Particle Attached to Spring Elastic Collision with Rigid Wall

To solve the given problem, let us model the motion of the mass attached to the spring using simple harmonic motion (SHM).

1. Motion before the collision

Let the equilibrium position of the mass be at $x = 0$. The particle starts moving at $t = 0$ from $x = 0$ with an initial speed $u_0$ in the positive $x$-direction. The angular frequency of oscillation is: $\omega = \sqrt{\frac{k}{m}}$

The position $x(t)$ and velocity $v(t)$ of the particle before the collision are given by: $x(t) = A \sin(\omega t)$ $v(t) = A \omega \cos(\omega t)$

where $A$ is the amplitude of the SHM. Since the initial velocity is $v(0) = u_0$, we have: $u_0 = A \omega \implies A = \frac{u_0}{\omega}$

The particle undergoes an elastic collision when its speed becomes $0.5 u_0$. Let $t_1$ be the time of collision: $v(t_1) = A \omega \cos(\omega t_1) = 0.5 u_0 = 0.5 A \omega$ $\cos(\omega t_1) = \frac{1}{2} \implies \omega t_1 = \frac{\pi}{3} \implies t_1 = \frac{\pi}{3}\sqrt{\frac{m}{k}}$

At the instant of collision ($t = t_1$), the position of the particle (location of the rigid wall) is: $x(t_1) = A \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}A$

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2. Motion after the elastic collision

Since the rigid wall is stationary and the collision is perfectly elastic, the magnitude of the velocity remains unchanged while its direction reverses: $v(t_1^+) = -0.5 u_0 = -0.5 A \omega$ The position immediately after the collision remains $x(t_1^+) = \frac{\sqrt{3}}{2}A$.

For $t \ge t_1$, the equation of motion can be written as: $x(t) = A \sin(\omega t + \phi)$ $v(t) = A \omega \cos(\omega t + \phi)$

Substituting the boundary conditions at $t = t_1$: $\sin(\omega t_1 + \phi) = \frac{\sqrt{3}}{2}$ $\cos(\omega t_1 + \phi) = -\frac{1}{2}$

This implies: $\omega t_1 + \phi = \frac{2\pi}{3}$

Since $\omega t_1 = \frac{\pi}{3}$, the phase constant is $\phi = \frac{\pi}{3}$. Thus, for $t \ge t_1$: $x(t) = A \sin\left(\omega t + \frac{\pi}{3}\right)$ $v(t) = A \omega \cos\left(\omega t + \frac{\pi}{3}\right)$

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3. Evaluation of the Options

Option (A): Speed upon returning to equilibrium position

At the equilibrium position, $x(t) = 0$: $\sin\left(\omega t + \frac{\pi}{3}\right) = 0 \implies \cos\left(\omega t + \frac{\pi}{3}\right) = \pm 1$

The magnitude of the velocity (speed) at $x = 0$ is: $|v| = A \omega = u_0$ Thus, Option (A) is correct.

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Option (B): First passage through the equilibrium position

The particle passes through $x = 0$ for the first time after $t = 0$ when: $\omega t + \frac{\pi}{3} = \pi$ $\omega t = \frac{2\pi}{3} \implies t = \frac{2\pi}{3} \sqrt{\frac{m}{k}}$ Thus, Option (B) is incorrect.

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Option (C): Time of maximum compression

Maximum compression occurs when $x(t) = -A$: $\sin\left(\omega t + \frac{\pi}{3}\right) = -1$ $\omega t + \frac{\pi}{3} = \frac{3\pi}{2}$ $\omega t = \frac{3\pi}{2} - \frac{\pi}{3} = \frac{7\pi}{6} \implies t = \frac{7\pi}{6} \sqrt{\frac{m}{k}}$ Thus, Option (C) is incorrect.

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Option (D): Second passage through the equilibrium position

The particle passes through $x = 0$ for the second time after $t = 0$ when: $\omega t + \frac{\pi}{3} = 2\pi$ $\omega t = \frac{5\pi}{3} \implies t = \frac{5\pi}{3} \sqrt{\frac{m}{k}}$ Thus, Option (D) is correct.

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Conclusion

The correct options are A and D.