A wave travelling along positive $X$-axis is given by $y = A \sin(\omega t - kx)$. If it is reflected from a…
A wave travelling along positive $X$-axis is given by $y = A \sin(\omega t - kx)$. If it is reflected from a rigid boundary such that $80\%$ amplitude is reflected, then equation of reflected wave is
$y = A \sin(\omega t + 0.8 kx)$
$y = - 0.8 A \sin(\omega t + kx)$
$y = A \sin(\omega t + kx)$
$y = 0.8 A \sin(\omega t + kx)$
Solution
On getting reflected from a rigid boundary, the wave suffers reflection.
Hence, if $y_{\text{incident}} = A \sin(\omega t - kx)$
Then, $y_{\text{reflected}} = (0.8A) \sin \{\omega t - k(-x) + \pi\}$
$= -0.8A \sin (\omega t + kx)$
an additional phase change of $\pi$.