Re: [Phys-l] question about coupled oscillators

[Jack Uretsky <jlu@hep.anl.gov>]

Hi Carl-
	The general statement goes something like this (in the 2 normal 
mode case):
	There are 2 coupled amplitudes, call them X1 and X2.  Your system 
is such that you can find 2 normal modes, with frequencies fA and fB. 
Corresponding to each frequency there is an eigenvector that is a linear 
combination of X1 and X2.  The eigenvectors, call them XA and XB are 
orthogonal to each other in the usual case.  The eigenmodes can be excited 
independently of each othe (that's what makes them "eigen"), so there 
should be no relationship between the phases of the excitations.
	The simplest case to work out, that I know of, is two pendulums 
connected by a spring.
					Regards,
						Jack




On Fri, 3 Nov 2006, Carl Mungan wrote:

> In writing down the assumed form for the displacement of each
> oscillator i in a coupled system, in general one includes a phase
> constant d_i in the argument of the cosine:
>
> x_i (t) = A_i cos (w*t + d_i)
>
> However, in all the examples I can think of, the relative phases of
> the oscillators *in a single normal mode* always turn out to be
> either 0 or pi. Why can't we get other values?
>
> In particular, the "weak coupling" problem (where you pull just one
> oscillator aside and then get beats) looks like an example of pi/2
> relative phase difference between the two oscillators. Except I don't
> have a single normal mode; I have two normal modes equally excited.
> But there's a part of me that wants to call this something like
> another normal mode since each particle is oscillating in a nicely
> repetitive manner. (Okay, maybe I need to require the two normal mode
> frequencies to be commensurate for this statement to be exactly
> correct.) That is, I'm wanting to define "normal mode" as any
> smoothly repetitive pattern, but apparently that's too loose a
> definition to be right.
>
> Please help clarify my thinking. -Carl

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