A particle is moving along the axis under the influnence of a force given by `F = -5x + 15`. At time `t = 0`, the particle is located at `x = 6` and having zero velocity it take `0.5` second to reach the origin for the first time. The equation of motion of the particle can be respected by

Options

A. `x = 3 + 3 cospit`
B. `x = 3 cospit`
C. `x = 3 + 3 sinpit`
D. `x = 3 + 3 cos(2pit)`

Explanation

To solve the problem step by step, we will analyze the motion of the particle under the influence of the force given and derive the equation of motion.

### Step 1: Analyze the Force
The force acting on the particle is given by:
\[ F = -5x + 15 \]
This force can be rewritten as:
\[ F = -5(x - 3) \]
This indicates that there is a restoring force that acts towards the position \( x = 3 \). Thus, \( x = 3 \) is the equilibrium position (mean position) of the particle.

### Step 2: Determine the Amplitude
At \( t = 0 \), the particle is at \( x = 6 \) with zero velocity. Since the particle is at its extreme position (maximum displacement), the amplitude \( A \) can be calculated as:
\[ A = |6 - 3| = 3 \]
So, the amplitude of the motion is \( 3 \).

### Step 3: Determine the Time Period
The problem states that it takes \( 0.5 \) seconds to reach the origin (which is \( x = 0 \)) for the first time. This means that the time taken to go from the maximum displacement (6) to the mean position (3) and then to the origin is half of the time period \( T \):
\[ 0.5 = \frac{T}{2} \]
Thus, the time period \( T \) is:
\[ T = 1 \text{ second} \]

### Step 4: Calculate Angular Frequency
The angular frequency \( \omega \) is given by:
\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{1} = 2\pi \]

### Step 5: Write the General Equation of Motion
The general equation of motion for a simple harmonic oscillator can be written as:
\[ x(t) = A \cos(\omega t + \phi) + x_0 \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant. Since we have:
- Amplitude \( A = 3 \)
- \( x_0 = 3 \) (mean position)

The equation becomes:
\[ x(t) = 3 + 3 \cos(2\pi t + \phi) \]

### Step 6: Determine the Phase Constant
At \( t = 0 \), the position \( x(0) = 6 \):
\[ 6 = 3 + 3 \cos(\phi) \]
Subtracting \( 3 \) from both sides gives:
\[ 3 = 3 \cos(\phi) \]
Thus:
\[ \cos(\phi) = 1 \]
This implies:
\[ \phi = 0 \]

### Final Equation of Motion
Substituting \( \phi = 0 \) into the equation gives:
\[ x(t) = 3 + 3 \cos(2\pi t) \]

### Conclusion
The equation of motion of the particle is:
\[ x(t) = 3 + 3 \cos(2\pi t) \]