Andy Buffler Department of Physics University of Cape Town

University of Cape Town Department of Physics PHY014F Vibrations and Waves Part 3 Travelling waves Boundary conditions Sound Interference and diffraction covering (more or less) French Chapters 7 & 8 Andy Buffler Department of Physics University of Cape Town andy.buffler@uct.ac.za 1

Problem-solving and homework Each week you will be given a take-home problem set to complete and hand in for marks... In addition to this, you need to work through the following problems in French, in you own time, at home. You will not be asked to hand these in for marks. Get help from you friends, the course tutor, lecturer,... Do not take shortcuts. Mastering these problems is a fundamental aspect of this course. The problems associated with Part 3 are: 7-1, 7-, 7-3, 7-4, 7-5, 7-6, 7-7, 7-8, 7-9, 7-10, 7-11, 7-15, 7-17, 7-18, 7-19, 7-1 You might find these tougher:

Travelling waves For a string clamped at both ends the displacement of the n th normal mode is nπ x yxt (, ) = Asin cosω n t ω n = L Since L n λ nπ vt π vt = then = L λ Then write: And using nπ x π vt ψ ( xt, ) = Asin cos L λ ( ) ( ) sinα sin β = sin α + β + sin α β 1 { } A π x πvt A πx πvt ψ ( xt, ) = sin + + sin λ λ λ λ A π A sin ( x vt) sin π = + + ( x vt) λ λ nπ v L 3

Travelling waves... A π A π ψ ( xt, ) = sin ( x vt) + sin ( x+ vt) λ λ A wave travelling in the +ve x-direction with speed v A wave travelling in the ve x-direction with speed v The speed v is called the phase velocity and is the speed of a crest (or any other point of specified phase) x v = t 4

Travelling waves...3 We have shown that a standing wave nπ x π vt ψ ( xt, ) = Asin cos L λ can be regarded as the superposition of two travelling waves: A π ψ1( xt, ) = sin x+ vt λ ( ) and A π ψ ( xt, ) = sin x vt λ ( ) We can think of the standing wave being made up of a travelling wave being reflected back and forth from the two ends. The quantity v introduced in the standing wave treatment acquires a definite physical meaning: phase velocity 5

Formation of standing waves two waves having the same amplitude and frequency, travelling in opposite directions, interfere with each other. 6

Travelling waves...4 The travelling waves we have considered are sinusoidal in shape. Other shapes (wave pulses etc) are possible and can be represented as the superposition of sinusoidal waves. Any functions f( x vt) or gx ( + vt) are solutions of the wave equation and represent travelling waves. f wave at time t 1 wave at time t x1 x f( x vt ) = f( x vt ) 1 1 x vt = x vt x1 x v = t t 1 1 1 x 7

Travelling waves...5 Travelling waves can be represented by sine or cosine functions: or π ψ ( xt, ) = Asin ( x vt) λ π ψ ( xt, ) = Acos ( x vt) λ x v = Acos π t λ λ x = Acos π ft λ wave number π x t k = = Acos π λ λ T = Acos{ kx ωt} (French: k = 1 λ! ) 8

Travelling waves...6 Sinusoidal travelling waves: or or ψ ( xt, ) = Asin{ kx ωt+ φ} ψ ( xt, ) = Acos kx ωt+ φ' { } ψ ( xt, ) = Asin{ kx+ ωt+ φ} ψ ( xt, ) = Acos kx+ ωt+ φ' { } waves in +ve x-direction waves in ve x-direction λ ω Phase velocity v phase = fλ = π f = π k 9

Different types of travelling waves transverse longitudinal... and water waves? 10

French page 09 Wave speeds in specific media Transverse waves on a stretched string: Longitudinal waves in a thin rod: v = v = Y ρ T µ Liquids and gases: mainly longitudinal waves Liquids: v B = B: bulk modulus ρ Gases: v γ p = = ρ γrt M γ p : ratio of specific heats : pressure 11

Wave pulses... not covered in detail. Read French 16-30 1

French page 8 Superposition of wave pulses 13

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French page 53 heavy spring Reflection of wave pulses light spring very heavy spring 16

Reflection of wave pulses heavy string light string 17

Reflection of wave pulses Phase change of π (i.e. inversion) on reflection from fixed end. No inversion on reflection from free end. 18

Dispersion The waves we are most familiar with (e.g. sound waves in air) are non-dispersive i.e. waves of different frequency travel at the same speed. Travelling wave on a string: π yxt (, ) = Asin ( x vt) λ T For a continuous string: v f n = µ For a lumpy, beaded string: = nf1 f n nπ = f0 sin ( N + 1) high frequency waves may have a different velocity than low frequency. 19

Dispersion No dispersion of light in a vacuum. But light is dispersed into colours by a prism: Snell s law: sini c = n( λ) = sin r v( λ) refractive index important for optic fibre communications 0

French page 13, 3 Dispersion 3 Take two sinusoidal travelling waves of slightly different frequencies: { ω } { ω } ψ1( xt, ) = Acos kx 1 1t ψ ( xt, ) = Acos kx t Then ψ ( xt, ) = ψ1( xt, ) + ψ( xt, ) where Acos{ kx 1 ω1t} Acos{ kx ωt} ( kx ωt) + ( kx ω t) ( kx ωt) ( kx ω t) = + Acos cos k1+ k ω1+ ω k1 k ω1 ω = Acos x t cos x t 1 1 = Acos kx ωt cos kx ωt 1 1 1 1 = k k { } { } + k ω + ω 1 1 = ω = 1 k = k k ω = ω1 ω 1

Dispersion 4 1 1 { } { } ψ ( xt, ) = Acos kx ωt cos kx ωt Average travelling wave Envelope { ωt} cos{ 1 kx 1 ωt} cos kx ω v phase = ω k Group velocity = v group = k = ω k

Dispersion 5 Phase velocity: Group velocity: v phase = v group = ω k ω = k dω dk the wave packet moves at v g so does transport of energy When analyzing an arbitrary pulse into pure sinusoids if these sinusoids have different characteristic speeds, then the shape of the disturbance must change over time a pulse that is highly localized at t = 0 will become more and more spread out as it moves along. 3

Dispersion 6 Consider surface waves on liquids For long wavelength waves (λ ~ m) on deep water (gravity waves): ω = gk then v g ω vφ = = k dω 1 = = dk g k g k v = φ 1 v g For short wavelength waves (λ ~ mm) on deep water (capillary waves or ripples): ω Sk S vφ = = ω = k 3 then k ρ ρ v 3 φ = v dω 3 Sk vg = = dk ρ S = surface tension g 4

Phase and group velocities of a wave and a pulse with = φ v g v 5

French page 37 The energy in a mechanical wave Consider a small segment of a string carrying a wave: ds T dy Mass of segment = µ dx T dx y If u = y x x+ dx x t dk 1 y µ then kinetic energy per unit length = = dx t Potential energy = Tds ( dx) y 1 y where ds = dx + dy = dx 1+ = dx 1 + +... x x du 1 y Then potential energy per unit length = = T 6 dx x

The energy in a mechanical wave For a travelling wave on a string: yxt (, ) = Asin π f t y x Then uy ( x, t) = = π facos π f t t v cos x = u0 π f t v y π fx π x At t = 0: uy ( x) = = u0cos = u0cos t v λ then dk 1 y 1 π µ µ u0 cos x = = dx t λ x v and λ 1 π x 1 K = µ u cos dx= λµ u λ 4 0 0 0 in one wavelength 7

The energy in a mechanical wave 3 Similarly for potential energy: At t = 0: y π A π x = cos x λ λ then and π cos du A T π x = dx λ λ λ π A T πx π A T λ U = cos dx = in one wavelength λ λ λ 0 1 U = π A µ f λ = λµ u u0 = π fa 0 4 Over one wavelength, total energy: 1 E = K + U = λµ u 0 8

French page 41 θ F Transport of energy y 0 by a wave x yxt (, ) = Asin π f t v A long string has one end at x = 0 and is driven at this point by a driving force F equal in magnitude to the tension T and applied in a direction tangent to the string. At x = 0 : y(0, t) = Asin π ft y π fa and Fy = Tsinθ T = T cos π ft x x= 0 v Then work done: π fat W = Fydy0 = cos π ft d Asin π ft v ( π fa) T ( = cos π ft) dt 9 v ( ) ( ) x

Transport of energy by a wave For one complete cycle: from t = 0 to t = 1/f : 1 f ut 0 ut 0 1 cycle = (1+ cos4 π ) = = λµ 0 v vf 0 W ft dt u Then mean power input P = W ut 1 v 1 0 cycle = = µ 0 f u v thusp = energy per unit length velocity energy flows along medium at velocity v and at v g if the medium is dispersive 30

Doppler effect... frequency changes of traveling waves due to motion of source and/or detector... S: source of waves (sound) λ D: detector S D v S : speed of source v D : speed of detector v v : speed of travelling wave = f λ When both vs and vd are zero,then the number of wavelengths vt passing D in time t is λ S v D D 31

Doppler effect... For the case of stationary S and moving D: For moving D we must add (or subtract) vtλ D wavelengths. S v D v D vt λ ± vdt Observed frequency: f ' = t v± v = D λ Source frequency: f = v λ λ f ' = f v ± v v D 3

Doppler effect...3 For the case of moving S and stationary D: In time period T, S moves a distance vt S. Wavelength at D is therefore shortened (or lengthened) by vt. Observed wavelength: λ ' = λ v v vs v vs = = f ' f f f v f ' = f v v S S vs For both source and detector moving: v± vd f ' = f v v S f S v S D D 33

Doppler effect...4 At supersonic speeds v > v, this relationship no longer applies: All wavefronts are buched along a V-shaped envelope in 3D... called a Mach cone... a shock wave exists alongs the surface of this cone since the bunching of the wavefronts causes an abrupt rise and fall of air pressure as the surface passes any point... causes a sonic boom S vs = v vs = v vs > v vt θ vt sinθ = = vt S v v S vt S Mach number 34

Doppler effect...5 supersonic aircraft travelling bullet bow wave of a boat... get a sonic boom whenever the speed of the source of the waves is greater than the speed of waves in that medium. Cherenkov radiation 35

Doppler effect...6 36

Physical optics Electromagnetic radiation can be modelled as a wave or a beam of particles... all observed phenomena can be described by either model, although the treatment my be easier with one, or the other. EM radiation propagates in vacuo, and may be thought of as E and B fields in phase with each other, and propagating at right angles to each other and to the direction of propagation. Velocity of EM radiation in vacuum = constant, c In a medium of refractive index n, v= c n EM wave from a point source: E is in phase around each circle... vacuum c... get a coherent plane wave at a large distance from point source 37

Interference If we use two in-phase sources to generate waves on the surface of a liquid it is easy to observe interference effects. At certain points the waves are in phase and add constructively, and at other points they are out of phase, interfere destructively, giving zero amplitude.... this is not an everyday observation with light sources... the wavelength of light is small (400-800 nm)... ordinary light sources are enormously larger than this... and are not monochromatic. 38

Try this...? filter double slit Interference... screen... no interference fringes observed... light from different portions of source is incoherent... no fixed phase relationship Thomas Young performed a classic interference experiment in 1801:... interference fringes observed on screen... filter pinhole double slit screen 39

Interference...3... but why do we see fringes at all... doesn t light travel in straight lines? Huyghens showed that one could regard each point on a wavefront as being a source of secondary wavelets... the envelope of these secondary wavelets form a subsequent wavefront... wavefronts 40

Huyghens Principle screen plane wave travelling in this direction Aperture large compared with a wavelength... some spreading of wave Aperture similar size to a wavelength... large angular spread from small aperture 41

Young s double slit experiment back to Young s experiment...... the two slits in the second screen act as coherent sources 4

Young s double slit experiment... d θ y D screen Path difference = d sinθ Maxima on screen when Minima on screen when dsinθ = mλ 1 ( m ) dsinθ = + y = Dsinθ for small θ mλ y = D for maxima d λ m = 0, ±1, ±,... 43

Young s double slit experiment...3... a more detailed treatment d r r θ P r 1 Suppose a plane wave illuminates the slits... the waves from the two slits are in phase at the slits. Since the distances r 1 and r differ, the waves will have a phase difference at P. E1 = E0cos( kr1 ω t) Electric fields of the two waves: E = E cos( kr ωt) 0... both waves have the same amplitude at P... not quite right since r 1 and r are different distances... but ok. 44

Young s double slit experiment...4 Total field: E = E1+ E = E0cos( kr1 ωt) + E0cos( kr ωt) ( + ) ( ) k r1 r k r1 r = E0 cos ωt cos k = E0 cos{ kr ωt} cos dsinθ π d = E0 cos{ kr ωt} cos sinθ λ We now take (time average of ) I = E E Thus for one slit only: similarly: E 0 1 = 0cos ( 1 ω ) = I E kr t I = E 0 45

Young s double slit experiment...5 Write 1 I = I = I = E 0 1 0 π d = 4 0 cos cos sinθ λ π d φ I = 4I0cos sinθ = 4I0cos λ π d So I will have maxima when cos sinθ = 1 λ π d or when sinθ = mπ m = 0, ±1, ±,... λ For both slits: I E { kr ωt} or dsinθ = mλ This approach gives us additional information... 46

Young s double slit experiment...6 Double slit interference pattern I two sources (coherent) two sources (incoherent) single source π d = I θ λ ( θ ) 4 0 cos sin I 4I 0 I 0 I 0 m = π d sinθ = λ 3 1 0 1 3 3π π π 0 π π 3π 47

Young s double slit experiment...7... can also use phasors... Ψ = Ae 1 0 Ψ jkr ( ωt) ( ωt) jkr 1 Ψ = 0 δ = Ae Ae 0 ( ωt+ δ ) jkr 1 Ψ Ψ Ψ 1 Ψ 1 Ψ Ψ Ψ( δ ) A 0 Ψ Ψ 1 Ψ 0 π π δ Ψ = 0 Ψ Ψ 1 48

Interference patterns from thin films Black (i.e. destructive intereference) from very thin film indicates that there is a phase change of π at one of the reflections. t λ For wedges having small angle α, bright fringes correspond to an increase of λ in thickness t : For nearly normal incidence, x t λ path difference = t dark Minima occur when the π out of phase t = λ 4 path difference = nλ bright π out of phase dark 3π out of phase t = λ For minima, t = nλ or x n α = nλ Distance between fringes: 49 x n 1 x = + n λ α

To obtain fringes on a visible scale a very small angle is necessary... such fringes may be seen on viewing the light reflected from a soap film held vertically as it drains... For other interference effects, look up for yourself: Lloyd s mirror Fresnel bi-prism S 1 S fringes S fringes S S 1 Michelson interferometer fringes Newton s rings fringes 50

Fresnel diffraction Around 1818 Augustin Fresnel applied Huyghen s priciple to the problem of diffraction of light by apertures and obstacles... took into account the realtive phases of the secondary wavelets consequent upon their having to travel diferent distances to the point of observation... analytical treatment is complicated... look at the results of a few special cases... Shadow of a straight edge (cast by a point source S)... no sharp edge to the shadow... illumination diminishes smoothly into the shadow... with fringes outside the shadow. S 51

Shadow of a narrow slit: centre of pattern at P may either be a relative maximum or minimum, depending on ratio of slit width to wavelength. Shadow of a narrow parallel-sided obstacle: centre always a relative maximum... but very faint unless obstacle is very narrow. S S Shadow of a circular obstacle: wavelets from all round circumference all reinforce at centre of shadow. S Poisson bright spot (found by Arago) 5

Fraunhofer diffraction... special (limiting) cases of Fresnel diffraction... when both the distances to source and screen both tend to infinity... treatment is easier since the paths of all the wavelets to any point of interest are parallel. It is practical sometimes to use a lens... what would be seen at infinity is then found in its focal plane. or S 53

Fraunhofer diffraction by a single slit Let slit width be a... picture below very much distorted in scale... incident wave y a dy θ r r 0 P a ysinθ To find the intensity of the wave at P we add the contributions at P of waves originating from all points of the aperture. a a jkr ω jkr t jkysinθ Ψ ( P) = = ( t) ( ω ) 0 Ae dy A e dy a a a jkr t jky ( ω ) = 0 sinθ Ae e dy a 54

Fraunhofer diffraction by a single slit... Ψ ( P) = a jkr ( ωt) 0 jky a 0 sinθ A e e dy ( ω ) 1 jk sinθ jkr t jky 0 sinθ = Ae 0 e = Ψ ( P) = a a jk sinθ jk sinθ jkr ( 0 ωt) e e Ae 0 jk sinθ use sinφ = Then Ae I =Ψ ( P) a a 1 ( ω ) sin( sinθ ) jkr t ka 0 a 0 1 I( θ ) e jφ I e j jφ sinα α = 0 kasinθ where sinα α 1 α = kasinθ 55 α

I( θ ) Fraunhofer diffraction by a single slit...3 I sinα α = 0 1 where α = kasinθ I 0 I( θ ) α = 3π π π 0 π π 3π λ λ λ sinθ = 3 λ λ λ 3 a a a a a a Minima at or α = mπ 1 kasinθ = mπ where or m = ±1, ±,... m π λ sinθ = = m a k a 56

Fraunhofer diffraction by a single slit...4 Using phasors... consider a single slit as N sources, each of amplitude A 0... in the example below, N = 10. The path difference between any two adjacent sources is d sinθ where d = a N and the phase difference is δ = dk sinθ. At the central maximum point at θ = 0, the waves from the N sources add in phase, giving the resultant amplitude A = NA. max 0 At the first minimum, the N phasors form a closed polygon... the phase difference between adjacent sources is then δ = π N When N is large, the waves from the first and last sources are approximately in phase A 0 A = NA max 0 δ = 57 π N

Fraunhofer diffraction by a single slit...5 At a general point... where the waves from two adjacent sources differ in phase by δ... The phase difference between the first and last wave is φ. r r φ φ A φ As N increases, the phasor diagram approximates the arc of a circle... the resultant amplitude A is the length of the chord of this arc. ( ) The length of the arc is A ( = NA ) 1 From the figure: sin φ = A r or A = r sin φ A max max 0 Therefore φ = or r r = Amax φ 58

Fraunhofer diffraction by a single slit...6 1 A 1 max sin 1 φ Combining... A= rsin φ = sin φ = Amax φ φ Then I I φ = = φ 1 A sin 1 0 A max 1 or I 1 sin φ = I0 1 φ...where φ is the phase difference between the first and last waves = π λ asinθ = kasinθ ( ) The second maximum occurs when the N 1 phasors complete circles: 1 A... and the resultant amplitude is the diameter of the circle... C 3 Amax Amax 4Amax The diameter A = = = then A = π π 3 π 9π 4 1 and I = I0 = I 0 9π. 59

Interference and diffraction from a double slit When there are two or more slits, the intensity pattern on a screen far away is a combination of the single slit diffraction pattern and the multiple slit interference pattern. sinα = I0 I( θ ) 4 cos α β π a α = sinθ λ π d β = sinθ λ 60

Interference and diffraction from a double slit... intensity pattern for two slits if a 0 : intensity pattern for one slit of width a : intensity pattern for two slits each of width a : sinθ = a λ λ a λ 3 d d λ λ d 0 λ d λ d λ a λ 3 d 61 a λ

Interference patterns from multiple slits (diffraction gratings) d... many slits, with inter-slit spacing d, illuminated by a plane wave... As with the double slit, each slit acts as a source of waves. Consider each pair of slits as a Young s double slit. Slits are coherent, in phase sources for light at an angle θ to the n th n bright fringe. For constructive interference d sinθ n = nλ n = 0, ±1, ±,... If m is the number of slits, then amplitude at θn is A = ma where a is the amplitude form a single slit Then I A m a (...many slits give intense fringes) 6

Interference patterns from multiple slits... d sinθ θ r 1 r r 3 r 4 to P r = r1+ dsinθ r3 = r + dsinθ r4 = r1+ 3dsinθ etc. Total field at P: E = E1+ E + E3 +... = E cos( kr ωt) + E cos( kr ωt) + E cos( kr ωt) +... 0 1 0 0 3 Use complex numbers: ( ω ) ( ω ) jkr ( ωt) E = E e + E e + E e + jkr1 t jkr t 3 0 0 0... 63

Interference patterns from multiple slits...3 ( ω ) ( ω ) jkr ( ωt) E = E e + E e + E e + jkr1 t jkr t 3 0 0 0... ( ω ) ( ω + sin θ ) ( ω + sin θ ) = Ee + Ee + Ee + j kr1 t j kr1 t kd j kr1 t kd 0 0 0... ( ) { } 1 ω sinθ sin θ ( 1) sinθ jkr t jkd jkd N jkd = Ee 0 1 + e + e +... + e... for N slits E = = E e geometric progression ( jkd sinθ 1 e ) ( ωt) jkr1 0 jkd sinθ Ee 1 e N N N N jkd sinθ ( jkd sinθ jkd sinθ e e e ) ( ωt) jkd θ ( jkd θ jkd θ e e e ) jkr1 0 sin sin sin 1 1 1 jφ jφ e e and using sinφ =... j 64

Interference patterns from multiple slits...4 E = E e E e 1 N 1 ( ) ( ) 1 ω jkd sinθ sin sinθ sin( kd sinθ ) jkr t Nkd 0 0 1 = Ee 0 N { ( 1+ 1 sin ) } sin sin β jkr jkd θ ωt Nβ Write E = E e 0 { ω } sin β sin β jkr t N where π d sin sin λ 1 β = kd θ = θ Then, as before, I( θ ) I sin Nβ sin β = 0 65

Interference patterns from multiple slits...5 I( θ ) I sin Nβ sin β = 0 β = π d sinθ λ I N = N = 3 N = 4 π d sinθ = 3π π π 0 π π 3π λ 66

Interference patterns from multiple slits...6 Phasors for a 10 slit grating (N = 10)... δ = 0,π δ = π N = 36 Ψ( δ ) 10A 0 δ = 4π N = 7 δ = 3π N = 54 δ = 5π N = 90 0 π π δ δ = 6π N = 108 δ = 7π N = 16 67

Circular apertures Different shaped apertures will give different diffraction patterns. A commonly encountered aperture in optics is the circular aperture.... diffraction pattern described by a Bessel function, with the first minimum at λ sinθ = 1. d... where d is the diameter To resolve two objects, 1 λ where θr = sin 1. d Rayleigh criterion θ > θ R θ θ < θ > θr θ = θr R 68

Interference and diffraction with water waves See French 80-94 69