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(adsbygoogle = window.adsbygoogle || []).push({});

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

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

With [itex]k_1=5,\; k_2=10,\; m_1=20,\; m_2=50 [/itex] I get [itex]\omega_1=0.2365,\;\omega_2=0.9456[/itex]

This comes from the equation [itex]\begin{pmatrix}

m_1 \omega^2+k_1+k_2 & -k_2 \\

-k_2 & m_2 \omega^2+k_2

\end{pmatrix}

\begin{pmatrix}

X_1 \\

X_2

\end{pmatrix}=\begin{pmatrix}

0 \\

0

\end{pmatrix}[/itex]

I have formed the impression (which must be wrong) that my values of [itex]\omega[/itex] should be eigen values with eigen vectors of [itex]\begin{pmatrix}

1 \\

1

\end{pmatrix}[/itex] and [itex]\begin{pmatrix}

1 \\

-1

\end{pmatrix}[/itex] Which describe the first two principle modes of vibration.

I expect to be able to check by substituting my values of [itex]\omega[/itex] into the matrix equation and get the zero matrix at the RHS when I used the eigen vectors for [itex]\begin{pmatrix}

X_1 \\

X_2

\end{pmatrix}[/itex]

However when I multiply out the matrix and the eigen vectors with my values of [itex]\omega[/itex] I get nothing like [itex]\begin{pmatrix}

0 \\

0

\end{pmatrix}[/itex]

Where is my understanding going wrong with this?

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

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