A wave representing by the equation `y = a cos(kx - omegat)` is suerposed with another wave to form a stationary wave such that point `x = 0` is a node. The equation for the other wave isA. `y=asin(kx-omegat)`B. `y=-acos(kx+omegat)`C. `y=-acos(kx-omegat)`D. `y=-asin(kx-omegat)`

A wave representing by the equation `y = a cos(kx - omegat)` is suerpo

1.	A wave representing by the equation `y = a cos(kx - omegat)` is suerposed with another wave to form a stationary wave such that point `x = 0` is a node. The equation for the other wave isA. `y=asin(kx-omegat)`B. `y=-acos(kx+omegat)`C. `y=-acos(kx-omegat)`D. `y=-asin(kx-omegat)`
Answer» Correct Answer - B 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 `(lamda)/(2)` So, if `y_(Incident)=acos(kx-omegat)` `rArr y_("reflected")=acos(-kx-omegat+pi)` `=-acos(omegat+kx)`.

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A wave representing by the equation `y = a cos(kx - omegat)` is suerposed with another wave to form a stationary wave such that point `x = 0` is a node. The equation for the other wave isA. `y=asin(kx-omegat)`B. `y=-acos(kx+omegat)`C. `y=-acos(kx-omegat)`D. `y=-asin(kx-omegat)`

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