Vibrations of air column in a pipe open at both ends

Now consider a pipe OPEN at both ends, again excited by a tuning fork held near one end. Even with both ends open, the air inside the pipe is still confined by the tube's walls and is slightly denser than the free air outside, so an incident wave still undergoes PARTIAL reflection at the far open end; this partially reflected wave superposes with the still-arriving incident wave (which is itself reflected again, a second time, at the near open end where the source sits) and, under the right conditions, builds up a genuine stationary wave, exactly as in the closed-pipe case but now with the different boundary condition that BOTH ends -- not just one -- must be antinodes, since both ends are equally free.

The fundamental mode has an antinode at EACH of the two open ends with exactly one node in between them -- HALF a wavelength fits the air column, $L=\lambda/2$, i.e. $\lambda=2L$; from $v=n\lambda=2nL$ (Eq. 6.22), the fundamental frequency is $n=\dfrac{v}{2L}=\dfrac{v}{2(l+2e)}$ (Eq. 6.23) -- the FIRST HARMONIC, and the lowest frequency this pipe can produce.

The next mode has three antinodes (both ends plus one more in the middle) and two nodes: ONE full wavelength fits the column, $L=\lambda_1$ (Eq. 6.24), giving from $v=n_1\lambda_1$ a frequency $n_1=\dfrac{v}{L}=\dfrac{v}{l+2e}=2n$ (Eq. 6.25) -- the SECOND HARMONIC, and the FIRST OVERTONE.

The mode after that has four antinodes and three nodes: one-and-a-half wavelengths fit the column, $L=\dfrac{3\lambda_2}{2}$ (Eq. 6.26), giving $n_2=\dfrac{3v}{2L}=3n$ (Eq. 6.27) -- the THIRD HARMONIC, and the SECOND OVERTONE. In general, the p-th overtone has frequency $n_p=(p+1)n$ (Eq. 6.28), for $p=0,1,2,3,\ldots$ ($p=0$ recovering the fundamental). …