## UnkleRhaukus one year ago $\begin{vmatrix}2-\xi^2&-1&0\\-1& 2-\xi^2&-1\\0&-1&2-\xi^2\end{vmatrix}=0$

1. anonymous

(' )-( .) what

2. anonymous

@THEHELPER123

3. UnkleRhaukus

FIND THE determinant

4. UnkleRhaukus

, solve for xi

5. beginnersmind

Just expand it out and solve for $\xi$ = 0. There's only 2 non-zero terms.

6. beginnersmind

BTW, you can use the substitution $u = 2 - \xi^{2}$ $det A = u^{3} - u$ $u^{3} - u=0$ to simplify the algebra.

7. 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?

8. UnkleRhaukus

hmm i guess the u-sub is a good idea

9. UnkleRhaukus

i'll try it again with the $$u$$-substitution,

10. beginnersmind

$det A = u^{3} - 2u$ is correct, not $det A = u^{3} - u$

11. 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}

12. UnkleRhaukus

hmm so u = 0 or u^2-2 =0

13. UnkleRhaukus

u = 0 or ±√2

14. UnkleRhaukus

that's three solutions for u, and three for xi xi = √(2-{-√2,0,√2})

15. UnkleRhaukus

so the normal node frequencies $$\omega$$, where : $$\omega/\omega_0 = \xi$$ are....

16. beginnersmind

Should be 6 roots for $$\xi$$. 2 for each u.

17. UnkleRhaukus

*the frequencies must be positive oh wait still in angular frequencies,

18. UnkleRhaukus

$\nu = \omega/2\pi =... \textit{physics}$

19. beginnersmind

Ok, that makes sense. I don't think I can help too much with the phyics though. Are these coupled springs?

20. UnkleRhaukus

yes

21. UnkleRhaukus

$\omega_0^2 = k/m = 6\,[\text{N/m}]/0.0905\,[\text{kg}]$so $\nu_1 = 0.99\,[\text{Hz}],\quad \nu_2 = 1.3\,[\text{Hz}], \qquad \nu_1 =2.34\,[\text{Hz}]$

22. UnkleRhaukus

what a rigmarole, so many variable, so few numbers

23. 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.

24. UnkleRhaukus

one trig function, is better than two