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 given the initial conditions, we can follow these steps:

### Step 1: Write down the equations for displacement, velocity, and acceleration
For a simple harmonic oscillator, the displacement \( x(t) \), velocity \( v(t) \), and acceleration \( a(t) \) can be expressed as:
1. \( x(t) = A \sin(\omega t + \phi) \)
2. \( v(t) = A \omega \cos(\omega t + \phi) \)
3. \( a(t) = -A \omega^2 \sin(\omega t + \phi) \)

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