2.4
Section 2.4 p1
Harmonic Oscillator &
Particle in a Box
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.4 Modeling of Free Oscillations
of a Mass—Spring System
Setting Up the Model (continued 1)
The simple (linear) harmonic oscillator consists of a body
moving in a straight line under the influence of a force
(1)
F = −kx
(Hooke’s law)
whose magnitude is proportional to the displacement x of the
body from the fixed point O, the point of equilibrium, and
whose direction is towards this point
k (> 0): the force constant.
“‒” a restoring force to ensure the force acts in the direction
opposite to that of the displacement, i.e. to pull it back to x = 0.
k ↑ → Stiffness ↑
Section 2.4 p2
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.4 Modeling of Free Oscillations
of a Mass—Spring System
Setting Up the Model (continued 2)
By Newton’s second law of motion, the acceleration
experienced by the body is
d 2x
F =m 2
(2)
dt
and simple harmonic motion is therefore described by the
differential equation
d 2x
m 2 = − kx
dt
Section 2.4 p3
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.4 Modeling of Free Oscillations
of a Mass—Spring System
Setting Up the Model (continued 3)
F
d 2 x1
d 2 x2
m1 2 + m2 2
dt
dt
Lagrangian mechanics
Section 2.4 p4
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.4 Modeling of Free Oscillations
of a Mass—Spring System
ODE of the H.O. System
d 2x k
+ x=
0
2
dt
m
ω2
This is a homogeneous linear ODE with constant
coefficients. A general solution is obtained, namely
(4)
x(t) = d1 cos ωt + d2 sin ωt
Section 2.4 p5
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.4 Modeling of Free Oscillations
of a Mass—Spring System
ODE of the H.O. System
Let the displacement and velocity at time t = 0 be
x(0) = A, x′(0) = 0
x(0) = A = d1 cos 0 + d2 sin 0 = d1
x′(0) = ‒ d1 ω sin 0 + d2 ω cos 0 = d2 ω = 0
x(t) = A cos ωt
Section 2.4 p6
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.4 Modeling of Free Oscillations
of a Mass—Spring System
ODE of the H.O. System
The maximum displacement of the body from equilibrium is the
amplitude A
The wavelength of the representative curve is the period τ =
2π/ω, the time taken for one complete oscillation
Section 2.4 p7
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.4 Modeling of Free Oscillations
of a Mass—Spring System
ODE of the H.O. System
The maximum displacement of the body from equilibrium is the
amplitude A
The wavelength of the representative curve is the period τ =
2π/ω, the time taken for one complete oscillation
The inverse of the period, ν =1/τ = ω/2π, is called the frequency
of oscillation, the number of oscillations in unit time
The quantity ω = 2πν is called the angular frequency
k
ω=
m
1
ν=
2π
k
m
Section 2.4 p8
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.4 Modeling of Free Oscillations
of a Mass—Spring System
ODE of the H.O. System
Energies
dV
F= −
dx
1 2
V=
x
( ) kx + c
2
Let V(0) = 0 → The potential energy is zero at the equilibrium
position
1 2
V ( x ) = kx
2
Section 2.4 p9
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.4 Modeling of Free Oscillations
of a Mass—Spring System
ODE of the H.O. System
Energies
x(t) = A cos ωt
1 2 1 2
V=
kx
kA cos 2 ωt
( x) =
2
2
1
T = kA2 sin 2 ωt
2
The total energy is therefore the constant
1 2
E = T + V = kA
2
Section 2.4 p10
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.4 Modeling of Free Oscillations
of a Mass—Spring System
ODE of the H.O. System
Energies
The relation between the potential and kinetic energies
Section 2.4 p11
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.4 Modeling of Free Oscillations
of a Mass—Spring System
ODE of the H.O. System
Energies
At maximum displacement from equilibrium, x = A, V has its
maximum value, V = E, and T = 0
As the body approaches the equilibrium position, V is converted
into T
at x = 0, V = 0 and the kinetic energy is a maximum, T = E.
Section 2.4 p12
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.4 Modeling of Free Oscillations
of a Mass—Spring System
ODE of the H.O. System
Bending
r → θ (bond length → bond angle)
Ebend
1
2
= kθ (θ − θ 0 )
2
Section 2.4 p13
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.4 Modeling of Free Oscillations
of a Mass—Spring System
ODE of the H.O. System
Harmonic Approximation
1
2
E=
k r ( r − ro )
stretch
2
Section 2.4 p14
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.2 Homogeneous Linear ODEs
with Constant Coefficients
EXAMPLE
The particle in a one-dimensional box
The ‘particle in a one-dimensional box’ is the name given to the
system consisting of a body allowed to move freely along a line
of finite length.
In quantum mechanics, this is one of the simplest systems that
demonstrate the quantization of energy.
Section 2.2 p15
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.2 Particle in a 1-D box
EXAMPLE
The particle in a one-dimensional box
Ĥψ = Eψ
For a particle of mass m moving in the x-direction
2
d
ψ ( x)

−
+ V ( x )ψ ( x ) =
Eψ ( x )
2
2m dx
2
kinetic
potential
total
ψ: the wave function
Section 2.2 p16
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.2 Particle in a 1-D box
EXAMPLE
The particle in a one-dimensional box
0 for 0 < x < l
V ( x) = 
∞ for x ≤ 0 and x ≥ l
V=∞
V=∞
V=0
x
0
Section 2.2 p17
l
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.2 Homogeneous Linear ODEs
with Constant Coefficients
EXAMPLE
The particle in a one-dimensional box
The constant value of V inside the box → no force acts on the
particle in this region;
setting V = 0 → the energy E is the (positive) kinetic energy of
the particle
The ∞ value of V at the ‘walls’ and outside the box → the
particle cannot leave the box; in quantum mechanics this means
that ψ = 0 at the walls and outside the box.
Section 2.2 p18
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.2 Homogeneous Linear ODEs
with Constant Coefficients
EXAMPLE
The particle in a one-dimensional box
For the particle within the box, we therefore have the boundary
value problem
2
2 d ψ ( x )
−
=
Eψ ( x )
2
2m dx
with boundary conditions
ψ(0) = ψ(l) = 0
Section 2.2 p19
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.2 Homogeneous Linear ODEs
with Constant Coefficients
EXAMPLE
The particle in a one-dimensional box
Let
2mE
ω = 2

d 2ψ
2
ω
ψ=
0
+
2
dx
2
=
ψ ( x ) d1 cos ω x + d 2 sin ω x
Section 2.2 p20
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.2 Homogeneous Linear ODEs
with Constant Coefficients
EXAMPLE
The particle in a one-dimensional box
Applying the boundary conditions
ψ ( 0=
) d=1 0
=
ψ ( l ) d=
0
2 sin ωl
It follows, because the sine function is zero only when its
argument is a multiple of π
i.e. ωl = nπ
where n is an integer
Section 2.2 p21
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.
2.2 Homogeneous Linear ODEs
with Constant Coefficients
EXAMPLE
The particle in a one-dimensional box
nπ
ω=
, n = 0, ± 1, ± 2,
l
nπ x
=
ψ n ( x ) d=
sin
, n 1, 2,
2
l
n = 0 → discarded ∵the trivial solution ψ0(x) = 0 is not a
physical solution
negative values of n have been discounted because ψ−n(x) =
−ψn(x) is merely ψn with a change of sign
n2h2
En =
8ml 2
Section 2.2 p22
Advanced Engineering Mathematics, 10/e by Edwin Kreyszig
Copyright 2011 by John Wiley & Sons. All rights reserved.