A particle is executing vertical SHM about the highest point of a projectile. When the particle is at the mean position, the projectile is fired from the ground with velocity u at an angle `theta` with the horizontal.The projectile hits the oscillating particle .Then, the possible time period of the particle is

Options

A. `( u sin theta)/( g)`
B. `( 2 u sin theta)/(g)`
C. `( 2 u sin theta )/( 3g)`
D. all the above

Answer

To solve the problem, we need to analyze the motion of both the projectile and the particle executing vertical Simple Harmonic Motion (SHM). Here’s a step-by-step solution:

### Step 1: Understand the Motion of the Projectile
The projectile is launched from the ground with an initial velocity \( u \) at an angle \( \theta \) with the horizontal. We can break this velocity into its horizontal and vertical components:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

### Step 2: Calculate the Time of Flight of the Projectile
The total time of flight \( T \) for a projectile launched at an angle \( \theta \) is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
where \( g \) is the acceleration due to gravity.

### Step 3: Determine the Time to Reach the Highest Point
The time taken to reach the highest point (where the vertical component of velocity becomes zero) is half of the total time of flight:
\[ t_{up} = \frac{T}{2} = \frac{u \sin \theta}{g} \]

### Step 4: Analyze the SHM of the Particle
The particle is executing vertical SHM about the highest point of the projectile. The mean position of the SHM corresponds to the equilibrium position where the particle oscillates.

### Step 5: Condition for Collision
For the projectile to collide with the particle executing SHM, the particle must be at its mean position when the projectile reaches its highest point. This means that the time taken for the projectile to reach its highest point must equal the time taken for the particle to complete half of its oscillation.

### Step 6: Relate the Time of Flight to the Time Period of SHM
Let \( T_p \) be the time period of the particle executing SHM. For the particle to be at the mean position when the projectile reaches its highest point, we have:
\[ t_{up} = \frac{T_p}{2} \]
Substituting \( t_{up} \):
\[ \frac{u \sin \theta}{g} = \frac{T_p}{2} \]

### Step 7: Solve for the Time Period of the Particle
Rearranging the equation gives:
\[ T_p = \frac{2u \sin \theta}{g} \]

### Step 8: Consider Other Possible Cases
The problem states that there can be multiple scenarios for the collision:
1. The projectile hits the particle at the mean position.
2. The projectile hits the particle when it is at the positive extreme.
3. The projectile hits the particle when it is at the negative extreme.

For these cases, the time period can also be expressed as:
- For the first case, \( T_p = \frac{2u \sin \theta}{g} \)
- For the second case, \( T_p = \frac{2u \sin \theta}{3g} \)
- For the third case, \( T_p = \frac{3u \sin \theta}{2g} \)

### Conclusion
The possible time periods of the particle can be:
1. \( T_p = \frac{2u \sin \theta}{g} \)
2. \( T_p = \frac{2u \sin \theta}{3g} \)
3. \( T_p = \frac{3u \sin \theta}{2g} \)