Passage IX) For SHM to take place force acting on the body should be proportional to -x or F = -kx. If A be the amplitude then energy of oscillation is `1/2`k`A^(2)`
Force acting on a block is `F=(-4x +8)`. Here F is in Newton and x the position of block on x-axis in meter

Options

A. motion of the block is periodic but not simple harmonic
B. Motion of the block is not periodic
C. motion of the block is simple harmonic about origin, x=0
D. motion of the block is simple harmonic about x=2m.

Answer

To solve the problem, we will analyze the force acting on the block and determine if it exhibits simple harmonic motion (SHM).

### Step-by-Step Solution:

1. **Identify the Force Equation**: The force acting on the block is given by:
\[ F = -4x + 8 \]

2. **Rearranging the Force Equation**: We can rearrange the force equation to express it in a standard form:
\[ F = -4x + 8 \implies F = -4(x - 2) \]
This shows that the force is proportional to the displacement from the position \(x = 2\).

3. **Determine the Nature of the Motion**: For SHM, the force must be proportional to the negative of the displacement from an equilibrium position. Here, the equilibrium position is at \(x = 2\). The force can be rewritten as:
\[ F = -4(x - 2) \]
This indicates that the force is indeed proportional to \(-x\) (specifically, \(-4(x - 2)\)), confirming that the motion is SHM about the point \(x = 2\).

4. **Calculate the Time Period**: The time period \(T\) of SHM is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Here, \(k = 4\) (the coefficient of \(x\) in the force equation). We need to know the mass \(m\) of the block to find the time period. Since the mass is not provided, we will express the time period in terms of \(m\):
\[ T = 2\pi \sqrt{\frac{m}{4}} = \pi \sqrt{\frac{m}{1}} \]

5. **Conclusion**: The block performs SHM about the position \(x = 2\) with a time period dependent on its mass.

### Final Answers:
- The motion is SHM.
- The equilibrium position is at \(x = 2\).
- The time period is \(T = \pi \sqrt{m}\).