two masses, each of mass *m*, are connected by a spring of resting length *L* and spring constant *k*_{2}. These are placed between fixed walls spaced a distance 2*L* apart, and each mass is attached to a separate wall by springs of resting length and spring constant *k*_{1}. Thus in the equilibrium position, the springs are at their resting lengths, and form a straight line.

Now, suppose we give the left mass a small displacement along this line. What is the subsequent motion of the masses?

The motion of the masses is one-dimensional; let the position of the masses be given by *x*_{1} and *x*_{2}, with each being zero at the initial positions, and positive values for rightward displacement. Then the force on mass 1 due to the left spring is . Similarly, the force on mass 2 due to the right spring is . Lastly, the central spring is compressed by an amount *x*_{1}–*x*_{2} when *x*_{1}>*x*_{2}, and extended by *x*_{2}–*x*_{1} when *x*_{1}<*x*_{2}. Thus, it exerts a force on mass 1 and the opposite force on mass 2. Thus, using *ΣF*=*ma*, we have equations of motion:

and

.

These are coupled second order linear differential equations. We can make them symbolically simpler by letting and . Using matrices to combine the equations, we have:

.

Let us try a solution of the form .

Then , and we have:

.

This holds for all *t* if and only if:

,

which is to say, that is an eigenvector of the matrix with eigenvalue *r*^{2}.

Now, to find the eigenvalues *x* of *A*, we have:

.

A quick check will find that the corresponding eigenvectors for *x*_{+} and *x*_{–} are constant multiples of and , respectively.

Noting that , we see that

and thus the values of are all imaginary.

Let and . Then our possible values for *r* are and .

Combining the imaginary exponentials that are conjugate to make the result real, we obtain and , giving the general solution:

.

Note that the pure *ω*_{1} solution corrensponds to the two masses oscillating together, with our central spring making no change in length; similarly, the higher-frequency *ω*_{2} solution corrensponds to the two masses oscillating exactly out of phase.

In phase motion:

Out of phase motion:

For our original problem’s initial conditions give at *t*=0. The velocity conditions give , the initial condition on *x*_{2} tells us that *C*_{1}=*C*_{2}, and the initial displacement on *x*_{1} tells us . Thus we have:

and .

Note that if *k*_{2} is small compared to *k*_{1}, so that the central spring is loose, and that the fundamental frequencies *ω*_{1} and *ω*_{2} are close, the oscillation will begin on mass 1 and will slowly transfer until it is mass 2 that is mostly oscillating, and then back to mass 1, and so on:

Horizontal axis is time, vertical axis is displacement (in units of *x*_{0}).

Red is *x*_{1}, blue is *x*_{2}, graph is done with .

Tags: Coupled Oscillator, Eigenvalues, Friday Physics, Oscillator, physics, Vibrational Modes

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