A particle of mass $m$ is attached to one end of a mass-less spring of force constant $k,$ lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time $t=0$ with an initial velocity $u_0$. When the speed of the particle is $0.5u_0,$ it collides elastically with a rigid wall. After this collision

The correct option is C The time at which the particle passes through the equilibrium position for the second time is $t=\frac{5\pi }{3}\sqrt{\frac{m}{k}}$

(A) is correct due to conservation of energy.
(B) $x=A\sin \omega t$
$\frac{dx}{dt}=A\omega \cos \omega t=\frac{u_0}{2}=u_0\cos \omega t$
$\Rightarrow \cos \omega t=\frac{1}{2}$
$\Rightarrow \omega t=\frac{\pi }{3}$
$\Rightarrow t=\frac{\pi }{3\omega }$
$\Delta t=\frac{2\pi }{3\omega }=\frac{2\pi }{3}\sqrt{\frac{k}{m}}$, its wrong
(C) Maximum compression will occur at time
$=\frac{2\pi }{3}\sqrt{\frac{k}{m}}+\frac{\pi }{2}\sqrt{\frac{k}{m}}=\frac{7\pi }{6}\sqrt{\frac{k}{m}},$ hence it's wrong.
(D) Second time at equilibrium at
$=\frac{2\pi }{3}\sqrt{\frac{k}{m}}+\pi \sqrt{\frac{k}{m}}=\frac{5\pi }{3}\sqrt{\frac{k}{m}},$ so correct.