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Lecture 33 – Coupled oscillators
Text: Fowles and Cassiday, Chap. 11
Coupled oscillators in one dimension
The previous lecture introduced the equations of motion for a vibrating chain, a discrete set of points linked together and subject to a uniform tension. Let's now introduce a Hooke's law potential for the elements of the chain. The solution of this system is generally determined using matrices (linear algebra is not a prerequisite for this course) so we will just solve a specific example of longitudinal motion of two oscillators.
A B
K
The end-points of the chain are fixed, and the two vertices A and B are free to move horizontally. There are three springs, with inequivalent spring constants and K , as indicated.
Now, imagine that mass m B at location x B is initially displaced to x B(0) = 1 (arbitrary units) and then released. Take x A(0)=0.
If the central spring were missing (or K = 0), m B would oscillate freely with angular frequency
o = (^ / m B)
1/
as usual.
However, in the presence of the central spring, energy is transferred between the vertices, and the motion of m B is gradually damped out as m A gains energy. The motion looks something like
x B
t
The motion of position A is similar, except that it starts at x A(0) = 0, resembling a sine function rather than a cosine.
We have seen this kind of waveform before when considering beats, where there is a high frequency component modulated by a low frequency (the beat frequency) amplitude. Mathematically, this behaviour can be obtained by a linear combination of two sinusoidal functions. Define Q 1 ( t ) and Q 2 ( t ) by Q 1 ( t ) = 2-1/2^ cos 1 t Q 2 ( t ) = 2-1/2^ cos 2 t
and take the linear combination x B( t ) = 2-1/2^ ( Q 1 + Q 2 ) = (1/2) (cos 1 t + cos 2 t ).
After a trig identity,
x B ( t ) = cos 1
t
cos^
1 −^2
t
Clearly this form satisfies the condition x B(0) = 1, and has the characteristics: first term varies rapidly - frequency equal to the mean of the two components second term varies slowly at the beat frequency.
The motion of position A is similar to that of B , except that it is shifted in phase by π/2. Taking the other linear combination x A( t ) = 2-1/2^ ( Q 1 - Q 2 ) = (1/2) (cos 1 t - cos 2 t ). we obtain
x A ( t ) = sin 1
t
sin^
1 −^2
t
This form satisfies x A(0) = 0, and has the usual beat frequency etc.
Normal modes
Just as there are some special oscillation frequencies associated with transverse waves on a string (because of the boundary conditions), there are some special modes associated with coupled oscillators. These are called normal modes , and correspond to situations in which there is no exchange of energy between vertices. There are two normal modes for the system investigated here:
- Displace both masses equally in the same direction x A(0) = x B(0) = 1 (to the right, here) v A(0) = v B(0) = 0 (released from rest)
The phenomena of normal modes appears everywhere in the vibrations of atoms and molecules. For example, with carbon dioxide
stretching
bending
There is another bending mode perpendicular to the page.
O C O
O C O
O C O
