Physics 430: Lecture 25
Coupled Oscillations
Dale E. Gary
NJIT Physics Department
11.1 Two Masses and Three Springs





We covered oscillations in Chapter 5, but here we add complexity by
considering a system with two oscillators coupled together. The coupling
can be strong or weak.
We will limit the discussion to oscillators obeying Hooke’s Law, and without
friction. It is a special case, but one with a wide application.
Consider the situation shown
in the figure at right. There are
two cars of masses m1 and m2,
and three springs of spring
constants k1, k2 and k3, and we
want to obtain the equations of motion for the two cars.
We could use the Lagrangian formalism, but let’s use the Newtonian
approach first. The equilibrium positions of the two cars are shown by the
lines, and we will use coordinates x1 and x2 relative to those.
The forces on m1 are k1x1 to the left, and k2(x2 -x1) to the right, so its
equation of motion is m1 x1  -k1 x1  k2 ( x2 - x1 ) Likewise: m2 x2  k2 x1 - (k2  k3 ) x2 .
 -(k1  k2 ) x1  k2 x2 ,
December 03, 2009
Two Masses and Three Springs-2

The two coupled equations of motion:
m1 x1  -(k1  k2 ) x1  k2 x2
m2 x2  k2 x1 - (k2  k3 ) x2 ,
can be written more compactly using matrix notation, as Mx  -Kx, where



-k2 
 k1  k2
 x1 
 m1 0 
x   ,
M
.
 , and K   -k
k

k
x
0
m
2
2
3
2
 2


Notice that this is a generalization of the single oscillator, which you can see
by setting k2 and k3 = 0. You then get a single equation as in Chapter 5.
Note also that if the coupling spring, k2 = 0, then the two equations become
uncoupled and describe two separate oscillators.
As we did in Chapter 5, we will find complex solutions z(t) = aeiwt, but you can
imagine that we might have more than one frequency of oscillation, w  k / m ,
since we have two ms and 3 ks. It turns out that we only need to assume
one frequency initially, but we will arrive at an equation for w that is satisfied
by more than one frequency. z (t )
 1   a1  iwt
 a1   1e-i1 
iwt
Let’s try the solutions: z(t )  
  a  e  ae , where a  a    -i 2  .
z
(
t
)
 2   2
 2   2 e 
December 03, 2009
Two Masses and Three Springs-3
Recall that when we write the solution in terms of complex exponentials, we
will ultimately have to keep only the real part of the solution, i.e. x(t) = Re z(t).
 When we substitute z(t) into Mx  -Kx,we find the relation
-w 2 Maeiwt  -Kaeiwt ,
so the following equation must be satisfied:

(K - w 2 M)a  0.

Clearly, if we ignore the trivial solution a = 0 (no motion at all), we must have
det(K - w 2 M)  0.
This is a generalization of the eigenvalue problem that was introduced in
Chapter 10.
 Since these are two x two matrices, you can see that the determinant will
give a quadratic equation for w2, with two solutions (two roots).
 This implies that there are two frequencies at which our system will oscillate,
and these are called normal modes of the system. We could find them for
this system with two different masses and three different spring constants,
but it is complicated and not very interesting. Let’s look at simpler systems.
December 03, 2009
11.2 Identical Springs and Equal
Masses



For this case, the matrices M and K become:
m 0 
 2k
M
,
and
K


 -k
 0 m

-k 
.
2k 
We then need to find the determinant of
 2k - mw 2
-k 
2
K -w M  
,
2
2k - mw 
 -k
which is
det(K - w 2 M)  (2k - mw 2 )2 - k 2  (k - mw 2 )(3k - mw 2 ),
Setting this to zero, we find two solutions (the two normal mode frequencies):
k
3k
w
 w1 , and w 
 w2 .
m
m
Now that we have the frequencies, we must still solve the equation
(K - w 2 M)a  0.
 We do this twice, once for each frequency, to obtain the motion x(t) for the
two oscillating carts.

December 03, 2009
First Normal Mode

First insert w1  k / m , so that the relation
 k -k 
K - w12 M  
.

 -k k 
As a check, you can see that the determinant of this matrix is zero. The
solutions are then
a1 - a2  0
 1 -1  a1 
2
(K - w M)a  k 
  a   0 or -a  a  0 .
1
1

 2
1
2
 These are both the same equation, and simply says that a1 = a2 = Ae-i. Since
 z1 (t )   a1  iw1t  A i (w1t - )
z(t )  
,
   a  e   A e
z
(
t
)
 
 2   2

we finally have

 x (t )   A
x(t )   1     cos(w1t -  ).
 x2 (t )   A
That is, both carts move in unison: x1 (t )  A cos(w1t -  )
x2 (t )  A cos(w1t -  )
[first normal mode].
December 03, 2009
First Normal Mode-2

The motion is shown in the figure below. Notice that the spring between the
two carts does not stretch or contract at all.

When we plot the motion, it looks like this (identical motions, in phase).
December 03, 2009
Second Normal Mode

Now insert w2  3k / m , so that the relation
 -k -k 
K - w12 M  
.

 -k -k 
As a check, you can see that the determinant of this matrix is zero. The
solutions are then
a1  a2  0
1 1  a1 
2
(K - w M)a  -k 
  a   0 or a  a  0 .
1
1

 2
1
2
 These are again both the same equation, and simply says that a1 = -a2 = Ae-i.
Since
 z1 (t )   a1  iw2t  A  i (w2t - )
z(t )  
,
   a  e   - A e
z
(
t
)
 
 2   2

we finally have

 x (t )   A 
x(t )   1     cos(w2 t -  ).
 x2 (t )   - A
x1 (t )  A cos(w2 t -  )
That is, both carts move oppositely:
x2 (t )  - A cos(w2 t -  )
[2nd normal mode].
December 03, 2009
Second Normal Mode-2

The motion is shown in the figure below. Notice that the spring between the
two carts stretches and contracts, contributing to the higher force, and hence,
higher frequency.

When we plot the motion, it looks like this (identical motions, in phase, but at
a higher frequency w2).
December 03, 2009
General Motion

It is important to realize that, although these are the only two normal modes
for the oscillation, the general oscillation is a combination of these two
modes, with possibly different amplitudes and phases depending on initial
conditions.
1
1
x(t )  A1   cos(w1t - 1 )  A2   cos(w2 t -  2 ).
1
 -1

The resulting motion is surprisingly complicated, but deterministic. Because
w2  3w1
is an irrational ratio, the motion never repeats itself, except in the case that
either A1 or A2 = 0.
Motion for
A1 = 1, 1 = 0
A2 = 0.7, 2 = p/2
December 03, 2009
Normal Coordinates

If the motion seems complicated, you should realize that there is an
underlying simplicity that is masked by our choice of coordinates. We can
just as easily choose for our coordinates the so-called normal coordinates
1  12 ( x1  x2 )
 2  12 ( x1 - x2 ).

Using these coordinates, as you can easily check, the two normal modes are
no longer mixed, but instead we have:
1 (t )  A cos(w1t -  ) 
[first normal mode]
.
 2 (t ) 
0

1 (t ) 
0

.
 2 (t )  A cos(w2 t -  ) 
[second normal mode]
December 03, 2009
11.3 Two Weakly Coupled
Oscillators

Let’s consider an interesting special case of two coupled oscillators with weak
coupling. We can arrange this by making the coupling spring weak compared
to the figure below, with k1 = k3 = k, and k2 << k.
As you can guess, this will have the same normal modes as before, with
-k 2 
 k  k2
m 0 
M
,
and
K

.



k  k2 
 0 m
 - k2
 Solving det(K - w 2 M )  ( k  k2 - mw 2 ) 2 - k22  ( k - mw 2 )( k  2k2 - mw 2 ), we
conclude that
k  2k2
k
w1 
, and w2 
.
m
m


The difference here is that k2 is small, so w1  w2.
December 03, 2009
Two Weakly Coupled Oscillators-2
To take account of this closeness in the two frequencies, we will write the
result in terms of the average frequency
w  w2
wo  1
.
2
 To emphasize how close the two frequencies are to this average, we can also
write
w1  wo - e , and w2  wo  e .

Thus, e is half the difference between the two frequencies.
 The two normal modes are then (i) the two cars moving in phase (with the
middle spring not involved at all) and (ii) the two cars moving exactly out of
phase:
1
1
z (t )  C1   ei (wo -e )t , and z(t )  C2   ei (wo e )t .
1
 -1
where the complex constants C1 and C2 are determined by initial conditions.
 Writing the general solution as the superposition of these, and factoring the
 1
1 
exponential, we have
Slowly varying
z(t )  C1   e-ie t  C2   eie t  eiwot .
envelope
 -1 
 1
December 03, 2009
Two Weakly Coupled Oscillators-3

In general, the constants are complex constants, but let’s look at the case
where they are equal and real, i.e. C1 = C2 = A/2, so the solution becomes
 cos e t  iwot
A e-ie t  eie t  iwot
z(t )   -ie t
e  A
e .
i
sin
e
t
2 e - eie t 



We take the real part of this to get
x1 (t )  A cos e t cos wo t 
.
x2 (t )  A sin e t sin wo t 

The result is the following, displaying “beat” frequencies.
Envelope is cos et
Rapid oscillation
is cos wot
December 03, 2009