OxfordAQA International AS and A-level Physics: Superposition and stationary waves (9630)

This guide follows the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, for exams from May/June 2020 onwards. It covers section 3.5.5 (Principle of superposition of waves and formation of stationary waves), which is shared International AS and A-level content, so it applies to both qualifications. Use it with the course hub and the printable checklist.

Companion pages: the revision notes and the practice questions. To check your starting point, try the free diagnostics first.

What 3.5.5 asks you to do

Specification statement	What you must be able to do
Stationary waves; nodes and antinodes on strings	Describe the pattern and identify nodes and antinodes
Formation by two same-frequency waves moving in opposite directions	Explain this, with a graphical explanation
Strings, microwaves and sound	Apply the ideas to each, including finding wavelength
Harmonics; f = (1/2l)√(T/μ) for the first harmonic	Use the equation and the harmonic series
Experiments on frequency against length, tension and mass per unit length	Describe, analyse and evaluate them

The specification says the terms “fundamental” and “overtone” will not be used. Say “first harmonic”, “second harmonic” and so on.

The principle of superposition

When two or more waves pass through the same point, you find the resultant displacement by adding the separate displacements as vectors at that instant. Displacements in opposite directions have opposite signs.

Worked example. At one instant, wave A gives +3.0 mm at a point and wave B gives −1.8 mm. The resultant is +3.0 + (−1.8) = +1.2 mm. If B were +1.8 mm, the resultant would be +4.8 mm. After the waves have passed through each other, each carries on unchanged.

How a stationary wave forms

A stationary wave is the pattern produced when two progressive waves with the same frequency (and similar amplitude) overlap while moving in opposite directions. In practice the second wave is usually the first wave reflected at a boundary.

Graphical explanation

Take wave 1 moving right and wave 2 moving left, each with amplitude A = 2.0 mm. Draw both on the same axes and add them point by point.

• t = 0. The waves lie exactly on top of each other. Every point has twice the single-wave displacement, so the resultant has amplitude 4.0 mm.
• t = T/4. Each wave has moved a quarter of a wavelength, one to the right and one to the left. The two curves are now mirror images, so at every point the displacements are equal and opposite. The resultant is a flat line.
• t = T/2. Each wave has moved half a wavelength. The curves coincide again, but the resultant is the t = 0 pattern turned upside down.

Points where the waves are always in antiphase never move. These are nodes. Points where they are always in phase oscillate with the largest amplitude. These are antinodes. Adjacent nodes are λ/2 apart, and a node and the next antinode are λ/4 apart.

Displacement (mm) of five points over half a cycle, taking x from a node:

Position	t = 0	t = T/8	t = T/4	t = T/2
x = 0 (node)	0	0	0	0
x = λ/8	2.8	2.0	0	−2.8
x = λ/4 (antinode)	4.0	2.8	0	−4.0
x = 3λ/8	2.8	2.0	0	−2.8
x = λ/2 (node)	0	0	0	0

Read across: every point reaches its maximum at the same time and passes through zero at the same time. Read down: amplitude varies with position, from zero at a node to 4.0 mm at an antinode.

Stationary and progressive waves compared

Feature	Progressive wave	Stationary wave
Amplitude	Same for every point	Zero at nodes, maximum at antinodes
Phase	Changes steadily along the wave	Same within a loop; opposite in neighbouring loops
Energy	Carried along the wave	Not transferred along the wave
Wavelength	Distance between points in phase	Twice the node-to-node distance

Stationary waves on strings and harmonics

Both fixed ends of a string are nodes. The lowest-frequency pattern, the first harmonic, is one loop with an antinode in the middle, so l = λ/2 and λ = 2l.

Harmonic	Loops	Wavelength	Frequency
First	1	2l	f₁
Second	2	l	2f₁
Third	3	2l/3	3f₁
nth	n	2l/n	nf₁

Combining v = fλ with λ = 2l, and the wave speed v = √(T/μ) for a stretched string, gives the equation in the specification:

f = (1/2l) √(T/μ) for the first harmonic

where T is the tension (N), μ is the mass per unit length (kg m⁻¹) and l is the vibrating length (m). It shows f ∝ 1/l, f ∝ √T and f ∝ 1/√μ.

Worked example. A string vibrates over l = 0.72 m. T = 58 N and μ = 3.2 × 10⁻³ kg m⁻¹. Find the first and third harmonic frequencies, and the tension needed for a first harmonic of 110 Hz.

1. T/μ = 58 / 3.2 × 10⁻³ = 18 125 m² s⁻², so v = 134.6 m s⁻¹.
2. f₁ = v/2l = 134.6 / 1.44 = 93.5 Hz.
3. f₃ = 3 × 93.5 = 280 Hz.
4. Rearrange: T = μ(2lf₁)² = 3.2 × 10⁻³ × (1.44 × 110)² = 80.3 N.

Scaling shortcuts, using the 93.5 Hz string: quadruple T and f₁ doubles to 187 Hz. Quadruple μ and f₁ halves to 46.7 Hz. Make l 1.5 times longer and f₁ falls to 62.3 Hz.

Stationary waves with microwaves

A microwave transmitter faces a metal plate. The reflected wave travels back and superposes with the outgoing wave. A probe receiver moved along the line shows signal minima (nodes) and maxima (antinodes). The metal plate is a node. Minima are λ/2 apart. Measuring across several gaps reduces the percentage uncertainty in λ.

Worked example. Nine successive minima span 11.4 cm. Find the wavelength and frequency, taking c = 3.00 × 10⁸ m s⁻¹.

1. Nine minima give 8 gaps, so λ/2 = 11.4 / 8 = 1.425 cm.
2. λ = 2.85 cm = 0.0285 m.
3. f = c/λ = 3.00 × 10⁸ / 0.0285 = 1.05 × 10¹⁰ Hz.

The minimum is often not exactly zero because the reflected wave is slightly weaker than the outgoing wave, so cancellation is incomplete.

Stationary waves with sound

A loudspeaker facing a flat reflecting board sets up a stationary sound wave. A microphone connected to an oscilloscope, moved along the line, finds positions of minimum and maximum signal. Sound is longitudinal, so a node is a point where the air does not oscillate. Its displacement is always zero, while the pressure varies most there.

Worked example. A loudspeaker emits 2.00 kHz. Take the speed of sound as 340 m s⁻¹. Find the distance between adjacent minima, and from a minimum to the next maximum.

1. λ = v/f = 340 / 2000 = 0.170 m.
2. Adjacent minima: λ/2 = 0.085 m.
3. Minimum to maximum: λ/4 = 0.043 m (0.0425 m).

Experiments on a string

You need to know how frequency of the first harmonic depends on length, tension and mass per unit length.

Apparatus. A string or wire runs from a vibration generator, over a bridge (fixing the length l) and then over a pulley to a hanging mass. The tension is T = mg. A signal generator drives the vibrator.

Method.

1. Vary one quantity at a time and keep the other two fixed.
2. Set the length, then slowly raise the signal frequency until a single large loop forms. The first harmonic is the lowest frequency that does this.
3. Find the frequency of maximum amplitude carefully, by going just above and below it and taking the middle.
4. Length: vary l with the bridge. Tension: add masses to the hanger. μ: use strings of different thickness, or twist two strings together. Find μ by weighing a long measured length and dividing mass by length.
5. Repeat each setting and take the mean.

Graphs that give straight lines.

Variation	Plot	Gradient
Length	f against 1/l	½√(T/μ)
Tension	f² against T	1/(4l²μ)
Mass per unit length	f² against 1/μ	T/(4l²)

Worked data analysis. A string with T fixed gives these first harmonic readings:

l / m	0.35	0.45	0.55	0.65	0.85
f / Hz	88.6	68.9	56.4	47.7	36.5
1/l / m⁻¹	2.86	2.22	1.82	1.54	1.18

A plot of f against 1/l is a straight line through the origin, which confirms f ∝ 1/l. Using the end points, gradient = (88.6 − 36.5) / (2.86 − 1.18) = 52.1 / 1.68 ≈ 31 m s⁻¹. Since gradient = v/2, v = 62 m s⁻¹. If 3.00 m of this string has mass 7.8 g, μ = 2.6 × 10⁻³ kg m⁻¹, and T = μv² = 2.6 × 10⁻³ × 62² = 10 N. That is a hanging mass of about 1.0 kg.

Sources of uncertainty. The main one is judging the frequency of maximum amplitude, since resonance is a range. Measure l against a metre rule from the node at the bridge to the node at the pulley, and note that the end at the vibrator is not quite a node. A string that slips or stretches changes T and μ.

Common errors

• Writing “fundamental” or “overtone” when the specification asks for harmonics.
• Saying the wave “stops” or “does not move”. The medium oscillates, but there is no net energy transfer along the wave.
• Taking the distance between adjacent nodes as λ. It is λ/2.
• Forgetting that the ends of a fixed string are nodes, and using l = λ.
• Leaving out the mass-to-weight step: T = mg, not the mass in kg.
• Treating each loop as one wavelength. Each loop is half a wavelength.
• In a graphical explanation, drawing only one instant. Show at least t = 0 and t = T/4.

Next steps

Try the practice questions, then condense your notes with the revision notes. For the earlier wave content in the same section, see the course hub.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), specification Version 4.4, for exams from May/June 2020 onwards, published by OxfordAQA. This guide covers section 3.5.5.