Q.Show that only odd harmonics are present in the vibrations of air column in a pipe closed at one end.

[!TLDR] In a closed pipe the fundamental has $L=\lambda/4$; each higher mode adds one more quarter-wavelength segment, so $L=(2p+1)\lambda/4$ always -- an ODD number of quarters -- giving $n_p=(2p+1)n$, i.e. only odd harmo …

A pipe closed at one end always has a NODE at the closed end and an ANTINODE at the open end (section 6.7.2), since the closed end constrains particle motion completely while the open end leaves it comparatively free. The simplest (fundamental) standing-wave pattern that satisfies BOTH conditions simultaneously fits exactly one quarter-wavelength into the air column: $L=\lambda/4$, giving $n=v/(4L)$ -- the first harmonic. Every successive allowed mode must ADD one full extra half-wavelength segment (one more complete node-to-antinode-to-node unit) while still ending on an antinode at the open end and starting on a node at the closed end -- which is only possible by adding exactly TWO more quarter-wavelength segments each time (never just one, since a single extra quarter-wavelength would flip the open end back to a node, violating the boundary condition). So the allowed lengths run $L=\dfrac{\lambda}{4},\dfrac{3\lambda}{4},\dfrac{5\lambda}{4},\dfrac{7\lambda}{4},\ldots$ -- always an ODD number of quarter-wavelengths, $L=(2p+1)\dfrac{\lambda_p}{4}$ for $p=0,1,2,\ldots$. Since $v=n_p\lambda_p$ and $\lambda_p=4L/(2p+1)$, the corresponding frequency is $n_p=\dfrac{v}{\lambda_p}=\dfrac{(2p+1 …

Apply the node-at-closed-end, antinode-at-open-end boundary conditions to find that only an odd number of quarter-wavelengths can fit the air column, then conver …

Assuming an EVEN number of quarter-wavelengths could also satisfy the boundary conditions, without checking that this would place a NODE (not an anti …