Module 4 · Electrons, Waves and Photons

Stationary Waves Part 2: Open–Open and Open–Closed Systems

Specification: OCR A H556 | Section: 4.4.4 Stationary waves | This page: a focused follow-on from Part 1, concentrating on air columns, allowed harmonics, phase behaviour and exam traps

By the end of this topic you should be able to…

compare open–open and open–closed systems without re-covering all of Part 1
recognise the node/antinode boundary conditions at each end of a pipe
derive the allowed wavelengths and frequencies in both systems
understand phase relationships in the loops of each stationary wave pattern
spot common OCR exam traps, especially about odd harmonics and end conditions

Big idea: once you know what each end of the tube must be — node or antinode — the whole harmonic pattern becomes predictable.

Part 1

Boundary conditions: what each end must do

You already met nodes and antinodes in Part 1. The main job here is to use them correctly in air columns.

Open end

An open end is a displacement antinode. Air can vibrate most freely there.

Closed end

A closed end is a displacement node. The air at the boundary cannot move back and forth.

Why this matters

These end conditions decide which harmonics are allowed and which are impossible.

OCR-friendly wording

For air columns, it is safest to say open end = antinode and closed end = node. That is the language most useful for stationary wave diagrams and harmonic questions.

Part 2

Open–open systems

An open pipe has an antinode at both ends. So the simplest possible pattern fits half a wavelength inside the pipe.

Open–open fundamentalL = λ/2 so λ₁ = 2L

Higher harmonics fit whole numbers of half-wavelengths into the length of the pipe.

Open–open harmonicsλ_n = 2L / n and f_n = nv / 2L where n = 1, 2, 3, 4 …

So an open–open pipe supports all harmonics: 1st, 2nd, 3rd, 4th, and so on.

Pattern memory trick

Open–open behaves like a string fixed at both ends in the mathematics of allowed harmonics, even though the physical end conditions are antinode–antinode rather than node–node.

Part 3

Open–closed systems

An open–closed pipe must have a node at one end and an antinode at the other. So the smallest possible pattern is a quarter wavelength.

Open–closed fundamentalL = λ/4 so λ₁ = 4L

The next allowed pattern must still keep one end as a node and the other as an antinode. That means only odd multiples of a quarter wavelength fit.

Open–closed harmonicsλ = 4L, 4L/3, 4L/5, …
f = v/4L, 3v/4L, 5v/4L, …

So an open–closed pipe supports odd harmonics only: 1st, 3rd, 5th, 7th …

High-frequency exam trap

Students often say “the second harmonic of a closed pipe”. In standard OCR A treatment, there is no second harmonic for an open–closed pipe. The next allowed mode after the fundamental is the third harmonic.

Comparison

Open–open vs open–closed

Feature	Open–open pipe	Open–closed pipe
End conditions	Antinode at both ends	Node at closed end, antinode at open end
Fundamental pattern	Half wavelength in the pipe	Quarter wavelength in the pipe
Fundamental wavelength	λ₁ = 2L	λ₁ = 4L
Fundamental frequency	f₁ = v / 2L	f₁ = v / 4L
Allowed harmonics	All: n = 1, 2, 3, 4 …	Odd only: n = 1, 3, 5 …
General form	f_n = nv / 2L	f_n = nv / 4L for odd n only

Fast comparison statement

For the same length L, the fundamental frequency of an open–open pipe is twice that of an open–closed pipe.

Interactive

Pipe mode explorer

Try switching to open-closed and dragging the harmonic slider to 2 or 4. The simulator will show why these are not allowed.

Allowed? yes

Pattern: fundamental

Wavelength: 2L

Frequency: v / 2L

Phase note: one loop only

This animation is meant to help you see which patterns are allowed, what the end conditions force the wave to do, and why even harmonics fail in an open–closed pipe.

Part 4

Phase relationships

Phase questions are easy to mishandle unless you focus on the stationary wave loops.

All points between two adjacent nodes oscillate in phase.
Points in neighbouring loops are in antiphase.
The same principle works in both open–open and open–closed systems.

For example, in the second harmonic of an open–open pipe, the two sections of the wave are in antiphase. In the third harmonic of an open–closed pipe, the two moving sections on either side of the internal node are in antiphase.

Misconception alert

Do not say that all points in the whole stationary wave are in phase. They are only in phase within the same loop between adjacent nodes.

Part 5

End correction: when it matters

For many OCR A harmonic questions, you can ignore end correction unless it is mentioned explicitly. But in practical measurements with resonance tubes, the displacement antinode at an open end is slightly outside the physical end of the tube.

This means the effective length of the air column is a little longer than the measured tube length.

Practical-use version

Include end correction only when dealing with measured resonance lengths or when the question explicitly hints at it. For standard harmonic sketching and most theoretical frequency questions, the simple formulas are usually enough.

Worked Examples

Worked examples

Worked example 1

An open pipe has length 0.85 m. The speed of sound is 340 m s⁻¹. Find the first three harmonic frequencies.

For an open pipe, f₁ = v / 2L = 340 / (2 × 0.85) = 200 Hz

All harmonics are allowed, so f₂ = 2f₁ = 400 Hz

f₃ = 3f₁ = 600 Hz

Worked example 2

A pipe closed at one end has length 0.85 m. The speed of sound is again 340 m s⁻¹. Find the first three allowed frequencies.

For an open–closed pipe, f₁ = v / 4L = 340 / (4 × 0.85) = 100 Hz

Only odd harmonics are allowed, so the next is the 3rd harmonic: f₃ = 3 × 100 = 300 Hz

Next allowed is the 5th harmonic: f₅ = 5 × 100 = 500 Hz

Worked example 3

A student claims that the second harmonic of a closed pipe of length L has wavelength 2L. Diagnose the error.

A standard open–closed pipe does not support even harmonics.

So “second harmonic” is not an allowed mode in this model.

The next allowed wavelength after 4L is 4L/3, corresponding to the 3rd harmonic.

Knowledge

Knowledge Check

What is the displacement condition at an open end of an air column?

1 mark

Open end = displacement antinode

Which harmonics are allowed in an open–closed pipe?

1 mark

Odd harmonics only

For the same pipe length, which system has the higher fundamental frequency: open–open or open–closed?

1 mark

Open–open

How are points in adjacent loops related in phase?

1 mark

They are in antiphase / 180° out of phase

Exam Practice

Exam-Style Questions

Explain why an open–closed pipe supports only odd harmonics.

4 marks

The closed end must be a node
The open end must be an antinode
Only odd numbers of quarter wavelengths satisfy these boundary conditions
Even harmonics would require an incorrect end condition at one end

A tube open at both ends has length 0.68 m. The speed of sound is 340 m s⁻¹.

(a) Calculate the fundamental frequency. [2 marks]
(b) State the frequency of the third harmonic. [1 mark]
(c) Determine the wavelength of the third harmonic. [2 marks]

5 marks

(a) f₁ = v / 2L = 340 / (2 × 0.68) = 250 Hz
(b) f₃ = 3 × 250 = 750 Hz
(c) λ₃ = 2L/3 = 1.36 / 3 = 0.453 m

A student writes: “The next frequency after the fundamental in a closed pipe is the second harmonic.” Explain fully why this is incorrect.

3 marks

Even harmonics are not allowed in an open–closed pipe
The next allowed mode after the fundamental is the third harmonic
This is because the pattern must keep a node at one end and antinode at the other

A resonance tube behaves as an open–closed pipe. Its first resonance occurs when the air column length is 0.170 m and the speed of sound is 340 m s⁻¹.

(a) Show that the tuning fork frequency is 500 Hz. [3 marks]
(b) State the next allowed resonant frequency for the same air column. [1 mark]

4 marks

(a) For first resonance, L = λ/4 so λ = 4L = 0.680 m
f = v/λ = 340 / 0.680 = 500 Hz
(b) Next allowed harmonic is the third, so frequency = 1500 Hz

Compare the phase relationships in the second harmonic of an open pipe with those in the third harmonic of a closed pipe.

4 marks

Within each loop, points oscillate in phase
Adjacent loops are in antiphase
The second harmonic of an open pipe has two loops
The third harmonic of a closed pipe has three quarter-wave sections / two adjacent loop changes giving alternating phase regions

A flute can be modelled as an open–open pipe and a clarinet approximately as an open–closed pipe. Both have the same air-column length.

Describe two important differences in their harmonic series.

4 marks

The flute supports all harmonics, whereas the clarinet supports only odd harmonics
The fundamental frequency of the flute is twice that of the clarinet for the same length
The overtone pattern is therefore more closely spaced in the flute
This contributes to different timbre / tone quality

Topic Summary

Open end

Displacement antinode.

Closed end

Displacement node.

Open–open pipe

All harmonics allowed; f_n = nv / 2L.

Open–closed pipe

Odd harmonics only; f = v / 4L, 3v / 4L, 5v / 4L …

open–open: λ_n = 2L / n

open–closed: λ = 4L, 4L/3, 4L/5 …

adjacent loops are in antiphase

same length: f(open–open) = 2f(open–closed)