Undamped 2 DOF vibration. What should the eigen vectors be here

[nerak99]

Normal modes of vibration, two masses, two spring, arranged vertically with m2 at the top, m1 underneath arranged (top to bottom) m2, k2, m1, k1, rigid support

I have solved the first part of an undamped coupled spring problem to give

$m_1m_2 \omega ^ 4 + ((m_1+m_2)k_2+m_2k_1)\omega ^2 +k_1k_2=0$ Since this is a show that Q, I know this is correct.

With $k_1=5,\; k_2=10,\; m_1=20,\; m_2=50$ I get $\omega_1=0.2365,\;\omega_2=0.9456$

This comes from the equation $\begin{pmatrix}<br /> m_1 \omega^2+k_1+k_2 & -k_2 \\<br /> -k_2 & m_2 \omega^2+k_2<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> X_1 \\<br /> X_2<br /> \end{pmatrix}=\begin{pmatrix}<br /> 0 \\<br /> 0<br /> \end{pmatrix}$

I have formed the impression (which must be wrong) that my values of $\omega$ should be eigen values with eigen vectors of $\begin{pmatrix}<br /> 1 \\<br /> 1<br /> \end{pmatrix}$ and $\begin{pmatrix}<br /> 1 \\<br /> -1<br /> \end{pmatrix}$ Which describe the first two principle modes of vibration.

I expect to be able to check by substituting my values of $\omega$ into the matrix equation and get the zero matrix at the RHS when I used the eigen vectors for $\begin{pmatrix}<br /> X_1 \\<br /> X_2<br /> \end{pmatrix}$
However when I multiply out the matrix and the eigen vectors with my values of $\omega$ I get nothing like $\begin{pmatrix}<br /> 0 \\<br /> 0<br /> \end{pmatrix}$

Where is my understanding going wrong with this?

 

[AlephZero]

I have formed the impression (which must be wrong) that my values of $\omega$ should be eigen values with eigen vectors of $\begin{pmatrix}<br /> 1 \\<br /> 1<br /> \end{pmatrix}$ and $\begin{pmatrix}<br /> 1 \\<br /> -1<br /> \end{pmatrix}$ Which describe the first two principle modes of vibration.

I expect to be able to check by substituting my values of $\omega$ into the matrix equation and get the zero matrix at the RHS when I used the eigen vectors for $\begin{pmatrix}<br /> X_1 \\<br /> X_2<br /> \end{pmatrix}$
However when I multiply out the matrix and the eigen vectors with my values of $\omega$ I get nothing like $\begin{pmatrix}<br /> 0 \\<br /> 0<br /> \end{pmatrix}$

Where is my understanding going wrong with this?

Your "impression" about the correct answer is wrong.

To see why it's wrong, imagine a two springs with equal stiffness, with a large mass at the bottom, and a small mass at the mid point. This is almost the same as a single mass at the end of the spring, and the eigenvector for the lowest mode will be approximately $\begin{pmatrix}<br /> 0.5 \\<br /> 1<br /> \end{pmatrix}$ not $\begin{pmatrix}<br /> 1 \\<br /> 1<br /> \end{pmatrix}$

The eigenector for the second mode will be close to $\begin{pmatrix}<br /> 1 \\<br /> 0<br /> \end{pmatrix}$, though that is a bit harder to "see" intuitively.

In general the eigenvectors depend on all the mass and stiffness properties. To calculate them, substitute the numbers for k m and $\omega$ into your equation $\begin{pmatrix}<br /> m_1 \omega^2+k_1+k_2 & -k_2 \\<br /> -k_2 & m_2 \omega^2+k_2<br /> \end{pmatrix}<br /> \begin{pmatrix}<br /> X_1 \\<br /> X_2<br /> \end{pmatrix}=\begin{pmatrix}<br /> 0 \\<br /> 0<br /> \end{pmatrix}$
If you calculated $\omega$ correctly, the system of equations will be singular, and you can solve them for the ratio of $X_1$ to $X_2$.

Repeat with the other value of $\omega$ to find the other eigenvector.

 

[nerak99]

Thank you very much for your prompt answer. I might even have ended up understanding eigenvectors in this context. It was well worth ploughing through the latex to get your answer.