8.1 Stationary waves

Syllabus
9702–2028–2029
Topic
8.1
Level
AS

Learning objectives

Superposition means overlapping waves add their instantaneous displacements

When waves overlap, the resultant displacement at each point is the algebraic sum of the individual displacements: y=y₁+y₂.

Keep signs and phases, then add the contributions at the same position and time. The waves continue afterwards rather than permanently merging.

Two equal pulses with the same sign reinforce to double displacement; opposite pulses can cancel temporarily.

Superposition does not mean one wave destroys the other or that energy disappears when displacement is zero.

Three experiments reveal stationary waves through fixed patterns

System Set-up and adjustment What reveals the stationary pattern Interpretation
microwaves transmitter faces a metal reflector; move a detector between them alternating signal minima and maxima incident and reflected microwaves superpose; minima are nodes and maxima antinodes
stretched string vibration generator drives a tensioned string with reflection at a fixed end; vary frequency, vibrating length or tension until resonance stable loops with fixed zero-motion points fixed points are displacement nodes; centres of loops are antinodes
air column loudspeaker or tuning fork drives air in a tube; vary frequency or air-column length until resonance large response; fine powder may gather where air motion is least closed end is a displacement node; open end is a displacement antinode

In each experiment, a wave and its reflection of the same frequency travel in opposite directions and superpose. A clear stationary pattern appears at resonance when the boundary conditions fit an allowed mode.

Moving a microwave detector or locating successive string/air displacement nodes gives a repeated spatial pattern: adjacent nodes or adjacent antinodes are λ/2 apart. Measure across several intervals where possible to reduce percentage uncertainty.

Node and antinode refer here to displacement amplitude. Pressure nodes and antinodes in an air column are reversed relative to displacement. Treat an open end as a displacement antinode and a closed end as a displacement node; end corrections are neglected and their theory is not required.

Add opposite-travelling profiles to build fixed nodes and antinodes

Take two coherent waves of the same type, frequency, wavelength, speed and amplitude travelling along the same line in opposite directions. At every position, add their signed displacements. Repeating this graphical addition at later times produces a stationary pattern rather than a travelling profile.

y1=Asin(kxωt),y2=Asin(kx+ωt),y=y1+y2=2Asin(kx)cos(ωt)y₁ = A sin(kx − ωt), y₂ = A sin(kx + ωt), y = y₁ + y₂ = 2A sin(kx) cos(ωt)

Time Result of adding the two profiles
0 maximum profile on one side of equilibrium
T/4 every non-node point passes through equilibrium
T/2 same maximum profile inverted
3T/4 every non-node point passes through equilibrium again
T original profile returns

Nodes are fixed positions where the two waves always cancel and resultant amplitude is zero: x = nλ/2. Antinodes lie midway between nodes, where resultant amplitude is maximum, 2A: x = (2n + 1)λ/4. An antinode is not always at maximum displacement; its particle passes through equilibrium every half-cycle.

All particles between the same pair of adjacent nodes oscillate in phase. Particles in neighbouring loops oscillate 180° out of phase. Every oscillating particle has the same frequency, but amplitude changes continuously from zero at a node to maximum at an antinode.

The envelope does not move along the medium, and there is no net energy transfer along a stationary wave. Fixed nodes do not mean every particle is permanently at rest—only particles exactly at nodes have zero amplitude.

Find wavelength from node or antinode spacing in a stationary wave

In a stationary wave, adjacent nodes or adjacent antinodes are separated by λ/2; a node to the nearest antinode is λ/4.

Measure over several intervals when possible, divide by the number of half-wavelength gaps, then multiply by two.

Four adjacent node gaps spanning 12 cm give λ=2×(12/4)=6.0 cm.

Counting positions rather than gaps gives an off-by-one error; the end-to-end distance must be divided by the number of intervals.