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Exponential decays and oscillations can be modelled with the second order linear differential equation with constant coefficients
$$ ax''+bx'+c=0. $$ Assuming that an object of mass \(m\) experiences a spring force \(F_k=-kx\) and a resisting damping force of the form \(F_R=-Rx'\), the total force experienced by the object is $$ \sum F=F_k+F_R=-kx-Rx'. $$ By Newton's third law, this is equal to mass times acceleration \(mx''\) and we obtain $$ mx''=-kx-Rx', $$ that is $$ mx''+Rx'+kx=0. $$ By assuming that \(x(t)=e^{rt}\) (or replacing \(x''\to r^2\), \(x'\to r\), \(x\to 1\)) we obtain the characteristic equation $$ mr^2+Rr+k=0, $$ which has the solution $$ r=\frac{-R\pm\sqrt{R^2-4mk}}{2m}. $$In case \(R\) is sufficiently large, the discriminant \(D=R^2-4mk>0\) and the solutions of the characteristic equation are real $$ r_1=\frac{-R+\sqrt{R^2-4mk}}{2m} $$ and $$ r_2=\frac{-R-\sqrt{R^2-4mk}}{2m}. $$ Because \(m,k,R\gt0\), we see that $$ -R+\sqrt{R^2-4mk}<-R+\sqrt{R^2}=0. $$ Thus both \(r_1,r_1<0\) and the solution is a linear combination of two exponential decays $$ x(t)=Ae^{r_1t}+Be^{r_2t}. $$ This is an overcritical damping with no oscillation.
In case \(D=R^2-4mk=0\), we have $$ r=-\frac{R}{2m} $$ and we obtain a single exponential decay $$ x(t)=(A+tB)e^{-\frac{R}{2m}t}. $$ If the values \(m,k,R\) are chosen in random, this case is unprobable. However, in this case \(x(t)\to 0\) in the fastest possible way; the equation is critically damped.In case \(R\) is sufficiently small, then \(D=R^2-4mk<0\) and we obtain $$ r_1=\frac{-R+j\sqrt{-D}}{2m} $$ and $$ r_2=\frac{-R-j\sqrt{-D}}{2m}. $$ This leads to the solution $$ x(t)=Ae^{-\frac{R}{2m}t}\sin\left(\underbrace{\frac{\sqrt{4mk-R^2}}{2m}}_{=\omega}+\varphi_0\right). $$ The system oscillates and the damping is undercritical.
If \(R=0\), we obtain the undamped oscillation with characteristic equation \(r^2+\omega^2=0\), with \(\omega^2=\frac{k}{m}\) and \(r=\pm\omega j\) and $$ x(t)=\sin(\omega t+\varphi_0). $$