$\sqrt{3} A =\sqrt{ A ^{2}+ A ^{2}+2 A ^{2} \cos \phi}$
$3 A ^{2}=2 A ^{2}+2 A ^{2} \cos \phi$
$\cos \phi=\frac{1}{2}$
$\therefore \phi=60^{\circ}$
$\therefore \text { Phase difference }=60 \text { degree }$

If the position and velocity of the particle at $t=0\, {s}$ are $2\, {cm}$ and $2\, \omega \,{cm} \,{s}^{-1}$ respectively, then its amplitude is $x \sqrt{2} \,{cm}$ where the value of $x$ is ..... .