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It took me a while but I think I see where the problem is. Following
Carl's initial wording I used the word "phase" incorrectly in connection
with the ratio of components of each eigenvector. For the simplest case
where the 2 oscillators have the same mass and the same
natural frequency ( omega squared = k/m) one finds two modes of
oscillation: One where the two masses move "in phase" with each other
because the ratio of the amplitudes of the components of the eigenvector
is 1 ( omega squared = k/m for this one), and one where
the two masses move in "opposite" phase because the ratio of the
amplitudes is -1 (omega squared = k/m + kc/m where kc is the spring
constant of
the spring coupling the two masses) . My understanding was that Carl was
referring to the ratio of 1 as phase =0 and -1 as phase = pi.
The actual motion of the masses is of course a linear superposition of
the two modes with magnitudes and relative phases that depend on the
initial conditions.
Now, unless I goofed in my calculation, if the masses are not equal and
or the springs tying each mass to a wall (in my example, but in general
this would mean that the free oscillations of each
oscillator have different frequencies) are different, the ratios of the
amplitudes of the components for each eigenvector will not be the usual
+1 or -1. The spring + mass case is rather messy but
the case of two pendulums of same lengths and different masses is
cleaner. One finds that in the lower frequency normal mode (omega
squared = g/l) the ratio of amplitudes (angles in this case) is 1. In
the higher frequency mode ( omega square = g/l + k/mu where k =
coupling spring constant and mu = reduced mass) the ratio of the
amplitudes is - m1/m2
(negative of the ratio of the masses which goes to -1 if m1 = m2) .
Karim Diff
Jack Uretsky wrote:
Hi all-
The coupled oscillator problem is discussed at length in Olenick,
et al. <The Mechanical Universe> (NOT the advanced edition), Ch 22.
The problem of 2 coupled oscillators can be worked out easily,
TKarim's example is OK (though the conclusion is erroneous). The paradigm
equation is (' stands for time derivative, f^2 means f squared)
x'' + f1^2 x +k(x-y) =0
y'' + f2^2 y +k(y-x) =0
The eigenvalue equation is a quadratic in the square of the
eigenfrequencies; f1, f2, k are real constants. It is easy to show that
both squared eigenfrequencies are real and positive.
The eigenvectors are two dimensional column vectors, one for each
(squared) eigenfrequency. The eigenvalue condition gives 1 relation
between the two components of each eigenvector. The normalization
condition requires the squared magnitudes of the two components to sum to
unity. Since only the magnitudes are involved, there is an undetermined,
arbitrary phase for each eigenvector. The relative phases may be
determined from the starting conditions on x and y. Different phases may
refer to different starting conditions, but 0 and/or pi are certainly not
mandated in general.
Regards,
Jack
On Wed, 8 Nov 2006, Karim Diff wrote:
I did not work it out in detail yet but what about this:
2 unequal masses & 3 horizontal springs (different k's) between 2 walls.
In crude ASCII
Wall |/\/\/\/\[ m 1 ]/\/\/\[m 2 ]/\/\/\/| Wall
The phases are 0 & pi if the two masses and the springs are the same
(basic coupled oscillators)
but I believe the phase relationship becomes more complicated in the
general case (2 different masses and 3 different k's).
Karim Diff
Carl Mungan wrote:
So if the eigenvectors can be complex even when one component is_______________________________________________
chosen to be real, then that implies that the oscillators in a normal
mode need NOT pass through the equilibrium position at the same
instant (ie. relative phases other than 0 or pi are possible).
In that case, can someone rig up a *simple soluble* example of such
an eigenvalue problem?
To make my request clear for those who have forgotten the beginning
of this thread. I want an example of a system of coupled oscillators,
as simple as possible, where the relative phases between oscillators
in a normal mode are neither 0 nor pi. All the usual textbook
examples I can think only have in-phase or exactly-out-of-phase
relative motions of the oscillators.
The correct statement, that you seem to be striving for, is that the
eigenvalues of a Hermitean matrix (includes real, symmetric
matrices) are real. This says nothing about the scaling of the
eigenvectors, which are usually scaled by normalizing them to a
(squared) norm of unity - which leaves them with an undetermined
phase. That phase is just the subject of the present discussion.
Regards,
Jack
Is that true even if I insist that one component arbitrarily have the
real value 1, to remove the indeterminacy in the overall scaling of
an eigenvector? This way d_i is the relative phase, which is what I
meant.
The argument is erroneous. The reality condition is on the
eigenvalues, not the eigenvectors.
Jack
Having thought about it a little more, this is what I come up with:
Newton's third law says that oscillators i and j exert equal
magnitude forces on each other. This in turn will lead to elements
K_ij and K_ji in the force-constant matrix K to be equal. In turn,
this means K is symmetric (and of course it's real); and so is the
mass matrix M. But this means the eigenvectors must be real, even if
I write the oscillator displacement vector x in complex form where
the i-th component is A_i exp (i d_i). But this means d_i can only be
0 or pi.
What do you think of this argument? -Carl
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