Standing Waves In Pipes

Resonance in air columns, described in NCERT Class 11, Chapter 15 (Waves), requires a displacement node at a closed end and a displacement antinode at an open end. For a pipe closed at one end, the fundamental fits a quarter wavelength, $L=\lambda /4$. The next allowed modes must again place a node at the closed end and an antinode at the open end, which happens only for $L=3\lambda /4,5\lambda /4,$ and so on, giving frequencies in the ratio $1:3:5:\dots$. Thus only odd harmonics are present. The option of only even harmonics is wrong because even multiples would require an antinode at the closed end, which is forbidden. The option of all integer harmonics describes a pipe open at both ends, not a closed pipe. The option that no harmonics exist beyond the fundamental contradicts the existence of the $3\lambda /4$ and higher modes. Quantitatively, the resonant frequencies of a closed pipe are $f_n=(2n-1)\frac{v}{4L}$ for $n=1,2,3,\dots$, generating the sequence $\frac{v}{4L},\frac{3v}{4L},\frac{5v}{4L}$, exactly the odd multiples of the fundamental. A further consequence is that a closed pipe of length $L$ has a fundamental one octave lower than an open pipe of the same length, since the closed pipe fits only a quarter wavelength while the open pipe fits half. A consistency check: the missing even harmonics are exactly why a closed pipe such as a clarinet sounds tonally different from an open pipe of the same fundamental, and the ratio $1:3:5$ confirms the odd-only spectrum.