 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 $x_s=|x_0(0)|\simeq 0.10\mathrm{mm}$, $x_p\simeq 1.0\mathrm{mm}$ and $f_r\simeq 150\mathrm{Hz}$. These are graph estimates, not exact supplied measurements.

Use oscillator parameters from static and resonant displacement. The static spring constant isFor light damping, ${\omega }_0\simeq {\omega }_r=2\pi (150)\simeq 9.42\times 10^2{\mathrm{s}}^{-1}$, givingNear resonance, $x_p\simeq F_0/(b{\omega }_0)$, soIndependently, the resonance enhancement $x_p/x_s\simeq 10$ estimates the same quality factor. The half-power displacement level is $x_p/\sqrt{2}\simeq 0.71\mathrm{mm}$; its intersections are approximately $142$ and $157\mathrm{Hz}$, giving bandwidth $\Delta f\simeq 15\mathrm{Hz}$ and $Q\simeq f_0/\Delta f\simeq 10$.

The exact displacement-peak frequency is ${\omega }_r^2={\omega }_0^2-b^2/(2m^2)$. At $Q\simeq 10$ 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.