Chapter 5 French

7/30/2019 Chapter 5 French

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Chapter V

Coupled Oscillations



Two, or more, oscillators coupled together

Examples

1. Two pendulums, coupled by a spring



2. A molecule

Molecule2CO



3. A crystalline solid



~

4. Coupled electrical oscillators



5. An elastic medium, like a vibrating string

A wave in a medium is a continuum of

coupled oscillators



Two Simple Pendulums Coupled by a Spring

1x 2x

m mk

Equations of Motion :

)xx(k x

gmdt

xdm 12

1

2

1

2

)xx(k x

gmdt

xdm 12

2

2

2

2



0)xx(

m

k x

g

dt

xd1212

1

2

0)xx(m

k x

g

dt

xd1222

2

2

The above two differential equations are

coupled.

Adding and subtracting the above equations :Decoupling

0qdt

qd

1

2

02

1

2

0qdt

qd

; 2

2

2

2

2



122211 xxq;xxq

Where,

m

k 2;

g 2

0

22

0

The solutions for are :21 q&q

)t(cosAq 1011

)tcos(Aq 222



21 x&x can then be obtained as :

)qq(2

1x 211 )qq(

2

1x; 212

21 q&q are known as the normal

coordinates and are known as normal

frequencies.

&0

Initial Conditions on are obtained

from those of 21 x&x21 q&q



Normal Coordinates : New coordinates, that

are linear combinations of the original, inwhich the equations of motion are

decoupled.

Normal Frequencies : Frequencies of

oscillation of the normal coordinates.



Let the initial conditions be such that

remains zero forever.2

q

0)0(q)0(q 22

Or, )0(x)0(x;)0(x)0(x 2121

Normal Mode Vibrations

Thus, 0)t(q&)tcos(A)t(q 21011

For this to happen, the initial conditions

must be such that



0)t(q&tcosA)t(q 201

tcos2

A)t(x)t(x 021 And,

Taking :

0)0(x)0(x;2A)0(x)0(x 2121



This mode of vibration is called a Normal

Mode Vibration

Normal Mode Vibration : A mode of vibration inwhich only one normal coordinate is excited,

the other normal coordinates remaining zero.

First Mode for the Coupled Pendulums

0t 4Tt 0



2Tt 0 4T3t 0

0Tt



Second normal mode vibration :

0)t(q1

)0(x)0(x&)0(x)0(x 2121

Required initial conditions :

0)0(q;0)0(q 11

Or,



Complete solution :

tcosA)t(q2

1)t(x

21

0)0(x)0(x;A)0(x)0(x 2112

0)0(q;A2)0(q 22

tcosA2)t(q2

tcosA)t(q2

1)t(x 22



0t 4Tt

2Tt 4T3t



Tt



General Motion (Both Modes Excited)

A

Initial Conditions :

A)0(x;0)0(x 21

0)0(x)0(x 21

A)0(q)0(q 21 0)0(q)0(q 21



tcosA)t(q;tcosA)t(q 201

Solutions for 21 q&q

Solutions for 21 x&x

]tcost[cos2

A)t(x 01

t2sint2sinA00

t

2

cost

2

cosA)t(x 002



)t(x2

t

)t(x1

t



Summary

1. There exist normal coordinates, which

are such that, the equations of motion in

them, are decoupled

2. Each normal coordinate behaves likea simple harmonic oscillation with its own

frequency, the normal frequency

3. With appropriate initial conditions, onecan excite only one normal coordinate, the

other remaining dormant. Such vibrational

modes are called normal mode vibrations.



4. In a normal mode vibration, each mass in

the coupled system, executes a SHO with

the same frequency, the corresponding

normal frequency. The amplitudes of motion

of the different masses, and their phases

are in general different.

5. In the most general motion, which results

from arbitrary initial conditions, the motion of each mass is rather complicated. There is no

definite frequency of vibration. However, the

motion is a superposition of SHMs



No Class on Saturday

Combined Class for Both Sections

On

Sunday (25/11)

At

10.00 AM

InRoom No. 5102



General Approach for Finding Normal Modes

Back to Coupled Pendulums :

0)xx(x

dt

xd12

2

s1

2

02

1

2

0)xx(xdt

xd12

2

s2

2

02

2

2

Since in a normal mode vibration, all

masses execute SHM of a common

frequency, put :

mk ;g 2

s20



tcosAx;tcosAx 2211

0AA)( 2

2

s1

22

s

2

0

0A)(A 2

22

s

2

01

2

s

In the matrix form :

0A

A

2

1

22

s

2

0

2

s

2s

22s

20



For non-trivial solutions for

( ), the determinant of

the matrix must be zero :

21 A&A

zeronotA&ABoth 21

04

s

222

s

2

0

Or,2

s

2

00 2or

First Normal Mode

1A

A

2

101



tcosAx;tcosAx 1211

Second Normal Mode

1A

A2

2

12

s

2

02

tcosAx;tcosAx 2221

Normal Coordinates

0xx 21

First Mode : 0xx 21

Second Mode :



Since, in the second normal mode vibration,

the first normal coordinate is identically zero

211 xxq

Since, in the first normal mode vibration, thesecond normal coordinate is identically zero

122 xxq



Prob. 5.10

m

m

k

k

Consider the vertical motion

of the system.

a) Find the normal modefrequencies and the ratio of the

amplitude of the two masses in

each mode



1x

2x

m

m


)xx(k xk td

xdm 1212

12

)xx(k td

xdm 1222

2

Or,

0)xx2(td

xd21

2

02

1

2

0)xx(td

xd12

2

02

2

2



Substituting : tcosAx;tcosAx 2211

0AA)2( 2

2

01

22

0

0A)(A 2

22

01

2

0

Writing the above as a matrix equation and

setting the determinant to zero, we get :

02

22

0

2

0

2

0

22

0



Or,2

0

2

2

53

We have : 22

0

2

0

2

1

2A

A

First Mode :15

2

A

A

2

1

Second Mode :15

2

A

A

2

1



2 cm

1.23 cm

2 cm

3.23 cm

First Mode

Second Mode



b*) Find the normal coordinates

First Mode : 0x2x15 21

Second Mode :

211 x2x15q

212 x2x15q

0x2x1521



c*) If the lower mass is pulled down a

distance A, while the upper mass is held

fixed, and the system released, describe thesubsequent motion.

0)0(q)0(q;A2)0(q;A2)0(q 2121

)t(cosA)t(q 1111

We have :

)t(cosA)t(q2222

Initial Conditions :



)tcos(B2)tcos(A2)t(x 22111

)tcos(B)15()tcos(A)15()t(x 22111

One can directly write down the complete

solutions for , without requiring to

obtain the normal coordinates .

21 x&x

21 q&q

When B is zero, the system oscillates in

the first mode with the requisite ratio of

amplitudes, and, so is the other way

around.



tcosA2)t(q 11

tcosA2)t(q 22

Now,

)qq(52

1x 211

212 q)15(q)15(54

1x



Prob. 5.9 The carbon

dioxide molecule can be

likened to a central

mass connected to two

other identical masses,

by identical springs of spring constant k.

a) Set up the equations of motion, find thenormal frequencies and ratios of the

amplitudes in the normal modes

2m

1m1m




)xx(k dt

xdm 122

1

2

1

)xx(k )xx(k dt

xdm 231222

2

2

)xx(k

dt

xdm 232

3

2

1

1x2x 3x



Or,0)xx(

m

k

dt

xd21

1

2

1

2

0)xx2x(

m

k

dt

xd321

2

2

2

2

0)xx(m

k

dt

xd23

1

2

3

2

Substituting :

tcosAx;tcosAx;tcosAx 332211



0Am

k A

m

k 2

1

1

2

1

0Am

k A

m

k 2A

m

k 3

2

2

2

2

1

2

0Am

k A

m

k 3

2

1

2

1

Setting the determinant of the coefficient

matrix to zero, the normal frequencies are

obtained as :



21

3

1

21m

2

m

1k ;

m

k ;0

From the equations for the coefficients :

1

2

11

321m

k :

m

k :

m

k A:A:A

First mode

1:1:1A:A:A 321

No oscillations, rigid shift of the molecule



Second Mode

1:0:1A:A:A 321

AAA



Third Mode

1:m

m2:1A:A:A

2

1321

1:7.2:1



Coupled and Driven Oscillators

tcos

m

F)xx(x

dt

dx

dt

xd 012

2

s2

2

02

2

2

2

0)xx(xdtdx

dtxd 12

2s1

20

121

2



Adding and subtracting the equations

tcosm

Fq

dt

dq

dt

qd 01

2

01

2

1

2

tcosmFq

dtdq

dtqd 0

222

22

2

Steady-state solutions of the above

equations are :

)t(cos)(A)t(q 111 )t(cos)(A)t(q 222



)t(cos)(A)tcos()(A2

1)t(x 22111

)t(cos)(A)tcos()(A2

1)t(x 22112

)(A1 )(A2

0



When the driving frequency matches one of

the two frequencies , the

displacements of the two pendulums

become large.

&0

There are, thus, two resonant frequencies,corresponding to the two normal frequencies

at which the system can oscillate

At the lower resonance frequency, the

two pendulums are in phase. At the

higher one, they are totally out of phase.



Prob. 5.12 Two identical masses are

connected to three identical springs on a

frictionless surface as shown

The free end is driven with a displacement :

tcosXX 0

Find and draw the graphs of the displacements

of the two masses.

X 1x 2x

mm kkk



X 1x 2x

mm kkk


)xx(k )Xx(k dt

xdm

1212

1

2

2122

2

2

xk )xx(k dt

xdm

Or, tcosXxx2dt

xd0

2

02

2

01

2

02

1

2

0x2xdt

xd

2

2

01

2

02

2

2



Addition and subtraction of the two equations

gives us :

tcosm

Fq

dt

qd 01

2

02

1

2

tcosm

Fq3

dt

qd 02

2

02

2

2

Where, 00212211 Xk F,xxq,xxq



Steady-state solution :

tcos)](A)(A[2

1)t(x 211

tcos)](A)(A[2

1)t(x 212

where,

22

0

0

2

0222

0

0

2

01

3

X)(A;

X)(A



tcos)3()(

2X)t(x

22

0

22

0

22

00

2

01

tcos

)3()(

1X)t(x

22

0

22

0

0

4

02

At a driving frequency , the mass

nearer the end being driven, becomesstationary

02

Indefinite Number of Coupled Oscillators



Indefinite Number of Coupled Oscillators

N Light Beads Connected by Massless Rods

Equilibrium

General State of Vibration

m

T



Equations of Motion

p P+1P-1

py1py

1py

1

2

)sinsin(Tdt

ydm 122

p

2

)]yy()yy[(T

1ppp1p

Or,0)yy2y(

dt

yd

1pp1p

2

02

p

2

TT

m

T2

0



0)yy2y(dt

yd1pp1p

2

02

p

2

N.........,,2,1p

It is assumed that there are two mass points,

at and , which are

permanently fixed.

Normal Modes

Substitute : tcosAy pp 0AA)2(A 1p

2

0p

22

01p

2

0

0p 1Np

Boundary Conditions : 0AA 1N0

0)t(y)t(y 1N0



2

0

22

0

p

1p1p 2

A

AA

The amplitudes must depend upon the

discrete index in such a manner that the

ratio on the left above is independent of p

and the boundary condition is satisfied.

pA

p

A clever guess :

0AA 1N0

)tstancons,&C(psinCAp

cos2A

AA

p

1p1p



For the boundary conditions to be satisfied,

we must have :

0)1N(sin N........,,2,1n)1N(

n

Each of the above possible values for

leads to a normal mode frequency :

2sin2 0

)1N(2

nsin2 0n N........,,2,1n



The amplitudes of the mass point in

the normal mode is :

1N

npsinCApn

thpthn

N,........,1n;N.......,,1p

px)1N(

xnsinC

pxL

xnsinC

L

nxsinC)x(A,where),p(AA nnpn



The First Mode

)1N(2

sin2 01

L

xsinC)x(A1



The Second Mode

Lx2sinC)x(A2

)1N(sin2 02



Three Particles



General Solution

N

1n

nnnp )t(cos1N

npsinC)t(y N.......,,1p



Prob. 5.16 (Resonance in N coupled

Oscillators)

Consider a system of N coupled oscillators

driven at a frequency . Forcing is done at

the extreme end such that

tcosh)t(y 1N

Find the resulting amplitudes of the particles.



In the steady state, each particle will oscillate

with the driving frequency.

tcos)(Ay pp

Here, is the driving frequency, andhence, given

Substituting the above into the equations of

motion, we get :

2

0

22

0

p

1p1p 2

A

AA

Boundary Conditions : hA;0A 1N0



Solution : psinCAp

Second boundary condition requires :

h)1N(sinC

)1N(sin

hC

2

0

22

0

p

1p1p 2cos2

A

AA

)1N(sin

)psin(h)(Ap 2

0

22

0

2

2cos

, Where,



The amplitudes blow up whenever

0)1N(sin

Or,1N

n

This happens when the driving frequency is

such that :

2

0

22

0

2

2

1N

ncos

Or,

)1N(2

nsin2 0

A li d f ill i b l h



Amplitudes of oscillations become large when

the driving frequency is one of the normal

mode (natural) frequencies

N

1n

nnnp )t(cos1N

npsinC)t(y

Most general solution of the driven oscillators :

tcos)1N(sin

)psin(h

Transient Steady State

Chapter 5 French

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