Absorbed-power bandwidth of a viscously damped oscillator ($\Delta \Omega =b/m$)

For a constant-amplitude force on $m\ddot{x}+b\dot{x}+kx=F_0\cos \Omega t$, the cycle-averaged absorbed power is $b{F}_0^2{\Omega }^2/[2((k-m{\Omega }^2{)}^2+b^2{\Omega }^2)]$. It peaks at ${\omega }_0=\sqrt{k/m}$. Its half-power frequencies are ${\Omega }_{\pm }=(\sqrt{b^2+4mk}\pm b)/(2m)$, hence their separation is exactly $b/m$ and $Q={\omega }_0/\Delta \Omega$. A displacement-amplitude resonance has different half-power frequencies away from weak damping; the measured observable must be specified.

Converting the two half-power frequencies using $\lambda =2\pi c/\omega$ gives $\Delta \lambda =2\pi c(1/{\omega }_{-}-1/{\omega }_{+})$. Since ${\omega }_{-}{\omega }_{+}={\omega }_0^2$, this reduces exactly to the displayed expression. The wavelength endpoints are not symmetric around $2\pi c/{\omega }_0$; a symmetric narrow-line sketch is an approximation. This conversion concerns the same measured response versus a reparametrized axis, not a transformation between spectral density conventions.