## Physics Friday 26

two masses, each of mass m, are connected by a spring of resting length L and spring constant k2. These are placed between fixed walls spaced a distance 2L apart, and each mass is attached to a separate wall by springs of resting length  and spring constant k1. 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 x1 and x2, 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 x1x2 when x1>x2, and extended by x2x1 when x1<x2. 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 r2.
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 ω1 solution corrensponds to the two masses oscillating together, with our central spring making no change in length; similarly, the higher-frequency ω2 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 x2 tells us that C1=C2, and the initial displacement on x1 tells us . Thus we have:
 and .
Note that if k2 is small compared to k1, so that the central spring is loose, and that the fundamental frequencies ω1 and ω2 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 x0).
Red is x1, blue is x2, graph is done with .

### 3 Responses to “Physics Friday 26”

1. Physics Friday 27 « Twisted One 151’s Weblog Says:

[…] Friday 27 Last week, we discussed the one-dimensional problem of two identical masses connected to each other by a spring of constant […]

2. Physics Friday 28 « Twisted One 151’s Weblog Says:

[…] Friday 28 In the previous two posts (here and here), we looked at a collection of masses (2 and 3,respectively) connected in a line by […]

3. Physics Friday 110 « Twisted One 151's Weblog Says:

[…] with a friend of mind, and which we both initially missed. Here, I will modify the situation from an earlier post to display this. Given two masses, m1 and m2, connected to each other by a spring of spring […]