Chapter 2 Solutions. Chapter 2 Solutions 1. 4( z t) ( z t) z t

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1 Chapter Solutions 1 Chapter Solutions.1. 1 z t ( z t) z z ( z t) t t It s a twice differentiable function of ( z t ), where is in the negative z direction. 1 y t ( y, t) ( y4 t) ( y 4 t) y y 8( y 4 t) t 3 t Thus, 4, 16, and, 1 y 16 t The velocity is 4 in the positive y direction..3 Starting with: A (, z t) ( zt) 1 1 z t ( ) z t A z [( zt ) 1] ( zt) 1 A 3 z [( zt) 1] [( zt) 1] 4( zt) ( zt) 1 A 3 3 [( zt) 1] [( zt) 1] 3( zt) 1 A [( z 3 t) 1]

2 Chapter Solutions ( zt) A t [( zt) 1] ( zt) A t t [( z t ) 1] 4 ( zt) A ( zt) 3 [( z t) 1] [( z t) 1] [( zt) 1] 4 ( zt) A 3 [( zt) 1] [( zt) 1] 3( zt) 1 A 3 [( zt) 1] Thus since 1 z t The wave moves with velocity in the positive z direction..4 c 8 c 3 10 m/ s m.5 Starting with: ( y, t) Aexp[ a( byct) ] ( y, t) Aexp[ a( byct) ] Aexp[ a( byct) ] Aa c c y texp[ a( byct) ] t b b b 4Aa c c y t exp[ a( byct) ] 4 t b b b Aa c y texp[ a( byct) ] y b b 4Aa c y t exp[ a( byct) ] 4 y b b 14 Hz Thus ( y, t) Aexp[ a( byct ) ] is a solution of the wave equation with cb / in the + y direction..6 number of waves 131, c, 9 (0.003) ( / ) 8 10 c / 310 /10, 3cm. Waves extend 3.9 m c / 3 10 / m 0.6 m / m 5 10 km m/s.

3 Chapter Solutions 3.9 The time between the crests is the period, so 1/s; hence 1/.0 Hz. As for the speed Lt / 4.5 m/1.5 s 3.0 m/s. We now know,, and and must determine. Thus, / 3.0 m/s/.0 Hz 1.5 m..10 = = m/s = (4.3 m); = 0.81 khz..11 = = 1498 m/s = (440 HZ) ; = 3.40 m..1 = (10 m)/.0 s) = 5.0 m/s; = / = (5.0 m/s)/(0.50 m) = 10 Hz..13 (/ ) and so ( /)..14 q / /4 0 /4 / 3/4 sin q 1 / 0 / 1 / cos q 0 / 1 / 0 / sin(q /4) / 1 / 0 / 1 sin(q /) 0 / 1 / 0 / sin(q 3 /4) / 0 / -1 / 0 sin(q /) 0 / 1 / 0 / q 5/4 3/ 7/4 sin q 0 / 1 / 0 cos q 1 / 0 / 1 sin(q /4) / 0 / 1 / sin(q /) 1 0 / 1 sin(q 3 /4) / 1 / 0 / sin(q /) 1 / 0 / 1 sin q leads sin(q p/) x / / 4 0 /4 / 3/4 kx x / 0 / 3/ cos( kx /4) / / / / cos( kx 3 /4) / / / / / / / t / /4 0 /4 / 3 /4 t ( / ) t / 0 / 3/ sin( t /4) / / / / / sin( /4 t) / / / / / / /

4 4 Chapter Solutions.17 Comparing y with Eq. (.13) tells us that A 0.0 m. Moreover, / 157 m 1 and so /(157m l ) m. The relationship between frequency and wavelength is, and so = / = (1. m/s)/ m 30 Hz. The period is the inverse of the frequency, and therefore l/ s..18 (a) ( ) m 4.0 m (b),so 0 m/s 5.0 Hz 4.0 m (c) ( xt, ) Asin( kxt ) From the figure, A = 0.00 m 1 k 0.5 m ; (5.0 Hz) 10 rad/ s 4.0 m ( xt, ) 0.00 msin x10t 0.00 cos x10t.19 (a) ( ) cm 30.0 cm. (c), so / (100 cm/s)/(30.0 cm) 3.33 Hz.0 (a) ( ) s 0.0 s. (b) 1/ 1/(0.0 s) 5.00 Hz. (c), so / (40.0 cm/s)/(5.00 s 1 ) 8.00 cm..1 A sin (k x t), 1 4sin(0.x 3t). (a) 3, (b) 1/0., (c) 1/3, (d) A 4, (e) 15, (f) positive x A sin(kx t), (1/.5) sin(7x 3.5t). (a) 3.5/, (b) / 7, (c) /3.5, (d) A 1/.5, (e) 1/, (f) negative x. From of Eq. (.6) (x, t) A sin(kx t) (a), so / (0.0 rad/s)/, (b) k /, so /k /(6.8 rad/m) 1.00 m, (c) 1/, so 1/ 1/(10.0/ Hz) 0.10s, (d) From the form of, A 30.0 cm, (e) /k (0.0 rad/s)/(6.8 rad/m) 3.18 m/s, (f) Negative sign indicates motion in x direction..3 (a) 10, (b) c Hz, (c) m, (e).0 10 s, (f) y direction (d) m/s,.4 / x k and / t k. Therefore x t k k / (1/ ) / ( ) 0..5 / x k ; / t ; / ( v) / ( / ) k ; therefore, / x (1/ ) / t ( k k ) 0..6 (x, t) A cos(kx t (/)) A{cos(kx t) cos(/) sin(kx t) sin(/)} A sin(kx t).7 y A cos(kx t ), a y y. Simple harmonic motion since a y y.

5 Chapter Solutions s; therefore 1/ Hz;, / m and k / m 1. (x, t) (10 3 V/m) cos[ m 1 (x (m/s)t)]. It s cosine because cos y(x, t) C/[ (x t) ]..30 (0, t) A cos(kt ) A cos(kt) A cos(t), then (0, /) A cos(/) A cos () A, (0, 3/4) A cos(3/4) A cos(3/) Since (y, t) (y t) A is only a function of (y t), it does satisfy the conditions set down for a wave. Since / y / t 0, this function is a solution of the wave equation. However, (y, 0) Ay is unbounded, so cannot represent a localized wave profile..3 k m 1, Hz, /k m/s..33 / (.0 m/s)(1/4 s) 0.5 m z t ( zt, ) (0.00 m)sin 0.50 m 1/4 s 1.5 m. s ( zt, ) (0.00 m)sin 0.50 m 1/4 s (z, t) (0.00 m) sin ( ) (z, t) (0.00 m) sin (11.8) (z, t) (0.00 m) sin 3.6 (z, t) (0.00 m) (0.9511) (z, t) m.34 d / dt ( / x)( dx/ dt) ( / y)( dy/ dt ) and let y t whereupon d / dt / x( ) / t 0 and the desired result follows immediately..35 d / dt ( / x)( dx/ dt) / t 0 k( dx/ dt) k and this is zero provided dx/dt, as it should be. For the particular wave of d Problem.3, / y( ) / t ( ) dt and the speed is m/s..36 a(bx + ct) ab (x + ct/b) g(x + t) and so c/b and the wave travels in the negative x-direction. Using Eq. (.34) / t / / x x t [ A( a)( bx ct) c exp[ a( bx ct) ]]/[ A( a)( bx ct) bexp[ a( bx ct) ]] c/ b; the minus sign tells us that the motion is in the negative x-direction..37 (z, 0) A sin(kz ); ( /1, 0) A sin( /6 ) 0.866; (/6, 0) A sin(/3 ) 1/; (/4, 0) A sin (/ ) 0. A sin (/ ) A(sin / cos cos / sin ) A cos 0, /. A sin(/3 /) A sin(5/6) 1/; therefore A 1, hence (z, 0) sin (kz /)..38 Both (a) and (b) are waves since they are twice differentiable functions of z t and x t, respectively. Thus for (a) a (z bt /a) and the velocity is b/a in the positive z-direction. For (b) a (x bt/a c/a) and the velocity is b/a in the negative x-direction.

6 6 Chapter Solutions.39 (a) (y, t) exp[ (ay bt) ], a traveling wave in the y direction, with speed /k b/a. (b) not a traveling wave. (c) traveling wave in the x direction, a/b, (d) traveling wave in the x direction, (x, t) 5.0 exp[ ax ( bat / ) ], the propagation direction is negative x; ba / 0.6 m/s. (x, 0) 5.0 exp(5x )..41 / m; 10.0 cm is a fraction of a wavelength viz. (0.100 m)/(0.300 m) 1/3; hence /3.09 rad corresponds to /1 or 4nm (x, t) A sin (x/ ± t/), 60 sin (x/ t/ ), 400 nm, / m/s. (1/1.33) Hz, s..44 exp[ i]exp[ i] (cos isin )(cos i sin ) (cos cos sin sin ) i(sin cos cossin ) cos( ) isin( ) exp[ i ( )] * A exp[it] A exp[ it] A * ; A. In terms of Euler s formula * A (cost i sin t)(cos t i sin t) A (cos t sin t) A..45 If z x iy, then z * x iy and z z * yi..46 z 1 x1 iy1 z x iy z1 z x1 x iy1 iy Re( z1 z) x1 x Re( z ) Re( z ) x x z 1 x1 iy1 z x iy Re( z1) Re( z) x1x Re( z z ) Re( x x ix y ix y y y ) x x y y Re( z ) Re( z ) Re( z z ) Thus A exp i(k x x k y y k z z), k x k, k y k, k z k, 1/ 1/ k [( k) ( k) ( k) ] k( )..49 Consider Eq. (.64), with / x f, / y f, /, /. z f t f Then t f whenever (1/ ) / ( 1) 0 1. *************************<<INSERT MATTER OF.50 IS MISSING>>***********************

7 Chapter Solutions 7.51 Consider the function: (z, t) Aexp[(a z b t abzt)]. Where A, a, and b are all constants. First factor the exponent: (a z b t abzt) (az bt) 1 b z t. a a 1 b Thus, (,) z t Aexp z t a a. This is a twice differentiable function of (z t), where ba /, and travels in the z direction..5 (h/m) /6(1) m..53 k can be constructed by forming a unit vector in the proper direction and multiplying it by k. The unit vector is ˆ [(40) iˆ(0) ˆj(10) k]/ 4 1 (4iˆ ˆj kˆ)/ 1 and ˆ ˆ ˆ k k(4i j k) / 1. r xiˆyj ˆzkˆ, hence ( x, y, z, t) Asin[(4 k / 1) x( k/ 1) y( k/ 1) z t]..54 ˆ ˆ ˆ k (1i 0 j 0 k), r xˆi yˆj zkˆ, so, Asin( kr t) Asin( kxt) where k / (could use cos instead of sin)..55 ( r1, t) [ r ( r r1), t] ( kr1, t) [ kr k( r r1), t] ( kr, t) ( r, t) since k( r r1) 0.56 Aexp[ i( kr t)] Aexp[ i( kxxkyykzzt)] The wave equation is: 1 v t where, x y z ( kx ky kz ) Aexp[ i( kxxkyykzzt)] Aexp ikx x ky y kz z t t where k k k k x y z k kx ky kz then, k Aexp[ i( kxxkyykzzt )] This means that is a solution of the wave equation if / k /. k

8 8 Chapter Solutions θ / /4 0 /4 / 3 /4 5 / 4 3/ 7/4 sin θ 1 1/ 0 1/ 1 1/ 0 1/ 1 1/ 0 sin θ sin θ 3 3/ 0 3/ 3 3/ 0 3/ 3 3/ 0 / /4 0 /4 / 3/4 5 / 4 3/ 7/4 sin 1 1/ 0 1/ 1 1/ 0 1/ 1 1/ 0 sin( /) 0 1/ 1 1/ 0 1/ 1 1/ 0 1/ 1 sin sin( /) Note that the amplitude of {sin() sin( /)} is greater than 1, while the amplitude of {sin() sin( 3/4) is less than 1. The phase difference is /8..60 x / /4 0 /4 / 3/4 kx / 0 / 3/ cos kx cos (kx ) cos kx cos (kx )