Closed Pipe Harmonics Calculator

Calculate resonant frequencies for a pipe closed at one end (odd harmonics)

Parameters

mⓘ
m/sⓘ
(odd)ⓘ
Show Trail

Controls

xⓘ

Calculated Values

Resonant Frequency:
171.50;Hz171.50;Hz
Harmonic n (odd):
1.00;1.00;

Examples

Closed pipe fundamental

L = 0.5 m, n = 1.

    Third harmonic

    n = 3.

      Visualization

      Resonance in a Closed Organ Pipe

      A closed organ pipe is open at one end and closed at the other. The closed end is a rigid boundary: displacement node (air cannot move into the wall). The open end is a displacement antinode. Only certain quarter-wavelength patterns fit in the length L.

      Allowed wavelengths satisfy L = n(λ_n/4) but only odd n = 1, 3, 5, … produce valid standing waves (node at closed end, antinode at open end). Hence f_n = nv/(4L) with n odd only.

      Fundamental (n = 1): λ₁ = 4L — one quarter of a wavelength fits in the pipe. Example: L = 0.5 m, v = 343 m/s → f₁ = 343/(4×0.5) = 171.5 Hz. The same physical length as an open pipe would give f₁(open) = 343 Hz — one octave higher.

      The first overtone is n = 3 (third harmonic), not n = 2 — even harmonics are missing. This gives clarinets and stopped organ pipes a hollow, “hollow wood” timbre compared to flutes.

      End correction at the open end slightly lengthens the effective column. Closed end has less correction. Bottles blown over the lip are often modeled as closed pipes with large end correction.

      Pressure and displacement are 90° out of phase in standing sound waves: at a displacement node, pressure variation is maximum (closed end of a pipe is often a pressure antinode).

      Class 12 contrasts open and closed pipes in tables. JEE favorites: ratio of frequencies, which harmonic is heard, and combining with string formulas.

      Key Concepts

      • f_n = nv/(4L), n = 1, 3, 5, …
      • Node at closed end, antinode at open
      • λ₁ = 4L
      • Missing even harmonics
      • f₁(closed) = ½ f₁(open) same L
      • First overtone is n = 3

      Real-World Applications

      • Clarinet and stopped organ pipes
      • Bottle resonance and “blowing across” demos
      • Closed resonance tube experiments
      • Subwoofer and duct acoustics (some modes)
      • Class 12–JEE closed pipe problems

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      Physics Equations

      Closed Pipe:
      fn=nv4Lf_n = \frac{nv}{4L}

      Step-by-Step Solution

      See how the main results are calculated.

      1

      Step 1: Closed Pipe Formula

      Equation:

      fn=nv4Lf_n = \frac{nv}{4L}

      Explanation:

      n = 1, 3, 5, … (odd harmonics only); closed end is a node.

      2

      Step 2: Given

      Result:

      L=0.5m,v=343m/s,n=1(odd)L = 0.5 m, v = 343 m/s, n = 1 (odd)
      3

      Step 3: Calculate

      Calculation:

      fn=1×3434×0.5=171.5000Hzf_n = \frac{1 \times 343}{4 \times 0.5} = 171.5000 Hz

      Result:

      fn=171.5000Hzfₙ = 171.5000 Hz
      4

      Step 4: Fundamental

      Calculation:

      f1=v/(4L)—quarter−waveinpipef₁ = v/(4L) — quarter-wave in pipe

      Explanation:

      Effective length ≈ L + 0.3d for end correction.

      5

      Step 5: Compare Open Pipe

      Same L: f₁(closed) = ½ f₁(open)

      Explanation:

      Closed pipe fundamental is one octave lower for same length.

      6

      Step 6: Application

      Clarinet behaves approximately as a closed cylindrical pipe.

      Explanation:

      Even harmonics suppressed — timbre differs from open pipe.

      Frequently Asked Questions (FAQ)

      Why no even harmonics?

      An even number of quarter-waves would force a displacement antinode at the closed end, which is impossible for a rigid closure.

      Is a clarinet exactly a closed pipe?

      Approximately for the lowest register; tone holes and bell change effective length and harmonic pattern at higher notes.

      What is the second resonant frequency?

      For n = 3: f₃ = 3v/(4L) = 3f₁ — not 2f₁.

      Pressure node or antinode at closed end?

      Closed end: displacement node, pressure antinode (air piles up). Open end: opposite.

      Can I use n = 2 in the formula?

      No for ideal closed pipe — n must be odd. Using n = 2 gives a mode that does not satisfy boundary conditions.

      Practice MCQs

      1. Closed pipe:
      2. Harmonics:
      3. Closed end is:
      4. n=3 is:
      5. Same L as open pipe, f₁ is:
      6. λ₁ =