A mass $m$ on a spring with stiffness $k$, slowed by a damper with coefficient $c$, is the standard example of a second-order equation. The complex roots and repeated roots you have studied are exactly the different ways it can move.
Key idea
$mx'' + cx' + kx = 0$, with characteristic roots $r = \frac{-c \pm \sqrt{c^{2} - 4mk}}{2m}$.
Undamped ($c = 0$): $x = R\cos(\omega_0t - \delta)$ with natural frequency $\omega_0 = \sqrt{k/m}$ and period $\frac{2\pi}{\omega_0}$.
$c^{2} < 4mk$: underdamped (oscillates with shrinking amplitude); $c^{2} = 4mk$: critically damped; $c^{2} > 4mk$: overdamped (returns without oscillating).
Undamped ($c = 0$): $x = R\cos(\omega_0t - \delta)$ with natural frequency $\omega_0 = \sqrt{k/m}$ and period $\frac{2\pi}{\omega_0}$.
$c^{2} < 4mk$: underdamped (oscillates with shrinking amplitude); $c^{2} = 4mk$: critically damped; $c^{2} > 4mk$: overdamped (returns without oscillating).
Why it works
Measure $x$ from the equilibrium position. The spring pulls back with force $-kx$ (Hooke's law) and the damper resists motion with force $-cx'$, so Newton's law $mx'' = -kx - cx'$ gives the equation. The sign of the discriminant $c^{2} - 4mk$ decides whether the roots are complex, repeated or real. For the undamped case, $c_1\cos\omega_0t + c_2\sin\omega_0t$ $= R\cos(\omega_0t - \delta)$ with $R = \sqrt{c_1^{2} + c_2^{2}}$: expand $\cos(\omega_0t - \delta)$ and match coefficients.
Worked example: An undamped spring
A 0.5 kg mass stretches a spring 0.098 m at rest ($g = 9.8$ m/s²). It is pulled to $x(0) = 0.2$ m and given $x'(0) = 1$ m/s. Find $x(t)$, the amplitude and the period.
- At rest $k(0.098) = 0.5(9.8)$, so $k = 50$ N/m and $\omega_0 = \sqrt{50/0.5} = 10$.
- $x = c_1\cos 10t + c_2\sin 10t$: $c_1 = 0.2$ and $10c_2 = 1$, so $c_2 = 0.1$.
- Amplitude $\sqrt{0.04 + 0.01} = \sqrt{0.05}$ $\approx 0.224$ m; period $\frac{2\pi}{10} \approx 0.628$ s.
Answer: $x = 0.2\cos 10t + 0.1\sin 10t$
Worked example: Three kinds of damping
For $m = 1$ and $k = 9$, classify $c = 2$, $c = 6$ and $c = 10$.
- $4mk = 36$.
- $c = 2$: $4 < 36$, underdamped. $c = 6$: $36 = 36$, critically damped. $c = 10$: $100 > 36$, overdamped.
Answer: under, critical, over
Worked example: A decaying oscillation
Solve $x'' + 2x' + 10x = 0$, $x(0) = 1$, $x'(0) = 0$.
- $r = -1 \pm 3i$, so $x = e^{-t}(c_1\cos 3t + c_2\sin 3t)$.
- $c_1 = 1$; $x'(0) = -c_1 + 3c_2 = 0$ gives $c_2 = \frac{1}{3}$.
Answer: $x = e^{-t}\left(\cos 3t + \frac{1}{3}\sin 3t\right)$: it swings with angular frequency 3 inside the envelope $e^{-t}$
Common mistake
Comparing $c$ with $4mk$ instead of $c^{2}$ with $4mk$ (equivalently, $c$ with $2\sqrt{mk}$). Another slip is giving the natural frequency as $\frac{k}{m}$ or $\sqrt{km}$; check units: $\sqrt{k/m}$ has units $\sqrt{(\text{N/m})/\text{kg}} = 1/\text{s}$.