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Given is a mechanical multiple degree of freedom system described by the following matrices and equation:

  • mass matrix ${\bf{m}} = \left[\begin{matrix} m & 0 & 0 \\ 0 & m & 0 \\ 0 & 0 & m/2 \end{matrix}\right]$,

  • stiffness matrix ${\bf{k}} = \left[\begin{matrix} 2k & -k & 0 \\ -k & 2k & -k \\ 0 & -k & k \end{matrix}\right]$,

  • displacements ${\bf{u}} = \left[\begin{matrix}u_1(t) \\ u_2(t) \\ u_3(t)\end{matrix}\right]$,

  • external force ${\bf{p}} = \left[\begin{matrix}0\\0\\p_0\sin(\omega t)\end{matrix}\right]$, and

  • equation of motion ${\bf{m\ddot{u}}}+{\bf{ku}} = \bf{p}$.

The natural frequencies $\omega_i$ have been derived from the eigenvalue problem $\det({\bf{k}}-\omega_i^2{\bf{m}})=0$ which led to:

$\omega_1^2=(2-\sqrt{3})\frac{k}{m}\,,\,\omega_2^2=2\frac{k}{m}\,,\,\omega_3^2=(2+\sqrt{3})\frac{k}{m}$.

I know that the steady-state solution for $\bf{u}$ is

${\bf{u}}=\frac{1}{\det({\bf{k}}-\omega^2{\bf{m}})}\rm{adj}({\bf{k}}-\omega^2{\bf{m}})\,\bf{p}$,

so I have to calculate the determinant and adjugate. I actually have no problem doing that, but the textbook solution looks much more elegant than mine and I don't know how to get there.

My solution:

$\det({\bf{k}}-\omega^2{\bf{m}}) = \frac{1}{2}(2k-\omega^2m)^3-3k^3$,

${\rm{adj}}({\bf{k}}-\omega^2{\bf{m}}) = \left[\begin{matrix} \dots & \dots & k^2 \\ \dots & \dots & k(2k-\omega^2m) \\ \dots & \dots & (2k-\omega^2m)^2-k^2 \end{matrix}\right]$ (only the third column is relevant to the solution)

Textbook solution:

$\det({\bf{k}}-\omega^2{\bf{m}}) = \frac{1}{2}m^3(\omega_1^2-\omega^2)(\omega_2^2-\omega^2)(\omega_3^2-\omega^2)=k^3(1-\frac{\omega^2}{\omega_1^2})(1-\frac{\omega^2}{\omega_2^2})(1-\frac{\omega^2}{\omega_3^2})$,

${\rm{adj}}({\bf{k}}-\omega^2{\bf{m}}) = \left[\begin{matrix} \dots & \dots & 1 \\ \dots & \dots & 2(1-\omega^2/\omega_2^2) \\ \dots & \dots & 4(1-\omega^2/\omega_2^2)^2-1 \end{matrix}\right]k^2$

My question is: How can I get the textbook solution which is written in terms of the natural frequencies? It seems to me like there is some way to express ${\bf{k}}-\omega^2{\bf{m}}$ mostly in terms of the natural frequencies and then I could go from there, but I don't know how I should do that.

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    $\begingroup$ This is a mathematics problem, not a physics problem. Flagging to migrate to math.stackexchange.com $\endgroup$
    – Kyle Kanos
    Dec 10, 2013 at 13:47
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    $\begingroup$ I disagree. This is the internal mechanics of (the mathematics that describes) a piece of physics. $\endgroup$ Dec 10, 2013 at 20:46
  • $\begingroup$ which textbook? $\endgroup$ Apr 26, 2020 at 15:10

2 Answers 2

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The determinant is fairly easy to calculate. You know already, essentially, the eigenvalues of the stiffness matrix; more accurately, you know the eigenvalues of the matrix $\mathbf{m}^{-1}\mathbf{k}$, because the $\omega_i$ are zeros of the equation $$0=\det(\mathbf{m}^{-1}\mathbf{k}-\omega^2).$$ (The more aesthetically minded would replace $\mathbf{m}^{-1}\mathbf{k}$ with $\mathbf{m}^{-1/2}\mathbf{k}\,\mathbf{m}^{-1/2}$ to get a hermitian matrix, but no matter.) If you express the second determinant in the corresponding eigenbasis, you get $$ \det(\mathbf{k}-\mathbf{m}\,\omega^2)=\det(\mathbf{m})\det(\mathbf{m}^{-1}\mathbf{k}-\omega^2) =\frac{m^3}2\det\begin{pmatrix} \omega_1^2-\omega^2 & 0 & 0 \\ 0 & \omega_2^2-\omega^2 & 0 \\ 0 & 0 & \omega_3^2-\omega^2 \end{pmatrix}, $$ which gives your textbook's expression. More generally, this is an expression of the principle that a matrix's determinant is the product of its eigenvalues.


The adjunct, on the other hand, does not (to my knowledge) satisfy any such nice relation; in any case it is a nasty beast to deal with and I think few people judiciously substituting in the definition $k=m\omega_2^2/2$ of $\omega_2$ instead of $k$.

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I suspect you are overwhelmed by the plethora of variables that obscure the fundamental cyclometric symmetry of the problem. You may scale all but one of them out of the symmetric matrix M, part of whose (symmetric) inverse you are seeking, really, by redefining $${\bf{M}} \equiv {\bf{k}}-\omega^2 {\bf{m}}= -k \left[\begin{matrix} 2(x-1) & 1 & 0 \\ 1 & 2(x-1) & 1 \\ 0 & 1 & x-1 \end{matrix}\right], $$ where $x\equiv\frac{\omega^2 m}{2k}$.

The cyclometric (Mercedes-Benz) symmetry of its eigenvalues is then evident from its simple determinant, once you recognize the triple-angle formula for the sine in it, $\det {\bf M}= - k^3 (x-1)(4(x-1)^2-3)= - k^3 (x-1)(x-1-\frac{\sqrt{3}}{2}) (x-1+\frac{\sqrt{3}}{2})$, displaying the sines of the 3 roots of unity. This result is only useful in factoring the cubic polynomial out of the inverse of M, which you know is possible since the adjugate, adjM = detM M$^{-1}$ is a polynomial in x, and not a ratio of polynomials.

Nevertheless, it is easier to just find the 3rd column of the inverse (which is all you need for your solution) directly, by solving for the three trivial conditions of just the 3rd column of ${\bf M ~ M}^{-1}={\bf I}$. You know from the above discussion that your answer must have an inverse factor of $(x-1)(x-1-\frac{\sqrt{3}}{2}) (x-1+\frac{\sqrt{3}}{2})$ in it, summarizing the resonances of the system, which it is checked to have; and the transpose of this 3rd column of the adjugate is simply $k^2 (1, 2(1-x), 4(1-x)^2-1)$, which is just your textbook's expression, and, of course yours, once the roots of the determinant, the zeroes, are plugged in. The last entry is $k^2((x-1-\frac{\sqrt{3}}{2}) (x-1+\frac{\sqrt{3}}{2}) +2)$, of course.

Even though, as Kyle Kanos argues in the comment, this is strictly a linear algebra problem, still the symmetry techniques employed and the methodology utilized are the bread and butter of physics, and one might well argue physicists are normally quicker if not better at handling those.

Academic overkill PS: If, beyond the scope of your problem, you were interested in the full inverse, useless here, you only need exploit the trigonometric observation above, and change variables one last time, $x-1\equiv \sin \phi $, the one essentially dictated by the cyclometricity of the problem. Then it is straightforward to observe that the adjugate is actually elegant, $${\bf{I}} = \left[\begin{matrix} 2\sin \phi & 1 & 0 \\ 1 & 2\sin \phi & 1 \\ 0 & 1 & \sin \phi \end{matrix}\right] \left[\begin{matrix} -\cos (2\phi) & -\sin \phi & 1 \\ -\sin \phi & 2\sin^2 \phi & -2\sin \phi \\ 1 & -2\sin \phi & 1-2\cos(2\phi) \end{matrix}\right] ~ {\Large /}\sin (3\phi), $$ which of course contains the above answer.

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