Two sinusoidal waves of same amplitude of 1 cm  propagate in opposite directions along x-axis. Wavelength of each wave is 6 cm speed of each wave is 0.5 cm/s. Then at any time t, displacement of a particle from its equilibrium position is zero at 

Frequency $\mathrm{f}=\frac{\mathrm{V}}{\mathrm{\lambda }}=\frac{0.5}{6}\mathrm{Hz}=\frac{1}{12}\mathrm{Hz}$

Wave number $\mathrm{K}=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}=\frac{2\mathrm{\pi }}{6}{\mathrm{cm}}^{-1}=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.{\mathrm{cm}}^{-1}$

Equation of wave-1, ${\mathrm{y}}_1=1\sin \left(\frac{\mathrm{\pi x}}{3}-2\mathrm{\pi }.\frac{1}{12}\mathrm{t}\right)$ and equation of wave-2, ${\mathrm{y}}_2=1.\sin \left(\frac{\mathrm{\pi x}}{3}+2\mathrm{\pi }.\frac{1}{12}.\mathrm{t}\right)$

$\therefore$ Equation of resultant wave is 

$\mathrm{y}={\mathrm{y}}_1+{\mathrm{y}}_2=2.\sin \frac{\mathrm{\pi x}}{3}.\cos \frac{\mathrm{\pi t}}{6}\mathrm{cm}$

For the particles which are permanently at rest, y = 0

$\begin{array}{l}\therefore \sin \frac{\mathrm{\pi x}}{3}=0\Rightarrow \frac{\mathrm{\pi x}}{3}=\mathrm{n\pi }\left(\mathrm{n}=1,2,3,.....\right)\\ \Rightarrow \mathrm{x}=3\mathrm{n}\mathrm{cm}\end{array}$