Chapter 2 Solutions. ψ 1 ψ. 16 t. ( z υt ), where υ is in the negative z direction. = 2 A. Chapter 2 Solutions = 32

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1 Chaper Soluions 1 Chaper Soluions.1 υ ( z+ υ) ψ υ( z+ υ) ψ υ I s a wice differeniable funcion of ( z υ ), where υ is in he negaive z direcion.. υ ψ ( y,) ( y 4) ( y 4 ) ψ ( y 4 ) ψ 3 Thus, υ 4, υ 16, and, 16 The velociy is ν 4 in he posiive y direcion..3 Saring wih: A ψ (, z ) ( z υ) + 1 υ ( υ ) z A z [( z υ ) + 1] ψ ( z υ) 1 A + 3 [( z υ) + 1] [( z υ) + 1] 4( z υ) ( z υ) + 1 A [( z υ) + 1] [( z υ) + 1] 3( z υ) 1 A [( z 3 υ ) + 1]

2 Chaper Soluions ( z υ) Aυ [( z υ) + 1] ψ ( z υ) Aυ [( z υ ) + 1] ν 4 ν( z υ) Aυ + ( z υ) 3 υ + υ + [( z ) 1] [( z ) 1] ν[( z υ) + 1] 4 ν( z υ) Aυ + 3 [( z υ) + 1] [( z υ) + 1] 3( z υ) 1 Aυ 3 [( z υ) + 1] Thus since υ The wave moves wih velociy υ in he posiive z direcion..4 c ν λ 3 10 m/s m Hz 474 THz.5 Saring wih: ψ ( y, ) Aexp[ a( by c) ] ψ ( y, ) Aexp[ a( by c) ] Aexp[ a( by c) ] Aa c c y exp[ a( by c) ] b b b ψ 4Aa c c y exp[ a( by c) ] 4 b b b Aa c y exp[ a( by c) ] b b ψ 4Aa c y exp[ a( by c) ] 4 b b Thus ψ ( y, ) Aexp[ a( by c ) ] is a soluion of he wave equaion wih υ cb / in he + y direcion..6 Number of waves, N d/λ ( )/( ) 1. D Nλ Nc/ν 1(3 10 )/( ) 3 m.7 c 3 10 λ v km. c 3 10 For WiFi, λ 6 cm. 9 v 5 10 c 3 10 For ligh used in he definiion of candela, λ nm. 14 v υ 343. λ 66 nm. This is comparable wih ligh waves. ν 5 10

3 Chaper Soluions s 0 waves period. Hence, τ 0.5 s and from ha ν Hz. If wo full wavelenghs fi ino 5 m, hen λ 5/.5 m. The speed is c λν 5 m/s..10 υ νλ m/s λ(94 Hz); λ» 0.4 m..11 For air: λ υ/ν 343/440; λ» 7 cm. For waer: λ υ/ν 1500/440; λ» 3.41 m..1 υ (6 m)/(1. s) 5 m/s; λ υ/ν (5 m/s)/(0 Hz) 0.5 m. Number of wavelenghs, Ν 6/0.5 4 wavelenghs..13 υ νλ (ω/π ) λ and so ω (π /λ)υ..14 q π / π /4 0 π /4 π / 3π /4 sin q 1 / 0 / 1 / cos q 0 / 1 / 0 / sin(q π /4) / 1 / 0 / 1 sin(q π /) 0 / 1 / 0 / sin(q 3π /4) / 0 / -1 / 0 sin(q + π /) 0 / 1 / 0 / q π 5π /4 3π / 7π /4 π sin q 0 / 1 / 0 cos q 1 / 0 / 1 sin(q π /4) / 0 / 1 / sin(q π /) 1 0 / 1 sin(q 3π /4) / 1 / 0 / sin(q + π /) 1 / 0 / 1 sin q leads sin(q p/) x λ/ λ/ 4 0 λ/4 λ/ 3λ/4 λ π kx x π π / 0 π / π 3π /π λ cos( kx π /4) / / / / cos( kx + 3 π /4) // / /// / τ / τ /4 0 τ /4 τ / 3 τ /4 τ ω ( πτ / ) π π / 0 π / π 3π /π sin( ω + π/4) / / / // sin( π/4 ω) /// / ///

4 4 Chaper Soluions.17 λ π /k 4 m, τ λ/ν 1.6 s, ν 1/τ 0.65 Hz and A 0.1 m..1 (a) λ ( ) m 4.0 m (b) υ νλ, so υ 0 m/s ν 5.0 Hz λ 4.0 m (c) ψ( x, ) Asin( kx ω + ε) From he figure, A 0.00 m π π 1 k 0.5π m ; ω πν π (5.0 Hz) 10π rad/ s λ 4.0 m π π π ψ( x, ) [ 0.00 m] sin x 10π 0.00 cos x 10π.19 (a) λ ( ) cm 30.0 cm. (c) υ νλ, so ν υ /λ (100 cm/s)/(30.0 cm) 3.33 Hz.0 (a) τ ( ) s 0.0 s. (b) ν 1/τ 1/(0.0 s) 5.00 Hz. (c) υ νλ, so λ υ/ν (40.0 cm/s)/(5.00 s 1 ).00 cm..1 Firs wave is of he form ψ A sin π (k x + ν). So, (a) ν, (b) λ 1/ , (c) τ 1/ 0.5, (d) A 5, (e) υ 5, and (f) negaive x. Second wave is of he form ψ A sin(kx ω). So, (a) ν 1.5/π, (b) λ π /5, (c) τ π /1.5, (d) A, (e) υ 1.5/5, and (f) posiive x.. (a) ν 0 Hz, (b) τ 1/ s, (c) A 0. m, (d) λ 1/4 0.5 m, (e) υ 5 m/s, and (f) posiive x.3 (a) λ m, (b) τ 0.05 s, (c) ν 1/ Hz, (d) A 0.5 m, (e) υ λ/τ 40 m/s, and (f) negaive y direcion..4 ψ/ x k ψ and Therefore ψ/ k υψ. ψ x υ ψ k + k ψ / (1/ ) / ( ) 0..5 ψ/ x k ψ; ψ/ ω ψ; ω / υ ( πv) / υ ( π/ λ) k ; herefore, ψ/ x (1/ υ ) ψ/ ( k + k ) ψ 0..6 From rigonomery, we know ha sin(α + π /) cosα. Therefore, he same funcion can be expressed as ψ (x, ) A sin(kx ω + π /)..7 υ y ω A cos(kx ω + ε), a y ω y. Simple harmonic moion since a y y.. τ s; herefore ν 1/τ Hz; υ νλ, λ υ /ν m and k π /λ m 1. ψ (x, ) (10 3 V/m) cos[ m 1 (x (m/s))]. I s cosine because cos y(x, ) C/[ + (x + υ ) ]..30 ψ (0, ) A cos(kυ + π) A cos(kυ ) A cos(ω), hen ψ (0, τ/) A cos(ωτ/) A cos (π) A, ψ (0, 3τ/4) A cos(3ωτ/4) A cos(3π/) None of he presened funcions can be differeniaed wice in a nonrivial way (he second derivaives are jus zero in all he cases). So, hey canno be valid wave funcions.

5 Chaper Soluions The phase is ϕ( z) π ( z ), Is parial derivaives are: ( ϕ/ ) π6 10 z 1 4 and ( ϕ/ z ) π 10 ; and as per Eq. (.3), he phase velociy is / ( ϕ/ ) ( ϕ/ z) 3 10 m/s..33 The wavelengh is λ vτ m y ψ( y, ) (0.050 m)sin π + λ τ.5 m 3.6 s ψ( y, ) (0.050 m)sin π + λ τ ψ( y, ) (0.050 m)sin π ψ( y, ) (0.050 m) sin π ( ) ( ) ψ( y, ) (0.050 m)sin π 3.09 ψ ( y, ) 0.07 m.34 dψ/ d ( / x)( dx/ d) + ( / )( dy/ d ) and le y whereupon dψ/ d / x( ± υ) +/ 0 and he desired resul follows immediaely..35 dϕ/ d ( ϕ/ x)( dx/ d) + ϕ/ k( dx/ d) k υ. This is zero when dx/d υ which is wha we mean o prove. Using i on Problem.3, we have 4 1 ( ϕ/ z)( υ) + ϕ/ 0; and hus, π 10 ( υ) π and from here υ 3 10 m/s..36 a(bx + c) ab (x + c/b) g(x + υ ) and so υ c/b and he wave ravels in he negaive x-direcion. Using Eq. (.34) ( / ) /( / x) x [ A( a)( bx + c) c exp[ a( bx + c) ]]/[ A( a)( bx + c) b exp[ a( bx + c) ]] c/ b ; he minus sign ells us ha he moion is in he negaive x-direcion..37 ψ (z, 0) A sin(kz + ε); ψ ( λ /1, 0) A sin( π /6 + ε) 0.66; ψ (λ/6, 0) A sin(π/3 + ε) 1/; ψ (λ/4, 0) A sin (π/ + ε) 0. A sin (π / + ε) A(sin π / cos ε + cos π / sin ε) A cos ε 0, ε π /. A sin(π /3 + π /) A sin(5π /6) 1/; herefore A 1, hence ψ (z, 0) sin (kz + π /)..3 Boh (a) and (b) are waves since hey are wice differeniable funcions of z υ and x + υ, respecively. Thus for (a) ψ a (z b /a) and he velociy is b/a in he posiive z-direcion. For (b) ψ a (x + b/a + c/a) and he velociy is b/a in he negaive x-direcion..39 (a) ψ (y, ) exp[ (ay b) ], a raveling wave in he +y direcion, wih speed υ ω/k b/a. (b) no a raveling wave. (c) raveling wave in he x direcion, υ a/b, (d) raveling wave in he +x direcion, υ ψ (x, ) 5.0 exp[ a( x + b/ a ) ], he propagaion direcion is negaive x; υ ba / 0.6 m/s. ψ (x, 0) 5.0 exp( 5x )..41 ω πv π(440) 763 rad/s. k π / λ πν / υ 3.14 (440/343).06 rad/m. ϕ( x, ) ( kx ω) (.06x 763 ). [ ] [ ] ϕ ϕ( x+ 0., ) ϕ( x, ).06( x+ 0.) x rad. z

6 6 Chaper Soluions x x.4 10 πrad; ϕ( x, ) ( kx ω) π v πυ v. λ c We are looking for d such ha ϕ( x+ d,) ϕ(,) x π. Afer subsiuing, we obain x d x d πυ v + πυ v π. From here, πυ v πor c c c c 3 10 d m 300 nm..43 ψ (x, ) A sin π (x/λ ± /τ), ψ 60 sin π (x/ / ), λ 400 nm, υ / m/s. ν (1/1.33) Hz, τ s..44 exp[ iα]exp[ iβ] (cosα + isin α)(cos β + i sin β) (cosαcos β sinαsin β) + i(sinαcos β + cosαsin β) cos( α + β) + isin( α + β) exp[ i ( α + β)] ψψ * A exp[iω ] A exp[ iω ] A * ; ψψ A. In erms of Euler s formula ψψ * A (cos ω + i sin ω )(cos ω i sin ω ) A (cos ω + sin ω ) A..45 If z x + iy, hen z * x iy and z z * yi..46 z 1 x1 + iy1 z x + iy z + z x + x + iy + iy Re( z + z ) x + x 1 1 Re( z ) + Re( z ) x + x.47 z 1 x1 + iy1 z x + iy 1 1 Re( z ) Re( z ) xx 1 1 Re( z z ) Re( xx + ixy + ixy yy) xx yy Thus Re( z 1) Re( z ) Re( z 1 z )..4 ψ A exp i(k xx + k yy + k zz), k x kα, k y kβ, k z kγ, 1/ 1/ k [( kα) + ( kβ) + ( kγ) ] k( α + β + γ )..49 Consider Eq. (.64), wih ψ/ x α f, ψ/ y β f, / z f, / f. ψ γ ψ υ Then (1/ ) / ( + + 1) f 0 whenever ψ υ ψ α β γ α + β + γ 1. *************************<<INSERT MATTER OF.50 IS MISSING>>***********************

7 Chaper Soluions 7.51 Consider he funcion: ψ (z, ) Aexp[ (a z + b + abz)]. Where A, a, and b are all consans. Firs facor he exponen: 1 (a z + b + abz) (az + b) b z. + a a 1 b Thus, ψ ( z, ) Aexp z + a a. This is a wice differeniable funcion of (z υ ), where υ ba /, and ravels in he z direcion..5 λ h m /( υ) ( ) 4 10 m ν υλ / 500 (4 10 ) Hz. Boh resuls are orders of magniude beyond any known wave phenomena in fac, he wavelengh is wihin an order of magniude of he so-called Planck's lengh he shores disance ha modern physics can conceive of..53 k can be consruced by forming a uni vecor in he proper direcion and muliplying i by k. The uni vecor is ˆ [(4 0) iˆ+ ( 0) ˆj+ (1 0) k]/ (4iˆ+ ˆj+ kˆ)/ 1 and ˆ ˆ ˆ k k(4i + j + k) / 1. r xiˆ+ yj ˆ+ zkˆ, hence ψ( x, y, z, ) Asin[(4 k / 1) x+ ( k/ 1) y+ ( k/ 1) z ω]..54 ˆ ˆ ˆ k (1i + 0 j + 0 k), r xˆi + yˆj + zkˆ, so, ψ Asin( k r ω+ ε) Asin( kx ω+ ε) where k π /λ (could use cos insead of sin)..55 ψ( r1,) ψ[ r ( r r1),] ψ( k r1,) ψ[ k r k ( r r1),] ψ( k r,) ψ( r,) since k ( r r1) 0.56 ψ Aexp[ ik ( r+ ω+ ε)] Aexp[ ikx ( + ky+ kz+ ω+ ε)] x y z The wave equaion is: 1 ψ ψ v where, + + x ψ ( k + k + k ) Aexp[ ikx ( + ky+ kz+ ω+ ε)] x y z x y z ψ ω Aexp ( ω + ε) i kx x ky y kz z where hen, k k + k + k x y z k k + k + k x y z ψ k Aexp[ ikx ( + ky+ kz+ ω + ε)] x y z This means ha ψ is a soluion of he wave equaion if ν ω / k ν ω/. k

8 Chaper Soluions.57 θθ ππ 3ππ/4 ππ/ ππ/4 0 ππ/4 ππ/ 3ππ/4 ππ 5ππ/4 3ππ/ 7ππ/4 ππ sin θθ sin(θθ + ππ/) sin θθ + sin(θθ + ππ/) θθ ππ 3ππ/4 ππ/ ππ/4 0 ππ/4 ππ/ 3ππ/4 ππ 5ππ/4 3ππ/ 7ππ/4 ππ sin θθ sin(θθ 3ππ/4) sin θθ + sin(θθ 3ππ/4) The new funcion will be described by ψ() ψ1() + ψ(). Using simple rigonomery, we can easily find ha ψ( ) Asin( ω + ϕ/) cos ϕ/. Wih a lile rearranging, ψ( ) (Acos ϕ/)sin( ω + ϕ/). The ampliude is A cos ϕ/ and he phase shif is ϕ /..60 x λ / λ /4 0 λ /4 λ / 3λ /4 λ kx π π / 0 π / π 3π /π cos kx cos (kx + π) cos kx + cos (kx + π)