the transfer function is negative, and the maximum pendulum displacement is now
smaller than the top’s maximum displacement. At very high driving frequencies
(
!0), the limit of the transfer function is zero, and the angular amplitude of
the pendulum motion is X0/l: the pendulum mass stays near the equilibrium point,
while the top moves back and forth.
The energy (for small oscillations) is
E =
1
2ml2˙2 +
1
2mgl2 + ml ˙ ˙X +
1
2mV 2
=
1
2ml(l
2 + g)2
0 cos2(
t) + ml
20X0 sin2(
t) +
1
2m
2X2
0 cos2(
t)
=
1
2mX2
0
cos2(
t)
(
2 + !2
0)
2
!2
0 −
2 +
2
+ sin2(
2t)
2
2
!2
0 −
2
and the power absorbed by the system is
dE
dt
= mlV ¨ + m˙R · ¨R
= mlX00
2 sin
t cos
t + mX2
0
2 sin
t cos
t
=
1
2mX2
0
2
2
!2
0 −
2 + 1
sin 2
t
=
1
2mX2
0
2!2
0
!2
0 −
2 sin 2
t
Since dE/dt is periodic with frequency 2
, the system absorbs energy during half of
the cycle of the top motion, and “returns” the energy in the other half. Which half is
which depends on the driving frequency being larger or smaller than the pendulum
energy, i.e., the pendulum moving in phase or out of phase with the support. The peak
power is proportional to a factor X2
0!2
0
2/(!2
0−
2): this is very large near resonance;
small and proportional to X0
2 at low driving frequencies; and independent of driving
frequency at large driving frequencies, / X2
0!2
0.
• Vertical periodic motion: Y = Y0 cos(
t), X = 0.
The equation of motion is −l¨ = l(g− ¨ Y ) sin = l(g+
2Y0 cos
t) sin . The motion
of the support adds an oscillating component to the gravitational acceleration. This
is not a driven oscillator like the previous case, because the oscillatory driving force
is vertical but the natural oscillatory pendulum motion is horizontal. If Y0
2 g,
we can use a perturbative approach to find a solution starting with the pendulum
solution; if Y0
2 g, we can also use a perturbative approach to find a solution
starting with the driven pendulum solution.
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