AP Physics 1: Open and Closed Pipe Resonance

The resonant hum of an organ pipe and the crisp note from a flute are not just sounds but physics in action. Understanding standing waves in air columns is crucial for AP Physics 1 because it connects fundamental wave principles to real-world acoustic systems, from musical instruments to architectural acoustics. This topic requires you to master how boundary conditions—whether an end is open or closed—dictate the possible standing wave patterns, wavelengths, and consequently, the series of notes an instrument can produce.

Fundamental Concepts: Reflection and Boundary Conditions

To grasp pipe resonance, you must first understand how a sound wave behaves at the end of a tube. A sound wave is a longitudinal pressure wave. When it hits a boundary, it reflects. The nature of that reflection depends on the boundary condition.

At a closed end, the air molecules cannot oscillate; this point becomes a displacement node (and a pressure antinode). Think of it like a fixed end on a string: the wave inverts upon reflection. At an open end, the air molecules are free to move maximally, creating a displacement antinode (and a pressure node). This is analogous to the free end of a string; the wave reflects without inversion. These conditions are the absolute rules that generate all possible standing wave patterns inside the tube.

Standing Wave Patterns in Pipes

A standing wave is the stable pattern formed by the interference of two identical waves traveling in opposite directions. In pipes, we visualize these patterns by drawing the displacement of air molecules from their equilibrium positions. The allowed patterns are those that fit the boundary conditions at both ends of the pipe.

For an open-open pipe (both ends open), both ends must be displacement antinodes. The simplest pattern, the fundamental frequency or first harmonic, has a single node in the center. For a pipe of length L, this pattern is exactly half a wavelength long, so $L=\frac{1}{2}\lambda$. The next possible pattern adds two more nodes, fitting a full wavelength into the pipe ($L=\lambda$). This pattern continues, allowing any integer number of half-wavelengths to fit: $L=n\frac{{\lambda }_n}{2}$, where $n=1,2,3,...$.

For an open-closed pipe (one end open, one end closed), the open end is an antinode and the closed end is a node. The fundamental pattern has a node at the closed end and an antinode at the open end. This is only one-quarter of a wavelength: $L=\frac{1}{4}\lambda$. The next possible pattern adds a node and an antinode, fitting three-quarters of a wavelength ($L=\frac{3}{4}\lambda$). This constraint means only odd numbers of quarter-wavelengths are allowed: $L=n\frac{{\lambda }_n}{4}$, where $n=1,3,5,...$ (odd integers only).

Determining Resonant Frequencies and Wavelengths

The resonant frequency is the frequency at which a standing wave forms. It is determined by the wavelength of the standing wave and the speed of sound v in the air column, using the universal wave equation $v=f\lambda$.

For an open-open pipe:

1. Start with the wavelength condition: $L=n\frac{{\lambda }_n}{2}$.
2. Solve for ${\lambda }_n$: ${\lambda }_n=\frac{2L}{n}$, where $n=1,2,3,...$
3. Find the frequency: $f_n=\frac{v}{{\lambda }_n}=n\frac{v}{2L}$.

This equation shows all integer harmonics are present. The frequencies are $f_1$, $2f_1$, $3f_1$, $4f_1$, etc.

For an open-closed pipe:

1. Start with the wavelength condition: $L=n\frac{{\lambda }_n}{4}$, where $n=1,3,5,...$
2. Solve for ${\lambda }_n$: ${\lambda }_n=\frac{4L}{n}$.
3. Find the frequency: $f_n=\frac{v}{{\lambda }_n}=n\frac{v}{4L}$.

Critically, n is only odd. This means the resonant frequencies are $f_1$, $3f_1$, $5f_1$, $7f_1$, etc. The even-numbered harmonics ($2f_1$, $4f_1$, ...) are missing.

Example Calculation: A pipe of length 0.5 m is used with a speed of sound of 340 m/s.

• As an open-open pipe, its fundamental frequency is $f_1=\frac{340}{2(0.5)}=340$ Hz. Its next resonance (first overtone) is at $f_2=2\times 340=680$ Hz.
• As an open-closed pipe, its fundamental is $f_1=\frac{340}{4(0.5)}=170$ Hz. Its next resonance is at $f_3=3\times 170=510$ Hz, not 340 Hz.

Why the Harmonic Series Differs

The physical reason open-closed pipes produce only odd harmonics stems directly from the asymmetric boundary conditions. An open end must be an antinode (A) and a closed end a node (N). The simplest pattern is N-A, which is 1/4 of a wave. To create the next possible standing wave, you must add a node and an antinode as a pair (N-A-N-A), which always adds half a wavelength. Adding half a wavelength to a quarter wavelength gives three-quarters. This process only ever yields odd multiples of the initial quarter-wavelength. In contrast, open-open pipes have symmetric boundaries (A-A). Adding a node-antinode pair (A-N-A) places you at half a wavelength for the fundamental. The next pattern adds another half wavelength, leading to all integer multiples.

This difference is audible. An open-closed pipe (like a clarinet) has a "hollower" or "darker" timbre because its sound lacks the even harmonics that open-open pipes (like a flute) possess.

Common Pitfalls

1. Confusing displacement and pressure nodes/antinodes. At a closed end, displacement is a node but pressure variation is an antinode (it's maximum). At an open end, displacement is an antinode but pressure is a node (atmospheric pressure). When drawing diagrams for AP Physics, you are almost always drawing the displacement of air molecules. Remember: Closed End = Displacement Node; Open End = Displacement Antinode.
2. Mislabeling harmonics for open-closed pipes. The most common error is calling the 170 Hz tone in our example the "first harmonic" and the 510 Hz tone the "second harmonic." This is incorrect. They are the first and third harmonics, respectively. The harmonic number n is always the integer in the frequency formula $f_n=n\frac{v}{4L}$. For open-closed pipes, n can only be odd. The 510 Hz tone corresponds to $n=3$.
3. Using the wrong length in calculations. For a pipe that is open at both ends, L is simply the physical length. However, the effective acoustic length is slightly longer due to the end correction (the antinode forms just beyond the physical open end). For most AP problems, you use the given physical length unless told otherwise. Just be conceptually aware that the actual resonant frequencies are slightly lower than those calculated with L.
4. Applying the open-open formula to an open-closed pipe. Always identify the boundary conditions first. If one end is closed, you must use the open-closed wavelength condition ($L=n\lambda /4$) and remember the odd-integer restriction. Automatically using $L=n\lambda /2$ will yield incorrect wavelengths and frequencies.

Summary

• Boundary conditions rule: A closed end is a displacement node; an open end is a displacement antinode. These conditions determine all possible standing waves.
• Open-open pipes support standing waves where $L=n({\lambda }_n/2)$, producing all integer harmonics: $f_n=n(v/2L)$ for $n=1,2,3,...$
• Open-closed pipes support standing waves where $L=n({\lambda }_n/4)$ for odd n only, producing only odd harmonics: $f_n=n(v/4L)$ for $n=1,3,5,...$
• The absence of even harmonics in open-closed pipes is a direct result of the asymmetric node/antinode requirement at the boundaries.
• Always use the wave equation $v=f\lambda$ to connect the calculated wavelength to the resonant frequency.
• Avoid common mistakes by carefully labeling harmonics based on the integer n in the frequency equation, not the order in which you hear the overtones.