Damped oscillator: Spring motion

Question:
A 0.400-kg object is attached to a spring of its force constant $k=200$ N/m and this motion is subject to a resistive force, $-bv$, which is proportional to the velocity. If the damped frequency is 99.5 % of the undamped frequency, find the value of $b$.

Answer:
The equation of motion gives:
\[
\sum F = -kx - b\frac{dx}{dt} = m\frac{d^2x}{dt^2}
\]
Let us put $\frac{dx}{dt}=x'\equiv Dx$ and $\frac{d^2x}{dt^2}=x''\equiv D^2x$. The differential equation becomes
\begin{eqnarray}
x''+\frac{b}{m}x'+\frac{k}{m}x &=& 0  \quad \mathrm{or} \\
\left(D^2+\frac{b}{m}D+\frac{k}{m}\right)x &=& 0
\end{eqnarray}
Let us also put $\omega_0 = \sqrt{\frac{k}{m}}$, which is called the rational frequency. This system is a damped oscillation, so the discriminant must be negative. Namely,
\begin{equation}
\left(\frac{b}{m}\right)^2 - 4\omega^2 < 0 \ \rightarrow \ \left(\frac{b}{m}\right)^2 < 4\omega^2
\end{equation}
The solution should be
\begin{equation}
D = -\frac{b}{2m} \pm \sqrt{\omega^2_0 - \frac{1}{4}\frac{b^2}{m^2}}i
\end{equation}
Therefore,
\begin{equation}
x = Ae^{-\frac{m}{2m}t}\cos(\omega t + \phi)
\end{equation}
where $\omega=\sqrt{\omega^2_0 - \frac{1}{4}\frac{b^2}{m^2}}$, which is called the damped frequency. This frequency is 99.5% of the undamped, so
\begin{eqnarray}
& & 0.995\omega_0 = \sqrt{\omega^2_0 - \frac{1}{4}\frac{b^2}{m^2}}  \\
& & 0.010\omega^2_0 = \frac{1}{4}\frac{b^2}{m^2}  \\
& & b = \sqrt{0.010 \times 4 \frac{k}{m} m^2}   \\
& & b = \sqrt{0.040 km}
\end{eqnarray}
Therefore,
\[
b = 1.789 \ \mathrm{kg/s}
\]

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