A wave travelling along positive x- axis is given by y=A sin (ω t-kx). If it is reflected from a rigid boundary such that 80% amplitude is reflected, then equation of reflected wave is

Question

A wave travelling along positive x- axis is given by y=A sin (ω t-kx). If it is reflected from a rigid boundary such that 80% amplitude is reflected, then equation of reflected wave is (A)  y=A sin (ω t+0.8kx)  (B)  y=-0.8A sin (ω t+kx)  (C)  y=A sin (ω t+kx)  (D)  y=0.8A sin (ω t+kx)

Tardigrade Admin · Accepted Answer

On getting reflected from a rigid boundary the wave suffers. Hence, if yincident=A sin (ω t-kx) Then, yreflected=(0.8A) sin ω t-k(-x)+π =-0.8A sin (ω t+kx) an additional phase change of π .

Q. A wave travelling along positive $x -$ axis is given by $y = A sin (ω t - k x) .$ If it is reflected from a rigid boundary such that 80% amplitude is reflected, then equation of reflected wave is

$y = A sin (ω t + 0.8 k x)$

$y = - 0.8 A sin (ω t + k x)$

$y = A sin (ω t + k x)$

$y = 0.8 A sin (ω t + k x)$

On getting reflected from a rigid boundary the wave suffers. Hence, if
$y_{in c i d e n t} = A sin (ω t - k x)$
Then, $y_{re f l ec t e d} = (0.8 A) sin {ω t - k (- x) + π}$
$= - 0.8 A sin (ω t + k x)$
an additional phase change of $π .$

Q. A wave travelling along positive x− axis is given by y=Asin(ωt−kx). If it is reflected from a rigid boundary such that 80% amplitude is reflected, then equation of reflected wave is

y=Asin(ωt+0.8kx)

y=−0.8Asin(ωt+kx)

y=Asin(ωt+kx)

y=0.8Asin(ωt+kx)