Resolução Marion [Mec.Clássica] - Capítulo 12

Resolução Marion [Mec.Clássica] - Capítulo 12

(Parte 1 de 7)

CHAPTER 12 Coupled Oscillations


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


1 κωω ω εκ


For the initial conditions [Eq. 12.2)],

Using (3), we can write


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



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):


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

(Parte 1 de 7)