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Dividing Newton's equation by the mass gives the stated model: $\ddot{z}$ is the acceleration, $\gamma \dot{z}$ represents viscous damping, $(k/m)z$ is the spring's restoring acceleration, and the right side is the applied driving force per unit mass. The complex notation is a convenience; the physical displacement is the real part. Put ${\omega }_0^2=k/m$. The homogeneous characteristic roots areFor $m,k>0$ and damping $\gamma \ge 0$, oscillations occur when ${\gamma }^2<4{\omega }_0^2$, the underdamped case of the damped harmonic oscillator.

For a steady response $z=Z{e}^{i\omega t}$, substitution givesTaking the modulus and phase givesThe phase is between zero and $\pi$ for positive damping and frequency; an unqualified arctangent loses the correct quadrant above resonance.

With the real drive $F_0\cos \omega t$ and real response $r\cos (\omega t-\varphi )$, the average driving power equals the dissipated viscous power:For $\gamma >0$, write this asIt is largest when the squared term vanishes, givingThis power peak is at ${\omega }_0$ even when the displacement-amplitude peak is shifted. The frequency bandwidth of viscous oscillator power absorption follows from the half-power condition $|{\omega }_0^2-{\omega }^2|=\gamma \omega$:In fact the bandwidth identity is exact in this model. Sharp resonance additionally gives ${\omega }_{1,2}\simeq {\omega }_0\mp \gamma /2$. With zero damping at exact resonance, there is no bounded steady harmonic response to which these finite-power formulas apply.

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 ${\omega }_0$.