How Ocarinas Work (Part 1)

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Ocarina Physics

Dear Ocarinist,

On this page and in part 2, I strive to thoroughly explain how ocarinas work right down to their fundamental physics.

Ocarinas have never been well explained by anyone in the ocarina community, while questions related to ocarina physics come up time and time again. I’ve seen many people express the wish that someone with expertise in physics wrote about this topic and settle it once and for all.

Being both a physicist and an ocarinist, I’ve decided to do this here and leave nothing out. This page is meant for laymen as much as for people savvy in physics, so that everybody gets something out of it.


Questions I answer in this article

  • How does the physics of an ocarina work?
  • Why do the finger holes have different sizes?
  • Why are there large and small ocarinas?
  • When and why does an ocarina change its pitch?
  • When and why does the loudness of an ocarina change?



Below, in section 1, I give you my derivation of the ocarina equation.
In section 2, at the end of this page, I quickly discuss the solution of the ocarina equation.
In section 3, in part 2 of this article, I explain what we can learn from it. Among them are some things I have never seen anyone discuss before.


Ocarina Physics

Section 1 – The Ocarina Equation

Whether you are savvy in physics or not, you should be able to get something out of this. I try to explain it in such a way that you can follow along, even if the equations mean nothing to you. If not, make sure you read part 3 one page 2 of this article, where I discuss what we learn from the results.



Air Column

Imagine an ocarina without holes except for the aperture, or mouth hole.

If we blow into the windway, the air column within it will start to move inwards. By blowing air inside, we increase the pressure in the chamber. This will push the air back out again, until the pressure inside is back normal. However, the air needs a moment to react to changes, so it won’t stop immediately and keeps on moving, until more air has left the chamber than we blew in.

Since the pressure outside is now larger than inside, some air is pushed back into the chamber, but it will overshoot again. This motion continues to go back and forth – it is an oscillation in the air pressure, which is exactly how sound is defined.

How fast the air column in the windway is moving depends on the ocarina. If it moves fast, we have a high frequency or high pitch. If it moves slowly, then the pitch is low.

Let’s call the displacement of the column at any point in time x(t). We need to determine this function, because if we know how the column moves, we know what sound an ocarina makes.


The Derivation

Ocarina Sketch

The air inside the ocarina occupies the volume V at a pressure p, which depends on time. The length of the windway is l, and the area of the hole is A.

Notice how the available volume inside (at the bottom of the animation) is changing due to the motion of the column. In general, we have

(1)   \begin{equation*} V=V_0 + A\cdot x(t) \end{equation*}

where the inner chamber has the constant volume V_0, and the column gives or occupies the volume A\cdot x(t) as it moves back and forth. (Note: volume = area times height)

Next, we need a law that governs the gas itself. More precisely, we need a law that governs how the volume of the gas is connected to its pressure – this is called an equation of state. For that, we need to recognize what kind of process is happening here.

Since we are describing sound, we know that one vibration of the air column takes only a fraction of a second. We can assume the process is too fast for heat to be exchanged between the air particles. Any thermodynamical process where no heat is exchanged is called adiabatic. An adiabatic process is governed by the following equation of state:

(2)   \begin{equation*} p\cdot V^\kappa = const. \end{equation*}

where \kappa is just a number, depending on what kind of gas we have. For dry air at 20°C or 68°F, we have \kappa=1.402

Now comes the technical part where I have to work with this equation. What I’ll do is reformulate this into an equation that describes the motion of the column – meaning it gives us x(t). Only the mathematically adept people will be able to follow this part, as I don’t want to simplify and thus take away from it. Anyone else, just don’t worry about it or scroll down :)

Take the time derivative of equation (2) to get rid of the constant on the right hand side. Then divide by V^\kappa to get

    \[ \dot{p} + \frac{p \kappa}{V} \dot{V} = 0 \]

Note: In physics, time derivatives are written as dots on top of the symbols.
Replacing \dot{V} by A\cdot\dot{x}(t) from (1) gives

    \[ \dot{p}+ \frac{p \kappa}{V} A\cdot\dot{x}(t) = 0 \]

This equation isn’t linear and its solution can’t be expressed as a function – it can only be solved by a computer algorithm. So as we always do in physics, we make a simplification to linearize this equation.

Instead of p and V in the second term, we  take only the pressure and volume p_0 and V_0 at rest. In other words, we neglect the small changes due to the motion of the column. Simplifying it this way doesn’t even produce an error in the end result. I’ve tested both.

So, we now have our linearized equation

    \[ \dot{p}+ \frac{p_0 \kappa}{V_0} A\cdot\dot{x}(t) = 0 \]

which can be integrated over time to find the relationship between pressure and air column position

(3)   \begin{equation*} p + \frac{p_0 \kappa}{V_0} A\cdot x(t) = 0 \end{equation*}

We want an equation that just has x(t) in it, so we need to get rid of the pressure entirely. We can do this by expressing it in terms of the force producing it.
Pressure is defined as a force F acting on a surface, meaning p = F / A.

Force is defined as mass m times the acceleration a of that mass. Acceleration is the second time derivative of the position x(t)

    \[F = m a = m \ddot{x}(t)\]

And the mass of the air column is the air density \rho times the volume. The volume of the whole column is l\cdot A. Putting all of this together gives

(4)   \begin{equation*} p = \rho l \ddot{x}(t) \end{equation*}

where A cancels. Finally, we can put (4) in for the pressure in (3) to get

(5)   \begin{equation*} \ddot{x}(t)+ \frac{p_0 \kappa A}{V_0 \rho l} \cdot x(t) = 0 \end{equation*}

This is the ocarina equation.


Section 2 – The Solution Of The Ocarina Equation

Anyone having done undergraduate physics will recognize the ocarina equation as the equation of an harmonic oscillator. An harmonic oscillator is a system that, if displaced, experiences a force trying to bring it back to its original rest position. Such a system could be a log bobbing on water, a tree branch waving in the wind, a pendulum, or the air column in an ocarina :)

Solving this type of differential equation is straightforward and the first thing a physics student learns at university, but I won’t bother you with it. Instead, I’ll give you the solution and rewrite the ocarina equation in a simple form

(6)   \begin{equation*} \ddot{x}(t) + \omega^2 x(t) = 0 \end{equation*}

where \omega is the angular frequency at which the column oscillates.

The solution of the above equation is the function x(t). It is

(7)   \begin{equation*} x(t) = \frac{v}{\omega}\cdot sin(\omega t) \end{equation*}

where v is the velocity of the air going in – that is how hard you are blowing. The factor v/\omega in front of the sine function is the amplitude (loudness) of the ocarina.

Normal frequency f and angular frequency are connected in this way: \omega = 2 \pi f

The frequency is what we wanted all along, because it is the pitch of the ocarina. Comparing (5) and (6) we find

    \[ f = \frac{1}{2 \pi} \sqrt{\frac{p_0 \kappa A}{V_0 \rho l}} \]

With this we have determined both the pitch and loudness of the ocarina.


Move on to part 2 of this article to find out what all of this means and what we learn from it about ocarinas.

Don’t miss it,



Ocarina Physics

6 Responses to “How Ocarinas Work (Part 1)”

  1. Tanguay Desgagné says:

    Good morning Allen !

    Great topic !

    I am greatly interested to read in whole this excellent article “How Ocarinas Work”, as I think there are huge analogies with how labial (or pucker) whistling resonance works (I am a whistler), eventhought their basic sound generating “devices” (ocarina mouth hole versus lips) and principles are very different.

    HOWEVER, unfortunately, I cannot read any equations on this web page
    Nothing appears, only blanks !

    So, could you please help me to see the full article with equations? Have I to do something to have access to these equation ?

    Also, I tried to send my inquiry to your “contact” tab, but they ask to put a code… but they don’t show the code we have to put in !

    Thank you very much for your attention,


    Tanguay Desgagne

    • Allen says:

      Hi Tanguay,

      That’s very strange. I’ve tested it with several PCs and in several browsers. Since you are also unable to use the contact form, you may have a script blocker installed. Perhaps it works if you try a different browser :)


  2. Matthew says:

    Hi Allen,

    I’m a high school student who is currently taking AP Physics. Everyone has to do an end-of-year physics project. I’m making an ocarina, and this article is really helpful and interesting. I’m going to look for the area of the frequency relationship experimentally.


  3. Megan says:

    Thank you so much for writing this! I’ve been trying to research the underlying physics of ocarinas and was frustratingly coming up empty handed. I’ve onl taken a little physics but I was able to follow all (or most :)) of the work which was wonderful! Thanks again!!!

  4. Andrew says:

    Hi Allen,

    where did the v come from? The units are right, but the solution of the harmonic oscillator in (6) should really be just sin(w t); of course this is checked by taking two derivatives and obtaining -w^2 sin(w t).

    • Allen says:

      Hi Andrew, I’m jumping straight to the solution of equation (6) to make it shorter, which is why the v appears without explanation. To solve a differential equation you need to add up all of its partial solutions. Equation (6) has two partial solutions: sin(wt) and cos(wt), but they can also have constants in front. The general solution is therefore: x(t) = A*sin(wt) + B*cos(wt), with constants A and B.

      This encompasses ALL possible solutions for eq. (6). To get the solution we are interested in, you need to enter the initial conditions of our particular situation: x(0) = 0 and x'(0) = v
      In words: at the beginning t=0, the air column is in position 0 and starts moving at speed v, since you start blowing into it.

      x(t) = A*sin(wt) + B*cos(wt)
      x'(t) = Aw*cos(wt) – Bw*sin(wt)

      Using the conditions gives:

      x(0) = A*sin(0) + B*cos(0) = B = 0
      x'(0) = Aw*cos(0) – Bw*sin(0) = A w = v, thus A=v/w

      Therefore: x(t) = v/w*sin(wt)

      I hope this helps :)

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