In a S.H.M., the path length is 4 cm and the maximum acceleration is `2pi^(2) cm//s^(2)`. The periodic time of S.H.M. is

Options

A. `(1)/(2)s`
B. `(1)/(4)s`
C. 2 s
D. 4 s

Explanation

To solve the problem, we need to find the periodic time (T) of the simple harmonic motion (S.H.M.) given the path length and maximum acceleration.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Path length (which is twice the amplitude, \(2a\)) = 4 cm
- Maximum acceleration (\(a_{max}\)) = \(2\pi^2\) cm/s²

2. **Calculate the Amplitude (a):**
- Since the path length is \(2a\), we can find the amplitude:
\[ 2a = 4 \text{ cm} \implies a = \frac{4}{2} = 2 \text{ cm} \]

3. **Relate Maximum Acceleration to Amplitude and Angular Frequency:**
- The formula for maximum acceleration in S.H.M. is given by:
\[ a_{max} = a \omega^2 \]
- Substituting the known values:
\[ 2\pi^2 = 2 \cdot \omega^2 \]

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

5. **Calculate the Periodic Time (T):**
- The relationship between the angular frequency and the periodic time is:
\[ T = \frac{2\pi}{\omega} \]
- Substituting the value of \(\omega\):
\[ T = \frac{2\pi}{\pi} = 2 \text{ seconds} \]

### Final Answer:
The periodic time of the S.H.M. is \(2\) seconds.

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