In this case where the driving force is also some periodic function such as: $\huge \begin{align} F_{\text{ext}} &= F_{0}\cos (\omega t)\\ \end{align}$ The [[../Math/Differential Equations|Differential Equation]] for the system would be: $\huge \begin{align} mx'' &= -k\vec x - b \vec x' + F_{0}\cos(\omega t)\\ \vec x'' + \frac{b}{m}\vec x' + \frac{k}{m} \vec x &=\frac{F_{0}}{m} \cos(\omega t) \end{align} $ Note that this is the same as before except the equation equals the force divided by the mass instead of 0. $\huge \begin{align} \vec x(t) &= A_{0}\sin (\omega t + \phi_{0}) \\ A_{0} &= \frac{F_{0}}{m}\pa{(\omega^2 - \omega_{0}^2)^2 + \frac{b^2\omega^2}{m^2}}^{-1/2} \\ \phi_{0} &= \arctan\pa{ \frac{\omega_{0}^2-\omega^2}{\omega \frac{b}{m}} } \end{align} $ Where $\omega_{0}$ is the [[Natural Frequency]] of the system and $\omega$ is the [[Frequency]] of the driving force $\vec F_{\text{ext}}$ and $\phi_{0}$ is the phase offset of the system. When discussing these problems we use a measure calleed $Q$ ([[Q-Factor]]) ### Energy $\huge\begin{align} E_{0} &= \frac{1}{2}k A_{0}^2 = \frac{1}{2} m \omega^2A_{0}^2\\ E(t) &= E_{0}e^{ -\frac{t}{\tau} } \\ E'(t) &= -\tau^{-1} E_{0} e^{-\frac{t}{\tau}}\\ &=-\frac{E(t)}{\tau} \end{align} $ $\huge \begin{align} \deriv E t &= -\frac{E}{\tau} = \deriv E \tau = \frac{\Delta E}{T} \\ \frac{\Delta E}{E} &= -\frac{2\pi}{Q} \\ T &= \frac{2\pi}{w_{0}}\\ Q &= \omega_{0} \tau \end{align}$ Where $Q$ is the [[Q-Factor]].