Worked examples — Damped oscillations — underdamped, critically damped, overdamped

Before anything, let us re-state the physical law itself so nothing appears "out of nowhere". A mass $m$ on a spring of stiffness $k$, dragged by a resistive force $-b\dot{x}$ proportional to velocity, obeys Newton's second law:

$m\ddot{x}+b\dot{x}+kx=0.$

Here $x$ is the displacement, $\dot{x}$ its velocity, $\ddot{x}$ its acceleration; $b>0$ is the damping constant (N·s/m). Divide through by $m$ to reach the standard form

$\ddot{x}+2\gamma \dot{x}+{\omega }_0^{2}x=0,$

whose characteristic roots are $\lambda =-\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}.$ The quantity under the root, ==${\gamma }^2-{\omega }_0^2$== (call it the discriminant $\Delta$), is the whole story.

• $\gamma$ = damping coefficient $=\frac{b}{2m}$ (how hard the syrup pulls) — this is where $b$ and $m$ enter.
• ${\omega }_0$ = natural frequency $=\sqrt{\frac{k}{m}}$ (how eager the spring is to swing) — this is where $k$ and $m$ enter.

So every time you see $\gamma$ or ${\omega }_0$ below, remember they are just bundled versions of the physical $b$, $m$, $k$ from that original ODE.

The general solution in each regime — state it once, up front

Each sign of $\Delta$ produces a different shape of solution. Memorise these three forms now; every example below just plugs numbers into one of them.

Picturing the discriminant: roots in the complex plane

Where do these roots live? Plot each root $\lambda$ as a dot on a plane whose horizontal axis is the real part (the decay rate, always $\le 0$ here) and whose vertical axis is the imaginary part (the oscillation rate). This plane is the honest picture behind the loose word "quadrant": it is the complex plane, and the sign of $\Delta$ decides whether the root dots sit off the horizontal axis (oscillation) or on it (pure decay).

Read the figure like this: as damping $\gamma$ grows from $0$, the two red root-dots start as a conjugate pair up/down the imaginary axis (underdamped — they have height, so they oscillate), slide left as decay increases, then collide on the real axis at $\gamma ={\omega }_0$ (critical — the two dots merge into one), and finally split apart along the real axis (overdamped — both real, no height, so no oscillation). "Off the axis ⇒ wobble; on the axis ⇒ crawl" is the entire topic in one moving picture. Whenever an example asks which regime, you are really asking where do the red dots sit.

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The scenario matrix

Every question in this topic lands in exactly one of these cells. Each example below is tagged with the cell(s) it covers.

Cell	Case class	Sign of $\Delta ={\gamma }^2-{\omega }_0^2$	What happens	Example
A	Underdamped	$\Delta <0$	oscillates in shrinking envelope	Ex 1, 6
B	Critically damped	$\Delta =0$	fastest clean stop	Ex 2
C	Overdamped	$\Delta >0$	two decaying exponentials, sluggish	Ex 3
D	Degenerate: $\gamma =0$	$\Delta <0$, no decay	pure SHM, rings forever	Ex 4
E	Limiting: $\gamma \to \infty$	$\Delta >0$, one root $\to 0$	one term lingers, very slow	Ex 5
F	Boundary crossing (find $b$)	set $\Delta =0$	classify above/below	Ex 2, 7
G	Energy / $Q$ decay	underdamped	how fast energy drains	Ex 8
H	Word problem (real device)	any	translate words to $m,k,b$	Ex 9
I	Exam twist (measure ${\omega }_d$, infer $\gamma$)	underdamped	invert the formulas	Ex 10

Each row corresponds to a position of the red root-dots in the figure above: rows A/D put them off the real axis, row B on it (merged), rows C/E split along it.

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Worked examples

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Recall Which cell is which — quick self-test

Sign of $\Delta <0$ means ::: underdamped (oscillates); root-dots off the real axis. Sign of $\Delta =0$ means ::: critically damped (fastest clean stop); dots merged on the real axis. Sign of $\Delta >0$ means ::: overdamped (two decaying exponentials); dots split along the real axis. $\gamma =0$ reduces the solution to ::: pure SHM, $x=A\cos ({\omega }_{0}t+\varphi )$, no decay; dots on the imaginary axis. As $\gamma \to \infty$ the slow root ${\lambda }_1$ tends to ::: $0^{-}$ (motion becomes ever slower). Formula for critical damping constant ::: $b_{crit}=2\sqrt{km}$. Per-cycle amplitude decay factor in terms of $Q$ ::: $e^{-2\pi /Q}$.

See also Second-order linear ODEs for the general theory behind the characteristic equation, and RLC circuits where $L,R,C$ play the roles of $m,b,k$ exactly.