Step 1 — Boundary conditions for a closed pipe: A closed end is a rigid boundary → displacement node (particles cannot move). An open end is a free boundary → displacement antinode (particles move freely).

Step 2 — Standing wave pattern must fit these conditions: The wave pattern must have a NODE at the closed end and an ANTINODE at the open end.

Step 3 — Allowed lengths: The distance from a node to the nearest antinode is always λ/4. So the minimum pipe length is L = λ/4 → fundamental: λ = 4L → $f_{1}$ = $\frac{v}{4L}$.

Step 4 — Next allowed pattern: To add another node-to-antinode segment: L = 3λ/4 → λ = 4L/3 → f = 3$\frac{v}{4L}$ = 3$f_{1}$ (3rd harmonic).

Step 5 — Pattern of allowed harmonics: L = (2n−1)λ/4 for n = 1, 2, 3… → $f_n$ = (2n−1)$\frac{v}{4L}$ = $f_{1}$, 3$f_{1}$, 5$f_{1}$… These are the 1st, 3rd, 5th… harmonics — ODD harmonics only.

Step 6 — Why no even harmonics? Even harmonics would require the pattern to end with a node at both ends or antinodes at both ends — violating the closed-open boundary conditions.

Conclusion: The asymmetric boundary condition (node-antinode) of a closed pipe strictly forbids even harmonics. This is a consequence of geometry, not material properties.