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 mnotion of the particle can be respected by

Options

A. `x = 3 + 3 cos pi t`
B. `x = 3 cos pi t`
C. `x = 3 + 3 sin pi t`
D. `x = 3 + 3 cos (2pi t)`

Explanation

To solve the problem, we need to find the equation of motion of the particle under the influence of the given force \( F = -5x + 15 \). We'll follow these steps:

### Step 1: Identify the acceleration
Using Newton's second law, we know that \( F = ma \), where \( m \) is the mass of the particle and \( a \) is its acceleration. We can express acceleration as:
\[ a = \frac{F}{m} \]
Substituting the expression for force:
\[ a = \frac{-5x + 15}{m} \]

### Step 2: Set up the differential equation
Since acceleration is the second derivative of position with respect to time, we can write:
\[ m \frac{d^2x}{dt^2} = -5x + 15 \]
This can be rearranged to form a standard second-order linear differential equation:
\[ \frac{d^2x}{dt^2} + \frac{5}{m}x = 15/m \]

### Step 3: Solve the homogeneous part
The homogeneous equation is:
\[ \frac{d^2x}{dt^2} + \frac{5}{m}x = 0 \]
The characteristic equation is:
\[ r^2 + \frac{5}{m} = 0 \]
This gives us:
\[ r = \pm i\sqrt{\frac{5}{m}} \]
Thus, the general solution for the homogeneous part is:
\[ x_h(t) = A \cos\left(\sqrt{\frac{5}{m}} t\right) + B \sin\left(\sqrt{\frac{5}{m}} t\right) \]

### Step 4: Solve the particular solution
To find a particular solution, we can assume a constant solution \( x_p = C \). Substituting into the non-homogeneous equation:
\[ 0 + \frac{5}{m}C = \frac{15}{m} \]
This gives:
\[ C = 3 \]
Thus, the particular solution is:
\[ x_p = 3 \]

### Step 5: Combine solutions
The general solution of the motion is:
\[ x(t) = x_h(t) + x_p = A \cos\left(\sqrt{\frac{5}{m}} t\right) + B \sin\left(\sqrt{\frac{5}{m}} t\right) + 3 \]

### Step 6: Apply initial conditions
We know that at \( t = 0 \), \( x(0) = 6 \) and \( \frac{dx}{dt}(0) = 0 \):
1. From \( x(0) = 6 \):
\[ A + 3 = 6 \implies A = 3 \]
2. From \( \frac{dx}{dt}(0) = 0 \):
\[ \frac{dx}{dt} = -A\sqrt{\frac{5}{m}} \sin\left(\sqrt{\frac{5}{m}} t\right) + B\sqrt{\frac{5}{m}} \cos\left(\sqrt{\frac{5}{m}} t\right) \]
At \( t = 0 \):
\[ B\sqrt{\frac{5}{m}} = 0 \implies B = 0 \]

### Final Equation of Motion
Thus, the equation of motion simplifies to:
\[ x(t) = 3 \cos\left(\sqrt{\frac{5}{m}} t\right) + 3 \]

### Step 7: Determine the mass
Given that the particle takes \( 0.5 \) seconds to reach the origin for the first time, we set \( x(0.5) = 0 \):
\[ 3 \cos\left(\sqrt{\frac{5}{m}} \cdot 0.5\right) + 3 = 0 \]
This leads to:
\[ \cos\left(\sqrt{\frac{5}{m}} \cdot 0.5\right) = -1 \]
Thus:
\[ \sqrt{\frac{5}{m}} \cdot 0.5 = \pi \implies \sqrt{\frac{5}{m}} = \frac{2\pi}{1} \implies \frac{5}{m} = 4\pi^2 \implies m = \frac{5}{4\pi^2} \]

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