seconds after the
coincident chimes
and the second bell rings at 4,8,12,16, seconds after the
coincident chimes. Therefore the next
pair of coincident chimes occurs at 12 seconds. Alternatively (and quite
unnecessarily unless you
are in an ODE modeling course), let the chimes of the first bell be
described by
, where we agree that a chime sounds whenever
with n>0 an integer. Similary, the second bell is
described by 
. The phase difference
satisfies 
, an ODE with a
solution 
( 
). Clearly 
(first set of coincident chimes)
when t=12 seconds.
radians represent one
complete revolution of the clock hands, then the minute hand is governed
by the ODE 
(angular frequency is radians per hour)
and the hour hand is governed by the ODE 
(angular
frequency is 
radians per hour). Letting t=0 be 1:00, the
initial conditions are m(0)=0 and 
. The
solutions of the ODEs are 
and 
.
Setting m(t)=h(t), we find that the hands are coincident at t=1/11
hours after 1:00.
Alternatively and without ODEs, we may proceed in two different ways.
Let 
be the fraction of the circle that the hour hand moves
between 1:00 and the time of coincidence. Let T be the time, in hours,
between 1:00 and the time of coincidence. The hour hand travels at a
rate of 
radians per hour. The minute hand travels at a
rate of radians per hour. The angular distance traveled by the
hour hand is 
. The angular distance traveled by
the minute hand is 
. Eliminating
from both of these equations and solving for T, we find that the time
between 1:00 and the time of coincidence is T = 1/11 hours.
We can also view the problem as a tortoise and hare race. Assume we
start at 1:00: the minute hand starts at the 12 and the hour hand starts
at the 1. It takes 1/12 of an hour for the minute hand to get to the 1.
But in that time the hour hand has advanced 
of a
revolution. It takes the minute hand of an hour to get
to this new position. In that time, the hour hand has advanced 
of a revolution. It takes the minute hand of an hour to get to this new position.
Continuing with this pattern, the minute hand will catch the hour hand
after an infinite number of steps; but this does not mean an infinite
amount of time (the resolution of Zeno's paradox). The time elapsed in
hours is given by the geometric series
When 
(the modulation of the frequency is very weak), we see
that 
which is the frequency of the uniform
oscillator. As 
(strong modulation of the
frequency), the period becomes arbitrarily large.
with 
. (You can use any interval of length .)
Equilibrium points are
the roots of 
or equivalently the intersection points of
the curves
and 
. We see that 
and 
are equilibrium points for all values of
. When 
, there are two additional equilibrium points
at 
. Several pictures of f
and the flows on the circle determine the stability of these points.
Either from the pictures or by linearized stability analysis, the equilibrium points behave as follows:
is stable for
and unstable otherwise.
is stable for
and unstable otherwise.
exist only for
and are unstable.
