Sound in a narrow pipe can form standing waves when reflections at the ends reinforce particular patterns. The allowed resonances are determined by the boundary conditions at each end.

For a longitudinal sound wave, it is useful to distinguish two oscillating quantities:

• the air-particle displacement or velocity;
• the acoustic pressure variation relative to ambient pressure.

Where particle displacement has a node, pressure has an antinode, and vice versa.

Boundary conditions at pipe ends

At an ideal closed end, the air cannot move through the wall, so particle displacement has a node there. The pressure variation can be large, so the closed end is a pressure antinode.

At an ideal open end, the pressure must remain close to atmospheric pressure, so the acoustic pressure variation has a node there. Particle displacement has an antinode.

These complementary conditions determine which wavelengths fit the pipe.

Pipe open at both ends

For a pipe of length $L$ open at both ends, displacement antinodes occur at both ends. The allowed standing waves satisfy

$$L=n\frac{\lambda_n}{2},\qquad n=1,2,3,\ldots$$

so

$$\lambda_n=\frac{2L}{n}.$$

Using $f=v/\lambda$,

$$\boxed{f_n=n\frac{v}{2L}}.$$

The fundamental frequency is

$$f_1=\frac{v}{2L},$$

and all integer harmonics are allowed:

$$f_1,\ 2f_1,\ 3f_1,\ldots$$

An ideal flute-like air column is approximately described by this family.

Pipe closed at one end and open at the other

Now let one end be closed and the other open. The closed end must be a displacement node and the open end a displacement antinode. The shortest pattern fitting those conditions contains one quarter of a wavelength:

$$L=\frac{\lambda_1}{4}.$$

More generally,

$$L=(2n-1)\frac{\lambda_n}{4},\qquad n=1,2,3,\ldots$$

so

$$\boxed{f_n=(2n-1)\frac{v}{4L}}.$$

Only the odd members of the harmonic series occur in the ideal one-dimensional model:

$$f_1,\ 3f_1,\ 5f_1,\ldots$$

with

$$f_1=\frac{v}{4L}.$$

A clarinet-like air column is often approximated by this kind of boundary condition.

Worked example

Take the speed of sound to be

$$v=343,\mathrm{m/s}$$

and a pipe of length

$$L=0.50,\mathrm m.$$

If the pipe is open at both ends,

$$f_1=\frac{343}{2(0.50)}=343,\mathrm{Hz}.$$

Its next two resonances are

$$686,\mathrm{Hz}$$

and

$$1029,\mathrm{Hz}.$$

If the same pipe is closed at one end,

$$f_1=\frac{343}{4(0.50)}\approx171.5,\mathrm{Hz}.$$

The next resonances are

$$3f_1\approx514.5,\mathrm{Hz}$$

and

$$5f_1\approx857.5,\mathrm{Hz}.$$

Changing only the boundary condition therefore changes both the fundamental frequency and the harmonic sequence.

Effective length and real instruments

The simple formulas treat pressure nodes and displacement antinodes as if they occur exactly at an open geometric end. In a real pipe, the oscillating air extends slightly outside the opening, producing an end correction and an effective acoustic length somewhat greater than the physical length.

Real wind instruments also contain flared bells, tone holes, mouthpieces, reeds, and nonuniform bores. These modify the ideal spectrum, but the underlying principle remains: the resonant frequencies are the normal modes allowed by the acoustic boundary conditions.

Opening or closing holes changes the effective length of the resonating air column, which is why wind instruments can change pitch without changing the speed of sound itself.