Small oscillations: diagonal matrix

[bznm]

Homework Statement

I'm solving an exercise about small oscillations.

I name $T$ the kinetic matrix and $H$ the hessian matrix of potential.

The matrix $\omega^2 T- H$ is diagonal and so find the auto-frequencies is easy! But I have a problem with normal modes. The lagrangian coordinates are two angles, $\theta$ and $\phi$.

$$\omega^2T-H(\theta, \phi)=\begin{pmatrix}m\omega^2-m \Omega &&&0 \\
0&&&M\omega^2-k
\end{pmatrix}$$

Normal modes are given by splitted oscillations of the two coordinates. Is it correct? Are they given by:
$\theta(t)=A_1 \cos(\Omega t+ \alpha_1)$ and $\phi(t)=A_2(\cos \frac{k}{M} t+\alpha_2)$? ($A_1, A_2$= constants depending on initial conditions)

And is the general solution of motion given by $$\theta(t)+\phi(t)=A_1 \cos(\Omega t+ \alpha_1)+A_2(\cos \frac{k}{M} t+\alpha_2)$$?

 

[BruceW]

looks pretty good. Except one slight mistake. hint: check the values of omega.

 

[bznm]

looks pretty good. Except one slight mistake. hint: check the values of omega.

Thanks for your answer!
I have forgotten the square for ω..

But I have a doubt:
If I find the eigenvectors, I obtain:

- If $\omega^2=\Omega$
the eigenvector is (μ_1,0)

- if $\omega^2=k/M$
the eigenvector is (0, μ_2)

where μ_1 and μ_2 belong to ℝ.

So, the correct solution of motion, introduced the q_i coordinates that shift the origin of the system in the point of equilibrium, should be:

${\bf q}(t)=A_1 cos (\sqrt(\omega)t+\alpha_1) \begin{pmatrix}\mu_1\\0 \end{pmatrix}+A_2 cos (\sqrt(k/M)t+\alpha_2) \begin{pmatrix}0\\μ_2 \end{pmatrix}$

is it correct? thanks again

 

[BruceW]

yeah, that all agrees with the matrix equation you wrote in the first post. Except one thing. You have written the eigenvectors the wrong way around. The whole idea of the normal mode method is that the matrix you wrote, when acting on an eigenvector, should give a zero vector. So when omega squared equal k/M, then what should the eigenvector be?

Edit: woops, ah sorry sorry. You did write them the correct way around. I didn't take the time to look at the matrix product carefully enough. So all your work is correct.

 

[bznm]

Thank you so much!

 

[BruceW]

no worries! you had pretty much done it already.