Step-by-step Solution

1. Analyze the initial motion (before collision)

The particle of mass m is attached to a spring with force constant k. It starts from the equilibrium position (x=0) at t=0 with an initial velocity $u_0$. This is a Simple Harmonic Motion (SHM).

The angular frequency of the oscillation is $\omega = \sqrt{k/m}$.

The equations for position x(t) and velocity v(t) for an SHM starting from the equilibrium position with a positive velocity are: $x(t) = A \sin(\omega t)$ $v(t) = A\omega \cos(\omega t)$

At t=0, $v(0) = u_0$. From the velocity equation, $v(0) = A\omega \cos(0) = A\omega$. Therefore, the amplitude of the motion is $A = u_0 / \omega = u_0 \sqrt{m/k}$.

The equations of motion before the collision are: $x(t) = A \sin(\omega t)$ $v(t) = u_0 \cos(\omega t)$

2. Determine the instant of collision

The collision occurs when the speed of the particle is $0.5 u_0$. Since the particle is moving away from the equilibrium position for the first time, its velocity is positive. $v(t_{coll}) = 0.5 u_0$ $u_0 \cos(\omega t_{coll}) = 0.5 u_0$ $\cos(\omega t_{coll}) = 0.5$

The smallest positive value for $\omega t_{coll}$ is $\pi/3$. So, the time of collision is $t_{coll} = \frac{\pi}{3\omega} = \frac{\pi}{3} \sqrt{\frac{m}{k}}$.

At this time, the position of the particle is: $x_{coll} = x(t_{coll}) = A \sin(\omega t_{coll}) = A \sin(\pi/3) = A \frac{\sqrt{3}}{2}$.

The velocity just before the collision is $v_{before} = 0.5 u_0$.

3. Analyze the motion after the collision

The collision is elastic with a rigid wall. This means the velocity of the particle is reversed, while its position remains the same at the instant of collision. Velocity just after collision: $v_{after} = -v_{before} = -0.5 u_0$. Position just after collision: $x_{after} = x_{coll} = A \frac{\sqrt{3}}{2}$.

The total mechanical energy of the system is $E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$. Let's check the energy just after the collision: $E_{after} = \frac{1}{2}m(v_{after})^2 + \frac{1}{2}k(x_{after})^2$ $E_{after} = \frac{1}{2}m(-0.5 u_0)^2 + \frac{1}{2}k(A \frac{\sqrt{3}}{2})^2$ Since $k = m\omega^2$ and $A = u_0/\omega$ (so $k A^2 = m u_0^2$): $E_{after} = \frac{1}{2}m(\frac{u_0^2}{4}) + \frac{1}{2}k A^2 (\frac{3}{4}) = \frac{1}{8}mu_0^2 + \frac{3}{8}m u_0^2 = \frac{4}{8}m u_0^2 = \frac{1}{2}m u_0^2$. This is the same as the initial energy. Since the total energy $E = \frac{1}{2}kA'^2$ is conserved, the amplitude of oscillation A' after the collision remains the same as the initial amplitude, A' = A.

4. Evaluate the options

A: The speed of the particle when it returns to its equilibrium position is u0. At the equilibrium position (x=0), the potential energy is zero, and all the energy is kinetic: $E = \frac{1}{2}mv_{eq}^2$. Since the total energy of the system remains $\frac{1}{2}mu_0^2$ after the collision: $\frac{1}{2}mv_{eq}^2 = \frac{1}{2}mu_0^2$ $v_{eq}^2 = u_0^2 \implies |v_{eq}| = u_0$. So, the speed at the equilibrium position is $u_0$. Option A is correct.

To analyze the remaining options, let's find the equation of motion for $t \ge t_{coll}$. We can define a new SHM starting from $t=t_{coll}$ with initial conditions $x = A\sqrt{3}/2$ and $v = -0.5 u_0 = -A\omega/2$. Let the new motion be described by $x(t) = A \cos(\omega(t - t_{coll}) + \delta)$. $v(t) = -A\omega \sin(\omega(t - t_{coll}) + \delta)$. At $t = t_{coll}$: $x(t_{coll}) = A \cos(\delta) = A\sqrt{3}/2 \implies \cos(\delta) = \sqrt{3}/2$. $v(t_{coll}) = -A\omega \sin(\delta) = -A\omega/2 \implies \sin(\delta) = 1/2$. These two conditions give $\delta = \pi/6$. So, for $t \ge t_{coll}$, the position is given by $x(t) = A \cos(\omega(t - t_{coll}) + \pi/6)$.

B: The time at which the particle passes through the equilibrium position for the first time is $t = \pi \sqrt{m/k}$. The particle was at equilibrium at t=0. The first time it returns to equilibrium is after the collision. We need to find the smallest $t > t_{coll}$ for which x(t)=0. $A \cos(\omega(t - t_{coll}) + \pi/6) = 0$ $\omega(t - t_{coll}) + \pi/6 = \pi/2$ (the first time the phase reaches a point where cosine is zero after $\pi/6$) $\omega(t - t_{coll}) = \pi/2 - \pi/6 = \pi/3$ $t - t_{coll} = \frac{\pi}{3\omega}$ $t = t_{coll} + \frac{\pi}{3\omega} = \frac{\pi}{3\omega} + \frac{\pi}{3\omega} = \frac{2\pi}{3\omega} = \frac{2\pi}{3} \sqrt{\frac{m}{k}}$. This is not equal to $\pi \sqrt{m/k}$. Option B is incorrect.

C: The time at which the maximum compression of the spring occurs is $t = \frac{4\pi}{3}\sqrt{m/k}$. Maximum compression occurs at x = -A. $A \cos(\omega(t - t_{coll}) + \pi/6) = -A$ $\cos(\omega(t - t_{coll}) + \pi/6) = -1$ $\omega(t - t_{coll}) + \pi/6 = \pi$ (first time the phase reaches $\pi$) $\omega(t - t_{coll}) = \pi - \pi/6 = 5\pi/6$ $t - t_{coll} = \frac{5\pi}{6\omega}$ $t = t_{coll} + \frac{5\pi}{6\omega} = \frac{\pi}{3\omega} + \frac{5\pi}{6\omega} = \frac{2\pi + 5\pi}{6\omega} = \frac{7\pi}{6\omega} = \frac{7\pi}{6} \sqrt{\frac{m}{k}}$. This is not equal to $\frac{4\pi}{3} \sqrt{m/k}$. Option C is incorrect.

D: The time at which the particle passes through the equilibrium position for the second time is $t = \frac{5\pi}{3}\sqrt{m/k}$. The first time (after t=0) the particle passes through equilibrium is at $t_1 = \frac{2\pi}{3\omega}$. After passing through equilibrium, the particle completes half an oscillation to return to the equilibrium position again. The time for half an oscillation is $T/2 = (2\pi/\omega)/2 = \pi/\omega$. So, the second time it passes through equilibrium is: $t_2 = t_1 + T/2 = \frac{2\pi}{3\omega} + \frac{\pi}{\omega} = \frac{2\pi + 3\pi}{3\omega} = \frac{5\pi}{3\omega}$. $t_2 = \frac{5\pi}{3} \sqrt{\frac{m}{k}}$. Option D is correct.

Final check: The correct options are A and D.