I am new to StackExchange, and I have a question about a school project I'm working on involving constructing an instrument. I am building a xylophone-glockenspiel-mash kind of thing, but I wanted to save some money on materials and I need to know: If I have the right material, and I know the current length and the current pitch in Hz when struck, can I calculate the length it needs to be to reach a specific note (in this case, F4)? The current length is 3ft or 91.4cm, and currently it hits 218.2Hz when struck. I have found nothing in my searches on the internet so far, although this may be because I don't know the proper terminology to search for. Thank you for reading, and thanks in advance for helping.

  • 1
    It will depend entirely on the type of material used for the bars, a certain metal alloy or type of wood. And if it is wood, it will vary with the density of the grain of the wood, its age, or its moisture content. I don't think you can easily generalize. You'll certainly end up having to mill or cut blocks and carefully cut them to tune them by hand and by ear.
    – user1044
    Commented Mar 31, 2014 at 7:54
  • There are other criteria you may want to take into account, especially harmonicity. I’ll try to find my old musical acoustics notes, but I have little to no hope.
    – Édouard
    Commented Mar 31, 2014 at 17:00
  • 1
    Usually the length isn't even that important. In many instruments the lengths are chosen in an aesthetically pleasing way, but you can do a lot by tuning them, see e.g. here: youtube.com/watch?v=5PesHXkN2M8
    – flawr
    Commented Jun 10, 2017 at 9:39

3 Answers 3


I have corrected this ! The guy below gave the correct answer, but I've worked it out so it can be used practically.

Present length^2 / needed length^2 = desired frequency / known frequency

Working that out

needed length = Square root of (present length * present length * known frequency /desired frequency )

349.2 hz is desired freq 91.4 cm is present length 218.2 hz is known freq

91.4 * 91.4 * 218.2 / 349.2

= 5220

Get the Square root of 5220

= 72.2 cm

Cut it way longer than that! Then test it and shave it down gradually until you reach the right pitch.

BTW here is a link to the frequencies of the music notes: https://pages.mtu.edu/~suits/notefreqs.html Could be handy!

NOTE: from actually doing this several practical tips.

1) Outside of an octave or so you may get different harmonics coming to the fore, especially with cheap wood. So when cutting down a larger piece of wood, you really need to use trial and error. I had a piece of wood that sounded a fairly clear Ab, and another about half as long is a C. I think that's because a different harmonic is coming through on the longer piece. 2) If there are impurities in the wood such as knots or swirls, it tends to affect the pitch, apparently in a downward direction, sometimes by as much as a tone. On my makeshift example I have several bits of wood the same length that are nonetheless different pitches by as much as a tone, and I think it is because one piece of wood has straight grain and the other has a knot in the middle. 3) Because of this you really do need to cut the wood a lot longer than you need!!! Then gradually shave off pieces until you reach the pitch you want. 4) I used a rubber strip attached to nails as a sort of bed for the bars, and it makes a very clear sound. Might be good to do this at the start, for testing the pitch. Otherwise hold the piece of wood at the very end when you hit it. A photo of my makeshift example, which is a 'proof of concept' for some lessons I'm doing with my students on making xylophones Here's my makeshift example, 'proof of concept' for xylophone construction lessons at school. Note the lengths of the keys. F is actually slightly longer than E, when it should be shorter, but I think this is because of the big knot on the E bar. This is the 'bed' in which the xylophone rests. I bought the rubber at a hardware store. It's very basic, but sounds quite reasonable. Here is what the 'bed' looks like, in which the bar rests at both ends. It allows the bar to vibrate freely.

Here are the lengths of these bars and the frequencies. They conform to the above equation, roughly, if you want to check!

C 40cm 261 hz D 40cm 293 hz E 35cm 329 hz F 36cm 349 hz G 32cm 391 hz A 32cm 440 hz


Eq. 4.39 of H. Olsen, Music Physics and Engineering 1967 gives the equation for the fundamental frequency of a free bar. For this problem, where you have the same material and the same cross-sectional shape, the frequency is proportional to 1/(length squared)


I think that rote numbers will not work. Wood is not a material of homogenous resilience, and xylophone bars are hollowed out for best resonance (and as part of tuning). So you need to figure in some waste material for experiments.

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