OCR Specification focus:
‘Describe patterns for strings and air columns in open and closed tubes.’
Understanding how stationary wave patterns form on strings and within air columns reveals the physical principles behind musical instruments, resonance, and the quantisation of standing wave modes.
Stationary Waves and Resonance Patterns
When a wave reflects and interferes with itself in a confined medium, it can create a stationary wave (or standing wave). Unlike progressive waves, which transfer energy through the medium, stationary waves store energy within a fixed region, producing distinct points of no motion (nodes) and maximum motion (antinodes).
Stationary waves form when the reflected wave and incident wave are coherent—having the same frequency, amplitude, and wavelength—and travel in opposite directions.
Stationary Waves on Strings
A stretched string fixed at both ends is one of the simplest systems that produces stationary waves. The two fixed points cannot move, so they must always be nodes. Between these nodes, the string vibrates in distinct patterns called modes of vibration or harmonics.
Formation of Harmonics on Strings
At specific frequencies, the string vibrates in modes where an integer number of half-wavelengths fits exactly into the string length.
EQUATION
—-----------------------------------------------------------------
Wavelength on a String (λₙ) = 2L / n
L = Length of string (m)
n = Harmonic number (1, 2, 3, …)
—-----------------------------------------------------------------
This equation ensures that the boundary conditions—nodes at both ends—are satisfied.
Between these nodes, the string oscillates with large amplitude at antinodes, where constructive interference is maximum.
Harmonic patterns on a string fixed at both ends, showing the fundamental and successive modes. Nodes occur at the fixed ends, with additional internal nodes and antinodes appearing for higher harmonics. The diagram is minimal and labels the stationary-wave envelope clearly. Source.
Harmonic Patterns
First Harmonic (Fundamental):
The simplest mode of vibration. One-half of a wavelength fits into the string length (L = ½λ).
2 nodes (at the ends)
1 antinode (in the centre)
Second Harmonic (First Overtone):
Two half-wavelengths fit into the length (L = λ).
3 nodes
2 antinodes
Third Harmonic (Second Overtone):
Three half-wavelengths fit into the length (L = 1.5λ).
4 nodes
3 antinodes
Each higher harmonic corresponds to a frequency that is an integer multiple of the fundamental frequency.
EQUATION
—-----------------------------------------------------------------
Harmonic Frequency (fₙ) = n × f₁
f₁ = Fundamental frequency (Hz)
n = Harmonic number
—-----------------------------------------------------------------
Therefore, f₂ = 2f₁, f₃ = 3f₁, and so forth, creating a harmonic series.
Stationary Waves in Air Columns
Air columns in tubes behave similarly to strings, but the boundary conditions differ depending on whether the tube ends are open or closed. The displacement of air molecules determines the wave pattern inside the tube.
Displacement and Pressure Nodes
Displacement Node: A point where air molecules do not move (corresponds to a pressure antinode).
Displacement Antinode: A point where air molecules oscillate with maximum amplitude (corresponds to a pressure node).
At an open end, air molecules can move freely, so it must be a displacement antinode.
At a closed end, air cannot move, so it must be a displacement node.
Patterns in Open and Closed Tubes
Open–Open Tube
Both ends are open, so both ends act as displacement antinodes.
The simplest pattern (first harmonic) fits half a wavelength into the tube.
Successive harmonics occur when whole numbers of half-wavelengths fit into the tube.
EQUATION
—-----------------------------------------------------------------
Wavelength in an Open–Open Tube (λₙ) = 2L / n
L = Length of tube (m)
n = Harmonic number (1, 2, 3, …)
—-----------------------------------------------------------------
Thus, all harmonics are possible in open–open tubes (n = 1, 2, 3, …).
Harmonics for a system with two free ends (open–open tube analogue), showing antinodes at both ends and integer-multiple modes. The graphic depicts the first three harmonics with clearly marked nodes and antinodes. Any additional surrounding text discusses boundary conditions more broadly but the figure itself focuses on the allowed patterns. Source.
Examples: flutes, organ pipes, and some brass instruments.
Closed–Open Tube
One end is closed (node), and the other end is open (antinode).
This pattern allows only odd harmonics, since only an odd number of quarter-wavelengths can fit within the tube.
EQUATION
—-----------------------------------------------------------------
Wavelength in a Closed–Open Tube (λₙ) = 4L / n
L = Length of tube (m)
n = Odd harmonic number (1, 3, 5, …)
—-----------------------------------------------------------------
This pattern allows only odd harmonics, since only an odd number of quarter-wavelengths can fit within the tube.
Standing-wave modes in a closed–open pipe, with displacement node at the closed end and antinode at the open end. The diagram shows the fundamental and higher odd harmonics, matching the quarter-wave condition. Labels are concise and directly aligned with GCSE/A-level expectations. Source.
First Harmonic (n=1): Quarter of a wavelength fits inside.
Node at closed end
Antinode at open end
Third Harmonic (n=3): Three-quarters of a wavelength fit.
Two additional nodes and antinodes appear inside the tube.
Fifth Harmonic (n=5): Five-quarters of a wavelength fit, continuing the odd sequence.
Examples: clarinets and certain organ pipes that are closed at one end.
Comparison of String and Tube Patterns
While the physics of reflection and interference underpin both systems, their end conditions create distinctive harmonic sequences:
Strings (both ends fixed): All harmonics (n = 1, 2, 3, …) are possible.
Open–Open Tubes: Same as strings; all harmonics are present.
Closed–Open Tubes: Only odd harmonics (n = 1, 3, 5, …) are present.
The frequency spacing of harmonics differs accordingly, affecting the timbre of sounds produced by different instruments.
EQUATION
—-----------------------------------------------------------------
Wave Speed (v) = f × λ
v = Speed of sound in medium (m/s)
f = Frequency of the wave (Hz)
λ = Wavelength (m)
—-----------------------------------------------------------------
This equation links measurable quantities for any wave pattern, allowing the speed of sound or string tension to be determined experimentally.
Resonance and Energy Transfer
When a vibrating source (e.g. a tuning fork or speaker) drives a string or air column at one of its natural frequencies, resonance occurs. The amplitude of vibration increases dramatically, indicating that energy is efficiently transferred from the driver to the medium.
Resonance conditions depend on the wavelength relationships described earlier:
For strings: length must equal an integer multiple of half-wavelengths.
For open tubes: length equals an integer multiple of half-wavelengths.
For closed tubes: length equals an odd multiple of quarter-wavelengths.
Resonance patterns are visual or acoustic evidence of standing waves. In laboratory investigations, these are demonstrated using:
Microwave experiments (stationary field patterns).
Vibrating strings with a mechanical driver.
Resonance tubes filled with air or water to vary effective length.
Applications and Relevance
Understanding these patterns provides a foundation for the physics of musical acoustics, sound engineering, and vibration analysis. The relationship between wavelength, frequency, and boundary conditions explains how instruments produce specific pitches and how resonant cavities enhance sound.
By mastering the patterns on strings and in tubes, students gain insight into wave behaviour, resonance phenomena, and the quantisation of vibrational modes—key aspects of both classical and modern physics.
FAQ
At a fixed end, the medium cannot move, so displacement must be zero — creating a node. The reflected wave must invert upon reflection to maintain this condition.
At an open end, air particles move freely and pressure must match atmospheric pressure. This creates a displacement antinode and a pressure node, where maximum oscillation of air occurs.
In real tubes, the antinode does not form exactly at the physical open end. The wave extends slightly beyond the tube’s mouth — this is known as the end correction.
This effectively increases the acoustic length of the tube.
For a single open end: typically about 0.6 times the tube radius.
For an open–open tube: correction applies to both ends.
Neglecting end corrections can lead to small but measurable frequency errors in experiments.
Only frequencies that meet the boundary conditions (nodes and antinodes forming at the ends) will reinforce themselves through constructive interference.
Other frequencies produce partial interference and quickly die out due to energy dissipation.
Thus, each tube or string supports a discrete set of resonant frequencies, forming the harmonic series. These are sometimes referred to as the tube’s or string’s normal modes of vibration.
For a string, the pitch depends on:
Length (L): Shorter strings produce higher frequencies.
Tension (T): Greater tension increases frequency.
Mass per unit length (μ): Thicker or denser strings lower frequency.
For an air column, pitch depends on:
Tube length: Shorter tubes give higher frequencies.
Temperature: Higher temperatures increase sound speed and frequency.
A resonance tube experiment provides a clear demonstration.
A vibrating tuning fork is held above a vertical tube partially filled with water.
The air column length is gradually altered by moving the water level.
Loud sounds occur when the column resonates with the tuning fork frequency, corresponding to stationary wave formation.
The distances between resonant positions can be used to calculate the speed of sound or verify harmonic relationships in a closed–open system.
Practice Questions
Question 1 (2 marks)
A stationary wave is formed on a string fixed at both ends. The distance between two adjacent nodes is 0.40 m.
Determine the wavelength of the wave on the string.
Mark Scheme:
Recognises that the distance between adjacent nodes equals half a wavelength (1 mark)
Correctly calculates wavelength: λ = 2 × 0.40 = 0.80 m (1 mark)
Question 2 (5 marks)
A student investigates stationary waves in a tube that is closed at one end and open at the other. The tube has a length of 0.60 m.
(a) State the boundary conditions for the closed and open ends. (2 marks)
(b) Calculate the wavelength and frequency of the fundamental mode, given that the speed of sound in air is 340 m s⁻¹. (2 marks)
(c) Explain why only odd harmonics can exist in this type of tube. (1 mark)
Mark Scheme:
(a)
Closed end: displacement node (air cannot move) (1 mark)
Open end: displacement antinode (maximum movement of air) (1 mark)
(b)
Recognises that the fundamental mode in a closed–open tube has L = ¼λ (1 mark)
Calculates wavelength: λ = 4 × 0.60 = 2.4 m (1 mark)
Calculates frequency: f = v / λ = 340 / 2.4 = 142 Hz (to 3 s.f.) (1 mark)
(c)
Only odd harmonics fit because the tube must always have a node at the closed end and an antinode at the open end, allowing only odd multiples of the quarter-wavelength pattern (1 mark)
