# Preface

This is devoted to the fasinated topic---pendulum equations.

Introduction to Linear Algebra with Mathematica

# Pendulum Equations

A simple pendulum consists of a single point of mass m (bob) attached to a rod (or wire) of length $$\ell$$ and of negligible weight. We denote by θ the angle measured between the rod and the vertical axis, which is assumed to be positive in counterclockwise direction. By applying the Newton’s law of dynamics, we obtain the equation of motion
$m\ell^2 \ddot{\theta} + mg\ell \,\sin \theta =0 .$
It can be simplified by putting $$\omega = \sqrt{g/\ell} :$$
$\ddot{\theta} + \omega^2 \sin \theta =0 .$
Further simplification is available by normalization, which leads to $$\omega = 1$$ and we get
$\ddot{\theta} + \sin \theta =0 .$
In practice, it is easier to study an ordinary differential equation as a system of equations involving only the first derivatives.
$\dot{x} = y, \qquad \dot{y} = -\sin x .$
The variables x and y can be interpreted geometrically. Indeed, the angle x = θ corresponds to a point on a circle whereas the velocity $$y = \dot{\theta}$$ corresponds to a point on a real line. Therefore, the set of all states (x ,y) can be represented by a cylinder, the product of a circle by a line. More formally, the phase space of the pendulum is the cylinder $$S^1 \times \mathbb{R} ,$$ its elements are couples (position,velocity).

Thus, at each point (x ,y) in the phase space, there is an attached vector $$(\dot{x}, \dot{y} ) = (y, - \sin x) .$$ This can be geometrically represented as a vector field on the cylindrical phase space.

Example: The swinging mass m has a kinetic energy of $$m \ell^2 \left( {\text d}\theta /{\text d}t \right)^2 /2$$ and a potential energy of $$mg\ell\left( 1 - \cos \theta \right) ;$$ the potential energy is zero for θ = 0. Let θM denote the maximum amplitude of the pendulum. Since dθ/dt = 0 at θ = θM, conservation of energy gives

$\frac{1}{2} \,m\ell^2 \left( \frac{{\text d}\theta}{{\text d}t} \right)^2 = mg\ell\,\cos \theta - mg\ell\,\cos \theta_M .$
Solving for the velocity dθ/dt, we obtain
$\frac{{\text d}\theta}{{\text d}t} = \pm \left( \frac{2g}{\ell} \right)^{1/2} \left( \cos \theta - mg\ell\,\cos \theta_M \right)^{1/2} ,$
with the mass m canceling out. We take t to be zero, when θ = 0 and dθ/dt > 0. An integration from θ = 0 t0 θM yields
$$\int_0^{\theta_M} \left( \cos \theta - mg\ell\,\cos \theta_M \right)^{-1/2} {\text d}\theta = \left( \frac{2g}{\ell} \right)^{1/2} \int_0^t {\text d}t = \left( \frac{2g}{\ell} \right)^{1/2} t . \label{pendulum.1}$$
This is one-fourth of a cycle, and therefore the time t is one-fourth of the period, T. We note that θ < θM, and with a bit of clairvoyance we try the half-angle substitution
$\sin \frac{\theta}{2} = \sin \left( \frac{\theta_M}{2} \right) \sin \phi .$
With this, Eq.\eqref{pendulum.1} becomes
$$T = 2\pi \left( \frac{\ell}{g} \right)^{1/2} \int_0^{\pi} \left[ 1 - \sin^2 \left( \frac{\theta_M}{2} \right) \sin^2 \phi \right]^{-1/2} {\text d}\phi . \label{pendulum.2}$$
Although not an obvious improvement over Eq.\eqref{pendulum.1}, the integral \eqref{pendulum.2} now defines the complete elliptic integral of the first kind,$$K \left( \sin^2 \theta_M /2 \right) .$$ From the series expansion, the period of our pendulum may be developed as a power series—powers of sin θM/2:
$T = 2\pi \left( \frac{\ell}{g} \right)^{1/2} \left\{1 + \frac{1}{4}\,\sin^2 \frac{\theta_M}{2} + \frac{9}{64}\,\sin^4 \frac{\theta_M}{2} + \cdots \right\} .$
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# Pendulum Equation with Resistance

We convert the pendulum equation with resistance
$m\ell^2 \ddot{\theta} + c\,\dot{\theta} + mg\ell \,\sin \theta =0 .$
to a system of two first order equations by letting $$x= \theta \quad\mbox{and} \quad y = \dot{\theta} :$$
$\frac{{\text d} x}{{\text d}t} = y , \qquad \frac{{\text d} y}{{\text d}t} = -\omega^2 \sin x - \gamma \,y .$
Here $$\gamma = c/(m\,\ell ) , \ \omega^2 = g/\ell$$ are positive constants. Therefore, the above system of differential equations is autonomous. Setting $$\gamma = 0.25 \quad\mbox{and}\quad \omega^2 =1 ,$$ we ask Mathematica to provide a phase portrait for the pendulum equation with resistance:

# Spherical Pendulum

Consider the motion under gravity of a bob of mass m attached to a fixed point by an inextensible massless rod of length ℓ. The bob is free to move on a sphere of radius ℓ. Using spherical coordinates (angles) θ and ϕ as the generalized coordinates, the kinetic and potential energies become
\begin{align*} \mbox{K} &= \frac{m}{2}\,\ell^2 \left( \dot{\theta}^2 + \dot{\phi}^2 \sin^2 \theta \right) , \\ \Pi &= mg\ell \left( 1- \cos \theta \right) . \end{align*}
This allows us to derive the Euler--Lagrange equations for the corresponding Lagrangian:
$\frac{\text d}{{\text d}t} \left( \frac{\partial {\cal L}}{\partial \dot{\theta}} \right) - \frac{\partial {\cal L}}{\partial \theta} = 0 , \qquad \frac{\text d}{{\text d}t} \left( \frac{\partial {\cal L}}{\partial \dot{\phi}} \right) - \frac{\partial {\cal L}}{\partial \phi} = 0 .$
Substituting the expressions for the kinetic and potential energies, we obtain two coupled equations
$\begin{split} \ddot{\theta} + \frac{g}{\ell} \,\sin\theta - \frac{1}{2} \,\dot{\phi}^2 \sin 2\theta &= 0, \\ \ddot{\phi}\,\sin\theta + 2\dot{\phi} \dot{\theta}\,\cos\theta &= 0. \end{split}$
The last equation is a first order differential equation with respect to the derivative of ϕ, so it can be integrated
$\dot{\phi} \sin^2 \theta = c, \quad \mbox{a constant}.$
Using this equation, we obtain the following differential equation for nagle θ:
$\dot{\theta} + \frac{g}{\ell} \,\sin\theta - \frac{c^2 \cos\theta}{\sin^3 \theta} = 0.$

# Applications

Example 2: Consider a pendulum with constant torque T. Its motion is governed by the differential equation
$\ddot{\theta} + \frac{g}{\ell}\,\sin\theta = \frac{T}{m\ell^2} .$
Its equilibria are determined by solving the algebraic equation
$\sin\theta = \frac{T}{mg\ell} .$
For real roots, $$\displaystyle \frac{T}{mg\ell} < 1,$$ illustrating fold or saddle-node bifurcation Its first integral is
$\frac{y^2}{2} - \frac{g}{\ell}\,\cos x - \frac{T}{m \ell^2}\, x = \mbox{constant}.$
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Example 3: Consider a pendulum in a plane that is rotating about a vertical axis with angular velocity ω. Its motion is governed by the differential equation
$\ddot{\theta} + \frac{g}{\ell}\,\sin\theta - \omega^2 \sin x \,\cos x = 0.$
Equilibria: y = 0, x = 0, π, and also $$\displaystyle \cos x = \frac{g}{\omega^2 \ell} .$$ For real roots, $$\displaystyle \frac{g}{\omega^2 \ell} < 1,$$ illustrating a pitchfort bifurcation. The first integral:
$\frac{y^2}{2} - \frac{g}{\ell} \,\cos x + \frac{\omega^2}{4}\.\cos 2x = \mbox{constant}.$
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We consider a periodic movement described by the equation
$\dot{x} = -y^3 , \qquad \dot{y} = x^3 .$
F[x_, y_] := -y^3;
G[x_, y_] := x^3;
sol = NDSolve[{x'[t] == F[x[t], y[t]], y'[t] == G[x[t], y[t]],
x[0] == 1, y[0] == 0}, {x, y}, {t, 0, 3*Pi},
WorkingPrecision -> 20]
ParametricPlot[Evaluate[{x[t], y[t]}] /. sol, {t, 0, 3*Pi}]

Out[11]=
{{x -> InterpolatingFunction[{{0, 9.4247779607693797154}}, <>],
y -> InterpolatingFunction[{{0, 9.4247779607693797154}}, <>]}}
X[t_] := Evaluate[x[t] /. sol]
Y[t_] := Evaluate[y[t] /. sol]
fns[t_] := {X[t], Y[t]};
len := Length[fns[t]];
Plot[Evaluate[fns[t]], {t, 0, 3*Pi}]

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Its translation from the Russian in English by Lisa Shields with an introduction by Peter Lynch is available on the web:
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