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I’ve been trying to find a way to solve this function for the first positive root. Solving this analytically seems to be above my high school education. Any and all help is greatly appreciated :)

The function is as follows:

$$\omega_i=\sqrt{\frac{N}{E\cdot I_i}} \ \ \ \ \ i=1,2$$

$$\cos(\omega_1 L_1)\sin(\omega_2L_2)+\sqrt{\frac{I_1}{I_2}}\sin(\omega_1L_1)\cos(\omega_2L_2)=0$$

All variables are constant except $N$.

I’ve attempted to use similar looking trigonometric identities such as $\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)$ but I cant seem to find a way to take the square root into account.

Solving this function graphically is straight forward (Example here), but I need to implement it in excel. The built in goal seek function doesn’t automatically update and hits false positives often, thus I’ve decided to implement a root seeking function. I’m not allowed to find 3rd party libraries as per my IT departments demands.

Looking at the functions shape, I’m scared the Newton-Raphson method with find a different root. Is there a better root seeking algorithm for this problem?

Тyma Gaidash
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Pikasso
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  • Welcome to Math.SE! ... To reduce some notational clutter, define $$n:=\sqrt{N} \qquad k_i:=L_i/\sqrt{EI_i} \qquad r:=\sqrt{I_1/I_2}$$ so that $\omega_i=nk_i/L_i$. Then the equation reduces to $$\cos(nk_1)\sin(nk_2)+r\sin(nk_1)\cos(nk_2)=0$$ Dividing-through by cosines, we can write this as $$r\tan(nk_1)+\tan(nk_2)=0 $$ Solving this for $n$ will require numerical methods, but once you have it, then $N=n^2$. – Blue Oct 17 '24 at 13:45
  • @Blue. Why not to continue with $x=nk_1$ ? – Claude Leibovici Oct 17 '24 at 13:52
  • @ClaudeLeibovici: "Why not to continue with $x=nk_1$?" ... Slavish devotion to notational symmetry? (Combining $I_1$ and $I_2$ into $r$ gave me jitters.) :) ... Actually, I was considering framing things in terms of equating functions of the form $\sqrt{I_i},\tan(nk_i)$, then realized the signs didn't work out. ... blah, blah, blah ... But you're right: In accordance with my stated goal of reducing notational clutter, I should have just written, say, $$\tan(x)+p\tan(qx)=0$$ for appropriate constants $p$ and $q$. – Blue Oct 17 '24 at 14:10
  • @Blue. In fact, it could be inresting to know which one to take $x=nk_1$ or $x=nk_2$ because the choice could have a significant impact on the numerical method. Any idea ? In any manner, thanks for the discussion. Cheers :-) – Claude Leibovici Oct 17 '24 at 14:26
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    @ClaudeLeibovici: Once numerical methods enter the room, I tend to back away slowly. :) – Blue Oct 17 '24 at 14:55
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    @Blue. Just thé opposite for me ! – Claude Leibovici Oct 17 '24 at 15:20
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    $\tan(x)+p\tan(qx)=0$ is equivalent to $\cos(qx)\sin(x)+p\cos(x)\sin(qx)=0$. Using product to sum formulas gives $\sin((1-q)x)+\frac{1+p}{1-p}\sin((1+q) x)=0$ which is addressed in this question. – Тyma Gaidash Oct 17 '24 at 15:37

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