Phase Space Trajectories: Bounded and
unbounded motion
Prabha Mandayam
Indian Institute of Technology, Madras
Physics 1010 (Nov 2020)
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Topics covered in this lecture
Phase diagram for bounded and unbounded motion
Stable and unstable equilibrium points
Example 1: Phase diagram of a damped oscillator
Example 2: Phase diagram of plane pendulum
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Particle moving in an arbitrary potential
For a particle moving in one-dimension under a general potential U (x), the
phase trajectories can be obtained by solving the energy conservation
equation.
1
mẋ2 + U (x) = E.
2
If U (x) is known, it is easy to solve exactly for ẋ(x).
ẋ(x) =
2p
E − U (x).
m
In some cases, it maybe difficult to obtain the exact form of U (x). However,
it is possible to obtain a qualitative picture of the phase diagram for motion
in an arbitrary potential (see figure below).
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Phase diagram for an asymmetric potential
For example, unlike the symmetric (parabolic) SHO potential, we may
consider a particle moving in an asymmetric potential, corresponding to a
nonlinear force.
p
If there is no damping, the velocity ẋ ∝ E − U (x), leading to the phase
trajectories shown below.
Figure 1: (a) Asymmetric potential and (b) phase diagram for bounded motion
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Stable equilibrium and bounded motion
The three oval paths in the Fig. 1 correspond to three different values of
total energy E.
If the total energy is less than the height of the potential on either side of
x = 0, the particle is trapped in the potential well.
When E approaches close to the minimum of the potential at x = 0, the oval
trajectories approach an ellipse.
The point x = 0 is a point of stable equilibrium since d2 U (x)/dx2 > 0 at
x = 0. Small perturbations around such a point leads to locally bounded
motion.
If the system were damped, the oscillating particle will spiral down the
potential well and eventually come to rest at the equilibrium position x = 0.
(See Example 1 on slide 9)
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Inverted asymmetric potential: unbounded motion
Motion in the neighbourhood of the maximum of a potential is qualitatively
very different. Consider, for example, a particle moving in an inverted
asymmetric potential shown below.
Figure 2: (a) Inverted asymmetric potential and (b) phase diagram for unbounded
motion
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Unstable equilibrium and unbounded orbits
x = 0 in now a point of unstable equilibrium, since d2 U (x)/dx2 < 0 at x = 0.
Small perturbations around this point leads to locally unbounded motion.
If the inverted potential in Fig. 2 were symmetric, that is, a parabolic
potential of the form U = − 12 kx2 ,
The phase trajectory corresponding to total energy E0 (equal to the maximum
value of the potential U (x)) would be a straight line.
The phase trajectories corresponding to E1 and E2 would be hyperbolas.
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Damped Oscillations
Recall that the equation of motion for a one-dimensional damped oscillator
with a damping force of the form, F (x) = −bẋ is given by,
mẍ + bẋ + kx = 0 ⇔ ẍ + γ ẋ + ω02 x = 0.
p
Here, γ = b/m is the damping parameter and ω0 = k/m is the
characteristic frequency of the undamped oscillator.
The general form of the solution for such a damped oscillator is,
"
!
!#
r r γ 2
γ 2
− γ2 t
2
2
x(t) = e
A1 exp
− ω0 t + A2 exp −
− ω0 t
.
2
2
Unlike the simple harmonic oscillator, the energy of the damped oscillator is
not constant in time. The oscillator constantly loses energy to the damping
medium, at a rate proportional to square of the velocity.
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Example 1: Phase diagram of a damped oscillator
Q: Construct a general phase diagram analytically for the damped oscillator.
Then, using a computer make a plot for x and ẋ versus t and a phase diagram for
the following values: A = 1cm, ω0 = 1rad/s, γ2 = 0.2s−1 , and δ = π/2rad.
Solution:
First we write the expressions for the displacement and velocity:
x(t)
=
ẋ(t)
=
γ
Ae− 2 t cos(ω1 t − δ),
hγ
i
γ
−Ae− 2 t
cos(ω1 t − δ) + ω1 sin(ω1 t − δ) .
2
(1)
We introduce a change of variables: u = ω1 x, w = γ2 x + ẋ. Then, u(t) and
w(t) evolve as,
γ
γ
u(t) = ω1 Ae− 2 t cos(ω1 t − δ); w(t) = −ω1 Ae− 2 t sin(ω1 t − δ)
Rewriting in polar coordinates, we have,
p
ρ(t) = u2 + w2 ,
φ(t) = ω1 t
γ
Eliminating t, we get, ρ = ω1 Ae− 2 /ω1 φ .
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Equation of the phase trajectories
Since the transformation from (x(t), ẋ(t)) to the dynamical variables
(u(t), w(t)) is linear, the phase paths take the same shape in the u-w plane,
as in the x-ẋ plane.
Equation of the phase trajectory: ρ = ω1 Ae−(γ/2ω1 ) φ is the equation of a
logarithmic spiral.
The radius vector of a given point on the phase space is continuously
decreasing, indicating the damped motion of the oscillator.
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Phase diagrams of the damped oscillator
(a) The phase trajectories of the damped oscillator in the (w, u) coordinates;
(b) actual phase trajectories in the (x, ẋ) plane.
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Plots of x, ẋ for damped oscillator
x(t) and ẋ(t) have been plotted numerically as a function of time, using the
given values of A, γ2 , k, m and δ.
For δ = π/2, the amplitude has x = 0 at t = 0, but ẋ has a large positive
value. So x attains a maximum value of 0.7m at t = 2s.
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Example 2: Plane Pendulum
Q: Consider a particle of mass m constrained by a weightless, rigid rod to move in
a vertical circle of radius l. Construct the phase diagram of such a plane
pendulum.
Solution:
We can get the equation of motion of the plane pendulum by equating the
torque about its support axis to the product of angular acceleration and
moment of inertia (I) about the axis : I θ̈ = lF
Torque is provided by the component of gravitational force perpendicular to
the rod. Thus, I = ml2 and F = −mg sin θ, giving the equation of motion,
θ̈ + ω02 sin θ = 0, ω02 ≡
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Phase Space
g
.
l
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Equation of phase trajectories
To obtain the phase trajectories, we do not need to solve for θ(t) and θ̇(t);
instead, we simply use the fact that the plane pendulum is a conservative
system.
Conservation of energy implies, T + U = E = constant, where,
T =
1
1 2
Iω = ml2 θ̇2 , U = mgl(1 − cos θ).
2
2
We have taken the zero of the potential energy to be the lowest point
(θ = 0) on the circular path described by the particle.
If θ0 denotes the highest point of the motion, then, T (θ = θ0 ) = 0 and
U (θ = θ0 ) = E = mgl(1 − cos θ0 ).
Thus the conservation of energy gives,
PH 1010
1 2 2
ml θ̇
2
=
⇒ θ̇
=
θ0
θ
2
2
2mgl sin
− sin
2
2
r s
g
θ0
θ
2
2
sin
− sin
.
2
l
2
2
Phase Space
(2)
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Bounded orbits of the plane pendulum
If θ and θ0 are small angles, Eq. (2) can be approximated as,
s
l
θ̇
g
!2
+ θ2 ≈ θ02 .
Defining
the coordinates of the phase space of the plane pendulum to be
q
l
θ, g θ̇, we see that the phase trajectories close to θ = 0 are circles.
Thus, for small θ0 , the motion is approximately simple harmonic! Recall that
θ0 specifies the total energy of the system.
For −π < θ < π and E < 2mgl ≡ E0 , the situation is identical to a particle
bound in the potential U (θ) = mgl(1 − cos θ). Phase trajectories are
therefore closed curves for this region.
Since the potential is periodic in θ, we get the same phase trajectories for
π < θ < 3π, −3π < θ < −π, and so on.
The points θ = 0, ±2π, ±4π, . . . are positions of stable equilibrium. They
indeed correspond to the minima of the potential U (θ).
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Unstable equilibria and unbounded orbits
At E = E0 = 2mgl, θ0 = ±π. Then, the equation of the phase trajectories
(Eq. (2)) simplifies to,
r
g
θ
θ̇ = ±2
cos
.
l
2
Thus the phase trajectories for E = E0 = 2mgl are cosine functions.
When the total energy E > E0 , the motion is no longer oscillatory, leading to
unbounded orbits on phase space. However, the motion is still periodic: the
pendulum starts executing complete revolutions about its support axis.
Thus, the phase trajectories for E = E0 separate locally bounded motion
from locally unbounded motion. Such a path is called a separatrix. A
separatix always passes through a point of unstable equilibrium.
The points θ = ±π, ±3π, . . . are positions of unstable equilibrium. These
correspond to maxima of the potential U (θ).
We now plot the phase diagram of the plane pendulum, for one complete
cycle, that is, for the range −π < 0 < π.
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Phase diagram of the plane pendulum
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