One end of a spring of force constant K is fixed to a vertical wall and the other to a body of mass m resting on a smooth horizontal surface. There is another wall at a distance $x_{0}$ from the body. The spring is then compressed by $3 x_{0}$ and released. The time taken to strike the wall from the instant of release is (given $\sin^{- 1}$ (l/3) = (π/9))

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$\frac{π}{6} \sqrt{\frac{m}{K}}$

$\frac{2 π}{3} \sqrt{\frac{m}{K}}$

$\frac{π}{4} \sqrt{\frac{m}{K}}$

$\frac{11 π}{9} \sqrt{\frac{m}{K}}$

answer is D.

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Detailed Solution

When spring is compressed by $3 x_{0}$ . Amplitude, $A = 3 x_{0}$ . The time taken from extreme compressed position to mean position, $t_{1} = T / 4 .$
If time taken ( $t_{2}$ ) from mean position to $x = x_{0}$ is given by
$x = A \sin \frac{2 {πt}_{2}}{T} \Rightarrow x_{0} = 3 x_{0} \sin \frac{2 {πt}_{2}}{T} \sin \frac{2 {πt}_{2}}{T} = \frac{1}{3} \Rightarrow \frac{2 {πt}_{2}}{T} = \frac{π}{9} \Rightarrow t_{2} = \frac{T}{18} t_{1} + t_{2} = \frac{T}{4} + \frac{T}{18} = \frac{11}{18} T = \frac{11}{8} 2 π \sqrt{\frac{m}{K}} = \frac{11}{9} π \sqrt{\frac{m}{K}}$