JEE Advance - Physics (2022 - Paper 2 Online - No. 1)
A particle of mass $1 \mathrm{~kg}$ is subjected to a force which depends on the position as $\vec{F}=$ $-k(x \hat{\imath}+y \hat{\jmath}) \mathrm{kg}\, \mathrm{m} \mathrm{s}^{-2}$ with $k=1 \mathrm{~kg} \mathrm{~s}^{-2}$. At time $t=0$, the particle's position $\vec{r}=$ $\left(\frac{1}{\sqrt{2}} \hat{\imath}+\sqrt{2} \hat{\jmath}\right) m$ and its velocity $\vec{v}=\left(-\sqrt{2} \hat{\imath}+\sqrt{2} \hat{\jmath}+\frac{2}{\pi} \hat{k}\right) m s^{-1}$. Let $v_{x}$ and $v_{y}$ denote the $x$ and the $y$ components of the particle's velocity, respectively. Ignore gravity. When $z=0.5 \mathrm{~m}$, the value of $\left(x v_{y}-y v_{x}\right)$ is __________ $m^{2} s^{-1}$.
Answer
3
Explanation
$$
F_x=-x=m a_x
$$
So $a_x=\frac{d^2 x}{d t^2}=-x$
$$ \Rightarrow x=A_x \sin \left(\omega t+\phi_x\right) \quad(\omega=1 \mathrm{rad} / \mathrm{s}) $$
and $v_x=A_x \omega \cos \left(\omega t+\phi_x\right)$
at $t=0, x=\frac{1}{\sqrt{2}} \mathrm{~m}$ and $v_x=-\sqrt{2} \mathrm{~m} / \mathrm{s}$
So $\frac{1}{\sqrt{2}}=A_x \sin \phi_x$
and $-\sqrt{2}=A_x \cos \phi_x$
$$ \begin{aligned} & \Rightarrow \tan \phi_x=-\frac{1}{2} ......(1) \\ & \text { and } A_x=\sqrt{\frac{5}{2}} \mathrm{~m} ......(2) \end{aligned} $$
Similarly
$$ \begin{aligned} & F_y=-y=m a_y . \\\\ & \Rightarrow \frac{d^2 y}{d t^2}=-y \end{aligned} $$
So, $y=A_y \sin \left(\omega \mathrm{t}+\phi_{\mathrm{y}}\right) \quad(\omega=1 \mathrm{rad} / \mathrm{s})$ and
$v_y=A_y \omega \cos \left(\omega t+\phi_y\right)$
at $t=0,y=\sqrt{2} \mathrm{~m}$ and $v_y=\sqrt{2} \mathrm{~m} / \mathrm{s}$
So $\sqrt{2}=A_y \sin \phi$
and $\sqrt{2}=A_y \cos \phi$
$$ \Rightarrow \phi=\frac{\pi}{4} \text { and } A_y=2 $$
So,
$\left(x v_y-y v_x\right) $
$ =\sqrt{\frac{5}{2}} \sin \left(\omega t+\phi_x\right) \times 2 \cos \left(\omega t+\phi_y\right)-2 \sin \left(\omega t+\phi_y\right) \times \sqrt{\frac{5}{2}} \cos \left(\omega t+\phi_x\right) $
$ =\sqrt{\frac{5}{2}} \times 2\left(\sin \left(\omega t+\phi_x\right) \cos \left(\omega t+\phi_y\right)-\sin \left(\omega t+\phi_y\right) \times \cos \left(\omega t+\phi_x\right)\right. $
$ =\sqrt{10} \sin \left(\phi_x-\phi_y\right)$
$ =\sqrt{10}\left(\sin \phi_x \cos \phi_y-\cos \phi_x \sin \phi_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) $
$=3$
So $a_x=\frac{d^2 x}{d t^2}=-x$
$$ \Rightarrow x=A_x \sin \left(\omega t+\phi_x\right) \quad(\omega=1 \mathrm{rad} / \mathrm{s}) $$
and $v_x=A_x \omega \cos \left(\omega t+\phi_x\right)$
at $t=0, x=\frac{1}{\sqrt{2}} \mathrm{~m}$ and $v_x=-\sqrt{2} \mathrm{~m} / \mathrm{s}$
So $\frac{1}{\sqrt{2}}=A_x \sin \phi_x$
and $-\sqrt{2}=A_x \cos \phi_x$
$$ \begin{aligned} & \Rightarrow \tan \phi_x=-\frac{1}{2} ......(1) \\ & \text { and } A_x=\sqrt{\frac{5}{2}} \mathrm{~m} ......(2) \end{aligned} $$
Similarly
$$ \begin{aligned} & F_y=-y=m a_y . \\\\ & \Rightarrow \frac{d^2 y}{d t^2}=-y \end{aligned} $$
So, $y=A_y \sin \left(\omega \mathrm{t}+\phi_{\mathrm{y}}\right) \quad(\omega=1 \mathrm{rad} / \mathrm{s})$ and
$v_y=A_y \omega \cos \left(\omega t+\phi_y\right)$
at $t=0,y=\sqrt{2} \mathrm{~m}$ and $v_y=\sqrt{2} \mathrm{~m} / \mathrm{s}$
So $\sqrt{2}=A_y \sin \phi$
and $\sqrt{2}=A_y \cos \phi$
$$ \Rightarrow \phi=\frac{\pi}{4} \text { and } A_y=2 $$
So,
$\left(x v_y-y v_x\right) $
$ =\sqrt{\frac{5}{2}} \sin \left(\omega t+\phi_x\right) \times 2 \cos \left(\omega t+\phi_y\right)-2 \sin \left(\omega t+\phi_y\right) \times \sqrt{\frac{5}{2}} \cos \left(\omega t+\phi_x\right) $
$ =\sqrt{\frac{5}{2}} \times 2\left(\sin \left(\omega t+\phi_x\right) \cos \left(\omega t+\phi_y\right)-\sin \left(\omega t+\phi_y\right) \times \cos \left(\omega t+\phi_x\right)\right. $
$ =\sqrt{10} \sin \left(\phi_x-\phi_y\right)$
$ =\sqrt{10}\left(\sin \phi_x \cos \phi_y-\cos \phi_x \sin \phi_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) $
$=3$
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