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 $\frac{L}{2}$ 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 $F_l=-k_1x_1$ . Similarly, the force on mass 2 due to the right spring is $F_r=-k_1x_2$ . Lastly, the central spring is compressed by an amount x₁-x₂ when x₁>x₂, and extended by x₂-x₁ when x₁<x₂. Thus, it exerts a force $F_{c1}=k_2(x_2-x_1)$ on mass 1 and the opposite force $F_{c2}=k_2(x_1-x_2)$ on mass 2. Thus, using ΣF=ma, we have equations of motion:
$\begin{eqnarray}ma_1&=&F_l+F_{c1}\\&=&-k_1x_1+k_2(x_2-x_1)\\ma_1&=&-(k_1+k_2)x_1+k_2x_2\\\ddot{x_1}&=&-\frac{k_1+k_2}{m}x_1+\frac{k_2}{m}x_2\end{eqnarray}$
and
$\begin{eqnarray}ma_2&=&F_r+F_{c2}\\&=&-k_1x_2+k_2(x_1-x_2)\\ma_2&=&-(k_1+k_2)x_2+k_2x_1\\\ddot{x_2}&=&-\frac{k_1+k_2}{m}x_2+\frac{k_2}{m}x_1\end{eqnarray}$ .

These are coupled second order linear differential equations. We can make them symbolically simpler by letting $\omega^2\equiv\frac{k_2}{m}$ and $\lambda\equiv\frac{k_1+k_2}{m}$ . Using matrices to combine the equations, we have:
$\begin{pmatrix}\ddot{x_1}\\\ddot{x_2}\end{pmatrix}=\begin{pmatrix}-\lambda&\omega^2\\\omega^2&-\lambda\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}$ .

Let us try a solution of the form $\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}ae^{rt}\\be^{rt}\end{pmatrix}$ .
Then $\begin{pmatrix}\ddot{x_1}\\\ddot{x_2}\end{pmatrix}=\begin{pmatrix}r^2ae^{rt}\\r^2be^{rt}\end{pmatrix}$ , and we have:
$\begin{pmatrix}r^2ae^{rt}\\r^2be^{rt}\end{pmatrix}=\begin{pmatrix}-\lambda&\omega^2\\\omega^2&-\lambda\end{pmatrix}\begin{pmatrix}ae^{rt}\\be^{rt}\end{pmatrix}$ .
This holds for all t if and only if:
$r^2\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}-\lambda&\omega^2\\\omega^2&-\lambda\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}$ ,
which is to say, that $\begin{pmatrix}a\\b\end{pmatrix}$ is an eigenvector of the matrix $A=\begin{pmatrix}-\lambda&\omega^2\\\omega^2&-\lambda\end{pmatrix}$ with eigenvalue r².
Now, to find the eigenvalues x of A, we have:
$\det(A-xI)=0$
$\begin{eqnarray}\det\begin{pmatrix}-\lambda-x&\omega^2\\\omega^2&-\lambda-x\end{pmatrix}&=&0\\(-\lambda-x)^2-\omega^4&=&0\\(x+\lambda)^2&=&\omega^4\\x_{\pm}&=&-\lambda\pm\omega^2\end{eqnarray}$ .
A quick check will find that the corresponding eigenvectors for x₊ and x_- are constant multiples of $\begin{pmatrix}1\\1\end{pmatrix}$ and $\begin{pmatrix}1\\-1\end{pmatrix}$ , respectively.
Noting that $\omega^2\lt\lambda$ , we see that $x_{\pm}\lt0$
and thus the values of $r=\pm\sqrt{x_{\pm}}$ are all imaginary.
Let $\omega_1^2\equiv\lambda-\omega^2=\frac{k_1}{m}$ and $\omega_2^2\equiv\lambda+\omega^2=\frac{k_1+2k_2}{m}$ . Then our possible values for r are $\pm{i}\omega_1$ and $\pm{i}\omega_2$ .
Combining the imaginary exponentials that are conjugate to make the result real, we obtain $\begin{pmatrix}x_1\\x_2\end{pmatrix}=C_1\begin{pmatrix}1\\1\end{pmatrix}\cos(\omega_1{t}-\phi_1)$ and $\begin{pmatrix}x_1\\x_2\end{pmatrix}=C_2\begin{pmatrix}1\\-1\end{pmatrix}\cos(\omega_2{t}-\phi_2)$ , giving the general solution:
$\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}C_1\cos(\omega_1{t}-\phi_1)+C_2\cos(\omega_2{t}-\phi_2)\\C_1\cos(\omega_1{t}-\phi_1)-C_2\cos(\omega_2{t}-\phi_2)\end{pmatrix}$ .
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 $x_1=x_0,\;\;x_2=0,\;\;\dot{x_1}=0,\;\;\dot{x_2}=0$ at t=0. The velocity conditions give $\phi_1=\phi_2=0$ , the initial condition on x₂ tells us that C₁=C₂, and the initial displacement on x₁ tells us $C_1=C_2=\frac{x_0}{2}$ . Thus we have:
$x_1=\frac{x_0}{2}(\cos\omega_1{t}+\cos\omega_2{t})$ and $x_2=\frac{x_0}{2}(\cos\omega_1{t}-\cos\omega_2{t})$ .
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 $\frac{k_2}{k_1}=0.05$ .

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

This entry was posted on June 27, 2008 at 12:01 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

3 Responses to “Physics Friday 26”

Physics Friday 27 « Twisted One 151’s Weblog Says:
July 4, 2008 at 12:03 am | Reply
[...] Friday 27 Last week, we discussed the one-dimensional problem of two identical masses connected to each other by a spring of constant [...]
Physics Friday 28 « Twisted One 151’s Weblog Says:
July 11, 2008 at 12:01 am | Reply
[...] 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 [...]
Physics Friday 110 « Twisted One 151's Weblog Says:
February 26, 2010 at 12:10 am | Reply
[...] 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 [...]

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