Stationary Waves in Organ Pipes: MDCAT Physics notes

Standing waves in air columns

When sound waves are reflected at the ends of a pipe, the incident and reflected waves superpose to form stationary waves. The boundary conditions are:

• A closed end is always a node (air cannot move there).
• An open end is always an antinode (air moves freely).

The distance between neighbouring nodes (or antinodes) is $\lambda/2$; from a node to the next antinode is $\lambda/4$.

Pipe open at both ends

There are antinodes at both ends. The simplest mode has one node in the middle, so $L = \lambda_1/2$.

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

All harmonics are present: $f_1 : f_2 : f_3 = 1 : 2 : 3$.

Pipe closed at one end

There is a node at the closed end and an antinode at the open end. In the simplest mode $L = \lambda_1/4$.

$$f_n = \frac{n v}{4L}, \qquad n = 1, 3, 5, \ldots$$

Only odd harmonics are present: $f_1 : f_3 : f_5 = 1 : 3 : 5$.

Property	Open pipe	Closed pipe
Ends	Antinode, antinode	Node, antinode
Fundamental $f_1$	$v/2L$	$v/4L$
Fundamental wavelength	$2L$	$4L$
Harmonics present	All	Odd only
Gap between successive resonances	$v/2L = f_1$	$v/2L = 2f_1$

Closing an open pipe

For the same length, the closed pipe's fundamental is half that of the open pipe: $f_c = v/4L = \tfrac12 (v/2L) = \tfrac12 f_o$, so $f_o = 2f_c$. Closing one end lowers the pitch by an octave.

Identifying a pipe from its resonances

1. Find the gap $\Delta f$ between successive resonances.
2. Divide each frequency by the smallest possible fundamental. If they are consecutive whole numbers, the pipe is open with $f_1 = \Delta f$. If they are consecutive odd numbers, the pipe is closed with $f_1 = \Delta f/2$.
3. Find $L$ from $f_1 = v/2L$ (open) or $f_1 = v/4L$ (closed).

Worked example. Successive resonances are 300, 500 and 700 Hz, with $v = 340\ \mathrm{m\,s^{-1}}$. The gap is 200 Hz. Dividing by 100 Hz gives 3, 5, 7 (odd), so the pipe is closed with $f_1 = 100$ Hz. Then $L = v/4f_1 = 340/400 = 0.85$ m.

Worked example. Resonances at 400, 600, 800 Hz: gap 200 Hz, ratios 2, 3, 4, so an open pipe with $f_1 = 200$ Hz and $L = 340/400 = 0.85$ m.

Key formulas

• Open: $f_n = nv/2L$, $n = 1,2,3,\ldots$
• Closed: $f_n = nv/4L$, $n = 1,3,5,\ldots$
• $v = f\lambda$

Common MDCAT traps

• For a closed pipe the resonance gap is twice the fundamental; do not take the gap as $f_1$.
• Even harmonics never appear in a closed pipe.
• Covering one end halves the fundamental, so $2f_c = f_o$, not $f_c = 2f_o$.
• The closed end is a node of displacement, not an antinode.

Quick revision

• Open end: antinode; closed end: node.
• Open pipe: all harmonics, $f_1 = v/2L$.
• Closed pipe: odd harmonics, $f_1 = v/4L$.
• Same length: open fundamental = 2 × closed fundamental.
• Successive resonances in either pipe differ by $v/2L$.