Wave Fundamentals Investigate how wave speed, frequency, wavelength, and amplitude are related. Experiment with Different end conditions (fixed, loose, open) to observe standing waves and resonance.
What is a Wave? A wave is a disturbance that transfers energy through a medium or space without transferring matter.
Types of Waves Property Transverse Longitudinal Oscillation direction Perpendicular to propagation Parallel to propagation Examples Light, water surface waves, strings Sound, pressure waves Crests and troughs Yes Compressions and rarefactions Polarisation Can be polarised Cannot be polarised
Wave Terminology Term Symbol Definition Unit Displacement x x x y y y Distance from equilibrium m Amplitude A A A Maximum displacement m Wavelength λ \lambda λ Distance between two consecutive identical points m Frequency f f f Number of complete oscillations per second Hz Period T T T Time for one complete oscillation s Wave speed v v v Speed at which the wave propagates m/s Phase ϕ \phi ϕ Position in the cycle of oscillation rad
Relationships F = { 1 } { T } F = \frac\{1\}\{T\} F = 1 { } { T } V = f λ V = f\lambda V = f λ
A sound wave has frequency 440 H z 440\mathrm{ Hz} 440 Hz 343 m / s 343\mathrm{ m/s} 343 m/s
λ = { v } { f } = { 3 43 } { 440 } = 0.780 { m } \lambda = \frac\{v\}\{f\} = \frac\{343\}\{440\} = 0.780\mathrm\{ m\} λ = v { } { f } = 3 { 43 } { 440 } = 0.780 { m } The Wave Equation For a travelling wave:
Y ( x , t ) = A sin ( ω t − k x + ϕ ) Y(x, t) = A\sin(\omega t - kx + \phi) Y ( x , t ) = A sin ( ω t − k x + ϕ ) Where:
ω = 2 π f \omega = 2\pi f ω = 2 π f k = 2 π λ k = \dfrac{2\pi}{\lambda} k = λ 2 π ϕ \phi ϕ Key Relations V = { ω } { k } = f λ V = \frac\{\omega\}\{k\} = f\lambda V = ω { } { k } = f λ Intensity The intensity of a wave is the power per unit area:
I = { P } { A } I = \frac\{P\}\{A\} I = P { } { A } For a point source radiating equally in all directions:
I = { P } { 4 π r 2 } I = \frac\{P\}\{4\pi r^2\} I = P { } { 4 π r 2 } Intensity is proportional to amplitude squared:
I ∝ A 2 I \propto A^2 I ∝ A 2 Sound Waves Nature of Sound Sound is a longitudinal mechanical wave that requires a medium. It consists of compressions (high Pressure) and rarefactions (low pressure).
Speed of Sound Medium Speed (m/s) Air at 20 ° C 20\degree\mathrm C 20° C 343 Water at 20 ° C 20\degree\mathrm C 20° C 1482 Steel 5960 Glass 5640
The speed of sound in air depends on temperature:
V ≈ 331 + 0.6 T C { m / s } V \approx 331 + 0.6T_C \mathrm\{ m/s\} V ≈ 331 + 0.6 T C { m / s } Inverse Square Law For a point source of sound:
I ∝ { 1 } { r 2 } I \propto \frac\{1\}\{r^2\} I ∝ 1 { } { r 2 } Doubling the distance from a source reduces the intensity to one quarter.
Sound Intensity Level Measured in decibels (dB):
β = 10 log { 10 } ( { I } { I 0 } ) \beta = 10\log_\{10\}\!\left(\frac\{I\}\{I_0\}\right) β = 10 log { 10 } ( I { } { I 0 } ) Where I 0 = 10 − 12 W / m 2 I_0 = 10^{-12}\mathrm{ W/m}^2 I 0 = 1 0 − 12 W/m 2
Source Level (dB) Threshold of hearing 0 Whisper 30 Normal conversation 60 Busy traffic 80 Rock concert 120 Jet engine at 30 m 150
A 10 dB increase represents a tenfold increase in intensity. A 3 dB increase approximately doubles the intensity. Decibels are logarithmic, so you cannot add them.
Electromagnetic Spectrum Region Wavelength Range Frequency Range Radio waves > 0.1 m \gt 0.1\mathrm{ m} > 0.1 m < 3 × 10 9 H z \lt 3 \times 10^9\mathrm{ Hz} < 3 × 1 0 9 Hz Microwaves 0.1 m m 0.1\mathrm{ mm} 0.1 mm 0.1 m 0.1\mathrm{ m} 0.1 m 3 × 10 9 3 \times 10^9 3 × 1 0 9 3 × 10 12 H z 3 \times 10^{12}\mathrm{ Hz} 3 × 1 0 12 Hz Infrared 700 n m 700\mathrm{ nm} 700 nm 0.1 m m 0.1\mathrm{ mm} 0.1 mm 3 × 10 12 3 \times 10^{12} 3 × 1 0 12 4.3 × 10 14 H z 4.3 \times 10^{14}\mathrm{ Hz} 4.3 × 1 0 14 Hz Visible light 400 n m 400\mathrm{ nm} 400 nm 700 n m 700\mathrm{ nm} 700 nm 4.3 × 10 14 4.3 \times 10^{14} 4.3 × 1 0 14 7.5 × 10 14 H z 7.5 \times 10^{14}\mathrm{ Hz} 7.5 × 1 0 14 Hz Ultraviolet 10 n m 10\mathrm{ nm} 10 nm 400 n m 400\mathrm{ nm} 400 nm 7.5 × 10 14 7.5 \times 10^{14} 7.5 × 1 0 14 3 × 10 16 H z 3 \times 10^{16}\mathrm{ Hz} 3 × 1 0 16 Hz X-rays 0.01 n m 0.01\mathrm{ nm} 0.01 nm 10 n m 10\mathrm{ nm} 10 nm 3 × 10 16 3 \times 10^{16} 3 × 1 0 16 3 × 10 19 H z 3 \times 10^{19}\mathrm{ Hz} 3 × 1 0 19 Hz Gamma rays < 0.01 n m \lt 0.01\mathrm{ nm} < 0.01 nm > 3 × 10 19 H z \gt 3 \times 10^{19}\mathrm{ Hz} > 3 × 1 0 19 Hz
Key Properties All EM waves travel at c = 3.0 × 10 8 m / s c = 3.0 \times 10^8\mathrm{ m/s} c = 3.0 × 1 0 8 m/s They are all transverse waves. They can all travel through a vacuum. They can all be polarised. Superposition Principle of Superposition When two or more waves overlap, the resultant displacement at any point is the algebraic sum of the Individual displacements:
Y { { t o t a l } } = y 1 + y 2 + y 3 + ⋯ Y_\{\mathrm\{total\}\} = y_1 + y_2 + y_3 + \cdots Y { { t o t a l }} = y 1 + y 2 + y 3 + ⋯ Constructive Interference Waves arrive in phase (path difference = n λ = n\lambda = nλ n n n
{ P a t h d i f f e r e n c e } = n λ \mathrm\{Path difference\} = n\lambda { P a t h d i f f er e n ce } = nλ The resultant amplitude is A 1 + A 2 A_1 + A_2 A 1 + A 2
Destructive Interference Waves arrive out of phase (path difference = ( n + 1 2 ) λ = (n + \frac{1}{2})\lambda = ( n + 2 1 ) λ
{ P a t h d i f f e r e n c e } = ( n + { 1 } { 2 } ) λ \mathrm\{Path difference\} = \left(n + \frac\{1\}\{2\}\right)\lambda { P a t h d i f f er e n ce } = ( n + 1 { } { 2 } ) λ The resultant amplitude is ∣ A 1 − A 2 ∣ |A_1 - A_2| ∣ A 1 − A 2 ∣
Two-Source Interference For coherent sources (same frequency, constant phase relationship), interference produces a pattern of bright and dark fringes (for light) or loud and quiet regions (for sound).
For double-slit interference with slit separation d d d D D D
D sin θ = n λ ( { b r i g h t f r i n g e s } ) D\sin\theta = n\lambda \quad (\mathrm\{bright fringes\}) D sin θ = nλ ({ b r i g h t f r in g es }) D sin θ = ( n + { 1 } { 2 } ) λ ( { d a r k f r i n g e s } ) D\sin\theta = \left(n + \frac\{1\}\{2\}\right)\lambda \quad (\mathrm\{dark fringes\}) D sin θ = ( n + 1 { } { 2 } ) λ ({ d a r k f r in g es }) For small angles (sin θ ≈ tan θ ≈ x D \sin\theta \approx \tan\theta \approx \dfrac{x}{D} sin θ ≈ tan θ ≈ D x
X n = { n λ D } { d } X_n = \frac\{n\lambda D\}\{d\} X n = n { λ D } { d } Fringe spacing:
Δ x = { λ D } { d } \Delta x = \frac\{\lambda D\}\{d\} Δ x = λ { D } { d }
Light of wavelength 600 n m 600\mathrm{ nm} 600 nm 0.2 m m 0.2\mathrm{ mm} 0.2 mm 1.5 m 1.5\mathrm{ m} 1.5 m
Δ x = { λ D } { d } = { 6 00 × 10 { − 9 } × 1.5 } { 0.2 × 10 { − 3 } } = { 9 × 10 { − 7 } } { 2 × 10 { − 4 } } = 4.5 × 10 { − 3 } { m } = 4.5 { m m } \Delta x = \frac\{\lambda D\}\{d\} = \frac\{600 \times 10^\{-9\} \times 1.5\}\{0.2 \times 10^\{-3\}\} = \frac\{9 \times 10^\{-7\}\}\{2 \times 10^\{-4\}\} = 4.5 \times 10^\{-3\}\mathrm\{ m\} = 4.5\mathrm\{ mm\} Δ x = λ { D } { d } = 6 { 00 × 1 0 { − 9 } × 1.5 } { 0.2 × 1 0 { − 3 }} = 9 { × 1 0 { − 7 }} { 2 × 1 0 { − 4 }} = 4.5 × 1 0 { − 3 } { m } = 4.5 { mm } Standing Waves Standing waves form when two identical waves travelling in opposite directions superpose. This Occurs due to reflection at a boundary.
Nodes and Antinodes Node : point of zero displacement (destructive interference)Antinode : point of maximum displacement (constructive interference)Standing Waves on Strings For a string of length L L L
Harmonics :
Harmonic Wavelength Frequency 1st (fundamental) λ 1 = 2 L \lambda_1 = 2L λ 1 = 2 L f 1 = v 2 L f_1 = \dfrac{v}{2L} f 1 = 2 L v 2nd λ 2 = L \lambda_2 = L λ 2 = L f 2 = v L = 2 f 1 f_2 = \dfrac{v}{L} = 2f_1 f 2 = L v = 2 f 1 3rd λ 3 = 2 L 3 \lambda_3 = \dfrac{2L}{3} λ 3 = 3 2 L f 3 = 3 v 2 L = 3 f 1 f_3 = \dfrac{3v}{2L} = 3f_1 f 3 = 2 L 3 v = 3 f 1 n n n λ n = 2 L n \lambda_n = \dfrac{2L}{n} λ n = n 2 L f n = n v 2 L = n f 1 f_n = \dfrac{nv}{2L} = nf_1 f n = 2 L n v = n f 1
The wave speed on a string under tension T T T μ \mu μ
V = { { T } { μ } } V = \sqrt\{\frac\{T\}\{\mu\}\} V = { T { } { μ }} Standing Waves in Pipes Open pipe (open at both ends):
F_n = \frac`\{nv}`\{2L\}, \quad n = 1, 2, 3, \ldots Both ends are antinodes.
Closed pipe (closed at one end):
F_n = \frac`\{nv}`\{4L\}, \quad n = 1, 3, 5, \ldots \mathrm\{ (odd harmonics only)\} The closed end is a node, the open end is an antinode.
An open pipe has a fundamental frequency of 440 H z 440\mathrm{ Hz} 440 Hz
F 3 = 3 f 1 = 3 ( 440 ) = 1320 { H z } F_3 = 3f_1 = 3(440) = 1320\mathrm\{ Hz\} F 3 = 3 f 1 = 3 ( 440 ) = 1320 { H z } The Doppler Effect Definition The Doppler effect is the change in observed frequency of a wave when there is relative motion between the source and the observer.
Moving Source, Stationary Observer F ′ = { f } { 1 ± { v s } { v w } } F' = \frac\{f\}\{1 \pm \frac\{v_s\}\{v_w\}\} F ′ = f { } { 1 ± v { s } { v w }} Where v s v_s v s v w v_w v w
Source approaching: f ′ = f 1 − v s / v w f' = \dfrac{f}{1 - v_s/v_w} f ′ = 1 − v s / v w f Source receding: f ′ = f 1 + v s / v w f' = \dfrac{f}{1 + v_s/v_w} f ′ = 1 + v s / v w f Moving Observer, Stationary Source F ′ = f ( 1 ± { v o } { v w } ) F' = f\!\left(1 \pm \frac\{v_o\}\{v_w\}\right) F ′ = f ( 1 ± v { o } { v w } ) Observer approaching: f ′ = f ( 1 + v o / v w ) f' = f(1 + v_o/v_w) f ′ = f ( 1 + v o / v w ) Observer receding: f ′ = f ( 1 − v o / v w ) f' = f(1 - v_o/v_w) f ′ = f ( 1 − v o / v w ) F ′ = f ( { v w ± v o } { v w ∓ v s } ) F' = f\!\left(\frac\{v_w \pm v_o\}\{v_w \mp v_s\}\right) F ′ = f ( v { w ± v o } { v w ∓ v s } ) Upper signs when approaching, lower signs when receding.
Electromagnetic Doppler Effect For light:
F ′ = f { { 1 + β } { 1 − β } } F' = f\sqrt\{\frac\{1 + \beta\}\{1 - \beta\}\} F ′ = f { 1 { + β } { 1 − β }} Where β = v c \beta = \dfrac{v}{c} β = c v v ≪ c v \ll c v ≪ c
{ Δ f } { f } ≈ { v } { c } \frac\{\Delta f\}\{f\} \approx \frac\{v\}\{c\} Δ { f } { f } ≈ v { } { c } Redshift : source receding, observed wavelength increasesBlueshift : source approaching, observed wavelength decreasesApplications Application Description Radar guns Measure speed of vehicles Medical ultrasound Measure blood flow velocity Astronomy Measure speed of stars/galaxies (redshift) Weather radar Track storm systems
A fire engine with siren at 700 H z 700\mathrm{ Hz} 700 Hz 30 m / s 30\mathrm{ m/s} 30 m/s v s o u n d = 343 m / s v_{\mathrm{sound}} = 343\mathrm{ m/s} v sound = 343 m/s
F ′ = { f } { 1 − v s / v w } = { 7 00 } { 1 − 30 / 343 } = { 7 00 } { 0.9125 } = 767 { H z } F' = \frac\{f\}\{1 - v_s/v_w\} = \frac\{700\}\{1 - 30/343\} = \frac\{700\}\{0.9125\} = 767\mathrm\{ Hz\} F ′ = f { } { 1 − v s / v w } = 7 { 00 } { 1 − 30/343 } = 7 { 00 } { 0.9125 } = 767 { H z } Diffraction Definition Diffraction is the spreading of waves when they pass through an aperture or around an obstacle.
Conditions Diffraction is most significant when the wavelength is comparable to the size of the aperture or obstacle. For light (very small λ \lambda λ For sound (larger λ \lambda λ Single Slit Diffraction For light of wavelength λ \lambda λ a a a
Minima occur at:
A sin θ = n λ , n = ± 1 , ± 2 , … A\sin\theta = n\lambda, \quad n = \pm 1, \pm 2, \ldots A sin θ = nλ , n = ± 1 , ± 2 , … The central maximum is twice as wide as the secondary maxima.
Rayleigh Criterion Two sources are just resolvable when the central maximum of one diffraction pattern coincides with the first minimum of the other:
θ { min } = 1.22 { λ } { D } \theta_\{\min\} = 1.22\frac\{\lambda\}\{D\} θ { min } = 1.22 λ { } { D } Where D D D
Resolution The ability to distinguish between two closely spaced objects depends on:
Wavelength (shorter λ \lambda λ Aperture diameter (larger D D D This is why astronomical telescopes use large mirrors and electron microscopes use short wavelengths (electrons).
A telescope with a mirror of diameter 0.1 m 0.1\mathrm{ m} 0.1 m 550 n m 550\mathrm{ nm} 550 nm
θ { min } = 1.22 × { 5 50 × 10 { − 9 } } { 0.1 } = 6.71 × 10 { − 6 } { r a d } \theta_\{\min\} = 1.22 \times \frac\{550 \times 10^\{-9\}\}\{0.1\} = 6.71 \times 10^\{-6\}\mathrm\{ rad\} θ { min } = 1.22 × 5 { 50 × 1 0 { − 9 }} { 0.1 } = 6.71 × 1 0 { − 6 } { r a d } This is approximately 1.38 1.38 1.38
Polarisation Definition Polarisation is the restriction of the oscillation direction of a transverse wave to one plane.
Methods Method Description Polarising filter Allows only one plane of oscillation to pass Reflection Light reflected from a surface is partially polarised Birefringence Certain crystals split light into two polarised beams Scattering Light scattered by the atmosphere is partially polarised
Malus’s Law When polarised light of intensity I 0 I_0 I 0 θ \theta θ
I = I 0 cos 2 θ I = I_0\cos^2\theta I = I 0 cos 2 θ Brewster’s Angle When light hits a surface at Brewster’s angle, the reflected light is completely polarised:
tan θ B = { n 2 } { n 1 } \tan\theta_B = \frac\{n_2\}\{n_1\} tan θ B = n { 2 } { n 1 } IB Exam-Style Questions Question 1 (Paper 1 style) Light of wavelength 590 n m 590\mathrm{ nm} 590 nm 0.5 m m 0.5\mathrm{ mm} 0.5 mm 2 m 2\mathrm{ m} 2 m
X 3 = { 3 λ D } { d } = { 3 × 590 × 10 { − 9 } × 2 } { 0.5 × 10 { − 3 } } = { 3 .54 × 10 { − 6 } } { 5 × 10 { − 4 } } = 7.08 × 10 { − 3 } { m } = 7.08 { m m } X_3 = \frac\{3\lambda D\}\{d\} = \frac\{3 \times 590 \times 10^\{-9\} \times 2\}\{0.5 \times 10^\{-3\}\} = \frac\{3.54 \times 10^\{-6\}\}\{5 \times 10^\{-4\}\} = 7.08 \times 10^\{-3\}\mathrm\{ m\} = 7.08\mathrm\{ mm\} X 3 = 3 { λ D } { d } = 3 { × 590 × 1 0 { − 9 } × 2 } { 0.5 × 1 0 { − 3 }} = 3 { .54 × 1 0 { − 6 }} { 5 × 1 0 { − 4 }} = 7.08 × 1 0 { − 3 } { m } = 7.08 { mm } Question 2 (Paper 2 style) A string of length 0.8 m 0.8\mathrm{ m} 0.8 m 4 g 4\mathrm{ g} 4 g 50 N 50\mathrm{ N} 50 N
(a) Find the speed of waves on the string.
μ = { 0 .004 } { 0.8 } = 0.005 { k g / m } \mu = \frac\{0.004\}\{0.8\} = 0.005\mathrm\{ kg/m\} μ = 0 { .004 } { 0.8 } = 0.005 { k g / m } V = { { T } { μ } } = { { 5 0 } { 0.005 } } = { 10000 } = 100 { m / s } V = \sqrt\{\frac\{T\}\{\mu\}\} = \sqrt\{\frac\{50\}\{0.005\}\} = \sqrt\{10000\} = 100\mathrm\{ m/s\} V = { T { } { μ }} = { 5 { 0 } { 0.005 }} = { 10000 } = 100 { m / s } (b) Find the fundamental frequency and the first three harmonic frequencies.
F 1 = { v } { 2 L } = { 1 00 } { 1.6 } = 62.5 { H z } F_1 = \frac\{v\}\{2L\} = \frac\{100\}\{1.6\} = 62.5\mathrm\{ Hz\} F 1 = v { } { 2 L } = 1 { 00 } { 1.6 } = 62.5 { H z } F 2 = 125 { H z } , f 3 = 187.5 { H z } F_2 = 125\mathrm\{ Hz\}, \quad f_3 = 187.5\mathrm\{ Hz\} F 2 = 125 { H z } , f 3 = 187.5 { H z } (c) Draw the standing wave pattern for the second harmonic.
The second harmonic has one node at the centre and antinodes at each quarter point. There are 3 Nodes (including both ends) and 2 antinodes.
Question 3 (Paper 1 style) An ambulance with siren at 800 H z 800\mathrm{ Hz} 800 Hz 20 m / s 20\mathrm{ m/s} 20 m/s v s o u n d = 340 m / s v_{\mathrm{sound}} = 340\mathrm{ m/s} v sound = 340 m/s
F ′ = { f } { 1 + v s / v w } = { 8 00 } { 1 + 20 / 340 } = { 8 00 } { 1.0588 } = 756 { H z } F' = \frac\{f\}\{1 + v_s/v_w\} = \frac\{800\}\{1 + 20/340\} = \frac\{800\}\{1.0588\} = 756\mathrm\{ Hz\} F ′ = f { } { 1 + v s / v w } = 8 { 00 } { 1 + 20/340 } = 8 { 00 } { 1.0588 } = 756 { H z } Question 4 (Paper 2 style) Unpolarised light of intensity I 0 I_0 I 0 60 ° 60\degree 60°
(a) Find the intensity after the first filter.
I 1 = { I 0 } { 2 } I_1 = \frac\{I_0\}\{2\} I 1 = I { 0 } { 2 } (b) Find the intensity after the second filter.
I 2 = I 1 cos 2 60 ° = { I 0 } { 2 } × { 1 } { 4 } = { I 0 } { 8 } I_2 = I_1\cos^2 60\degree = \frac\{I_0\}\{2\} \times \frac\{1\}\{4\} = \frac\{I_0\}\{8\} I 2 = I 1 cos 2 60° = I { 0 } { 2 } × 1 { } { 4 } = I { 0 } { 8 } Question 5 (Paper 2 style) A car horn produces sound at 400 H z 400\mathrm{ Hz} 400 Hz 25 m / s 25\mathrm{ m/s} 25 m/s
(a) Find the frequency heard by the observer as the car approaches.
F ′ = { 4 00 } { 1 − 25 / 343 } = { 4 00 } { 0.927 } = 431 { H z } F' = \frac\{400\}\{1 - 25/343\} = \frac\{400\}\{0.927\} = 431\mathrm\{ Hz\} F ′ = 4 { 00 } { 1 − 25/343 } = 4 { 00 } { 0.927 } = 431 { H z } (b) Find the frequency heard as the car moves away.
F ′ = { 4 00 } { 1 + 25 / 343 } = { 4 00 } { 1.073 } = 373 { H z } F' = \frac\{400\}\{1 + 25/343\} = \frac\{400\}\{1.073\} = 373\mathrm\{ Hz\} F ′ = 4 { 00 } { 1 + 25/343 } = 4 { 00 } { 1.073 } = 373 { H z } (c) Calculate the change in frequency.
Δ f = 431 − 373 = 58 { H z } \Delta f = 431 - 373 = 58\mathrm\{ Hz\} Δ f = 431 − 373 = 58 { H z } Summary Quantity Formula Wave speed v = f λ v = f\lambda v = f λ Intensity I = P 4 π r 2 I = \dfrac{P}{4\pi r^2} I = 4 π r 2 P Sound level β = 10 log 10 ( I / I 0 ) \beta = 10\log_{10}(I/I_0) β = 10 log 10 ( I / I 0 ) Double-slit maxima d sin θ = n λ d\sin\theta = n\lambda d sin θ = nλ Single-slit minima a sin θ = n λ a\sin\theta = n\lambda a sin θ = nλ String harmonics f n = n v 2 L f_n = \dfrac{nv}{2L} f n = 2 L n v Doppler (source moving) f ′ = f 1 ∓ v s / v w f' = \dfrac{f}{1 \mp v_s/v_w} f ′ = 1 ∓ v s / v w f Malus’s law I = I 0 cos 2 θ I = I_0\cos^2\theta I = I 0 cos 2 θ Rayleigh criterion θ = 1.22 λ D \theta = 1.22\dfrac{\lambda}{D} θ = 1.22 D λ
For wave problems, always identify the type of wave and the relevant equations. For interference Problems, determine whether you need path difference or phase difference. For standing waves, identify whether the system is a string, open pipe, or closed pipe. For Doppler problems, Identify what is moving (source, observer, or both).
Wave Intensity and Amplitude Relationship Between Intensity and Amplitude For a wave, intensity is proportional to the square of the amplitude:
I ∝ A 2 I \propto A^2 I ∝ A 2 If the amplitude doubles, the intensity quadruples.
Intensity at a Distance from a Point Source I = { P } { 4 π r 2 } I = \frac\{P\}\{4\pi r^2\} I = P { } { 4 π r 2 } This means:
I 1 r 1 2 = I 2 r 2 2 I_1 r_1^2 = I_2 r_2^2 I 1 r 1 2 = I 2 r 2 2
At 10 m 10\mathrm{ m} 10 m 0.5 W / m 2 0.5\mathrm{ W/m}^2 0.5 W/m 2 25 m 25\mathrm{ m} 25 m
I 2 = I 1 × { r 1 2 } { r 2 2 } = 0.5 × { 1 00 } { 625 } = 0.08 { W / m } 2 I_2 = I_1 \times \frac\{r_1^2\}\{r_2^2\} = 0.5 \times \frac\{100\}\{625\} = 0.08\mathrm\{ W/m\}^2 I 2 = I 1 × r { 1 2 } { r 2 2 } = 0.5 × 1 { 00 } { 625 } = 0.08 { W / m } 2 Phase and Phase Difference Phase Difference Phase difference Δ ϕ \Delta\phi Δ ϕ
Δ ϕ = { 2 π Δ x } { λ } \Delta\phi = \frac\{2\pi \Delta x\}\{\lambda\} Δ ϕ = 2 { π Δ x } { λ } Where Δ x \Delta x Δ x
Phase Difference Description 0 , 2 π , 4 π , … 0, 2\pi, 4\pi, \ldots 0 , 2 π , 4 π , … In phase (constructive) π , 3 π , 5 π , … \pi, 3\pi, 5\pi, \ldots π , 3 π , 5 π , … Anti-phase (destructive) π / 2 \pi/2 π /2 90 ° 90\degree 90°
Coherence Two sources are coherent if they have:
The same frequency. A constant phase relationship. Only coherent sources produce a stable interference pattern.
Diffraction Gratings Equation For a diffraction grating with N N N d = 1 / N d = 1/N d = 1/ N
D sin θ = n λ D\sin\theta = n\lambda D sin θ = nλ The maximum number of orders visible:
N { max } = { d } { λ } N_\{\max\} = \frac\{d\}\{\lambda\} N { max } = d { } { λ } (rounded down to the nearest integer).
Advantages Over Double Slit Sharper, brighter fringes. Larger angular separation. More accurate measurement of wavelength.
A diffraction grating has 500 l i n e s / m m 500\mathrm{ lines/mm} 500 lines/mm 600 n m 600\mathrm{ nm} 600 nm
D = { 1 } { 500000 } = 2 × 10 { − 6 } { m } D = \frac\{1\}\{500000\} = 2 \times 10^\{-6\}\mathrm\{ m\} D = 1 { } { 500000 } = 2 × 1 0 { − 6 } { m } sin θ 1 = { λ } { d } = { 6 00 × 10 { − 9 } } { 2 × 10 { − 6 } } = 0.3 ⟹ θ 1 = 17.5 ° \sin\theta_1 = \frac\{\lambda\}\{d\} = \frac\{600 \times 10^\{-9\}\}\{2 \times 10^\{-6\}\} = 0.3 \implies \theta_1 = 17.5\degree sin θ 1 = λ { } { d } = 6 { 00 × 1 0 { − 9 }} { 2 × 1 0 { − 6 }} = 0.3 ⟹ θ 1 = 17.5° sin θ 2 = { 2 λ } { d } = 0.6 ⟹ θ 2 = 36.9 ° \sin\theta_2 = \frac\{2\lambda\}\{d\} = 0.6 \implies \theta_2 = 36.9\degree sin θ 2 = 2 { λ } { d } = 0.6 ⟹ θ 2 = 36.9° Maximum order: n max = 2 × 10 − 6 600 × 10 − 9 = 3.33 n_{\max} = \dfrac{2 \times 10^{-6}}{600 \times 10^{-9}} = 3.33 n m a x = 600 × 1 0 − 9 2 × 1 0 − 6 = 3.33
Additional IB Exam-Style Questions Question 6 (Paper 2 style) A string of length 0.6 m 0.6\mathrm{ m} 0.6 m 240 m / s 240\mathrm{ m/s} 240 m/s
(a) Calculate the fundamental frequency.
F 1 = { v } { 2 L } = { 2 40 } { 1.2 } = 200 { H z } F_1 = \frac\{v\}\{2L\} = \frac\{240\}\{1.2\} = 200\mathrm\{ Hz\} F 1 = v { } { 2 L } = 2 { 40 } { 1.2 } = 200 { H z } (b) Draw the standing wave pattern for the third harmonic and state its frequency.
The third harmonic has 3 half-wavelengths fitting on the string, with 4 nodes and 3 antinodes.
F 3 = 3 f 1 = 600 { H z } F_3 = 3f_1 = 600\mathrm\{ Hz\} F 3 = 3 f 1 = 600 { H z } (c) The tension in the string is doubled. Find the new fundamental frequency.
V = { { T } { μ } } ⟹ v ′ = { 2 } v V = \sqrt\{\frac\{T\}\{\mu\}\} \implies v' = \sqrt\{2\}v V = { T { } { μ }} ⟹ v ′ = { 2 } v F 1 ′ = { 2 } × 200 = 283 { H z } F_1' = \sqrt\{2\} \times 200 = 283\mathrm\{ Hz\} F 1 ′ = { 2 } × 200 = 283 { H z } Question 7 (Paper 2 style) Two loudspeakers are 3 m 3\mathrm{ m} 3 m 686 H z 686\mathrm{ Hz} 686 Hz 343 m / s 343\mathrm{ m/s} 343 m/s
(a) Calculate the wavelength.
λ = { v } { f } = { 3 43 } { 686 } = 0.5 { m } \lambda = \frac\{v\}\{f\} = \frac\{343\}\{686\} = 0.5\mathrm\{ m\} λ = v { } { f } = 3 { 43 } { 686 } = 0.5 { m } (b) A listener walks along a line parallel to the speakers, 4 m 4\mathrm{ m} 4 m
For constructive interference: path difference = n λ = n\lambda = nλ
Using geometry, the path difference Δ = d sin θ \Delta = d\sin\theta Δ = d sin θ θ \theta θ
3 sin θ = n × 0.5 3\sin\theta = n \times 0.5 3 sin θ = n × 0.5 For n = 1 n = 1 n = 1 sin θ = 1 / 6 \sin\theta = 1/6 sin θ = 1/6 θ = 9.6 ° \theta = 9.6\degree θ = 9.6° 4 tan 9.6 ° = 0.68 m 4\tan 9.6\degree = 0.68\mathrm{ m} 4 tan 9.6° = 0.68 m
For n = 2 n = 2 n = 2 sin θ = 1 / 3 \sin\theta = 1/3 sin θ = 1/3 θ = 19.5 ° \theta = 19.5\degree θ = 19.5° 4 tan 19.5 ° = 1.41 m 4\tan 19.5\degree = 1.41\mathrm{ m} 4 tan 19.5° = 1.41 m
Question 8 (Paper 1 style) Light of wavelength λ \lambda λ a a a
The first minimum occurs at a sin θ = λ a\sin\theta = \lambda a sin θ = λ a a a sin θ \sin\theta sin θ
Question 9 (Paper 2 style) Unpolarised light of intensity 200 W / m 2 200\mathrm{ W/m}^2 200 W/m 2 30 ° 30\degree 30° 60 ° 60\degree 60°
Find the intensity after each filter.
After filter 1: I 1 = 200 2 = 100 W / m 2 I_1 = \dfrac{200}{2} = 100\mathrm{ W/m}^2 I 1 = 2 200 = 100 W/m 2
After filter 2: I 2 = 100 cos 2 30 ° = 100 × 0.75 = 75 W / m 2 I_2 = 100\cos^2 30\degree = 100 \times 0.75 = 75\mathrm{ W/m}^2 I 2 = 100 cos 2 30° = 100 × 0.75 = 75 W/m 2
After filter 3: I 3 = 75 cos 2 ( 60 ° − 30 ° ) = 75 cos 2 30 ° = 75 × 0.75 = 56.25 W / m 2 I_3 = 75\cos^2(60\degree - 30\degree) = 75\cos^2 30\degree = 75 \times 0.75 = 56.25\mathrm{ W/m}^2 I 3 = 75 cos 2 ( 60° − 30° ) = 75 cos 2 30° = 75 × 0.75 = 56.25 W/m 2
For the A-Level treatment of this topic, see Wave Properties .
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Unit tests probe edge cases and common misconceptions. Integration tests combine Wave Properties with other physics topics to test synthesis under exam conditions.
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Intuition Wave properties describe how disturbances travel through space. The fundamental relationship speed equals frequency times wavelength is like saying a runner’s pace equals stride length times strides per second. Superposition means waves pass through each other without permanent damage, adding and subtracting their displacements. Standing waves emerge from reflection at boundaries, trapping energy in fixed patterns. The Doppler effect explains why an approaching ambulance sounds higher pitched and a receding one lower, because the relative motion compresses or stretches the arriving wavefronts.
Common Pitfalls Confusing wave speed, frequency, and wavelength. Remember v = f λ v = f\lambda v = f λ
Forgetting that waves transfer energy, not matter (except for matter waves in quantum mechanics).
Misidentifying nodes and antinodes in standing wave diagrams.
Using the wrong equation from the data sheet. Take time to read the full equation, including conditions and variable definitions.
Neglecting air resistance or assuming ideal conditions when the question specifies a real-world scenario.
Forgetting to include units in final answers, especially when working with derived units like N kg − 1 m 2 \text{N}\,\text{kg}^{-1}\,\text{m}^2 N kg − 1 m 2
Cross-References Topic Site Link [Wave Properties] A-Level View [Wave Properties] IB View [Wave Properties] DSE View
Worked Examples Common Mistakes Confusing transverse and longitudinal waves: Transverse waves oscillate perpendicular to direction (light, waves on string). Longitudinal waves oscillate parallel to direction (sound, slinky). Only transverse waves can be polarised, sound cannot be polarised.
Forgetting that frequency doesn’t change when waves enter a new medium: When a wave enters a different medium, speed and wavelength change, but frequency stays the same (determined by the source). Students often incorrectly assume frequency changes.
Misidentifying the normal: The normal is the line perpendicular to the surface at the point of incidence. All angles are measured from the normal, not from the surface. Students often measure from the surface, giving wrong angles.
Confusing the wave equation v = f λ v = f\lambda v = f λ The wave equation relates speed, frequency, and wavelength. It does not describe energy. Energy depends on amplitude and frequency, not wavelength directly.
Forgetting that intensity is proportional to amplitude squared: I ∝ A 2 I \propto A^2 I ∝ A 2
See Also 