p34_086 - 86. (a) From Eq. 34-1, 2E 2 = 2 [Em sin(kx t)] =...

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86. (a) From Eq. 34-1, 2 E ∂t 2 = 2 ∂t 2 [ E m sin( kx ωt )] = ω 2 E m sin( kx ωt ) , and c 2 2 E ∂x 2 = c 2 2 ∂x 2 [ E m sin( kx ωt )] = k 2 c 2 sin( kx ωt )= ω 2 E m sin( kx ωt ) . Consequently, 2 E ∂t 2 = c 2 2 E ∂x 2 is satisfed. Analogously, one can show that Eq. 34-2 satisfes 2 B ∂t 2 = c 2 2 B ∂x 2 . (b) From E = E m f ( kx ± ωt ), 2 E ∂t 2 = E m 2 f ( kx ± ωt ) ∂t 2 = ω 2 E m d 2 f du 2 ¯ ¯ ¯ ¯ u = kx ± ωt and c 2 2 E ∂x 2 = c 2 E m 2 f ( kx ± ωt ) ∂t 2 = c 2 E m k 2 d 2 f du 2 ¯ ¯ ¯ ¯ u = kx ± ωt . Since
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