A wave travelling along the positive x- axis | vedclass

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(B) When a wave reflects from a rigid boundary,it undergoes a phase change of $\pi$ radians.Given the incident wave is $y_{incident} = A\sin(\omega t

Vedclass · Accepted Answer

$y = - 0.8A\sin (\omega t + kx)$

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(B) When a wave reflects from a rigid boundary,it undergoes a phase change of $\pi$ radians.Given the incident wave is $y_{incident} = A\sin(\omega t - kx)$.The reflected wave travels in the opposite direction (negative $x-$ axis),so the sign of $kx$ changes to $+kx$.The amplitude of the reflected wave is $80\%$ of the original amplitude,which is $0.8A$.Including the phase change of $\pi$,the reflected wave equation becomes:$y_{reflected} = 0.8A\sin(\omega t + kx + \pi)$.Using the trigonometric identity $\sin(	heta + \pi) = -\sin(	heta)$,we get:$y_{reflected} = -0.8A\sin(\omega t + kx)$.

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

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