WAVES CP4 REVISION LECTURE ON. The wave equation. Traveling waves. Standing waves. Dispersion. Phase and group velocities.

Transcription

1 CP4 REVISION LECTURE ON WAVES The wave equation. Traveling waves. Standing waves. Dispersion. Phase and group velocities. Boundary effects. Reflection and transmission of waves.

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3 (a) uniform linear density ρ y T 2 θ 2 small displacements y θ 1 T 1 T 1 cosθ 1 = T 2 cosθ 2 for small θ, cosθ 1 T 1 = T 2 = T ρ δx 2 y t 2 = T sinθ 2 T sinθ 1 sinθ tanθ y x = ρ δx 2 y t 2 = T Thus ρ 2 y t 2 = T 2 y x 2 [ ( y x ) 2 ( y ] x ) 1 }{{} ( 2 y/ x 2 )δx+... x i.e., 2 y x 2 = 1 2 y v 2 t 2, v 2 T ρ wave equation

4 (b) x=a x=a +L Consider general solution with separated variables: y(x,t) = (Acosk(x a)+bsink(x a))(ccoskvt+dsinkvt), v = T/ρ Fixed ends : y(x = a,t) = y(x = a+l,t) = 0 = A = 0,k = nπ/l = y n (x,t) = sin[nπ(x a)/l](c n cos(nπvt/l)+d n sin(nπvt/l)) normal modes general solution : y(x,t) = n y n (x,t) Then, if the initial displacement is of the form y(x,0) = Csin[nπ(x a)/l] for some integer n, subsequent displacements y(x, t) retain the same shape, but with a different time-dependent normalisation f(t) = C cos(nπvt/l) + D sin(nπvt/l)

5 Normal modes of the string with fixed ends: y n (x,t) = sin[nπ(x a)/l](c n cos(nπvt/l)+d n sin(nπvt/l)), v = T/ρ (c) String initially at rest y/ t(x,0) = 0 D n = 0 Initial displacement y(x,0) = Asin[2π(x a)/l]cos[π(x a)/l] can be written as y(x,0) = 1 2 Asin[3π(x a)/l]+ 1 2 Asin[π(x a)/l]. Therefore the displacement of the string at subsequent times is given by y(x,t) = 1 2 Asin[3π(x a)/l]cos(3πvt/l)+ 1 2 Asin[π(x a)/l]cos(πvt/l)

6 June 2011

7 Standing waves with speed c: y(x,t) = (αcoskx+βsinkx)(γcosckt+δsinckt), ω = ck Fixed endpoints: y(x = 0,t) = 0 = α (γcosckt+δsinckt) = 0 for any t = α = 0 y(x = D,t) = 0 = βsinkd = 0 = kd = nπ, n integer = wavenumbers and frequencies: k n = nπ/d, ω n = nπc/d General solution : y(x,t) = n y n (x,t) where y n (x,t) = sin nπx D Two lowest modes: n = 1 : n = 2 : y 1 (x,t) = sin πx D y 2 (x,t) = sin 2πx D ( γ n cos nπct D +δ nsin nπct ) D ( γ 1 cos πct D +δ 1sin πct ) D ( γ 2 cos 2πct D +δ 2sin 2πct ) D

8 June 2011

9 Substituting the ansatz e i(ωt±kx) into the equation gives ω 2 +c 2 k 2 = µ 2 i.e., ω 2 = c 2 k 2 +µ 2 Phase velocity : v p = ω k = c2 k 2 +µ 2 Group velocity : v g = ω k = with k = v p v g = c 2 c 2 k c2 k 2 +µ 2 = v p > c, v g < c = c 1+µ 2 /(c 2 k 2 ) For k 0, v p µ k ; v g c2 k µ 0 For k, v p c; v g c c 1+µ2 /(c 2 k 2 )

10 September Consider the superposition of two travelling waves of equal amplitude with closely spaced angular frequencies and wave numbers, ω = ω 1 ω 2 and k = k 1 k 2, respectively. Show that the resultant wave exhibits beats, and sketch the waveform. Explain the significance of the phase and group velocities, v p and v g, respectively. [6] Show that for waves travelling through a dispersive medium, v g = v p λ dv p dλ, where λ is the wavelength in the medium. [6] Waves propagate through a medium and are characterized by the dispersion relation v 2 p = c2 + λ 2 ω 2 0, where ω 0 is a constant and c is the speed of light. Show that the product of the phase and group velocities is c 2, and comment on the physical significance of the values of v p and v g with respect to c. A wave travels with v g = 0.9c at λ = 350 nm. Calculate the change in group velocity when λ decreases by 0.02 nm. [8]

11 (a) Superposition of travelling waves y 1 (x,t) and y 2 (x,t): y 1 = Asin[(k+ k/2)x (ω+ ω/2)t], y 2 = Asin[(k k/2)x (ω ω/2)t] Using sinα+sinβ = 2sin[(α+β)/2]cos[(α β)/2], the resultant wave is y = y 1 +y 2 = 2Asin(kx ωt)cos[( k x ω t)/2] 1st factor sin varies with frequency ω and wave number k, i.e., close to the original waves y 1 and y 2, and corresponding speed v = ω/k (phase velocity). 2nd factor cos varies much more slowly, with frequency ω/2 and wave number k/2 amplitude modulation, moving at speed v g = ω/ k (group velocity). The modulating envelope encloses a group of short waves. For ω, k 0, v g = dω dk

12 (b) Waves travelling through a dispersive medium: v g = dω dk = d(vk) dk = v + 2π λ dv dλ = v +k dv dk = v + 2π λ ) dv ( λ2 = v λ dv dλ 2π dλ 1 dk/dλ (c) Suppose the dispersion relation v = v(λ) is given by v 2 = c 2 +λ 2 ω 2 0 Then 2v dv = 2λ dλ ω 2 0, i.e. dv dλ = λω2 0 v So v g = v λ dv dλ = v λ2 ω0 2 v, i.e. v g = v v2 c 2 v = c2 v = v g v = c 2

13 v g = c2 v δv g = c2 v 2 dv dλ v g v δλ = c2 v 2 λ δλ So δv g = x 2 c v g = xc v = c x ( x 1 ) δλ x λ = cx(1 x2 ) δλ λ x = 0.9, λ = 350nm, δλ = 0.02nm = δv g = c

14 June Find all possible solutions to the equation u u +c t x = 0 of the form u = g(t)f(x), where c is a real constant. u = g(t)f(x) = f g t +c g f x = 0 Then 1 g g t }{{} function of t f x }{{} function of x = c 1 f = K constant = g = g 0 e Kt, f = f 0 e Kx/c So u = g 0 f 0 }{{} A e K(t x/c)

15 June 2008

16 k 1 = ω / c 1 c 1 c 2 k 2 = ω / c 2 x=0 z I = Re e i(ωt k 1x) ; z T = Re ( te i(ωt k 2x) ) ; z R = Re (re i(ωt+k 1x) ) continuity of z at x = 0: z I +z R = z T = 1+r = t continuity of z/ x at x = 0 = ik 1 +ik 1 r = ik 2 t Hence ik 1 (1 r) = ik 2 (1+r) = r = k 1 k 2 k 1 +k 2 reflected amplitude t = 1+r = 2k 1 k 1 +k 2 transmitted amplitude

17 Note The continuity of z and z/ x at the boundary x = 0 in the previous problem determines the amplitudes of the reflected and transmitted waves in terms of the amplitude of the incident wave as a function of the wave numbers k 1 and k 2 : r = k 1 k 2 k 1 +k 2 ; t = 2k 1 k 1 +k 2 The energy transport across the boundary is described by the coefficients (see next problem) T R reflected flux incident flux = r2 transmitted flux incident f lux = k 2 k 1 t 2 reflection coefficient transmission coefficient The relation 1 = R+T expresses energy conservation.

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19 !!"#$%&''(%)*%+,-.%% y i =e i(k 1x!"t)! 1! 2 v 1,2 = T! 1,2 = " k 1,2 y r =re!i(k 1x+"t) x =0 y t =t e i(k 2x!"t)! k,2 =" # 1,2 T 9$"2.702%602.$42$%72%.$572:;%% / % % 1+r =t k 1 1!r y 1 ( x =0,t)= y 2 ( x =0,t) (String continuous)!y 1!t ( x =0,t)=!y 2!t ( )=k 2 t! r = 1!k 2 /k 1 ( x =0,t) (Forces continuous if no mass at join) = 1! " /" 2 1, t = 1+k 2 /k 1 1+ " 2 /" k 2 /k 1 = 2 1+ " 2 /" 1 +0<"5%$542.=7..702% P 1,2 = 1 2 T!k 1,2 A 2 " R= k 1 k 1 r 2 = % ' & # 1 $ # 2 # 1 + # 2 ( * ) 2 T = k 2 t 2 = # 2 4 k 1 # 1 1+ # 2 /# 1 ( ) = 4 # 1 # 2 2 ( # 1 + # 2 ) 2

20 !"#$%&''(%)*%+,-.%% y i =e i(k 1x!"t)! 1! 2 v 1,2 = T! 1,2 = " k 1,2 y r =re!i(k 1x+"t) x =0 y t =t e i(k 2x!"t)! k,2 =" # 1,2 T 9$"2.702%602.$42$%72%.$572:;%% / % % 1+r =t k 1 1!r y 1 ( x =0,t)= y 2 ( x =0,t) (String continuous)!y 1!t ( x =0,t)=!y 2!t ( )=k 2 t! r = 1!k 2 /k 1 ( x =0,t) (Forces continuous if no mass at join) = 1! " /" 2 1, t = 1+k 2 /k 1 1+ " 2 /" k 2 /k 1 = 2 1+ " 2 /" 1 +,4."%37<"5"26"%="$>""2%72673"2$%423%5"?"6$"3%>4@".% r! 1 "! 2 = r e i#, r is negative for! 1 <! 2 i.e. " =#! +! 2

21 June A long uniform string is stretched along the x-axis, has a mass density µ, and is held under tension T. A transverse wave of displacement y(x,t) travels along the string. (a) The wave equation for transverse waves propagating along the string may be written as 2 y(x,t) x 2 2 y(x,t) t 2. = µ T [ ] Show that a Gaussian pulse propagating along the string, A exp (x+vt) 2 /x 2 0, where A, v, and x 0 are constants, is a solution of the above differential equation. Sketch the pulse at two different locations corresponding to times t 1 and t 2 > t 1. [6] (b) Show that the wave energy density along the string is given by u = 1 2 { µ ( ) y 2 +T t ( ) } y 2 x Hence show that for a sinusoidal wave of amplitude A and angular velocity ω travelling along the string, the energy per wavelength is given by E λ = 1 2 µa2 ω 2. (c) Assume now that the string extends to infinity in the x direction, but the other end is terminated by a mass m at x = 0 which is free to move in the y-direction. Calculate the fraction of the energy reflected when a sinusoidal wave propagates in the +x direction. What happens to the rest of the energy? [4]. [5] [5]

22 (a) 2 y x 2 = µ T 2 y t 2 For y(x,t) = Aexp [ (x+vt) 2 /x 2 0] one has 2 y x 2 = 2 y x (x+vt) 2 y x y t 2 = 2v2 y x 2 0 and +4(x+vt) 2 v 2 y x 4 0 = v 2 2 y x 2. Thus y(x,t) is solution provided v 2 = T/µ. y t t 2 > t 1 1 v (t 2 t 1 ) x

23 (b) linear mass density µ tension T transverse displacements y y dx dy x Kinetic energy of element dx : dk = 1 ( ) 2 y 2 µ dx t dk kinetic energy density : dx = 1 ( ) 2 y 2 µ t ( Potential energy of element dx : dv = T (dx)2 +(dy) dx) 2 = T dx 1+( y/ x) 2 }{{} 1+(1/2)( y/ x) potential energy density : dx = 1 ( y 2 T dx x dv dx = 1 2 T ( ) 2 y x ) 2 Energy density u = dk dx + dv dx = 1 2 µ ( ) 2 y + 1 t 2 T ( ) 2 y x

24 Energy density u = dk dx + du dx = 1 2 µ ( ) 2 y + 1 t 2 T ( ) 2 y x For y(x,t) = Asin(ωt kx), v = ω/k = T/µ, one has y t = ωacos(ωt kx), y x = kacos(ωt kx) 1 λ λ 0 1 λ λ 0 kinetic energy per wavelength λ : dx dk dx = 1 2λ µ ω2 A 2 dx dv dx = 1 2λ T k2 A 2 So E λ = 1 λ λ 0 dx λ dx cos 2 (ωt kx) 0 } {{ } λ/2 potential energy per wavelength λ : λ ( dk dx + dv dx dx cos 2 (ωt kx) 0 } {{ } λ/2 ) = 1 4 µ ω2 A 2 = 1 4 T ω2 v 2 A2 = 1 4 µ ω2 A 2 = 1 4 µ ω2 A µ ω2 A 2 = 1 2 µ ω2 A 2

25 (c) Terminating mass m at x = 0: m 2 y y = T t2 x y(x,t) = e i(ωt kx) +re i(ωt+kx) = mω 2 (1+r) = ikt(1 r) i.e. r = kt imω2 kt +imω 2 R = r 2 = 1 all energy reflected with phase change 2φ, tanφ = mω 2 /Tk phase: + 1 if m = 0, e iπ = 1 if m