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Question 14

A wave represented by the equation $$y_1 = a\cos(kx - \omega t)$$ is superimposed with another wave to form a stationary wave such that the point $$x = 0$$ is a node. The equation for the other wave is

Solution

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 $$y_{incident} = a \ cos \ (kx - \omega t)$$ $$\implies \quad y_{incident} = a \ cos \ (-kx - \omega t + \pi)$$

$$= -a \cos \ (\omega t + kx)$$

Hence equation for the other wave is $$y = a \cos \ (kx + \omega t + \pi)$$

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