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(B) The resultant wave is formed by the superposition of two waves travelling in opposite directions: $Y_1 = A \sin(\omega t - kx)$ and $Y_2 = A \sin(

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$A$ standing wave having nodes at $x=\left(n+\frac{1}{2}\right) \frac{\lambda}{2}$,where $n=0, 1, 2, 3, \ldots$

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(B) The resultant wave is formed by the superposition of two waves travelling in opposite directions: $Y_1 = A \sin(\omega t - kx)$ and $Y_2 = A \sin(\omega t + kx)$.Using the trigonometric identity $\sin(C) + \sin(D) = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$,we get:$Y = Y_1 + Y_2 = 2A \sin(\omega t) \cos(kx)$.This represents a standing wave.Nodes occur where the amplitude is zero,i.e.,$\cos(kx) = 0$.This implies $kx = (2n+1) \frac{\pi}{2}$ for $n = 0, 1, 2, \ldots$.Substituting $k = \frac{2\pi}{\lambda}$,we get $\frac{2\pi}{\lambda} x = (2n+1) \frac{\pi}{2}$.Solving for $x$,we find $x = (n + \frac{1}{2}) \frac{\lambda}{2}$.

$A$ wave travelling along a uniform string represented by $Y=A \sin (\omega t-k x)$ is superimposed on another wave travelling along the same string represented by $Y=A \sin (\omega t+k x)$. The resultant is

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