Mathematics for Physics 3: Dynamics and Differential Equations
14
70
Lecture 14: Coupled oscillators and normal modes
So far we mainly studied dynamical systems in which only one body moves. The second Newton law was then applied
to that body, either in one or in three dimensions. We now wish to generalise the discussion to dynamical systems
that involve several different bodies, or components, each having its own mass. While each of these components
admits the second Newton law with its own forces, the forces may depend on the position of other components, thus
correlating their motion. We will see that so long as linearity of the equations of motion is preserved the solutions
become simple upon identifying collective modes, each correlating in a particular way the motion between different
components of the system. These collective modes are called Normal Modes.
Coupled oscillators: equations of motion
Let us start with a simple example, two bodies are connected by three springs; the springs are arranged longitudinally
between two walls, such that the central spring connects directly between the two masses, while the ones on the sides
connect between each body and the wall adjacent to it. The system is shown in fig. 13.
Figure 13: System of two bodies connected via three springs between two static walls .
Let us assume that each of the springs admits Hooke’s law with a corresponding coefficient ki and the bodies
have masses m1 and m2 , respectively. We assume that at t = 0 the system is perturbed away from its equilibrium
state, and we wish to solve for the position of the bodies y1 (t) and y2 (t) at any later time t.
Note that coordinates yi are measured to the right, from the left wall. It would be useful to first find the
equilibrium state and then define coordinates xi (see below) that measure the displacements away from that state.
Of course, to determine the equilibrium state one needs extra information, including the distance between the two
walls, which we denote by L, and the natural (non-extended) lengths of the springs, which we denote by li for
i = 1, 2, 3.
To start analysing the system, let us write down the forces which spring i applies on the body (of mass mi )
to its right. The positive direction is to the right (recall that the positions yi (t) are measured towards the right).
According to Hooke’s law it is:
�
�
Fi = −ki (yi − yi−1 ) − li
(211)
where (yi −yi−1 ) is the actual length of the spring while li is its natural (non-extended) length. In the case considered,
where we have three springs, y0 = 0 and y3 = L, so the forces are:
�
�
F1 = −k1 y1 − l1
�
�
F2 = −k2 y2 − y1 − l2
(212)
�
�
F3 = −k3 L − y2 − l3
We note that each of the bodies mi is affected by the force of two springs, the spring on its left contributing Fi ,
while the spring on its right contributing −Fi+1 . For example, in fig. 13 body m1 feels F1 by the spring on its left
and −F2 by the spring on its right. The fact that the spring asserts on body m1 the exact opposite of the force it
assets on body m2 is a consequence the third Newton law.
We can now readily write the second Newton law for each of the masses. We get:
mÿi = Fi − Fi+1
(213)
Mathematics for Physics 3: Dynamics and Differential Equations
71
namely,
m1 ÿ1 = F1 − F2
(214)
�
�
�
�
m1 ÿ1 = −k1 y1 − l1 + k2 y2 − y1 − l2
�
�
�
�
m2 ÿ2 = −k2 y2 − y1 − l2 + k3 L − y2 − l3 .
(215)
m2 ÿ2 = F2 − F3
or explicitly, using (212),
This is the set of equations of motion of the system. It is useful to write it in a matrix form:
�
��
� �
��
� �
�
m1 0
ÿ1 (t)
k1 + k2
−k2
y1 (t)
k 1 l 1 − k 2 l2
+
=
0 m2
ÿ2 (t)
−k2
k2 + k3
y2 (t)
k2 l2 + k3 (L − l3 )
(216)
Coupled oscillators: equilibrium and oscillations around it
Importantly, these equations are linear. It is an example of a linear system. We can formally write it as:
Ô[y] = c
Ô ≡ M D2 + K
(217)
where Ô is a linear functional operator acting on the function y(t), where as before, D is a differentiation operator,
generating differentiation with respect to time, and
�
�
�
�
�
�
�
�
ÿ1 (t)
m1 0
k1 + k2
−k2
k 1 l 1 − k 2 l2
y(t) ≡
,
M≡
,
K≡
,
c≡
ÿ2 (t)
0 m2
−k2
k2 + k3
k2 l2 + k3 (L − l3 )
(218)
Owing to linearity we can apply the tools we learnt in previous lectures, such as superposition. The fact that we
are dealing with matrices and vectors whose components corresponds to the coordinates of different bodies does not
change this fundamental property.
As we know, an immediate use linearity is the fact that the general solution can be written as a sum of the general
homogeneous solution and a particular solution of the non-homogeneous equation. In the case considered, the r.h.s
of (216), i.e. the part which makes it non-homogeneous is constant (time independent). It is therefore clear that a
particular non-homogeneous solution is a constant vector
�
�
�
�
y1 (t)
η1
y P (t) =
=
.
y2 (t) P
η2
Stated differently, one can remove the non-homogeneous term in (216) by simply shifting the coordinates as
follows:
�
� �
� �
�
y1 (t)
η1
x1 (t)
=
+
,
(219)
y2 (t)
η2
x2 (t)
where the constants η1 and η2 are defined such that
�
��
� �
�
k1 + k2
−k2
η1
k 1 l 1 − k 2 l2
=
.
−k2
k2 + k3
η2
k2 l2 + k3 (L − l3 )
Upon inverting the K matrix we can find the solution for η1 and η2 :
�
�
�
� �
�
η1
l1 + (L − l1 − l2 − l3 ) k2 k3 / k2 k3 + k1 k�2 + k3 k2
�
=
η2
l1 + l2 + (L − l1 − l2 − l3 ) k3 (k1 + k2 )/ k2 k3 + k1 k2 + k3 k2
(220)
(221)
This is the equilibrium point. At this point the forces on the r.h.s of eq. (214) vanish.
Coupled oscillators: homogeneous equations of motion
Now we can turn our attention to the dynamical part: the system is perturbed about the equilibrium point and
starts to oscillate. We see that solving this problem amounts to solving the homogeneous system:
�
��
� �
��
� �
�
m1 0
ẍ1 (t)
k1 + k2
−k2
x1 (t)
0
+
=
.
(222)
0 m2
ẍ2 (t)
−k2
k2 + k3
x2 (t)
0
Mathematics for Physics 3: Dynamics and Differential Equations
72
For simplicity let us now assume that the mass matrix is the unit matrix, m1 = m2 = 1 (if this is not the case one
can rescale the coordinates so as to effectively end up with the same equation; we shall do that in the next lecture).
With this assumption we get:
�
� �
��
� �
�
ẍ1 (t)
k1 + k2
−k2
x1 (t)
0
+
=
.
(223)
ẍ2 (t)
−k2
k2 + k3
x2 (t)
0
or, equivalently, in the more formal operator notation:
�
�
D2 + G [x(t)] = 0 .
(224)
where G = K (if the mass matrix is non-trivial, G will differ from K. We will see in the next lecture how). This
is the familiar form of simple harmonic motion, with the only remaining complication (compared to problems we
solved before) being the fact that we now have a set of coupled
for the two displacements x1 and x2 . This
� equations
�
x1 (t)
calls for change of basis in the linear vector space x(t) =
so as to find modes that decouple from each
x2 (t)
other. Before doing that it is useful to note that the coupling between the two bodies m1 and m2 in this particular
example is only due to the middle spring. Indeed, only k2 appears in the non-diagonal terms of the equation, and
in its absence the system will fall in two separate oscillators in the original basis x. For any nonzero k2 , it would be
convenient to change basis.
Diagonalizing the system: example
Let us now consider the example of (223). For simplicity we now assume extra symmetry. Let us take equal masses:
m1 = m2 = 1 and further assume that k3 = k1 so the mechanical system in fig. 13 is left-right symmetric. Note that
we do not assume that the initial conditions are symmetric. Indeed the system may be perturbed in an arbitrary
way.
Under these assumptions our (homogeneous) equation of motion system takes the form:
�
or, in a matrix notation,
ẍ1 (t)
ẍ2 (t)
�
+
�
�
k1 + k2
−k2
��
−k2
k2 + k1
��
G
�
x1 (t)
x2 (t)
�
=
�
0
0
ẍ + Gx = 0
�
.
(225)
(226)
Let us substitute the following trial function into (226):
x(t) = p cos(ωt − φ)
(227)
This gives:
−ω 2 p cos(ωt − φ) + Gp cos(ωt − φ) = 0 .
so for this to hold at any time t one needs:
�
�
G − ω2 I p = 0 .
where I is a unit matrix. This is an eigenvalue problem: for the ansatz of (227) to satisfy the equation, p must be
an eigenvector of the matrix G, with the eigenvalue λ = ω 2 . We shall see how the eigenvalue problem emerges from
more general considerations in (242) below.
For our particular example the eigenvalue equation is:
�
��
�
k1 + k2 − λ
−k2
p1
=0
(228)
−k2
k2 + k1 − λ
p2
To have a non-trivial solution for p the determinant of G − λI must vanish. This leads to the following quadratic
equation:
(k1 + k2 − λ)2 − k22 = 0
which is solved by
λ1 = k 1 ,
λ2 = k1 + 2k2 .
The corresponding eigenvectors can be obtained by returning to equation (242):
�
�
G − λj I p(j) = 0 .
Mathematics for Physics 3: Dynamics and Differential Equations
73
for j = 1 and 2. For j = 1 we have:
�
k1 + k2 − k1
−k2
−k2
k2 + k1 − k1
��
(1)
p1
(1)
p2
�
=0
1
1
�
=⇒
(1)
(1)
p1 = p2
(229)
so
1
p(1) = √
2
where the
√
�
(230)
�
�T
2 normalization was chosen such that p(j) p(j) = 1. For j = 2 we have:
�
k1 + k2 − (k1 + 2k2 )
−k2
−k2
k2 + k1 − (k1 + 2k2 )
��
(2)
p1
(2)
p2
�
=0
=⇒
(2)
(2)
p1 = −p2
(231)
so
1
p(2) = √
2
�
1
−1
�
Thus the general solution of the equation of motion (225) is:
��
�
��
�
x(t) = β (1) p(1) cos t λ1 − φ(1) + β (2) p(2) cos t λ2 − φ(2)
�
�
�
�
��
�
��
�
1
1
1
1
= β (1) √
cos t k1 − φ(1) + β (2) √
cos t k1 + 2k2 − φ(2)
1
−1
2
2
(232)
(233)
where β (1) and β (2) as well as φ(1) and φ(2) , four parameters in total, will be determined by the initial conditions.
Recall that there are two bodies, for each of which we need to supply a condition on the position and on the velocity
at time t = 0. This is precisely sufficient to uniquely fix these four parameters.
The physical interpretation of the two normal modes is intuitive:
• The first mode, corresponding to ω 2 = k1 , is the mode in which the two masses oscillate in phase with one
(1)
(1)
another: their displacements are identical x1 (t) = x2 (t). This means that the central spring connecting
them is always in the same length, and it thus does not influence the frequency of the oscillation: the latter is
indeed dictated by the two springs connected to the walls k1 = k3 .
• The second mode, corresponding to ω 2 = k1 + 2k2 , is the mode in which the two masses oscillate in exact anti(2)
(2)
phase with one another: their displacements are identical in magnitude but opposite in sign: x1 (t) = −x2 (t).
This means that the central spring extends (or contracts) twice as much as the others, and indeed, two-to-one
is the ratio through which the corresponding Hooke’s constants enter the relevant frequency ω 2 = k1 + 2k2 .
Our analysis shows that any motion of this system (with any given set of initial conditions), however complex it may
appear to be, can be seen as some linear combination of these two simple normal modes.