Simple and Physical Pendula
Simple Pendulum: “point” mass m on massless string L
 Howdowefindnaturalfrequency  ?
o UseNewton’s2ndlawtogetdifferentialequationforSHM
o Identify  fromdifferentialequation
 Forcesonhangingmass:
o Forstringat  fromvertical, Ftangential   mg sin   Tangentialacceleration:
o Reminder:forpointatdistanceRfromcentreofbodywithangular
d 2
acceleration   dt 2 ,tangentialaccelerationis a tangential  R o SoformassonstringoflengthL,tangentialaccelerationis atangential
(Remember:  mustbeinradianshere)
 Newton’s2ndLaw: Ftangential
d 2
 matangential gives  mg sin   mL 2 dt
d 2
g


sin  o Result: dt 2
L
o NotquiteSHMBUTforsmallangle, sin    (for  inradians)
d 2
 L 2 dt
d 2
g



 SimplePendulumresult:forsmall  ,have dt 2
L g
2


o LookslikeSHMwith
L  Note,doesnotdependonm!
 StrictlySHMforsmallangleonly.Slightlydifferentfromsinusoidal
forlargeramplitude.
2
L
T


2

o Periodis

g o Solutionis    max cos t     Canseeswingingmotionforobjectsthatare
muchmorecomplicatedthanamassona
string.
o Needmoregeneralresult!
Physical Pendulum:
 NOTa“point”massonamasslessstring.Lookatrotationofrigidbody.
 Newton’s2ndlawforrigidrotatingobjectwithmomentofinertiaIis   I o Restoringtorqueduetogravityatcofm
distancedfrompivotis:
    mgd sin  d
 clockwiseforpositive  d 2
o Angularaccelerationis   dt 2  Canrewrite   I asdifferentialequation:
d 2
mgd
sin  (notquiteSHM)


o dt 2
I
 Forsmallangle, sin    ,get
d 2
mgd



o dt 2
I
mgd
2


o SHMwith
I Small angle result for Physical Pendulum with moment of inertia I
mgd


 SHMwithangularfrequency
I I
2
T


2

 Periodis

mgd  AngularfrequencyandperioddependonshapeofswingingobjectANDon
locationofpivot(boththroughIandd)
o Because I  m ,masscancelsoutand  doesnotdependonm
 Checkforsimplependulum
2
o Forsimplependulum, I  mL and d  L mgd g
2

 asexpectedforsimplependulum

 So
I
L
Example: Physical Pendulum consisting of a
suspended from one end
 Mass=M,Length=L
1
I

ML2  Momentofinertia:
3
L
d

 Distancefrompivottocofm:
2
mgd
MgL / 2
3g




 Soangularfrequencyis
I
ML2 / 3
2 L o Independentofmass
rod
DAMPED OSCILLATION:
 Objectmovingthroughfluid(likeair)feelsspeed‐dependentresistiveforce:


o Dampingforce: R  bv (non‐conservative.Dissipatesmech.energy)
 Formassmonspringkmovingthroughfluidwithcoefficientb
o Newton’s2ndLaw:  Fx  kx  bv x  ma x d 2x
b dx k


 x
o Givesdifferentialequation: dt 2
m dt m
k


 Definenaturalfrequency 0
m
o FrequencyofoscillatorinABSENCEofdamping
b
 Solutiontodifferentialequationdependsonrelativesizeof 0 and 2m o i.e.comparisonofnaturalfrequencyandamountofdamping
b
UNDERDAMPED MOTION: small b so that 0  2m
or
b  2m 0
 Systemdoesnotoscillateatnaturalfrequency:
2
 b 





 o Shiftedto
m
2


o Besuretodistinguish  and 0 2
0
2m
 Amplitudeofoscillationdecaysexponentiallywithtimeconstant b  bt / 2 m
cos t     Completesolutionis: x  Ae
CRITICALLY DAMPED MOTION: see for b  bc  2m 0
 Systemreturnstoequilibrium
exponentiallywithnooscillation
b
OVERDAMPED MOTION: large b so that 0  2m
 Systemreturnstoequilibriumwithno
oscillationandmoreslowlythancritical
case
(See Dan Russell animations: link on web page)
or
b  2m0
FORCED OSCILLATIONS AND RESONANCE
 Dampedoscillatorcanbekeptatconstantamplitudebyaddingenergythrough
periodicforce–forcedoscillation
o Amplitudedependsonfrequency
 Example:Applyperiodicforce F t   F0 sin  t tomassmonspringkmoving
throughfluidwithcoefficientb
o Careful:Besuretodistinguish  and 0 dx
d 2x
o Newton’s2ndLawgives: F0 sin  t  b dt  kx  m dt 2 o Ifdrivingforceisappliedtosysteminitiallyatrest,initialmotionis
complicatedbuteventuallysettlesintosteadystatesolution
o Aftersystem“settlesdown”,getsteadystatesolution:
 x  A  cos t     oscillateswithfrequency  ,not 0  Amplitudedependsondrivingfrequencyanddrivingforce
F0 / m
A
2

 2   02  b / 2m 2
 maximumamplitudeat
resonance,   0 o amplitudeatresonance
dependsonb

