Given: \( K = 1200 \, N/m \), \( m = 3 \, kg \), \( x = 0.02 \, m \)
The angular frequency \( \omega = \sqrt{\frac{K}{m}} = \sqrt{\frac{1200}{3}} = 20 \, rad/s \)
The frequency \( f = \frac{\omega}{2\pi} = \frac{20}{2\pi} \approx 3.18 \, Hz \)
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