ME 240: Introduction to Dynamics and Vibrations
Mechanical Engineering Department
The University of Michigan
Computer Assignment #1
Supplemental Document
Originally prepared by Akira Saito and modified here by Todd Lillian
1
Introduction
In this document, we present how to solve a secondorder ordinary differential equation using
Mat
lab
. Sample
Matlab
codes are provided for solving a simple example problem that is similar to
computer assignment #1. Modifying the
Matlab
code in this document may be a good starting
point for working on your assignment.
Example Problem:
The equation of motion of a simple pendulum shown in Fig.
1
can be expressed as a secondorder,
nonlinear ordinary differential equation of the following form:
mL
¨
θ
+
mg
sin
θ
= 0
(1)
where
θ
is the angle defined in Fig.
1
,
¨
θ
denotes the second derivative of
θ
with respect to time
t
,
m
is the mass and is set to (1/9.8) kg,
g
is gravity and is set to 9.8 m/s
2
, and
L
is the length and
is set to 9.8 m. Considering the following initial conditions:
θ
(
t
= 0) =
π
4
,
˙
θ
(
t
= 0) = 0
(2)
answer the following questions by solving Eq. (
1
) using
Matlab
:
(a)
Obtain the value of
˙
θ
when the pendulum first hits
θ
= 0
.
(b)
Plot tension
T
as a function of
θ
.
T
is defined as
T
=
mg
cos
θ
+
mL
˙
θ
2
Light, Rigid rod
Figure 1: Simple Pendulum
1
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2
State Equations
In order to make Eq. (
1
) more suitable for a numerical solution, we transform this equation into
a special form called
state equations
. In general the state equations have the following form:
˙
x
=
f
(
x
,
u
)
(3)
where
x
contains all the
state variables
,
f
is a
state function
, and
u
contains
inputs
. If you have
n
state variables, Eq. (
3
) can be written more explicitly as,
˙
x
1
˙
x
2
.
.
.
˙
x
n
=
f
1
(
x
,
u
)
f
2
(
x
,
u
)
.
.
.
f
n
(
x
,
u
)
(4)
Now let’s transform Eq. (
1
) into the form of Eq. (
4
). We define
θ
as the first state variable,
x
1
,
and therefore
x
1
=
θ
. The natural choice for the second state variable is
˙
θ
, therefore
x
2
=
˙
θ
. We
note that that the second time derivative of
θ
is the first time derivative of
˙
θ
. Let us rewrite the
definitions using the new notation:
x
1
x
2
=
θ
˙
θ
(5)
Now considering the time derivative of Eq. (
5
) and using Eq. (
1
),
¨
θ
=

g
L
sin
θ
, we get the following:
˙
x
1
˙
x
2
=
˙
θ
¨
θ
=
x
2

g
L
sin
x
1
(6)
It is noted that we have two components in the state functions:
f
1
(
x
,
u
) =
f
1
([
x
1
;
x
2
]
,
0
) =
x
2
(7)
f
2
(
x
,
u
) =
f
2
([
x
1
;
x
2
]
,
0
) =

g
L
sin
x
1
(8)
The next step is to write a
Matlab
code to solve these equations.
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 Fall '09
 PERKINS
 Mechanical Engineering, Derivative, state equations, Akira Saito

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