Stationary waves in air columns

The two ends a pipe can have

A closed end is a displacement node and an open end is a displacement antinode. Blow across the top of a bottle, or into an organ pipe, and the note you hear is a stationary wave in the column of air inside. A sound wave travels down the tube, reflects at the far end, and superposes with the wave still arriving. The difference from a string is in the boundary conditions, because a pipe can have one end of each kind.

The air touching a solid wall cannot move back and forth along the tube, which is why a closed end is a node. The air at an open end is free, and moves further than anywhere else in the pipe, which is why an open end is an antinode. Those are displacement nodes and antinodes, which is how pipe diagrams are drawn; the pressure variation is the inverse, largest at the closed end and smallest at the open end.

The spacings are unchanged from the string: node to node or antinode to antinode is λ/2$\lambda/2$λ/2, and node to the nearest antinode is λ/4$\lambda/4$λ/4. Name the two ends first, then count quarters and halves between them.

Take a tube closed at one end and open at the other. It needs a node at one end and an antinode at the other, and the shortest pattern that manages that is a single quarter of a wavelength. So the first harmonic of a closed tube of length L has

which means the pipe sounds a wavelength four times its own length, twice as long as the longest wavelength a string of the same length can carry.

Describe the drawing

One tube closed at the left and open at the right, sounding its first harmonic, drawn twice on two baselines. The upper cyan pattern is the displacement of the air: it is pinned to zero at the closed end, which is a node marked N, and swings furthest at the open end, an antinode marked A, so a quarter of a wavelength fits the tube. The lower amber pattern is the pressure, which is the gradient of the displacement reversed and so is the exact inverse picture: full swing at the closed end and nothing at all at the open end, where the air is free to move and the pressure stays atmospheric. Dashed lines join each end of the tube to the same end below. Pipe diagrams are drawn in displacement by convention, which is why the end that looks quiet is the end where the pressure varies most.

Why a closed tube sounds only the odd harmonics

The next pattern that fits must still start at a node and finish at an antinode, so it adds half a wavelength: L = 3λ/4$L = 3\lambda/4$L = 3λ/4, then 5λ/4$5\lambda/4$5λ/4, and onwards. Every length the tube allows is an odd number of quarter wavelengths, so with n odd,

and the frequencies climb f₁$f_1$f_1, 3f₁$3f_1$3f_1, 5f₁$5f_1$5f_1, with the even harmonics missing altogether. The reason is the mismatch between the ends. An even number of quarter wavelengths is a whole number of half wavelengths. Half a wavelength always runs node to node or antinode to antinode, so fitting one would need the same kind of end at both ends of the tube. A closed tube has one of each and can never fit one. The gap is audible: it gives a stopped organ pipe and a clarinet their hollow tone.

A tube open at both ends needs an antinode at each end instead. Antinode to antinode is half a wavelength, so its first harmonic has λ₁ = 2L$\lambda_1 = 2L$λ_1 = 2L and every whole number of half wavelengths fits after that, λ_n = 2L/n$\lambda_n = 2L/n$λ_n = 2L/n with n taking every whole value. An open tube therefore supports all integer harmonics, f₁$f_1$f_1, 2f₁$2f_1$2f_1, 3f₁$3f_1$3f_1, the same series as the stretched string but built between two antinodes rather than two nodes. Compare the two pipes at equal length and the closed one starts from the longer wavelength, 4L against 2L, so it sounds an octave lower and on half as many harmonics.

Describe the drawing

Two air columns of the same length side by side, each drawn twice. The tube closed at one end has a displacement node at the closed end and an antinode at the open end, and holds one quarter wavelength in its first harmonic and three in its third; no even harmonic fits. The tube open at both ends has an antinode at each end and holds one half wavelength in its first harmonic and two in its second, so every harmonic sounds. Brackets along each tube mark the quarter and half wavelengths.

Measuring the speed of sound in a resonance tube

The closed pipe gives a way to measure the speed of sound with nothing more than a tuning fork and a metre rule. Stand a long tube upright with its lower end in a tall cylinder of water, so the water surface closes the bottom while the top stays open. Raising or lowering the tube changes the length of the trapped air column, which makes it a closed pipe you can tune.

Strike a tuning fork of known frequency f and hold it flat, a centimetre or so above the open top, taking care that it never touches the glass. Start with the air column short and lengthen it slowly. At one length the sound swells suddenly, because the column's first harmonic has come into step with the fork and the column is resonating. Record that length L₁$L_1$L_1, measured from the top of the tube down to the water surface. Keep going and a second, quieter resonance arrives at L₂$L_2$L_2, where the same column is running in its third harmonic.

Describe the drawing

A tube standing in water, drawn twice. At the first resonance the air column holds a quarter of a wavelength, with a displacement node at the water surface and an antinode at the open rim. At the second resonance the water has been lowered until the column holds three quarters of a wavelength, with a second node halfway up. The two column lengths therefore differ by half a wavelength.

Those two lengths are a quarter and three quarters of the same wavelength, so their difference is half a wavelength, λ = 2(L₂ − L₁)$\lambda = 2(L_2 - L_1)$λ = 2(L_2 - L_1), and the speed follows.

The second reading is there so that you can take the difference and need not rely on L₁$L_1$L_1 alone. The antinode actually forms slightly above the open end rather than exactly at it, so every length you measure is short by the same small amount, and subtracting one reading from the other cancels it out. CIE tells you to treat that offset as negligible, and this method makes that safe.

Two habits sharpen the result. Approach each resonance from both directions, once lengthening the column and once shortening it, and average the pair, because the loudness peak is broad and easy to overshoot. Then swap the fork for others of different frequency and repeat. Since L₁$L_1$L_1 is close to v/4f$v/4f$v/4f, a graph of L₁$L_1$L_1 against 1/f$1/f$1/f comes out straight with gradient v/4. Its small negative intercept is the offset you have just been cancelling.