Re: Explanation for Resonant Coupling?

[Maurice Barnhill <mvb@UDEL.EDU>]

Thomas M. Philip wrote:
> Thanks for the explanation, but a couple of more questions...
> ...
> Also, why can't both pendulums oscillate together so that their total
> energy would
> equal the inital energy?  Why does one slow down while the other picks up,
> and then after a certain period, the process reverses?

If the two pendula are identical, they will swing together with no
changes in the motion under two different conditions:
  1.  Both are started at the same time and from the same angle.
  2.  They are started at the same time and opposite angles.
(If the pendula are not identical there are still two stable motions but
both are more complicated.)

The frequency of motion when they are moving together is not the same as
the frequency when they are moving in opposite directions, and the two
types of motion are called the two normal modes.  The motion when the
first of the pendula is moving and the second is stationary at t=0 is
the sum of the two normal modes; the motion when the second pendulum is
moving at the beginning and the first is not is the difference of the
two normal modes [or the reverse depending on which is called #1 and
which is called #2].  Since the two normal modes have different
frequencies, they gradually get out of step, which leads to the first
pendulum slowing down and the second starting to move.  Eventually the
normal modes are exactly out of step, and only pendulum 2 is moving,
then after another interval the modes are back in step and only pendulum
1 moves, and so forth until friction stops the system.

<ALGEBRA MODE>
Let the pendula be x1 and x2, and let the modes be A and B.  The motion
when the first pendulum is started and the second stationary at t=0 is

x1(t) = G[ cos(fA t) + cos(fB t)]
x2(t) = G[ cos(fA t) - cos(fB t)]

in the absence of friction.  Using a trig identity

x1(t) = 2 G cos[ (fA+fB) t/2 ] cos[(fB-fA) t/2]
x2(t) = 2 G sin[ (fA+fB) t/2 ] sin[(fB-fA) t/2]

and in each case the first factor is the frequency of the pendulum and
the second factor gives the slower variation in the amplitude of the
motion.
</ALGEBRA MODE>

To get the values of the two frequencies fA and fB you must know details
of the system.
> > At 03:33 PM 1/11/99 -0800, you wrote:
> > > I was wondering if someone could explain the following:
> > >
> > > Tie a string horizontally above the ground.  Tie two strings with small
> > > balls attached to them (pendulums) to the horizontal string.  Set one
> > > pendulum in motion.  Gradually it will come to a rest and the second
> > > pendulum will begin oscillating.  Then the second stops, the first resumes
> > > and so on.
> > >
> > > I was told it had to do with Resonant Coupling, but I still don't fully
> > > understand it.
> > >
> > > --Thomas
> > > tmphilip@prontomail.com
> > Homer B. (Jim) James                    www.pitt.edu/~hbjst/homepage.htm
> > Adjunct Professor of Physics/Physical Science
> > Community College of Allegheny County
> > Dept. of Physics and Physical Science

--
Maurice Barnhill, mvb@udel.edu
http://www.physics.udel.edu/~barnhill/
Physics Dept., University of Delaware, Newark, DE 19716