# Creating a xylophone out of copper pipes

I have been trying to create a xylophone out of copper pipes for a school project and have been running into a few problems with calculating the length of each bar. I have been using this paper as reference, specifically equation (1). I have cut two pipes so far with lengths of 0.884m and 0.585m which played the notes B3 and A♭6 respectivly, which is far from the expected E2 and G♭3. For reference, my pipes have an inner radius of 0.25 in and an outer radius of 0.3125 in.

Even when I tried "reverse calculating" the speed of sound in the bar using (1) with both sets of data, I got answers both different from the expected 3700 m/s and from each other. Does anybody have any idea why this could be, either an error on my part or in my calculations? Is there another equation I should be using or is it more likly I am just making some silly error?

For reference, here are my calculations for the 0.884m bar:

K = 1/2 sqrt(0.25^2 + 0.3125^2) = 0.2 in = 0.0048 m

f = (pi * v * K * 3.011^2) / (8 * L^2) = (pi * 3700 m/s * .0048 m * 3.011^2) / (8 * (0.884 m)^2) = 80.9 Hz

This is around (a bit lower than) the frequency of E2, but when tested the tuner registers B3.

After some more tinkering around and reading through some comments, I am more lost than ever. I tried mounting the bars this time using rubber foam weather seal but the pitch was being registered even more spastic from the tuner. I have attached a video here, which hopefully can be of use.

• Are you using the pipes for resonators or as the vibrating bodies themselves? – phoog Jan 20 at 4:19
• I suspect that you're getting an overtone. B3, for example, is the third harmonic of E2. A♭6 is the 9th harmonic of G♭3. I don't quite see how those pipe lengths would give fundamental pitches so far apart, though, so I'm not sure you've calculated the expected pitches correctly. – phoog Jan 20 at 5:20
• Can you record the sound it is making and post it here? Just to rule out harmonics being read by the tuner, which happens frequently. Or are you really sure that's not the case? Can you do FFT images? – Von Huffman Jan 20 at 6:02
• How are you mounting these sections? If there's even a little damping or restraint you may find that the fundamental is being destroyed in favor of an overtone. In the meantime, if you can get some literature on vibraphone design (where the bar length and the resonant pipe lengths have to be matched up), that might help. – Carl Witthoft Jan 20 at 14:35
• For your analysis, I strongly recommend cutting one clean length, making sure you can hear a fundamental, and use that to estimate a single constant "Alpha" in the reduced equation frequency = Alpha/(length^2) . If a second bar 1/4 or 4X as long doesn't produce an octave, then you have to figure out how better to mount or strike your pipes. – Carl Witthoft Jan 20 at 14:39

Your radius is 0.25" but you use 0.5" in the calculations. I fixed that ran through your numbers and got 237Hz for n=1, which is not that far from B3=247Hz (actually it's about half way between B and Bb).

I was only working to two significant figures though. (Tip - it's less error prone to just put numbers in in the right units and THEN calculate ...)

EDIT : Here is my working at OP's request. Actually it was 236Hz, not 237Hz. But close enough for jazz (xylophone) ;-)

0.25" = 6.35mm

0.3125" = 7.94mm

K = sqrt(63.00 + 40.32) / 2 = 5.08mm = 0.00508m

f = [(pi * v * K) / (8 * L^2)] * m^2

If L = 0.884m

= [59.05 / (8 * 0.781)] * m^2

= (59.05 / 6.25) * m^2

= 9.44 * m^2

If n = 2, m = 5

f = 25 * 9.44 = 236Hz

(The frequency your tuner is picking up seems to be a harmonic.)

• Oops I think I accidentally typed 0.5in instead of 0.25in, thank you for catching that error. However, that 0.2in was still correct (I think) with 0.25in. How did you calculate 237 Hz? Is it possible that you could include your calculations? – COsborne Jan 20 at 13:04
• Done. If this solves your problem, would you please consider marking it as the answer? Thank you. – danmcb Jan 21 at 18:50