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A wave is a disturbance that transfers energy from one place to another without transferring matter. Understanding the fundamental properties of waves — and the distinction between transverse and longitudinal types — is essential to every topic in this course.
There are two fundamental categories of wave, classified by the direction of oscillation relative to the direction of energy transfer.
In a transverse wave, the oscillations are perpendicular to the direction of energy transfer (the direction of wave propagation).
Examples include:
A key property of transverse waves is that they can be polarised — the oscillations can be restricted to a single plane. This is explored in detail in Lesson 9.
In a longitudinal wave, the oscillations are parallel to the direction of energy transfer. The medium undergoes alternating compressions (regions of high pressure/density) and rarefactions (regions of low pressure/density).
Examples include:
Longitudinal waves cannot be polarised, because the oscillation direction is already constrained to be along the propagation direction.
Exam Tip: If asked how you can distinguish transverse from longitudinal waves experimentally, state that only transverse waves can be polarised. This is the definitive test.
Every periodic wave is described by a set of measurable quantities. These must be defined precisely for full marks in an exam.
| Quantity | Symbol | Definition | SI Unit |
|---|---|---|---|
| Amplitude | A | Maximum displacement of a point on the wave from its equilibrium (rest) position | m |
| Wavelength | λ | Minimum distance between two points oscillating exactly in phase (e.g., crest to crest) | m |
| Frequency | f | Number of complete oscillations passing a given point per unit time | Hz (s⁻¹) |
| Period | T | Time taken for one complete oscillation | s |
| Wave speed | v | Distance travelled by the wave per unit time | m s⁻¹ |
Frequency and period are reciprocals of each other:
f = 1/T and equivalently T = 1/f
Worked Example 1 — A sound wave has a period of 2.5 ms. Calculate its frequency.
T = 2.5 ms = 2.5 × 10⁻³ s
f = 1/T = 1/(2.5 × 10⁻³) = 400 Hz
The fundamental equation relating wave speed, frequency, and wavelength is:
v = f λ
This can be derived from first principles. In one complete period T, the wave advances by exactly one wavelength λ. Since speed = distance/time:
v = λ/T = λ × (1/T) = f λ
This equation applies to all waves — mechanical and electromagnetic, transverse and longitudinal.
Worked Example 2 — A radio wave has a frequency of 98.5 MHz. Calculate its wavelength.
v = c = 3.00 × 10⁸ m s⁻¹ (radio waves are electromagnetic)
f = 98.5 MHz = 98.5 × 10⁶ Hz
λ = v/f = (3.00 × 10⁸)/(98.5 × 10⁶) = 3.05 m
Worked Example 3 — A guitar string vibrates at 440 Hz and produces a sound wave in air with wavelength 0.773 m. Calculate the speed of sound in air.
v = f λ = 440 × 0.773 = 340 m s⁻¹ (to 3 s.f.)
Worked Example 4 — Ultrasound of frequency 2.0 MHz travels through soft tissue at 1540 m s⁻¹. Calculate the wavelength.
λ = v/f = 1540/(2.0 × 10⁶) = 7.7 × 10⁻⁴ m = 0.77 mm
Exam Tip: Always convert units before substituting into the wave equation. Frequencies in MHz or kHz must be converted to Hz; wavelengths in nm or mm must be converted to m.
The phase of a point on a wave describes its position within the oscillation cycle. Phase is measured in radians (rad) or degrees (°), where one complete cycle = 2π rad = 360°.
The phase difference between two points on a wave (or between two waves) describes how far one oscillation is ahead of or behind the other.
| Phase difference | In degrees | In radians | Meaning |
|---|---|---|---|
| In phase | 0° (or 360°) | 0 (or 2π) | Points oscillate together — same displacement at all times |
| Antiphase | 180° | π | Points have equal but opposite displacements at all times |
| Quarter cycle ahead | 90° | π/2 | One point leads the other by a quarter of a wavelength |
If two points on a wave are separated by a distance Δx along the direction of propagation, their phase difference Δφ is:
Δφ = (2π/λ) × Δx
Equivalently, the path difference in terms of wavelength determines the phase relationship:
Worked Example 5 — Two points on a wave of wavelength 0.60 m are separated by 0.45 m. What is their phase difference?
Δφ = (2π/λ) × Δx = (2π/0.60) × 0.45 = (2π × 0.75) = 1.5π rad
Convert: 1.5π rad = 270° (or equivalently 3π/2 rad)
The path difference is 0.75λ, which is neither a whole number nor a half-integer multiple of λ, so the points are neither in phase nor in antiphase.
Worked Example 6 — Two loudspeakers emit sound of wavelength 0.50 m in phase. A listener is 3.00 m from one speaker and 3.75 m from the other. What does the listener hear?
Path difference = 3.75 − 3.00 = 0.75 m
Path difference in wavelengths = 0.75/0.50 = 1.5λ
Since 1.5λ = (1 + ½)λ, this is a half-integer multiple of λ. The waves arrive in antiphase, producing destructive interference — the listener hears a quiet sound (a minimum).
Waves can be represented graphically in two important ways:
This is a snapshot of the wave at one moment in time, plotting displacement (y-axis) against position along the wave (x-axis). From this graph you can read:
This shows how the displacement of one point varies with time. From this graph you can read:
Common Misconception: Students often confuse wavelength and period. Wavelength is a spatial quantity (metres) read from a displacement-distance graph; period is a temporal quantity (seconds) read from a displacement-time graph.
The intensity of a wave is the power transmitted per unit area, measured perpendicular to the direction of energy transfer:
I = P/A
where I is intensity (W m⁻²), P is power (W), and A is the area (m²) through which the wave passes.
For a point source radiating equally in all directions, the power spreads over the surface of a sphere of radius r:
I = P/(4πr²)
This gives the inverse square law: intensity is inversely proportional to the square of the distance from the source.
Intensity is proportional to the square of the amplitude:
I ∝ A²
This means that doubling the amplitude quadruples the intensity.
Worked Example 7 — A point source emits sound with a power of 0.50 W. Calculate the intensity at a distance of 4.0 m from the source.
I = P/(4πr²) = 0.50/(4π × 4.0²) = 0.50/(4π × 16) = 0.50/201.1 = 2.5 × 10⁻³ W m⁻²
Worked Example 8 — At a distance of 2.0 m from a source, the intensity of a sound wave is 0.080 W m⁻². What is the intensity at 8.0 m?
Using the inverse square law: I₁r₁² = I₂r₂²
I₂ = I₁ × (r₁/r₂)² = 0.080 × (2.0/8.0)² = 0.080 × (0.25)² = 0.080 × 0.0625 = 5.0 × 10⁻³ W m⁻²
The distance quadrupled, so the intensity decreased by a factor of 16.