A particle of mass 1 kg is subjected to a force which depends on the position as $\overrightarrow{\mathrm{F}}=-\mathrm{k}(\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}}){\mathrm{kgms}}^{-2}$ with $\mathrm{k}=1{\mathrm{kgs}}^{-2}$. At time t = 0, the particle's position $\overrightarrow{\mathrm{r}}=\left(\frac{1}{\sqrt{2}}\widehat{\mathrm{ı}}+\sqrt{2}\widehat{\mathrm{ȷ}}\right)\mathrm{m}$ and its velocity $\overrightarrow{\mathrm{v}}=\left(-\sqrt{2}\widehat{\mathrm{ı}}+\sqrt{2}\widehat{\mathrm{ȷ}}+\frac{2}{\mathrm{\pi }}\widehat{\mathrm{k}}\right){\mathrm{ms}}^{-1}$. Let v_x and v_y denote the x and the y components of the particle's velocity, respectively. Ignore gravity. When z = 0.5 m, the value of $\left({\mathrm{xv}}_{\mathrm{y}}-{\mathrm{yv}}_{\mathrm{x}}\right)$ is _____${\mathrm{m}}^2{\mathrm{s}}^{-1}$.

$F_x=-x=m{a}_x$
so, $a_x=\frac{d^{2}X}{d{t}^2}=-X$
$\Rightarrow x=A_x\sin \left(\omega t+{\varphi }_x\right)(\omega =1\mathrm{rad}/\mathrm{s})$
and $v_x=A_x\omega \cos \left(\omega t+{\varphi }_x\right)$
at $t=0,x=\frac{1}{\sqrt{2}}\mathrm{m}\text{ and }{v}_x=-\sqrt{2}\mathrm{m}/\mathrm{s}$
$\begin{array}{r}\frac{1}{\sqrt{2}}=A_x\sin {\varphi }_x\\ -\sqrt{2}=A_x\cos {\varphi }_x\end{array}$
$\Rightarrow \tan {\varphi }_x=-\frac{1}{2}$........(1)
$A_x=\sqrt{\frac{5}{2}}m$........(2)
Similarly
$\begin{array}{l}{\mathrm{F}}_{\mathrm{y}}=-\mathrm{y}={\mathrm{ma}}_{\mathrm{y}}.\\ \Rightarrow \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}=-\mathrm{y}\\ \text{ So, }\mathrm{y}={\mathrm{A}}_{\mathrm{y}}\sin (\mathrm{\omega t}+\mathrm{\varphi y})(\mathrm{\omega }=1\mathrm{rad}/\mathrm{s})\\ \text{ and }{\mathrm{v}}_{\mathrm{y}}={\mathrm{A}}_{\mathrm{y}}\mathrm{\omega cos}\left(\mathrm{\omega t}+{\mathrm{\varphi }}_{\mathrm{y}}\right)\\ \text{ at }\mathrm{t}=0\mathrm{y}=\sqrt{2}\mathrm{m}\text{ and }{\mathrm{v}}_{\mathrm{y}}=\sqrt{2}\mathrm{m}/\mathrm{s}\\ \text{ So }\sqrt{2}={\mathrm{A}}_{\mathrm{y}}\sin \mathrm{\varphi }\\ \text{ and }\sqrt{2}={\mathrm{A}}_{\mathrm{y}}\cos \mathrm{\varphi }\\ \Rightarrow \mathrm{\varphi }=\frac{\mathrm{\pi }}{4}\text{ and }{\mathrm{A}}_{\mathrm{y}}=2 (3\mathrm{and} 4)\\ \text{ So, }\left({\mathrm{xv}}_{\mathrm{y}}-{\mathrm{yv}}_{\mathrm{x}}\right)=\sqrt{\frac{5}{2}}\sin \left(\mathrm{\omega t}+{\mathrm{\varphi }}_{\mathrm{x}}\right)\times 2\cos \left(\mathrm{\omega t}+{\mathrm{\varphi }}_{\mathrm{y}}\right)-2\sin \left(\mathrm{\omega t}+{\mathrm{\varphi }}_{\mathrm{y}}\right)\times \sqrt{\frac{5}{2}}\cos \left(\mathrm{\omega t}+{\mathrm{\varphi }}_{\mathrm{x}}\right)\\ =\sqrt{\frac{5}{2}}\times 2\left(\sin \left(\mathrm{\omega t}+{\mathrm{\varphi }}_{\mathrm{X}}\right)\cos \left(\mathrm{\omega t}+{\mathrm{\varphi }}_{\mathrm{y}}\right)-\sin \left(\mathrm{\omega t}+{\mathrm{\varphi }}_{\mathrm{y}}\right)\times \cos \left(\mathrm{\omega t}+{\mathrm{\varphi }}_{\mathrm{X}}\right)\right.\\ =\sqrt{10}\sin \left({\mathrm{\varphi }}_{\mathrm{x}}-{\mathrm{\varphi }}_{\mathrm{y}}\right)\\ =\sqrt{10}\left(\sin {\mathrm{\varphi }}_{\mathrm{x}}\cos {\mathrm{\varphi }}_{\mathrm{y}}-\cos {\mathrm{\varphi }}_{\mathrm{X}}\sin {\mathrm{\varphi }}_{\mathrm{y}}\right)\\ =\sqrt{10}\left(\frac{1}{\sqrt{5}}\times \frac{1}{\sqrt{2}}-\left(-\frac{2}{\sqrt{5}}\right)\times \frac{1}{\sqrt{2}}\right)\\ =3\end{array}$