= Solution
The response is quasistatic at low <frequency>, resonantly enhanced when the spring and inertia nearly cancel, and suppressed as $1/\omega^2$ at high <frequency>. The original graph gives approximate readings $x_s=|x_0(0)|\simeq0.10\,\mathrm{mm}$, $x_p\simeq1.0\,\mathrm{mm}$ and $f_r\simeq150\,\mathrm{Hz}$. These are graph estimates, not exact supplied measurements.
Use <oscillator parameters from static and resonant displacement>. The static spring constant is
$$
\boxed{k\simeq\frac{0.9\,\mathrm N}{1.0\times10^{-4}\,\mathrm m}=9.0\times10^3\,\mathrm{N\,m^{-1}}.}
$$
For light <damping>, $\omega_0\simeq\omega_r=2\pi(150)\simeq9.42\times10^2\,\mathrm{s^{-1}}$, giving
$$
\boxed{m\simeq\frac{k}{\omega_0^2}\simeq1.0\times10^{-2}\,\mathrm{kg}=10\,\mathrm g.}
$$
Near resonance, $x_p\simeq F_0/(b\omega_0)$, so
$$
\boxed{b\simeq\frac{0.9}{(9.42\times10^2)(1.0\times10^{-3})}
\simeq0.95\,\mathrm{kg\,s^{-1}},\qquad Q=\frac{m\omega_0}{b}\simeq10.}
$$
Independently, the resonance enhancement $x_p/x_s\simeq10$ estimates the same <quality factor>. The half-power displacement level is $x_p/\sqrt2\simeq0.71\,\mathrm{mm}$; its intersections are approximately $142$ and $157\,\mathrm{Hz}$, giving bandwidth $\Delta f\simeq15\,\mathrm{Hz}$ and $Q\simeq f_0/\Delta f\simeq10$.
The exact displacement-peak <frequency> is $\omega_r^2=\omega_0^2-b^2/(2m^2)$. At $Q\simeq10$ its difference from $\omega_0$ is only about $0.25\%$, well below the precision justified by this printed curve. Greater numerical precision for the inferred parameters would be misleading.
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