QJEEAdv22Ph82f8: JEE Advanced 2022 Physics - Class 11 JEEAdv22Ph82f8 Solution: A particle of mass is subjected to a force which depends on the …

Question

Verified Solution • 2025–26 Session moderate

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}

.

Accepted Answer

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}&amp; \Rightarrow \tan \phi_x=-\frac{1}{2} ......(1) \\&amp; \text { and } A_x=\sqrt{\frac{5}{2}} \mathrm{~m} ......(2)\end{aligned}

Similarly

\begin{aligned}&amp; F_y=-y=m a_y . \\\\&amp; \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

Step-by-Step Verified Solution