Fundamental frequency of open pipe 500Hz — find first three harmonics of closed pipe of same length

Question

An open pipe has a fundamental frequency of 500 Hz. Find the first three harmonics produced by a closed pipe of the same length.

Solution — Step by Step

For an open pipe, both ends are antinodes. The fundamental mode has half a wavelength fitting in the pipe length:

L = \frac{\lambda}{2} \Rightarrow \lambda = 2L

Fundamental frequency of open pipe:

f_{\text{open}} = \frac{v}{\lambda} = \frac{v}{2L} = 500 \text{ Hz}

Therefore: $\dfrac{v}{2L} = 500$ Hz, so $\dfrac{v}{L} = 1000$ Hz (we’ll use this).

For a closed pipe (one end closed, one end open):

Closed end = node, open end = antinode
Only ODD harmonics are produced
Fundamental: $L = \lambda/4 \Rightarrow f_1 = \frac{v}{4L}$
Harmonics: $f_n = n \cdot \frac{v}{4L}$ for $n = 1, 3, 5, 7, \ldots$

f_{\text{closed, fundamental}} = \frac{v}{4L} = \frac{1}{2} \times \frac{v}{2L} = \frac{500}{2} = 250 \text{ Hz}

The harmonics of a closed pipe are at odd multiples of the fundamental:

1st harmonic (fundamental): $f_1 = \frac{v}{4L} = 250$ Hz
2nd harmonic produced = 3rd overtone = $3 \times 250 = \mathbf{750 \text{ Hz}}$
3rd harmonic produced = 5th overtone = $5 \times 250 = \mathbf{1250 \text{ Hz}}$

(Note: “harmonics” in a closed pipe means the odd-numbered ones — 1st, 3rd, 5th, etc.)

Why This Works

The constraint in a closed pipe is that the closed end must be a node (no displacement) and the open end must be an antinode. This forces only odd multiples of the fundamental to fit in the pipe — even harmonics would require a node at both ends or an antinode at both ends, which contradicts the boundary conditions.

An open pipe, with antinodes at both ends, can support all harmonics (both odd and even). That’s why the open pipe’s harmonic series is richer (more tones) than the closed pipe’s.

Relation between fundamental frequencies of same-length pipes:

f_{\text{closed}} = \frac{1}{2} f_{\text{open}}

The fundamental of a closed pipe is always half that of an open pipe of the same length.

Alternative Method — Counting Wavelengths

Mode	Open pipe	Closed pipe
Fundamental	$L = \lambda/2$	$L = \lambda/4$
2nd (if exists)	$L = \lambda$	$L = 3\lambda/4$
3rd (if exists)	$L = 3\lambda/2$	$L = 5\lambda/4$

For the closed pipe with $L$ such that $v/(4L) = 250$ Hz:

$f_1 = 250$ Hz, $f_3 = 750$ Hz, $f_5 = 1250$ Hz.

Common Mistake

Students often include even harmonics for a closed pipe. A closed pipe only produces odd harmonics: 1st, 3rd, 5th (250, 750, 1250 Hz). The “2nd harmonic” at 500 Hz and the “4th harmonic” at 1000 Hz are absent for a closed pipe. This is a defining property of closed pipes and a very common exam trap.

Remember: closed pipe = odd harmonics only; open pipe = all harmonics. This explains why clarinets (approximated as closed pipes) have a more “hollow” tone than flutes (open pipes) — the absence of even harmonics changes the timbre. This fact appears in JEE context as a conceptual question about musical instruments.

Question

Solution — Step by Step

Why This Works

Alternative Method — Counting Wavelengths

Common Mistake

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