Damped oscillations — underdamped, critically damped, overdamped: Exercises

A handful of symbols run through everything, so let us pin them down in plain words before any problem uses them:

Figure — Damped oscillations — underdamped, critically damped, overdamped

How to read this figure (the map for the whole page): the horizontal axis is the damping $γ$ (with $ω_{0}$ held fixed at $1$ ); the vertical axis is the discriminant $Δ = γ^{2} - ω_{0}^{2}$ . The slate curve is $Δ$ itself. Where the curve sits below the dashed zero line (lavender band, $γ < ω_{0}$ ) you are underdamped; where it crosses zero (mint line, $γ = ω_{0}$ ) you are critical; where it rises above zero (coral band, $γ > ω_{0}$ ) you are overdamped. Pick your $γ$ relative to $ω_{0}$ , read off the sign of $Δ$ , and you know the regime before doing anything else.

Level 1 — Recognition

Exercise 1.1

A mass–spring–damper has $m = 4 kg$ , $k = 36 N/m$ , $b = 6 N\cdots/m$ . Which regime?

Recall Solution

WHAT we compute: the two frequencies, then compare them via the discriminant.

$ω_{0} = k / m = 36/4 = 9 = 3 rad/s$ . (WHY: natural frequency is $k / m$ .)
$γ = b /2 m = 6/ (2 \cdot 4) = 6/8 = 0.75 s^{- 1}$ . (WHY: $γ = b /2 m$ .)
Discriminant $Δ = γ^{2} - ω_{0}^{2} = 0.5625 - 9 < 0$ , equivalently $γ = 0.75 < ω_{0} = 3$ .

Answer: underdamped. Damping is far weaker than the spring, so the mass swings back and forth inside a shrinking envelope.

Exercise 1.2

For $m = 1 kg$ , $k = 9 N/m$ , state the value of $b$ that puts the system exactly on the boundary between wobbling and crawling.

Recall Solution

WHAT: the boundary is critical damping, $γ = ω_{0}$ , i.e. the discriminant $Δ = 0$ .

$ω_{0} = 9/1 = 3 rad/s$ .
Set $γ = ω_{0}$ : $b /2 m = ω_{0} \Rightarrow b = 2 m ω_{0} = 2 \cdot 1 \cdot 3 = 6 N\cdots/m$ .

Answer: $b_{cr i t} = 6 N\cdots/m$ . Below this you underdamp; above it you overdamp. (WHY: $γ = ω_{0}$ makes the discriminant $Δ = γ^{2} - ω_{0}^{2}$ hit exactly zero — the definition of the borderline.)

Level 2 — Application

Exercise 2.1

Underdamped system: $m = 2 kg$ , $k = 200 N/m$ , $b = 8 N\cdots/m$ . Find $ω_{0}$ , $γ$ , and the damped frequency $ω_{d}$ .

Recall Solution

$ω_{0} = 200/2 = 100 = 10 rad/s$ .
$γ = 8/ (2 \cdot 2) = 8/4 = 2 s^{- 1}$ . Since $2 < 10$ : underdamped, good — $ω_{d}$ is real.
$ω_{d} = ω_{0}^{2} - γ^{2} = 100 - 4 = 96 \approx 9.80 rad/s$ .

WHY $ω_{d} < ω_{0}$ : drag slows each swing slightly, so the observed rhythm is a touch slower than the ideal undamped one.

Exercise 2.2

The same system starts at $x (0) = 0.05 m$ released from rest. The amplitude envelope is $A e^{- γ t}$ with $A = 0.05 m$ (approximately, for light damping). What fraction of the initial amplitude remains after $t = 1 s$ ? What fraction of the initial energy?

Recall Solution

WHAT: evaluate the decaying envelope, then square it for energy.

Amplitude fraction: $\frac{A ( 1 )}{A ( 0 )} = e^{- γ t} = e^{- 2 \cdot 1} = e^{- 2} \approx 0.135$ → about 13.5% left.
Energy fraction: energy $\propto$ amplitude $^{2}$ , so $\frac{E ( 1 )}{E ( 0 )} = e^{- 2 γ t} = e^{- 4} \approx 0.0183$ → about 1.8% left.

WHY the factor 2 for energy: $E \sim \frac{1}{2} k A^{2}$ , and squaring $e^{- γ t}$ gives $e^{- 2 γ t}$ . Energy always leaks twice as fast (in the exponent) as amplitude.

Level 3 — Analysis

Exercise 3.1

Underdamped oscillator with $ω_{0} = 5 rad/s$ , $γ = 0.5 s^{- 1}$ . Compute the quality factor $Q = ω_{0} /2 γ$ , and estimate how many full oscillation cycles occur before the amplitude falls to $1/ e$ of its start.

Recall Solution

$Q = \frac{ω _{0}}{2 γ} = \frac{5}{2 \cdot 0.5} = \frac{5}{1} = 5$ .
Amplitude reaches $1/ e$ when $e^{- γ t} = e^{- 1}$ , i.e. at $t = 1/ γ = 1/0.5 = 2 s$ .
Number of cycles in that time: period $T = 2 π / ω_{d}$ . Here $ω_{d} = 25 - 0.25 = 24.75 \approx 4.975 rad/s$ , so $T \approx 2 π /4.975 \approx 1.263 s$ .
Cycles $= t / T = 2/1.263 \approx 1.58$ cycles. (WHY divide $t$ by $T$ : one period $T$ is the time for exactly one complete cycle, so the elapsed time $t$ contains $t / T$ of those periods — that count IS the number of cycles.)

Sanity check via $Q$ — where " $Q / π$ " comes from: the number of cycles to reach $1/ e$ is $N = \frac{t _{1/ e}}{T} = \frac{1/ γ}{2 π / ω _{d}} = \frac{ω _{d}}{2 π γ} .$ For light damping $ω_{d} \approx ω_{0}$ , so $N \approx \frac{ω _{0}}{2 π γ} = \frac{1}{π} \cdot \frac{ω _{0}}{2 γ} = \frac{Q}{π}$ . Here $Q / π = 5/ π \approx 1.59$ — matches the $1.58$ above. (WHY it works: $Q = ω_{0} /2 γ$ already packages "oscillation speed over decay speed"; dividing by $π$ just converts that ratio into whole cycles.)

Exercise 3.2

Overdamped system: $ω_{0} = 3 rad/s$ , $γ = 5 s^{- 1}$ . Find both real roots $λ_{1, 2}$ , and say which term dominates the long-time behaviour and why.

Recall Solution

WHY roots at all: the equation of motion $m \overset{x}{¨} + b \overset{x}{˙} + k x = 0$ (divided by $m$ , giving $\overset{x}{¨} + 2 γ \overset{x}{˙} + ω_{0}^{2} x = 0$ ) is solved by guessing $x = e^{λ t}$ . Each derivative just multiplies by $λ$ , so the ODE collapses into the characteristic equation $λ^{2} + 2 γ λ + ω_{0}^{2} = 0$ . Solving that quadratic with the quadratic formula gives the roots below — that is the whole reason $λ_{1, 2}$ appear.

$λ = - γ \pm γ^{2} - ω_{0}^{2} = - 5 \pm 25 - 9 = - 5 \pm 16 = - 5 \pm 4$ .
So $λ_{1} = - 5 + 4 = - 1 s^{- 1}$ and $λ_{2} = - 5 - 4 = - 9 s^{- 1}$ .
Solution shape: $x (t) = A e^{- t} + B e^{- 9 t}$ , where $A$ and $B$ are integration constants fixed by the initial conditions $x (0)$ and $\overset{x}{˙} (0)$ — they set how much of each exponential is present, but not the decay rates.
Dominant at long times: the $e^{- t}$ term. Its decay rate is smallest in magnitude ( $∣ λ_{1} ∣ = 1$ ), so $e^{- 9 t}$ vanishes almost instantly while $e^{- t}$ lingers.

WHY overdamped feels sluggish: the slow root $λ_{1}$ sits close to $0$ , so a slow tail dominates the return to rest. More damping → slower crawl, not faster.

The two exponentials and their sum are plotted below — notice how the coral fast root ( $e^{- 9 t}$ ) is gone almost immediately while the lavender slow root ( $e^{- t}$ ) sets the pace of the whole dashed curve:

Look at the dashed slate curve (the sum $x (t)$ ): after $t \approx 0.5 s$ it tracks the lavender slow root almost exactly — visual proof that $λ_{1} = - 1$ governs the tail.

Level 4 — Synthesis

Exercise 4.1

A car suspension has effective $m = 250 kg$ (per wheel) and spring $k = 40000 N/m$ . Engineers want critical damping (fastest settle, no bounce). Find the required damping constant $b_{cr i t}$ , and confirm the corresponding $Q$ .

Recall Solution

WHAT: critical damping needs $γ = ω_{0}$ , i.e. $b_{cr i t} = 2 k m$ .

$b_{cr i t} = 2 k m = 2 40000 \cdot 250 = 2 1 0^{7}$ .
$1 0^{7} = 10 \cdot 1 0^{3} \approx 3162.3$ , so $b_{cr i t} \approx 6324.6 N\cdots/m$ .
Check $Q$ : at critical, $γ = ω_{0}$ , so $Q = ω_{0} /2 γ = 1/2 = 0.5$ . Critical damping is exactly $Q = \frac{1}{2}$ .

WHY $2 k m$ : set $γ = ω_{0}$ → $b /2 m = k / m$ → $b = 2 m k / m = 2 k m$ . The $m$ 's combine cleanly.

Exercise 4.2

Take the parent-note system $m = 2$ , $k = 50$ and compare settling behaviour at three damping values: $b = 8$ (underdamped), $b = 20$ (critical), $b = 40$ (overdamped). For each, give $γ$ , name the regime, and identify the slowest decay rate that governs the tail.

Recall Solution

$ω_{0} = 50/2 = 5 rad/s$ throughout.

$b$	$γ = b /2 m$	Regime	Slowest tail rate
8	$2$	underdamped ( $2 < 5$ )	envelope decays at $γ = 2$
20	$5$	critical ( $5 = 5$ )	$e^{- 5 t}$ (times a $t$ factor)
40	$10$	overdamped ( $10 > 5$ )	$λ_{1} = - 10 + 100 - 25 = - 10 + 75 \approx - 1.34$

Underdamped tail rate: $γ = 2 s^{- 1}$ .
Critical tail rate: $γ = 5 s^{- 1}$ — fastest overall envelope.
Overdamped slow root: $λ_{1} \approx - 1.34 s^{- 1}$ — slower than both others, confirming the crawl.

WHERE the " $t$ factor" comes from (critical case): when $γ = ω_{0}$ the discriminant $Δ = 0$ , so the two roots merge into a single repeated root $λ = - γ$ . A 2nd-order ODE always needs two independent solutions, but the merge leaves only one, $e^{- γ t}$ . The standard remedy for a repeated root supplies the missing second solution as $t e^{- γ t}$ (you can verify it satisfies $\overset{x}{¨} + 2 γ \overset{x}{˙} + ω_{0}^{2} x = 0$ when $γ = ω_{0}$ ). Hence the general form $x (t) = (A + B t) e^{- γ t},$ with $A, B$ again fixed by the initial conditions. Effect on the envelope: the lone factor of $t$ grows only linearly while $e^{- γ t}$ shrinks exponentially, so the product still decays to zero — just with one brief rise-then-fall bump rather than a pure exponential. That is why critical damping can settle so cleanly yet quickly.

Punchline: critical ( $γ = 5$ ) has the fastest-decaying dominant behaviour. Push past it to $b = 40$ and a $- 1.34 s^{- 1}$ tail appears — slower than even the underdamped case. This is why critical damping is the engineering sweet spot.

The three responses are plotted together below (all normalised to start at $x = 1$ , released from rest), so you can see the sweet spot rather than just read it:

Trace each curve: the lavender underdamped one dips below zero (it wobbles), the mint critical curve slides to zero fastest without crossing, and the coral overdamped curve crawls above the mint one the whole way — the mint curve reaching zero soonest is the geometric meaning of "critical is the fastest clean stop."

Level 5 — Mastery

Exercise 5.1

An underdamped oscillator's amplitude drops from $12 cm$ to $3 cm$ over $4$ full oscillations. The observed damped period is $T = 0.5 s$ . Find (a) the damping coefficient $γ$ , and (b) the quality factor $Q$ (use $ω_{d} \approx ω_{0}$ for the light-damping estimate).

Recall Solution

WHAT: use the envelope ratio across a known time.

Time elapsed: $t = 4 T = 4 \cdot 0.5 = 2 s$ .
Envelope ratio: $\frac{A ( t )}{A ( 0 )} = e^{- γ t} = \frac{3}{12} = \frac{1}{4}$ .
Take logs: $- γ t = ln (1/4) = - ln 4$ , so $γ = \frac{ln 4}{t} = \frac{ln 4}{2} = \frac{1.3863}{2} \approx 0.693 s^{- 1}$ .
(b) $ω_{d} = 2 π / T = 2 π /0.5 = 4 π \approx 12.566 rad/s$ . With light damping $ω_{0} \approx ω_{d}$ , so $Q \approx \frac{ω _{0}}{2 γ} \approx \frac{12.566}{2 \cdot 0.693} \approx 9.07$ .

WHY logs: exponential decay becomes linear in log-space, turning "ratio over time" into a clean solve for $γ$ . This logarithmic decrement trick is the standard experimental way to measure damping.

Exercise 5.2

Same physical damper is dropped into an RLC circuit analogy: $L = 0.5 H$ , $C = 200 μ F$ , $R = 20 Ω$ . Using the mechanical–electrical dictionary $m \to L$ , $b \to R$ , $1/ k \to C$ , classify the circuit's transient response.

Recall Solution

WHAT: map to the same $γ$ vs $ω_{0}$ test. For RLC, $ω_{0} = 1/ L C$ and $γ = R /2 L$ .

$ω_{0} = \frac{1}{L C} = \frac{1}{0.5 \cdot 200 \times 1 0 ^{- 6}} = \frac{1}{1 0 ^{- 4}} = \frac{1}{0.01} = 100 rad/s$ .
$γ = \frac{R}{2 L} = \frac{20}{2 \cdot 0.5} = \frac{20}{1} = 20 s^{- 1}$ .
Compare: $γ = 20 < ω_{0} = 100$ → underdamped — the current/charge rings and decays, just like the mechanical spring.
Bonus $Q = ω_{0} /2 γ = 100/40 = 2.5$ : a modestly ringing circuit.

WHY the same test works: the governing ODE $L \overset{q}{¨} + R \overset{q}{˙} + q / C = 0$ is identical in form to $m \overset{x}{¨} + b \overset{x}{˙} + k x = 0$ . Same maths → same three regimes → same $γ$ -vs- $ω_{0}$ decision. See Second-order linear ODEs.

Recall Quick self-check (reveal answers)

Regime test in one line? ::: Check the sign of the discriminant $Δ = γ^{2} - ω_{0}^{2}$ ; negative → under, zero → critical, positive → over. Amplitude vs energy exponents? ::: $A \propto e^{- γ t}$ , $E \propto e^{- 2 γ t}$ . Critical damping in terms of $b$ ? ::: $b_{cr i t} = 2 k m$ , equivalently $Q = \frac{1}{2}$ . Why overdamped is slow? ::: Its slow root $λ_{1} = - γ + γ^{2} - ω_{0}^{2}$ sits near $0$ . Why the extra $t$ at critical damping? ::: Repeated root gives only one solution $e^{- γ t}$ ; the second independent one is $t e^{- γ t}$ .