To ascertain the particle's displacement from its equilibrium position when its velocity is \(2 \, \text{cm/sec}\), we apply the principles of Simple Harmonic Motion (SHM).Given: Maximum velocity at the equilibrium position, \(v_{\text{max}} = 6 \, \text{cm/sec}\) Amplitude of SHM, \(A = 4 \, \text{cm}\) Velocity at position \(x\), \(v = 2 \, \text{cm/sec}\)Step 1: Determine Angular FrequencyThe maximum velocity in SHM is related to angular frequency \(\omega\) and amplitude \(A\) by the equation \(v_{\text{max}} = \omega A\).Solving for \(\omega\):\[\omega = \frac{v_{\text{max}}}{A} = \frac{6 \, \text{cm/sec}}{4 \, \text{cm}} = 1.5 \, \text{sec}^{-1}\]Step 2: Utilize the Velocity-Position EquationThe velocity \(v\) at a position \(x\) in SHM is described by \(v = \omega \sqrt{A^2 - x^2}\).Rearranging to solve for \(x\):\[v = \omega \sqrt{A^2 - x^2}
\Rightarrow \sqrt{A^2 - x^2} = \frac{v}{\omega}
\Rightarrow A^2 - x^2 = \left( \frac{v}{\omega} \right)^2
\Rightarrow x^2 = A^2 - \left( \frac{v}{\omega} \right)^2
\Rightarrow x = \sqrt{A^2 - \left( \frac{v}{\omega} \right)^2}\]Step 3: Substitute and CalculateSubstitute the given values: \(A = 4 \, \text{cm}\), \(v = 2 \, \text{cm/sec}\), and \(\omega = 1.5 \, \text{sec}^{-1}\):\[x = \sqrt{4^2 - \left( \frac{2}{1.5} \right)^2}
x = \sqrt{16 - \left( \frac{4}{2.25} \right)}
x = \sqrt{16 - \frac{16}{9}}
x = \sqrt{\frac{144}{9} - \frac{16}{9}}
x = \sqrt{\frac{128}{9}}
x = \frac{\sqrt{128}}{3}
\]\[x = \frac{8 \sqrt{2}}{3} \approx 3.77 \, \text{cm}\]Rounding to the nearest whole number:\[x \approx 4 \, \text{cm}\]Conclusion:The particle's position from the mean position when its velocity is \(2 \, \text{cm/sec}\) is:\[\boxed{4 \, \text{cm}}\]