COUPLED OSCILLATORS
A real physical object can be regarded as a large number of simple oscillators coupled
together (atoms and molecules in solids). The question is: how does the coupling affect
the behavior of each of the individual oscillators?
Two identical pendulums
We have two identical pendulums (length L) for which we consider small oscillations. In
order to find what is the simplest motion, we imagine two experiments:
1) If we draw the two masses aside some distance and release them simultaneously from
rest, they will swing in identical phase with no relative change in position. The
spring will remain unstretched (or uncompressed) and will exert no force on either
mass.
θ1
θ2
We call this vibration pattern the first mode of vibration of the system.
2) The other obvious way of starting a symmetric oscillation will be to stretch the spring
from both ends. If we release the masses from rest simultaneously, we may notice that:
a) The spring now exerts forces during motion
b) From symmetry of motions of A and B, their positions are mirror images of each
other
mg
mg
We call this vibration pattern the second mode of vibration of the system.
Note: each pendulum in the one of the modes above oscillates with the same frequency:
the normal oscillation frequency.
The two oscillating patterns are called normal modes.
Both are SHM of constant angular frequency and amplitude.
General motion as superposition of normal modes
We take two coupled pendulums, identical, each starting from rest. Any motion of the
system, showing no special symmetry may be described as a combination of the two
normal modes of oscillation.
1
2
θ1
θ2
T1
T2
Fe
mgsinθ1
mg
Fe
mgsinθ2
mg
We assume small displacements from equilibrium: x1 , x2 .
Each pendulum swings because of the combined force of gravity mg and the string
tension T. The combined force is:
mg
mg sin θ 1 ≅ mg θ 1 =
x1 for mass 1,
L
but force along the spring exerted by mass 1 is F1 ~ mgθ 1 cosθ1 ~ mgθ 1 ~ mgx1 /L
mg
mg sin θ 2 ≈ mg θ 2 =
x 2 for mass 2,
L
but force along the spring exerted by mass 2 is F2 ~ mgθ 2 cosθ2 ~ mgθ 2 ~ mgx2 /L
Note Displacements from equilibrium are very small, angles θ 1 and θ 2 are very small:
sinθ 1 ,2 ~ θ1,2 ; cosθ1,2 ~ 1. Displacements are given by: x1 ≅ Lθ1 , x2 ≅ Lθ2
If we consider x1 ≠ x2 , the spring is stretched by x2 - x1 and the elastic restoring force in
the spring will be Fe = k(x2 - x1 )
The total restoring force on mass 1 is
The total restoring force on mass 2 is:
-[mgx1 /L - k(x2 - x1 )]
-[mgx2 /L + k(x2 - x1 )]
Equations of motion can be written for each of the masses by using Newton's second law:
 d 2 x1
mg
x1 + k ( x 2 − x 1)
m 2 = −
 dt
L

2
mg
 d x2
 m dt 2 = − L x 2 − k ( x 2 − x1 )
(1)
( 2)
The Symmetry Method to solve the system of second order differential
equations
The two equa tions are symmetric. We add and subtract them:
 d 2 ( x1 + x 2 )
mg
=
−
( x1 + x 2 )
m
L

dt 2

2
 m d ( x1 − x 2 ) = − m g + 2 k  ( x − x )

 1
2

dt 2
L m
(3)
We obtained equations that look like the SHM. We have obtained independent
oscillations in (x1 + x2 ) and (x1 - x2 ). We solve for (x1 + x2 ), (x1 - x2 ):
x1 + x2 = A cosω pt
x1 - x2 = B cosωs t
where ωp =
g
L
and
ωs =
g
k
+2
L
m
ωp and ωs are called the normal frequencies.
(4)
In writing the solutions, we need to apply the initial conditions. For this we have to
distinguish between the two oscillating patterns:
(1) parallel oscillation:
θ1
θ2
mg
mg
and (2) symmetric oscillation:
mg
mg
1) Parallel oscillation:
Let B = 0, x1 - x2 = 0
The two pendulums are moving in parallel. The spring does nothing (as if it didn't
exist). They both oscillate with ωp = the natural frequency of free pendulum without
coupling. This is the first (lower) normal mode of oscillation.
2) Symmetric Oscillation:
Let A = 0,
x1 + x2 = 0
The spring gets expanded/shrunk by twice the movement of each pendulum.
g
k
ωs =
+2
is determined by both the pendulum and spring. Each pendulum
L
m
oscillates with frequency ωs but they are out of phase by π. This is the second (higher)
normal mode.
General solution of the system of differential equations is a linear
combination of normal modes:
A
B
cosωp t + cos ωs t
2
2
A
B
x 2 ( t ) = cos ωp t − cos ωs t
2
2
x1 ( t ) =
(5)
We need a solution that satisfies an initial condition. Let's use an example :
a
 x 1 (0 ) = a

Initial conditions for the two masses are: 
dx1
x
&
(
0
)
=
1

dt

a =

0 =

A
+
2
A
−
2
B
2 

B
2 
è
t =0
=0
and
x 2 (0 ) = 0
x& 2 (0 ) = 0
A=a
 ⇒ A= B=a
A = B
which results if the following solution:
a
x1 ( t ) = (cosωp t + cosωs t )
2
a
x 2 ( t ) = (cosωp t − cosωs t )
2
The two normal modes are:
x1 + x 2 = a cos ωp t
first
x1 − x 2 = a cos ωs t
sec ond ( higher )
(lower ) mod e
mod e
We define by q1 = (x1 + x2) and q2 = (x1 - x2) the normal coordinates of
the system
α +β
α −β
cos
2
2
α+β
α −β
cos α − cos β = −2 sin
sin
2
2
If we use:
cos α + cos β = 2 cos
 ωp + ωs   ωp − ωs 
x1 ( t ) = a cos
t  cos
t  are the two individual oscillations
2
2

 

 ωp + ωs   ωp − ωs 
x 2 ( t ) = −a sin 
t  sin 
t 
2
2

 

If ωp − ωs << ωp + ωs , the spring constant is very small and the coupling between the
two pendulums is very weak.
ωp =
g
;
L
ωs =
g
k
+2
L
m
are the two normal frequencies.
We notice that in each normal mode, the individua l oscillators oscillates with the same
normal frequency
Observation. Up to now, we have studied only coupled oscillations of the same angular
frequency. If the two frequencies are different, we obtain beats: modulations of
amplitude produced by two oscillations of slightly different frequencies.