Mecflu - oscila??es juntadas, Exercícios de Meteorologia
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Mecflu - oscila??es juntadas, Exercícios de Meteorologia

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1-1

CHAPTER 12 Coupled Oscillations

12-1.

m1 = M

k1

x1

k12 k2

m2 = M

x2 The equations of motion are

( )

( )

1 1 12 1 12 2

2 2 12 2 12 1

0

0

Mx x x

Mx x x

κ κ κ

κ κ κ

+ + − =  + + − = 

(1)

We attempt a solution of the form

( )

( )

1 1

2 2

i t

i t

x t B e

x t B e

ω

ω

=  = 

(2)

Substitution of (2) into (1) yields

( )

( )

2 1 12 1 12 2

2 12 1 2 12 2

0

0

M B B

B M B

κ κ ω κ

κ κ κ ω

+ − − =  

− + + − = 

(3)

In order for a non-trivial solution to exist, the determinant of coefficients of and must vanish. This yields

1B 2B

( ) ( )2 21 12 2 12 12M Mκ κ ω κ κ ω κ+ − + − = 2   (4) from which we obtain

( )21 2 12 1 2 12 2 1

4 2 2M M

κ κ κ2 2ω κ κ κ + +

= ± − + (5)

This result reduces to ( )2 12 12 Mω κ κ κ= + ± for the case 1 2κ κ κ= = (compare Eq. (12.7)].

397

398 CHAPTER 12

If were held fixed, the frequency of oscillation of m would be 2m 1

(201 1 12 1 M

)ω κ κ= + (6)

while in the reverse case, would oscillate with the frequency 2m

(202 2 12 1 M

)ω κ κ= + (7)

Comparing (6) and (7) with the two frequencies, ω+ and ω− , given by (5), we find

( )22 21 2 12 1 2 12 1

2 4 2M

κ κ κ κ κ+  = + + + − +  

ω κ

( ) ( ) 21 2 12 1 2 1 12 0 1 1

2 2M M

κ κ κ κ κ κ κ ω + + − = + =  1> + (8)

so that

01ω ω+ > (9)

Similarly,

( )22 21 2 12 1 2 12 1

2 4 2M

κ κ κ κ κ−  = + + − − +  

ω κ

( ) ( ) 21 2 12 1 2 2 12 0 1 1

2 2M M

κ κ κ κ κ κ κ ω + − − = + =  2< + (10)

so that

02ω ω− < (11)

If , then the ordering of the frequencies is 1κ κ> 2

01 02ω ω ω ω+ −> > > (12)

12-2. From the preceding problem we find that for 12 1 2,κ κ κ

1 12 2 121 2;M M κ κ κ

ω ω + +

≅ ≅ κ

(1)

If we use

101 02;M M 2κ κω ω= = (2)

then the frequencies in (1) can be expressed as

COUPLED OSCILLATIONS 399

( )

( )

12 1 01 01 1

1

12 2 02 02 2

2

1 1

1 1

κ ω ω ω ε

κ

κ ω ω ω ε

κ

 = + ≅ + 

  

= + ≅ +  

(3)

where

12 121 2 1 2

; 2 2 κ κ

ε ε κ κ

= = (4)

For the initial conditions [Eq. 12.22)],

( ) ( ) ( ) ( )1 2 1 20 , 0 0, 0 0, 0 0x x x= = = ,=x D (5)

the solution for ( )1x t is just Eq. (12.24):

( ) 1 2 1 21 cos cos2 2 D t tx t

ω ω ω ω+ −  =        (6)

Using (3), we can write

( ) ( )1 2 01 02 1 01 2 02 2 2

ω ω ω ω ε ω ε ω

ε+ +

+ = + + +

≡ Ω + (7)

( ) ( )1 2 01 02 1 01 2 02 2 2

ω ω ω ω ε ω ε ω

ε− −

− = − + −

≡ Ω + (8)

Then,

( ) ( ) ( )1 cos cosx t D t t t tε ε+ + − −= Ω + Ω + (9) Similarly,

( )

( ) (

1 2 1 2 2 sin sin2 2

sin sin

x t D t t

D t t t+ + − −

+ −

)t

  =      

= Ω + Ω +

ω ω ω ω

ε

ε (10)

Expanding the cosine and sine functions in (9) and (10) and taking account of the fact that ε+ and ε− are small quantities, we find, to first order in the ε’s,

( )1 cos cos sin cos cos sinD t t t t t t t tε ε+ − + + − − +≅ Ω Ω − Ω Ω − Ω Ω  x t (11) −

( )2 sin sin cos sin sin cosD t t t t t t t tε ε+ − + + − − +≅ Ω Ω + Ω Ω + Ω Ω  x t (12) − When either ( )1x t or ( )2x t reaches a maximum, the other is at a minimum which is greater than zero. Thus, the energy is never transferred completely to one of the oscillators.

400 CHAPTER 12

12-3. The equations of motion are

2 1 2 0 1

2 2 2 0 2

0

0

m x x x

M

m x x x

M

ω

ω

+ + =    + + = 

(1)

We try solutions of the form

( ) ( )1 1 2 2;i t i tx t B e x t B eω ω= = (2)

We require a non-trivial solution (i.e., the determinant of the coefficients of B and equal to zero), and obtain

1 2B

( ) 2

22 2 4 0 0

m M

ω ω ω  − −    = (3)

so that

2 2 20 m M

ω ω ω− = ± (4)

and then

2

2 0

1 m M

ω ω =

± (5)

Therefore, the frequencies of the normal modes are

2 0

1

2 0

2

1

1

m M

m M

ω ω

ω ω

= +

= −

(6)

where 1ω corresponds to the symmetric mode and 2ω to the antisymmetric mode.

By inspection, one can see that the normal coordinates for this problem are the same as those for the example of Section 12.2 [i.e., Eq. (12.11)].

12-4. The total energy of the system is given by

( ) ( ) ( )22 2 2 21 2 1 2 12 2 11 1 12 2 2

E T U

M x x x x x xκ κ

= +

= + + + + − (1)

Therefore,

COUPLED OSCILLATIONS 401

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

1 1 2 2 1 1 2 2 12 2 1 2 1

1 1 12 2 1 1 2 2 12 2 1

1 12 1 12 2 1 2 12 1 12 2 2

dE M x x x x x x x x x x x x

dt

Mx x x x x Mx x x x x

Mx x x x Mx x x x

= + + + + − −

    = + − − + + + −    

  + − + − + +  

κ κ

κ κ κ κ

κ κ κ κ κ κ

2

= + (2)

which exactly vanishes because the coefficients of and are the left-hand sides of Eqs. (12.1a) and (12.1b).

1x 2x

An analogous result is obtained when T and U are expressed in terms of the generalized coordinates 1η and 2η defined by Eq. (12.11):

( )2 21 214T M η η= + (3)

( )2 21 2 121 14 2U 2 1κ η η κ η= + + (4)

Therefore,

( ) [1 12 1 1 2 2 2 dE

M M dt

]2 2η κ κ η η η κη η ⋅ = + + + +  (5)

which exactly vanishes by virtue of Eqs. (12.14).

When expressed explicitly in terms of the generalized coordinates, it is evident that there is only one term in the energy that has as a coefficient (namely, 12κ

2 12 1κ η

1

), and through Eq. (12.15) we see that this implies that such a term depends on the ’s and 1C ω , but not on the ’s and 2C 2ω .

To understand why this is so, it is sufficient to recall that 1η is associated with the anitsymmetrical mode of oscillation, which obviously must have 12κ as a parameter. On the other hand, 2η is associated to the symmetric mode, ( ) ( )1 2x t x t= , ( ) ( )1 2x t x t= , in which both masses move as if linked together with a rigid, massless rod. For this mode, therefore, if the spring connecting the masses is changed, the motion is not affected.

12-5. We set . Then, the equations of motion are 1 2 12κ κ κ= = ≡κ

1 1 1 2

2 2 2 1

2 0

2 0

m x x x

m x x x

κ κ

κ κ

+ − =   + − = 

(1)

Assuming solutions of the form

( )

( )

1 1

2 2

i t

i t

x t B e

x t B e

ω

ω

=  = 

(2)

we find that the equations in (1) become

402 CHAPTER 12

( )

( )

2 1 1 2

2 1 2 2

2 0

2 0

m B B

B m B

κ ω κ

κ κ ω

− − =  

− + − = 

(3)

which lead to the secular equation for 2ω :

( ) ( )2 21 22 2m m 2κ ω κ ω κ− − = (4) Therefore,

2 1 2

3 1 1

m m κ µ

ω µ  

= ± − +  (5)

where ( )1 2 1 2m m m mµ = + 1 2=

is the reduced mass of the system. Notice that (5) agrees with Eq.

(12.8) for the case m m and M= 12κ κ= . Notice also that 2ω is always real and positive since

the maximum value of ( )1 m+ 23 mµ is 3 4 . (Show this.) Inserting the values for 1ω and 2ω into either of the equations in (3), we find

1

1 11 21 1 2

3 2 1 1

m a a

m m µ

µ

   − + − =  +   

 (6)

and

112 22 1 2

3 2 1 1

m a

m m µ

µ

   = − − −  +   

a (7)

Using the orthonormality condition produces

11 1

1 a

D = (8)

( )

1 2 1 2 2

2 1 2 21

1

3 2 1 1

m m m m m m m

D a

 +  − + −

+  = (9)

where

( ) ( ) ( ) 2

2 21 1 2 1 1 2 2

2 2 1 2

32 2 2 1

m m m m m

m m m m ≡ − + + − −

+ 1 2mD m (10)

The second eigenvector has the components

( )

1 2 1 2 2

2 1 2 12

2

3 2 1 1

m m m m m m m

D a

 +  − − −

+  = (11)

1 Recall that when we use 1ω ω= , we call the coefficients ( )1 1 aβ ω ω 11= = and ( )2 1 aβ ω ω= = 21 , etc.

COUPLED OSCILLATIONS 403

22 2

1 a

D = (12)

where

( )

2 2 2 1 1 1

2 1 2 1 22 2 2 2 2 2 1 2

3 3 2 1 1 2 1 1

m m m m m m

m m m m m

      ≡ + + − + − −     

+      1 2mD m (13)

The normal coordinates for the case in which ( )0jq 0= are

( ) ( )

( ) ( )

1 1 11 10 2 21 20

2 1 12 10 2 22 20

cos

cos

t m a x m a x t

t m a x m a x

1

2t

η ω

η ω

= +

= + (14)

12-6.

m

m k

k

x1 x2 If the frictional force acting on mass 1 due to mass 2 is

( )1 2f x xβ= − − (1) then the equations of motion are

( )

( )

1 1 2 1

2 2 1 2

0

0

mx x x x

mx x x x

+ − + =  + − + = 

β κ

β κ (2)

Since the system is not conservative, the eigenfrequencies will not be entirely real as in the previous cases. Therefore, we attempt a solution of the form

( ) ( )1 1 2 2;tx t B e x t B e tα α= = (3)

where iα λ ω= + is a complex quantity to be determined. Substituting (3) into (1), we obtain the following secular equation by setting the determinant of the coefficients of the B’s equal to zero:

( )22m 2 2α βα κ β α+ + = (4) from which we find the two solutions

( )

1 1

2 2

;

1

i m m

m m

κ κα ω

α β β κ

 = ± = ± 

  = − ± − 

(5)

The general solution is therefore

404 CHAPTER 12

( ) ( )2 21 11 11 12 12i m t i m t m t m m t mt mB e B e e B e B eκ κ β κ β κβ − −+ − − + −= + + +x t (6) and similarly for ( )2x t .

The first two terms in the expression for ( )1x t are purely oscillatory, whereas the last two terms contain the damping factor te β− . (Notice that the term ( )212 expB m tβ κ+ − increases with time if

2 mβ κ> , but is not required to vanish in order to produce physically realizable motion because the damping term, exp(–βt), decreases with time at a more rapid rate; that is

12B +

2β β− + 0κ <m− .)

To what modes do 1α and 2α apply? In Mode 1 there is purely oscillating motion without friction. This can happen only if the two masses have no relative motion. Thus, Mode 1 is the symmetric mode in which the masses move in phase. Mode 2 is the antisymmetric mode in which the masses move out of phase and produce frictional damping. If 2 mβ κ< , the motion is one of damped oscillations, whereas if 2 mβ κ> , the motion proceeds monotonically to zero amplitude.

12-7.

m

k

m

k

x1

x2

We define the coordinates and as in the diagram. Including the constant downward gravitational force on the masses results only in a displacement of the equilibrium positions and does not affect the eigenfrequencies or the normal modes. Therefore, we write the equations of motion without the gravitational terms:

1x 2x

1 1 2

2 2 1

2 0

0

mx x x

mx x x

κ κ

κ κ

+ − =   + − = 

(1)

Assuming a harmonic time dependence for ( )1 tx and ( )2x t in the usual way, we obtain

( )

( )

2 1 2

2 1 2

2 0

0

m B B

B m B

κ ω κ

κ κ ω

− − =  

− + − = 

(2)

Solving the secular equation, we find the eigenfrequencies to be

COUPLED OSCILLATIONS 405

2 1

2 2

3 5 2

3 5 2

m

m

κ ω

κω

+ =

− =

(3)

Substituting these frequencies into (2), we obtain for the eigenvector components

11 21

12 22

1 5 2

1 5 2

a a

a a

− = 

 

+ = 

(4)

For the initial conditions ( ) ( )1 20 0x x= = 0 , the normal coordinates are

( )

( )

1 11 10 20

2 12 10 20

1 5 cos

2

1 5 cos

2

t ma x x t

t ma x x t

η ω

η ω

 − = +  

 + = +  

1

2

(5)

Therefore, when , 10 201.6180x x= − ( )2 0tη = and the system oscillates in Mode 1, the antisymmetrical mode. When , 10 20.6180x x= 0 ( )1 0tη = and the system oscillates in Mode 2, the symmetrical mode.

When mass 2 is held fixed, the equation of motion of mass 1 is

1 12mx x 0κ+ = (6)

and the frequency of oscillation is

10 2 m κ

ω = (7)

When mass 1 is held fixed, the equation of motion of mass 2 is

2 2 0mx xκ+ = (8)

and the frequency of oscillation is

20 m κ

ω = (9)

Comparing these frequencies with 1ω and 2ω we find

1 1

2 2

3 5 2 2 1.1441

4

3 5 0.6180

4

m m

m m

κ κ ω ω

κ κ ω ω

+ = = 0

0

>    − = = < 

406 CHAPTER 12

Thus, the coupling of the oscillators produces a shift of the frequencies away from the uncoupled frequencies, in agreement with the discussion at the end of Section 12.2.

12-8. The kinetic and potential energies for the double pendulum are given in Problem 7-7. If we specialize these results to the case of small oscillations, we have

(2 2 21 2 1 21 2 22T m )φ φ φ φ= + + (1)

( )2 21 21 22U mg φ φ= + (2)

where 1φ refers to the angular displacement of the upper pendulum and 2φ to the lower pendulum, as in Problem 7-7. (We have also discarded the constant term in the expression for the potential energy.)

Now, according to Eqs. (12.34),

,

1 2 jk j kj k

T m q= ∑ q (3)

,

1 2 jk j kj k

U A q= ∑ q (4)

Therefore, identifying the elements of { }m and { }A , we find

{ } 2 2 1 1 1

m  

=    

m (5)

{ } 2 0 0 1

mg  

=    

A (6)

and the secular determinant is

2 2

2 2

2 2 0

g

g

ω ω

ω ω

− − =

− − (7)

or,

2 2 42 2 g g

ω ω ω    0− − − =       (8)

Expanding, we find

2

4 24 2 g g

ω ω   0− +    = (9)

which yields

COUPLED OSCILLATIONS 407

( )2 2 2 gω = ± (10) and the eigenfrequencies are

1

2

2 2 1.848

2 2 0.765

g g

g g

ω

ω

= + =

= − =

(11)

To get the normal modes, we must solve

( )2 0jk r jk jr j

A m aω− =∑

For k = 1, this becomes:

( ) ( )2 211 11 1 21 21 2 0r r r rA m a A m aω ω− + − = For r = 1:

( ) ( )2 211 212 2 2 2 2 2g gmg m a m a− + − + =  

0  

Upon simplifying, the result is

21 112a a= −

Similarly, for r = 2, the result is

22 122a a=

The equations

1 11 1 12

2 21 1 22

x a a

x a a

2

2

η η

η η

= +

= +

can thus be written as

1 11 1 22 1 2

x a a 2η η= +

1 11 1 22 22x a aη η= − +

Solving for 1η and 2η :

1 2 11 2 2211

2 2 ;

22 2 x x x x

aa η η

− + = = 2

408 CHAPTER 12

2 1 2 1

2 2 1 1

occurs when 0; i.e. when 2

occurs when 0; i.e. when 2

x x

x x

η η

η η

= =

= =

Mode 2 is therefore the symmetrical mode in which both pendula are always deflected in the same direction; and Mode 1 is the antisymmetrical mode in which the pendula are always deflected in opposite directions. Notice that Mode 1 (the antisymmetrical mode), has the higher frequency, in agreement with the discussion in Section 12.2.

12-9. The general solutions for ( )1x t and ( )2x t are given by Eqs. (12.10). For the initial conditions we choose oscillator 1 to be displaced a distance D from its equilibrium position, while oscillator 2 is held at , and both are released from rest: 2 0x =

( ) ( ) ( ) ( )1 2 1 20 , 0 0, 0 0, 0x x x= = = 0=x D (1)

Substitution of (1) into Eq. (12.10) determines the constants, and we obtain

( ) (1 1cos cos2 D

x t t t)2ω ω= + (2)

( ) (2 2cos cos2 D

x t t t)1ω ω= − (3)

where

121 2 2

M M κ κ κ

ω ω +

> = (4)

As an example, take 1 21.2ω ω= ; ( )1x t vs. ( )2x t is plotted below for this case.

It is possible to find a rotation in configuration space such that the projection of the system point onto each of the new axes is simple harmonic.

By inspection, from (2) and (3), the new coordinates must be

1 1 2 1cosx x x D tω≡ − =′ (5)

2 1 2 2cosx x x D tω≡ + =′ (6)

These new normal axes correspond to the description by the normal modes. They are represented by dashed lines in the graph of the figure.

COUPLED OSCILLATIONS 409

0.2

0.2

–0.5

0.4

0.6

0.8

1.0 0.4 0.6 0.8 1.0

ω π

2 5 2

t =

ω π

2 3 2

t =

ω π

2 3 t =

ω π

2 2 t =

7π 2

x2(t)/D x2′

x1′

x1(t)/D

ω2t = π

ω2t = 2π

ω1 = 1.2 ω2

ω2t = 0

12-10. The equations of motion are

( )

( )

1 1 12 1 12 2 0

2 2 12 2 12 1

cos

0

mx bx x x F t

mx bx x x

κ κ κ ω

κ κ κ

+ + + − =  + + + − = 

(1)

The normal coordinates are the same as those for the undamped case [see Eqs. (12.11)]:

1 1 2 2 1;x x x x2η η= − = + (2)

Expressed in terms of these coordinates, the equations of motion become

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

2 1 2 1 12 2 1 12 2 1 0

2 1 2 1 12 2 1 12 2 1

2 cos

0

m b F

m b

tη η η η κ κ η η κ η η ω

η η η η κ κ η η κ η η

+ + + + + + − − =  − + − + + − − + = 

(3)

By adding and subtracting these equations, we obtain the uncoupled equations:

012 1 1 1

0 2 2 2

2 cos

cos

Fb t

m m m

Fb t

m m m

κ κ η η η ω

κη η η ω

+ + + =    + + = 

(4)

With the following definitions,

410 CHAPTER 12

2 12 1

2 2

0

2

2

b m

m

m

F A

m

β

κ κ ω

κω

=   + =   =    = 

(5)

the equations become

2 1 1 1 1

2 2 2 2 2

2 c

2 c

os

os

A t

A t

η βη ω η ω

η βη ω η ω

+ + =

+ + = (6)

Referring to Section 3.6, we see that the solutions for ( )1 tη and ( )2 tη are exactly the same as that given for x(t) in Eq. (3.62). As a result ( )1 tη exhibits a resonance at 1ω ω= and ( )2 tη exhibits a resonance at 2ω ω= .

12-11. Taking a time derivative of the equations gives ( )q I=

11 2 0 I

LI MI C + + =

22 1 0 I

LI MI C + + =

Assume 1 1 i tI B e ω= , 2 2

i tI B e ω= ; and substitute into the previous equations. The result is

2 21 1 2 1

0i t i t i tLB e B e M B e C

ω ω ωω ω− + − =

2 22 2 1 1

0i t i t i tLB e B e M B e C

ω ω ωω ω− + − =

These reduce to

( )2 21 21 0B L B MC ω ω  − + − =  

( )2 21 2 1 0B M B LCω ω  − + − =  

This implies that the determinant of coefficients of B and must vanish (for a non-trivial solution). Thus

1 2B

COUPLED OSCILLATIONS 411

2 2

2 2

1

0 1

L M C

M L C

ω ω

ω ω

− − =

− −

( ) 2

22 21 0L M C ω ω − − =  

2 2 1

L M C ω ω− = ±

or

( )

2 1 C L M

ω = ±

Thus

( )

( )

1

2

1

1

C L M

C L M

ω

ω

= +

= −

12-12. From problem 12-11:

1 1 2 1

0LI I MI C + + = (1)

2 2 1 1

0LI I MI C + + = (2)

Solving for in (1) and substituting into (2) and similarly for I , we have 1I 2

2

1 1 2

2

2 2 1

1 0

1 0

M M L I I I

L C CL

M M L I I I

L C CL

  − + − =    

  

− + − =    

(3)

If we identify

2

12

1 1

M m L

L

M LC

M C L

κ

κ

 = − 

  =    = −    

(4)

412 CHAPTER 12

then the equations in (3) become

( )

( )

1 12 1 12 2

2 12 2 12 1

0

0

mI I I

mI I I

κ κ κ

κ κ κ

+ + − =  + + − = 

(5)

which are identical in form to Eqs. (12.1). Then, using Eqs. (12.8) for the characteristic frequencies, we can write

( )

( )

1 2

2 2

1 1

1 1

M L M C L MC L L

M L M C L MC L L

ω

ω

+ = =

  −−  

− = =

  +−  

(6)

which agree with the results of the previous problem.

12-13.

I1

C1

q1

I2

C2

q2

L1 L12 L2

The Kirchhoff circuit equations are

( )

( )

1 1 1 12 1 2

1

2 2 2 12 2 1

2

0

0

q L I L I I

C

q L I L I I

C

+ + − =    + + − = 

(1)

Differentiating these equations using q I= , we can write

( )

( )

1 12 1 1 12 2 1

2 12 2 2 12 1 2

1 0

1 0

L L I I L I C

L L I I L I C

+ + − =    + + − = 

(2)

As usual, we try solutions of the form

( ) ( )1 1 2 2;i t i tI t B e I t B eω ω= = (3)

which lead to

COUPLED OSCILLATIONS 413

( )

( )

2 2 1 12 1 12 2

1

2 2 12 1 2 12 2

1

1 0

1 0

L L B L B C

L B L L B C

ω ω

ω ω

  + − − =  

     

− + + − =     

(4)

Setting the determinant of the coefficients of the B’s equal to zero, we obtain

( ) ( )2 21 12 2 12 1 1 2

1 1 L L L L L

C C ω ω    

+ − + − =     

4 2 2ω  (5)

with the solution

( ) ( ) ( ) ( )

( ) ( )

2 2 1 12 1 2 12 2 1 12 1 2 12 2 12 1 22

2 1 2 1 12 2 12 12

4

2

L L C L L C L L C L L C L C C

C C L L L L L ω

 + + + ± + − + + =  + + − 

(6)

We observe that in the limit of weak coupling ( )12 0L → and 1 2L L L= = , C C , the frequency reduces to

1 2 C= =

1 LC

ω = (7)

which is just the frequency of uncoupled oscillations [Eq. (3.78)].

12-14.

I1

C1 C12 I2 C2

L1 L2

The Kirchhoff circuit equations are (after differentiating and using q I= )

1 1 1 2 1 12 12

2 2 2 1 2 12 12

1 1 1 0

1 1 1 0

L I I I C C C

L I I I C C C

  + + − =     

  

+ + − =     

(1)

Using a harmonic time dependence for ( )1I t and ( )2 tI , the secular equation is found to be

2 21 12 2 121 2 2 1 12 2 12 12

1C C C C L L

C C C C C ω ω

  + + − −

  

 = 

  (2)

Solving for the frequency,

( ) ( ) ( ) ( ) 2 2 21 1 2 12 2 2 1 12 1 1 2 12 2 2 1 12 1 2 1 22 1 2 1 2 12

4

2

C L C C C L C C C L C C C L C C C C L L

L L C C C ω

 + + + ± + − + + = (3)

414 CHAPTER 12

Because the characteristic frequencies are given by this complicated expression, we examine the normal modes for the special case in which 1 2L L L= = and C C1 2 C= = . Then,

2 12 1

12

2 2

2

1

C C LCC

LC

ω

ω

+ =

=

(4)

Observe that 2ω corresponds to the case of uncoupled oscillations. The equations for this simplified circuit can be set in the same form as Eq. (12.1), and consequently the normal modes can be found in the same way as in Section 12.2. There will be two possible modes of oscillation: (1) out of phase, with frequency 1ω , and (2) in phase, with frequency 2ω .

Mode 1 corresponds to the currents and oscillating always out of phase: 1I 2I

;

I1 I2 I1 I2

Mode 2 corresponds to the currents and oscillating always in phase: 1I 2I

;

I1 I2 I1 I2

(The analogy with two oscillators coupled by a spring can be seen by associating case 1 with Fig. 12-2 for 1ω ω= and case 2 with Fig. 12-2 for 2ω ω= .) If we now let and , we do not have pure symmetrical and antisymmetrical symmetrical modes, but we can associate

1L L≠ 2 21C C

2ω with the mode of highest degree of symmetry and 1ω with that of lowest degree of symmetry.

12-15.

I1

C1

I2

C2

L1 R L2

Setting up the Kirchhoff circuit equations, differentiating, and using q I= , we find

( )

( )

1 1 1 2 1 1

2 2 2 1 2 2

1 0

1 0

L I R I I I C

L I R I I I C

+ − + =    + − + = 

(1)

Using a harmonic time dependence for ( )1I t and ( )2 tI , the secular equation is

2 21 2 1 2

1 1 0L i R L i R R

C C ω ω ω ω ω    

− − − − + =     2 2

  (2)

COUPLED OSCILLATIONS 415

From this expression it is clear that the oscillations will be damped because ω will have an imaginary part. (The resistor in the circuit dissipates energy.) In order to simplify the analysis, we choose the special case in which 1 2L L L= = and 1 2C C C= = . Then, (2) reduces to

2

2 1 0L i R R C

ω ω ω  2 2− − +   = (3)

which can be solved as in Problem 12-6. We find

1

2 2

1 LC

i L R R

L C

ω

ω

= ±

  = ± −   

(4)

The general solution for is ( )1I t

( ) 2 21 1

1 11 11 12 12 i LC t i LC t R L C t L R L C t LRt LB e B e e B e B e− − −−+ − + −I t − = + + +  

(5)

and similarly for . The implications of these results follow closely the arguments presented in Problem 12-6.

( )2I t

Mode 1 is purely oscillatory with no damping. Since there is a resistor in the circuit, this means that and flow in opposite senses in the two parts of the circuit and cancel in R. Mode 2 is the mode in which both currents flow in the same direction through R and energy is dissipated. If

1I 2I

2 C<R L , there will be damped oscillations of and , whereas if 1I 2I 2R L C> , the currents will

decrease monotonically without oscillation.

12-16.

O

y

x

R

P

R

Mg Q(x,y)

Mg

θ

φ

Let O be the fixed point on the hoop and the origin of the coordinate system. P is the center of mass of the hoop and Q(x,y) is the position of the mass M. The coordinates of Q are

( )

( )

sin sin

cos cos

x R

y R

θ φ

θ φ

= +  = − + 

(1)

The rotational inertia of the hoop through O is

416 CHAPTER 12

(2) 2O CM 2I I MR MR= + = 2

The potential energy of the system is

( )

hoop mass

2 cos cos

U U U

MgR θ φ

= +

= − + (3)

Since θ and φ are small angles, we can use 2s 1 2x x≅ −co . Then, discarding the constant term in U, we have

( )2 21 2 2

U MgR θ φ= + (4)

The kinetic energy of the system is

( )

hoop mass

2 2 2 O

2 2 2 2 2

1 1 2 2

1 2

2

T T T

I M x y

MR MR

= +

= + +

 = + + + 

θ

θ θ φ θφ

(5)

where we have again used the small-angle approximations for θ and φ. Thus,

2 2 2 1

3 2 2

T MR θ φ θφ = + +  (6)

Using Eqs. (12.34),

,

1 2 jk j kj k

T m q= ∑ q (7)

,

1 2 jk j kj k

U A q= ∑ q (8)

we identify the elements of { }m and { }A :

{ } 2 3 1 1 1

MR  

=    

m (9)

{ } 2 0 0 1

MgR  

=    

A (10)

The secular determinant is

2 2

2 2

2 3 0

g R

g R

ω ω

ω ω

− − =

− − (11)

from which

COUPLED OSCILLATIONS 417

2 2 42 3 g g R R

ω ω ω    0− − − =       (12)

Solving for the eigenfrequencies, we find

1

2

2

2 2

g R

g R

ω

ω

=

=

(13)

To get the normal modes, we must solve:

( )2 0jk r jk jr j

A m aω− =∑

For k = r = 1, this becomes:

2 211 212 2 3 2 g g

mgR mR a mR a R R

0 − − =   

or

21 112a a= −

For k = 1, r = 2, the result is

12 22a a=

Thus the equations

1 11 1 12

2 21 1 22

x a a

x a a

2

2

η η

η η

= +

= +

can be written as

1 11 1 22 2

2 11 1 222

x a a

x a a 2

η η

η η

= +

= − +

Solving for 1η , 2η

1 2 11 1 11 22

2 ;

3 3 x x x x

a a η η 2

− + = =

1η occurs when the initial conditions are such that 2 0η = ; i.e., 10 20 1 2

x x= −

This is the antisymmetrical mode in which the CM of the hoop and the mass are on opposite sides of the vertical through the pivot point.

2η occurs when the initial conditions are such that 1 0η = ; i.e., 10 20x x=

418 CHAPTER 12

This is the symmetrical mode in which the pivot point, the CM of the hoop, and the mass always lie on a straight line.

12-17.

kk m km km

x1 x2 x3 Following the procedure outlined in section 12.6:

2 21 2 1 1 1 2 2 2

T mx mx m= + + 23x

( ) ( )222 21 2 1 3 2

2 2 2 1 2 3 1 2 2 3

1 1 1 1 2 2 2 2

U kx k x x k x x kx

k x x x x x x x

= + − + − +

 = + + − − 

3

Thus

0 0

0 0 0 0

m

m

m

   =     

m

2 0

2 0 2

k k

k k k

k k

−   = − −   − 

A

Thus we must solve

2

2

2

2 0 2 0

0 2

k m k

k k m k

k k m

ω ω

ω

− − − − −

− − =

This reduces to

( ) ( )32 2 22 2 2k m k k mω ω 0− − − = or

( ) ( )22 2 22 2 2k m k m kω ω − − −  0= If the first term is zero, then we have

1 2k m

ω =

If the second term is zero, then

22 2k mω− = ± k

COUPLED OSCILLATIONS 419

which leads to

( ) ( )

2 3

2 2 2 2 ;

k k

m m

+ − = =ω ω

To get the normal modes, we must solve

( )2 0jk r jk jr j

A m aω− =∑

For k = 1 this gives:

( ) ( )2 1 22 0r r rk m a k aω− + − = Substituting for each value of r gives

( )

( )

( )

11 21 21

12 22 22 12

13 23 23 13

1 : 2 2 0 0

2 : 2 0 2

3 : 2 0 2

r k k a ka a

a ka a

r k a ka a a

= − − = → =

= − − = → = −

= − = → =

r k a

Doing the same for k = 2 and 3 yields

11 31 21

12 32 22 32

13 33 23 33

0

2

2

a a a

a a a a

a a a a

= − =

= = −

= =

The equations

1 11 1 12 2 13

2 21 1 22 2 23

3 31 1 32 2 33

x a a a

x a a a

x a a a

3

3

3

η η η

η η η

η η η

= + +

= + +

= + +

can thus be written as

1 11 1 22 2 33 3

2 22 2 33 3

3 11 1 22 2 33

1 2

2

1 2

x a a a

x a a

x a a a 3

η η η

η η

η η η

= − +

= +

= − − +

We get the normal modes by solving these three equations for 1η , 2η , 3η :

420 CHAPTER 12

1 3 1

11

1 2 2

22

2

2 2 2

x x a

x x a

η

η 3 x

− =

− + − =

and

1 23 33

2 4

x x a

η 3 x+ +

=

The normal mode motion is as follows

1 1

2 2

3 2 1 3

:

: 2

: 2

x x

x x

x x x

η

η

• → • ← • = −

← • • → ← • = − = −

• → • → • → = =

3

1 32

2

xη

12-18.

x1 y1

m

b

x

M

θ

1 1

1 1

sin ; cos

cos ; sin

x x b x x b

y b b y b

θ θ θ

θ θ θ

= + = +

= − =

Thus

( )

( )

( )

2 2 2 1 1

2 2 2 2

1

1 1 2 2

1 1 2 cos

2 2

1 cos

T Mx m x y

Mx m x b b x

U mgy mgb

θ θ θ

θ

= + +

= + + +

= = −

For small θ, 2

cos 1 2 θ

θ − . Substituting and neglecting the term of order 2θ θ gives

COUPLED OSCILLATIONS 421

( ) ( )2 2 2

2

1 1 2

2 2

2

T M m x m b b x

mgb U

θ θ

θ

= + + +

=

Thus

2 M m mb

mb mb

+  =    

m

0 0 0 mgb   =    

A

We must solve

( )2 2

2 2 0 M m mb

mb mgb mb

ω ω ω ω

− + − 2 =− −

which gives

( )( )

( )

2 2 2 4 2

2 2 2

0

0

M m mb mgb m b

Mb mgb m M

ω ω ω

ω ω

+ − −

 − + = 

2 =

Thus

( )

1

2

0

g M m

mb

ω

ω

=

= +

( )2 0jk r jk jr j

A m aω− =∑

Substituting into this equation gives

( )

( ) ( )

21

12 22

0 2

2, 2

a k

bm a a k

m M

= =

= − = = +

, 1r

r

=

Thus the equations

11 1 12 2

21 1 22 2

x a a

a a

η η

θ η η

= +

= +

become

( )11 1 22 2

mb x a a

m M η η= −

+

422 CHAPTER 12

22 2aθ η=

Solving for 1η , 2η :

( )

2 22

1 11

a

bm x

m M n

a

θ η

θ

=

+ +

=

( )

1 2

2 1

occurs when 0; or 0

occurs when 0; or

n n

bm n n x

m M

θ

θ

= =

= = − +

12-19. With the given expression for U, we see that { }A has the form

{ } 12 13

12 23

13 23

1 1

1

ε ε ε ε ε ε

− −   = − −   − − 

A (1)

The kinetic energy is

( )2 2 21 2 312T θ θ θ= + + (2) so that { }m is

{ } 1 0 0 0 1 0 0 0 1

   =     

m (3)

The secular determinant is

2 12 13

2 12 23

2 13 23

1 1

1

ω ε ε ε ω ε ε ε ω

− − − 0− − − =

− − − (4)

Thus,

( ) ( ) ( )32 2 2 2 212 13 23 12 13 231 1 2ω ω ε ε ε ε ε ε− − − + + − = 0 (5) This equation is of the form (with 1 2 x− ≡ω )

3 2 23 2x x 0− − =α β (6)

which has a double root if and only if

( )3 22 2α β= (7)

COUPLED OSCILLATIONS 423

Therefore, (5) will have a double root if and only if

3 22 2 2

12 13 23 12 13 233

ε ε ε ε ε ε

 + + = 

  (8)

This equation is satisfied only if

12 13 23ε ε ε= = (9)

Consequently, there will be no degeneracy unless the three coupling coefficients are identical.

12-20. If we require , then Eq. (12.122) gives 11 212a a= 31 213a a= − , and from Eq. (12.126) we

obtain 21 1 14a = . Therefore,

1 2 1 3

, , 14 14 14

 = −   a (1)

The components of can be readily found by substituting the components of above into Eq. (12.125) and using Eqs. (12.123) and (12.127):

2a 1a

2 4 5 1

, , 42 42 42

− =    a (2)

These eigenvectors correspond to the following cases:

a1 a2

12-21. The tensors { }A and { }m are:

{ }

1 3

3 2 3

3 1

1 0

2 1 2 2

1 0

2

κ κ

1 κ κ κ

κ κ

       =        

A (1)

{ } 0 0

0 0 0

m

m

m

0    =     

m (2)

thus, the secular determinant is

424 CHAPTER 12

2 1 3

2 3 2 3

2 3 1

1 0

2 1 1

0 2 2

1 0

2

m

m

m

κ ω κ

κ κ ω κ

κ κ ω

− =

(3)

from which

( ) ( ) ( )22 2 2 21 2 3 11 02m m mκ ω κ κ ω− − − −κ ω (4) =

In order to find the roots of this equation, we first set ( ) 23 11 2 2κ κ κ= and then factor:

( ) ( ) ( )

(κ ω (5) ) ( )

( ) ( )

2 2 2 1 1 2 1 2

2 2 4 2 1 1 2

2 2 2 1 1 2

0

0

0

m m m

m m m

m m m

κ ω κ ω κ ω κ κ

ω κ κ ω

κ ω ω ω κ κ

 − − − − 

 − − + = 

 − − + = 

=

Therefore, the roots are

1 1

1 2 2

3 0

m

m

κ ω

κ κ ω

ω

=

+ =

=

(6)

Consider the case 3 0ω = . The equation of motion is

23 3 3 0+ =η ω η (7)

so that

3 0η = (8)

with the solution

( )3 t at bη = + (9)

That is, the zero-frequency mode corresponds to a translation of the system with oscillation.

COUPLED OSCILLATIONS 425

12-22. The equilibrium configuration is shown in diagram (a) below, and the non- equilibrium configurations are shown in diagrams (b) and (c).

1 2

4 3

x1

x2

x3

O

2A 2B

(a)

1 2

x3 x

x3

O

OA

A Aθθ

2

  

}

(b)

1 4

x3 x

x3

O

OB

B

Bφ φ

2

  

{

(c)

The kinetic energy of the system is

2 23 1 2 1 1 1 2 2 2

T Mx I I 2θ φ= + + (1)

where ( ) 21 1 3I M= A and ( ) 22 1 3I M= B . The potential energy is

( ) ( ) ( ) ( )

( )

2 2 2 2 3 3 3 3

2 2 2 2 2 3

1 2

1 4 4 4

2

U x A B x A B x A B x A B

x A B

 = − − + + − + + + + − + 

= + +

κ θ φ θ φ θ φ θ

κ θ φ

φ

(2)

Therefore, the tensors { }m and { }A are

{ } 2

2

0 0 1

0 3

1 0 0

3

M

MA 0

MB

       =        

m (3)

{ } 2 2

4 0 0 0 4 0 0 0 4

A

B

κ κ

κ

   =     

A (4)

The secular equation is

426 CHAPTER 12

( )2 2 2 2 2 2 21 14 4 4 3 3

M A MA B MBκ ω κ ω κ ω  − − −     0  =

(5)

Hence, the characteristic frequencies are

1

2

3 2

2

3 2

3 2

M

M

M

κ ω

κ ω

κ ω ω

=

=

= =

(6)

We see that 2 3ω ω= , so the system is degenerate.

The eigenvector components are found from the equation

( )2 0jk r jk jr j

A m aω− =∑ (7)

Setting to remove the indeterminacy, we find 32 0a =

21 2 3 2

01 0 0 ; 3 ; 0 0 0 3

M

MA

MB

         = = =             

aa a (8)

The normal coordinates are (for ( ) ( ) ( )3 0 0 0x θ φ 0= = = )

( )

( )

( )

1 30 1

0 2

0 3 3

cos

cos 3

cos 3

t x M t

A M t

B M t t

η ω

θ 2tη ω

φ η ω

=

=

=

(9)

Mode 1 corresponds to the simple vertical oscillations of the plate (without tipping). Mode 2 corresponds to rotational oscillations around the axis, and Mode 3 corresponds to rotational oscillations around the -axis.

1x

2x

The degeneracy of the system can be removed if the symmetry is broken. For example, if we place a bar of mass m and length 2A along the of the plate, then the moment of inertia around the is changed:

2 -axisx

1 -axisx

( ) 21 1 3

I M m= +′ A (10)

The new eigenfrequencies are

COUPLED OSCILLATIONS 427

1

2

3

2

3 2

3 2

M

M m

M

κ ω

κ ω

κ ω

=

= +

=

(11)

and there is no longer any degeneracy.

12-23. The total energy of the r-th normal mode is

2 21 1

2 2

r r r

r r

E T U

2 rη ω η

= +

= + (1)

where

ri tr re= ωη β (2)

Thus,

ri tr r ri e= ωη ω β (3)

In order to calculate T and U , we must take the squares of the real parts of r r rη and rη :

( ) ( ) ( ) 222

2

Re Re cos sin

cos sin

r r r r r r r

r r r r r r

i i t i

t t

t = = + + 

−  

η η ω µ ν ω ω

ω ν ω ω µ ω= − (4)

so that

221 cos sin

2r r r r r r T tω ν ω µ ω= + t   (5)

Also

( ) ( ) ( ) 222

2

Re Re cos sin

cos sin

r r r r r r

r r r r

i t i

t t

t = = + + 

  

η η µ ν ω ω

µ ω ν ω= − (6)

so that

221 cos

2r r r r r r U tω µ ω ν ω= − xin t   (7)

Expanding the squares in T and U , and then adding, we find r r

428 CHAPTER 12

( )2 2 21

2

r r r

r r r

E T U

ω µ ν

= +

= +

Thus,

22 1 2r r r

E ω β= (8)

So that the total energy associated with each normal mode is separately conserved.

For the case of Example 12.3, we have for Mode 1

( )1 10 20 cos2 M

x x 1tη ω= − (9)

Thus,

( )1 1 10 20 sin2 M

x x 1tη ω= − − ω (10)

Therefore,

2 21 1 1 1 1 2 2

E 21η ω η= + (11)

But

2 121 2

M κ κ

ω +

= (12)

so that

( ) ( )

( ) ( )

2 22 212 12 1 10 20 1 10 20

2 12 10 20

2 21 1 sin cos

2 2 2 2

1 2

4

M M E x x t x x

M M

x x

+ + = − + −

= + −

κ κ κ κ 1tω ω

κ κ (13)

which is recognized as the value of the potential energy at t = 0. [At t = 0, , so that the total energy is

1 2 0x x= = ( )1 0=U t .]

12-24. Refer to Fig. 12-9. If the particles move along the line of the string, the equation of motion of the j-th particle is

( ) ( )1j j j jmx x x x xκ κ−= − − − − 1j+ (1) Rearranging, we find

( )1 2j j j jx x x xm 1 κ

−= − + + (2)

which is just Eq. (12.131) if we identify mdτ with mκ .

COUPLED OSCILLATIONS 429

12-25. The initial conditions are

( ) ( ) ( )

( ) ( ) ( )

1 2 3

1 2 3

0 0 0

0 0 0 0

q q q a

q q q

= = =   = = = 

(1)

Since the initial velocities are zero, all of the rν [see Eq. (12.161b)] vanish, and the rµ are given by [see Eq. (12.161a)]

3

sin sin sin 2 4 2 4r a r r rπ π π

µ  = + +   (2)

so that

1

2

3

2 1 2

0

2 1 2

a

a

µ

µ

µ

+ = 

 

=   −

=  

(3)

The quantities ( )si are the same as in Example 12.7 and are given in Eq. (12.165). The displacements of the particles are

n 1jr nπ + 

( ) ( ) ( )

( ) ( ) (

( )

)

( ) ( )

1 1 3 1

2 1 3 1

3 1 3 1

1 2 cos cos cos cos

2 4

2 1 cos cos cos cos

2 2

1 2 cos cos cos cos

2 4

q t a t t a t t

a t t a t t

q t a t t a t t

ω ω ω ω

ω ω ω ω

ω ω ω ω

= + + −

= − + +

= − + +

3

3

3

q t (4)

where the characteristic frequencies are [see Eq. (12.152)]

2 sin , 1, 2 8r

r r

md τ πω  =   

, 3= (5)

Because all three particles were initially displaced, there can exist no normal modes in which any one of the particles is located at a node. For three particles on a string, there is only one normal mode in which a particle is located at a node. This is the mode 2ω ω= (see Figure 12-11) and so this mode is absent.

12-26. Kinetic energy ( ) [ ] 2

2 2 2 2 2 1 2 3

22

mb mb

mb

T m mb

   = + + ⇒ =    + 

θ θ θ

Potential energy

430 CHAPTER 12

( ) ( ) ( ) ( ) ( )

( ) ( )

[ ]

2 22 1 2 3 2 1 3

2 2 2 2 2 2 2 1 2 3 1 2 3 1 2 2 3

2 2

2 2 2

2 2

1 cos 1 cos 1 cos sin sin sin sin 2

2 2 2 2 2

0

0

k U mgb b

mgb kb

mgb kb kb

A kb mgb kb kb

kb mgb kb

2   = − + − + − + − + −   

≈ + + + + + − −

 + −  ⇒ = − + −   − + 

θ θ θ θ θ θ

θ θ θ θ θ θ θ θ θ θ

θ

The proper frequencies are solutions of the equation

[ ] [ ]( ) ( )

( ) ( )

2 2 2 2

2 2 2 2 2

2 2

0

0 Det Det 2

0

mgb kb mb kb

A m kb mgb kb mb kb

kb mgb kb mb

 + − −  

= − = − + − −   

− + −  

ω

ω ω

ω

2

2 2

We obtain 3 different proper frequencies

2 1 1

2 2 2

2 3 3

4.64 rad/s

3 3 4.81 rad/s

4.57 rad/s

mg kb mg kb mb mb

mg kb mg kb mb mb

mg g mb f

+ + = ⇒ = =

+ + = ⇒ = =

= ⇒ = =

ω ω

ω ω

ω ω

Actually those values are very close to one another, because k is very small.

12-27. The coordinates of the system are given in the figure:

L1

m1

m2

θ1

θ2 L2

Kinetic energy:

( )( )

( )

2 2 2 2 2 2 2 1 1 1 2 1 1 2 2 1 2 1 2 1 2

2 2 2 2 2 1 1 1 2 1 2 2 2 2 1 2 1 2

1 1 2 cos

2 2

1 1 1 2 2 2 jk j kjk

T m L m L L L L

m L m L m L m L L m

= + + − −

≈ + + − = ∑

θ θ θ θ θ θ θ

θ θ θ θ θ θ

COUPLED OSCILLATIONS 431

( ) 21 2 1 2 1 2

2 2 1 2 2 2

jk

m m L m L L m

m L L m L

 + − ⇒ =    − 

Potential energy:

( ) ( ) ( )

( )

1 1 1 2 1 1 2 2

2 2 1 2

1 2 1 2 2

1 cos 1 cos 1 cos

1 2 2 2 jk j kjk

U m gL m g L L

m m gL m gL A

 = − + − + − 

≈ + + = ∑

θ θ θ

θ θ θ θ

( )1 2 1

2 2

0 0jk

m m gL A

m gL

 +  ⇒ =   

 

Proper oscillation frequencies are solutions of the equation

[ ] [ ]( )2Det 0A m− =ω

( ) ( ) ( ) ( ) ( )2 221 2 1 2 1 2 1 1 2 2 1 2

1,2 1 1 22

m m g L L m m g m L L m L L

m L L

 + + + + − + + ⇒ =ω

The eigenstate corresponding to 1ω is 11

21

a

a    

where

( ) ( ) ( ) ( ) ( ) ( )

1 2 1 1 2 21 112 221 2

1 2 1 2 1 2 1 1 2 2 1 2

2 1

m m L gm L a a

m L m m g L L m m g m L L m L L

  +  = −  + + + + − + +   

×

The eigenstate corresponding to 2ω is 12

22

a

a    

where

( ) ( ) ( ) ( ) ( ) ( )

1 2 1 1 2 22 122 221 2

1 2 1 2 1 2 1 1 2 2 1 2

2 1

m m L gm L a a

m L m m g L L m m g m L L m L L

  +  = −  + + − + − + +   

×

These expressions are rather complicated; we just need to note that and have the same

sign

11a 21a

11

21

0 a a  

>   

while and have opposite sign 12a 22a 11

21

0 a a  

<  .  

The relationship between coordinates ( )1 2,θ θ and normal coordinates 1 2,η η are

12 1 1

1 11 1 12 2 22

2 21 1 22 2 11 2 1

21

~

~

a a a a a a a

a

 −= +  ⇔ = +   − 

2

2

η θ θ θ η η θ η η

η θ θ

432 CHAPTER 12

To visualize the normal coordinate 1η , let 2 0=η . Then to visualize the normal coordinate 2η ,

we let 1 0=η . Because 11

21

0 a a > and 12

22

a a

0< , we see that these normal coordinates describe two

oscillation modes. In the first one, the two bobs move in opposite directions and in the second, the two bobs move in the same direction.

12-28. Kinetic energy: [ ] 2

22 2 2 2 1 1 2 2 2

2

01 1 2 2 0

m b T m b m b m

m b

  = + ⇒ =  

  θ θ

Potential energy: ( ) ( ) ( )

[ ]

1 1 2 2 1

2 2 1

2 2 2

1 cos 1 cos sin sin 2 k

U m gb m gb b b

m gb kb kb A

kb m gb kb

= − + − + −

 + − ⇒ ≈  − + 

2θ θ θ θ

Solving the equation, [ ] [ ]( )2Det 0A m−ω = , gives us the proper frequencies of oscillation,

2 21ω 25 (rad/s) g b = = 2 22

1 2

25.11(rad/s) g k k b m m = + + =ω

The eigenstate corresponding to 1ω is 11

22

a

a    

with 21 117.44a a=

The eigenstate corresponding to 2ω is 12

22

a

a    

with 22 128.55a a=

From the solution of problem 12-27 above, we see that the normal coordinates are

121 1 2 1 22

~ 0 a a − = + 2.12η θ θ θ θ

112 1 2 1 21

~ 0 a a − = + 2.13η θ θ θ θ

Evidently 1η then characterizes the in-phase oscillation of two bobs, and 2η characterizes the out-of-phase oscillation of two bobs.

Now to incorporate the initial conditions, let us write the most general oscillation form:

( )

( )

( )

1 1 2 2

1 1 2 2

1 1 2 2

1 11 12

2 21 22

11 12

Re

Re

Re 7.44 8.35

i t i i t i

i t i i t i

i t i i t i

a e a e

a e a e

a e a e

− −

− −

− −

= +

= +

= −

ω δ ω δ

ω δ ω δ

ω δ ω δ

θ α α

α

α α

θ α

where α is a real normalization constant. The initial conditions helps to determine parameters α’s, a’s, δ’s.

COUPLED OSCILLATIONS 433

( )( ) ( )( )

1 11 1 12 2

2 11 1 12

Re 0 7 cos cos 0.122 rad

Re 0 0 7.44 cos 8.35 cos 0

t a a

t a a

° = = − ⇒ + = − 

= = ° ⇒ + =

θ α δ α δ

θ α δ α 2δ

=

1 1sin sin 0⇒ =δ δ . Then

( )

1 11 1 12 2 1 2

2 11 1 12 2 2

cos cos 0.065 cos 0.057 cos

7.44 cos 8.35 cos 0.48 cos cos

a t a t t

a t a t t

= + = − −

= + = − 1t

θ α ω α ω ω ω

θ α ω α ω ω ω

where 1 5.03 rad/s=ω , 2 4.98 rad/s=ω (found earlier)

Approximately, the maximum angle 2θ is 0.096rad and it happens when

2 1

cos 1 cos 1

t

t

=  = −

ω ω

which gives

( ) 2 1

1 2

2 2 1 2 1 2

t n k t k n

=  + ⇒ == + 

ω π ω ω π ω

because 1 2

101 100 =

ω ω

we finally find 50k n= = and 2

100 63 st = =

π ω

.

Note: 2 max 0.96 rad=θ and at this value the small-angle approximation breaks down, and

the value 2maxθ we found is just a rough estimate.

434 CHAPTER 12

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