# 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...

• Is the pin to be excited longitudinally or transversely? Is the pin fixed at a some point or is it totally free to move? What is the orientation of the pin relative to the vertical? Does the weight of the pin have to be taken into account? – Farcher Sep 27 '19 at 7:10
• Assume it is suspended vertically by a thread. I just really need the equations and reasoning to arrive to the correct answer. – Drunken Code Monkey Sep 27 '19 at 11:39
• Related for one end fixed and the rod undergoing transverse oscillations Derive equation for a cantilever in SHM and Fundamental frequency of a material and its Young's modulus – Farcher Sep 27 '19 at 15:55
• Also, is the pin solid or hollow? Is it a completely straight cylinder, or is it tapered? In general, what is the specific geometry we're talking about here? – probably_someone Sep 27 '19 at 17:11
• If the notation is standard, it's likely that $\rho$ (not a p, pronounced "rho") is the density of the material and $E$ is its Young's modulus – user2723984 Sep 27 '19 at 18:41

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.