COUPLED OSCILLATIONS

 

 

We will study coupled oscillations of a linear chain of identical non-interacting bodies connected to each other and to fixed endpoints by identical springs.

 

First, recall Newton’s second Law of Motion:

 

 

Newton’s second Law of Motion Everyone unconsciously knows this Law. Everyone knows that heavier objects require more force to move the same distance than do lighter objects. The Second Law, however, gives us an exact relationship between force, mass, and acceleration:

 

In the presence of external forces, an object experiences an acceleration directly proportional to the net external force and inversely proportional to the mass of the object.

This Law is widely known with the following equation:

 

F = ma 

 

 

 

                                                                                            

 

Where F is the net force, m is the mass of the object which the force F acts upon and a is the acceleration of the object. Since the acceleration is the second derivative of the distance with respect to time, the above law can be stated as

 

 

where   stands for the second derivative of x with respect to time t.

 

 The velocity, force, and acceleration have both a magnitude and a direction associated with them. Scientists and mathematicians call this vector quantity (magnitude plus direction). The equation shown above is actually a vector equation and can be applied in each of the component directions.

 

A second law that we will need is Hooke’s law

 

Hooke’s law discovered by the English scientist Robert Hooke in 1660--states that the force f exerted by a coiled spring is directly proportional to its extension${\mit\Delta} x$. The constant of proportionality is called the spring constant. The extension of the spring is the difference between its actual length and its natural length (i.e., its length when it is exerting no force). The force acts parallel to the axis of the spring. Obviously, Hooke's law only holds if the extension of the spring is sufficiently small. If the extension becomes too large then the spring deforms permanently, or even breaks. Such behavior lies beyond the scope of Hooke's law.

 

 

 

 

 

 

 

Let us consider first the simple case of one mass attached to a spring one end of which is attached to a vertical wall:

 

 

If x(t) is the position of the mass m from the equilibrium position at time t and k is the spring constant, then Newton’s second law of motion together with Hooke’s law give:

 

 

or equivalently,

 

 

This is one of the most famous equations of all physics. It is known as the harmonic equation. Let

 

 

then (w0 is called the frequency of the oscillation) it is widely known that the solution to the harmonic equation is

 

 

where A0 is a positive real number representing the maximum value of  x(t). One can also show that the solution can be written as

 

 

where and are the values of the position of the mass and its velocity at time t=0 respectively. The period of the oscillation described by formula (1) is

 

 

and the quantity

 

 

 

is called the natural frequency of the oscillation.

 

In what follows we treat some more complicated cases of oscillations.

 

1) Case of two masses Consider two identical bodies joined up with identical springs on a frictionless track as follows:

 

 

Here A and B represent the equilibrium positions of the two masses. Let x1(t) and x2(t) be the distances from the equilibrium positions of the two masses at time t and let k be  the spring constant.

 

The force acting on the first body has two parts by Hooke’s law: the first part is –kx1 due to the leftmost spring and the second part is k(x2-x1) due to the center spring. The net force acting on the first mass is then

 

 

Similarly, the net force acting on the second mass is

 

 

Applying the second Newton’s of motion gives the following system of differential equations:

 

 

 

 

 

This can be written in matrix form as follows:

 

 

 

 

Let us now find the eigenvalues of the matrix

 

 

 

 

that appears in the above equation.

 

Recall that the eigenvalues of A are those values of λ satisfying the equation:

 

 

where I is the identity matrix of the same size as A:

 

 

Let us now find corresponding eigenvectors. If  is an eigenvector of A corresponding to the eigenvalue 1, then one would have, or (A- I)X=0:

 

 

 

 

which means that a=b. So

 

 


is a basis for the eigenspace corresponding to 1. Similarly, one can show that the vector

 

 

is a basis for the eigenspace corresponding to 3.

 

Now let

 

 

then  where

 

 

is the “diagonal form” of A. Note that  so equation (*) above becomes

 

 

 

 

Now, let us consider the following change of variables:

 

 

so,

 

 

Taking the second derivatives gives:

 

 

so equation (**) above becomes

 

 

 

after simplification. This gives the following system of harmonic equations:

 

 

that  we know how to solve by the above simple case of a single mass attached to a spring.

 

So, what is the physical interpretation of all this?

 

Well, it is not difficult to see that there are two special kinds of motions that one can easily describe:

1.      Let us look again at the eigenvector

 

 

corresponding to the eigenvalue 1. The fact that the components are equal tells us that x1 and x2 are always equal. Consequently, the system oscillates back and forth but the middle spring is never stretched. It is as if we had two masses, each attached to a spring of constant k. It easy to see then that the frequency is given by

 

 

2.      In the case of the second eigenvector

 

 

 

x1 and x2 are always equal but have opposite directions. As you can guess, this gives an “in and out” motion type. The frequency of the system is also predictable in this case: each mass is attached to a spring compressed by distance x1 and to another stretched by a distance of 2 x1. It is as if the mass is attached to one single spring of constant 3k. We know that the frequency in this case is

 

 

Note that

 

 

is an eigenvector corresponding to the eigenvalue 3, this is why the  appears in the frequency above.

 

These two particular cases are called the normal modes of the system. As you ca guess, they have the property that if the system starts out in one of these modes, it will remain in this mode.

 

Of course, the above problem involving two masses can be dealt without talking about eigenvalues and eigenvectors. The benefit of using that algebraic technique is more apparent in the complicated cases of more than two masses.

 

2) Case of three masses Let us consider the case of three masses:

 

 

 Repeating the same argument as in the previous case gives the following system

 

 

 

with

 

 

as usual. One can show that (please work out the details) that the eigenvalues of the matrix:

 

 

 

are

 

 

and that

 

 

are corresponding eigenvectors.

 

One motion easy to describe is the one corresponding to the eigenvalue 2: The mass in the middle is not moving and the other two are moving in opposite directions. Each one of these masses has two springs attached to it; this explains the eigenvalue 2. The other two motions are little harder to describe.

 

 

Let us look now at a vibration example described by the following diagram:

 

 

Using

 

 

the system can be represented by the matrix equation:

 

 

where, as usual, the symbol represents the second derivative of x with respect to time.

 

Let us take this time the matrix A to be:

 

 

then its eigenvalues are and , and the corresponding eigenvectors are:

 

 

So, the normal frequencies of the vibration are and the normal modes of vibration are as follows: