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3|SimpleHarmonicMotion
87
derivativeoftotalenergy,is
dE=dt
=
F
x;
frict
v
x
bv
2
x
.Thatis
dE
dt
=
d
dt
1
2
mv
2
x
+
1
2
kx
2
=
mv
x
a
x
+
kxv
x
=(
ma
x
+
kx
)
v
x
bv
2
x
= 2
m v
2
x
= 2
m
2
v
2
0
!
02
e
2
t
sin
!
0
t
+cos
!
0
t
2
(3
:
22)
= 2
m
2
v
2
0
!
02
e
2
t
[1 sin2
!
0
t
]
Caution:Donotusecomplexexponentialsincomputingenergy.Youmustrstconverteverything
intothereal(cosine,sine)form.Thesquareof
e
i!t
hascrosstermsthatdonotbelong. Therealpart
of
e
i!t
2
iscos2
!t
andthatisnothinglikecos
2
!t
.
3.4OtherOscillators
Anystableequilibriumwillprovideaharmonicoscillatorforsmallmotions.Well,almostany;seesection
3.12foronethatdoesn’t.
m
^
z
^r
^
Fig.3.4
Pendulum
The pendulum is one of theeasiest oscillatorstostudy and forthisreason it was
probablytherst(Galileo). Theidealmodelofapendulumhasapointmassonthe
endofalight and inelasticstring. . Tond d its equation ofmotion,useplanepolar
coordinateswiththeangle
measuredfromthevertical.pendulum
d
2
dt
2
g
sin
(3
:
23)
Youcanderivethiseitherbyusingthetorqueequation,
~
=
I~
,orbyusing
theequations(0.41)describingaccelerationinpolarcoordinates,orbyusingenergy
conservation,
dE=dt
=0. Takethesecondchoicerst,solookbackatsection0.6.
Heretheradius
r
isaconstant,
,soitsderivativesarezeroandtheaccelerationis
~a
=^
r
r
_
2
+
^
r
+2_
r
_
=^
r
_
2
+
^
Eq.(0.41)
Theforceon
m
comesfromgravityandthestring,taking
z
positivedownward.
~
F
^
rF
string
+
mg
^
z
^
rF
string
+^
rmg
cos
^
mg
sin
Combinetheseandyouhave
^
rF
string
+^
rmg
cos
^
mg
sin
=
m
^r
_
2
+
^

!
F
string
+
mg
cos
m‘
_
2
and
mg
sin
=
m‘
(3
:
24)
The last equation is the one of interesthere, as it is the dierential equation for
versus
t
. The
problemisthatit’snotasimpleharmonicoscillator,sowhat’sitdoinginthischapter? Itsaysthatthe
secondderivativeof
withrespectto
t
isasineof
,not
itself. Thesimpleformsofsolutionsuch
ascos
!t
or
e
t
arenotright. Tryitandsee. Itiscertainlynotalinear,constantcoecientdierential
equation.
Thereishoweveranapproximationthatworkssurprisinglywell. Iftheangleofoscillationissmall
thenyoucanusetheseriesexpansionofthesineandget
d
2
dt
2
g
sin
g
1
6
3
+
1
120
5

 
g
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3|SimpleHarmonicMotion
88
Howgoodisthisapproximation?ThetermomittedshouldbealotlessthanwhatI’mkeeping.
1
6
3
soiftheratiois1%then
3
<:
06
;
or
<
p
:
06=
:
24=14
Amoredetailedcalculation,section4.6,showsthattheerrorincalculatingthefrequencyisabout0
:
4%
when
swingsbackandforth14
.Thisissmallerthan1%becausetheerrorinestimatingthesine
reachesitsworstvalueonlyattheendsoftheswing. Formostofthependulum’smotion,theerroris
smallerthanitsmaximumvalue.
Withthissmallangleapproximation,theequationofmotionisthatofasimpleharmonicoscillator
d
2
dt
2
g
=)
(
t
)=
A
sin(
!
0
t
+
)
;
where
!
0
=
r
g
(3
:
25)
Thatthefrequencyofoscillationis (nearly) independentoftheamplitude(
A
) oftheoscillationwas
asurprise to Galileoand others of his time (ca.1600). . A A consequence of this observationwas the
developmentofthependulumclock.Inthenextapproximation,keepingthe
3
term,thefrequencyis
approximately
!
0
r
g
1
16
2
max
(3
:
26)
Seesection4.6andproblem4.44forthedetailsofthis.
EnergyMethod
Anothermethod to derive the dierential equation of motion uses conservation of energy, and this
way doesn’t t require e knowing g the e equation (0.41) for r acceleration n in polar r coordinates. . Write e the
totalenergy. Thekinetic c energy is
mv
2
=
2=
m‘
2
_
2
=
2,andthepotentialenergyisthegravitational
mgh
=
mg‘
(1 cos
). Here e Itookthezero of potential atthe bottom of the swing. . Energy y is
conserved,soitstime-derivativeiszero.
h
E
=
1
2
m‘
2
_
2
+
mg‘
(1 cos
)
;
then
dE
dt
=
1
2
m‘
2
d
_
2
d
_
d
_
dt
mg‘
d
cos
d
d
dt
=
m‘
2
_
d
2
dt
2
+
mg‘
sin
d
dt
=0
(3
:
27)
The chain rule is theonly mathematical tool youneed todothese manipulations. . The e factorof
_
cancels,andthe resultisEq. (3.23) again, butwithless eort. . Do o you lose anythingby using this
simplerapproach?Yes,butjustthepartofEq.(3.24)forthetensioninthestring.
ElectricCircuits
Thesimplestelectriccircuitswillexhibitoscillations.Allthatyouneedisaninductorandacapacitor.
I
L
q
C
R
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3|SimpleHarmonicMotion
89
L
dI
dt
+
q
C
=0
;
and
I
=
dq
dt
=)
L
d
2
q
dt
2
+
q
C
=0
Putinaresistorandyouhave
L
d
2
q
dt
2
+
R
dq
dt
+
q
C
=0
Addanoscillatingvoltagesourceandyoureplacethezeroontherightsideby
V
0
cos
!t
.
You don’t t have to study y anything new in n order r to o solve these e equations; they’re the same
equationsyou’vebeenreadingaboutinthelastseveralpagesexceptthatthesymbolshavechanged.
IndescribingmechanicalfrictionImadeapointabout the sortof modellingthatyouhave todoto
describeit. Thesamequestionarisesheretoo. . Canyoureallysaythattheelectricalresistanceis
IR
withaconstant
R
? Sometimesandinsomeapproximationyoucan,butnotalways. Foranordinary
incandescentlightbulbyouwouldprobablyexpectthistobevalid,butittooisjustanapproximation.
Theresistanceinthetungstenlamentdependsonitstemperature,andthatinturndependsonthe
sizeofthecurrentgoingthroughit. Thatreplacesthepotentialdierence
IR
withaslightlynonlinear
functionof
I
. Ifyouusea uorescentbulb,thesituationisnotevenclosetolinear.Therelationbetween
currentandvoltageinthatcaseisnotevensingle-valued. Itisnotafunction,andthevoltage-current
relationevenhasanegativeslope(
dI=dV <
0)insomeregions. That’s s thereasonthat uorescent
xturesusea ballast to keep p the system stable. . Thatballast t couldbe justanotherresistor, butin
practicethatwouldwastetoomuchenergysootherdevicesareusedinstead.
3.5ForcedOscillations
Whenyouhaveanysortofharmonicoscillatoryoucanalsohaveotherforcesbesidesthespringandthe
damping(orthecapacitanceandtheresistance). Iftheoscillatorisachildonaswing(apendulum)
youmaybepushingperiodically. Ifit’samassonaspringyoumaybepullingontheendofthespring,
varyingtheforceonthemassbychangingthelengthofthespring.Inanycase,thedierentialequation
willbesomethinglike
m
d
2
x
dt
2
b
dx
dt
kx
+
F
external
(
t
)
(3
:
28)
Thisisnotalinear,homogeneous,constantcoecientdierentialequation,andthesumoftwoofits
solutionsisnotasolution,butitcomesprettyclose.
Terminology:\Homogeneous"meansthatifyoumultiplythedependentvariable
x
byaconstant
theneverytermintheequationismultipliedbythatconstant.* Equation(3.14)satisesthis,but(3.28)
doesn’t. Itisaninhomogeneousequation.
\Linear"meansthatthedependentvariableanditsderivativesappearonlytotherstorzerothpower.
Bothoftheseequationssatisfythis.
Thereisaplanofattackforhandlinginhomogeneousproblemssuchasthese.
1. Tossouttheinhomogeneoustermandndthegeneralsolutionofthishomogeneouspart,complete
withallthearbitraryconstants.
2.Findanyonesolutiontothefull,inhomogeneousequation,anditdoesn’tneedtohaveanyarbitrary
constants.
3. Addtheresultsofsteps1and2.
Thiswillprovideasolutiontothefullequationcompletewiththearbitraryconstantsneededtotthe
initialconditions.Toshowthatthisistrue,simplyplugtheresultintotheoriginalequation.
* orapowerofthatconstant
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3|SimpleHarmonicMotion
90
Example
x
Putamassontheendofaspringandhangthespringvertically. Theequationofmotionis
m
d
2
x
dt
2
=
F
x
kx
+
mg
(3
:
29)
Stepone: Tossoutthe
mg
termandyoualreadyknowthesolutiontobe
x
hom
(
t
)=
A
sin
!
0
t
+
B
cos
!
0
t
.
Steptwo:Themostecientwaytondasolutiontoanequationisguesswork.Whatfunction
doesittakesothatacombinationofthefunctionanditssecondderivativegivesaconstant?Aconstant
ofcourse. TherearemoresystematicwaysthatI’llgettoinsection3.11,butthiswilldoforthesimple
cases.
x
inh
=
C
=
mg=k
Stepthree:
x
=
x
hom
+
x
inh
=
Ae
i!
0
t
+
Be
i!
0
t
+
mg=k
(3
:
30)
IftheinitialconditionsarethatIreleasethemassfromrestattheorigin,
x
(0)=0=
A
+
B
+
mg
k
_
x
(0)=0=
Ai!
0
Bi!
0
giving
A
=
B
mg
2
k
x
(
t
)=
mg
k
[1 cos
!
0
t
]
(3
:
31)
Asusual,whatdoesthissay?Thedimensionscheckeasily. Also,it’sreal. Duringthederivationofthe
answer,alltheequationswerecomplex,sothatitcomesoutrealsaysthatIhaven’tmadetoomany
mistakes.Now,whatdoesitlooklikeforsmalltime?
small
t
=)
x
mg
k
h
1
2
!
2
0
t
2
+
i
=
1
2
mg
k
!
2
0
t
2
=
1
2
gt
2
Therstthingitdoesisdrop. Ofcourse.
Afterthatitoscillatesbetween
x
=0(whencos
!
0
t
=1)andthebottompoint
x
=2
mg=k
(when
cos
!
0
t
= 1). . Theequilibriumpoint,wherethetotalforceiszero,ishalfwaybetweenthesetwo.
t
x
Example
Applyaconstantforcetoanoscillator,but foranite time. . Thenshutito. Assumethemass
startsattheoriginatrest,
m
x
kx
+
F
(
t
)
F
(
t
)=
F
0
(0
<t<T
)
0
(otherwise)
0
T
F
0
(3
:
32)
Fortheinterval0
<t<T
thedierentialequationis
m
x
+
kx
=
F
0
,andthesolutiontothisis
x
(
t
)=
A
cos
!
0
t
+
B
sin
!
0
t
+
F
0
=k
Usetheinitialconditions.
x
(0)=0=
A
+
F
0
=k;
x_(0)=0=B
whichimply
x
(
t
)=
F
0
k
[1 cos
!
0
t
]
(
t<T
)
(3
:
33)
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3|SimpleHarmonicMotion
91
Atthe endofthe
T
-interval,startagainwith thenal conditions attheend of 0
< t<T
asthe
initialconditionsfor
T <t<
1. (Iwillsetthisupthehardwayrst,thenshowaneasierway.) ) The
dierentialequationnolongerhasthe
F
0
term,sothesolutionis
for
t>T;
x
(
t
)=
C
cos
!
0
t
+
D
sin
!
0
t;
from(3.33):
x
(
T
)=
F
0
k
[1 cos
!
0
T
]
;
_
x
(
T
)=
F
0
k
!
0
sin
!
0
T
match
x
and _
x
:
x
(
T
)=
C
cos
!
0
T
+
D
sin
!
0
T
=
F
0
k
[1 cos
!
0
T
]
_
x
(
T
)= 
C!
0
sin
!
0
T
+
D!
0
cos
!
0
T
=
F
0
k
!
0
sin
!
0
T
Thelasttwolinesaretwoequationsforthetwounknowns
C
and
D
. Straight-forwardtosolve,butthis
isnottheeasywaytodotheproblem.Thisisthesamesortofdicultythatoccurredintheexample
atEq.(2.4),andyousimplifythealgebrainthesameway. Lookbacktothatequationandseewhat
occurredthere.
Insteadofgrindingthroughthisalgebra,jumptotheeasiermethod
:::
Planaheadandchooseyourcoordinatescarefully.Thisincludeswhereyoustartyourclock.The
naturaloriginforthesecondhalfoftheproblemis
T
,notzero. Thatsuggestswritingthesolutionas
x
(
t
)=
C
cos
!
0
(
t
T
)+
D
sin
!
0
(
t
T
)
(
t>T
)
then
C
=
x
(
T
)=
F
0
k
[1 cos
!
0
T
] and
D
_
x
(
T
)
=!
0
=
F
0
k
sin
!
0
T
(3
:
34)
Anytimethatyoundyourselfwith analgebraicmess,looktosee ifyoucouldhavesetitupmore
skillfully. Ifnot,thenshownofearandplungeahead,buthereyouseewhatalittleinsightwilldo.Put
thisalltogetherand
x
(
t
)=
8
<
:
0
(
t<
0)
F
0
k
[1 cos
!
0
t
]
(0
<t<T
)
F
0
k
(1 cos
!
0
T
)cos
!
0
(
t
T
)+(sin
!
0
T
)sin
!
0
(
t
T
)
(
T<t
)
(3
:
35)
Examinetheresults.
(a) If
!
0
T
=2
allmotion stopsafter
t
=
T
. Thathappensbecauseatthetimejustbefore
T
the
mass comes into agentlestop atthe origin. . If f you turnotheforce atthatinstantthen it stays
stopped.
(b) If f the force lasts s just a a short time (short compared d to what?
!
0
T
 1), then n the motion n is
approximately
x
(
t
)
F
0
k
1
2
(
!
0
T
)
2
cos
!
0
t
+
!
0
T
sin
!
0
t
Thesecond(sine)termisthedominantonenow,
F
0
k
!
0
T
sin
!
0
t
=
F
0
T
m
.
m
k
!
0
sin
!
0
t
=
F
0
T
m
!
0
!
2
0
sin
!
0
t
=
v
initial
!
0
sin
!
0
t
and
v
initial
=
F
0
T=m
isthevelocityjustaftertheimpulse.
(c)YoucanndatrigonometricidentitytosimplifytheappearanceofEq.(3.35). Doesithelp?
Aretherestilleasierandmoresystematicways? Yes;gotosection3.11tosee.
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3|SimpleHarmonicMotion
92
3.6HarmonicForcing
Anothertime-dependentforce,onethatappearsveryoftenandinmanycontexts.IfIholdaspringby
oneendwiththemasshangingfromtheotherandIthenmovemyhandupanddown,thatwillprovide
anoscillatingforceon
m
. If f achild is on aswing and Ipush periodically, that’sa timedependent
force.Thesimplestmathematicalformforsuchapushisacosineorsine.Thegeneralizationtoamore
complicatedperiodicforcecanwait.Callthefrequencyofthisforce (capital
!
),then
m
d
2
x
dt
2
kx
b
dx
dt
+
F
external
(
t
)
where
F
external
(
t
)=
F
0
cos
t
(3
:
36)
Ifthedampingtermisabsent,you canguess asolutiontothis. . You u wantafunctionsuchthatthe
secondderivativeplusthefunctionitselfresultinacosineof
t
. Acosineitselfwilldothat. Youhave
toadjustthecoecient,butthat’seasytodo. Seeproblem3.14. Thedampingtermmakesinspired
guessingmorediculttodobecausetherstderivativechangesacosineintoasine.
Hereiswherethecomplexexponentialformcomesintoitsown.Uptonowthere’slittleI’vedone
thatIcouldn’tdojustwithsimplesinesandcosines,butthedampingtermmakesthosemuchharder
todealwith. Therighttechniqueistosaythatthecosineistherealpartofthecomplexexponential.
Solveeverythingwiththeexponentialformandthentaketherealpartoftheresult.
m
d
2
x
dt
2
+
b
dx
dt
+
kx
=
F
0
e
i
t
(realpartunderstood)
(3
:
37)
Amoreexplicitwaytosaythisisthatinsteadofsolvingoneequation,solvetwo.Twodierentforcing
functions (cosine e and sine) ) give e twodierent
x
’s (
x
r
and
x
i
) thatsatisfy twodierent dierential
equations.
m
d
2
x
r
dt
2
+
b
dx
r
dt
+
kx
r
=
F
0
cos
t
and
m
d
2
x
i
dt
2
+
b
dx
i
dt
+
kx
i
=
F
0
sin
t
Nowtaketherstoftheseequationsandadditto
i
timesthesecond.
m
d
2
(
x
r
+
ix
i
)
dt
2
+
b
d
(
x
r
+
ix
i
)
dt
+
k
(
x
r
+
ix
i
)=
F
0
(cos
t
+
i
sin
t
)
IfIlet
x
=
x
r
+
ix
i
,thisisEq.(3.37).
Nowif you wantan exponentialoutput onthe righthand side,put anexponentialin. . Usea
solutionthat’saconstanttimes
e
i
t
andyougetasolution.
x
inh
(
t
)=
Ce
i
t
 
mC
2
e
i
t
+
ibC
e
i
t
+
kCe
i
t
=
F
0
e
i
t
Theexponentialcancelsandyouhavetheequationfor
C
.
C
=
F
0
m
2+
ib
+
k
;
and
x
inh
(
t
)=
F
0
e
i
t
m
2+
ib
+
k
(3
:
38)
Before worrying aboutadding thesolution tothe homogeneous part of the equation, I’lllookmore
closelyat thisinhomogeneoussolution. . Ittakesabitofmanipulation n toput thisintoasimpleand
useful form,andthekey isto rememberthatwhileadditionandsubtractionofcomplexnumbersis
moreeasily doneintermsofrectangularcomponents,multiplicationand divisionworkmoreeasilyin
polarform.
3|SimpleHarmonicMotion
93
A
B
p
A
2+
B
2
Thedenominatoris
m
2
+
ib
+
k
=
A
+
iB
=
p
A
2+
B
2
A
+
iB
p
A
2
+
B
2
(3
:
39)
The reason forthis multiplication and division by the samefactoris
thatitmakesthenalfractionhavemagnitudeone. Thatallowsme
towriteitasanexponential,
e
i
.Fromthepicture,thecosineandthe
sineoftheangle
arethetwotermsinthefraction.
A
+
iB
=
p
A
2
+
B
2
(cos
+
i
sin
)=
p
A
2
+
B
2
e
i
and
tan
=
sin
cos
=
B
A
(3
:
40)
There’saslightlytrickypointaboutthis. Ifyousay
=tan
1
(
B=A
)thenwhathappenswhen
A
is
zero?Orifitchangessign?Doyouget
=
2or 
=
2?Whatistan
1
( 1)?Therealwaytodothisis
tolookbackattherectangularform,
A
+
iB
.Knowingthequadrantsthat
A
and
B
areinwilltellyou
whatangleyouaredealingwith.Thearctangentismultiplevaluedandbyspecifyingbothrectangular
componentsyouremovetheambiguity.* Thebestwayremains: lookatthepicture.Whereare
A
and
B
inthecomplexplane? Thatanswersthequestion.Thisway y tan
1
(
y=x
)!tan
1
(1
=
1)=3
=
4
andtan
1
( 1
=
1)= 
=
4or+7
=
4.
ThedenominatorinEq.(3.38)isnow
p
m
2+
k
)2+
b
2 2
e
i
;
where
tan
=
sin
cos
=
b
k
m
2
(3
:
41)
Thisinhomogeneoussolutionisthen
x
inh
(
t
)=
F
0
e
i
t
p
m
2
+
k
)
2
+
b
2
2
e
i
=
F
0
e
i
(
t
)
p
m
2
+
k
)
2
+
b
2
2
(3
:
42)
Fortherealpartofthis,change
e
i
(
t
)
!cos(
t
). Itisalsoeasiertointerprettheresultifyou
dividenumeratoranddenominatorby
m
andwrite
!
2
0
insteadof
k=m
.Theangle
iscalledthephase
lag,appearingasitdoesin(
t
).
x
inh
(
t
)=
(
F
0
=m
)
e
i
(
t
)
q
2+
!
2
0
2
+
b
2 2
=m
2
(3
:
43)
Fortherealpart,thenumeratorof
x
r
(
t
)becomes(
F
0
=m
)cos(
t
).Thisisworthmorestudy
allbyitselfbeforecombiningitwiththehomogeneoussolution. Supposethatthedampingcoecient
b
issmall,andexaminethedenominatorofthelastequation. Thesquarerootfactorisafunctionof
theappliedfrequency ,andas variesfromzerotoinnitytherstterm,(
!
2
0
2
)
2
,variesfrom
!
4
0
toinnity,butnotice: itvanishesinbetween. . When
2
=
!
2
0
=
k=m
thisparenthesisiszeroand
allthat’sleftinthedenominatorof(3.43)isthedampingfactor
b
=m
andIsaidI’massumingthat
b
issmall.Whenthisdenominatorissmalltheinhomogeneoussolutionislarge. Thisisthephenomenon
ofresonance.
* Ifyouwriteacomputerprogram m thatcallson theinverse tangentfunction,look foralibrary
versionthattakestwoarguments(
A
and
B
)andnotjustone.
3|SimpleHarmonicMotion
94
Inthesameanalysis, whentheforcingfrequencyvaries,thenwhathappenstotheangle
in
Eq.(3.41)?As !0,theanglegoestozero. As increasesthedenominatorgoestozerosotan
goestoinnity.Nowwhat? Theanswerisreallymucheasiertondbylookingattheoriginalcomplex
form,Eq.(3.39).
m
2
+
ib
+
k
/
e
i
Inthisrectangularform,small makesthispositiverealandthen
!0asstated.As increasesyou
pickupapositiveimaginarypart,carrying
upto
=
2when
m
2
=
k
.Finally,forverylarge thisis
mostlynegativereal( 
2
)plusacomparativelysmallamountofpositiveimaginary(
i
),so
!
.
Thenextthreegraphsshowtheamplitudeandthephase
asafunctionofthedrivingfrequency
. Therearesix x plots in eachcase,forvariousvalues of
b=m!
0
from0
:
5to0
:
02. Therstisthe
amplitude(the1
=
p
factorinEq.(3.42)).Thesecondplotsthephaseforthesameparameters.The
thirdplotsthecomplexnumberthatisthewholecoecientof
F
0
e
i
t
inthesameequation. Itisworth
seeinghowvariouspartsofthethirdgraphcorrelatewiththersttwographs.
!
0
:
5!
:
4!
:
3!
:
2!
b=m!
0
=
:
1!
"0
:
02
Fig.3.5(a)
!
0
0
:
5!
:
02
Fig.3.5(b)
Inthesecondgraphyoucanseethephase
asitstartsfromzero,meaningthatwhenyoupush
withaslowoscillationthemasswillfollowyourpush. Ifhowevertheforcingfrequency ishigh,then
!
andtheresponseapproaches180
outofphasefromtheappliedforce.Thesephasedierences
haveamajoreectonthestructureofoceantides,discussedinchapterveonpage170. Thegraph
showinghowthephaseapproaches
whentheforcingfrequencyis wellabovethenaturalfrequency
helps explain why the highest high tides can occurnearthe time that theMoon is either r risingor
setting,notwhenitisoverhead.(Didyouknowthat?)
:
5
:
1
:
02
Fig.3.5(c)
1
m
2+
ib
+
k
Thenextlogicalstepwouldbetoaddthefullsolutionofthehomogeneousequationandtoapply
initialconditions.Iwon’t.Itisnotveryinstructiveanditismoderatelycomplicated.I’llexpectyouto
doitfortheundampedcase,problem3.14,becauseinthatspecialcasethereisenoughpayoforthe
relativelymodesteortinvolved.
Terminology:TheinhomogeneoustermthatIjustderivediscalledthesteady-stateterm,andthe
homogeneoustermcomputedinsection3.3isthetransient. Thatissimplybecausethehomogeneous
solutionhasanegativeexponentialinitanditeventuallydiesout. Thesteady-statetermkeepsgoing.
3|SimpleHarmonicMotion
95
Isthevalue
2
=
k=m
=
!
2
0
thepositionofthepeakintherstofthesegraphs? That’sthe
onethatshows theamplitudeoftheresponsetotheforce. . Noitisnot. Thepeakoccurswhenthe
denominatorinEq.(3.43)isaminimum,andtondthatyouhavetodierentiatewithrespectto .
Betteryet,dierentiatewithrespectto
2
.
d
d
2
2
+
!
2
0
2
+
b
2
2
=m
2
=2
2
!
2
0
+
b
2
=m
2
=0
then
2
=
!
2
0
b
2
=
2
m
2
Youseefromthisthattheresonancepeakreally occursslightlybelowthenaturalfrequency. . Forthe
mostcommoninterestingcasesthedampingis smallandthisdistinction isnotlarge,butitisthere
andyoucanseeitintheamplitudegraph,Figure3.5(a).
MechanicalResonanceExample
Attachamasstotheendofaspringandletithangdownfromyourhand. Nowmoveyourhandup
anddowninalowfrequencyoscillation. Themasswillfollowyourhandinessentiallythesamemotion.
Graduallyincreasethefrequencyof yourhand’s motionandtheamplitude of the mass’smotionwill
rise,eventually becominglargeenoughthatthe massmay y otheendofthespring. . Thisfollows
theleftsideofgures Fig.3.5(a) andFig.3.5(b). . The e amplitudeof
m
’s motionincreases,whileit
more-or-lessstaysinphasewiththemotionofyourhand: 0
<<=
2.
Now repeat the e experiment but start with a high frequency oscillation n of f your hand. . Then
gradually lower the frequency and d watch the e mass. . Now w you u are following the e same two o gures,
but coming g in from the e right. . Again, , the response of f the mass will increase dramatically, , but t its
directionofmotionwillbeoppositetothatofyourhand,correspondingtothemathematicalstatement
=
2
<<
.
OpticalResonanceExample
Lighthitsanatom,andasaconsequenceitappliesanoscillatingforcetotheatom. Ifthefrequency
ofthelightmatchesoneofthe naturaltransitionfrequenciesintheatom,yougetalargeresponse,
andthelightiseitherstronglyabsorbedorstronglyscattered. Inthesunthisphenomenoncausesthe
Fraunhoferlines. The e lightfromthehotsun passesthroughthe(relatively) coolersolaratmosphere
whereun-ionizedatomscanexist,andthespeciccolorsoflightwhosefrequenciesmatchtheresonant
frequenciesofsomeoftheatomsisabsorbedorscattered,leavingadarklineinthespectrum.
400
450
500
550
600
650
700
750
wavelengthinnm
656.2
656.4
656.6
656.8
657.0
657.2
657.4
657.6
657.8
Fig.3.6
Thelinelabeled\C"intherstpictureistheH
lineduetothepresenceofexcitedhydrogen
atoms.Thesecondpictureisagreatlymagniedversioninwhichthislineappearsasthelargeblobat
3|SimpleHarmonicMotion
96
awavelengthjustbelow656.3nm.Thelargewidthofthislineispartlycausedbythenaturalwidthof
theresonanceasinthegraphontheprecedingpage,buttheatomalsosuersmanycollisionswithits
neighbors,andthisaccountsformostofthespreadinginthiscase.
Detailedunderstandingofthisphenomenonrequiresquantummechanics,butitissurprisingjust
howmuchoftheintuitionthatyougetfromtheclassicalharmonicoscillatorwillextendtotheadvanced
treatment.
Example
A fewlinesbackIsaidthatIdidn’twanttoaddthefullhomogeneoussolutiontotheproblem. . I
willmakeanexceptionhere.Intheundampedcase,whatdoesthesolutionlooklikewithanoscillating
externalforce?
F
x
kx
+
F
0
sin
t
=
m
x
!
x
hom
=
A
cos
!
0
t
+
B
sin
!
0
t
(3
:
44)
Fortheinhomogeneouspart,youwantacombinationof
x
and
x
toproduceasin
t
. That’seasy: a
sin
t
doesthejob.
x
inh
(
t
)=
C
sin
t
!
m
x
inh
kx
inh
=
C
m
2
+
k
sin
t
=
F
0
sin
t
Thatdeterminestheconstant
C
andthewholesolutionisnow
x
(
t
)=
A
cos
!
0
t
+
B
sin
!
0
t
+
F
0
m
2+
k
sin
t
Usetheinitialconditionthatthemassstartsattheoriginwithzerovelocity.
x
(0)=0=
A;
_
x
(0)=0=
B!
0
+
F
0
m
2+
k
Putthesetogetheranduse
!
2
0
=
k=m
toget
x
(
t
)=
F
0
=m
!
2
0
2
h
sin
t
!
0
sin
!
0
t
i
(3
:
45)
First,inordertoseeifthismakessense,whatmustitlooklikeforsmalltime?Thespringstartswith
noforceandtheaddedforcestartsatzero,soitmuststartwithzeroacceleration. Expandit:
x
(
t
)=
F
0
=m
!
2
0
2
h
t
1
6
3
t
3
+ 
!
0
!
0
t
1
6
!
3
0
t
3
+
i
=
1
6
F
0
m
t
3
+
Is this acceleration right? ? Twoderivativesgive 
x
=(
F
0
=m
)
t
+, and thatis s exactlywhatthe
startingequation(3.44)says.Itis,isn’tit?
Whatdoesthissolutionlooklikeatlatertimes?Theequation(3.45)hastwooscillationsoftwo
dierentfrequencies. Sometimestheyadd;sometimestheysubtract. Agraphwillhelp.
Fig.3.7
This graph uses two fairly random frequencies (  = 7
:
97and
!
0
= 11
:
3in some equally arbitrary
units).Theindividualfrequenciesarealreadyhardtodiscerninthisgraph.
Thereisaninterestingexampleofwhathappenstotheseresonancesinacircumstanceonlyslightly
more complicated thanthese: : aforced d pendulum is almost a forced d simple harmonic oscillator,but
not quite. . This s leads s to some very surprising results, , and d there e is a qualitative description of the
phenomenoninsection11.1. It’ssomethingthatyou’renotusedtointhisbook|pageswithoutany
equations.
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