At t=0, the displacement x(0) of the block in a linear oscillator like that of Fig. 15.5 is `-8.50cm`. (Read x(0) as ''x at time zero''). The block's velocity v(0) then is `-0.920m//s`, and its acceleration a(0) is `+47.0 m//s^(2)`.
What is the angular frequency `omega` of this system?

Explanation

To find the angular frequency \( \omega \) of the linear oscillator, we can use the relationships between displacement, velocity, and acceleration in simple harmonic motion (SHM).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Displacement at \( t = 0 \): \( x(0) = -8.50 \, \text{cm} = -0.0850 \, \text{m} \)
- Velocity at \( t = 0 \): \( v(0) = -0.920 \, \text{m/s} \)
- Acceleration at \( t = 0 \): \( a(0) = +47.0 \, \text{m/s}^2 \)
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