problem15_09

University Physics with Modern Physics with Mastering Physics (11th Edition)

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Unformatted text preview: 15.9: a) ∂y = − Ak sin( kx + ωt ) ∂x ∂y = − Aω sin (kx + ωt ) ∂t ∂2 y = − Ak 2 cos(kx + ωt ) 2 ∂x ∂2 y = − Aω2 cos(kx + ωt ), 2 ∂t and so b) ∂2 y ∂x 2 = 2 k2 ∂ y 2 ω ∂t 2 , and y ( x, t ) is a solution of Eq. (15.12) with v = ω k . ∂2 y = − Ak 2 sin( kx + ωt ) 2 ∂x ∂2 y = − Aω2 sin (kx + ωt ), ∂t 2 ∂y = + Ak cos(kx + ωt ) ∂x ∂y = + Aω cos(kx + ωt ) ∂t , and y ( x, t ) is a solution of Eq. (15.12) with v = ω k . c) Both waves are moving in the − x -direction, as explained in the discussion preceding Eq. (15.8). 2 d) Taking derivatives yields v y ( x, t ) = −ωA cos (kx + ωt ) and a y ( x, t ) = −ω A sin (kx + ωt ). and so ∂x 2 ∂2 y = 2 k2 ∂ y 2 ω ∂t 2 ...
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This document was uploaded on 02/05/2008.

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