# 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.

> IB DP Physics Higher Level (2023 syllabus) — C.4 Standing waves and resonance · Explain · 4 marks

## Mark scheme (4 marks)

1. A standing wave requires a node at a closed end (zero displacement) and an antinode at an open end.
2. For the closed pipe, the length L must equal an odd number of quarter-wavelengths (L = λ/4, 3λ/4, 5λ/4 …), because a node-to-antinode distance is always a quarter wavelength.
3. For the open pipe, both ends are antinodes, so the length must contain a whole number of half-wavelengths (L = λ/2, λ, 3λ/2 …), allowing both odd and even multiples of the fundamental.
4. Therefore the closed pipe's allowed frequencies are f = v/4L, 3v/4L, 5v/4L … (odd harmonics only), missing the even harmonics that the open pipe (f = v/2L, 2v/2L, 3v/2L …) produces, because the asymmetric boundary condition removes the even-harmonic solutions.

## Key terms

- [harmonic](https://www.gradenine.co.uk/glossary/harmonic)

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