Introduction to Stationary Waves Revision Notes | A Level AQA Physics

1. Why Stationary Waves Matter

Pluck a guitar string and you hear a clear, sustained note. The string does not send a wave travelling down to the bridge and disappear — instead, the wave bounces back and forth, and the two travelling waves combine to create something entirely different: a stationary wave. Understanding this is central to explaining musical instruments, microwave standing-wave experiments, and resonance in pipes.

Stationary waves sit at the heart of AQA A Level waves, and examiners test them in three distinct ways: graphical explanation of formation, description of nodes and antinodes, and application to strings, microwaves, and sound. This lesson builds each of those skills in turn.

2. Formation of Stationary Waves

Picture two identical waves on a string — same frequency, same amplitude, same wavelength — but travelling in opposite directions. At every point along the string, the displacements from both waves add together according to the principle of superposition. The result is not a travelling wave; the pattern stays fixed in space.

The most common way to produce a stationary wave in the lab is to send a wave towards a fixed boundary. The reflected wave travels back along the same path, and the incident and reflected waves superpose. Because both waves have the same frequency (the reflection does not change frequency), the conditions for a stationary wave are met automatically.

2a. Graphical Explanation of Formation

The graphical argument is one the AQA specification explicitly requires, so it is worth working through carefully. Consider two sinusoidal waves of equal amplitude $A$ and wavelength $\lambda$, one moving to the right (Wave 1) and one to the left (Wave 2). At any instant, you can draw both waves on the same axis and add their displacements point by point.

• At $t = 0$: the two waves are exactly in phase. Their displacements add constructively at every point, giving a resultant with amplitude $2A$.
• At $t = T/4$ (quarter period later): each wave has moved a quarter wavelength in its own direction. The waves are now exactly out of phase — every crest of one aligns with a trough of the other. The resultant displacement is zero everywhere.
• At $t = T/2$: the waves are back in phase but the pattern is inverted compared to $t = 0$. The resultant again has amplitude $2A$, but the crests and troughs have swapped positions.
• At $t = 3T/4$: the resultant is again zero everywhere.
• The cycle repeats: the pattern oscillates between $+2A$ and $-2A$ at fixed positions, while certain points remain at zero displacement throughout.

Those permanently stationary points are the nodes. The points that oscillate between $+2A$ and $-2A$ are the antinodes. The key insight from the graphical argument is that nodes and antinodes are fixed in space — the pattern does not move along the medium, which is why we call it a stationary wave.

Notice from the chart that the positions of zero displacement (nodes) remain at $x = 0$, $x = \lambda/2$, and $x = \lambda$ at every instant. The antinodes at $x = \lambda/4$ and $x = 3\lambda/4$ swing between $+2A$ and $-2A$ but never move along the string. This is the graphical proof the examiner expects.

3. Nodes and Antinodes

Now that the graphical picture is clear, the definitions of nodes and antinodes follow directly from it. These two terms carry significant mark weight in AQA questions, so precision matters.

Between every pair of adjacent nodes lies an antinode — a point of maximum amplitude. All particles between two adjacent nodes oscillate in phase with each other, but they are in antiphase with particles in the next segment (between the next pair of nodes). This in-phase/antiphase pattern is a key feature that distinguishes stationary waves from progressive waves.

4. Stationary Waves on Strings and Harmonics

A stretched string fixed at both ends is the classic stationary wave system. When the string vibrates, the fixed ends must be nodes (they cannot move). The stationary wave pattern that fits between the two fixed ends depends on the frequency of vibration — only certain frequencies produce a stable pattern. These are the harmonics.

The AQA specification requires you to describe stationary waves on strings in terms of harmonics. The term fundamental and the term overtone are not used in this specification — use first harmonic, second harmonic, and so on throughout your answers.

• First harmonic: the simplest pattern — one antinode in the middle, nodes at each fixed end. The string length $L$ equals half a wavelength: $L = \lambda_1/2$, so $\lambda_1 = 2L$.
• Second harmonic: two antinodes, three nodes (including the two fixed ends). The string length equals one full wavelength: $L = \lambda_2$, so $\lambda_2 = L$.
• Third harmonic: three antinodes, four nodes. The string length equals three half-wavelengths: $L = 3\lambda_3/2$, so $\lambda_3 = 2L/3$.
• In general, for the $n$th harmonic: $L = n\lambda_n/2$, giving $\lambda_n = 2L/n$.

Because wave speed $v = f\lambda$, and the wave speed on a given string under given tension is fixed, a shorter wavelength means a higher frequency. The first harmonic has the longest wavelength and therefore the lowest frequency — this is the lowest frequency at which a stationary wave can form on that string.

Worked example: Finding the First Harmonic Frequency of a String

A string of length 0.60 m is fixed at both ends. The speed of transverse waves on the string is 240 m/s. Calculate the frequency of the first harmonic. [3 marks]

1. $f_1 = \frac{v}{2L}$ — [M1] Correct formula for first harmonic. The first harmonic has one antinode, so $L = \lambda/2$, giving $\lambda = 2L$. Then $f = v/\lambda = v/(2L)$.
2. $f_1 = \frac{240}{2 \times 0.60}$ — [M1] $v = 240$ m/s, $L = 0.60$ m. Units: m/s divided by m gives Hz. Check: denominator = $2 \times 0.60 = 1.20$ m.
3. $2 \times 0.60 = 1.20$ m — Expanding the multiplication explicitly before dividing avoids arithmetic errors.
4. $f_1 = \frac{240}{1.20} = 200$ Hz — [A1] $240 \div 1.20 = 200$. Units: m/s ÷ m = Hz. Sanity check: 200 Hz is in the audible range, which is physically reasonable for a 0.60 m string.

Answer: $f_1 = 200$ Hz

Each harmonic is a distinct mode of vibration. In practice, a plucked or bowed string vibrates in a mixture of harmonics simultaneously — the relative amplitudes of those harmonics determine the timbre (tone quality) of the instrument. The first harmonic sets the pitch you hear.

5. Stationary Waves with Microwaves and Sound

The same physics that governs a vibrating string applies to any wave that can reflect and travel back along its original path. AQA requires you to consider stationary waves formed with microwaves and with sound waves, not just strings.

In a microwave stationary wave experiment, a microwave transmitter directs waves towards a metal reflecting plate. The reflected wave travels back towards the transmitter, and the two waves superpose to form a stationary wave pattern in the air between them. A small microwave detector (probe) moved along the line between transmitter and reflector detects alternating maxima (antinodes) and minima (nodes). The distance between successive nodes is $\lambda/2$, so measuring this spacing allows the microwave wavelength to be determined directly.

For sound waves, a stationary wave can be set up in a closed or open pipe. A loudspeaker at one end sends sound waves down the pipe; the waves reflect from the closed end (or from the open end due to partial reflection at the boundary between the pipe and open air). The incident and reflected sound waves superpose to form a stationary wave. At a closed end, the air cannot move, so a closed end is always a node. At an open end, the air is free to move, so an open end is treated as an antinode (in practice, the antinode is located slightly beyond the open end due to the end correction, but for AQA A-level purposes the open end is taken as an antinode).

A practical demonstration uses a resonance tube (closed at one end): the tube length is adjusted while a loudspeaker of fixed frequency is held at the open end. The length is increased until resonance is heard — a sudden increase in loudness — indicating that a stationary wave has formed. For a pipe closed at one end, resonance occurs when an odd number of quarter-wavelengths fits the tube length (i.e. $L = \lambda/4$, $3\lambda/4$, $5\lambda/4$, …). For an open pipe, resonance occurs when a whole number of half-wavelengths fits the tube length (i.e. $L = \lambda/2$, $\lambda$, $3\lambda/2$, …). This resonance condition is the pipe equivalent of the harmonic condition for strings.

6. Comparing Stationary and Progressive Waves

Examiners frequently ask you to compare stationary and progressive waves. The comparison requires you to address both similarities and differences — a common source of lost marks is giving only one side.

7. Exam Technique and Common Mistakes

Stationary wave questions appear in multiple formats on AQA papers: graphical description, definition recall, calculation of harmonic frequencies, and extended comparison questions. The following guidance targets the most common mark losses.

For extended-response questions asking you to explain stationary wave formation graphically, structure your answer in three beats: describe the two component waves, describe the superposition at a specific instant (e.g. $t = 0$ and $t = T/4$), then identify the fixed positions of nodes and antinodes and explain why they are fixed. This three-beat structure reliably covers the marking points.

Across all three contexts — strings, microwaves, and sound — the underlying physics is identical: two waves of equal frequency travelling in opposite directions superpose to produce a fixed pattern of nodes and antinodes. Mastering the graphical argument and the harmonic language for strings gives you the tools to handle any stationary wave question the examiner sets.