The maximum displacement of a particle executing S.H.M. is 1 cm and the maximum acceleration is `(1.57)^(2)`cm per `sec^(2)` . Then the time period is

Options

A. 0.25 sec
B. 4.00 sec
C. 1.57 sec
D. `(1.57)^(2)` sec

Answer

To solve the problem, we need to find the time period of a particle executing Simple Harmonic Motion (S.H.M.) given the maximum displacement (amplitude) and the maximum acceleration.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Maximum displacement (Amplitude, A) = 1 cm
- Maximum acceleration (A_max) = \( (1.57)^2 \) cm/s²

2. **Use the Formula for Maximum Acceleration:**
The maximum acceleration in S.H.M. is given by the formula:
\[ A_{max} = A \cdot \omega^2 \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency.

3. **Substitute the Known Values:**
Substitute the values of \( A \) and \( A_{max} \) into the equation:
\[ (1.57)^2 = 1 \cdot \omega^2 \]
This simplifies to:
\[ \omega^2 = (1.57)^2 \]

4. **Calculate \( \omega \):**
Taking the square root of both sides, we find:
\[ \omega = 1.57 \text{ rad/s} \]

5. **Use the Formula for Time Period:**
The time period \( T \) of S.H.M. is related to the angular frequency by the formula:
\[ T = \frac{2\pi}{\omega} \]

6. **Substitute \( \omega \) into the Time Period Formula:**
Substitute \( \omega = 1.57 \) rad/s into the time period formula:
\[ T = \frac{2\pi}{1.57} \]

7. **Calculate the Time Period:**
Using the approximation \( \pi \approx 3.14 \):
\[ T \approx \frac{2 \times 3.14}{1.57} \approx 4 \text{ seconds} \]

### Final Answer:
The time period \( T \) is approximately **4 seconds**.

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