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

Options

A. the speed to the particle when its returns to its equilibrium is `u_(0)`.
B. the time is at which paricle passes through the equilirbium position for the first time is `t = pisqrt((m)/(k))`.
C. the time at which the maximum compression of the spring occurs is `t = (4pi)/(3)sqrt((m)/(k))`.
D. the time at which the particles passes throuhout the equilirbium position for the second time is `t = (5pi)/(3)sqrt((m)/(k))`.

Explanation

To solve the problem step-by-step, we will analyze the motion of the particle attached to the spring and its interaction with the wall.

### Step 1: Understand the Initial Conditions
- The particle of mass \( m \) is attached to a spring with spring constant \( k \).
- It starts from the equilibrium position with an initial velocity \( u_0 \) at \( t = 0 \).

**Hint:** Remember that the equilibrium position is where the spring is neither compressed nor stretched.

### Step 2: Determine the Time of Collision
- The particle moves in simple harmonic motion (SHM). The velocity of a particle in SHM can be expressed as:
\[ v(t) = u_0 \cos(\omega t) \]
where \( \omega = \sqrt{\frac{k}{m}} \).
- We need to find the time \( t_1 \) when the speed of the particle is \( 0.5 u_0 \):
\[ 0.5 u_0 = u_0 \cos(\omega t_1) \]
Simplifying gives:
\[ \cos(\omega t_1) = 0.5 \]
This implies:
\[ \omega t_1 = \frac{\pi}{3} \]
Therefore:
\[ t_1 = \frac{\pi}{3 \omega} \]

**Hint:** Use the properties of cosine to find angles corresponding to given values.

### Step 3: Determine the Time of Return to Equilibrium
- The particle returns to the equilibrium position after a complete cycle of SHM. The time taken to return to equilibrium after the first collision is \( t_2 = 2t_1 \):
\[ t_2 = 2 \cdot \frac{\pi}{3 \omega} = \frac{2\pi}{3 \omega} \]

**Hint:** The motion of the particle is periodic; each complete cycle takes a certain time.

### Step 4: Calculate the Time of the Second Pass Through Equilibrium
- The time \( t_3 \) when the particle passes through the equilibrium position for the second time is:
\[ t_3 = t_2 + t_1 = \frac{2\pi}{3 \omega} + \frac{\pi}{3 \omega} = \frac{5\pi}{3 \omega} \]

**Hint:** Keep track of the time intervals for each segment of the motion.

### Step 5: Analyze the Collision with the Wall
- The collision with the wall is elastic, meaning the speed of the particle will reverse direction but maintain its magnitude. Thus, after the collision, the speed will still be \( 0.5 u_0 \) but in the opposite direction.

**Hint:** In elastic collisions, both momentum and kinetic energy are conserved.

### Step 6: Determine the Speed When Returning to Equilibrium
- After the collision, the particle will continue its motion and eventually return to the equilibrium position. The speed at this point will be equal to its initial speed \( u_0 \) because the energy in the system remains conserved.

**Hint:** The total mechanical energy in SHM remains constant, allowing you to deduce speeds at various points.

### Final Conclusion
- The speed of the particle when it returns to the equilibrium position after the collision will be \( u_0 \).

**Final Answer:** The speed of the particle when it returns to equilibrium is \( u_0 \).