Maximum acceleration of a particle in SHM is `16 cm//s^(2)` and maximum uelocity is `8 cm//s`.Find time period and amplitude of oscillations.

Explanation

To solve the problem, we need to find the time period and amplitude of a particle in Simple Harmonic Motion (SHM) given the maximum acceleration and maximum velocity.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Maximum acceleration, \( a_{\text{max}} = 16 \, \text{cm/s}^2 \)
- Maximum velocity, \( v_{\text{max}} = 8 \, \text{cm/s} \)

2. **Use the Formulas for SHM**:
- The maximum velocity in SHM is given by:
\[ v_{\text{max}} = A \omega \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.
- The maximum acceleration in SHM is given by:
\[ a_{\text{max}} = A \omega^2 \]

3. **Set Up the Equations**:
- From the maximum velocity equation:
\[ 8 = A \omega \quad \text{(1)} \]
- From the maximum acceleration equation:
\[ 16 = A \omega^2 \quad \text{(2)} \]

4. **Divide Equation (2) by Equation (1)**:
- This gives:
\[ \frac{16}{8} = \frac{A \omega^2}{A \omega} \]
- Simplifying this, we get:
\[ 2 = \omega \quad \Rightarrow \quad \omega = 2 \, \text{rad/s} \]

5. **Substitute \( \omega \) Back to Find Amplitude \( A \)**:
- Substitute \( \omega \) into equation (1):
\[ 8 = A \cdot 2 \quad \Rightarrow \quad A = \frac{8}{2} = 4 \, \text{cm} \]

6. **Calculate the Time Period \( T \)**:
- The time period \( T \) is given by:
\[ T = \frac{2\pi}{\omega} \]
- Substituting \( \omega = 2 \):
\[ T = \frac{2\pi}{2} = \pi \, \text{s} \]

### Final Answers:
- Amplitude \( A = 4 \, \text{cm} \)
- Time Period \( T = \pi \, \text{s} \)