Standing waves in organ pipes — open and closed pipe harmonics

medium CBSE NEET JEE-MAIN NEET 2023 3 min read
Tags Waves

Question

An organ pipe of length 50 cm produces sound waves with speed 340 m/s. Find the fundamental frequency and the first three harmonics for (a) an open pipe and (b) a closed pipe.

(NEET 2023, similar pattern)


Solution — Step by Step

An open pipe has antinodes at both ends (the air is free to vibrate maximally at open ends). The condition for standing waves is:

L=nλ2,n=1,2,3,L = n \cdot \frac{\lambda}{2}, \quad n = 1, 2, 3, \ldots

All harmonics (even and odd) are present.

Fundamental (n=1n = 1): f1=v2L=3402×0.5=340 Hzf_1 = \frac{v}{2L} = \frac{340}{2 \times 0.5} = \mathbf{340 \text{ Hz}}

Second harmonic (n=2n = 2): f2=2f1=680 Hzf_2 = 2f_1 = \mathbf{680 \text{ Hz}}

Third harmonic (n=3n = 3): f3=3f1=1020 Hzf_3 = 3f_1 = \mathbf{1020 \text{ Hz}}

Pattern: f1,2f1,3f1,4f1,f_1, 2f_1, 3f_1, 4f_1, \ldots — all integer multiples of the fundamental.

A closed pipe has a node at the closed end (air cannot vibrate) and an antinode at the open end. The condition is:

L=nλ4,n=1,3,5, (odd only)L = n \cdot \frac{\lambda}{4}, \quad n = 1, 3, 5, \ldots \text{ (odd only)}

Only odd harmonics are present. This is the crucial difference from an open pipe.

Fundamental (n=1n = 1): f1=v4L=3404×0.5=170 Hzf_1 = \frac{v}{4L} = \frac{340}{4 \times 0.5} = \mathbf{170 \text{ Hz}}

Third harmonic (n=3n = 3): f3=3f1=510 Hzf_3 = 3f_1 = \mathbf{510 \text{ Hz}} (this is the 1st overtone)

Fifth harmonic (n=5n = 5): f5=5f1=850 Hzf_5 = 5f_1 = \mathbf{850 \text{ Hz}} (this is the 2nd overtone)

Pattern: f1,3f1,5f1,7f1,f_1, 3f_1, 5f_1, 7f_1, \ldots — only odd multiples.


Why This Works

Standing waves form when sound reflects back and forth inside the pipe. At a closed end, the air molecule is fixed (node) — it can’t move because of the rigid wall. At an open end, pressure equals atmospheric pressure, so displacement is maximum (antinode).

The “fit” condition determines which wavelengths survive. An open pipe needs half-wavelengths to fit (λ/2\lambda/2 per segment between two antinodes). A closed pipe needs quarter-wavelengths (λ/4\lambda/4 between a node and the nearest antinode). Since only odd numbers of quarter-wavelengths fit in a closed pipe, even harmonics are absent.

The fundamental of a closed pipe is half that of an open pipe of the same length: v/(4L)v/(4L) vs v/(2L)v/(2L). A closed pipe sounds lower.


Alternative Method — Quick Comparison

PropertyOpen PipeClosed Pipe
EndsAntinode — AntinodeNode — Antinode
Fundamentalv/(2L)v/(2L)v/(4L)v/(4L)
Harmonics presentAll (1,2,3,1, 2, 3, \ldots)Odd only (1,3,5,1, 3, 5, \ldots)
1st overtone2nd harmonic (2f12f_1)3rd harmonic (3f13f_1)
Sound qualityRicher (more harmonics)Hollow (missing evens)

NEET loves this question: “The ratio of frequencies of the 3rd harmonic of an open pipe to the 2nd harmonic of a closed pipe of the same length is?” Open 3rd harmonic = 3v/(2L)3v/(2L), closed 2nd harmonic = 3v/(4L)3v/(4L). Ratio = 2:1. Practice these ratio questions — they appear almost every year.


Common Mistake

The trickiest trap: “2nd overtone of a closed pipe.” Students write the 2nd harmonic — WRONG. For a closed pipe, the 1st overtone is the 3rd harmonic, the 2nd overtone is the 5th harmonic. The nthn^\text{th} overtone of a closed pipe is the (2n+1)th(2n+1)^\text{th} harmonic. Always convert overtone to harmonic number first, then calculate.

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