Vibrations In Air Columns

Question

A narrow organ pipe closed at one end and open at the other has a length of 0.25 m, and sound travels in air at 340 m per second. What is the fundamental frequency produced by this pipe?

RankGuru · Accepted Answer

In a pipe closed at one end, the closed end must be a displacement node while the open end is an antinode, so the fundamental mode fits only a quarter wavelength of the resonant standing wave into the pipe length, giving $\lambda = 4L$. This makes the fundamental frequency $f_1 = v/(4L)$. Substituting $v = 340	ext{ m/s}$ and $L = 0.25	ext{ m}$ gives $f_1 = \frac{340}{4 	imes 0.25} = \frac{340}{1.0} = 340	ext{ Hz}$. The value $680	ext{ Hz}$ uses $\lambda = 2L$, which is the rule for an open pipe, not a closed one. The value $170	ext{ Hz}$ wrongly uses $\lambda = 8L$, doubling the pipe relation. The value $425	ext{ Hz}$ has no valid harmonic correspondence for a closed pipe. This applies the NCERT analysis of air columns, where a closed pipe sounds only odd harmonics built on $f_1 = v/4L$. As a plausibility check, a closed pipe of given length produces a lower fundamental than an open pipe of the same length, and the next allowed mode here is the third harmonic at 1020 Hz, confirming the odd-harmonic series consistent with 340 Hz.

Vibrations In Air Columns

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