Organ Pipes — Open vs Closed Resonance

Opening — Why Organ Pipes Confuse Students

Open and closed organ pipes look almost identical on paper, but their resonance frequencies are completely different. Students memorize formulas and trip up on whether to put a 2 or a 4 in the denominator, or whether the third harmonic of a closed pipe is the third or the fifth multiple. We’re going to fix this once and for all by going through the physics — not just the formulas.

The trick is to draw the standing wave pattern. End conditions force the wave: at a closed end, air can’t move (displacement node), and at an open end, air moves freely (displacement antinode). Match the wavelength to these end conditions, and the frequencies follow.

This topic is a NEET favourite — almost guaranteed every other year — and shows up in JEE Main as a one-marker.

Key Terms & Definitions

Open organ pipe: open at both ends. Examples — flute, recorder. Both ends are displacement antinodes.

Closed organ pipe: closed at one end, open at the other. Examples — clarinet, organ pipe with stopped end. Closed end is a displacement node, open end is an antinode.

Fundamental frequency (first harmonic): lowest frequency the pipe can resonate at.

Overtone: any resonance frequency above the fundamental. The first overtone is the second harmonic of an open pipe but the third harmonic of a closed pipe — students often confuse these.

End correction: in real pipes, the antinode lies slightly outside the open end. We ignore it for theoretical problems.

Methods / Concepts

Step 1 — Draw the Standing Wave

Always start with a sketch. For open pipe: antinode at both ends (curve at maximum amplitude there). For closed pipe: node at closed end, antinode at open end.

The fundamental fits the longest possible wavelength inside the pipe consistent with these boundaries.

Open pipe fundamental: half a wavelength fits — antinode-node-antinode pattern across length $L$.

$L = \frac{\lambda}{2} \implies \lambda = 2L \implies f_1 = \frac{v}{2L}$

Closed pipe fundamental: a quarter wavelength fits — antinode-quarter wave-node.

$L = \frac{\lambda}{4} \implies \lambda = 4L \implies f_1 = \frac{v}{4L}$

Step 2 — Get Higher Harmonics

For open pipe, we add half wavelengths: $L = n\lambda/2$, so $\lambda_n = 2L/n$ and $f_n = nv/(2L)$. All integers $n$ work.

For closed pipe, we add full half-wavelengths to the fundamental quarter-wavelength: $L = (2n-1)\lambda/4$, so $\lambda_n = 4L/(2n-1)$ and $f_n = (2n-1)v/(4L)$. Only odd multiples of the fundamental.

Step 3 — Compare to Open Pipe

A closed pipe of length $L$ has fundamental $v/(4L)$, which is half the fundamental of an open pipe of the same length. So closed pipes sound an octave lower than open pipes of equal length. This is why a clarinet (closed at the reed end) sounds lower than a flute (open at both ends) of similar length.

Solved Examples

Example 1 (Easy — CBSE)

An open pipe of length 50 cm produces a fundamental frequency of 340 Hz. Find the speed of sound.

$f_1 = \frac{v}{2L} \implies v = 2L f_1 = 2 \cdot 0.5 \cdot 340 = 340 \text{ m/s}$

Example 2 (Medium — NEET 2022)

A closed organ pipe and an open organ pipe of the same length produce fundamentals in the ratio 1:2. Verify, and find the ratio of their second overtones.

Fundamentals: closed = $v/4L$, open = $v/2L$. Ratio = $1:2$. Confirmed.

Second overtone: for closed pipe, overtones are 1st (3rd harmonic), 2nd (5th harmonic). So second overtone of closed = $5v/(4L)$.

For open pipe, overtones are 1st (2nd harmonic), 2nd (3rd harmonic). So second overtone of open = $3v/(2L) = 6v/(4L)$.

Ratio of second overtones = $5:6$.

Example 3 (Hard — JEE Advanced flavour)

A closed pipe is found to have a frequency 100 Hz that is the same as the third harmonic of an open pipe of length 1.5 m. Find the length of the closed pipe. Take $v = 340$ m/s.

Third harmonic of open pipe: $f = 3v/(2L_{\text{open}}) = 3 \cdot 340/(2 \cdot 1.5) = 340$ Hz. But the question states 100 Hz — let me re-read with the assumption that the third harmonic is 100 Hz.

Setting $3v/(2L_{\text{open}}) = 100$: $L_{\text{open}} = 3 \cdot 340/200 = 5.1$ m. Now the closed pipe has some harmonic at 100 Hz: closed harmonics are odd multiples of $v/(4L_c)$. If 100 Hz is the fundamental: $L_c = 340/(4 \cdot 100) = 0.85$ m. If it’s the third harmonic: $L_c = 3 \cdot 340/(4 \cdot 100) = 2.55$ m. Question wording determines which. Length depends on which harmonic.

Exam-Specific Tips

Common Mistakes to Avoid

Practice Questions

Q1. An open pipe is 1 m long. Find the first three resonance frequencies. ($v = 340$ m/s)

Q2. A closed pipe is 0.5 m long. Find its fundamental and first two overtones. ($v = 340$ m/s)

Q3. What length of closed pipe will produce a fundamental of 256 Hz? ($v = 340$ m/s)

Q4. A pipe produces frequencies 200, 600, 1000, 1400 Hz. Open or closed? Find length.

Q5. Two open pipes 1 m and 1.01 m long are sounded together. How many beats per second? ($v = 340$ m/s)

FAQs

Q: Why do closed pipes only have odd harmonics? Because the standing wave must have a node at the closed end and an antinode at the open end. Only odd quarter-wavelengths fit this asymmetry.

Q: What is end correction and when do we use it? End correction $e \approx 0.6r$ accounts for the antinode lying slightly outside the open end. Used when high precision is needed; ignored in school-level problems.

Q: Can a pipe be closed at both ends? Theoretically yes, but it wouldn’t radiate sound efficiently. Both ends become nodes, fundamental at $L = \lambda/2$, all harmonics present.

Q: Why does a flute (open both ends) sound brighter than a clarinet (closed one end)? The flute’s all-harmonics spectrum is rich and bright; the clarinet’s odd-only spectrum is hollow and woody.

Q: How does temperature affect organ pipe frequencies? Speed of sound $v \propto \sqrt{T}$, so $f \propto \sqrt{T}$ for fixed $L$. Pipes go slightly sharp when warm.

Q: Does pipe diameter affect frequency? Mainly through end correction — wider pipes have larger end correction and slightly lower resonance frequency.

Q: How does this relate to the resonance tube experiment in Class 11 lab? The resonance tube is exactly a closed pipe with adjustable length. By finding two consecutive resonance lengths, you get $\lambda/2 = L_2 - L_1$, and hence $v = f \cdot \lambda$.