10. the harmonic oscillator

7/31/2019 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

)(


94/147

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


101/147

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


103/147

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


105/147

Light D am ping


106/147

Resonance cu rve o r loren t zian


107/147

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


109/147

A is con st ant for a given freq uen cy

2

1

20m2Q

11

1/22222oo

mRA

1


110/147

22

1tano

2


111/147

St eady st ate

22

2

2/

1

8

1

o

o

m

FE


112/147

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


113/147

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


114/147

tcosAx 221

o

o

m

FA

w h e r e

tsinAx v

t2sin2

AF-tcostsinAFP 00 Fv

tcosFF 0 Driving for ce

&

p


115/147

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


116/147

p

)(Ax tcos

2/12222 ])()[(

1

o

o

m

FA

)(Av tsin

)(sintcos tAFFvP 0

)i (P AFF


117/147

)sin(cosP 0 ttAFFv

tcos)sinAF(

tcostsin)cosAF(

2

0

0

Average value is zero

sinAF21P 0

1


118/147

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


119/147

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


120/147


121/147

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


122/147

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


123/147

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


124/147

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


125/147

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


126/147

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


127/147

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


128/147

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


129/147

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


130/147

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


131/147

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


132/147

Is on ly an appr oxim at ion

It w il l ho ld i f M


133/147

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


134/147

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


135/147

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


136/147

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


137/147

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


138/147

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


139/147

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


140/147

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


141/147

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


142/147

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


143/147

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


144/147

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


145/147

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


146/147

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 .


147/147

http: / /www.acoust ics.sal ford.ac.uk/feschools /waves/shm4.htm

10. the harmonic oscillator

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