Explain why the resonant frequencies of a pipe closed at one end form a series containing only odd harmonics, whereas a pipe open at both ends produces both odd and even harmonics.

A standing wave needs a node at the closed end and an antinode at the open end. The distance between a node and an antinode is one‑quarter of a wavelength, so the pipe length must be an odd multiple of λ/4:
L = λ/4, 3λ/4, 5λ/4 …
Thus the resonant frequencies are f = v/4L, 3v/4L, 5v/4L … – only odd harmonics.

For a pipe open at both ends both ends are antinodes. The distance between two antinodes is half a wavelength, so the length can contain any integer number of half‑wavelengths:
L = λ/2, λ, 3λ/2 …
Hence the resonant frequencies are f = v/2L, 2v/2L, 3v/2L … – both odd and even harmonics are present.

The closed pipe therefore lacks the even harmonics because the asymmetric boundary condition (node–antinode) forces the length to be an odd multiple of λ/4, whereas the symmetric open–open condition allows all integer multiples of λ/2.