UnkleRhaukus
  • UnkleRhaukus
\[\begin{vmatrix}2-\xi^2&-1&0\\-1& 2-\xi^2&-1\\0&-1&2-\xi^2\end{vmatrix}=0\]
Linear Algebra
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
(' )-( .) what
anonymous
  • anonymous
@THEHELPER123
UnkleRhaukus
  • UnkleRhaukus
FIND THE determinant

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UnkleRhaukus
  • UnkleRhaukus
, solve for xi
beginnersmind
  • beginnersmind
Just expand it out and solve for $\xi$ = 0. There's only 2 non-zero terms.
beginnersmind
  • beginnersmind
BTW, you can use the substitution \[u = 2 - \xi^{2}\] \[det A = u^{3} - u\] \[u^{3} - u=0\] to simplify the algebra.
UnkleRhaukus
  • UnkleRhaukus
\[ \begin{align*} \begin{vmatrix} 2-\xi^2&-1 &0 \\ -1 & 2-\xi^2&-1 \\ 0 &-1 &2-\xi^2 \end{vmatrix} &= 0 \\ \\[1ex] (2-\xi^2)\Big[(2-\xi^2)^2-(-1)^2\Big] -(-1)\Big[(-1)(2-\xi^2)-0\Big] +0 &= 0 \\ (2-\xi^2)\Big[(2-\xi^2)^2-1\Big]-(2-\xi^2) &= 0 \\ (2-\xi^2)\left(\Big[(2-\xi^2)^2-1\Big]-1\right) &= 0 \\ (2-\xi^2)\Big((2-\xi^2)^2-2\Big) &= 0 \\ (2-\xi^2)\Big((2-\xi^2)^2-2\Big) &= 0 \\ \end{align*}\] oh , where's the 2?
UnkleRhaukus
  • UnkleRhaukus
hmm i guess the u-sub is a good idea
UnkleRhaukus
  • UnkleRhaukus
i'll try it again with the \(u\)-substitution,
beginnersmind
  • beginnersmind
\[det A = u^{3} - 2u \] is correct, not \[det A = u^{3} - u \]
UnkleRhaukus
  • UnkleRhaukus
\[\begin{align}\begin{vmatrix} u&-1 &0 \\ -1 & u&-1 \\ 0 &-1 &u \end{vmatrix} &= 0 \\ \\[1ex] u\Big[u^2-(-1)^2\Big] -(-1)\Big[(-1)(u)-0\Big] +0 &= 0 \\ u\Big[u^2-1\Big]-u &= 0 \\ u^3-2u &= 0 \end{align}\]
UnkleRhaukus
  • UnkleRhaukus
hmm so u = 0 or u^2-2 =0
UnkleRhaukus
  • UnkleRhaukus
u = 0 or ±√2
UnkleRhaukus
  • UnkleRhaukus
that's three solutions for u, and three for xi xi = √(2-{-√2,0,√2})
UnkleRhaukus
  • UnkleRhaukus
so the normal node frequencies \(\omega\), where : \(\omega/\omega_0 = \xi\) are....
beginnersmind
  • beginnersmind
Should be 6 roots for \(\xi\). 2 for each u.
UnkleRhaukus
  • UnkleRhaukus
*the frequencies must be positive oh wait still in angular frequencies,
UnkleRhaukus
  • UnkleRhaukus
\[\nu = \omega/2\pi =... \textit{physics}\]
beginnersmind
  • beginnersmind
Ok, that makes sense. I don't think I can help too much with the phyics though. Are these coupled springs?
UnkleRhaukus
  • UnkleRhaukus
yes
1 Attachment
UnkleRhaukus
  • UnkleRhaukus
\[\omega_0^2 = k/m = 6\,[\text{N/m}]/0.0905\,[\text{kg}]\]so \[\nu_1 = 0.99\,[\text{Hz}],\quad \nu_2 = 1.3\,[\text{Hz}], \qquad \nu_1 =2.34\,[\text{Hz}]\]
UnkleRhaukus
  • UnkleRhaukus
what a rigmarole, so many variable, so few numbers
beginnersmind
  • beginnersmind
I actually prefered variables to numbers when I studied physics for the most part. I'd much rather write \(\omega^{2}\) than have to square 1.3823 every time. The only difference was when the teacher decided it was a good idea to take a general solution cos(ax+phi1) + sin(ax+phi2) and express it as a single trigonometric function. Except it wasn't ax+phi1 but a more complicated expression of the input variables.
UnkleRhaukus
  • UnkleRhaukus
one trig function, is better than two

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