A particle of mass 1 kg is subjected to a force which depends on the position as F-kx-y- kg ms-2 with k-1 kgs-2- At time t-0- the particles-s position r- 1-2-2- m and its velocity v- -2-2-2-k- ms-1- Let Vx and Vy denote the x and the y components of the particle-s velocity respectively- Ignore gravity- When z - 0-5 m- the value of xvy-yvx is m2s-1-

Given: mass m=1 kg F= k(xî+yĵ) kg ms^ 2 k=1 kgs^ 2. at nbsp;t=0, r= ( 1√(2)î+√(2)ĵ )m v= ( √(2î)+√(2)ĵ+2πk̂ )ms^ 1 z = 0.5 m, We know that, for F_x= kx, x= ...

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Byju's Answer

Standard XII

Physics

SHM expression

A particle of...

Question

A particle of mass $1 kg$ is subjected to a force which depends on the position as $\to F = - k (x^i + y^j) k g m s^{- 2}$ with $k = 1 k g s^{- 2}$ . At time $t = 0$ , the particles's position $\to r = (\frac{1}{\sqrt{2}}^i + \sqrt{2}^j) m$ and its velocity $\to v = (- \sqrt{2^i} + \sqrt{2}^j + \frac{2}{π}^k) m s^{- 1}$ . Let $V_{x}$ and $V_{y}$ denote the $x$ and the $y$ components of the particle's velocity respectively. Ignore gravity. When $z = 0.5 m$ , the value of $(x v_{y} - y v_{x})$ is _________ $m^{2} s^{- 1}$ .

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Solution

Given:
mass $m = 1 kg$
$\to F = - k (x^i + y^j) k g m s^{- 2}$
$k = 1 k g s^{- 2}$ .
at $t = 0$ ,
$\to r = (\frac{1}{\sqrt{2}}^i + \sqrt{2}^j) m$
$\to v = (- \sqrt{2^i} + \sqrt{2}^j + \frac{2}{π}^k) m s^{- 1}$
$z = 0.5 m$ ,
We know that, for
$F_{x} = - k x,$
$x = A_{1} sin (w t + ϕ)$
and $V_{x} = A_{1} ω cos (ω t + ϕ_{1})$
$where ω = \sqrt{\frac{k}{m}} = 1, at t = 0 \Rightarrow V_{x} = - \sqrt{2}, x = \frac{1}{\sqrt{2}} \Rightarrow \frac{1}{\sqrt{2}} = A_{1} sin ϕ_{1} a n d - \sqrt{2} = A_{1} cos ϕ_{1}$
After solving we get $A_{1} = \frac{\sqrt{5}}{\sqrt{2}}, sin ϕ_{1} = \frac{1}{\sqrt{5}}, {cos}_{1} = - \frac{2}{\sqrt{5}}$
Now for given value of $z$ $t = \frac{S_{z}}{V_{z}} = \frac{0.5}{\frac{2}{π}} = \frac{π}{4}$
Therefore at $t = \frac{π}{4}, x = \frac{\sqrt{5}}{\sqrt{2}} sin (\frac{π}{4} + ϕ_{1}) x = \frac{\sqrt{5}}{\sqrt{2}} (sin \frac{π}{4} cos ϕ_{1} + cos \frac{π}{4} sin ϕ_{1})$
$x = \frac{\sqrt{5}}{\sqrt{2}} [(\frac{1}{\sqrt{2}} \times (- \frac{2}{\sqrt{5}}) + (\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{5}})] x = - \frac{1}{2}$
$V_{x} = \frac{\sqrt{5}}{\sqrt{2}} cos (\frac{π}{4} + ϕ_{1}) V_{x} = \frac{\sqrt{5}}{\sqrt{2}} [(\frac{1}{\sqrt{2}}) (- \frac{2}{\sqrt{2}}) - (\frac{1}{\sqrt{2}}) (\frac{1}{\sqrt{5}})] V_{x} = - \frac{3}{2}$

$Similarly for , F_{y} = - k y, at t = 0 \Rightarrow V_{y} = \sqrt{2}, y = \sqrt{2}$
$y = A_{2} sin (ω t + ϕ_{2}) \Rightarrow V_{y} = A_{2} ω cos (ω t + ϕ_{2}) where ω = 1$
$\Rightarrow \sqrt{2} = A_{2} sin (ϕ_{2}), \sqrt{2} = A_{2} cos ϕ_{2}$
On solving we get
$A_{2} = 2, ϕ_{2} = \frac{π}{4}$
Therefore for $t = \frac{π}{4}, y = 2 sin (\frac{π}{4} + \frac{π}{4}) = 2, V_{y} = 2 cos (\frac{π}{4} + \frac{π}{4}) = 0$
Now,
$x V_{y} - y V_{x} = (- \frac{1}{2}) (0) - (2) (- \frac{3}{2}) = 3$

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