Q. A wave travelling along positive $x$-axis is given by $y = A\sin{(\omega t - kx)}.$ If it is reflected from rigid boundary such that 80% amplitude is reflected, then equation of reflected wave is
A) $y = A\sin{(\omega t + kx)}$
B) $y = - 0.8A\sin{(\omega t + kx)}$
C) $y = 0.8A\sin{(\omega t + kx)}$
D) $y = A\sin{(\omega t + 0.8kx)}$
Explanation
(b) On getting reflected from a rigid boundary the wave suffers
Hence if $y_{incident} = A\sin(\omega t = kx)$
Then $y_{reflected} = (0.8A)\sin\left\lbrack \omega t - k( - x) + \pi \right\rbrack$
$= - 0.8A\sin{(\omega t + kx)}$ an additional phase change of $\pi$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.