Superposition and stationary waves flashcards

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Principle of superposition
When two or more waves meet at a point, the resultant displacement is the vector sum of the individual displacements.
Coherent sources
Sources with a constant phase difference between them (so they have the same frequency).
Path difference
The difference between the distances travelled by two waves from their sources to the point where they meet.
Phase difference
How far one oscillation is ahead of or behind another, as a fraction of a cycle; 2π rad (360°) is one whole cycle.
Wavefront
A line or surface joining neighbouring points on a wave that are in phase, e.g. along one crest.
Interference
The superposition of waves from coherent sources, giving a steady pattern of maxima and minima.
Node
A point on a stationary wave where the amplitude is always zero.
Antinode
A point on a stationary wave where the amplitude is a maximum, midway between adjacent nodes.
Fundamental (first harmonic)
The lowest frequency at which a stationary wave forms on a string or in a pipe.
Mass per unit length μ
Mass of a string ÷ its length, in kg m⁻¹. For a wire, μ = ρA.
Phase difference from path difference
phase difference = 2π × path difference / λ (radians). (No — learn it)
Conditions for maximum and minimum (sources in phase)
Maximum: path difference = nλ. Minimum: path difference = (n + ½)λ. (No — learn it)
Speed of a transverse wave on a string
v = √(T/μ) (Yes)
Fundamental frequency of a stretched string
f = (1/2L)√(T/μ), from v = fλ with λ = 2L. (No — derive it)
Harmonics on a string fixed at both ends
L = nλ/2, so fₙ = nv/2L = nf₁ (all whole numbers n). (No — derive it)
Fundamental of a pipe closed at one end
L = λ/4, so f₁ = v/4L; then 3f₁, 5f₁ … only. (No — derive it)
Fundamental of a pipe open at both ends
L = λ/2, so f₁ = v/2L; then 2f₁, 3f₁ … (all harmonics). (No — derive it)
Distance between adjacent nodes (or adjacent antinodes)
λ/2. Node to the nearest antinode is λ/4.
Phase of points on a stationary wave
All points between two adjacent nodes are in phase; points in adjacent loops are in antiphase.
Displacement–distance snapshots of a stationary wave at different times
Same nodes every time; the shape grows and shrinks inside a fixed envelope, flat twice per period.
End conditions in pipes
Closed end: displacement node. Open end: displacement antinode.
Core Practical 5: which graph for f against length?
f against 1/L — a straight line through the origin with gradient ½√(T/μ).
How does a stationary wave form?
A wave reflects; incident and reflected waves of the same frequency travel in opposite directions and superpose — cancelling at nodes, reinforcing at antinodes.
Stationary vs progressive wave: energy
A progressive wave transfers energy along its direction of travel; a stationary wave has no net energy transfer.
Why is destructive interference not always total silence?
The resultant amplitude is the difference of the two amplitudes; it is zero only if they are equal.
Why does a pipe closed at one end have only odd harmonics?
It needs a node at one end and an antinode at the other, so only odd numbers of λ/4 fit its length.
Core Practical 5: how is resonance found?
Increase the signal-generator frequency slowly until the string vibrates in one loop with largest amplitude; check from above too and take the mean.
Core Practical 5: how is μ of each string found?
Weigh a measured length of the string on a top-pan balance; μ = mass ÷ length.
In-phase sources, λ = 0.30 m, path difference 0.45 m. Maximum or minimum?
0.45 ÷ 0.30 = 1.5λ → antiphase → minimum.
Path difference 0.10 m, λ = 0.40 m. Phase difference?
2π × 0.10 ÷ 0.40 = π/2 = 1.57 rad (90°).
Nodes 15 cm apart on a string vibrating at 200 Hz. Wave speed?
λ = 2 × 0.15 = 0.30 m; v = 200 × 0.30 = 60 m s⁻¹.
Pipe closed at one end, 0.55 m long, v = 330 m s⁻¹. Fundamental and next resonance?
λ = 4 × 0.55 = 2.2 m; f₁ = 330 ÷ 2.2 = 150 Hz; next is 3f₁ = 450 Hz.
String: T = 90 N, μ = 2.5 × 10⁻³ kg m⁻¹. Wave speed?
v = √(90 ÷ 2.5 × 10⁻³) = √36 000 = 190 m s⁻¹ (189.7).
A string's fundamental is 196 Hz. The tension is doubled. New fundamental?
f ∝ √T: 196 × √2 = 277 Hz.
Using 2π × distance/λ for two points on a stationary wave
Wrong — that is the progressive-wave rule. On a stationary wave, same loop = in phase, adjacent loops = antiphase.
Node spacing taken as the wavelength
Adjacent nodes are λ/2 apart: double the node spacing to get λ.
Sources in antiphase
The conditions swap: whole wavelengths of path difference give minima, half-odd wavelengths give maxima.
Units in v = √(T/μ)
T in N (not kN); μ in kg m⁻¹ (convert g → kg, mm → m before finding the area).

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