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 $\mathrm{t}=0 \mathrm{~s}$ are $2 \mathrm{~cm}$ and $2 \omega \mathrm{cm} \mathrm{s}^{-1}$ respectively, then its
amplitude is $x \sqrt{2} \mathrm{~cm}$ where the value of $x$ is
$\mathrm{x}(\mathrm{t})=\mathrm{A} \sin (\omega \mathrm{t}+\phi)$
$\mathrm{v}(\mathrm{t})=\mathrm{A} \omega \cos (\omega \mathrm{t}+\phi)$
$2=A \sin \phi$ ...(1)
$2 \omega=A \omega \cos \phi$ .....(2)
From (1) and (2)
$\begin{aligned} \tan \phi &=1 \\ \phi &=45^{\circ} \end{aligned}$
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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