# How Do You Derive and Analyze the Matrix for 3 Coupled Oscillators?

• Lambda96
In summary: No. If \lambda < 0 then \mathbf{x} = \mathbf{v}e^{\pm \omega_0\sqrt{|\lambda|}t} and we have exponential growth or decay. Thus we must have \lambda \geq 0.
Lambda96
Homework Statement
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Hi,

I am not sure if I have derived the matrix correctly, because of my results in task b

I solved task 1 as follows, I assumed that all three particles move to the right

$$m \dot{x_1}=-k(x_1 - x_2)$$
$$2m \dot{x_2}=-k(x_2-x_2)-3k(x_2-x_3)$$
$$3m \dot{x_3}=-3k(x_3-x_2)$$

Then I simply divided all three equations by the masses and got the following form

$$\dot{x_1}=-\frac{k}{m}(x_1 - x_2)$$
$$\dot{x_2}=-\frac{k}{2m}(x_2-x_2)-\frac{3k}{2m}(x_2-x_3)$$
$$\dot{x_3}=-\frac{k}{m}(x_3-x_2)$$

Then I set up the 3 equations in the required matrix form:$$\frac{d}{dt^2} \vec{x}=\left( \begin{array}{rrr} -\frac{k}{m} & \frac{k}{m} & 0 \\ \frac{k}{2m} & -(\frac{k}{2m}+\frac{3k}{2m}) & \frac{3k}{2m} \\ 0 & \frac{k}{m} & -\frac{k}{m} \\ \end{array}\right) \left( \begin{array}{rrr} x_1 \\ x_2 \\ x_3 \\ \end{array}\right)$$

For task part b, I simply put ##\vec{x}(t)=e^{i \omega t} \vec{v}## into the matrix above and then divided out the exponential term on both sides.$$-\omega^2 \vec{v}= \left( \begin{array}{rrr} -\frac{k}{m} & \frac{k}{m} & 0 \\ \frac{k}{2m} & -(\frac{k}{2m}+\frac{3k}{2m}) & \frac{3k}{2m} \\ 0 & \frac{k}{m} & -\frac{k}{m} \\ \end{array}\right) \vec{v}$$

After that I just got everything on one site and used that ##\frac{k}{m}=\omega_0## is

$$\left( \begin{array}{rrr} 0 \\ 0 \\ 0 \\ \end{array}\right)= \left( \begin{array}{rrr} -\frac{k}{m}+\omega^2 & \frac{k}{m} & 0 \\ \frac{k}{2m} & -(\frac{k}{2m}+\frac{3k}{2m})+\omega^2 & \frac{3k}{2m} \\ 0 & \frac{k}{m} & -\frac{k}{m}+\omega^2 \\ \end{array}\right) \vec{v}$$

Now, to determine ##\omega##, I simply formed the determinant of the matrix and got the following:

$$\frac{3 \omega_0^6}{2}+\frac{3 \omega_0^4 \omega^2}{2}-2 \omega_0^2 \omega^4- \omega^6=0$$

If I now solve the equation for ##\omega##, I get the following values

$$\omega_1=\pm \omega_0$$
$$\omega_2=\pm \sqrt{\frac{1}{2}(3+ \sqrt{3})}\sqrt{-\omega_0^2}$$
$$\omega_3=\pm \frac{\sqrt{(\sqrt{3}-3) \omega_0^2}}{\sqrt{2}}$$

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PhDeezNutz
You can simplify your calculations by writing the system as $$\ddot{\mathbf{x}} + \omega_0^2 \begin{pmatrix} 1 & -1 & 0 \\ -\tfrac12 & 2 & -\tfrac32 \\ 0 & -1 & 1 \end{pmatrix}\mathbf{x} = 0$$ where $\omega_0^2 = k/m$. Then after substituting $\mathbf{x} = \mathbf{v}e^{i\omega t}$ you can set $\omega^2 = \lambda \omega_0^2$ to obtain $$\omega_0^2\begin{pmatrix} 1 - \lambda & -1 & 0 \\ -\tfrac12 & 2 - \lambda & -\tfrac 32 \\ 0 & -1 & 1 - \lambda \end{pmatrix}\mathbf{v} = 0.$$ The determinant is then (EDIT: This is incorrect; please see below) $$(1 - \lambda)((2 - \lambda)(1 - \lambda) + \tfrac32) - \tfrac12 (1 - \lambda) = (1- \lambda)(\lambda^2 - 3\lambda + 3).$$ Thus we have $\lambda = 1, \sqrt{3}e^{\pm i\pi/6}$. Therefore $$\frac{\omega}{\omega_0} = \pm 1, 3^{1/4}e^{\pm i\pi/12}, 3^{1/4}e^{\pm i 5 \pi /12}.$$ We can immediately see that the roots occur in complex conjugate pairs, as expected.

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Lambda96 and PhDeezNutz
I've not explicitly checked the calculation, but how can you get complex eigenvalues for a non-dissipative system of harmonic oscillators?

Lambda96 and pasmith
vanhees71 said:
I've not explicitly checked the calculation, but how can you get complex eigenvalues for a non-dissipative system of harmonic oscillators?

Yes, I see I have made a sign error; the determinant should be $$(1 - \lambda)((2-\lambda)(1 - \lambda) - \tfrac32 ) - \tfrac12(1 - \lambda) = (1 - \lambda)(\lambda^2 - 3\lambda).$$

Lambda96 and vanhees71
That looks good!

Lambda96
I should also note that we expect a zero eigenvalue: The centre of mass does not accelerate, so $$m_i \ddot x_i = m_i M_{ij} x_j = 0$$ for every $x_j$. But this requires that $$m_i M_{ij} = 0$$ so $m_i \neq 0$ is a left eigenvector with eigenvalue zero.

Lambda96 and vanhees71
Thank you pasmith and vanhees71 for your help , sorry I'm only getting back to you now, the last few weeks have been a bit stressful

Now I have understood how I can solve the task much easier, thank you very much

One quick question, does the sign of the eigenvalues actually tell me in which direction the masses are moving?

Lambda96 said:
One quick question, does the sign of the eigenvalues actually tell me in which direction the masses are moving?

No. If $\lambda < 0$ then $\mathbf{x} = \mathbf{v}e^{\pm \omega_0\sqrt{|\lambda|}t}$ and we have exponential growth or decay. Thus we must have $\lambda \geq 0$. The zero eigenvalue corresponds to a constant drift of the centre of mass; the other eigenvalues give the frequencies of oscillation about the centre of mass. The corresponding eigenvectors will show you the relative motion of the masses.

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