Transcript Sect. 4.4

Elliptic Integrals
Section 4.4 & Appendix B
• Brief math interlude:
– Solutions to certain types of nonlinear oscillator
problems, while not expressible in closed form in
terms of “elementary” functions (trig functions, etc.),
they are expressible in terms of Elliptic Integrals
– There is nothing mysterious about these! They are
just special functions which have been studied
completely & thoroughly by mathematicians 150
or more years ago.
– All properties are known (derivatives, Taylor’s series,
integrals, etc.) & tabulated for common values of the
arguments, …..
• Here, because of the application to the Plane
Pendulum problem, we are interested in the
Elliptic Integral of the 1st Kind: F(k,φ).
• From Appendix B:
F(k,φ)  ∫dθ[1- k2 sin2θ]-½
(limits: 0 < θ < φ),
(k2 < 1)
Or, with z = sinθ,
F(k,x) = ∫dz [(1- z2)(1- k2z2)]-½
(limits: 0 < z < x),
(k2 < 1)
Plane Pendulum
Section 4.4
• Pendulum: A mass m, constrained by a massless,
extensionless rod to move in a vertical circle of radius .
• The gravitational force acts downward, but the
component of this force influencing the
motion is  to the support rod.
• This is a nonlinear oscillator system with a
symmetric restoring force.
– Only for very small angular displacements is this is
linear oscillator!
Figure of Plane Pendulum Motion
• Component of the
gravitational force
involved in the
motion =
F(θ) = - mg sinθ
• Equation of motion
(rotational version of
Newton’s 2nd Law):
Torque about the axis
= (moment of inertia)  (angular acceleration)
N = I(d2θ/dt2) =  F(θ);
I = m2

m2 (d2θ/dt2) = - mg  sinθ
• Or
(d2θ/dt2) + (g/) sinθ = 0
Define:
ω02  (g/)
So:
Or:
(d2θ/dt2) + ω02 sinθ = 0
θ + ω02 sinθ = 0
(“natural frequency”)
• If & only if the angular displacement θ is small, then
sinθ  θ & the equation of motion becomes:
θ + ω02θ  0
• This is simple harmonic motion for the angular
displacement θ.
Frequency ω0  (g/)½ Period τ  2π (/g)½
• General equation of motion for the plane pendulum:
θ + ω02 sinθ = 0
(1)
A VERY nonlinear differential eqtn!
A VERY nonlinear oscillator eqtn!
• We could try to solve (1) directly. However,
instead, follow the text & use energy methods!
• The restoring force for the motion:
F(θ) = - mg sinθ (upper figure). This is
a conservative system  A potential
energy function U(θ) exists (lower figure).
Taking the zero of energy at bottom of
the path at θ = 0 & using:
 F(θ) = - (dU/dθ)  U(θ) = mg (1- cosθ)
U(θ) = mg (1- cosθ)
• Kinetic energy: T = (½)I(dθ/dt)2 = (½)m2(θ)2
• Total energy E = T + U is conserved
• Let the highest point of the motion (determined by initial
conditions!) be θ  θ0. θ0 is the amplitude of the oscillatory
motion.
– By definition, T(θ0)  0.
Also, U(θ0) = E = mg(1- cosθ0) = 2mgsin2[(½)θ0] (trig identity)
– Similarly a trig identity gives: U(θ) = 2mgsin2[(½)θ]
• Conservation of total energy 
E = 2mgsin2[(½)θ0) = T + U = (½)m2(θ)2 + 2mgsin2[(½)θ]
• So 2mgsin2[(½)θ0] = 2mgsin2[(½)θ]+ (½)m2(θ)2
m cancels out!
(2)
• Solving (2) for θ = θ(θ) (gives the phase diagram of
the pendulum!) & using the frequency for small
angles: ω02  (g/)
(dθ/dt) =  2ω0{sin2[(½)θ0] - sin2[(½)θ]}½ (3)
• We could integrate (3) & get t(θ) rather than θ(t)
using the period for small angles: τ0 = (2π/ω0)  2π(/g)½
dt = [τ0/(4π)]{sin2[(½)θ0] - sin2[(½)θ]}-½ dθ (4)
• Instead of t(θ), use (4) to get the period τ. Using the fact that
the motion is symmetric & also the definition of the period:

τ = (τ0/π)∫{sin2[(½)θ0] - sin2[(½)θ]}-½dθ
(limits 0  θ  θ0)
(5)
τ = (τ0/π)∫{sin2[(½)θ0] - sin2[(½)θ]}-½dθ
(5)
(limits 0  θ  θ0)
• (5) is an Elliptic Integral of the 1st Kind: F(k,x = 1)
τ  (τ0/π) F(k,1)
F(k,1) = ∫dz [(1- z2)(1- k2z2)]-½, (limits: 0 < z < 1), (k2 < 1)
where k  sin[(½)θ0], z  {sin[(½)θ]/sin[(½)θ0]}
This is tabulated in various places.
• For oscillatory motion, we must have |θ0| < π or
-1< k < 1; {k  sin[(½)θ0]}; (k2 < 1)
• Why? What happens if |θ0| = π?
τ  (τ0/π) F(k,1)
F(k,1) = ∫dz [(1- z2)(1- k2z2)]-½, (limits: 0 < z < 1), (k2 < 1)
where k  sin[(½)θ0], z  {sin[(½)θ]/sin[(½)θ0]}
• Consider small displacements from equilibrium (but
not necessarily so small that sinθ = θ!) (small kz < 1):
– Expand the (1-k2z2)-½ part of the integrand in a Taylor’s
series, & integrate term by term:
(1-k2z2)-½  1 + (½)k2z2 + (3/8)k4z4+ ...

τ  (τ0/π)∫dz(1-z2)-½[1+ (½)k2z2 + (3/8)k4z4 ..]
Using tables (& skipping steps) gives:
τ  τ0[1 + (¼)k2 + (9/64)k4 + ..]
• The period is (approximately):
τ  τ0[1 + (¼)k2 + (9/64)k4 + ..]
(6)
• We had: k  sin[(½)θ0], θ0= amplitude of the oscillations (max
angular displacement). In terms of the amplitude, the period is:
 τ  τ0{1 + (¼)sin2[(½)θ0]+(9/64)sin4[(½)θ0] +..} (7)
If k is large, we need many terms for an accurate result. For
small k, this rapidly converges. k = sin( θ0) is determined by
the initial conditions!!!
• PHYSICS: Unlike the simple pendulum (where sinθ  θ),
the period for a real pendulum depends
STRONGLY on the amplitude!
– For the simple pendulum, the period τ0 = 2π(/g)½ is
“isochronous” (independent of amplitude)
• The period is (approximately; τ0 = 2π(/g)½)
τ  τ0{1+ (¼)sin2[(½)θ0] + (9/64)sin4[(½)θ0] +..} (8)
• For small k = sin[(½)θ0] we can also make the small
θ0 approximation & expand sin[(½)θ0] for small θ0:
sin[(½)θ0]  (½)θ0 - (1/48)(θ0)3
Put this into (8) & keep terms through 4th order in θ0
τ  τ0[ 1 + (1/16)(θ0)2 + (11/3072)(θ0)4 + .. ]
Finally the period as a function of amplitude θ0 for small θ0:
τ  τ0[ 1 + (0.0625)(θ0)2 + (0.00358)(θ0)4 + .. ]
Phase Diagram for the Plane Pendulum
• From conservation of energy, we had:
E = 2mgsin2[(½)θ0] = T + U = (½)m2(θ)2 + 2mgsin2[(½)θ]
So 2mgsin2[(½)θ0] = 2mgsin2[(½)θ] + (½)m2(θ)2 (2)
• Solving (2) for θ = θ(θ) (gives the phase diagram of
the pendulum!)
– Using the frequency for small angles: ω02  (g/)
(dθ/dt) =  2ω0{sin2[(½)θ0] - sin2[(½)θ]}½ (3)
(dθ/dt) =  2ω0{sin2[(½)θ0] - sin2[(½)θ]}½ ;
ω02  (g/) defining E0  2mg
Phase Diagram for the Plane Pendulum
•
Qualitative Discussion
Energy Eqtn: 2mgsin2[(½)θ0] = 2mgsin2[(½)θ] + (½)m2(θ)2 (2)
ω02  (g/) (dθ/dt) =  2ω0{sin2[(½)θ0] - sin2[(½)θ]}½
(3)
• For θ & θ0 small, eqtn (2) becomes: (θ/ω0)2 + θ2  (θ0)2 Using
coordinates (θ/ω0) = (/g)½θ & θ, phase paths are ellipses (this is a SHO!).
• For general –π < θ < π, we have
E < 2mg  E0. m is bound in a well:
U(θ) = mg(1 - cosθ)
Phase paths are closed curves given by (3)
• U(θ) is periodic in θ, so we only need to plot
– π < θ < π. From (dU/dθ) = 0 & looking at
(d2U/dθ2), points θ =  2nπ, 0 are
positions of stable equilibrium. Also, when damping exists (as in a
real pendulum) these points become attractors (for long times, the phase
paths will spiral towards these points).
• If E > 2mg  E0, the motion no longer oscillatory, but it is
still periodic! This corresponds ------------------------- E
to the pendulum making complete
(circular) revolutions about the
support axis. We still have
U(θ) = mg(1 - cosθ) but the particle has enough energy to move
from one periodic valley to the next (“over the hill”; see figure).
• We still have conservation of energy
E = T + U = (½)m2(θ)2 + 2mgsin2[(½)θ]
But, now, θ0 is not defined! Instead, E is just
some constant determined by initial conditions.
The phase paths are open curves, still given by:
(dθ/dt) =  2ω0{sin2[(½)θ0) - sin2[(½)θ)]}½ (3)
• If E = 2mg  E0  θ0 =  π (mass initially vertical!)
The phase path eqtn (dθ/dt) =  2ω0{sin2[(½)θ0] - sin2[(½)θ]}½
Becomes:
(dθ/dt) =  2ω0cos[(½)θ]
The phase paths in this case are 2
simple cosine functions (the heavy
curves in the figure)
• The phase paths with E = E0 don’t
represent actual continuous motions
of the pendulum! These are paths of
unstable equilibrium.  If the pendulum were at rest
with θ0 =  π, any small disturbance would cause it to move
on some path E = E0 + δ (δ very small).
• If the motion were on the phase path
E = E0, the pendulum would reach θ = nπ
with 0 velocity {(dθ/dt) =  2ω0cos[(½)nπ] = 0}
but only after an infinite time! Proof:
We had the period (limits 0  θ  θ0)
τ = (τ0/π)∫{sin2[(½)θ0] - sin2[(½)θ]}-½dθ
Set θ0 = π & get τ  
• A phase path separating locally bounded motion
from locally unbounded motion (like E = E0 for the
pendulum) is called a “SEPARATRIX”
– A separatrix always passes through a point of unstable equilibrium.
Motion in the vicinity of a separatrix is extremely sensitive to the
initial conditions. Points on either side of the separatrix have very
different trajectories (like pendulum case just described!)