M40007: Introduction to Applied Mathematics
Introduction to Applied Mathematics
Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: THEend8_
M40007: Introduction to Applied Mathematics
1 Spring-mass systems
Consider 3 masses connected by 2 springs as shown in Figure 1. The masses are
labelled 1 , 2 and 3 . The springs are labelled a© and b©. In the upper row
the masses are in their equilibrium positions; in the lower row the masses are in
strained positions and non-zero tensions Ta and Tb exist in the springs joining the
masses. For now, the only forces acting on the masses are due to the springs.
x1 x2 x3
xˆ1 xˆ2 xˆ3
x
x
φ1 φ2 φ3
1 2 3
equilibrium
displaced
a© b©
Ta Tb
Figure 1: A spring-mass system. If any two masses connected by a spring are
displaced from equilibrium a tension exists in the spring connecting them.
The equilibrium positions of the masses are
x1, x2, x3 (1)
and their positions once displaced are
xˆ1, xˆ2, xˆ3. (2)
The important quantities are the displacements from equilibrium which are then
given by
φ1 = xˆ1 − x1, φ2 = xˆ2 − x2, φ3 = xˆ3 − x3. (3)
Hooke’s law: According to Hooke’s law, the tension in spring a© is
Ta = ca(φ2 − φ1), (4)
where ca > 0 is the spring constant. The quantity φ2 − φ1 is the extension of the
spring. As drawn in Figure 1 this tension will be positive since φ2 > φ1. This means
that the masses will be pulled towards each other by spring a©. Such a spring is
sometimes called a linear spring because of the linear relationship between the force
2
and the extension of the spring.
Similarly, Hooke’s law says that the tension in spring b© is
Tb = cb(φ3 − φ2), (5)
where cb > 0 is its spring constant. As drawn in Figure 1 this tension will be
negative since φ3 < 0 and φ2 > 0. This means that the masses will be pushed away
from each other by spring b©.
2 Spring-mass system as a graph
Any spring-mass system can be modelled as a graph with the nodes being the
masses and the displacement of each node being the node potential. The edges of
the graph are the springs, each having a spring constant, and the tension in each
spring being the edge variable.
Consider the spring-mass system shown in Figure 2.
φ1 φ2 φ3
1 2 3
a© b©
Ta Tb
Figure 2: A typical spring-mass system.
There are n = 3 masses, or nodes to the graph, and m = 2 springs, or edges to
the graph. In order to write down an incidence matrix for this graph, we need to
make a choice of direction on each edge, i.e., pick the arrows. However, since the
edges are aligned along an x-axis, say, the most natural choice is to pick all arrows
to point to the right in Figure 2, along the positive x direction. The incidence matrix
of the graph is then
A =
1 2 3( )−1 1 0 edge a©
0 −1 1 edge b© (6)
We will use
Φ =
φ1φ2
φ3
(7)
3
to denote the vector of node potentials, i.e., the vector of the displacements of the
masses. Then,
AΦ =
(
φ2 − φ1
φ3 − φ2
)
. (8)
This is the vector of the extensions of the springs. If we introduce the diagonal
matrix C where
C =
(
ca 0
0 cb
)