 Solution

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.