A particle executes simple harmonic motion represented by displacement function as $x(t)=A \sin (\omega t+\phi)$
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 ..... .
JEE MAIN 2021, Medium
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