Principle of Superposition and Stationary Waves

Understanding how waves combine and create standing wave patterns

SPEC 3.3.1.3 • PROGRESSIVE AND STATIONARY WAVES

Principle of Superposition

Definition

When two or more waves meet at a point, the resultant displacement at that point equals the vector sum of the individual displacements due to each wave at that point.

Unlike particles, waves can pass through each other. When they overlap, they combine momentarily then continue unchanged.

Key Concept

The superposition is instantaneous - after passing through each other, waves continue unchanged.

Types of Interference

Constructive

Waves in phase (phase difference = 0)

Path difference = nλ (whole wavelengths)

Displacements add → larger amplitude

Destructive

Waves in antiphase (phase difference = π)

Path difference = (n + ½)λ (half wavelengths)

Displacements cancel → zero amplitude

Exam Tip

For total destructive interference, waves must have equal amplitudes and be exactly in antiphase (π radians or 180°).

Formation of Stationary Waves

Stationary Wave

A wave pattern formed when two progressive waves of the same frequency, wavelength, and amplitude travel in opposite directions in the same plane and superpose.

Requirements for Formation

Two progressive waves travelling in opposite directions

Same frequency (and therefore same wavelength)

Same amplitude (or similar)

Important

No energy is transferred by a stationary wave - energy is stored, oscillating between kinetic and potential forms.

Worked Example01

Two waves meet at a point. Wave A has displacement +3.0 cm and wave B has displacement +5.0 cm. Calculate the resultant displacement. What if wave A had displacement -3.0 cm instead?

1

Apply superposition: resultant = vector sum of displacements

2

Case 1 (same direction): (+3.0) + (+5.0) = +8.0 cm

3

Case 2 (opposite): (-3.0) + (+5.0) = +2.0 cm

Answer+8.0 cm (constructive) and +2.0 cm (partial cancellation)

Nodes and Antinodes

Node (N)

Point of zero displacement at all times

Formed by destructive interference

Particles at nodes do not move

Antinode (A)

Point of maximum amplitude

Formed by constructive interference

Particles oscillate with maximum displacement

Key Relationships

Distance between adjacent nodes = λ/2 | Distance between adjacent antinodes = λ/2 | Node to nearest antinode = λ/4

Worked Example02

A stationary wave has a distance of 0.40 m between adjacent nodes. Calculate: (a) the wavelength, (b) the distance from a node to the nearest antinode.

1

Distance between adjacent nodes = λ/2

2

Part (a): λ/2 = 0.40 m → λ = 2 × 0.40 = 0.80 m

3

Part (b): Node to antinode = λ/4 = 0.80/4 = 0.20 m

Answer(a) λ = 0.80 m, (b) 0.20 m

Stationary Waves on Strings

A string fixed between two points produces stationary waves when plucked. Waves travel along the string and reflect at fixed ends.

Phase Change on Reflection

When a wave reflects at a fixed end, there is a phase change of π radians (180°). The reflected pulse is inverted.

Conditions for Stationary Wave Formation

Two waves must have: same frequency (and wavelength), same amplitude, and be travelling in opposite directions.

Boundary Condition

Fixed ends must be nodes (zero displacement) - the string cannot move at fixed points.

Frequency of Vibrating String

The frequency of a stationary wave on a string depends on three factors:

First Harmonic Frequency

f = \frac{1}{2 L} \frac{T}{μ}

SymbolQuantityUnit

fFrequencyHz

LLength of stringm

TTension in stringN

μ (mu)Mass per unit lengthkg m⁻¹

Typical Mistake

Using the total mass instead of mass per unit length (μ). Always check: μ must be in kg m⁻¹, not kg!

Harmonics on a String

Key Formulas

λ = \frac{2 L}{n}

f_{n} = n f_{0}

f_{n} = \frac{n}{2 L} \frac{T}{μ}

n = harmonic number (1, 2, 3...), f₀ = fundamental frequency

Worked Example03

A guitar string of length 0.65 m vibrates in its fundamental mode. The speed of waves on the string is 330 m s⁻¹. Calculate the fundamental frequency.

1

For fundamental (n = 1): λ = 2L = 2 × 0.65 = 1.30 m

2

Use v = fλ, rearrange: f = v/λ

3

f = 330/1.30 = 254 Hz

Answerf₀ = 254 Hz

Worked Example04

A 70 cm long stretched string vibrates in its second harmonic mode. Tension = 200 N, mass per unit length = 1.2 g m⁻¹. Calculate the frequency.

1

Convert: L = 0.70 m, μ = 1.2 × 10⁻³ kg m⁻¹

2

Use:

f = \frac{n}{2 L} \frac{T}{μ}

with n = 2

3

f = \frac{2}{2 \times 0.70} \frac{200}{1.2 \times 1 0 ^{- 3}} = \frac{1}{0.70} 166667

4

f = (1/0.70) × 408 = 583 Hz

Answerf = 583 Hz

Worked Example05

A string has fundamental frequency 200 Hz. Calculate the 2nd and 3rd harmonic frequencies. How many nodes in the 3rd harmonic?

1

Harmonic frequencies: fₙ = n × f₀

2

2nd harmonic: f₂ = 2 × 200 = 400 Hz

3

3rd harmonic: f₃ = 3 × 200 = 600 Hz

4

Nodes in nth harmonic = n + 1 → 3rd harmonic: 4 nodes

Answerf₂ = 400 Hz, f₃ = 600 Hz, 4 nodes

Stationary Waves in Pipes

Sound waves form stationary waves in air columns. The pattern depends on open/closed ends.

Closed End

Must be a NODE - air cannot move

Open End

Must be an ANTINODE - air free to vibrate

Pipe TypeFundamental λHarmonics

Closed (one end)λ = 4LOdd only (1, 3, 5...)

Open (both ends)λ = 2LAll (1, 2, 3...)

Worked Example06

A closed pipe has length 0.50 m. Speed of sound = 340 m s⁻¹. Calculate: (a) fundamental frequency, (b) frequency of next possible harmonic.

1

Closed pipe fundamental: λ = 4L = 4 × 0.50 = 2.0 m

2

Part (a): f₀ = v/λ = 340/2.0 = 170 Hz

3

Part (b): Closed pipes have ODD harmonics only

4

Next = 3rd harmonic: f₃ = 3 × 170 = 510 Hz

Answer(a) 170 Hz, (b) 510 Hz

Experimental Demonstrations

Stationary Microwaves

Setup: Microwave transmitter reflects off metal plate

Detection: Move probe between transmitter and plate

Result: Signal varies: max at antinodes, min at nodes

Measure: Distance between nodes = λ/2

Stationary Sound (Kundt's Tube)

Setup: Speaker at one end of glass tube

Detection: Sprinkle powder (cork dust) along tube

Result: Powder collects at nodes, shaken from antinodes

Use: Find speed of sound: v = fλ

Real-World Application

A motorist driving between two transmitters notices radio signal varying in strength. If maxima occur every 5 s at 30 m s⁻¹:

λ = 2 × (30 × 5) = 300 m (distance between maxima = λ/2)

Stationary vs Progressive Waves

PropertyProgressive WaveStationary Wave

Energy TransferEnergy transferred along waveNo net energy transfer

AmplitudeSame for all particlesVaries: zero (nodes) to max (antinodes)

PhaseVaries continuously along waveAll particles between nodes in phase

WavelengthDistance between in-phase pointsTwice distance between adjacent nodes

FrequencyAll particles same frequencyAll particles same frequency

Exam Focus

A common question asks you to compare these wave types. Key differences: energy transfer, amplitude variation, phase relationships.

Practice

Question 1: A string of length 1.2 m is fixed at both ends and vibrates in its 3rd harmonic. Calculate: (a) the wavelength, (b) the number of antinodes, (c) the frequency if wave speed is 180 m s⁻¹.

a)

λ = 2L/n = 2(1.2)/3 = 0.80 m

b)

Number of antinodes = n = 3

c)

f = v/λ = 180/0.80 = 225 Hz

Practice

Question 2: A closed pipe and open pipe have the same length L. If the open pipe's fundamental frequency is 400 Hz, what is the closed pipe's fundamental frequency?

1

Open pipe: λ = 2L → f_open = v/(2L)

2

Closed pipe: λ = 4L → f_closed = v/(4L)

3

Ratio: f_closed/f_open = (v/4L)/(v/2L) = 1/2

4

f_closed = 400 × (1/2) = 200 Hz

Summary

Key Concepts

ConceptKey Point

SuperpositionResultant = vector sum of individual displacements

Stationary waveTwo progressive waves, same f, λ, A, opposite directions; no energy transfer

Node separationλ/2 between adjacent nodes or antinodes

String formula

f = \frac{1}{2 L} \frac{T}{μ}

Harmonicsfₙ = nf₀; λ = 2L/n for string fixed both ends

Closed pipeλ = 4L; odd harmonics only

Open pipeλ = 2L; all harmonics

Key Exam Skills

Use f = (1/2L)√(T/μ) for string calculations with tension

Distinguish open vs closed pipe harmonics

Remember phase change of π at fixed boundaries