Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm sure you have tried sometime to make a sound by blowing in an empty bottle. Of course, the tone/frequency of the sound modifies if the bottle changes its shape, volume, etc.

I am interested in understanding how the tone/frequency of this sound can be (approximately) calculated from the shape of the bottle. Is there some formula? If you have any references about this please post them in your answers.

Being a mathematician I keep hearing that the principal tone is given by the first eigenvalue of the Laplace operator, and I'd like to understand the physical explanation.

share|cite|improve this question
up vote 4 down vote accepted

Yes. This is called Helmholtz resonance or cavity resonance which is an important application in acoustics. You could find larger sets of sample frequencies recorded in coke bottles at Hyperphysics...

In my basic understanding: The air in the bottle exhibits a single resonant frequency. When additional volume of air is blown into the closed cavity, the air will overflow out causing the pressure to decrease inside the bottle. Due to the newly produced low pressure, air outside rushes in. Thus, the air will oscillate into and out of the container for a few cycles at some natural frequency.

                    Cavity Resonance

Thus, The frequency of sound in such a closed bottle is determined to be $$f_c=\frac{v}{2\pi}\sqrt{\frac{A}{VL}}$$

Edit for symmetry and water add-in: The above formula could be used for an air cavity (as per your question). But in case of a water-filled bottle (If you require an appropriate answer), the experiment becomes somewhat complicated because we've to take the properties of sound in water into account. For example, $v_{air}$ is barely 340 m/s whereas $v_{water}$ is as high as 1484 m/s. It's easy to do if we write down some frequencies using different volumes of water (using the same bottle), and concluding a general relation between frequency & volume. This could be easily achieved through a graph...

                    Graph plot

This is the $f$ vs. $1/\sqrt{V}$ graph plotted by a bunch of good guys (for a 0.6 liter coke bottle). This curve gives the equation of a straight line which says that $$f=\frac{5184.93}{\sqrt{V}}-30.4$$

share|cite|improve this answer
Thank you for your answer. It seems to me though that the sound made by a bottle half filled with water (for example) changes if I do not keep the bottle vertically, while obviously none of the terms of the above formula changes. It is possible that different sounds have the same frequency? – Beni Bogosel Nov 19 '12 at 20:23
@BeniBogosel: Hello Beni, I think I can't understand your comment. $V$ is the volume of cavity which decreases as you fill water.. A water-filled bottle has less volume than a normal bottle cavity. Hence you get a higher frequency of sound. Is it what you're asking? – Waffle's Crazy Peanut Nov 20 '12 at 1:33
You didn't understand what I said. :) At the start the bottle is half filled with water. If I blow while the bottle is in a vertical position then I get one sound. If I hold the bottle at an angle of 45 degrees and blow, then the sound is slightly different. Is it possible that the formula holds only for rotational symmetric bottles? – Beni Bogosel Nov 20 '12 at 11:11
@BeniBogosel: Hello Beni. No, The formula holds good for all (clarified further for symmetry and water add-in). The pathway for air (neck of bottle) makes the air volume inside it to oscillate (jump in & out). As you incline your bottle, the frequency remains the same. Or else, a sub-woofer (Helmholtz resonator) won't give the same sound when you change its orientation. I haven't heard of something like that ("inclining changes the frequency"). – Waffle's Crazy Peanut Nov 20 '12 at 14:00
@BeniBogosel I think that the phenomenon you describe happens because the different angulation causes a different area of the opening port to be exposed to your blowing. – usumdelphini Jul 9 '14 at 15:20

The bottle behaves as a closed cylinder, so the distance between the opening of the bottle and it's base, $d$, is simply related to the wavelength of the fundamental sound by $\lambda = d/4$.

The shape of the bottle affects the tone because it causes energy to be transferred into harmonics of the fundamental frequency. However I don't know of any easy way to calculate the relative intensities of the harmonics from the shape of the bottle.

share|cite|improve this answer
Suggestion to the answer (v1): Change the term closed cylinder to cylinder with one open and one closed end. – Qmechanic Nov 19 '12 at 15:16
It's a good point about Helmholtz resonance, as suggested by Bjorn and Crazy Buddy. I'd guess that for most bottles it's just the length of the bottle that matters because you set up the usual closed pipe standing wave. However if the bottle is a sphere there won't be any strong reflection of the sound from the bottom of the bottle and you'll need to analyse the problem using Helmholtz resonance. It would be interesting to take a selection of bottles and see which effect is most important. Sadly I'm unlikely to have the time to do this. – John Rennie Nov 20 '12 at 10:13
""However if the bottle is a sphere there won't be any strong reflection of the sound from the bottom of the bottle and you'll need to analyse the problem using Helmholtz resonance. "" No. For a typical Bottle the resonant frequency is so low, that Helmholz always is the method of choice. The form is not really important. In case You have a cylindric tube with a round (semispherical) bottom, you have a pipe organ and the resonance is due to length then. – Georg Jan 22 '15 at 18:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.