Q:

Calculating the resonant frequency of a metal pin

Is there a way to calculate the required diameter of a cylindrical metal (let's say aluminum) pin of a fixed length so that it has a specific resonant frequency? Let's say I have an aluminum pin that is 4 inches long, and I want it to have a resonant frequency of 256Hz (middle C), how would I go about calculating its diameter?
Edit:
Thanks to Farcher for this link, I think this is the equation I am looking for, however not being a physicist I'm not too sure what each term represents, or the units used. Can someone help me break this down:
$$f = \frac{v}{2L} = \sqrt{\frac{E}{4L^2\rho}}\\
E = \rho\; \left(\;2\;L\;f\;\right)^2$$
I get that L is the length, and F the frequency, but what are p and E? Also I do not see where the diameter or any dimension other than the length comes into effect... Am I just mistaken, and the resonant frequency is only dependent on the material and length, no matter the other dimensions? Somehow that doesn't seem right to me...

A:

There are three types of waves in a thin rod - longitudinal, bending (transverse), torsional. The velocity of propagation of longitudinal waves in the rods is equal
$$c=\sqrt {\frac {E}{\rho}}$$
The dispersion equation of longitudinal waves is 
$$\omega =ck$$
In this case, the frequency does not depend on the diameter of the cylinder. For bending waves of thin rods, we have the following dispersion law:
$$\omega_x=k^2\sqrt {\frac {EI_y}{\rho S}},\omega_y=k^2\sqrt {\frac {EI_x}{\rho S}} $$
In the case of a cylindrical rod, we have $S=\pi r^2, I_x=I_y=\frac {\pi}{4}r^4$. In this case, the frequency depends on the radius of the cylinder as
$$\omega =\frac {rk^2}{2}\sqrt {\frac {E}{\rho}}$$
For aluminum, we have $\rho = 2500 kg/m^3, E=70 GPa$. The value of k depends on the method of fixing the rod. For a rod with fixed ends, we have the equation 
$$\cos(kL)\cosh (kL)=1$$
Equation solution
$$k_1L=4.7300, k_2L=7.8532, k_3L=10.9956,...$$
In our case, we have $L=0.0254*4 m, k_1=46.5551 m^{-1},\omega =2 \pi 256 rad/s$. Hence $r=0.00028 m$ it's too thin.
For a rod with one free end, we have the equation 
$$\cos(kL)\cosh (kL)=-1$$
Equation solution $k_1L=1.8751,k_1=18.4557$. Hence $r=0.00178487 m$ it is quite suitable for generating sounds.
References
L. D. Landau, E. M. Lifshitz (1970). Theory of Elasticity. Vol. 7 (2nd ed.). Pergamon Press. ISBN 978-0-08-006465-9.

