If the incident wave be represented by y(x,t) = a sin (kx – ωt), then, for reflection at a rigid boundary the reflected wave is represented by?

Question

If the incident wave be represented by y(x,t) = a sin (kx – ωt), then, for reflection at a rigid boundary the reflected wave is represented by?

TuteeHUB · Accepted Answer

y(x,t) = -a sin(kx + ωt + π/2)

TuteeHUB · Answer

y(x,t) = a sin (kx + ωt)

TuteeHUB · Answer

y(x,t) = -a sin(kx + ωt)

TuteeHUB · Answer

y(x,t) = a sin (kx + ωt - π/2)

1.	If the incident wave be represented by y(x,t) = a sin (kx – ωt), then, for reflection at a rigid boundary the reflected wave is represented by?
A.	y(x,t) = a sin (kx + ωt)
B.	y(x,t) = -a sin(kx + ωt)
C.	y(x,t) = -a sin(kx + ωt + π/2)
D.	y(x,t) = a sin (kx + ωt - π/2)
Answer» C. y(x,t) = -a sin(kx + ωt + π/2)

Discussion