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 amplitude and time period of a particle undergoing simple harmonic motion (SHM) given its maximum acceleration and maximum velocity.

### Step-by-Step Solution:

1. **Identify the given values:**
- Maximum acceleration, \( A_{max} = 16 \, \text{cm/s}^2 \)
- Maximum velocity, \( V_{max} = 8 \, \text{cm/s} \)

2. **Use the formula for maximum acceleration in SHM:**
\[ A_{max} = \omega^2 A \]
where \( \omega \) is the angular frequency and \( A \) is the amplitude.

3. **Use the formula for maximum velocity in SHM:**
\[ V_{max} = \omega A \]

4. **From the second equation, express \( A \) in terms of \( \omega \):**
\[ A = \frac{V_{max}}{\omega} = \frac{8}{\omega} \]

5. **Substitute \( A \) from step 4 into the first equation:**
\[ A_{max} = \omega^2 \left(\frac{8}{\omega}\right) \]
Simplifying this gives:
\[ A_{max} = 8\omega \]

6. **Now, substitute the value of \( A_{max} \):**
\[ 16 = 8\omega \]
Solving for \( \omega \):
\[ \omega = \frac{16}{8} = 2 \, \text{radians/second} \]

7. **Substitute \( \omega \) back into the equation for \( A \):**
\[ A = \frac{V_{max}}{\omega} = \frac{8}{2} = 4 \, \text{cm} \]

8. **Now, calculate the time period \( T \):**
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi \, \text{seconds} \approx 3.14 \, \text{seconds} \]

### Final Answers:
- Amplitude \( A = 4 \, \text{cm} \)
- Time period \( T \approx 3.14 \, \text{seconds} \)