Explanation

(d) Equation of stationary wave is
$y_{1} = a\sin{kx\cos{\omega t}}$
And equation of progressive wave is
$y_{2} = a\sin(\omega t - kx)$
$= a(\sin{\omega t\cos{kx - \cos{\omega t\sin{kx}}}})$
At $x_{1} = \frac{\pi}{3k}\ and\ x_{2} = \frac{3\pi}{2k}$
$\sin{kx_{1}\ or\ \sin{kx_{2}}}$is zero.
$\therefore$neither$x_{1}$ nor $x_{2}$ is node.
$\mathrm{\Delta}x = x_{1} - x_{2} = \frac{3\pi}{2k} - \frac{\pi}{3k} = \frac{7\pi}{6k}$
As $\mathrm{\Delta}x = \frac{7\pi}{6k},$ therefore, $\frac{2\pi}{k} > \mathrm{\Delta}x > \frac{\pi}{k}$
But $\frac{2\pi}{k} = \lambda,$ so, $\lambda > \mathrm{\Delta}x > \frac{\lambda}{2}$.
In case of a stationary wave, phase difference between any two points is either zero or $\pi.$
$\therefore\ \ \ \phi_{1} = \pi\ and\ \ \phi_{2} = k\mathrm{\Delta}x = k\frac{7\pi}{6k} = \frac{7}{6}\pi$
$\therefore\ \ \frac{\phi_{1}}{\phi_{2}} = \frac{\pi}{\frac{7}{6}\pi} = \frac{6}{7}$