Open: all harmonics. Closed: odd harmonics only. Why closed pipe sounds different.

Organ Pipes — Open vs Closed End Harmonics

Question

An organ pipe of length $L$ is open at both ends. Another pipe of the same length $L$ is closed at one end. Compare the fundamental frequencies and list all harmonics produced by each pipe. Why does the closed pipe produce only odd harmonics?

Solution — Step by Step

For an open pipe, both ends are open — so both ends must be displacement antinodes (maximum vibration). For a closed pipe, the closed end is a displacement node (zero movement) and the open end is an antinode.

These boundary conditions are the entire reason the two pipes behave differently. Everything follows from here.

For open-open: we need antinodes at both ends. The simplest standing wave that satisfies this has exactly half a wavelength fitting in length $L$ :

\frac{\lambda_1}{2} = L \implies \lambda_1 = 2L

f_1 = \frac{v}{\lambda_1} = \frac{v}{2L}

We can fit any integer number of half-wavelengths: $\frac{n\lambda}{2} = L$ , giving $\lambda_n = \frac{2L}{n}$ .

f_n = \frac{nv}{2L} = n \cdot f_1 \quad \text{where } n = 1, 2, 3, 4, \ldots

Open pipe produces all harmonics: 1st, 2nd, 3rd, 4th…

For closed-open: node at closed end, antinode at open end. The simplest wave needs one-quarter wavelength in length $L$ :

\frac{\lambda_1}{4} = L \implies \lambda_1 = 4L

f_1' = \frac{v}{4L}

Notice this is half the fundamental of the open pipe of the same length. The closed pipe sounds an octave lower.

We need a node at the closed end and an antinode at the open end. The wavelengths that satisfy this are: $\frac{\lambda}{4}, \frac{3\lambda}{4}, \frac{5\lambda}{4}, \ldots$

So: $(2n-1)\frac{\lambda}{4} = L$ , giving:

f_n = \frac{(2n-1)v}{4L} \quad \text{where } n = 1, 2, 3, \ldots

This gives frequencies: $\frac{v}{4L},\ \frac{3v}{4L},\ \frac{5v}{4L},\ \ldots$ — only odd harmonics.

Why This Works

The key is the node-antinode constraint at the closed end. A node means zero displacement, which forces the wave to fit an odd quarter-wavelength in the pipe. Even harmonics would require an antinode at both ends — which only an open pipe can satisfy.

Think of it this way: an even harmonic of the closed pipe would need a node at both ends (closed) and antinodes at both ends (open) simultaneously — that’s impossible with one closed end. The geometry simply forbids it.

This is why a clarinet (closed-pipe equivalent, roughly) sounds fundamentally different from a flute (open pipe). The missing even harmonics give the clarinet its characteristic hollow, woody tone. This is not a trivial point — JEE Main 2024 directly asked about the tonal difference in terms of harmonics present.

Alternative Method — Using Mode Diagrams

Draw the standing wave patterns visually.

For the open pipe, the modes look like: $\frac{1}{2}\lambda$ , $\lambda$ , $\frac{3}{2}\lambda$ fitted inside $L$ — each gives a new harmonic, and since both ends are free, any integer multiple works.

For the closed pipe, sketch the allowed patterns: node on left, antinode on right. You’ll see only $\frac{1}{4}\lambda$ , $\frac{3}{4}\lambda$ , $\frac{5}{4}\lambda$ fit — the pattern skips even ones every time.

In JEE/NEET MCQs, the quickest check: closed pipe → odd multiples of $\frac{v}{4L}$ , open pipe → all multiples of $\frac{v}{2L}$ . Memorise the ratio: fundamental of closed = $\frac{1}{2}$ × fundamental of open (same length).

Common Mistake

Students often write $f_n = \frac{nv}{4L}$ for the closed pipe — applying the open pipe formula with just a $4L$ denominator. This is wrong. The correct formula is $f_n = \frac{(2n-1)v}{4L}$ . The harmonic number is $(2n-1)$ , not $n$ . So the 3rd harmonic of a closed pipe is $\frac{5v}{4L}$ , not $\frac{3v}{4L}$ . In JEE objective questions, this exact trap appears — they give you $n = 3$ and expect the correct $f = \frac{5v}{4L}$ , not $\frac{3v}{4L}$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Mode Diagrams

Common Mistake

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