y¼0,decomposedintotheXandYmodes.The

higherfrequencyoftheYmodeensuresthatafteranumberofoscillationstheYmodewill

havegained

halfavibration(aphaseof%rad)ontheXmode;thisisshowninFigure4.5.

ThecombinationoftheXandYmodesthen

givesythevalueof2aandx¼0,andthe

processisrepeated.WhenYgainsanother

halfvibrationthenxequals2aagain.The

pendulumsmay

exchangeenergy;thenormalmodesdonot.

Toreinforcetheimportance

ofnormalmodes

andtheircoordinates

letusreturn

to

equations(4.3a)and(4.3b).Ifwemodifyournormalcoordinatestoread

Xq¼

m 2

!

1=2ðxþyÞ

and

Yq¼

m 2

!

1=2

ðx!yÞ

tt

y displacementx displacement

2a

2a

00

Figure

4.4

Beh

aviourwith

time

ofindividual

pen

dulums,

showing

complete

energy

exch

ange

betwee

nthepen

dulumsas

xdecreases

from

2ato

zero

whilst

ygrowsfrom

zero

to2

a

x = 0

a2a

a

YX

a− a

+

−

Figure

4.5

Thefaster

vibrationofthe

Ymoderesultsin

aphasegainof%radove

rthe

Xmode

of

vibration,to

give

y¼

2a,

whichis

shownhereas

aco

mbinationofthenorm

almodes

X!

Y

84

CoupledOscillations

then

wefindthatthekineticenergyinthoseequationsbecomes

Ek¼T¼a_ XX2þc_ YY2¼1 2_ XX2 qþ1 2_ YY2 q

ð4:4aÞ

andthepotentialenergy V¼bX2þdY2¼1 2

g l

!

X2 qþ1 2

g lþ2s m

"#

Y2 q

¼1 2!2 0X2 qþ1 2!2 sY2 q;

ð4:4bÞ

where!2 0¼g=land!2 s¼g=lþ2s=m.

NotethatthecoefficientsofX2 qandY2 qdependonlyonthemodefrequenciesandthatthe

propertiesofindividualpartsofthesystem

arenolongerexplicit.

Thetotalenergyofthesystem

isthesumoftheenergiesofeach

separateexcitedmode

fortherearenocrossproductsXqYqintheenergyexpressionofourexample,i.e.,

E¼TþV¼

1 2_ XX2 qþ1 2!2 0X2 q

"#

þ1 2_ YY2 qþ1 2!2 sY2 q

"#

Atomsinpolyatomicmoleculesbehaveas

themassesofourpendulums;thenormal

modes

oftwotriatomicmoleculesCO2andH2Oareshownwiththeirfrequencies

in

Figure4.6.Normalmodesandtheirvibrationswilloccurfrequentlythroughoutthisbook.

O OC

O

C C

O

HH

HH

OO

HH

O OO

H2O

105°

ω1 = 11 × 1013 sec−1

ω2 = 11.27 × 1013 sec−1

ω3 = 4.78 × 1013 sec−1

ω3 = 2 × 1013 sec−1

ω2 = 7.05 × 1013 sec−1

ω1 = 4.16 × 1013 sec−1

CO

2

Figure

4.6

NormalmodesofvibrationfortriatomicmoleculesCO2andH2O

NormalCoordinates,DegreesofFreedomandNormalModesofVibration

85

TheGeneral

MethodforFindingNorm

alModeFrequencies,

Matrices,Eigenvectors

andEigenvalues

Wehavejustseen

thatwhen

acoupledsystem

oscillatesinasinglenormalmodeeach

componentofthesystem

willvibratewithfrequencyofthatmode.Thisallowsustoadopt

amethodwhichwillalwaysyieldthevalues

ofthenormalmodefrequencies

andthe

relativeamplitudesoftheindividualoscillatorsateach

frequency.

Supposethatoursystem

ofcoupledpendulumsinthelastsectionoscillatesinonlyone

ofitsnormalmodesoffrequency!.

Then,intheequationsofmotion

m€ xxþmgðx=lÞþsðx!yÞ¼0

and

m€ yyþmgðy=lÞ!sðx!yÞ¼0

Ifthependulumsstartfromtest,wemay

assumethesolutions

x¼Aei!t

y¼Bei!t

whereAandBarethedisplacementamplitudesofxandyatthefrequency!.Usingthese

solutions,theequationsofmotionbecome

½!m!2Aþðmg=lÞAþsðA!BÞ'ei!t¼0

½!m!2Bþðmg=lÞB!sðA!BÞ'ei!t¼0

ð4:5Þ

Thesumoftheseexpressionsgives

ðAþBÞð!m!2þmg=lÞ¼0

whichissatisfiedwhen!2¼g=l,thefirstnormalmodefrequency.Thedifferencebetween

theexpressionsgives

ðA!BÞð!m!2þmg=lþ2sÞ¼0

whichissatisfied

when!2¼g=lþ2s=m,thesecondnormalmodefrequency.

Insertingthevalue!2¼g=l

inthepairofequationsgivesA¼B(the‘inphase’

condition),whilst!2¼g=lþ2s=mgivesA¼!B(theantiphaseconditon).

Thesearetheresultswefoundintheprevioussection.

Wemay,however,bydividingthroughbymei!t ,rewriteequation(4.5)inmatrixform

as

!2 0þ!2 s

!!2 s

!!2 s

!2 0þ!2 s

$%

A B

$%

¼!2A B

$%

ð4:6Þ

where

!2 0¼g l

and!2 s¼s m

86

CoupledOscillations

Thisiscalled

aneigenvalueequation.Thevalueof!

2forwhichnon-zerosolutionsexist

arecalled

theeigenvaluesofthematrix.Thecolumnvectorwithcomponents

AandBisan

eigenvectorofthematrix.

Equation(4.6)may

bewritten

inthealternativeform

ð!2 0þ!

2 s"!

2Þ

"!

2 s

"!

2 sð!

2 0þ!

2 s"!

2Þ

!

A B

!

¼0

ð4:7Þ

andtheseequationshaveanon-zerosolutionifandonly

ifthedeterminantofthematrix

vanishes;that

is,if

ð!2 0þ!

2 s"!

2Þ2

"!

4 s¼

0

or

ð!2 0þ!

2 s"!

2Þ¼

%!

2 s

i.e.

!2 1¼!

2 0or

!2 2¼!

2 0þ2!

2 s

asweexpect.

Thesolution!

2 1¼!

2 0in

equation(4.6)yieldsA

¼Bas

previouslyand!

2 2¼!

2 0þ2!

2 s

yieldsA

¼"B.

Because

thesystem

startedfrom

rest

wehavebeenable

toassumesolutionsofthe

simple

form

x¼

Aei!t

y¼

Bei!t

When

thependulumshavean

initialvelocity

att¼

0,theboundaryconditionsrequire

solutionsoftheform

x¼

Aei

ð!tþ"xÞ

y¼

Bei

ð!tþ"yÞ

whereeach

norm

almodefrequency!has

itsownparticularvalueofthephaseconstant".

Thenumber

ofadjustableconstantsthen

allowsthesolutionsto

satisfythearbitrary

values

oftheinitialdisplacementsandvelocities

ofboth

pendulums.

(Problems4.1,4.2,4.3,4.4,4.5,4.6,4.7,4.8,4.9,4.10,4.11)

Mas

sorIn

ductan

ceCo

upling

Inalaterchapterweshalldiscuss

thepropagationofvoltageandcurrentwaves

alonga

transm

issionlinewhichmay

beconsidered

asaseries

ofcoupledelectrical

oscillators

havingidenticalvalues

ofinductance

andofcapacitance.Forthemomentweshallconsider

theenergytransfer

betweentwoelectrical

circuitswhichareinductivelycoupled.

Mass

orInductanceCoupling

87

Amutual

inductance

(shared

mass)

exists

betweentwoelectrical

circuitswhen

the

magnetic

fluxfrom

thecurrentflowingononecircuit

threadsthesecondcircuit.Any

changeoffluxinducesavoltagein

both

circuits.

Atransform

erdependsuponmutual

inductance

foritsoperation.Thepower

sourceis

connectedto

thetransform

erprimarycoilofnpturns,over

whichis

woundin

thesame

sense

asecondarycoilofnsturns.Ifunitcurrentflowingin

asingleturn

oftheprimarycoil

producesamagnetic

flux#,then

thefluxthreadingeach

primaryturn

(assumingnoflux

leakageoutsidethecoil)is

np#andthetotalfluxthreadingallnpturnsoftheprimaryis

Lp¼

n2 p#

whereL

pis

theselfinductance

oftheprimarycoil.Ifunitcurrentin

asingle

turn

ofthe

secondarycoilproducesaflux#,then

thefluxthreadingeach

secondaryturn

isns#andthe

totalfluxthreadingthesecondarycoilis L

s¼

n2 s#;

whereL

sis

theselfinductance

ofthesecondarycoil.

Ifallthefluxlines

from

unitcurrentin

theprimarythread

alltheturnsofthesecondary,

then

thetotalfluxlines

threadingthesecondarydefines

themutualinductance

M¼

nsðn

p#Þ¼

ffiffiffiffiffiffiffiffiffiffi

LpL

s

p

Inpractice,

because

offluxleakageoutsidethecoils,

M<

ffiffiffiffiffiffiffiffiffiffi

LpL

s

p

andtheratio

Mffiffiffiffiffiffiffiffiffiffi

LpL

s

p¼

k;thecoefficientofcoupling:

IftheprimarycurrentI p

varieswithei!

t ,achangeofI p

gives

aninducedvoltage

"L

pdI p=dt¼

"i!

LI p

intheprimaryandan

inducedvoltage"M

dI p=d

t¼

"i!

MI p

inthe

secondary.

Ifweconsider

now

thetworesistance-freecircuitsofFigure

4.7,whereL1andL2are

coupledbyfluxlinkageandallowed

tooscillate

atsomefrequency

!(thevoltageand

currentfrequency

ofboth

circuits),then

thevoltageequationsare

i!L1I 1

"i

1

!C

1

I 1þi!

MI 2

¼0

ð4:8Þ

C2

L 2

L1

C 1

M = Mutual Inductance

M

Figu

re4.7

Inductively(m

ass)

coupledLC

circuitswithmutual

inductance

M

88

CoupledOscillations

and

i!L2I 2

"i

1

!C

2

I 2þi!

MI 1

¼0

ð4:9Þ

whereM

isthemutual

inductance.

Multiplying(4.8)by!=iL

1gives

!2I 1

"I 1

L1C

1

þM L1

!2I 2

¼0

andmultiplying(4.9)by!=iL2gives

!2I 2

"I 2

L2C

2

þM L2

!2I 1

¼0;

wherethenaturalfrequencies

ofthecircuit!

2 1¼

1=L

1C

1and!

2 2¼

1=L

2C

2give

ð!2 1"!

2ÞI

1¼

M L1

!2I 2

ð4:10Þ

and

ð!2 2"!

2ÞI

2¼

M L2

!2I 1

ð4:11Þ

Theproduct

ofequations(4.10)and(4.11)gives

ð!2 1"!

2Þð!

2 2"!

2Þ¼

M2

L1L2

!4¼

k2!

4;

ð4:12Þ

wherekisthecoefficientofcoupling.

Solvingfor!

gives

thefrequencies

atwhichenergyexchangebetweenthecircuits

allowsthecircuitsto

resonate.Ifthecircuitshaveequalnaturalfrequencies!

1¼!

2¼!

0,

say,

then

equation(4.12)becomes

ð!2 0"!

2Þ2

¼k2!

4

or

ð!2 0"!

2Þ¼

%k!

2

that

is

!¼

%!

0ffiffiffiffiffiffiffiffiffiffiffi

1%

kp

Thepositivesigngives

twofrequencies

!0 ¼

!0

ffiffiffiffiffiffiffiffiffiffiffi

1þ

kp

and

!00¼

!0

ffiffiffiffiffiffiffiffiffiffiffi

1"

kp

atwhich,ifweplotthecurrentam

plitudeversusfrequency,twomaxim

aappear(Figure4.8).

Mass

orInductanceCoupling

89

Inloose

couplingkandM

aresm

all,and!

0 (!

00(!

0,so

that

both

system

sbehave

almost

independently.

Intightcouplingthefrequency

difference

!00"!

0increases,

the

peakvalues

ofcurrentaredisplacedandthedip

betweenthepeaksismore

pronounced.In

thissimpleanalysistheeffectofresistance

has

beenignored.In

practicesomeresistance

is

alwayspresentto

limittheam

plitudemaxim

um.

(Problems4.12,4.13,4.14,4.15,4.16)

CoupledOscillationsofaLo

aded

String

Asafinalexam

pleinvolvingalargenumber

ofcoupledoscillatorsweshallconsider

alight

stringsupportingnequalmassesmspaced

atequaldistance

aalongitslength.Thestringis

fixed

atboth

ends;ithas

alength

ðnþ1Þa

andaconstanttensionTexistsat

allpointsand

alltimes

inthestring.

Smallsimple

harmonic

oscillationsofthemassesareallowed

inonly

oneplaneandthe

problem

isto

findthefrequencies

ofthenorm

almodes

andthedisplacementofeach

mass

inaparticularnorm

almode.

Thisproblem

was

firsttreatedbyLagrange,itsparticularinterestbeingtheuse

itmakes

ofnorm

almodes

and

thelightit

throwsupon

thewavemotion

and

vibration

ofa

continuousstringtowhichitapproxim

ates

asthelinearseparationandthemagnitudeofthe

massesareprogressivelyreduced.

Figure

4.9

showsthedisplacementyroftherth

masstogether

withthose

ofitstwo

neighbours.Theequationofmotionofthis

massmay

bewritten

by

consideringthe

components

ofthetensiondirectedtowardstheequilibrium

position.Therth

massis

pulled

downwardstowardstheequilibrium

positionbyaforceTsin'1,dueto

thetension

Current amplitude

Coupling

(a)k large

(b) k intermediate

(c)k small

ω0

ω

(a)

(b)(c)

Figu

re4.8

Variationofthecurrentam

plitudein

each

circuitneartheresonantfrequency.Asm

all

resistance

prevents

theam

plitudeat

resonance

from

reachinginfinitevalues

butthis

has

been

ignoredin

thesimpleanalysis.Flatteningoftheresponse

curvemaximumgives

‘frequency

bandpass’

coupling

90

CoupledOscillations

onitsleftandaforceTsin'2dueto

thetensiononitsrightwhere

sin'1¼

yr"

yr"

1

a

and

sin'2¼

yr"

yrþ

1

a

Hence

theequationofmotionisgiven

by

md2yr

dt2

¼"T

ðsin'1þsin'2Þ

¼"T

yr"

yr"

1

aþ

yr"

yrþ

1

a

$%

so

d2yr

dt2

¼€ yyr¼

T maðy

r"1"2yrþ

yrþ

1Þ

ð4:13Þ

If,in

anorm

almodeofoscillationoffrequency

!,thetimevariationofyris

simple

harmonicabouttheequilibrium

axis,wemay

writethedisplacementoftherth

massin

this

modeas

yr¼

Arei!

t

whereA

risthemaxim

um

displacement.Sim

ilarly

yrþ

1¼

Arþ

1ei!tandyr"

1¼

Ar"

1ei!

t .

Usingthesevalues

ofyin

theequationofmotiongives

"!

2A

rei!

t¼

T maðA

r"1"2A

rþ

Arþ

1Þe

i!t

or

"A

r"1þ

2"

ma!

2

T

&'

Ar"

Arþ

1¼

0ð4:14Þ

Thisisthefundam

entalequation.

m

m

m

aa

yr+ 1

yr− 1

yr

yr− yr − 1

yr− yr+ 1

1θ

2θ

Figu

re4.9

Displacements

ofthreemassesonaloaded

stringunder

tension

Tgivingequationof

motionm€ yy r

¼Tðy

rþ1"2yrþ

yr"

1Þ=

a

CoupledOscillationsofaLoadedString

91

Theprocedure

nowisto

startwiththefirstmassr¼

1andmovealongthestring,writing

outthesetofsimilar

equationsas

rassumes

thevalues

r¼

1;2;3;.

..;n

remem

beringthat,

because

theendsarefixed

y0¼

A0¼

0and

ynþ1¼

Anþ1¼

0

Thus,when

r¼

1theequationbecomes

2"

ma!

2

T

&'

A1"

A2¼

0ðA

0¼

0Þ

When

r¼

2wehave

"A

1þ

2"

ma!

2

T

&'

A2"

A3¼

0

andwhen

r¼

nwehave "A

n"1þ

2"

ma!

2

T

&'

An¼

0ðA

nþ1¼

0Þ

Thus,wehaveasetofnequationswhich,when

solved,willyield

ndifferentvalues

of!

2,

each

valueof!beingthefrequency

ofanorm

almode,thenumber

ofnorm

almodes

being

equal

tothenumber

ofmasses.

Theform

alsolutionofthissetofnequationsinvolves

thetheory

ofmatrices.However,

wemay

easily

solvethesimple

casesforoneortwomassesonthestring(n

¼1or2)and,

inadditon,itispossible

toshow

what

thecomplete

solutionfornmassesmustbewithout

usingsophisticatedmathem

atics.

First,when

n¼

1,onemassonastringoflength

2a,weneedonly

theequationforr¼

1

wherethefixed

endsofthestringgiveA

0¼

A2¼

0.

Hence

wehave

2"

ma!

2

T

&'

A1¼

0

giving

!2¼

2T

ma

asingle

allowed

frequency

ofvibration(Figure

4.10a).

When

n¼

2,stringlength

3a(Figure

4.10b)weneedtheequationsforboth

r¼

1and

r¼

2;that

is

2"

ma!

2

T

&'

A1"

A2¼

0

92

CoupledOscillations

and

"A

1þ

2"

ma!

2

T

&'

A2¼

0ðA

0¼

A3¼

0Þ

Eliminating

A1orA

2showsthat

thesetwoequationsmay

besolved

(are

consistent)

when

2"

ma!

2

T

&'

2

"1

¼0

that

is

2"

ma!

2

T"1

&'

2"

ma!

2

Tþ1

&'

¼0

Thus,therearetwonorm

almodefrequencies

!2 1¼

T ma

and

!2 2¼

3T

ma

aa

m (a)

(b)

mm

m

m

A1 A1

A2

A1 = −A2

A2

ω22

3T

ma

=

A1 = A2

ω12

T ma

=

ω2

2T

ma

=

n =

1

n =

2

Figu

re4.10

(a)Norm

alvibrationofasingle

massm

onastringoflength

2aat

afrequency

!2¼

2T=m

a.(b)Norm

alvibrationsoftwomassesonastringoflength

3ashowingtheloose

coupled

‘inphase’modeoffrequency!

2 1¼

T=m

aandthetightercoupled‘outofphase’modeoffrequency

!2 2¼

3T=m

a.Thenumber

ofnorm

almodesofvibrationequalsthenumber

ofmasses

CoupledOscillationsofaLoadedString

93

Usingthevalues

of!

1in

theequationsforr¼

1andr¼

2gives

A1¼

A2theslow

‘in

phase’

oscillationofFigure

4.10b,whereas!

2gives

A1¼

"A

2thefaster

‘anti-phase’

oscillationresultingfrom

theincreasedcoupling.

Tofindthegeneral

solutionforanyvalueofnletusrewrite

theequation

"A

r"1þ

2"

ma!

2

T

&'

Ar"

Arþ

1¼

0

intheform

Ar"

1þ

Arþ

1

Ar

¼2!

2 0"!

2

!2 0

where

!2 0¼

T ma

Weseethat

foranyparticularfixedvalueofthenorm

almodefrequency

!ð!

jsay)the

righthandsideofthisequationisconstant,independentofr,so

theequationholdsforall

values

ofr.What

values

canwegiveto

Arwhichwillsatisfythis

equation,meetingthe

boundaryconditionsA

0¼

Anþ1¼

0at

theendofthestring?

Let

usassumethat

wemay

express

theam

plitudeoftherthmassat

thefrequency!jas

Ar¼

Ceir'

whereCisaconstantand'issomeconstantangle

foragiven

valueof!j.Thelefthand

sideoftheequationthen

becomes

Ar"

1þ

Arþ

1

Ar

¼Cðe

iðr"1Þ'

þeiðr

þ1Þ'Þ

Ceir'

¼ðe

"i'

þei'Þ

¼2cos'

whichisconstantandindependentofr.

Thevalueof' j

(constantat!j)iseasily

foundfrom

theboundaryconditions

A0¼

Anþ1¼

0

which,usingsinr'

from

eir'gives

A0¼

Csinr'

¼0

ðautomatically

atr¼

0Þ

and

Anþ1¼

Csinðn

þ1Þ'

¼0

when

ðnþ1Þ'

j¼

j(for

j¼

1;2;.

..;n

94

CoupledOscillations

Hence

'j¼

j(

nþ1

and

Ar¼

Csinr'

j¼

Csin

rj(

nþ1

whichistheam

plitudeoftherth

massat

thefixed

norm

almodefrequency

!j.

Tofindtheallowed

values

of!

jwewrite

Ar"

1þ

Arþ

1

Ar

¼2!

2 0"!

2 j

!2 0

¼2cos'j¼

2cos

j(

nþ1

giving

!2 j¼

2!

2 01"cos

j(

nþ1

!

ð4:15Þ

wherejmay

takethevalues

j¼

1;2;.

..;n

and!

2 0¼

T=m

a.

Note

that

thereisamaxim

um

frequency

ofoscillation!j¼

2!

0.Thisiscalled

the‘cut

off’frequency

andsuch

anupper

frequency

limitischaracteristicofalloscillatingsystem

s

composedofsimilar

elem

ents(themasses)

repeatedperiodically

throughoutthestructure

ofthesystem

.Weshallmeetthisin

thenextchapteras

afeature

ofwavepropagationin

crystals.

Tosummarize,

wehavefoundthenorm

almodes

ofoscillationofncoupledmasseson

thestringto

havefrequencies

given

by

!2 j¼

2T

ma

1"cos

j(

nþ1

!

ðj¼

1;2;3

...nÞ

Ateach

frequency

!jtherth

masshas

anam

plitude

Ar¼

Csin

rj(

nþ1

whereCis

aconstant.

(Problems4.17,4.18,4.19,4.20,4.21,4.22)

TheWav

eEq

uation

Finally,in

thischapter,weshowhowthecoupledvibrationsin

theperiodicstructure

ofour

loaded

stringbecomewaves

inacontinuousmedium.

TheWaveEquation

95