Vibrations of Air Column

What this figure shows. A photograph or drawing of a clarinet is shown, a long slender wind instrument held vertically with a reed mouthpiece visible at its upper end and a flared bell opening at its lower end, its body dotted with the round finger-holes and metal keys players use to change pitch. The figure identifies the clarinet as a concrete, real-world instance of a closed organ pipe -- despite having finger holes along its side, its basic acoustic behaviour is that of a tube effectively closed at the reed end (where the vibrating reed nearly blocks air motion, forcing a node there) and open at the bell end, which is exactly the closed-pipe boundary condition the text uses to derive the pip …

What this figure shows. A tube of length L, closed at its lower end and open at its upper end, is drawn in its fundamental (lowest) mode of vibration, with a node (labelled N) marked exactly at the closed lower end and an antinode (labelled A) marked at the open upper end, and the smooth curve of the standing-wave displacement pattern drawn bulging outward from the tube's central axis to show that displacement is zero at the closed end and maximum at the open end. A brace alongside the tube marks its length L as exactly one quarter of the wavelength lambda-one, i.e. $L=\lambda_1/4$. The figure is the geometric basis for the closed-pipe fundamental-frequency formula $f_1=v/(4L)$ derived in the surrounding text: because the closed end must always be a node and the open end always an antinode, the shortest standing-wave pattern that can fit this boundary condition spans exactly one quart …

What this figure shows. The same closed tube of length L is now drawn in its second allowed mode of vibration, showing a total of two nodes (one still at the closed end, one partway along the tube) and two antinodes (one at the open end, one partway along), with the standing-wave curve now showing two bulging lobes along the tube's length instead of the single lobe of the fundamental mode. A brace marks the tube's length L as three-quarters of the wavelength lambda-two for this mode, $L=3\lambda_2/4$. The figure shows visually why the closed pipe skips straight from the fundamental to a mode three times its frequency (the first overtone, called the third harmonic since $f_2=3f_1$): the next standing-wave pattern that still satisfies node-at-closed-end, antinode-at-open-end must fit three quarter-wavelengths into the same fixed length L, not two, because a node-antinode bou …

What this figure shows. The closed tube of length L is drawn in its third allowed mode of vibration, now showing three nodes and three antinodes distributed along its length, with the standing-wave curve showing three bulging lobes. A brace marks the tube's length L as five quarter-wavelengths of this mode's wavelength lambda-three, $4L=5\lambda_3$. The figure continues the pattern established by the previous two figures, confirming that successive allowed closed-pipe modes require 1, 3, 5, ... quarter-wavelengths (always an odd count) to fit the fixed length L, which is exactly why a closed organ pipe's second overtone has a frequency five times the fundamental (the fifth harmonic, $f_3=5f_1$) and why closed pipes support only …

What this figure shows. A photograph or drawing of a flute is shown, a long slender wind instrument held horizontally, open at both ends (an embouchure hole near one end where the player blows across the opening, and an open far end), with a row of finger-holes and keys visible along its body. The figure identifies the flute as a concrete, real-world instance of an open organ pipe -- air is free to move at both ends of its tube -- which is exactly the open-pipe boundary condition the text uses to derive the pipe's fundamental frequency (twice that of an equal-length closed pipe) and its a …

What this figure shows. A tube of length L, open at both ends, is drawn in its fundamental mode of vibration, with antinodes (labelled A) marked at both the left and right open ends and a single node (labelled N) marked exactly at the midpoint of the tube, and the standing-wave curve drawn bulging outward at both ends and pinching to zero displacement in the middle. A brace marks the tube's length L as exactly half the wavelength lambda-one for this mode, $L=\lambda_1/2$. The figure is the geometric basis for the open-pipe fundamental-frequency formula $f_1=v/(2L)$ derived in the surrounding text: because both ends of an open pipe must be antinodes, the shortest standing-wave pattern that can satisfy this double boundary condition spans exactly half a wavelength, with a single node fo …

What this figure shows. The same open tube of length L is drawn in its second allowed mode, showing three antinodes (at both open ends and one additional antinode in the middle of the tube) and two nodes located symmetrically between them, with the standing-wave curve now showing two full bulging lobes along the tube. A brace marks the tube's length L as exactly one full wavelength of this mode, $L=\lambda_2$. The figure shows why, unlike the closed pipe, the open pipe's very next mode above the fundamental is simply twice the fundamental frequency (the first overtone, called the second harmonic since $f_2=2f_1$): with antinodes required at both ends, the tube's length can accommodate any integer number of half-wavelengths, so the sequence of allowed lengths increases by simple …

What this figure shows. The open tube of length L is drawn in its third allowed mode, showing four antinodes (including both open ends) and three nodes distributed between them, with the standing-wave curve now showing three full bulging lobes along the tube's length. A brace marks the tube's length L as three-halves of the wavelength lambda-three for this mode, $L=3\lambda_3/2$. The figure completes the open-pipe harmonic sequence shown across Figures 11.41-11.43: successive open-pipe modes require 1, 2, 3, ... half-wavelengths (every integer, none skipped) to fit the fixed length L, confirming that an open organ pipe supports every harmonic of its fundamental, in the ratio $f_1:f_2:f_3:\ldots=1:2:3:\ldots$, unlike th …

Worked out. A flute (an open pipe) sounds a fundamental note of 450 Hz, and the task is to find the frequencies of its second, third and fourth harmonics, and then to find the lowest three harmonics of a clarinet (a closed pipe) sounding at the same fundamental 450 Hz. Since an open pipe supports every harmonic in the ratio $1:2:3:4:\ldots$, the flute's harmonics are simply integer multiples of 450 Hz: $f_2=2\times450=900\ \text{Hz}$, $f_3=3\times450=1350\ \text{Hz}$, $f_4=4\times450=1800\ \text{Hz}$. Since a closed pipe instead supports only odd harmonics in the ratio $1:3:5:7:\ldots$, the clarinet's lowest three harmonics use odd multipliers instead of consecutive integers: its second (actual) harmonic present is $f_2=3\times450=1350\ \text{Hz}$, its third present harmonic is $f_3=5\times450=2250\ \text{Hz}$, and its fourth present harmonic is $f_4=7\times450=3150\ \text{Hz}$ -- vividly illustrating why a closed pipe's harmon …

Worked out. A closed organ pipe of length 30 cm has its third harmonic frequency exactly equal to the fundamental frequency of a separate open organ pipe, and the task is to find the length of that open pipe. The closed pipe's third harmonic (using $f_n=(2n-1)f_1$ with $n=2$, since the third harmonic is the first overtone) is $f=3v/(4l_1)$ with $l_1=30\ \text{cm}$. The open pipe's fundamental is $f=v/(2l_2)$. Setting these equal, $3v/(4l_1)=v/(2l_2)$, the common speed v cancels, leaving $3/(4\times30)=1/(2l_2)$, which rearranges to $l_2=(2\times30)/3=20\ \text{cm}$. The example demonstrates how matching two different pipes' resonant frequencies -- one at a harmonic other than its fundamental -- lets an unknown pipe length be found purely from the ratio of the two pipes' resonance-condition formu …