The acceleration and velocity maximum of simple harmonically oscillating system are `8m//sec^(2) and `8m//sec`respectively. What is the angular frequency? (in rad//sec)

Explanation

To solve the problem, we need to find the angular frequency (ω) of a simple harmonic motion (SHM) system given the maximum acceleration (a_max) and maximum velocity (v_max).

### Step-by-step solution:

1. **Identify the given values:**
- Maximum acceleration, \( a_{max} = 8 \, \text{m/s}^2 \)
- Maximum velocity, \( v_{max} = 8 \, \text{m/s} \)

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

3. **Set up the equations:**
- From the first equation, we can express \( A \) in terms of \( a_{max} \) and \( \omega \):
\[ A = \frac{a_{max}}{\omega^2} \]
- From the second equation, we can express \( A \) in terms of \( v_{max} \) and \( \omega \):
\[ A = \frac{v_{max}}{\omega} \]

4. **Equate the two expressions for amplitude \( A \):**
\[ \frac{a_{max}}{\omega^2} = \frac{v_{max}}{\omega} \]

5. **Cross-multiply to eliminate \( A \):**
\[ a_{max} \cdot \omega = v_{max} \cdot \omega^2 \]

6. **Rearrange the equation to solve for \( \omega \):**
\[ \omega^2 = \frac{a_{max}}{v_{max}} \]
\[ \omega = \sqrt{\frac{a_{max}}{v_{max}}} \]

7. **Substitute the known values:**
\[ \omega = \sqrt{\frac{8 \, \text{m/s}^2}{8 \, \text{m/s}}} \]
\[ \omega = \sqrt{1 \, \text{sec}^{-1}} = 1 \, \text{rad/s} \]

8. **Final answer:**
The angular frequency \( \omega \) is \( 1 \, \text{rad/s} \).