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A closed pipe (closed at both ends) resonates at f_n = nv/(2L). An open pipe (open at both ends) also resonates at f_n = nv/(2L). A pipe open at one end and closed at the other resonates at f_n = (2n-1)v/(4L), containing only odd harmonics. These differences arise from boundary conditions: closed ends require nodes (zero velocity), open ends require antinodes.
You already know from standing waves that resonance occurs when a wave reflects back on itself and the reflected wave reinforces the original — the two waves superpose constructively at every point. In a string, the fixed endpoints force displacement nodes there. In an air column, the physics is analogous but the boundary conditions differ depending on whether the end is open or closed.
At a closed end, air molecules cannot move — the wall stops them. This forces a displacement node (zero molecular motion) at that position. At an open end, air pressure must match the atmosphere outside, which means the pressure variation drops to zero there. Since pressure and displacement are 90° out of phase in a sound wave, zero pressure variation at an open end means maximum displacement — an antinode. These two boundary conditions are the only physics you need to derive all the resonance formulas.
For an open-open pipe, both ends require antinodes. The simplest standing wave pattern that satisfies this has antinodes at both ends with a node in the middle — that is exactly half a wavelength fitting in the pipe: L = λ/2, so λ = 2L. Higher harmonics fit additional half-wavelengths: L = nλ/2, giving f_n = nv/(2L) for all integers n = 1, 2, 3, … The full harmonic series is present. For a closed-closed pipe, both ends need nodes — the math works out identically and all harmonics are present.
For a closed-open pipe (like a clarinet), one end has a node and the other an antinode. The simplest pattern that satisfies this is a quarter wavelength: L = λ/4, so λ = 4L, and f₁ = v/(4L). The next pattern must have a node at the closed end and antinode at the open end with one more half-cycle in between — that requires L = 3λ/4, giving f = 3v/(4L). The pattern continues as L = (2n−1)λ/4, yielding only odd harmonics: f_n = (2n−1)v/(4L). This is why a clarinet (closed-open) sounds darker than a flute (open-open) at the same fundamental pitch — the clarinet's timbre lacks the even harmonics that would add brightness. The tube geometry is acoustic destiny: boundary conditions dictate the overtone spectrum, which dictates the instrument's voice.