# 10.8.3 Pipe Resonance

In the set-up below, the speaker is emitting a monotone. The length of a closed pipe is adjusted by sliding the extension in or out of the pipe. When the pipe is at some “special” lengths, the pipe becomes “possessed” and hums loudly in chorus with the speaker. Is this black magic or what?

Let’s try to visualize what’s happening in the pipe. The speaker is emitting a sound wave, a fraction of which is transmitted into the pipe. This sound wave is now kind of trapped in the pipe, undergoing repeated reflections at both ends of the pipe[1].

Since the speaker is transmitting sound wave into the pipe continuously, it means that the pipe is accumulating incident and reflected sound waves. And all these waves superpose with one another. At most frequencies, the interference is mostly destructive, and the resultant wave in the pipe has negligible amplitude. At resonant frequencies, however, constructive interference occurs. The resultant wave is a standing wave whose amplitude is the sum of the amplitude of all the incident and reflected sound waves in the pipe. The pipe is in resonance.

The H2 syllabus only requires you to work out the resonant frequencies of a given pipe by drawing. And the guiding principle to our drawings is that

a closed end must be a displacement node/pressure antinode, and

an open end must be a displacement antinode/pressure node.

Open Pipe

An open pipe is a pipe that is open at both ends. So a displacement antinode must be formed at both ends. To solve for the resonant frequencies, we only have to visualize the wavelengths that will “fit” the pipe and have displacement antinodes at both ends.

With some thought, we realize that in order to have two antinodes (A and A) at both ends, we can do ANA, then ANANA, then ANANANA and so on. Basically, we start out with one single “Χ” segment, and keep squeezing in one more “Χ” segment to progress to the next harmonic.

Since each “Χ” segment corresponds to half-a-wavelength, we are basically fitting integer number of half-wavelengths into the pipe. So

$\displaystyle n\frac{{{{\lambda }_{n}}}}{2}=L,\text{ }n=1,2,3,...\text{ }$

This means that the resonant wavelengths are

$\displaystyle {{\lambda }_{n}}=\frac{{2L}}{n},\text{ }n=1,2,3,...\text{ }$

Since $v=f\lambda$, the resonant frequencies are

$\displaystyle {{f}_{n}}=n\frac{v}{{2L}},\text{ }n=1,2,3,...\text{ }$

The first 5 harmonics of an open pipe of length L are tabulated below. Notice how the wavelengths of the overtones are $\displaystyle \frac{1}{2},\text{ }\frac{1}{3},\text{ }\frac{1}{4},\text{ }\frac{1}{5}...$ of the fundamental wavelength, and the frequencies of the overtones are 2, 3, 4, 5,… times the fundamental frequency.

Closed Pipe

First of all, let me clarify that a closed pipe does not refer to a pipe with two closed ends. Instead, a closed pipe refers to a pipe that is closed at one end and open at the other end. A closed pipe must have a displacement antinode at the open end and a displacement node at the closed end.

In order to have a node and an antinode (N and A) at either end, we can do a NA, then NANA, then NANANA and so on. Basically, we start out with one single “<” segment, and keep squeezing in one more “>”  and one more “<” segment to progress to the next harmonic.

Since each “<” or “>” segment corresponds to a quarter-wavelength, we are basically fitting odd number of quarter-wavelengths into the pipe. So

$\displaystyle n\frac{{{{\lambda }_{n}}}}{4}=L,\text{ }n=1,3,5,7...\text{ }$

This means that the resonant wavelengths are

$\displaystyle {{\lambda }_{n}}=\frac{{4L}}{n},\text{ }n=1,3,5,7...\text{ }$

Since $v=f\lambda$, the resonant frequencies are

$\displaystyle {{f}_{n}}=n\frac{v}{{4L}},\text{ }n=1,3,5,7...\text{ }$

The first 5 harmonics of a closed pipe of length L are tabulated below. Notice how the wavelengths of the overtones are $\displaystyle \frac{1}{3},\text{ }\frac{1}{5},\text{ }\frac{1}{7},\text{ }\frac{1}{9}...$ of the fundamental wavelength, and the frequencies of the overtones are 3, 5, 7, 9,… times the fundamental frequency.

Demonstration

Applet

Walter Fendt (Standing Longitudinal Waves)

Concept Test

2063

Interesting

Ruben’s Tube

Kundt’s Tube

Water Standing Waves

[1] You may be wondering why a sound wave is reflected even by an open end? If you’re interested, the appendix provides some discussion.