A wave represented by the equation y = A cos | vedclass

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Question

(B) Let the first wave be $y_1 = A \cos (kx - \omega t)$.Let the second wave be $y_2 = A \cos (kx + \omega t + \phi)$.The resultant wave is $y = y_1 +

Vedclass · Accepted Answer

$-A \cos (kx + \omega t)$

Vedclass · Answer

(B) Let the first wave be $y_1 = A \cos (kx - \omega t)$.Let the second wave be $y_2 = A \cos (kx + \omega t + \phi)$.The resultant wave is $y = y_1 + y_2 = A [\cos (kx - \omega t) + \cos (kx + \omega t + \phi)]$.Using the identity $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$,we get:$y = 2A \cos (kx + \frac{\phi}{2}) \cos (\omega t + \frac{\phi}{2})$.For $x = 0$ to be a node,the amplitude at $x = 0$ must be zero for all $t$.$2A \cos (k(0) + \frac{\phi}{2}) = 0 \implies \cos \frac{\phi}{2} = 0$.This implies $\frac{\phi}{2} = \frac{\pi}{2}$,so $\phi = \pi$.Substituting $\phi = \pi$ into the equation for $y_2$:$y_2 = A \cos (kx + \omega t + \pi) = -A \cos (kx + \omega t)$.

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

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