The following equations represent transverse waves

$\begin{array}{r}\left(a\right)z_1=A\cos (kx-\omega t)\\ \\ \left(b\right)z_2=A\cos (kx+\omega t)\\ \left(c\right)z_3=A\cos (ky-\omega t)\end{array}$

Identify the combination (s) of the waves which will produce (i) standing wave (s) and (ii) a wave travelling in the direction making an angle of $45°$ with the positive x and positive y axes. In each case, find the positions at which the resultant intensity is always zero. 

The given waves are represented by

$\begin{array}{l}{z}_1=A\cos (kx-\omega t) \left(1\right) \\ z_2=A\cos (kx+\omega t) \left(2\right)\\ z_3=A\cos (ky-\omega t) \left(3\right)\end{array}$

(i) Standing waves are produced when two waves travelling along the same straight line but in opposite direction superpose. Equation (1) represents a wave travelling along the positive x direction and Eq. (2) represents a wave travelling along the negative x direction. Hence they will produce standing waves when they superpose. From the superposition principle, the resultant wave is given by 

$\begin{array}{l}z=z_1+z_2\\ =A\cos (kx-\omega t)+A\cos (kx+\omega t)\\ =2A\cos \left\{\frac{(kx-\omega t)+(kx+\omega t)}{2}\right\}\times \\ \cos \left\{\frac{(kx+\omega t)-(kx-\omega t)}{2}\right\}\\ =2A\cos kx\cos \omega t\text{ or }z=A_r\cos \omega t\end{array}$

where $A_r=2 A \cos kx$ is the resultant amplitude. The resultant intensity is 

$I\propto A_{\mathrm{r}}^2\propto 4A^2{\cos }^{2}kx$

Therefore, $I$ will be zero if 

${\cos }^{2}kx=0\text{ or }\cos kx=0$

or $kx=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\dots$

  $=\left(n+\frac{1}{2}\right)\pi ;n=0,1,2,3,\dots$

or $x=\left(n+\frac{1}{2}\right)\frac{\pi }{k}=\frac{\pi }{2k},\frac{3\pi }{2k},\frac{5\pi }{2k},\dots$etc.

The intensity is zero at these values of $x.$ These values correspond to nodes.

(ii) When two waves of equal amplitudes, one travelling along the positive x direction and the other travelling along the positive y direction, both having displacements in the same straight line, superpose, the resultant wave travels in a direction making an angle $45°$ with the positive x and positive y directions. Equation (1) represents a transverse wave of amplitude A travelling along the positive x direction with particle displacements $z_1$ along the z direction. Equation (3) represents a transverse wave of amplitude A travelling along the positive y direction with particle displacements $z_3$ along the z direction. When these two waves superpose, the particle displacements in the resultant wave are given by 

$\begin{array}{l}z=z_1+z_3\\ =A\cos (kx-\omega t)+A\cos (ky-\omega t)\\ =2A\cos \left\{\frac{(kx-\omega t)+(ky-\omega t)}{2}\right\}\times \cos \left\{\frac{(kx-\omega t)-(ky-\omega t)}{2}\right\}\\ =2A\cos \left[\frac{k(x+y)}{2}-\omega t\right]\times \cos \left[\frac{k(x-y)}{2}\right]\end{array}$

or  $z=A_{\mathrm{r}}\cos \left[\frac{k(x+y)}{2}-\omega t\right]$

where $A_{\mathrm{r}}=2A\cos \left[\frac{k(x-y)}{2}\right]$ is the resultant amplitude.

The resultant intensity is 

$I\propto A^2\propto 4A^2{\cos }^2\left[\frac{k(x-y)}{2}\right]$

Therefore, $l$ will be zero if

${\cos }^2\left[\frac{k(x-y)}{2}\right]=0\text{ or }\cos \left[\frac{k(x-y)}{2}\right]=0$

or $\frac{k(x-y)}{2}=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},.....$etc.

$=\left(n+\frac{1}{2}\right)\pi ;n=0,1,2,\dots$etc.

or $(x-y)=(2n+1)\frac{\pi }{k};n=0,1,2,\dots$etc.

$=\frac{\pi }{k},\frac{3\pi }{k},\frac{5\pi }{k},\dots$etc.

Hence the intensity of the wave will be zero at points whose $x$ and $y$ coordinates satisfy the above condition.