Values of the acceleration `A` of a particle moving in simple harmonic motion as a function of its displacement `x` are given in the table below.
`|{:(A(mm s^(-2)),16,8,0,-8,-16),(x(mm),-4,-2,0,2,4):}|`
The pariod of the motion is

Options

A. `1/(pi)s`
B. `(2)/(pi)s`
C. `(pi)/(2)s`
D. `pis`

Explanation

To find the period of the motion for a particle moving in simple harmonic motion (SHM) based on the given acceleration and displacement values, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the relationship between acceleration and displacement**:
In simple harmonic motion, the acceleration \( A \) is related to the displacement \( x \) by the equation:
\[ A = -\omega^2 x \]
where \( \omega \) is the angular frequency.

2. **Using the given data**:
We have the following values from the table:
- When \( x = -4 \, \text{mm} \), \( A = 16 \, \text{mm/s}^2 \)
- When \( x = -2 \, \text{mm} \), \( A = 8 \, \text{mm/s}^2 \)
- When \( x = 0 \, \text{mm} \), \( A = 0 \, \text{mm/s}^2 \)
- When \( x = 2 \, \text{mm} \), \( A = -8 \, \text{mm/s}^2 \)
- When \( x = 4 \, \text{mm} \), \( A = -16 \, \text{mm/s}^2 \)

3. **Choosing a pair of values**:
Let's use the first pair where \( A = 16 \, \text{mm/s}^2 \) and \( x = -4 \, \text{mm} \):
\[ 16 = -\omega^2 (-4) \]

4. **Solving for \( \omega^2 \)**:
Rearranging the equation gives:
\[ 16 = 4\omega^2 \]
Dividing both sides by 4:
\[ \omega^2 = \frac{16}{4} = 4 \]

5. **Finding \( \omega \)**:
Taking the square root of both sides:
\[ \omega = \sqrt{4} = 2 \, \text{rad/s} \]

6. **Calculating the period \( T \)**:
The period \( T \) is given by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{2} = \pi \, \text{seconds} \]

7. **Conclusion**:
Therefore, the period of the motion is:
\[ T = \pi \, \text{seconds} \]

### Final Answer:
The period of the motion is \( \pi \) seconds.