A point mass m moves along the x -axis under an elastic…
A point mass $m$ moves along the $x$-axis under an elastic (restoring) force $-kx$ and a friction force $-r\dot x$ proportional to velocity, where $k>0$ and $r\ge0$ are constants. Set up the equation of motion and classify the resulting behavior according to the strength of the friction. $m\ddot x + r\dot x + kx = 0$.
Explanation of Method
Newton's second law $m\ddot x+r\dot x+kx=0$ has characteristic roots $\lambda=-n\pm\sqrt{n^2-\omega_0^2}$; the sign of $r^2-4mk$ splits the motion into damped oscillation, critical damping, and overdamped decay, with $x(t)\to0$ in every frictional case.
Step 1: Standard form
Divide by $m$ and introduce the damping coefficient $n=r/(2m)\ge0$ and the natural frequency $\omega_0=\sqrt{k/m}$ (so $\omega_0^2=k/m$): \[\ddot x + 2n\dot x + \omega_0^2x = 0. \tag{1}\]
Step 2: Characteristic equation
\[ \lambda^2+2n\lambda+\omega_0^2=0 \;\Longrightarrow\; \lambda=-n\pm\sqrt{n^2-\omega_0^2}. \tag{2} \]
Since $n^2-\omega_0^2=\dfrac{r^2-4mk}{4m^2}$, the sign of the discriminant in (2) is the sign of $r^2-4mk$; the behavior of the system falls into three cases.
Step 3: Case I — light friction (underdamped)
$r^2<4mk$, i.e. $n<\omega_0$.
The discriminant in (2) is negative, giving complex conjugate roots $\lambda=-n\pm i\omega$, where $\omega=\sqrt{\omega_0^2-n^2}>0$. The general solution is:
\[ x(t)=e^{-nt}\bigl(C_1\cos\omega t+C_2\sin\omega t\bigr)=A\,e^{-nt}\cos(\omega t-\varphi_0), \] where $A=\sqrt{C_1^2+C_2^2}$ and $\varphi_0=\arctan(C_2/C_1)$. This is a damped oscillation: the motion crosses $x=0$ infinitely often with quasi-period $T=2\pi/\omega=2\pi/\sqrt{\omega_0^2-n^2}$, while the amplitude $A\,e^{-nt}$ decays to $0$ as $t\to\infty$.
Step 4: Case II — critical friction
$r^2=4mk$, i.e. $n=\omega_0$.
The discriminant vanishes and (2) has the repeated real root $\lambda_1=\lambda_2=-n$. The general solution is: $x(t)=(C_1+C_2t)\,e^{-nt}.$
This is critical damping: the system returns to equilibrium as fast as possible without oscillating — $x(t)$ can cross zero at most once (depending on the sign of $C_1,C_2$) and then approaches $0$ monotonically, with $x(t)\to0$ as $t\to\infty$.
Step 5: Case III — heavy friction (overdamped)
$r^2>4mk$, i.e. $n>\omega_0$.
The discriminant is positive, giving two distinct real roots $\lambda_{1,2}=-n\pm\sqrt{n^2-\omega_0^2}$; both
are negative, since $\sqrt{n^2-\omega_0^2} \[ x(t)=C_1e^{\lambda_1t}+C_2e^{\lambda_2t}. \] This is overdamped, non-oscillatory motion: $x(t)$ approaches equilibrium at least as
slowly as in the critical case (the slower of the two exponentials, $e^{\lambda_1t}$ with
$\lambda_1=-n+\sqrt{n^2-\omega_0^2}$, dominates), and again $x(t)\to0$ as $t\to\infty$.
Step 6: Special subcase — no friction
$r=0$, i.e. $n=0$.
Equation (1) reduces to $\ddot x+\omega_0^2x=0$, with purely imaginary roots $\lambda=\pm i\omega_0$ and general solution
\[ x(t)=C_1\cos\omega_0t+C_2\sin\omega_0t=A\cos(\omega_0t-\varphi_0). \]
This is undamped simple harmonic motion of constant amplitude $A$ and period $T_0=2\pi/\omega_0$, independent of $t$ — the limiting case of Case I as $n\to0^+$.
Step 7: Summary
\[ \begin{array}{lccl} \textbf{Condition} & \textbf{Discriminant} & \textbf{Motion} & \textbf{Solution} \\ \hline r=0 & n^2-\omega_0^2=-\omega_0^2<0 & \text{undamped SHM} & x=A\cos(\omega_0t-\varphi_0) \\ r^2<4mk & n^2-\omega_0^2<0 & \text{underdamped (oscillatory decay)} & x=Ae^{-nt}\cos(\omega t-\varphi_0) \\ r^2=4mk & n^2-\omega_0^2=0 & \text{critically damped} & x=(C_1+C_2t)e^{-nt} \\ r^2>4mk & n^2-\omega_0^2>0 & \text{overdamped (non-oscillatory)} & x=C_1e^{\lambda_1t}+C_2e^{\lambda_2t} \end{array} \]
Result
Final answer:
\[\boxed{\,x(t)=\begin{cases} A\,e^{-nt}\cos(\omega t-\varphi_0), & r^2<4mk \quad (\omega=\sqrt{\omega_0^2-n^2}) \\[4pt] (C_1+C_2t)\,e^{-nt}, & r^2=4mk \\[4pt] C_1e^{\lambda_1t}+C_2e^{\lambda_2t}, & r^2>4mk \quad \bigl(\lambda_{1,2}=-n\pm\sqrt{n^2-\omega_0^2}\bigr) \end{cases}\,}\]
Remark
General remark. Whenever friction is present ($r>0$, i.e. $n>0$), $x(t)\to0$ as $t\to\infty$ in every case: friction always eventually returns the system to equilibrium. What changes with the strength of the friction is only how it gets there — oscillating through equilibrium (Case I, $r^2<4mk$) versus approaching it directly, without oscillation (Cases II and III, $r^2\ge4mk$). The oscillatory character present in the frictionless case is therefore preserved only as long as the damping stays below the critical threshold $r^2=4mk$.