I have a bit of a conundrum I'm trying to figure out. I have some spare time on my hands and I decided I wanted to custom make a set of wind chimes for my parents, and have been doing some research on how best to tune the pipes. My goal is to tune them to a specific scale (I'm a hobby musician) and not just create a "pleasant enough" sounding set.
My research brought me to several different places, not the least of which was wikipedia which has a section specifically on the math of wind chimes. So I took this equation from wikipedia, regarding the math of wind chimes (https://en.wikipedia.org/wiki/Wind_chime#Mathematics_of_tubular_wind_chimes): $$v_1 =\frac{\beta^2_1}{2\ \pi} \sqrt{\frac{EI}{\mu}} = \frac{22/3733}{2\pi L^2}\sqrt{\frac{EI}{\mu}},$$
where $L$ is the length of the tube, $E$ is the Young's modulus for the tube material, $I$ is the second moment of area of the tube, and $\mu$ is the mass per unit length of the tube. Young's modulus $E$ is a constant for a given material. If the inner radius of the tube is $r_i$ and the outer radius is $r_0$, then the second moment of area for an axis perpendicular to the axis of the tube is: $$I = \frac{\pi}{4}(r_o^4 - r_i^2).$$
The mass per unit length is: $$\mu = \pi \rho(r_o^2 - r_i^2),$$ where $\rho$ is the density of the tube material. The frequency is then $$f =\frac{22.3733}{4\cdot\pi} \sqrt{\frac{E}{2\rho}} \sqrt{D^2 + W^2},$$ where $W$ is the wall thickness and $D$ is the average diameter. Using 68 for young's modulus of aluminum, and 2.7 for density, 12.7 for $D$, and 1.65 for $W$, I believe the result becomes: $$f =\frac{22.3733}{4\cdot \:\pi} \sqrt{\frac{68}{2\:\rho}} \sqrt{12.7^2 + 1.65^2}.$$
Using symbolab.com to solve for L, given that I am using aluminum, and I know what frequencies I want to achieve (f = 523.25hz, C5 as an example), and the corresponding variables for the material I came to find that: $$L=\sqrt{\frac{44.7466\sqrt{2788.268617}}{2093\sqrt{5.4}\pi }}.$$
Which gave me a measurement of 393.24mm of length. After cutting the tube to the exact dimension, I discovered that the frequency was significantly off. Closer to approximately 432hz.
Now what is interesting to me is that, from the wikipedia page for acoustic resonance (https://en.wikipedia.org/wiki/Acoustic_resonance#Open_at_both_ends) I find the equation:
$f = \frac{nv}{2L}$ where $n$ is the primary harmonic I'm trying to find, $v$ is the velocity of sound in air and $L$ is my tube length, it confirms that 393.24mm should indeed be closer to 432hz, rather than the 523.25hz I was expecting from the first equation.
My question, therefore, is what did I do wrong with the more complex equation from the wind chimes article? The real-world results do not mesh with the expected math, so I assume I did the math incorrectly. I'm hoping to be able to use both equations to cross-check each other to confirm that I do indeed have the correct measurements to cut down on lost time and wasted material, if at all possible.
I'm excited to hopefully receive feedback from the community, and look forward to improving my understanding of the more complex math than I'm used to dealing with.