辅导案例-AME 525
AME 525 - Fall 2019 - Final Project
Due: Mon 12/9.
Late projects will not be accepted.
PART I: Harmonic Oscillator. Consider a one-dimensional mass-spring system where the spring force
is proportional to the distance x measured from an origin taken to coincide with location of the un-stretched
spring: F (x) = −kx. Our goal is to write the equation of motion of the mass and understand its behavior.
The equation of motion is given by
mx¨+ kx = 0 (1)
The energy E = 12mx˙
2 + 12kx
2 is conserved. The value of E is dictated by the initial conditions x(0) = xo,
x˙(0) = x˙o. That is,
E =
1
2
mx˙2 +
1
2
kx2 =
1
2
mx˙2o +
1
2
kx2o = constant
m
k
x
Figure 1: spring mass system
Solve the equations of motion. The solution to (1) is given by
x(t) = a cos(ωt+ φ), (2)
where
ω =

k
m
, a =

x2o +
x˙2o
ω2
, tanφ = − x˙o
ωxo
Here, ω is the frequency, a is the amplitude of oscillations and φ as the phase. φ is relevant only when
comparing different trajectories. Note that if xo = 0 and x˙o = 0, then one has x = 0 and x˙ = 0 for all time.
The oscillator is said to be in a state of equilibrium, or simply in equilibrium.
We will now present an alternative way of looking at all possible solutions, which in the case of the
harmonic oscillator are either the mass is at rest, also called in equilibrium, or the mass is undergoing periodic
oscillations.
Phase space (x, x˙). For mx¨+kx = 0, the solution curves in the phase space are concentric ellipses. There
are two ways to see this:
• Look at the solution x = a cos(ωt+ φ) and its time derivative x˙ = aω sin(ωt+ φ). One has
x = a cos(ωt+ φ)
x˙ = aω sin(ωt+ φ)
 =⇒
x
a
= cos(ωt+ φ)


= − sin(ωt+ φ)
 =⇒ x
2
a2
+
x˙2
a2ω2
= 1
It is easy to see that this equation represents a family of similar ellipses. Through each point of the
phase plane (x, x˙) there passes one and only one ellipse; it corresponds to a given value of a. In other
words, each ellipse corresponds to a class of initial conditions, namely, to a set of initial conditions
that have the same total energy.
1
• Look at the energy of the system E = 1
2
mx˙2 +
1
2
kx2 which is conserved. That is, E = constant. Thus,
all solutions must satisfy
x˙2 + ω2x2 =
2E
m
=⇒ x˙
2
2E
m
+
x2
2E
mω2
= 1
Again, this equation represents a family of similar ellipses. Through each point of the phase plane
(x, x˙) there passes one and only one ellipse; it corresponds to one value of the total energy or one
energy level.
PART II: Simple Pendulum. Consider the simple planar pendulum consisting of a point mass m at-
tached to an inextensible string as shown in the diagram. Our goal is to write down the equation of motion
of the mass and understand its behavior. The equation of motion for the pendulum is given by:
θ¨ = −g
l
sin θ (3)
We will focus on understanding the behavior of equation (3). This is a non-linear 2nd-order ordinary
differential equation. We can’t use the tools we know from linear ODEs to solve this equation. Actually,
the solution to this equation for general initial conditions θo and θ˙o is not readily available in terms of a
simple analytic function. Yet, we can understand a lot about the behavior of the system even without solving
explicity for θ(t).
e1
e2 θ
l
m
g
Figure 2: simple planar pendulum
Before we proceed note that equation (3) conserves energy. A simpler way to arrive at energy conser-
vation is to multiply both side of the equation by θ˙ and integrate over time to get
E =
1
2
θ˙2 − g
l
cos θ = constant
Phase space. The phase space (θ, θ˙) is 2-dimensional. But, given that (3) is periodic in θ with period
2pi, one only needs to understand the behavior of (3) over one segment of the phase space corresponding to
one period of θ, say θ ∈ [−pi, pi], θ˙ ∈ R. A depiction of the phase velocity field and phase portrait over two
periods in θ is shown in Figures 3. The periodicity in θ is apparent.
It is useful to replace (3), which is a second-order differential equation, by two equivalent first-order
differential equations. This can be done by defining the coordinates θ1 and θ2
θ1 = θ
θ2 = θ˙
(4)
Taking the time derivative of the above relations and using (3), one gets
θ˙1 = θ˙ = θ2
θ˙2 = θ¨ = −g
l
sin(θ) = −g
l
sin(θ1)
(5)
2
−6 −4 −2 0 2 4 6
−5
0
5
x1
x 2
−6 −4 −2 0 2 4 6
−3
−2
−1
0
1
2
3
Figure 3: (left)Phase velocity field for the simple pendulum θ¨ + sin θ = 0 (where g/l = 1). The origin is a stable center while
the point (±pi, 0) is a saddle. (right) Phase portrait for the simple planar pendulum. The parameter values are g = 1 and l = 1.
That is, equation (3) can be rewritten as 2 first-order differential equations
θ˙1 = θ2
θ˙2 = −g
l
sin(θ1)
(6)
Phase Portrait. The phase portrait is a collection of trajectories in the (θ, θ˙). In general, one needs to
integrate the second-order equation (3) in order to find θ(t) and θ˙(t). But since energy is an integral of
motion and since this is a one-degree of freedom system (the degree of freedom is θ). Then, the system is
integrable and the trajectories in the phase space can be obtained from the conservation of energy. In other
words, Energy conservation,
E =
1
2
θ˙2 − g
l
cos θ = constant
describes solution curves in the phase space (θ, θ˙). Figure 3 shows the phase velocity and the phase portrait
of the simple planar pendulum.
Equilibrium Points or Fixed Points. We look for solutions for which the system is in equilibrium, that
is, the phase velocity is zero, θ˙1 = 0 and θ˙2 = 0. As we will see later, these solutions are extremely important
for understanding the behavior of the pendulum. We have two such solutions:
• θ = 0 and θ˙ = 0 is an equilibrium point of (3), that is, starting at (0, 0), one has θ = 0 and θ˙ = 0 for
all time. This corresponds to when we place the pendulum at its lowest point without initial velocity.
• θ = ±pi and θ˙ = 0 is also an equilibrium.This corresponds to when we place the pendulum at its highest
point (vertically upward) without initial velocity.
Linear Stability. One way to check if an equilibrium is stable is to look at the linearized equations at the
equilibrium:
• An equilibrium is said to be linearly stable if the eigenvalues (i.e., the roots of the characteristic
polynomial) of the linearized equations at the equilibrium have strictly negative real parts.
• An equilibrium is said to be neutrally stable if the eigenvalues (i.e., the roots of the characteristic
polynomial) of the linearized equations at the equilibrium are pure imaginary.
• An equilibrium is said to be unstable if at least one of the eigenvalues (i.e., the roots of the characteristic
polynomial) of the linearized equations at the equilibrium has a positive real part.
3
In the pendulum example, one has two equilibria (0, 0) and (±pi, 0). Let’s begin by examining the
stability of the equilibrium at (0, 0).
Linearized equations for the pendulum about the vertically down position. We linearize (3)
about (0, 0). This means that we begin by letting
θ1 = 0 + ∆θ1, θ2 = 0 + ∆θ2
where ∆θ1 and ∆θ2 are ‘small’. We then substitute into (6) to get
˙̂
∆θ1 = ∆θ2,
˙̂
∆θ2 = −g
l
sin(∆θ1)
but, since ∆θ1 is small, one has: sin(∆θ1) ≈ ∆θ1. To this end, the linearized equation is
˙̂
∆θ1 = ∆θ2,
˙̂
∆θ2 = −g
l
(∆θ1)
The eigenvalues (i.e., roots of the characteristic polynomial) of the above linear equations are pure imaginary
±i

g
l
. The equilibrium (0, 0) is said to be neutrally stable.
Linearized equations for the pendulum about the vertically up position. We linearize (3) about
(±pi, 0). This means that we begin by letting
θ = ±pi + ∆θ1, θ2 = 0 + ∆θ2
where ∆θ1 and ∆θ2 are ‘small’. We then substitute into (6) to get
˙̂
∆θ1 = ∆θ2,
˙̂
∆θ2 = −g
l
sin(±pi + ∆θ1)
we use
sin(±pi + ∆θ1) = −g
l
[sin(±pi) cos(∆θ1) + cos(±pi) sin(∆θ1)]
Remember that sin(±pi) = 0 and cos(±pi) = −1 and simplify
˙̂
∆θ1 = ∆θ2,
˙̂
∆θ2 =
g
l
sin(∆θ1)
Again, using sin(∆θ1) ≈ ∆θ1, the linearized equations become
˙̂
∆θ1 = ∆θ2,
˙̂
∆θ2 =
g
l
(∆θ1)
The eigenvalues (i.e., roots of the characteristic polynomial) of the above linear equations are real ±

g
l
with one positive real root. The equilibrium (±pi, 0) is a saddle.
DELIVERABLES
1. Write a computer program to generate the plots shown in figure 3.
2. For a harmonic oscillator with negative spring stiffness, i.e. k < 0,
(a) plot the phase portrait,
(b) identify equibrium points and comment on their stablity.
3. Show that the equations of the pendulum, when linearized about the vertically down position, are
equivalent to a simple harmonic oscillator with positive k
4. Show that the equations of the pendulum, when linearized about the vertically up position are equiv-
alent to a harmonic oscillator with negative k.
Write your own document with the above deliverables and a discussion of the most important results.
(approx. 3 pages of text and plots together). All in a single PDF document. This document is due on
Monday December 9 (no extension!)
4
51作业君 51作业君

扫码添加客服微信

添加客服微信: IT_51zuoyejun