# Musical instrument confusions.

1. May 22, 2010

### stepan

Hello everyone. I am currently a grade 11 High-School student and have embarked on making the Waterphone() for my musical instrument project. Fortunately, I have stumbled upon this how-to for something to start on http://www.em411.com/show/blog/7895/0/DIY_Waterphone.html [Broken]. I will make it look more like this however: http://www.fundomentalmusic.com/_Media/dscn0569.jpg [Broken]

Although I have a general idea of how to get the frequencies, I am still unsure and need a second consultation (My teacher said she didn't know unless she did research, and I don't think she will be using her time for that :P ).

For the Rods

I will use the table from http://www.phy.mtu.edu/~suits/notefreqs.html to get frequencies of different notes.

I will use the same rods, so the only determinate for frequency will be height.

Because the rods are closed-air columns, I know the height of the rod will be 1/4*wavelength

Therefore for C4
$$\int$$=261.63Hz
v=345m/s(at 22 degrees Celcius)

$$\int$$=v$$\lambda$$
$$\lambda$$= 0.76m

therefore rod height is 19cm .

Is this right? . This is the only normal solution i came up with, although there is something that just doesn't sit right with me.

For the Cylinder in the center

I am not sure what that is for the waterphone. I think it is open at both ends but I don't know what note that is supposed to make, if it's supposed to make a note at all. Any help from people acquainted with the waterphone on this issue would be greatly appreciated.

If it is open at both ends though would i count the space that the cylinder is attached to and say it is a closed air-column, or say that it is an open air column and count just the rod.

Any comments would be greatly appreciated!

Sincerely,
Stepan

Last edited by a moderator: May 4, 2017
2. May 23, 2010

### Stonebridge

Welcome to Physics Forums, Stepan. This is an interesting question.
Firstly,
are the rods hollow; or solid?
The instrument in the video sounds like the vibrations are due to the vibration of the metal rods themselves, rather than the air that might or might not be inside them. If this is the case (metal not air) your calculation is not valid because that uses the air vibration. You need to calculate the frequency of standing waves in the metal of the rod itself using the speed of sound in the metal, not the air.
If this is the case, the nearest case mathematically would be that of a tuning fork, where one end vibrates and the other is fixed.
Here is the theory and formula for the frequency
http://en.wikipedia.org/wiki/Tuning_fork
Secondly. It doesn't sound like the instrument is actually tuned accurately to the notes of the scale. The pitch of a note, to my ear at least, seems to change oddly after it has been bowed.
It's likely that the rods have a number of frequencies at which they can vibrate (this gives them their strange sound) and that these will consist of higher harmonics and non-harmonic frequencies. If the rods are hollow, you will get the air column frequencies and their harmonics as well. It all adds to the overall sound.
I also guess that there are harmonic vibrations set up in the other rods as they are all acoustically linked. It may prove very difficult to "tune" the instrument by calculating the theoretical length of the rod. You can be sure that the physics will not be quite so straight forward, the vibration far more complex than that. Probably the best you can achieve is to get the rods of different length to approximate to a scale.
As I said. I don't think the instrument in the video is accurately tuned to a particular scale. With such a weird sound it doesn't really matter...

Last edited: May 23, 2010
3. May 23, 2010

### stepan

Hello Stonebridge and thanks for the warm welcome. I was so focused on the textbook i would have never thought that the rods could be solid. As you have mentioned, because the hollow rods would produce additional frequencies, For the sake of simplicity, and the fact that my teacher requires the instrument to play an octave, I think it would be a good idea to stick with solid rods.
After some time perusing the internet, I found the patent for the Waterphone, which has helped me clear a lot of things. Your assumptions about the varying frequencies were right, as the waterphone uses the chamber as a resonator , and the movement of water , plus the flexible posterior changes tones. The large cylinder in the middle affects how low the sounds will be.

Because the resonator acts on all the rods equally, I think that a scale can be achieved, with error that is consistent throughout. My teacher is marking if the instrument can play to a scale, so if I do this will it be ok? I also have a tuner that I use for my guitar, so ounce I finish I can use that to tone each rod; Although I am not sure how hard it will be to tone it ( I will be cutting the rods to refine tone).

Because the instrument changes the tones, I think what I can do to show that it is toned is to initially have the water removed, and hit each rod to show the octave. After that, I can put the water back in

Thank you for linking me to the formula on how to calculate the frequencies of solids. To make sure I got it right, I’ll do an example for C4

speed of sound in aluminum is 5100 m/s, and the
I can find that the Young modulus for Aluminum is 70*10^10 Pa(N/m^2) (E)
I can find that the density of aluminum is 2700kg/m^3 (p)
If we say the radius of the prongs in m is 0.007m (R)

Calculating this gets

http://www.wolframalpha.com/input/?i=sqrt(0.004m*(sqrt((pi*7*10^10N/m^2)/2700kg/m^3)/263Hz
I definitely did something wrong :S

Last edited: May 23, 2010
4. May 23, 2010

### sophiecentaur

If you cut a rod and the note turns out to be too high, you can lower the note by filing the end near the support to make it a bit thinner. This is how they tune glockenspeils etc. (and quartz crystals).

5. May 23, 2010

### Stonebridge

Using the tuning fork formula for a note of frequency around the C above middle C, I get a length of 52cm for the rod if its radius is 5mm
This seems very reasonable. A thinner rod would not need to be so long. Higher frequencies would need a shorter rod.
The sort of frequencies this instrument produces are in the higher octaves, I would imagine around 1000Hz or more. By careful choice of rod radius you will get reasonable sized rods.
The formula seems to be saying that, given all other things constant (radius, material) the frequency of the note is proportional to 1/√length of rod.
To set up an octave (top note double the frequency of bottom note), you know that the ratio of the length of the longest to the shortest rod will be √2
You would not be far off a chromatic scale if you cut the intermediate rods each an equal difference so that they form a stepped set of 13 from the longest to the shortest.

Last edited: May 23, 2010
6. May 23, 2010

### stepan

Thank you for the help sophie :) I'll give that a try ounce i get everything together.
Thanks for all the assistance Stonebridge. The way to make notes, as you have outlined makes perfect sense. When I used the formula to try to replicate your answer, I ended up with the unit m^(3/4)N^(1/4)/(kg^(1/4)sqrt(Hz) . I must have done something tremendously wrong to get this answer. Do you have an idea of where I went wrong in the computation? Did i use the wrong formula? ( Computation on link)

Last edited: May 23, 2010
7. May 23, 2010

### Stonebridge

Using the formula on Wiki we get
for R=0.005m, frequency = 512Hz

L² = 0.005/512 x √((Pi.70x10^10)/2700)
L² = 0.28m
L=53cm
I'm not sure about your computation. There are 4 left brackets and only 2 right brackets so something is not quite right.
Sophie's suggestion is good.
My suggestion would be to get a rod of about the right length for your highest note - it can be quite a high frequency. Over 1000Hz if you want.
Then one for the octave below, which should be the length of the other one times √2.
Tune them in as best you can. Then the other notes in the chromatic scale will just form equal steps in length between the top and bottom. You will finish with 12 rods in gradually shorter lengths like steps.
If you don't want a chromatic scale but prefer a diatonic (major) scale, then just use those rods which give a note on the scale.
Assuming you have a chromatic (semitone/halftone) scale to start with and rods
1,2,3,4,5,6,7,8,9,10,11,12,13 to give an octave, for the major scale just leave out rods
2,4,7,9, and 11
I'm sure the maths is approximate and there would be no benefit in trying to make the rods to any calculated length.
It's better to get them near and then tune them manually.
You may also find other metals work better than aluminium. Give them a try if you can get them.

8. May 23, 2010

### stepan

The computation that you have shown gives the right digits, but It is the units that I am worried about. If L^2 were to be 0.28m , than wouldn't L be 0.52sqrt(m) ? Same thing happens with the remaining part, Where the Pa and the kg/m^3 don't really cancel each-other out.

9. May 23, 2010

### waterphoneman

This may work on other instruments but I doubt if it will work on a waterphone. Perhaps thinning the rod at the free end might work better.

10. May 23, 2010

### waterphoneman

(Although aluminum is a good metal for sound I think you would be better of using
stainless steel.)

11. May 23, 2010

### waterphoneman

This may work on other instruments but it would weaken the rod on waterphones. A better solution would be to thin down the free end of the rod which will lower the tone.

12. May 23, 2010

### waterphoneman

I don't want to get into much information about tuning waterphones as that is about the only thing that is keeping me a head of the knock offs but you are correct in saying that the waterphone is not tuned to a conventional scale. That said, there is much conflict in a newly constructed waterphone until adjustments are made to remove them or at least reduce them. There can be conflict between the various rods and conflict between the rods and the bottom pan. Once the water is introduce that changes everything and I doubt if your waterphone will maintain a scale at that point as any little movement with bend a tone.

13. May 23, 2010

### stepan

Thanks for the replys waterphoneman, or Mr. Richard Water ? :O. I have taken your suggestion on the type of metal to use, and have switched the rods with Stainless steel. As for your last post, would you agree that if I were to not introduce the water initially, as to show the teacher that I have set the tones, would work? I would then add water to demonstrate the pleasing, tone-bending effects of the waterphone. I am still unsure as to the formula that has been suggested, as I can't get rid of the plethora of units that I end up with. Any comments on that would be greatly appreciated.

14. May 23, 2010

### waterphoneman

Yes, I am Richard Waters. You need to switch the resonator material to stainless steel but the rods will be easier to deal with if they are bronze.
Yes, you can probably maintain a scale of sounds if you do not use water initially.
The formula for tuning I do not want to venture into for reasons already stated but you should be able to accomplish your goal using a tuning fork. Good luck.

15. May 24, 2010

### Stonebridge

typo in my post, sorry
L² = 0.28m²
L=0.53m
The units are fine.
In the formula
L² = R/f x √Pi.E/ρ the units are m² on both sides*
I don't think it's worth worrying too much over getting a formula that will accurately predict the frequency of the rods given their length, thickness and material. It will only ever be an approximation. When it does show itself to be approximate, you then find yourself in the awkward position of having to explain why it doesn't give the right answer.
Makers of musical instruments don't construct pianos, flutes etc based on mathematical formulas to get the dimensions exactly right. It's done by "ear".
Get the two octave rods tuned first and cut the others as I explained before.

Thanks also to Mr Waterphone. It's fascinating to learn about your instrument.
I'm pleased to hear that you confirmed my suspicions that this is not tuned to a particular scale, and that aluminium is probably not a good material. I remember trying to make some wind chimes out of hollow aluminium tubes many years ago. They would never behave according to my calculations!
BTW
How does the point on the rod where you bow it affect the sound?

*
RHS
ms.√[N.mˉ²/kgmˉ³]
ms.√[kgmsˉ².mˉ² / kgmˉ³]
ms.√[m²sˉ²]

Last edited: May 24, 2010
16. May 24, 2010

### stepan

Thank you taking the time to comment Mr. Waters and as Stonebridge said, you have created a fascinating instrument. I will call my partner and tell him to replace the pans today.

I think I will set the rods up the way you suggested Stonebridge. The reason i was looking for a formula was because the teacher required it; I agree that It would be better to build by the ear. Thank you for the explanation for the units!

17. May 24, 2010

### waterphoneman

Stepan, You are welcome and thank you all for the compliments.

18. May 25, 2010

### stepan

With everyones help, I was able to get the rod heights and set the resonator up.( I glued the two pieces together because a special type of welding torch was required, and my school did not have it). With the rods, however, I was told that they couldn't weld it on, and since it is a different material, It won't stick. Is there any way to stick the rods on the resonator?

19. May 25, 2010

### sophiecentaur

You could tap a thread on the ends of the rods and bolt them firmly on (washers and nuts each side of the resonator face). It would alter the pitch but you can compensate for that. The essential thing would be to get a really rigid join. at that point.

20. May 25, 2010

### stepan

Thanks for the help Sophie, that sounds like a great idea. How would i go about tapping the thread on the ends of the rod though?