**UFBA**

# Mecflu - oscila??es juntadas

(Parte **1** de 2)

CHAPTER 12 Coupled Oscillations

12-1.

m = M k x k k m = M x The equations of motion are

Mx x x Mx x x κκ κ

We attempt a solution of the form itit xt B e

In order for a non-trivial solution to exist, the determinant of coefficients of and must vanish. This yields 1B 2B

from which we obtain

If were held fixed, the frequency of oscillation of m would be 2m1

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

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

so that

so that

If we use

then the frequencies in (1) can be expressed as

COUPLED OSCILLATIONS 399

1 κωω ω εκ

where

For the initial conditions [Eq. 12.2)],

Using (3), we can write

Then,

sin sin xt D t t

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,

( )2 sin sin cos sin sin cosD t t t t t t t tε ε+ − ++ − − +≅Ω Ω + Ω Ω + Ω Ω xt (12) −

When either ()1xt or ()2xt 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.

12-3. The equations of motion are m x x

m x x

We try solutions of the form

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

so that

and then

Therefore, the frequencies of the normal modes are

m M

m M 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. (12.1)].

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

(1)

Therefore,

COUPLED OSCILLATIONS 401 dE M x x x x x x x x x x

Mx x x x x Mx x x x x

Mx x x x Mx x x x which exactly vanishes because the coefficients of and are the left-hand sides of Eqs.

An analogous result is obtained when T and U are expressed in terms of the generalized coordinates 1η and 2η defined by Eq. (12.1):

Therefore,

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

), and through Eq. (12.15) we see that this implies that such a term depends on the ’s and 1Cω, but not on the ’s and 2C2ω.

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

spring connecting the masses is changed, the motion is not affected.

mx x x

Assuming solutions of the form itit xt B e

we find that the equations in (1) become mB B

Bm B κω κ

which lead to the secular equation for 2ω:

Therefore, m κµω µ

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

Inserting the values for 1ω and 2ω into either of the equations in (3), we find1

and

Using the orthonormality condition produces

m m where

The second eigenvector has the components m m

COUPLED OSCILLATIONS 403

where

cos cos tm a x m a x t tm a x m a x 1 2t

12-6.

m k x If the frictional force acting on mass 1 due to mass 2 is

mx x x x

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

where iαλω=+ 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:

i m

m m

The general solution is therefore

im t i m t mt m m t mtmB e B e e B e B eκκ βκ βκβ − −+− − + −=+ + +xt (6)

The first two terms in the expression for ()1xt are purely oscillatory, whereas the last two terms contain the damping factor teβ−. (Notice that the term ()212expBmtβκ+− increases with time if

2mβκ>, 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+

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 2mβκ<, the motion is one of damped oscillations, whereas if 2mβκ>, the motion proceeds monotonically to zero amplitude.

12-7.

m k k x

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:

mx x x

mB B

Bm B κω κ

Solving the secular equation, we find the eigenfrequencies to be

COUPLED OSCILLATIONS 405

Substituting these frequencies into (2), we obtain for the eigenvector components 1 21

15 cos tm a x x t

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

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

Comparing these frequencies with 1ω and 2ω we find

4 m m

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

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),

and the secular determinant is g gωωωω

or,

Expanding, we find

which yields

COUPLED OSCILLATIONS 407

To get the normal modes, we must solve

For k = 1, this becomes:

For r = 1:

Upon simplifying, the result is

=+ can thus be written as

x x x

occurs when 0; i.e. when 2

occurs when 0; i.e. when 2 x x

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 ()1xt and ()2xt 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: 20x=

where

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

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

x (t)/D ωt = π

4π ωt = 2π ω = 1.2 ω ωt = 0

12-10. The equations of motion are mx bx x x F t mx bx x x κ κκ ω

The normal coordinates are the same as those for the undamped case [see Eqs. (12.1)]:

2c os0 mb Fmb tη ηη η κ κ η η κ η η ω

(3)

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

2 cos

cos

Fb t m m

Fb t m m

With the following definitions, b m

the equations become os os

A t A t η βη ω η ω η βη ω η ω ++ =

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

itIBeω=; and substitute into the previous equations. The result is

These reduce to

This implies that the determinant of coefficients of B and must vanish (for a non-trivial solution). Thus 1 2B

COUPLED OSCILLATIONS 411

or

Thus

12-12. From problem 12-1:

If we identify 2

M mL

412 CHAPTER 12 then the equations in (3) become mI I I mI I I κκ κ

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

which agree with the results of the previous problem. 12-13.

C q

C q q LI L I I

q LI L I I

Differentiating these equations using qI= , we can write

As usual, we try solutions of the form

COUPLED OSCILLATIONS 413

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

with the solution

(6)

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

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

1CC C C L

Solving for the frequency,

Because the characteristic frequencies are given by this complicated expression, we examine the normal modes for the special case in which 12LLL== and CC12C==. Then,

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: 1I2I

Mode 2 corresponds to the currents and oscillating always in phase: 1I2I

(The analogy with two oscillators coupled by a spring can be seen by associating case 1 with

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

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,

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

1 LC iL R

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

2C<RL, there will be damped oscillations of and , whereas if 1I2I2RLC>, the currents will decrease monotonically without oscillation.

O y x

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 cosxRyR θφ θ φ

The rotational inertia of the hoop through O is

The potential energy of the system is

hoop mass 2c os cos

Since θ and φ are small angles, we can use 2s12xx≅−co. Then, discarding the constant term in U, we have

The kinetic energy of the system is hoop mass

T T IM x y

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

Using Eqs. (12.34),

MgR

The secular determinant is

from which

COUPLED OSCILLATIONS 417

Solving for the eigenfrequencies, we find g R

To get the normal modes, we must solve:

For k = r = 1, this becomes:

or

xa a xa a 2

Solving for 1η, 2η

3 x x xaa

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.

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. k m km km x Following the procedure outlined in section 12.6:

2 2 2 Uk x k x x k x x kx

Thus m m

k k k

Thus we must solve

km k k m k k m

This reduces to

COUPLED OSCILLATIONS 419 which leads to

km +−==ωω

To get the normal modes, we must solve

For k = 1 this gives:

rk k a ka a a ka a rk a ka a a

xa a a xa a a xa a a

=+ + can thus be written as

xa a a xa a

xa a a 3 x a

x a and

The normal mode motion is as follows 1 x x x xηη

12-18.

y m x M sin ; cos cos ; sin x b x x b yb b y b θ θθ θθ θ =+ =+

1 2c os

1c os

TM x m x y

Mx m x b b x

Um gy mgb

For small θ, 2 cos 1

COUPLED OSCILLATIONS 421

TM m x m b bx

mgb

Thus

2 M m b

We must solve

Mm mb

which gives

M m mb mgb m b

Mb mgb m M ωω ωωω

Thus

g Mm

Substituting into this equation gives ( ) bm a k

=+ become

xa amM ηη=−+ bm x

mM n occurs when 0; or 0 occurs when 0; or bm n x

The kinetic energy is

The secular determinant is

Thus,

COUPLED OSCILLATIONS 423

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

This equation is satisfied only if

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

These eigenvectors correspond to the following cases:

κκ 1κ κ κ

thus, the secular determinant is m m m from which

m m m m m m

Therefore, the roots are

m m

so that

COUPLED OSCILLATIONS 425

12-2. The equilibrium configuration is shown in diagram (a) below, and the nonequilibrium configurations are shown in diagrams (b) and (c).

x x x x

(b) x x

(c)

The kinetic energy of the system is

Ux A B x A B x A B x A B xA B κθ φ φ (2)

The secular equation is

Hence, the characteristic frequencies are

We see that 23ωω=, so the system is degenerate. The eigenvector components are found from the equation

a (8)

cos cos 3 cos 3 tx M t

AM t

BM t θ 2tη ω

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

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

The new eigenfrequencies are

COUPLED OSCILLATIONS 427

M m M and there is no longer any degeneracy. 12-23. The total energy of the r-th normal mode is

2 r rr

2rη ωη where

Thus,

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

cos sin r r r r r r r r r r r i t itt ωνωωµω=− (4) so that

Also

cos sin r r r r r r r r it itt µωνω=− (6) so that

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

r r r r

Thus,

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

Thus,

Therefore,

But

M Ex x t x x κκ κκ 1tω ω

κκ (13) which is recognized as the value of the potential energy at t = 0. [At t = 0, , so that the

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

COUPLED OSCILLATIONS 429 q q a

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

so that

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

12 cos cos cos cos

21 cos cos cos cos

12 cos cos cos cos qt a t t a t t a t t a t t qt a t t a t t

3 qt (4) where the characteristic frequencies are [see Eq. (12.152)]

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 2ωω= (see Figure 12-1) and so this mode is absent.

mb mb

mb Tm mb

Potential energy

1c os 1c os 1c os sin sin sin sin 2 k Um gb b mgb kb

mgb kb kb

A kb mgb kb kb kb mgb kb

The proper frequencies are solutions of the equation mgb kb mb kb

(Parte **1** de 2)