Lesson 7: Wave Properties

Waves transfer energy without transferring matter, and a single mathematical model — Theme C.2 — covers ripples on water, sound, light and even matter waves. This lesson builds the model: the quantities that describe a wave, the two great families (travelling and standing), and what happens at boundaries (reflection, refraction, diffraction).

Describing waves

A wave is characterised by wavelength λ (distance between two adjacent in-phase points), frequency f (oscillations per second at a point), period T = 1/f, amplitude (maximum displacement from equilibrium; intensity ∝ amplitude²) and speed c given by the wave equation v = fλ — the single most-used wave formula in the course. Displacement–distance graphs are snapshots of the whole wave at one instant; displacement–time graphs show one point oscillating. Confusing these two graphs is the classic Paper 1 trap: λ is read from the first, T from the second.

Transverse waves (light, water surface, string) oscillate perpendicular to the energy transfer; longitudinal waves (sound) oscillate parallel to it, forming compressions and rarefactions. Electromagnetic waves need no medium and all travel at c = 3.00 × 10⁸ m s⁻¹ in vacuum; mechanical waves need a medium and usually travel faster in denser, stiffer media.

Reflection, refraction and diffraction

At a boundary, frequency never changes (it is fixed by the source); speed and wavelength may. Refraction follows Snell's law n₁ sin θ₁ = n₂ sin θ₂, with refractive index n = c/v. Entering a denser medium (larger n), the wave slows, λ shortens, and the ray bends towards the normal. Above the critical angle θc = arcsin(1/n) for light leaving glass or water, total internal reflection occurs — the physics of optical fibres. Diffraction — spreading at apertures and around obstacles — is significant when the aperture is comparable to λ; the first minimum for a single slit is at sin θ = λ/b (developed in Lesson 21).

Worked problem 1 — the wave equation and strings

A guitar string vibrates at 440 Hz. The speed of transverse waves on the string is 132 m s⁻¹. (a) Find the wavelength on the string. (b) The sound produced travels through air at 343 m s⁻¹; find its wavelength in air.

(a) λ = v/f = 132/440 = 0.30 m   (the string's fundamental: string length = λ/2 = 0.15 m? No —
    the wavelength of the travelling waves that superpose is 0.30 m; standing wave nodes are λ/2 apart)
(b) λ_air = 343/440 = 0.78 m

The frequency is the same in both media — pitch is a property of the source; the wavelength adapts to the speed. This is why a diver hears a different "size" of sound wave underwater but the same note.

Worked problem 2 — refraction into glass

Light of wavelength 589 nm in air strikes glass (n = 1.50) at 48° to the normal. Find (a) the refraction angle, (b) the wavelength and speed in the glass, (c) the critical angle for glass–air.

(a) sin θ₂ = sin 48° / 1.50 = 0.7431/1.50 = 0.495 → θ₂ = 29.7°
(b) v = c/n = 2.00 × 10⁸ m s⁻¹ ; λ = 589/1.50 = 393 nm
(c) sin θc = 1/1.50 → θc = 41.8°

Angles are measured from the normal, never from the surface — check the diagram in every question.

Practical worked example — measuring the speed of sound

A 512 Hz tuning fork is held over a vertical tube whose water level is lowered until resonance is first heard at an air column of length L = 0.163 ± 0.002 m. The closed end is a node and the open end an antinode, so the first resonance is λ/4:

λ ≈ 4L = 0.652 m ;  v = fλ = 512 × 0.652 = 334 m s⁻¹

The percentage uncertainty from L alone is 1.2%, but the accepted value is 343 m s⁻¹ at 20 °C — outside the range. The discrepancy is the end correction: the antinode lies slightly above the tube's mouth (≈ 0.3 × diameter). Recognising and correcting this is what turns a routine practical into a strong IA-style evaluation.

Nature of Science

The wave model and particle model of light competed for two centuries. Newton's corpuscles explained reflection; Huygens's waves explained refraction; Young's double-slit fringes (1801) seemed decisive for waves — until the photoelectric effect (Lesson 14) revived particles. Modern physics keeps both models, each with a domain: wave–particle duality is a lesson in how evidence, not authority, settles scientific disputes.

Real-world application

Medical ultrasound uses reflections at tissue boundaries (the same impedance-mismatch physics as partial reflection at a string joint). Frequencies of 2–18 MHz are chosen because shorter λ gives finer resolution, at the cost of shallower penetration — a genuine engineering trade-off between two wave properties.

Common mistakes (examiner style)

• Reading the period from a displacement–distance graph or the wavelength from a displacement–time graph.
• Changing the frequency when a wave refracts.
• Measuring angles from the surface instead of the normal.
• Using v = fλ with f in kHz or λ in cm without converting.
• Stating that in total internal reflection the angle of incidence is less than the critical angle (it must be greater).

IB Exam Experience

Paper 1-style questions. 1. A wave has frequency 5.0 Hz and speed 2.0 m s⁻¹. Its wavelength is: A. 0.40 m B. 2.5 m C. 10 m D. 0.10 m. Key: A — λ = v/f = 0.40 m. 2. Light passes from air into water. Which quantity is unchanged? A. speed B. wavelength C. frequency D. direction. Key: C. 3. Sound of frequency 170 Hz travels at 340 m s⁻¹. Two points 1.0 m apart along the wave are: A. in phase B. in antiphase C. a quarter cycle out of phase D. unrelated. Key: B — λ = 2.0 m, so 1.0 m is half a wavelength = antiphase.

Paper 2-style structured question (6 marks). A progressive wave on a rope is described by a displacement–distance graph at one instant: amplitude 0.030 m, wavelength 0.60 m; the wave speed is 1.8 m s⁻¹. (a) Calculate the frequency. [2] (b) A point P is at a peak; state the displacement of P exactly half a period later and justify your answer. [2] (c) The rope is joined to a thinner rope in which the speed is 2.7 m s⁻¹. Calculate the wavelength in the second rope. [2] *Marking notes:* (a) f = v/λ = 1.8/0.60 = 3.0 Hz. (b) −0.030 m (a trough): half a period is a phase change of π, peak becomes trough. (c) f is unchanged: λ₂ = 2.7/3.0 = 0.90 m. Award method marks for stating that frequency is constant across the boundary — the examiners' report singles this out as the discriminator.

IA idea: Measure how the speed of waves on a stretched string depends on tension, testing v ∝ √T with a log–log plot and full uncertainty treatment.

Standing waves: the other great wave family

Superposing two identical waves travelling in opposite directions produces a standing (stationary) wave: nodes (zero displacement at all times) spaced λ/2 apart, antinodes between them, and — crucially — no net energy transfer along the medium. Allowed patterns are quantised by the boundary conditions: a string fixed at both ends supports λₙ = 2L/n, giving harmonics fₙ = nv/(2L); a pipe open at both ends behaves the same way, while a pipe closed at one end supports only odd harmonics, fₙ = (2n − 1)v/(4L), with a node at the closed end and an antinode at the open end. Instruments are distinguished by these harmonic series: the clarinet (closed pipe) sounds hollow because its spectrum lacks even harmonics; the oboe (open pipe) has them all.

Worked extra — harmonics of a guitar string

A string of length 0.65 m has wave speed 130 m s⁻¹. The fundamental is f₁ = v/(2L) = 130/1.30 = 100 Hz; the overtones are 200, 300, 400 Hz. Plucking at the midpoint suppresses even harmonics (they would need an antinode where the string was displaced into a node-shaped start) — which is why plucking position changes the tone. Fingering the string at its midpoint halves L and doubles every frequency: the octave, the most consonant interval, is the physics of Lesson 6's ratios in sound form.

Intensity, amplitude and the decibel scale

Intensity is power per unit area and scales as amplitude²: doubling the amplitude quadruples the intensity. Because sources radiate spherically, I ∝ 1/r² (Lesson 15's law again). Sound levels are quoted in decibels, dB = 10 log₁₀(I/I₀) with I₀ = 10⁻¹² W m⁻²: a factor 10 in intensity is +10 dB, a doubling is +3 dB. A classroom at 60 dB and a busy road at 80 dB differ in intensity by a factor of 100 — logarithmic scales compress enormous ranges, at the price of intuition, which is why examiners test the conversions directly.

Progressive versus standing waves: the comparison examiners want

Progressive waves transfer energy, have all points oscillating (with amplitude constant, phase varying along the wave), and move their pattern at v = fλ. Standing waves store energy, have fixed nodes and antinodes (amplitude varying from point to point, all points between two nodes in phase with each other), and their pattern is motionless. Wavelength is defined in both, but "wave speed" applies only to the progressive parents of the standing wave. This six-point contrast — energy, amplitude profile, phase profile, frequency spread, pattern motion, and the role of boundaries — is the guaranteed content of at least one question in most HL papers and appears in SL as a shorter state-and-explain item.