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A wave represented by the equation $y_1=a \cos$ $(k x-\omega t)$ is superimposed with another wave to form a stationary wave such that the point $x-0$ is node. The equation for the other wave is
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Verified Answer
The correct answer is:
$a \cos (k x+\omega t+\pi)$
$a \cos (k x+\omega t+\pi)$
Since the point $x=0$ is a node and reflection is taking place from point $x=0$. This means that reflection must be taking place from the fixed end and hence the reflected ray must suffer an additional phase change of $\pi$ or a path change of $\frac{\lambda}{2}$.
So, if $y_{\text {incident }}=a \cos (k x-\omega t)$
$$
\begin{aligned}
\Rightarrow y_{\text {incident }} & =a \cos (-k x-\omega t+\pi) \\
& =-a \cos (\omega t+k x)
\end{aligned}
$$
Hence equation for the other wave
$$
y=a \cos (k x+\omega t+\pi)
$$
So, if $y_{\text {incident }}=a \cos (k x-\omega t)$
$$
\begin{aligned}
\Rightarrow y_{\text {incident }} & =a \cos (-k x-\omega t+\pi) \\
& =-a \cos (\omega t+k x)
\end{aligned}
$$
Hence equation for the other wave
$$
y=a \cos (k x+\omega t+\pi)
$$
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