The formula is:

$$ f = \frac{3.5161}{2\pi L^2}\sqrt{\frac{EI}{\rho A}} $$

$A$ = area, $\rho$ = density, $I$ = second moment of area cross section, $E$ = Young's modulus, and $L$ = length. Can anyone help transpose this equation so that A is the unknown subject?

I can't find my way around it, and the people I've asked are stumped?

So i have some sprung steel, I'm calculating the frequency to make a SPECIFIC note.... but i want to know what size I need my steel (rectangular rod) to be to create the desired note/frequency.

Am I looking at it all wrong?


closed as off-topic by tpg2114, Kyle Kanos, Yashas, David Hammen, sammy gerbil Mar 10 '17 at 1:52

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  • $\begingroup$ Please check that the edited formula is still correct. $\endgroup$ – Mark H Mar 9 '17 at 15:29
  • $\begingroup$ Perfect, thank you Mark. Sorry I'm new to this, and very rusty with the old engineering maths. $\endgroup$ – Dale Mass Mar 9 '17 at 15:32
  • $\begingroup$ Out of curiosity, what instrument is this for? Xylophone? $\endgroup$ – Mark H Mar 9 '17 at 15:40
  • $\begingroup$ I'm a craftsman, it's a mixture between a Kalimba (Mbira) and a Cajon drum. I want to be as accurate as possible making the Steel "keys". Unsure about the Length. I've been working on various tests all week. $\endgroup$ – Dale Mass Mar 9 '17 at 15:48

$$ f = \frac{3.5161}{2\pi L^2}\sqrt{\frac{EI}{\rho A}} $$ $$ f^2 = \left(\frac{3.5161}{2\pi L^2}\right)^2\frac{EI}{\rho A} $$ $$ Af^2 = \left(\frac{3.5161}{2\pi L^2}\right)^2\frac{EI}{\rho} $$ $$ A = \left(\frac{3.5161}{2\pi L^2}\right)^2\frac{EI}{\rho f^2} $$ or $$ A = \left(\frac{3.5161}{2\pi L^2f}\right)^2\frac{EI}{\rho} $$

  • $\begingroup$ Mark H, extremely helpful, the workings out makes it perfect to revise. You've really helped me out, thank you. $\endgroup$ – Dale Mass Mar 9 '17 at 15:40

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