Download - 10. the Harmonic Oscillator
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Part - I I
V ibrat ion s an d w ave sV. Satya Narayana M ur t hy
A217BITS Pilani Hyde rabadCampusHyderabad
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Top ics t o be co ve red
Kleppner & Ko lenkowCh 10 Harm o n ic oscil lat o r
A P FrenchCh 3 Free vib rat ions o f a physical syst em
Osci l lat ion s invo lving m assive sp rin gs
Ch 4 Fo rced vib rat ions and resonanceThe p ow er absor bed by a dr iven o sci llat or
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Ch 2 Sup er po si t ion of per iod ic m ot ions
Ch 5 Cou pled osci llat ors & no rm al m od es
Ch 6 Nor m al m od es o f cont inu ou ssystems
Free vibrat ion s o f a st ret ched st r in g
Sup erpo si t ion of m od es on a st r ingForced harm on ic v ibrat ions of a
st retched st r ing
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Ch 7 Progressive w avesNo rm al m od es and t ravelling w avesPro gressive w aves
Dispe rsion , phase and grou p velocit yThe en ergy in a m echanical w ave
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The H arm onic O scillat o r
Kleppner & Ko lenkow (CH 10)A P French (CH - 3 & 4)
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Top ics t o be co ve red
Per iod ic m ot ion s
Sim ple harm on ic m ot ion
Dam ped harm on ic oscil lat or
Fo rced h arm on ic osci llat o r
The p ow er absor bed by a dr iven o sci llat or
Osci l lat ion s invo lving m assive sp rin gs
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In ever y day l ife w e com e acro ss var io us
t h ings t hat m ove
The m o t ion o f p hysical syst em s can b e
classi f ied in t o 2 b ro ad cat ego ries
1 Translat ional m ot ion
2 V ibrat iona l m ot ion
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Per iodic m ot ion s
Vibratory / O scilla tory m ot ions
SH M
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Per iod ic m ot ion s
A m ovem ent t hat repeat s w it h p er iod icit y
Ex:
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The pat t ern t hat rep eat s m ay be sim ple or
compl icated
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V ibratory / oscillato ry m ot ion
W hat is th e d i fference bet w ee n o scilla to ry
and v ibra tory m ot ion?
A bod y in per iod ic m ot ion m oves back andfor t h over t he sam e path
In o sci llat ion t im e taken t o com plete o ne cycleis con st ant , in v ibrat ion i t m ay no t beOsci l lat ion s occu r in physical or b iolo gicalsystemsVibrat ion s occu r in m echan ical syst em s
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Every osci llat o r y m o t ion is per io d ic
bu t every per iod ic m ot ion n eed no t beosci l latory
EX. t he osci llat ion s o f a pendu lum
t he v ibrat ion s of a st r ing of a gui t ar
Uni form circu lar m ot ionis a per iod ic m ot ion, buti t is no t osci llat o ry
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Sim ple Harm onic M ot ion / Sinuso ida l M ot ion
Sim ple per iod ic m ot ion
In m any syst em s a sm all d isp lacem ent (x)f ro m t he equ i libr ium po si t ion set s up SHM
Rest o r ing fo rce = -kx
W here k is a const ant
(st i f f ness o r sp rin gconstant)
SHM No f r ict ion fo rce
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l
d l
x
In equ i lib r ium netforce act ing on t hemass, F = -k d l = M g
Now M is d isp lacedf rom equ i lib r iumpo sit ion by a
distance x
M g
M g
To t al net fo rce act ing on M isFnet = -k (d l + x) M g = -kx
Ve rt ica l sp ring m ass syste m
Fnet
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The equat ion for t he m ot ion o f SHM is:
)1( 0x
m
kx
kxxm
How t o so lve?
kxF
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The so lut ion for t h is equ at ion is o f t he form :
tCcosx o
Ano t her po ssib le solut ion is : tBsinx o
Therefore t he m ost general so lu t ion w il l be
)2( tCcostBsinx oo
)3( )tAcos(x 0
Equ . (2) can be w r i t t en in convenient for m as
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Rot at ing vecto r re pre se nt at ion
SHM can b e represent ed as geom et r icpro ject ion o f un ifo rm circu lar m ot ion
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Acosx If w e t ake cou nt er c lock w ise d i rect ion as +ve
then =0t +)3( )tAcos(x 0
The value o f is de term ined f rom t he value
of x at t =0
0
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tCcostBsinx oo )2(
)tAcos(x 0 )3(
AcosC
AsinB
o r
w h e r e
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)tcos(Ax 02
0
)4( 0xx2
0
Com paring eq. (1) & (4)
)5(m
k0
)tAcos(x 0 )3(
0xmkx )1(
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Exam p les o f SH M
Sim ple pen du lum
l
m
mgcos
s
mgsin
m g
For sm all angu lar d isp lacem ent s
sin
l
Sl S
Eq. Of m ot ion is
mgsinSm
(6)0gsinS
and
0Sl
gS )6(
l
g0
kxxm
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O scil lat ion s of f loat ing bo d ie s
W hen a body is in equ ilib r ium ,t he w eigh t is balanced by t hebuoyan t fo rce
xgxm A mAg
0
m
Displace t he bo dy f ro m i tsequ il ibr ium po si t ion b y anam o u n t x
t he extra bu oyancy force isgiven b y:
m g
Fb
x
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H arm o n ic o scillat ion s o f a n LC circu it
LC
K
source
vo lt age acro ss capacit o r
2
2
dt
qdL
dt
diLv
vol t age acro ss ind uct o r
C
qv
qLC
q
0qLC
1q
LC
10
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Nomencla ture
)tAcos(x 0 )3(
x = inst an t aneous d isp lacem ent o f t hep ar t icle at t im e t
A = am p lit u d e (m axim u m d isp lacem en t )o = an gu lar f req uen cy = 2/ T = p hase fact o r o r p hase an gle
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Ene rgy o f a n o scillato r
The to t al energy (po t ent ial +kinet ic) is acon stant fo r an un dam ped oscillat or
22 mv2
1kx
2
1KUE
tsinAm21tcoskA
21E 0
22200
22
2kA
2
1E
The ind ivid ual values o f P.E and K.E w illvary w it h t im e
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Tim e average value s
f ( t )
t 1 t 2
fW hat is t im e average?
dttf2
1
t
t
W hat is area und er t hecurve be tw een t
1and
t 2 ?
dttfttf2
1
t
t
12 dttftt1
f2
1
t
t12
o r
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Examples
0 1 2 3 4 5 6
-1.0
-0.5
0.0
0.5
1.0
Sin()
(radian)
0sin t
0 1 2 3 4 5 60.0
0.2
0.4
0.6
0.8
1.0
(radian)
Sin2()
21
sin2
t
2
1)(sin
2sin
2
0
22
dtttM at hem at ically
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W hat is t he t im e average values o f P.E. orK.E. over o ne p erio d?
2224
1sin
2
1.. kAtkAEP
2224
1cos
2
1.. kAtkAEK
.... EPEK
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Tim e average value
W hen fr ict ion is pr esent , t h is is no lon gert r u e
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Calculating 0 or T f rom E
Sp ring m ass syste mm
22xm
2
1kx
2
1KUE
Since E is const an t 0dt
dE
0xm
kx
m
k0
k
m2T
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Sim ple pen du lum
m
s
l-y
y
m
l
22
222
y2lys
ylsl
For sm all
sy
2l
sy
2
mgymv2
1
E2
2
2
dt
dsv
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22 sl
mg
2
1sm
2
1E
2l
sy
dt
ds
v2
2
2
Sin ce E is const an t 0dt
dE
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0dtdE 0s
lgs
l
g0
gl2T
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Recap
Per iod ic m ot ion
Vibrato ry / Osci llat or y m ot ion
Equ . o f m ot ion fo r d i f ferent SHO
How t o guess a so lut ion for secon d ord erdi f ferent ial equ. h aving con st ant coef f icient s
Tim e average values o f KE & PE
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Tod ay's t op ics
Com plex nu m bers
Dam ped harm on ic oscil lat or
Equ . o f m ot ionLight ly dam pedHeavily dam pedCri t ically dam ped
EnergyQu ali ty fact or
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Com plex num bers
)tAcos(x 0 So l. o f a SHM
)t(sinAx 00
)tcos(Ax 020
To sim pl i fy t he calculat ion s w e use com plex
numbers
W hat is t he use of com plex nu m bers inharm on ic osci llat or ?
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Com plex num bers are represent ed b y
z = x + iyx is t he real part and y is t he im aginary part
Graph ical represent at ion o f com plex num bers
y
xA
A cos
A sin
Im aginary axis
Real axis
z = x + iy = A (cos+ i sin )
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z = x + iy = A (cos+ i sin )
z = A ei
y
x
A
xy - co m p lex p lan evect or of lengt h A m akes an angle w i t ht he real axis
Geometr ical lyw hat is t hemeaning?
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OX
Y
)(i)(i
21112 eeAAz
t
Add vect or o f lengt h A 2at an gle (2- 1) to A1
Turn i t by an angle
(t + 1)
O X
A1
A212
1 t
A1
12
A2
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)tAcos(x 0
How to represen t in com plex fo rm ?
)tsin(Ay 0
Con sider t he im aginary com po nent
)tsin(Ai)t(cosAZ 00
)t(i 0eAZ Calculationbecomessimpler
Real part rep resen t st he eq u. of SHM
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)t(i 0eAZ
)t(i
00eiAZ
)t(sinAx 00 Real p ar t
)t(i2
00eAZ
)tcos(Ax 02
0 Real p ar t
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Da m pe d Ha rm on ic O scillato r
SHM No f r ict ion force
W hat is t he e f fect o f f r i ct ion on t he harm onicosci l lator?
Assum e a specia l for m of f r ict ion force viscous fo rce veloci ty f = - bv
Cond it ion : Viscous fo rce ar ises w hen anob ject m oves t hro ugh a f lu id at speedsw hich are no t so large t o cause t ur bu lence
b = coef f icient o f dam ping force
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To t al fo rce act ing on m is F = Fspring + f
bvkxF
xbkxxm
0xxx
0x
m
kx
m
bx
2
0
Equ. o f m ot ion is
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In com plex form
020 xxx
How t o so lve ?
To conver t in t o com plex for m use t he
com panion equat ion
02
0 yyy
02
0 zzz
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The so lu t ion w ill be o f t he form ,tezz 0
Sub st i tu t ing th e so lut ion b ack in to t heo riginal equ atio n gives us:
0202
0 )(ezt
02
0 zzz
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2
0
2
4
2
The m ost general solu t ion w i ll be:
t
B
t
A21 ezezz
Here zA and zB are co n st an t s an d 1 and 2are t he t w o roo t s
02
0
2
0 )(ezt
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2
0
2
21
42
,
t
B
t
A ezezz21
2
o
2
4
2
o
2
4
2o
2
4
Case (i) Case (i i) Case (ii i)
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2
o
2
4
Case (i) Light Dam ping
o rUnder Dam ping
22
4o is im aginary
1
22
o i2
4
i
2
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ti2ti1t/211 ezezez
Th e so lu t io n t o t h e d if feren t ial eq uat io n is:
tCsintBcosex 11t/2
Real part o f x is
tA(t)costcosAex 112t
o r
The solut ion is osci llat or y, bu t w i th a redu cedfrequ ency and t im e vary ing (expo nent ia llydecaying) ampl i tude
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4
22
o1
2
o
2
4
01
tA(t)tAext
112 coscos
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2
o
2
4
Case (ii)
Heavy Dam ping
o rOver Dam ping
2
o
2
4
is real
2
2
o
4
12
2
Bo t h ro o t s are n egat ive
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This rep resent s no n -osci llat o ry behavior
The act ual d isp lacem ent w i ll dep end u po nt he in i t ia l con di t ions
tt 21 BeAex
Real part of t he solut ion is
ttezezz 21 21
So lut ion is
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01
2
o
2
4
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2
o
2
4
Case (iii) Cri t ical Dam p ing
t/2Cex
The so l. t o a 2nd o rder d if feren t ial equ .shou ld have t w o independent const an t sw h ich are t o be f ixed by t he in it ial cond it ions
2
So l. is
The so lu t ion is incom p let e W hy?
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so lu t ion w i ll be o f t he fo rm teBtAx )2/(
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teBtAx )2/(tt 21 BeAex tAext
12 cos
AirThick
o ilWater
Light Heavy Critical
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Ene rgy of a D am pe d H arm on ic O scillato r
frictionW0EtE
From w ork energy th eorem
Wf r ic t ion= w ork do ne by th e f r ict ion fo rcef ro m t im e 0 t o t
f = -bvoppo ses t he m ot ion
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x(t)
x(0)
f fdxW t
0
2dtbv 0Frict ion fo rce dissipat es en ergy
E(t ) decreases w i th t im e
22
2
1
2
1xmkxK (t)U (t)E (t)
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Fo r t he ligh t l y dam ped oscillato r
)tt/2)cos(Aexp(x 1
)t(t)sinexp(2
mAmv
2
1K(t) 1
22
1
22
can b e n eglect ed
)tcos(2
)tsin(Aev 1
1
1
t2
1
12
1
2o
2
4
01
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)t(cosekA2
1kx
2
1tU 1
2t22
)t(kcos)t(sinmeA21
tE
1
2
1
22
1
t2
For light dam ping
m
k
2
0
2
1
t2ekA2
1tE
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At t =0 20 kA
2
1E
In general t0eEtE
0 1 2 3 4 5 6 7
0
1
2
3
4
5
E
time(s)
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The decay is charact er ized by a t im e ,dam p ing t im e, du r ing w hich t he energyfal ls t o e-1 of i t s in i t ia l va lue
t0eEtE 00 0.368E
e
EtE
W h e n 1
Tim e con sta nt
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Recap
Com plex nu m bersDam ped harm on ic oscil lat orEqu . o f m ot ion xbkxxm
2
0
2
2142
,
t
B
t
A ezezz21
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2
o
2
4
2o
2
4
2
o
2
4
Case (i) Case (i i) Case (i ii)
teBtAx )2/(tt 21 BeAex tAext
12 cos
Ligh t Heavy Cr it ical
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t0eEtE
0 1 2 3 4 5 6 7
0
1
2
3
4
5
E
time(s)
00 0.368EeEtE
1
Energy of a light ly dam ped harm on icoscil lator
Tim e con st ant ()
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To day s t o p ics
Q facto r o f DHO
Forced Harm onic osci llat o r
Und am ped FHOEqu . Of m ot ionSolut ionResonance+ve and ve aspects
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Q ua lity facto r
The d am pin g can b e specif ied by a
dim ension less param eter Q
radianperdissipatedenergy
oscillatortheinstoredenergyQ
Rate o f change of energy EeE
dt
dE t0
Energy d issipat ed in a t im e T is ETTdt
dE
E(t)
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T = 2/ 1 osci llat es t hr ough 2 radians
Energy d issipat ed per radian is1
E
E
EQ 01
1
Light dam ping Q>>1Heavy dam pin g Q is lowUn dam ped osci llat or Q is inf in i te
ETTdtdE Energy d issipat ed in 2
radians
01 Q
1
E l 10 2
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Exam ple 10 .2A m usician s t un ing for k r in gs at A ab ove
m idd le C, 440 Hz. A so un d level m et erindicat es t hat t he sou nd int ensi ty decreasesby a fact or o f 5 in 4 s. W hat is t he Q o f t he
t un ing fork?Solution
A440 o r C523.3 St and ard Pi tch / Concert
Pit ch - is a un iversa l f requ ency or no te t hatall inst ru m ent s are set t oTh is concert pi t ch enables m usicians t o p layinst rum ents to ge ther in h arm ony
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N ot e f (H z) (m )
A 3 2 20 .00 1 57 .
A#3 / Bb
3 2 33 .08 1 48 .
B3 2 46 .94 1 40 .
C4 2 61 .63 1 32 .
C#
4 / Db
4 2 77 .18 1 24 .D 4 2 93 .66 1 17 .
D # 4 / Eb
4 3 11 .13 1 11 .
E4 3 29 .63 1 05 .
F4 3 49 .23 9 8 .8
F#4 / Gb4 3 69 .99 9 3 .2
G4 3 92 .00 8 8 .0
G#4 / Ab
4 4 15 .30 8 3 .1
A 4 440 .00
A t un ing fork is no rm al ly used t o set t he p i tch
Q 1
Given - soundintensi ty
decreases bya fact or o f 5 in4 s
M idd le C
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sou nd in t ensi ty is pro po r t ion al to t he energyo f osci llat ion
4
4
(0)
eE(0)eE(0)e5
10.4sec4
ln5
Energy loss du e t o heat ing of m etals
7004.0
)440(2 Q1
t0eEtE
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In an exp er im ent , a p ap er w eigh t su sp en dedf rom a hef t y rubber band had a per iod o f
1.2 s and t he am p lit ude o f oscillat io ndecreased by a fact o r o f 2 af t er t h reeper iods. W hat is t he est im at ed Q o f t he
system?
2
t
AeA(t)
Solution
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3.6.6
(0)
Ae
Ae2 ln21.8
10.39s
is sam e bu t low er Q fo r t he rubber band
Th is is because o f t he h igher f requency o f
t h e t un ing fo rkIt goes t h rough m any cycles in a given t im ean d lo ses less o f it s en ergy p er cycle
131.2*0.39
2
Q 1
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Fo rce d H arm o nic O scil lat or
U nd am pe d Forced O scillato rEqu . of m ot ion o f a SHO
tcosFkxxm 0
kxxm
Driving for ce
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tcosFkxxm 0
How t o so lve?
Try t he solut ion
tcosAx
RHS o f eq u . has co st
LHS o f eq u . m ust also have co st
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tcosFkxxm 0
tcosAx
Equ . o f m ot ion
solut ion
tFtAkm o coscos2
2mk
FA o
22
1
o
o
m
FA
-
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The solut ion is tm
Fx
cos
122
0
0
Incom plete so lut ion ???
No arb it rary con stant s
M ust ab le to specify x0 and v0
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Com plete so lut ion is
)tBcos(cos
1
m
Fx 022
0
0
t
St eady st ate
solut ion
General solut ion ofund am ped oscil lato r
02
0 xx
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Resonance22
1
o
o
m
FA
0A
= 0 A is f in i te
A 0Resonance
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0 20 40 60 80 100
-0.010
-0.005
0.000
0.005
0.010
0
A
22
1
o
o
m
FA
00;A
00;A
Th e d isp lacem en t is o pp osit e t o t h e d irect io no f t he fo rce!
Th ere is a ph ase d if feren ce o f bet w een t hed isp lacem en t an d t h e ap p lied fo rce
0;A -ve A ?
The ph eno m eno n of reson ance has bo t h ve
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The ph eno m eno n of reson ance has bo t h +veand ve aspect s
+ve aspectsSm all dr iv ing for ce gives large am p l it udeTun ing radios t o t he desi red f req uen cy
-w ave ovenf ood w ith no w ate rcon t en t cannot beheated
Applied -w ave f requ ency is equ al to t heH2O m olecu les (no n zero d ipo le m om ent )natu ral f requen cy
ve aspects
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-ve aspects
To red uce response at resonance d issipat ivef r ict ion fo rce is needed - Forced Dam pedHarm on ic Osci llat o r
R
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Recap
Q 01
tcosFkxxm 0
solut ion tm
Fx
cos
122
0
0
For ced Un dam ped osci llat or
Q facto r o f DHO
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Com plete so lut ion is
)tBcos(cos
1
m
Fx 022
0
0
t
St eady st ate
solut ion
General solut ion ofund am ped oscil lato r
02
0 xx
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22
1
o
o
m
F
A
0 20 40 60 80 100
-0.010
-0.005
0.000
0.005
0.010
0
A
-ve A ?
d isp lacem ent is opposit e t o t he d irect ion o ft h e fo rce!
Th ere is a ph ase d if feren ce o f bet w een t hed isp lacem en t an d t h e ap p lied fo rce
tcosFkxxm 0
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To day s t o p ics
Forced dam ped harm on ic osci llat or
Equ . Of m ot ionSolut ionResonance
EnergyQu ali ty fact or
F d D d H i O ill
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Forced D am pe d H arm on ic O scillato r
Und am ped FHO drivingspring FFF
tcosFkxxm 0
SHM springFF kxxm
Act ual m ot ion is t he sup erpo sit ion ofosci llat ion s at t w o f requ encies and 0
Transient beh avior
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drivingviscousspring FFFF Dam ped FHO
tcosFbv-kxxm 0
In t he ini t ial st age t ransient st ate exist s
Af t er a suf f icient ly lon g t im e t he nat ura losci llat ion s dies ou t because of t he d am pin gforce
No w t he o sci llat or osci llat es at t he f requ ency
of t he d r iv ing force St eady st ate
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tmFxxx oo cos
2
Wi l l x =A cost sat isf y t h is d i f feren t ial equ .?No !
Th e veloci ty term gives sin t
tcosFbv-kxxm 0
tcosm
Fxm
kxm
bx 0
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t
m
Fxxx oo cos2
How to f ind t he so lu t ion?
W r ite t he above equat ion in com plex form
ti
em
F
zzz
02
0
Solu t ion w ill be o f t he form z = zoeit
i t
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Real part o f z = zoeit gives t he solut ion t o
Forced dam ped harm on ic osci llat or
Subst i tut ing z = zoeit in com plex equ at ion
imFz
em
Fiez titi
22
0
00
02
0
2
0
1
)(
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i
1
m
Fz
22
0
00
Inpolar
fo rm
2
0
2
1
2
1
2222
0
0*
00
0
tan
)()(
1
Re
m
FzzR
z i
2222
0
22
00
)()(
i(
m
F
)
2222
0
i
0
)()(
e
m
F
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Real p ar t
The com plete so lut ion is z = zoeit
titii ReeRez
)cos( tRx
1/22222oo
1
m
FRA
22
1tan
o
Phase d i f ference b et w eent he d r iving forceand t he d isp lacem ent
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1/2
2222
o
o
1
m
FRA
A is con st ant for a given freq uen cy
0dt
dA
2
1
20m 2Q
11
At = m ax
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For ligh t d am p in g, A is m axim um fo r = oand t he am p lit ude at resonance is:
o
oo
m
FA )(
The behavio r o f A and as funct ions o f ,d ep en ds o n t he rat io / o
1/22222oo
1
m
FRA
1F
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10
1
Q 01
1/22222oo
1
m
FA
2
1
20m 2Q
11
1
o 1FA
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As increases, t he m axim um am pl it ud e occursat a f requ ency less t han t he reson ant f req uen cy
1
0
2
20m 2Q
11
1/22222o m
A
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1
-
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22
1tano
Undam ped FHO Dam ped FHO
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0 20 40 60 80 100
-0.010
-0.005
0.000
0.005
0.010
0
A
Undam ped FHO Dam ped FHO
1/2
2222
o
o
1
m
FA
22
1tano
22
1
o
o
m
F
A
Energy
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gy
For st eady st at e m ot ion am pl i tu de is con st ant
in t im e tAx cos
tAv sin)sin(
2
1
2
1)( 222 tAmmvtK
)(cos2
1
2
1)( 222 tkAkxtU
22
4
1AmK
22
4
1AkU
1
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2222
222
)(41
o
oo
mFE
)(mA41E 20
22
22
4
1AmK
22
4
1AkU
St ead y st ate
Light D am ping
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Light D am ping
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Resonance cu rve o r loren t zian
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22
2
2/
1
8
1
o
o
m
F
E
max imum
height 2
4
Falls t o on e hal f m axim um
2
2
02
2
0
Recap
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For ced d am ped harm on ic osci llat or
tcosFbv-kxxm 0
Recap
Equ . Of m ot ion
tAx cosSt eady st ate so lut ion
1/22222oo
1
m
FRA
22
1tano
o 1FRA
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A is con st ant for a given freq uen cy
2
1
20m2Q
11
1/22222oo
mRA
1
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22
1tano
2
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St eady st ate
22
2
2/
1
8
1
o
o
m
FE
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Tod ay s t op ics
Pow er absor bed by an o sci llat orSim ilar it y betw een th e pow er andenergy cur vesQ fact or calcu lat ion f rom reson ancecurves
Oscil lat ion s invo lving m assive sp rin gs
Pow er ab sorb ed by an oscillat or
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y
How to m ain ta in t he am pl it ude o f a fo rcedharm on ic osci llat or con st ant ly?
Rat e at w hich energy is sup pl ied t o a dr ivenosci llat or t o m ainta in i t s am pl it ud econst ant ly is
Fvdt
dxF
dt
dwP
U n d a m p e d FH O
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tcosAx 221
o
o
m
FA
w h e r e
tsinAx v
t2sin2
AF-tcostsinAFP 00 Fv
tcosFF 0 Driving for ce
&
p
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t2sin
2
AFP 0 Fv
t2sinP
0P Energy is fed int o t he syst em in o ne hal fcycle and is t aken out again du rin g nexthalf cycle
D a m p e d FH O
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p
)(Ax tcos
2/12222 ])()[(
1
o
o
m
FA
)(Av tsin
)(sintcos tAFFvP 0
)i (P AFF
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)sin(cosP 0 ttAFFv
tcos)sinAF(
tcostsin)cosAF(
2
0
0
Average value is zero
sinAF21P 0
1
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sin
2
1 0AFP
22220
22
0
1
m
F
2
1P
For light dam ping 0 =
2
2
0
2
0
2
1mF
81P
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220
2
0
/2
1
m
F
8
1
P
222
2/1
81
oo
mFEAverage en ergy is
Resonance cu rve o r loren t zian
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max imumheight 2
4
Falls t o on e hal f m axim um
2
2
02
2
0
220
2
0
/2
1
m
F
8
1P
22
2
2/
1
8
1
o
o
m
FE
half m axim umEP
o r
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E o r P
2
0
0- +
= Ful l w idt h athal f
m axim u m /resonancew i d t h
ha lf m axim um2
W id t h o f t he cu rve
2
o r
Q ua lity facto r
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Q y
curveresonanceofwidthfrequency
frequencyresonance
oQ
Gives t he f requ ency select ive pro per t y of
an o scil lat o r
Sharp ness o f resonance
cu rve m eans t he syst emw il l no t respo nd u nlessdr iven very n ear i ts reson ance freq uen cy
Q = 10 is m ore select ive
Respo nse in t im e vs respo nse in f req ue ncy
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50 100 150
0.00
0.02
0.04
0.06
0.08
0.10
0.12
E
FWHM
P
Oscillat o rs w h ich are very f requency
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select ive also h ave w eak d am p in g.
So su ch an o scillat o r do es no t recover f ro m ad ist u r ban ce o r d o es n o t resp o nd q u ickly.
Th e d am p in g t im e an d t h e reso n an ce cu r vew id t h obey 1
Th is relat ion is closely relat ed t o one fo rm o ft he He isenberg Uncer tain ty p r incip le
1
O scillat ion s invo lving m assive sp rings
-
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mk M
To t al energy is a const ant
2
2
1kxU
E = K + U = Const an t
K = K spr ing + K mass
l
0dt
dE
How t o calculat e t he KE of t he spr ing ?
W hat is t he f requen cy of oscil lat ion ?
Assumptions
-
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The sp ring osci llat ion s are n o t so large t hat
t hey cause t he spr ing coi ls t o b um p int o eacho the rSt ret ch ing for ce is sam e at all p o int s along
t he spr ingAll th e po int s in t he spr ing un dergodisp lacem ent s pro po r t ional to t he i r d ist ances
fro m f ixed end St at ic ext ensionVeloci ty is t he sam e for a ll t he e lem ent s oft he spr ing
l
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l/ 3 l/ 3 l/ 3
m
m
x
3
2x
3
x
ms ds
dsl
MdM
2
2
1(dM)dvdK
dt
dx
l
sdv
xl
s
Displacemento f ds
2
d
-
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2
2
2
1
2
1
dt
dx
l
sds
l
M(dM)dvdK
dss
dt
dx
l
MdK
2
2
3
2
l
spring dssdt
dx
l
MK
0
2
2
32
2
6
dt
dxMKspring
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E = PE spr ing + KE spr ing + KE mass
22
2
2
1
62
1
dt
dxm
dt
dxMkxE
0dt
dE
3
Mm
k
Sup pose m = 0
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3Mm
k
p p
Mk3
Th e ab ove calcu lat io n is n o t exact W hy?Because of t he assum pt ions(i) Ext en sio n o f t h e sp r in g is p ro p o rt io n al
t o t he d istance f rom t he f i xed end(ii) Ve locit y (dx/ d t ) is th e sam e fo r all t h e
e lem ent s o f t he spr ing
-
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Is on ly an appr oxim at ion
It w il l ho ld i f M
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Recap
Pow er absor bed by an o sci llat or0P Und am ped FHO
220
2
0
/21
mF
81P
22
2
2/
1
8
1
o
o
m
FE
Average en ergy is
Dam ped FHO
Q fact or calcu lat ion f ro m reson ance cur ves
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Oscil lat ion s invo lving m assive sp rin gs
E o r P
0- +
= Ful l w idt h athalfm axim u m /
resonancew i d t h
mM ,k ,l
3
Mm
k
Tod ay s t op ics
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y p
Vibrat ion e l im inato r
Sup erp osi t io n pr incip le
Sup erp osi t io n o f v ibrat ion s of equ alf renquency
Sup erp osi t io n o f v ibrat ion s of d i f ferentf renquency
Exam p le 1 0 .5 V ib rat io n Elim in at o r
-
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Pist ons suppo rt edby air p ressu re
To red uce t he ef fect of f loo r v ibrat ion s on a
del icate app arat us such as sen sit ive b alanceo r o pt ical syst em
M any v ibrat ion el im inat ors use spr ings
-
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Aut om ob ile vib rat ion e l im inator
inst ead o f air suspen sion
The form of equ at ion o f m ot ion is sam e forair or spr ing v ibrat ion el im inat ors
Area o f t he air co lum n = A Pat m
-
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Area o f t he air co lum n = A
At st at ic equ i libr ium APMgAP atm0
h
M
P0
a t m
Mos t l y APMg atm MgAP0
Equ i libr ium heightM ass i t sup po rt s = M
Now con sider d isp lacem ent o f M
-
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Equ i libr ium height
Displacem ent of M
from equ i lib r ium
Disp lacem ent o f t he low er end o f tab le leg
h
x
y
M
Iner t ial f ram e
Equat ion o f m ot ion o f M MgPAxM
-
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Equat ion o f m ot ion o f M MgPAxM
Inst ant aneo us pr essu re
yh
gx
h
gx
Aft er sim pl i f icat ion
The f lo or v ibrates tyy cos0
-
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yy 0
tyh
gx
h
gx cos0
Un dam ped forced oscil lat or
So lu t io n o f t he eq uat io n is x = xo cost
h
gand
yx 022
0
2
000
2x
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The ob ject o f t he air suspension is t o m aket he rat io x
0/ y
0as sm all as possib le
22
o
o
o
o
y
x
For > 2oox
2 oo
x
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For >> o 2oy
Fo r t h e vib rat io n elim in at o r t o b e su ccessf u l -The reson ance freq uen cy m ust be lowcom pared t o t he dr iv ing f requ ency
h
g0 Table m ust have long legs!
22 ooy
Am pl it ud e of v ibrat ion s is redu ced
0 Vibrat ion am pl if ier222
oox
-
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To avo id t h is prob lem , dam p ing is requ iredby m eans o f a dashpo t
h
M
x
y
Dashpot
0 Vibrat ion am pl if ier22 ooy
Damping
t er m b vw here v isrelat ive
velocit y ofi ts end s
yxb
2
1
24 (x
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2
222
0
0
0
0
(
(
y
x
To m ake x0sm all relat ive
t o yo , reduce0 below
Pract ical airt ables have 0o f 1 Hz o r less
Co il Sp r ings are used int b il t i l t t h
-
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au t om ob iles t o iso lat e t he
chassis f rom road v ib rat ions.
Dam p ing is p rovided by Shock
absor bers (dashp ot s).
Fo r a sm o o t h r ide, one shou ld
have a m assive and w eakspr ings (sm all k) so t hat t heresonance f requency is low .
-
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