JEE Advanced 2022 — Paper 2 — PHY-I solution

JEE Advanced2022Paper 2PHY-I

A particle of mass 1 kg is subjected to a force which depends on the position as $F = - k (x \hat{i} + y \hat{j})$ kg ms $^{- 2}$ with $k = 1$ kg s $^{- 2}$ . At time $t = 0$ , the particle's position $r = (\frac{1}{2} \hat{i} + 2 \hat{j})$ m and its velocity $v = (- 2 \hat{i} + 2 \hat{j} + \frac{2}{π} \hat{k})$ 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 $(x v_{y} - y v_{x})$ is _______ m $^{2}$ s $^{- 1}$ .

Your answer

Solution

The force components give independent SHM along $x$ and $y$ with angular frequency $ω = 1$ rad/s, since $\overset{x}{¨} = - x$ and $\overset{y}{¨} = - y$ .

Write $x (t) = A_{x} sin (t + ϕ_{x})$ and $y (t) = A_{y} sin (t + ϕ_{y})$ . The combination $x v_{y} - y v_{x}$ for two such oscillators equals $A_{x} A_{y} sin (ϕ_{x} - ϕ_{y})$ , a constant of motion (independent of time and of $z$ ).

From the initial conditions at $t = 0$ : $x_{0} = \frac{1}{2}$ , $v_{x 0} = - 2$ give $A_{x} = x_{0}^{2} + v_{x 0}^{2} = \frac{1}{2} + 2 = \frac{5}{2}$ , with $sin ϕ_{x} = \frac{1}{5}$ and $cos ϕ_{x} = - \frac{2}{5}$ .

Similarly $y_{0} = 2$ , $v_{y 0} = 2$ give $A_{y} = 2$ with $sin ϕ_{y} = cos ϕ_{y} = \frac{1}{2}$ .

Therefore $x v_{y} - y v_{x} = A_{x} A_{y} (sin ϕ_{x} cos ϕ_{y} - cos ϕ_{x} sin ϕ_{y}) = 10 (\frac{1}{5} \cdot \frac{1}{2} + \frac{2}{5} \cdot \frac{1}{2}) = 10 \cdot \frac{3}{10} = 3$ .

$∴ x v_{y} - y v_{x} = 3 m^{2} s^{- 1}$ .

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