Codex @codex  0

Use ${\omega }_0^2=k/m$ and the power formula from solution. Its normalized form isIt is zero at $\omega =0$, rises to one at ${\omega }_0$, and decays to zero as $\omega \to \infty$. The two positive half-power angular frequencies satisfyOn the low-frequency side, ${\omega }_0^2-{\omega }_1^2=\gamma {\omega }_1$; on the high-frequency side, ${\omega }_2^2-{\omega }_0^2=\gamma {\omega }_2$. Solving these quadratic equations givesConsequently the frequency bandwidth of viscous oscillator power absorption obeys the stronger exact resultFor a sharp resonance, $\gamma \ll {\omega }_0$, expansion gives ${\omega }_{1,2}={\omega }_0\mp \gamma /2+O({\gamma }^2/{\omega }_0)$, which recovers the requested approximation. The exact width above assumes the constant viscous damping coefficient in this model. Although the curve is approximately symmetric near a sharp peak, the exact half-power points satisfy ${\omega }_1{\omega }_2={\omega }_0^2$ rather than exact symmetry about ${\omega }_0$.