Help with formula for musical instrument [closed]

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 gerbilMar 10 '17 at 1:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – tpg2114, Kyle Kanos, Yashas, David Hammen, sammy gerbil
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• Please check that the edited formula is still correct. – Mark H Mar 9 '17 at 15:29
• Perfect, thank you Mark. Sorry I'm new to this, and very rusty with the old engineering maths. – Dale Mass Mar 9 '17 at 15:32
• Out of curiosity, what instrument is this for? Xylophone? – Mark H Mar 9 '17 at 15:40
• 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. – Dale Mass Mar 9 '17 at 15:48
• – sammy gerbil Mar 10 '17 at 1:55

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