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
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
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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 =

m1 0 · · · 0 0
0 m2 · · · 0 0
0 0 · · · 0 0
. . . . .
. . . . .
0 0 0 mn−1 0
0 0 0 0 mn

. (30)
Since all masses are positive then, provided they are non-zero, M is a positive-
definite matrix.
5 Two masses between fixed walls
Consider the situation shown in Figure 4 where two masses, each of unit mass, are
connected by three springs labelled a©, b© and c© between 2 fixed walls. Let all
spring constants be unity.
a© b©
Ta Tb
c©
Tc
0 1 2 3
φ0 φ1 φ2 φ3
Figure 4: Two masses, three springs, between two fixed walls.
Let the walls and the masses each constitute a node in the graph and label them
0 , 1 , 2 and 3 as shown in Figure 4. Let the vector of displacements be
Φ =

φ0
φ1
φ2
φ3
 (31)
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The Laplacian matrix in this case is easy to find using the standard construction:
K =
0 1 2 3

1 −1 0 0 0
−1 2 −1 0 1
0 −1 2 −1 2
0 0 −1 1 3
(32)
The walls are fixed implying that
φ0 = φ3 = 0. (33)
It is therefore convenient to consider the reduced Laplacian
Kˆ =
(
2 −1
−1 2
)
(34)
since masses 1 and 2 move according to Newton’s second law, which says that
fˆ− KˆΦˆ = Id
2Φˆ
dt2
, (35)
where I is the 2-by-2 identity matrix – this is the matrix M when the masses are
both unity – and
Φˆ =
(
φ1
φ2
)
, fˆ =
(
f1
f2
)
, (36)
where fi is the external force on mass i for i = 1, 2.
Free oscillations: First let us assume there are no external forces on the masses
so that they are in free oscillation under the influence of the internal spring forces.
This means that
f1 = f2 = 0. (37)
Equation (35) simplifies to
−KˆΦˆ = Id
2Φˆ
dt2
. (38)
This is a system of linear second order differential equations, with constant coeffi-
cients, for the displacements. It is natural to seek solutions of the form
Φ = Φ0eiωt, (39)
where ω ∈ R is a real constant and i is the square root of −1. The vector Φ0 will be
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complex-valued. On substitution of (39) into (38) we find
−KˆΦˆ0eiωt = −ω2Φˆ0eiωt (40)
which simplifies to
KˆΦˆ0 = ω2Φˆ0. (41)
This is an eigenvalue problem for Kˆ. The vector Φˆ0 is an eigenvector and ω2 is the
corresponding eigenvalue.
The matrix Kˆ is real and symmetric which means that it has 2 eigenvalues and
they are both real. So too are the corresponding eigenvectors. The mathematical
approach based on complex numbers used here appears to generate complex eigen-
vectors. However, the required real eigenvectors can be constructed on making the
following three observations.
Note 1: Since both Kˆ and ω2 are real then, on taking a complex conjugate of
(41),
KˆΦˆ0 = ω2Φˆ0 (42)
which shows that Φˆ0 is also an eigenvector of Kˆ and having the same eigenvalue as
Φˆ0.
Note 2: Since (38) is invariant if ω 7→ −ω then if Φ0eiωt is a solution of (38) then
so too is Φ0e−iωt.
Note 3: Notes 1 and 2 mean that if Φ0eiωt is a solution of (38) then so too are the
linear combinations
Re
[
Φ0eiωt
]
=
1
2
[
Φ0eiωt +Φ0e−iωt
]
(43)
and
Im
[
Φ0eiωt
]
=
1
2i
[
Φ0eiωt −Φ0e−iωt
]
. (44)
Recall that the eigenspace of a matrix is a vector space so eigenvectors with the
same eigenvalue can be added together to form other eigenvectors with the same
eigenvalue.
Let us find the eigenvalues and eigenvectors for the 2-mass problem shown in
Figure 4. It is convenient to set
λ = ω2 (45)
and to solve the standard eigenvalue problem
KˆΦˆ0 = λΦˆ0. (46)
The existence of a non-trivial solution requires the following zero determinant con-
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dition to hold:
der
(
2− λ −1
−1 2− λ
)
= 0 (47)
or
(2− λ)2 − 1 = 0. (48)
This leads to
λ = 1, 3. (49)
The corresponding values of ω are
ω = ±1,±
√
3. (50)
It is common to call the values of ω found above to be the natural frequencies of
oscillation of the system. The corresponding eigenvectors – also known as the natural
modes of oscillation – are easy to find:
λ = 1, Φˆ1 =
(
1
1
)
, (51)
λ = 3, Φˆ2 =
(
1
−1
)
. (52)
It is easy to check that these two eigenvectors are orthogonal, i.e.,
ΦˆT1 Φˆ2 = 0, (53)
as expected from linear algebra results on the eigenvectors of real symmetric matri-
ces.
The general solution for the time-dependent vector Φ is therefore
Φ =
(
1
1
)
(c1eit + c1e−it) +
(
1
−1
)
(c2e
√
3it + c2e−
√
3it), (54)
where c1, c2 ∈ C are two complex constants. This general solution therefore de-
pends on 4 real constants; the real and imaginary parts of c1 and c2. Physically,
those 4 constants would be determined by the initial positions and initial velocities
of the 2 masses.
What are the reaction forces at the walls? These are the external forces at nodes
0 and 3 . Let the vector of external forces at the nodes be
f =

R0
0
0
R3
 . (55)
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Now we can invoke Newton’s second law in the form (29) which implies
f−KΦ =

R0
0
0
R3
−K

0
φ1(t)
φ2(t)
0
 =

0
−ω2φ1(t)
−ω2φ2(t)
0
 (56)
to deduce that
R0 = −φ1(t), R3 = −φ2(t). (57)
Exercise: Verify that if
c1 =
1
2
(A− iB), c2 = 12(C− iD) (58)
then the solution (54) can be written in the alternative form
Φ =
(
1
1
)
(A cos t+ B sin t) +
(
1
−1
)
(C cos
√
3t+ D sin
√
3t). (59)
6 Three masses between fixed walls
Now consider the situation shown in Figure 5 where three masses, each of unit
mass, are connected by four springs labelled a©, b©, c© and d© between 2 fixed
walls. Let all the spring constants be unity.
a© b©
Ta Tb
c©
Tc
0 1 2 3
φ0 φ1 φ2 φ3
d©
Td
4
φ4
Figure 5: Two masses, three springs, between two fixed walls.
The walls and the masses constitute nodes in a graph; label them 0 , 1 , 2 , 3
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and 4 as shown in Figure 5. Let the vector of displacements be
Φ =

φ0
φ1
φ2
φ3
φ4
 . (60)
The Laplacian matrix in this case is easy to find using the standard construction:
K =
0 1 2 3 4

1 −1 0 0 0 0
−1 2 −1 0 0 1
0 −1 2 −1 0 2
0 0 −1 2 −1 3
0 0 0 −1 2 4
(61)
The walls are fixed implying zero displacements:
φ0 = φ4 = 0. (62)
It is therefore convenient to consider the reduced Laplacian
Kˆ =
 2 −1 0−1 2 −1
0 −1 2
 (63)
and to consider only the motion of masses 1 , 2 and 3 under Newton’s second
law. This law says that
fˆ− KˆΦˆ = Id
2Φˆ
dt2
, (64)
where I is the 3-by-3 identity matrix – this is the matrix M when the masses are all
unity – and
Φˆ =
 φ1φ2
φ3
 , fˆ =
 f1f2
f3
 , (65)
where fi is the external force on mass i for i = 1, 2, 3.
Free oscillations: First we assume there are no external forces on the masses
so that they are in free oscillation under the influence of the internal spring forces.
This means that
f1 = f2 = f3 = 0, fˆ = 0. (66)
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Equation (64) simplifies to
−KˆΦˆ = Id
2Φˆ
dt2
. (67)
As before we seek solutions of this system of linear second order differential equa-
tions of the form
Φ = Φ0eiωt, ω ∈ R. (68)
On substitution of (68) into (67) we find
−KˆΦˆ0eiωt = −ω2Φˆ0eiωt (69)
which simplifies, on cancellation of common terms, to the eigenvalue problem
KˆΦˆ0 = λΦˆ0, λ = ω2. (70)
The existence of non-trivial solutions requires the zero determinant condition to
hold:
det
 2− λ −1 0−1 2− λ −1
0 −1 2− λ
 = 0. (71)
The solution of this characteristic equation leads to
λ = 2, 2±
√
2 (72)
leading to the natural frequencies
ω = ±
√
2,±
√
2±
√
2. (73)
The corresponding eigenvectors are easy to find:
λ = 2, Φˆ1 =
 10
−1
 , (74)
λ = 2 +
√
2, Φˆ2 =
 −1√2
−1
 , (75)
λ = 2−
√
2, Φˆ3 =
 1√2
1
 . (76)
It can be verified that these eigenvectors are mutually orthogonal, as expected from
results in linear algebra.
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The general solution for Φ can therefore be written as
Φ =
 10
−1
 (c1e√2it + c1e−√2it) +
 −1√2
−1
 (c2e√2+√2it + c2e−√2+√2it)
+
 1√2
1
 (c3e√2−√2it + c3e−√2−√2it),
(77)
where c1, c2 and c3 ∈ C are complex constants. This general solution depends on
6 real constants determined by the initial positions and initial velocities of the 3
masses.
The structure of this type of problem for the free oscillation of masses con-
nected to springs between two fixed walls should be clear from these two exam-
ples. Adding more masses and springs leads to higher dimensional eigenvalue
problems. It becomes increasingly challenging, however, to find the associated nat-
ural frequencies and natural modes of oscillation, at least, using this direct solution
scheme based on evaluating determinants. Some deeper mathematical understand-
ing of the structure of the Laplacian matrices underlying these oscillatory systems
is needed.
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