y1 = 0.1sin(3t + 4p x)

at t = 0, y1 = 0.1sin 4p x

at x = 4.5, y1 = 0.1sin (4p )(4.5) = 0.1sin 9(2p ) = 0
 

y2 = 0.2sin(3t - 5p x)

at t = 0, y2 = 0.2sin (-5p x) = 0.2sin[-(5/2)2p x]

at x = 4.5, y2 = 0.2sin [-(5/2)(2p )(4.5)] = 0.2sin [-11.25(2p )]

= 0.2sin[-0.25(2p )] = 0.2 sin(-p /4) = 0.2sin(-45p ) = -0.14
 

y = y1 + y2 = 0 - 0.14 = -0.14

Answer is D

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