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 of the particle when it returns to its equilibrium position is `u_0`.
B. the time at which the particle passes through the equilibrium position for the first time is `t = pi sqrt((m)/(k))`
C. the time at which the maximum compression of the spring occurs is `t = (4 pi)/(3) sqrt((m)/(k))`
D. the time at which the particle passes through the equilibrium position for the second time is `t = (5 pi)/(3) sqrt((m)/(k))`.

Explanation

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

### Step 1: Understand the motion of the particle
The particle starts from the equilibrium position with an initial velocity \( u_0 \). The motion of the particle can be described by simple harmonic motion (SHM) since it is attached to a spring.

### Step 2: Write the equation of motion
The position \( x \) of the particle at time \( t \) can be expressed as:
\[ x(t) = A \sin(\omega t) \]
where \( A \) is the amplitude of the motion and \( \omega \) is the angular frequency given by:
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the time period of the motion, given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

### Step 3: Determine the velocity of the particle
The velocity \( v \) of the particle can be derived from the position function:
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t) \]
At the moment of collision with the wall, the speed of the particle is given as \( 0.5u_0 \). Therefore, we set:
\[ A \omega \cos(\omega t) = 0.5 u_0 \]

### Step 4: Relate the maximum velocity to the initial conditions
The maximum velocity \( v_{max} \) occurs when the particle is at the equilibrium position (where \( x = 0 \)):
\[ v_{max} = A \omega = u_0 \]
From this, we can express \( A \) in terms of \( u_0 \):
\[ A = \frac{u_0}{\omega} \]

### Step 5: Substitute and solve for \( \cos(\omega t) \)
Substituting \( A \) into the velocity equation:
\[ \frac{u_0}{\omega} \cdot \omega \cos(\omega t) = 0.5 u_0 \]
This simplifies to:
\[ \cos(\omega t) = 0.5 \]
Thus, we find:
\[ \omega t = \frac{\pi}{3} \quad \text{(or } 60^\circ\text{)} \]

### Step 6: Calculate the time \( t \)
Using \( \omega = \frac{2\pi}{T} \):
\[ \frac{2\pi}{T} t = \frac{\pi}{3} \]
This gives:
\[ t = \frac{T}{6} \]

### Step 7: Determine the total time to return to equilibrium
The time taken to return to the equilibrium position after the collision is also \( \frac{T}{6} \). Therefore, the total time to return to the equilibrium position after the collision is:
\[ t_{total} = \frac{T}{6} + \frac{T}{6} = \frac{T}{3} \]

### Step 8: Substitute \( T \) into the equation
Substituting \( T = 2\pi \sqrt{\frac{m}{k}} \):
\[ t_{total} = \frac{2\pi \sqrt{\frac{m}{k}}}{3} \]

### Conclusion
The total time taken for the particle to return to its equilibrium position after colliding with the wall is:
\[ t_{total} = \frac{2\pi}{3} \sqrt{\frac{m}{k}} \]