A particle executes simple harmonic motion represented by displacement function as
$x(t)=A \sin (\omega t+\varphi)$
If the position and velocity of the particle at t = 0 s are 2 cm and $2 \omega \mathrm{~cm} \mathrm{~s}^{-1}$ respectively, then its amplitude is $x \sqrt{2} \mathrm{~cm}$ where the value of x is_____.