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A particle of mass m = 3 kg is constrained to move along a straight line. A and B are two fixed points on the line at a separation of L = $1 m .$ When the particle is at some point P between A and B, it is acted upon by two forces ${\vec{F}}_{1} = (200 \frac{N}{m}) \vec{P A}$ and ${\vec{F}}_{2} = (100 \frac{N}{m}) \vec{P B},$ where magnitudes of $\vec{P A}$ and $\vec{P B}$ are in metre. At time t = 0, the particle is projected from A towards B with speed $v = \frac{10}{\sqrt{3}}$ m/s. Then

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Particle reaches to B at $t = \frac{π}{30}$ sec

Oscillation energy of the particle is $\frac{100}{3} J$

Particle reaches to B at $t = \frac{π}{15}$ sec

Oscillation energy of the particle is $\frac{200}{3} J$

answer is B, C.

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Detailed Solution

$m \vec{a} = - 200 \vec{x} + 100 (\vec{L} - \vec{x}), ∴ a = - \frac{300}{3} (x - \frac{1}{3}) S o, ω = 10 r a d / s$

For amplitude, $\frac{10}{\sqrt{3}} = 10 \sqrt{A^{2} - \frac{1}{9}}, ∴ A = \frac{2}{3} m$

So, $t_{A B} = \frac{T}{12} + \frac{T}{4} = \frac{T}{3},$ where $T = \frac{2 π}{10} \sec$ and $E_{o s c} = \frac{1}{2} (3) (100) (\frac{4}{9}) = \frac{200}{3} J$

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