M40007: Introduction to Applied Mathematics
c Darren Crowdy (2021)
Imperial College London
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M40007 c Darren Crowdy (2021)
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
x 1 x 2 x 3
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
x 1, x 2, x 3. (2)
The important quantities are the displacements from equilibrium which are then
given by
φ1 = x 1 x1, φ2 = x 2 x2, φ3 = x 3 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
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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)
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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
)
, (9)
where the positive spring constants are on the diagonal, then Hooke’s law – encom-
passing (4) and (5) – can be written in matrix form as
T =
(
Ta
Tb
)
= CAΦ. (10)
It is clear that Hooke’s law for T in this mechanical spring-mass system is anal-
ogous to Ohm’s law for the currents w in the electric circuit problem.
Also in analogy with the electric circuit problem, where
f = ATw (11)
gives the divergence of the currents at the nodes, the quantity defined by
fI = ATT (12)
now gives the divergence of the spring tensions at the masses. Physically, the diver-
gence of the spring tensions at the masses gives the net internal forces on the masses
due to the tensions in the springs; the subscript on fI is added to emphasize that these
are internal spring forces. To see this, note that
ATT =
1 01 1
0 1
( Ta
Tb
)
=
TaTb Ta
Tb
. (13)
As is evident from Figure 2, the quantities in this vector are the total forces on each
mass 1 , 2 and 3 in the positive x-direction due to the springs.
Putting the pieces together, from (12) and (10) we have
fI = ATT = ATCAΦ. (14)
Equivalently, the vector of total internal spring forces on each mass is given by
fI = KΦ, K = ATCA. (15)
Here K is the weighted Laplacian matrix now weighted by the set of (positive)
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spring constants.
What is the analogue of the Kirchhoff current law at the nodes in this spring-
mass system That depends on which physical problem is of interest. If we seek
conditions for the masses to be in equilibrium – and therefore not moving – the
required condition is that the
total force on each mass must be zero. This is the
condition of force balance in mechanical equilibrium.
For additional generality, in addition to the internal spring forces, we will also
allow in our formulation for a set of external forces on each mass. These external
forces will be collected in the vector
f =
f1f2
f3
. (16)
Among the many possibilities, these external forces could be due to gravitational
forces caused by a gravitational field, or due to reaction forces if any given mass is
simply constrained not to move.
The constraint that the total force on each mass vanishes at mechanical equilib-
rium is then equivalent to the vector equation
f + fI = 0, (17)
where, to get the total force on each mass, we have added together the external
forces and the internal forces due to the springs. For mechanical equilibrium, the
total force on each mass must vanish.
On combining (15) and (17) we arrive at
f = fI = ATCAΦ = KΦ. (18)
Remarkably, this equation is exactly the same as that appearing in the electric
circuit problem; it is just the physical interpretation of the vectors that changes. In
the spring-mass application, the Φ is the vector of displacements of the masses at
the nodes and f is the vector of external forces on the masses. The spring-constant-
weighted Laplacian K encodes the influence of the springs.
3 Two masses hanging under gravity
Consider 2 masses attached to a ceiling, pulled down by the external force of gravity
and connected by 2 springs as shown in Figure 3. It is convenient to think of the
ceiling as node 1 and the two masses, of mass m2 and m3, as nodes 2 and 3 . The
springs are labelled a and b . Their spring constants are ca and cb respectively.
The challenge is to determine the equilibrium displacements, φ2 and φ3, of the
two masses, and also the reaction force R at the ceiling.
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1
2
a
b
3
φ2
φ3
φ1 = 0
m2
m3
x
Figure 3: Two masses hanging under gravity. The ceiling is viewed as mass 1 but
has zero displacement.
By the usual constructions, it is easy to find the spring-constant-weighted Lapla-
cian to be
K =
1 2 3 ca ca 0 1 ca ca + cb cb 2
0 cb cb 3
(19)
The external forces at the nodes are
f =
Rm2g
m3g
, (20)
where R is the reaction force at the ceiling. The force due to gravity, assumed to the
in the positive x-direction, is given by the mass multiplied by the acceleration due
to gravity g. The equations for equilibrium are therefore
f =
Rm2g
m3g
= K
0φ2
φ3
=
ca ca 0 ca ca + cb cb
0 cb cb
0φ2
φ3
. (21)
This can be readily solved –either directly, or, for example, using Schur comple-
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ments – to find
φ2 =
(
m2 +m3
ca
)
g, φ3 =
m3g
cb
+
(
m2 +m3
ca
)
g (22)
and
R = (m2 +m3)g. (23)
Since the displacement of node 1 has been fixed, which is akin to fixing a unit
voltage at the + node in a 2-point source/sink electric circuit problem, the reaction
force R can be thought of as akin to the effective conductance in the circuit problem.
Remark: In this course the masses in a spring-mass network should only be
taken to be subject to the external force of gravity if this is stated explicitly.
4 Newton’s second law of motion
The structure of the equations for equilibrium of a spring-mass system has been seen
to fit neatly into the same mathematical framework that worked successfully for the
electric circuit problem. However, the spring-mass system allows for an interesting
generalization to the non-equilibrium setting.
The physical law needed to study this is Newton’s second law of motion. It is one
of the most basic laws of mechanics. Suppose a mass m has position x(t) where we
now allow this position to vary with time t. The velocity v of this mass is defined
as the vector
v =
dx
dt
(24)
and its acceleration a as
a =
dv
dt
=
d2x
dt2
. (25)
If F denotes the total force acting on the mass, then Newton’s second law of motion
dictates that
F = ma. (26)
The idea now is to allow the masses in the spring-mass systems just consid-
ered to be displaced from equilibrium with their subsequent positions determined
according to Newton’s second law of motion.
Consider a spring-mass system with n movable masses. The total force on any
mass is
F = f + fI = f KΦ, (27)
where K is the n-by-n Laplacian weighted by any spring constants exerting forces
on the masses and f denotes any external forces. The vector of accelerations of the
masses is
d2Φ
dt2
. (28)
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Newton’s second law (26) therefore implies
F = f KΦ = Md
2Φ
dt2
, (29)
where we have introduced the n-by-n diagonal matrix M where
M =