Wavefronts, the principle of superposition, phase and phase difference

Coherent sources, path difference and two-source interference

Turning a path difference into a phase difference, and deciding between a maximum and a minimum

How a stationary wave forms; nodes, antinodes and displacement–distance graphs at different instants

Harmonics on a string fixed at both ends

Harmonics in pipes closed at one end and open at both ends

The speed of a transverse wave on a string, v = √(T/μ)

Core Practical 5: how length, tension and mass per unit length affect the frequency of a vibrating string

1. When two or more waves meet at a point, the resultant displacement is the vector sum of the individual displacements, and this single rule (superposition) explains interference, stationary waves and the harmonics of strings and pipes.

2. Coherent sources have a constant phase difference, so they have the same frequency. Only coherent waves give a steady interference pattern.

3. For sources in phase: path difference = nλ gives a maximum; path difference = (n + ½)λ gives a minimum.

4. Phase difference = 2π× (path difference)/λ, in radians.

5. A stationary wave forms when two waves of the same frequency travel in opposite directions and superpose. Adjacent nodes are λ/2 apart; antinodes lie midway between them.

6. String fixed at both ends: L = nλ/2, all harmonics. Pipe closed at one end: L = λ/4, 3λ/4, …, odd harmonics only. Pipe open at both ends: L = nλ/2, all harmonics.

7. Speed of a transverse wave on a string: v = √(T/μ), so the fundamental frequency is f = 1/2L√(T/μ).

Where two sets of ripples cross, the water does something neither set would do alone, yet afterwards each set carries on unchanged. One simple rule describes the overlap.

"Vector sum" here means that displacements carry a sign. A displacement of +3.0 mm from one wave and −2.0 mm from another add to give +1.0 mm. The waves themselves pass through each other unchanged; superposition only describes what the medium does while they overlap.

Constructive superposition: the waves arrive in phase, crest on crest, and the resultant amplitude is the sum of the amplitudes.

Destructive superposition: the waves arrive in antiphase, crest on trough, and the resultant amplitude is the difference of the amplitudes. It is zero only if the two amplitudes are equal.

Phase, phase difference and wavefronts

Phase describes where a point is in its cycle of oscillation; one whole cycle is 2π rad (360°). Phase difference is how far one oscillation is ahead of or behind another, as a fraction of a cycle, in radians or degrees.

In phase: phase difference 0, 2π, 4π, … The oscillations reach their peaks together.

Antiphase: phase difference π, 3π, 5π, … One is at a peak when the other is at a trough.

A wavefront is a line or surface joining neighbouring points that are in phase, such as a line drawn along one crest. The direction of travel is at right angles to the wavefronts, and adjacent wavefronts drawn crest to crest are one wavelength apart. Plane waves have straight, parallel wavefronts; a small source gives circular wavefronts spreading out from it.

If the phase difference between two overlapping waves keeps changing, maxima and minima move about too quickly to be seen or heard. A steady pattern needs sources that stay in step.

Two loudspeakers driven by one signal generator, or two dippers on one vibrating bar in a ripple tank, are coherent. Two separate lamps are not.

What you detect at a point depends on the path difference: the difference between the distances from the two sources to the point. Each extra wavelength of path puts one wave a whole cycle behind, back in step; an extra half wavelength puts crest on trough.

From path difference to phase difference

Because one wavelength of path is one whole cycle, 2π rad, the conversion is a simple proportion:

The phase difference is in radians, and the path difference and wavelength must be in the same unit. This relationship is not in the formulae list, so learn it. For two sources that are in phase:

path difference = nλ → phase difference 2nπ → maximum (constructive);

path difference = (n + ½)λ → phase difference (2n + 1)π → minimum (destructive), where n = 0, 1, 2, …

If the two sources are themselves in antiphase, the conditions swap: whole wavelengths of path difference give minima and half wavelengths give maxima. So always check how the sources are driven before naming a maximum or a minimum.

WORKED EXAMPLE · QUESTION

Two dippers fixed to the same vibrating bar touch the water in a ripple tank. The bar vibrates at 12 Hz and the ripples travel at 0.30 m s⁻¹. Point P is 16.0 cm from one dipper and 9.75 cm from the other. Point Q is 13.5 cm from one dipper and 21.0 cm from the other. Determine whether the water at P and at Q oscillates with a large or a small amplitude.

SOLUTION

Wavelength: λ= v/f = 0.30 ÷12 = 0.025 m = 2.5 cm.

Path differences: P: 16.0 - 9.75 = 6.25 cm = 2.5λ. Q: 21.0 - 13.5 = 7.5 cm = 3.0λ.

Phase differences: the dippers move together, so they are in phase. P: 2π×2.5 = 5π rad (antiphase). Q: 2π×3.0 = 6π rad (in phase).

P is a minimum (small amplitude);

Q is a maximum (large amplitude). The amplitude at P is small rather than exactly zero, because the ripple from the nearer dipper has spread out less and so has the larger amplitude.

WORKED EXAMPLE · QUESTION

Two loudspeakers S₁ and S₂, 1.80 m apart, are connected to the same signal generator set to 590 Hz. A listener stands at P, 7.32 m from S₁ and 6.46 m from S₂. The speed of sound is 340 m s⁻¹. (a) Calculate the phase difference between the two waves at P and describe what the listener hears. (b) Calculate the lowest frequency at which the listener would hear a maximum at P.

SOLUTION

Wavelength: λ= 340 ÷590 = 0.576 m.

(a) Path difference = 7.32 - 6.46 = 0.86 m, which is 0.86 ÷0.576 = 1.49 wavelengths. Phase difference = 2π×0.86 ÷0.576 = 9.38 rad. Removing one whole cycle: 9.38 - 2π= 3.09 rad ≈π. The waves are in antiphase at P, so the listener is in a quiet region (a minimum).

(b) A path difference of zero is not possible at P, so the lowest frequency for a maximum makes the path difference exactly one wavelength: λ= 0.86 m, f = 340 ÷0.86 = 395 Hz.

Send a continuous wave along a rope tied to a wall: it reflects, so the rope carries two waves of the same frequency moving in opposite directions, and their sum no longer travels.

At some points the two waves always arrive in antiphase and cancel. These are nodes, points of zero amplitude.

Midway between nodes the waves arrive in phase and the amplitude is greatest. These are antinodes.

Adjacent nodes are λ/2 apart, and so are adjacent antinodes. A node and the next antinode are λ/4 apart.

Here λ is the wavelength of the two travelling waves, so a measured node spacing must be doubled to give it.

Displacement–distance graphs of a stationary wave

A displacement–distance graph is a snapshot. For a stationary wave, every snapshot has the same nodes and fits inside the same envelope; only the size and sign of each loop change.

Twice in every period the whole string is momentarily straight (zero displacement everywhere). The energy is then all kinetic.

All points between two adjacent nodes oscillate in phase but with different amplitudes. Points in adjacent loops are in antiphase.

So on a stationary wave the phase difference between two points is either 0 or π, unlike a progressive wave, where it changes steadily with distance.

A string clamped at both ends has a node at each end, so a whole number of loops, each λ/2 long, must span the length L:

The lowest frequency, with one loop, is the fundamental or first harmonic. Every whole-number multiple of it is possible; these are the harmonics. Here v is the wave speed on the string in m s⁻¹, which does not change when you change the frequency. These results are not in the formulae list: derive them from "a whole number of λ/2 fits in L" together with v = fλ, which is in the list.

At any other frequency the reflected waves arrive out of step and do not reinforce, so only a small, irregular motion appears; at a harmonic, each reflection adds to the motion and a large steady amplitude builds up.

WORKED EXAMPLE · QUESTION

A rubber cord is stretched between two fixed points 1.80 m apart and driven near one end. At 24 Hz a steady stationary wave with three loops appears. (a) Calculate the wavelength and the speed of the waves on the cord. (b) Calculate the fundamental frequency and the frequency at which four loops would appear.

SOLUTION

(a) Three loops are three half wavelengths: λ= 2 ×1.80 ÷3 = 1.20 m. Then v = fλ= 24 ×1.20 = λ = 1.20 m, v = 28.8 m s⁻¹.

(b) The wave speed is unchanged. The fundamental has one loop, so λ= 2L = 3.60 m and f₁ = 28.8 ÷3.60 = 8.0 Hz. Four loops: 4 ×8.0 = 32 Hz. f₁ = 8.0 Hz; four loops at 32 Hz.

Quick check: 24 Hz is the third harmonic, so the fundamental must be 24 ÷3 = 8.0 Hz.

Sound reflects at both ends of a pipe, so a stationary wave forms in the air column. Sound is longitudinal, so this is a pattern of air displacement along the pipe. Diagrams draw the envelope sideways to show the size of the motion.

At a closed end the air cannot move: a displacement node.

At an open end the air moves freely: a displacement antinode. In reality it sits just beyond the end; ignore this end correction unless told otherwise.

Closed at one end: the shortest pattern is node to antinode, a quarter wavelength. Adding half a wavelength at a time gives L = λ/4, 3λ/4, 5λ/4, … Only odd harmonics are possible: f₁, 3f₁, 5f₁, … with f₁ = v/4L.

Open at both ends: antinode to antinode, so L = λ/2, λ, 3λ/2, … All harmonics are possible: f₁, 2f₁, 3f₁, … with f₁ = v/2L, the same rule as a string.

WORKED EXAMPLE · QUESTION

A plastic tube has length 0.320 m. The speed of sound in the air inside is 340 m s⁻¹. Ignore end corrections. (a) One end is closed with a cap. Calculate the fundamental frequency and the next two frequencies at which the tube resonates. (b) The cap is removed. Calculate the new fundamental frequency.

SOLUTION

(a) Closed at one end, the fundamental has L = λ/4: λ= 4 ×0.320 = 1.28 m, so f₁ = 340 ÷1.28 = 266 Hz (265.6 Hz). Only odd harmonics are possible: 3f₁ = 797 Hz and 5f₁ = 1330 Hz. 266 Hz, 797 Hz, 1330 Hz

(b) Open at both ends, the fundamental has L = λ/2: λ= 2 ×0.320 = 0.64 m, so f₁ = 340 ÷0.64 = 531 Hz.

Closing one end halves the fundamental and removes the even harmonics; the open tube has 531 Hz, 1060 Hz, 1590 Hz …

The speed of waves on a string is set by the tension and the mass per unit length.

Here v is the wave speed (m s⁻¹), T the tension (N) and μ the mass per unit length (kg m⁻¹). This equation is in the formulae list supplied in the exam. The mass per unit length is μ= mass ÷ length. For a wire of density ρ and cross-sectional area A, μ= ρA; learn this step, because it is not given.

A tighter string pulls each displaced section back more strongly, so the disturbance passes on faster: v ∝√T.

A heavier string responds more slowly to the same tension: v ∝1/√μ.

Combining v = √(T/μ) with the fundamental of a string (λ= 2L) and v = fλ gives the fundamental frequency of a stretched string:

This is not in the formulae list: derive it from v = fλ, λ= 2L and v = √(T/μ). It shows three proportionalities that the core practical tests: f ∝1/L at constant T and μ; f ∝√T at constant L and μ; f ∝1/√μ at constant L and T.

WORKED EXAMPLE · QUESTION

A steel piano wire has diameter 0.80 mm and a vibrating length of 0.95 m. The density of steel is 7800 kg m⁻³. The tension in the wire is 620 N. (a) Calculate the fundamental frequency of the wire. (b) Calculate the tension needed to raise the fundamental frequency to 220 Hz.

SOLUTION

Mass per unit length: radius = 0.40 ×10⁻³ m, so A = π(0.40 ×10⁻³)² = 5.03 ×10⁻⁷ m² and μ= ρA = 7800 ×5.03 ×10⁻⁷ = 3.92 ×10⁻³ kg m⁻¹.

Wave speed: v = √(620 ÷3.921 ×10⁻³) = 398 m s⁻¹.

(a) Fundamental, λ= 2L: f = 398 ÷(2 ×0.95) = 209 Hz.

(b) With L and μ fixed, f ∝√T, so T ∝f²: T = 620 ×(220 ÷209.3)² = 685 N.

Core Practical 5 tests the relationship f = 1/2L√(T/μ) by changing one variable at a time and finding the fundamental frequency each time.

Set-up: as in the figure; slotted masses on a hanger give T = mg.

Finding resonance: raise the frequency slowly until one loop has the largest amplitude; approach from above too and take the midpoint. Read f from the signal generator.

Length: keep T and the string fixed; move the vibration generator to change L (metre rule). Use at least five lengths.

Tension: keep L and the string fixed and change the hanging mass.

Mass per unit length: keep L and T fixed and use strings of different μ (weigh a measured length of each).

Graphs: f against 1/L, f against √T (or f² against T), and f against 1/√μ should each be a straight line through the origin.

Safety: keep feet clear of the masses (padding underneath) and wear eye protection in case the string snaps.

WORKED EXAMPLE · QUESTION

A student keeps the hanging mass at 0.400 kg and measures the fundamental frequency f for five lengths. Her results: L = 0.400 m, 71 Hz; 0.500 m, 58 Hz; 0.600 m, 47 Hz; 0.700 m, 41 Hz; 0.800 m, 36 Hz. The signal generator displays the frequency to the nearest 1 Hz. g = 9.81 N kg⁻¹. (a) Explain why a graph of f against 1/L should be a straight line through the origin. (b) Use the gradient of her graph to determine the mass per unit length of the string. (c) Calculate the percentage uncertainty in the frequency reading of 36 Hz.

SOLUTION

(a) Rearranging, f = ½√(T/μ) ×(1/L). With T and μ constant this has the form y = mx, so f against 1/L is a straight line through the origin with gradient ½√(T/μ).

(b) The 1/L values are 2.50, 2.00, 1.67, 1.43 and 1.25 m⁻¹. Tension T = 0.400 ×9.81 = 3.92 N. Gradient of the best-fit line = 28.6 Hz m. From gradient = ½√(T/μ): μ= T ÷(4 ×gradient²) = 3.924 ÷(4 ×28.6²) = 1.20 × 10⁻³ kg m⁻¹.

(c) Uncertainty = half the resolution of the display, ±0.5 Hz: 0.5 ÷36 ×100 = 1.4%.

Judging the frequency of largest amplitude is usually a bigger uncertainty than the display; repeat readings from both sides and average.

1 Progressive waves

Every calculation here starts from v = fλ, and the phase difference along a progressive wave (2π per wavelength) is the idea that path difference builds on.

4 Diffraction and the wave nature of electrons

A diffraction grating is many coherent sources at once. The condition for a bright line, nλ= dsin θ, is the "path difference = whole number of wavelengths" rule from this page.

Unit 1 forces and materials, and Unit 5 oscillations

The tension in the practical comes from a hanging weight, T = mg, and the mass per unit length of a wire from its density and area, μ= ρA. A string or pipe driven at one of its harmonic frequencies is an example of resonance (Unit 5): a driving frequency that matches a natural frequency gives a large amplitude.

1. What is meant by two sources being coherent?

2. Two in-phase sources emit waves of wavelength 0.30 m. At a point the path difference is 1.05 m. Is the point a maximum or a minimum?

3. Adjacent nodes on a stationary wave are 18 cm apart and the frequency is 250 Hz. Calculate the wave speed.

4. A pipe closed at one end has a fundamental frequency of 125 Hz. State the next two resonant frequencies.

5. The tension in a string is made four times larger, with the length and string unchanged. What happens to the fundamental frequency?

6. Two points on a stationary wave are 0.10 m and 0.30 m from a node at one end. The wavelength is 0.80 m. State the phase difference between them.

1. They emit waves with a constant phase difference between them, which means they must also have the same frequency.

2. 1.05 ÷0.30 = 3.5 wavelengths, a whole number plus a half, so the waves arrive in antiphase: a minimum.

3. λ= 2 ×0.18 = 0.36 m; v = fλ= 250 ×0.36 = 90 m s⁻¹.

4. Only odd harmonics are possible: 3 ×125 = 375 Hz and 5 ×125 = 625 Hz.

5. f ∝√T, so the frequency is multiplied by √4 = 2: it doubles.

6. Nodes are λ/2 = 0.40 m apart, at 0, 0.40 m, 0.80 m … Both points lie between the nodes at 0 and 0.40 m, so they are in the same loop and oscillate in phase: phase difference 0 (not 2π×0.20/0.80 = π/2, which is the progressive-wave rule).