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 performing Simple Harmonic Motion (SHM).

### Step-by-Step Solution:

1. **Understanding the Motion of the Projectile:**
The projectile is launched from the ground with an initial velocity \( u \) at an angle \( \theta \) with the horizontal. The time of flight \( T \) for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
where \( g \) is the acceleration due to gravity.

2. **Finding the Time to Reach Maximum Height:**
The time taken to reach the maximum height (where the projectile will hit the oscillating particle) is half of the total time of flight:
\[ t_{\text{max height}} = \frac{T}{2} = \frac{u \sin \theta}{g} \]

3. **Understanding the SHM of the Particle:**
The particle is executing vertical SHM about the highest point of the projectile. The mean position of the particle corresponds to the equilibrium position in SHM, and it oscillates about this point.

4. **Condition for Collision:**
The projectile will hit the particle when it reaches its maximum height. For the collision to occur, the time taken by the projectile to reach its maximum height must equal the time taken for the particle to reach its mean position.

5. **Relating Time Period of SHM to Time of Flight:**
The time period \( T_p \) of the SHM can be related to the time taken to reach the maximum height of the projectile:
\[ T_p = 2 \times t_{\text{max height}} = 2 \times \frac{u \sin \theta}{g} \]
Thus, the time period of the particle performing SHM is:
\[ T_p = \frac{2u \sin \theta}{g} \]

### Conclusion:
The possible time period of the particle is given by:
\[ T_p = \frac{2u \sin \theta}{g} \]