Question
The wave $y_1=a \sin (\omega t-k x)$ is superimposed by another wave $y_2=a \sin (\omega t+k x)$ then find the ampli-tude of the resultant wave and also explain what are the condition for the amplitude to be maximum and minimum?

Answer

According to the principle of superposition the amplitude of the resultant wave is
$y = y _1+ y _2$
$\therefore \quad y=a \sin (\omega t-k x)+a \sin (\omega t+k x)$
Therefore, $y =2 a \sin \omega t \cos kx$ or $y =(2 a \cos kx ) \sin \omega t$ so, resultant wave amplitude $A=(2 a \cos k x)$
For maximum amplitude
$\begin{aligned} \cos k x= \pm 1 \Rightarrow & k x= m \pi \\ & (\text { where } m =0,1,2, \ldots \ldots \ldots .)\end{aligned}$
$x=\frac{ m \pi}{ k }($ where $m =0,1,2, \ldots \ldots \ldots)$
For minimum amplitude
$\begin{aligned} \cos k x=0 \Rightarrow & k x=(2 m-1) \pi \\ & (\text { where } m =1,2,3, \ldots \ldots \ldots \ldots .)\end{aligned}$
$\begin{aligned} \therefore \quad & x=\frac{(2 m-1) \pi}{ k } \\ & (\text { where } m =1,2,3, \ldots \ldots \ldots \ldots)\end{aligned}$

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