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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 256 25 "The mathematical pendul
um" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "We
 follow Gerd Baumann's approach in Mathematica to determine the period
 of the mathematical pendulum." }}{PARA 0 "" 0 "" {TEXT -1 165 "First,
 we write down the kinetic and potential energies: the kinetic energy \+
is expressed in terms of the angular velocity and the moment of inerti
a for a point mass " }{TEXT 274 1 "m" }{TEXT -1 23 " suspended at dist
ance " }{TEXT 273 1 "l" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 43 "We call the angular velocity d/dt phi=omega" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 24 "Tkin:=1/2*m*l^2*omega^2;" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 77 "The potential energy follows from the height of the pen
dulum bob (point mass " }{TEXT 275 1 "m" }{TEXT -1 74 "). We choose th
e potential energy to be zero if the mass is at the bottom." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Upot:=m*g*l*(1-cos(phi));" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "For plotting we have to pick som
e parameter values:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "l:=1;
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "g:=10;" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 5 "m:=1;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 141 "plots[contourplot](Tkin+Upot,phi=-3/2*Pi..3/2*Pi,ome
ga=-10..10,contours=[1,5,9,13,17,21,25,29],axes=boxed,filled=true, col
oring=[red,green]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 547 "This phas
e space diagram is obviously periodic with period 2 Pi. For small angu
lar velocities one has a system approximated by a harmonic oscillator,
 while for larger velocites (or energies, i.e., higher contours) one h
as solutions that simply revolve (winding the pendulum; in the periodi
c phase space diagrams one leaves e.g., at the right to re-enter at th
e same angular velocity at the left. The motion is left to right in th
e top half (for the open curves), and right to the left in the bottom \+
half (look at the sign of the angular velocity)." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 115 "The regularity of the di
agram is very striking. The units of energy (chosen for the contour li
nes) are in MKSA, if " }{TEXT 259 1 "l" }{TEXT -1 1 "," }{TEXT 258 1 "
g" }{TEXT -1 1 "," }{TEXT 257 1 "m" }{TEXT -1 19 " are given in MKSA.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 358 "Ther
e is a critical energy where the separation occurs between oscillating
 solutions (closed contours, in the somewhat abstract phase space one \+
obtains elliptic or deformed elliptic 'trajectories'; translate those \+
into oscillations using your imagination, by watching the sign of omeg
a, passage through phi=0, i.e., the bottom position of the pendulum, e
tc.)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 229 "
The locations phi=+Pi, omega=0, or phi=-Pi, omega=0 correspond to the \+
unstable equilibrium (pendulum at the top at rest). One can show that \+
the phase-space 'trajectory' become a cosine curves of phi/2 [Marion-T
hornton, eq. 4.31]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 385 "For winding solutions we observe how their angular vel
ocity is modulated as the go through the (periodic) potential well. Fo
r sufficiently high energies (not shown) the phase space 'trajectories
' approach straigh lines, as the pendulum bob's kinetic energy is so h
igh that the variation in the potential energy (by being moved up and \+
down in the graviational field) becomes negligible." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 298 "Time-dependent perturb
ations of the pendulum by an external force can lead to a loss of regu
larity in the phase space diagram (CHAOS). The modulation can come fro
m a simple coupling to a second pendulum, which leads to a time-varyin
g support point (science museum demonstrations + classroom demo!)." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "The dete
rmination of the phase-space 'trajectories' was easy. The actual integ
ration of phi(t) is not!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 145 "The \+
total energy can be expressed through the potential energy at the turn
ing point (pendulum bob at rest), i.e., through the initial angle phi0
." }}{PARA 0 "" 0 "" {TEXT -1 77 "One introduces the energy parameter \+
(initial phase angle phi0) k=sin(phi0/2);" }}{PARA 0 "" 0 "" {TEXT -1 
135 "To recover the general expression for the potential energy we una
ssign the parameters for length, gravitational acceleration, and mass:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "l:='l': g:='g': m:='m':
 Upot;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "Upot:=algsubs(cos
(phi)=1-2*sin(phi/2)^2,Upot);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 26 "Etot:=subs(phi=phi0,Upot);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
61 "The total energy for a given trajectory thus depends only on " }
{TEXT 261 1 "k" }{TEXT -1 66 "=sin(phi/2). Therefore, the period is ca
lculated as a function of " }{TEXT 260 1 "k" }{TEXT -1 1 "." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 145 "To calculate t
he period for the oscillatory solutions, one uses the energy expressio
n which depends on phi and omega=diff(phi,t), isolates omega." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "Tkin+Upot=Etot;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "solve(%,omega);" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 41 "DE:=diff(phi(t),t)=subs(phi=phi(t),%[1]);" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "sol:=dsolve(\{DE,phi(0)=p
hi0\},phi(t));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 224 "Maple 5.0 come
s up with a formal solution for phi(t). It hints at the possibility of
 finding the solution through elliptic integrals. In Maple 6.0 no answ
er is returned after the program tries to find a solution for a while.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 "For t
he determination of the period " }{TEXT 262 1 "T" }{TEXT -1 117 " one \+
can proceed differently: the differential equation DE can be separated
 so that an expression for dt is obtained:" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 32 "subs(diff(phi(t),t)=dphi/dt,DE);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 12 "solve(%,dt);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 203 "If we integrate the right-hand side over phi from 0 to p
hi0, the left-hand side will give us a quarter of the period T (for th
e full period phi would have to continue through 0 to -phi0 and back t
o phi0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
71 "The integral to be carried out falls into a category known as elli
ptic." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "assume(k>0);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "A required integral can be express
ed as an elliptic function:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
40 "Int(1/sqrt((1-z^2)*(1-k^2*z^2)),z=0..1);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 9 "value(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 20 "omega[0]:=sqrt(g/l);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "T
o help Maple we restrict the inital phase to be positive, and less tha
n Pi." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "assume(phi0>0,phi0
<Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "T:=2/omega[0]*Int(
1/sqrt(sin(phi0/2)^2-sin(phi/2)^2),phi=0..phi0);" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 12 "T:=value(T);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 126 "For plotting purposes let us simplify matters by chosing
 the pendulum length and the gravitational acceleration in MKSA units:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "g:=10; l:=10;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "T;" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 67 "We now explore this answer as a function of the initial
 angle phi0:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(T,phi0
=0..Pi);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 108 "In the limit of the \+
intial phase going to Pi (unstable equilibrium at the top) we obtain a
n infinite period:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "limit
(T,phi0=Pi,left);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 108 "What do we get for small initial phase angles (we
 should recover isochronicity in the small-amplitude limit)" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plot(T,phi0=0..Pi/16);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "taylor(T,phi0=0,5);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "T1:=algsubs(sin(phi0/2)=k,T)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "series(T1,k=0,8);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "T1t:=convert(%,polynom);" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "This agrees with eq. 2.25 in Baum
ann (or analytical mechanics texts)." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "plot(\{Re(T1),T1t\},k=0..1);" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 50 "An alternative would be to introduce the variable " }
{TEXT 269 1 "k" }{TEXT -1 26 " directly in the integrand" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "T2:=2/omega[0]*int(1/sqrt(k^2-sin(p
hi/2)^2),phi=0..2*arcsin(k));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 13 "simplify(T2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "This does \+
not work in Maple5! " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "sim
plify(T2,symbolic);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 132 "Maple is \+
reluctant to carry out the cancellation of function and its inverse. I
n Maple6 T2 is directly calculated as 4*EllipticK(k)." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 75 "A further exploration \+
should look into the calculation of trajectories phi(" }{TEXT 270 1 "t
" }{TEXT -1 2 ")!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 376 "Before we proceed with this step we construct the phase \+
diagram (a plot of the time derivative of the dynamic variable [the ph
ase angle phi] against the dynamic variable itself. For this we need t
he energy expression (we will look at contour lines of equal height, i
.e. lines of constant energy, as the mechanical energy is conserved fo
r a frictionless mathematical pendulum)." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "l:='l': g:='g': m:='m':" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 11 "phi:='phi';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 3 "DE;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "Let us make use of Ge
rd Baumann's analytic work:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
20 "omega[0]:=sqrt(g/l);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "
Etot;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "E_s:=Etot/(m*l^2*o
mega[0]^2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 255 "The scaled energy
 parameter allows to distinguish easily the different types of motion:
 for E_s<2 we have oscillations, for E_s>2 rotations, and E_s=2 corres
ponds to the asymptotic solution where the pendulum bob remains in the
 unstable equilibrium point." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 55 "We express the scaled energy in terms of the pa
rameter " }{TEXT 263 1 "k" }{TEXT -1 13 "=sin(phi0/2):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "k0:=7/8;" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 12 "E_s:=2*k0^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 11 "phi:='phi':" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 99 "We define \+
the differential equation such that positive and negative angular velo
cities are allowed." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "DE:=
(diff(phi(t),t))^2=omega[0]^2*(2*(cos(phi(t))-1+E_s));" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 69 "Oscillatory solutions to the equation of \+
motion (E_s<2) are given by:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 49 "theta:=(k,t)->2*arcsin(k*JacobiSN(omega[0]*t,k));" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "g:=10; l:=1;" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 28 "subs(phi(t)=theta(k0,t),DE);" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 19 "evalf(subs(t=1,%));" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 83 "Without further work Maple cannot figure out that the exp
ression written for theta(" }{TEXT 271 1 "t" }{TEXT -1 134 ") solves t
he differential equation. Nevertheless, we can verify that within nume
rical accuracy the differential equation is satisfied." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "plot([theta(0.5,t),theta(0.8,t),th
eta(0.9,t),theta(0.999,t)],t=0..2*Pi,color=[red,blue,green,black],view
=[0..2*Pi,-Pi..Pi]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 181 "One obse
rves that for moderate energy values the solution still looks like a s
ine-curve, although the dependence of the period on the energy is appa
rent already. Only for values of " }{TEXT 264 1 "k" }{TEXT -1 164 " cl
ose to unity do we see a qualtiative change in the behaviour of the so
lution, i.e., a dramatic slowdown of the pendulum bob in the vicinity \+
of the turning point." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 99 "We can follow this up with an animation. To display th
e pendulum's motion we need to calculate the " }{TEXT 267 1 "x" }
{TEXT -1 5 " and " }{TEXT 266 1 "y" }{TEXT -1 36 " coordinates. We cho
ose a length of " }{TEXT 268 1 "l" }{TEXT -1 99 "=1 and define the coo
rdinates such that the height is zero at the suspension point of the p
endulum." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "x:=(k,t)->evalf
(sin(theta(k,t))): y:=(k,t)->-evalf(cos(theta(k,t))):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "x(0.99,1),y(0.99,1);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 50 "We draw lines from the origin to the pend
ulum bob:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 151 "PL:=(k,t)->pl
ot([[0,0],[x(k,t),y(k,t)]],style=line,color=magenta,title=cat(cat(\"k=
 \",convert(evalf(k,3),string)),\"time= \",convert(evalf(t,3),string))
):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "PLs:=seq(PL(0.995,it/
10),it=1..200):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plo
ts):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "display(PLs,inseque
nce=true,axes=boxed,scaling=constrained);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 231 "The slowdown of the pendulum bob as the top position is \+
approached is very obvious. Thus the lack of isochronicity, or the dep
endence of the period on the maximum amplitude (energy deposited into \+
the pendulum) is made very obvious." }}{PARA 0 "" 0 "" {TEXT -1 298 "I
n Maple 6 the title for each frame is actually displayed in the animat
ion. However, we found two errors for Maple 6.0 in this worksheet: 1) \+
the contourplot is wrong; 2) the numerical evaluation of the JacobiSN \+
functions is in error. We can only hope that a service pack will remed
y these problems." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 215 "While the demonstration of Maple's handling of el
liptic integrals has been impressive, we wish to demonstrate also a ge
neral approach that can be applied to any differential equation, namel
y the numerical solution. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
63 "DE:=k->(diff(phi(t),t))^2=omega[0]^2*(2*(cos(phi(t))-1+2*k^2));" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "omega[0]:=1;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "IC:=phi(0)=0;" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 43 "sol:=dsolve(\{DE(0.999),IC\},phi(t),numeric);
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "Bad news! We need to go back
 to the original Newton law. Maple 6 is, however, improved in this are
a." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "omega:='omega': # it \+
is not good to mix omega(t), and omega[0], which makes omega a table.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "DE:=diff(phi(t),t)=omeg
a(t),diff(omega(t),t)=-(g/l)*sin(phi(t));" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 94 "The initial condition is chosen to be the turning point, \+
where all energy is in the potential." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "phifun:=k->2*arcsin(k);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "phi0:=phifun(0.999);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 27 "IC:=omega(0)=0,phi(0)=phi0;" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 68 "sol:=dsolve(\{DE,IC\},[omega(t),phi(t)],numeric,
output=listprocedure);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "t
heta:=subs(sol,phi(t));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 125 "In th
e plot command we need to put quotes around theta(t), as the expressio
n can only be evaluated for numerical values of t." }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 27 "plot('theta(t)',t=0..2*Pi);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Omega:=subs(sol,omega(t));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plot('Omega(t)',t=0..2*Pi);
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 152 "It is evident that there are
 extended time segments when the pendulum bob moves with low velocitie
s. A useful exercise would be to reduce the parameter " }{TEXT 265 1 "
k" }{TEXT -1 137 ", and to observe when the relationship between angul
ar velocity and angle returns to that of a harmonic oscillator (sine/c
osine curves). " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "We can also lo
ok at the solutions where the pendulum bob rotates." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 40 "theta:=(t,k)->2*JacobiAM(sqrt(g/l)*t,k);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "g:=10; l:=1;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "plot(Re(theta(t,0.6*I)),t=0.
.4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "evalf(theta(1,5.5*I
));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "The imaginary values of " 
}{TEXT 272 1 "k" }{TEXT -1 84 " come from the following analytic conti
nuation of the energy-initial angle relation:" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 9 "E_s:=2.1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 28 "solve(1-E_s=cos(phi1),phi1);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 17 "k:=evalf(sin(%));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 141 "Exercise: Adapt the animation to display the motion in the cas
e where the pendulum bob has sufficient energy to rotate over the top \+
position." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 \+
0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }