The acceleration of a particle performing a linear S.H.M. is `16 cm//s^(2)`, when it is at a dis"tan"ce of 4 cm from the mean position. The period of S.H.M. is

Options

A. 6.28 sec
B. 1.57 sec
C. 5 sec
D. 3.14 sec

Explanation

To solve the problem, we need to find the period of a particle performing simple harmonic motion (S.H.M.) given its acceleration and distance from the mean position.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Acceleration (a) = 16 cm/s²
- Distance from the mean position (x) = 4 cm

2. **Use the Formula for Acceleration in S.H.M.:**
The acceleration of a particle in S.H.M. is given by the formula:
\[ a = -\omega^2 x \]
where \( \omega \) is the angular frequency and \( x \) is the displacement from the mean position.

3. **Substitute the Known Values:**
Since we are given the magnitude of acceleration, we can ignore the negative sign for this calculation:
\[ 16 = \omega^2 \cdot 4 \]

4. **Solve for Angular Frequency (\( \omega \)):**
Rearranging the equation gives:
\[ \omega^2 = \frac{16}{4} = 4 \]
Taking the square root:
\[ \omega = 2 \text{ rad/s} \]

5. **Calculate the Period (T):**
The period \( T \) of S.H.M. is related to the angular frequency by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{2} = \pi \text{ seconds} \]

6. **Convert to Decimal:**
Approximating \( \pi \):
\[ T \approx 3.14 \text{ seconds} \]

### Final Answer:
The period of the S.H.M. is approximately **3.14 seconds**.

---