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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As given
$v(t)=A \omega \cos (\omega t+\phi)$
$2=A \sin \phi.....(1)$
$2 \omega=A \omega \cos \phi.....(2)$
From $(1)$ and $(2)$
$\tan \phi=1$
$\phi=45^{\circ}$
Putting value of $\phi$ in equation $(1)$
$2=A\left\{\frac{1}{\sqrt{2}}\right\}$
$A=2 \sqrt{2}$
$x=2$
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