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A particle of mass 2 kg is moving on a straight line under the action of force $F = (8 - 2 x) N .$ The particle is released from a rest at $x = 6 m$ . For the subsequent motion match the following (All the values in the right column are in their S.I. units)
Column-I Column-II
P Equilibrium position is at x= 1 2
Q Amplitude of S.H.M is 2 $π / 2$
R Time taken to go directly from $x = 2$ to $x = 4$ 3 4
S Energy of SHM is 4 6

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$P - 4; Q - 2; R - 1; S - 3$

$P - 3; Q - 1; R - 2; S - 3$

$P - 2; Q - 3; R - 3; S - 4$

$P - 1; Q - 2; R - 4; S - 3$

answer is B.

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Detailed Solution

$F = (8 - 2 x)$
At eq. position
$F = 0$
So, $8 - 2 x = 0$
$x = 4$
(A) $x = 4$
B) $A m p . \Rightarrow A = 6 - 4$ Question Image
$A = 2 m$
C) Total time per $T = \frac{2 π}{ω} = 2 π ∵ ω = 1$
$T i m e t a k e x = 4 t o x = 2 i s \frac{T}{4} = \frac{2 π}{4} = \frac{π}{2}$
D) $a = 4 - x$

$\frac{v d v}{d x} = 4 - x$
$\int_{0}^{v} v^{2} d v = \int_{6}^{4} (4 - x) d x$
$\frac{v^{2}}{2} = {(4 x - \frac{x^{2}}{2}) |}_{6}^{4}$
$\frac{v^{2}}{2} = 2$
At equilibrium position, energy of S.H.M. = K. E. of parts
$= \frac{1}{2} m v^{2} = 2 \times 2 = 4 J$

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