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(A) The resultant displacement $Y$ is the sum of the two waves:$Y = A \sin(\omega t - kx) + A \sin(\omega t + kx)$Using the trigonometric identity $\s

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$A$ standing wave having nodes at $x = (n + 1/2) \lambda/2, n = 0, 1, 2$

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(A) The resultant displacement $Y$ is the sum of the two waves:$Y = A \sin(\omega t - kx) + A \sin(\omega t + kx)$Using the trigonometric identity $\sin(C) + \sin(D) = 2 \sin((C+D)/2) \cos((C-D)/2)$:$Y = 2A \sin(\omega t) \cos(kx)$This represents a standing wave.For nodes,the amplitude must be zero,so $\cos(kx) = 0$.$kx = (2n + 1) \pi/2$Since $k = 2\pi/\lambda$:$(2\pi/\lambda) x = (2n + 1) \pi/2$$x = (2n + 1) \lambda/4 = (n + 1/2) \lambda/2$,where $n = 0, 1, 2, \dots$Thus,the nodes are at $x = (n + 1/2) \lambda/2$.

$A$ travelling wave represented by $y = A \sin (\omega t - kx)$ is superimposed on another wave represented by $y = A \sin (\omega t + kx)$. The resultant is

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