Free Pearson Edexcel International A Level Physics lesson
Interference and stationary waves
Use phase to predict reinforcement, cancellation and resonance.
What you’ll learn
- Connect path difference and phase difference.
- Explain nodes and antinodes from opposite travelling waves.
- Test how string frequency depends on length, tension and mass per length.
Add displacement, not intensity
When small disturbances overlap in a linear medium, the resultant displacement at a point is the algebraic sum of the individual displacements at that same instant. This is superposition. Equal upward and downward pulses can momentarily give zero displacement without permanently destroying either pulse. When periodic waves overlap, the repeated pattern of reinforcement and cancellation is called interference. Their relative phase determines which outcome occurs at a chosen point.
Coherent sources have the same frequency and a stable phase relationship. If they start in phase and travel at the same speed, a path difference of a whole number of wavelengths produces an in-phase arrival. A half-odd-integer wavelength produces opposite phase. The phase difference corresponding to path difference Δx is 360°Δx/λ or 2πΔx/λ radians, with any initial source phase added separately.
Maximum reinforcement of two equal-amplitude waves doubles displacement amplitude; it does not mean that two intensities simply add everywhere. In a fixed medium intensity is proportional to amplitude squared, and energy is redistributed between bright and dark regions. Complete cancellation requires equal amplitudes as well as opposite phase. If source phase drifts, the interference pattern moves or washes out during an observation, even when the average frequencies are nearly equal.
Δφ = 360° Δx/λ = 2π Δx/λ rad; resultant displacement = y₁ + y₂
Symbols and units
- Δx is path difference in m; λ is wavelength in m; Δφ is additional phase difference in rad (or degrees).
- y₁ and y₂ are signed displacements in the same unit.
How the equation explains the idea
- A path increase of λ adds 2π radians, so the fractional cycle Δx/λ gives Δφ = 2πΔx/λ. Add any initial source phase separately.
- At one point and instant, add y₁ + y₂. Equal-amplitude opposite-phase waves cancel locally; opposite phase with unequal amplitudes leaves their amplitude difference.
Conditions. Two sinusoidal waves at the same frequency in a linear medium. Both displacement amplitudes are 2 mm. The resultant amplitude is √(A₁² + A₂² + 2A₁A₂ cos φ).
Relative phase / degrees; Resultant amplitude / mm.
At 0° or 360° the amplitude is 4 mm; at 180° it is zero. At 90° it is √8 ≈ 2.83 mm. This is resultant amplitude, not an instantaneous displacement trace or intensity.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
A path difference that matters modulo a cycle
- Two in-phase coherent waves have wavelength 0.60 m and path difference 0.90 m.
- Phase difference = 360° × 0.90/0.60 = 540°.
- 540° differs from 180° by one full cycle, so equal-amplitude waves cancel at that point.
Opposite phase can result from 1.5 wavelengths, not only 0.5 wavelength.
Keep this idea. Convert path difference into phase, include source phase, then add signed displacements.
A pattern whose nodes stay put
A stationary wave can form when a travelling wave and its reflection have equal frequency and travel in opposite directions along the same medium. Their superposition creates fixed positions with zero amplitude, called nodes, and positions of greatest amplitude, called antinodes. The pattern is stationary even though the medium oscillates. In the ideal equal-wave case there is no net time-averaged energy transfer along the pattern; energy repeatedly changes form locally.
Adjacent nodes are separated by half a wavelength; a node and its nearest antinode are separated by a quarter wavelength. All particles within one loop between adjacent nodes oscillate in phase, though their amplitudes differ. Adjacent loops oscillate in opposite phase. A snapshot can have every point at zero displacement simultaneously; this does not turn every point into a node, because node refers to amplitude being zero for the entire cycle.
Boundary conditions select allowed modes. An ideal string fixed at both ends has nodes at both ends, so a length L contains an integer number n of half wavelengths. The fundamental contains one loop, with λ = 2L, while higher modes contain more loops. For sound in air columns, distinguish displacement nodes from pressure nodes: a displacement node corresponds to a pressure antinode. Always state which quantity a stationary-wave diagram represents.
Fixed–fixed string: L = nλ/2; fₙ = nv/(2L); adjacent-node separation = λ/2
Symbols and units
- L is fixed–fixed string length in m; n is positive integer mode number; λ is wavelength in m.
- v is wave speed in m s⁻¹ and fₙ is mode frequency in Hz.
How the equation explains the idea
- Each loop spans half a wavelength. Fitting n loops gives L = nλ/2, hence λ = 2L/n.
- Insert this wavelength into f = v/λ: fₙ = nv/(2L). At fixed L and v, doubling n doubles frequency and halves wavelength. A node stays at zero displacement throughout every cycle.
Conditions. Ideal equal counter-propagating waves form the second harmonic on a 1 m fixed–fixed string. Two snapshots half a period apart are shown, with maximum antinode displacement 2 mm.
Position along string / m; Displacement / mm.
At 0, 0.5 and 1 m both curves have zero displacement, and these positions stay nodes at all times. Adjacent loops have opposite signs in either snapshot. The snapshots are not two different modes.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
A three-loop string
- A 0.90 m string fixed at both ends vibrates with three equal loops.
- Its length contains three half wavelengths: λ = 2 × 0.90/3 = 0.60 m.
- At wave speed 120 m s⁻¹, f = 120/0.60 = 200 Hz.
Three loops correspond to the third harmonic, not three full wavelengths.
Keep this idea. Identify boundaries and count half wavelengths; fixed zero amplitude defines a node.
Core practical 5: what sets a string’s frequency?
The speed of a transverse wave on a thin stretched string is √(T/μ), where T is tension and μ is mass per unit length. Combining this with the fundamental fixed-end condition gives f = √(T/μ)/(2L). Tension increases the restoring influence; greater mass per length increases inertia. The relationship assumes small transverse displacement, a flexible uniform string and tension that stays approximately constant during vibration.
Use a signal generator and vibration generator to drive a string with a controlled vibrating length. A hanging mass over a low-friction pulley can provide tension approximately mg when the mass is stationary. Vary length at fixed tension and string, tension at fixed length and string, then linear density using different strings while controlling the other quantities. Measure a long sample’s mass and length to determine μ rather than attempting to weigh a tiny segment.
Find a clear fundamental resonance, reduce drive amplitude if nonlinear motion or excessive stretching occurs, and repeat the frequency search from above and below. Test f against 1/L, f² against T, or f² against 1/μ rather than assuming every curved raw graph disproves the theory. A pulley with friction or a heavy hanging section can make mg an imperfect model of tension. Keep hanging masses low over a catch tray and secure the generator and supports.
v = √(T/μ); μ = m/Lsample; f₁ = (1/2L)√(T/μ)
Symbols and units
- T is string tension in N (here not period); μ is mass per length in kg m⁻¹; L is vibrating length in m.
- f₁ is fundamental frequency in Hz; m and Lsample are sample mass in kg and sample length in m.
How the equation explains the idea
- Find μ = m/Lsample, converting grams to kilograms first. The small-amplitude uniform-string model gives v = √(T/μ), then f₁ = v/(2L).
- Square both sides to obtain f₁² = T/(4L²μ). At fixed L and μ, f₁² versus T has gradient 1/(4L²μ); quadrupling tension doubles frequency.
Conditions. Fundamental of a uniform fixed–fixed string: L = 0.50 m, μ = 0.0040 kg m⁻¹, small-amplitude model. The zero-tension point is the mathematical extrapolation; a slack string does not support this model.
Tension / N; Frequency squared / Hz².
The line f² = T/(4L²μ) has slope 250 Hz² N⁻¹. At 16 N, f² = 4000 Hz², hence f ≈ 63.2 Hz. Frequency itself is proportional to √T, not to T.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
A heavier string at the same tension
- A string with T = 16 N and μ = 0.0040 kg m⁻¹ has speed √4000 = 63.2 m s⁻¹.
- For a vibrating length 0.80 m, fundamental frequency is 63.2/(2 × 0.80) = 39.5 Hz.
- Replacing it with four times the linear density halves wave speed and frequency if tension and length are unchanged.
The square-root dependence matters: four times μ gives half the frequency.
Keep this idea. Change one factor at a time and linearise the square-root model before judging the evidence.
Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.
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