Stationary Waves in Air Columns

At a rigid closed end the air cannot move back and forth freely, so it is a displacement node. At an open end air motion is large, giving an approximate displacement antinode just outside the opening. The pressure pattern is opposite: a displacement node corresponds to a pressure antinode.

Use a labelled displacement diagram consistently. Drawing a node at an open end because atmospheric pressure is fixed mixes pressure and displacement. A stationary wave forms by superposition of incident and reflected waves of the same frequency. Energy is not carried along the column as in one progressive wave.

A tube open at both ends has displacement antinodes at both ends; its fundamental contains half a wavelength. A tube closed at one end has a node and an antinode, so its fundamental contains a quarter wavelength.

In the ideal closed–open tube only odd multiples of the fundamental occur. The next resonance is the third harmonic, not the second harmonic. Successive allowed resonant lengths at a fixed frequency differ by half a wavelength for either arrangement.

Use a tuning fork or suitable sound source over a tube whose air-column length can be varied, for example by changing a water level. Record two successive positions of strong resonance at the same frequency. Their separation gives λ/2, so v=2fΔL.

A real open-end antinode lies slightly outside the tube. Using a length difference largely cancels a constant end correction. Repeat resonance judgements, avoid excessive sound levels and measure a consistent reference point. Air temperature affects sound speed and should be recorded.

Try each problem before opening the worked solution. State the physical model, show substitutions with consistent units, and explain what your answer means.