Damped oscillations — underdamped, critically damped, overdamped

WHAT are we describing?

WHY velocity-proportional? Because viscous drag (a body moving slowly through fluid/air) really does scale with speed — slow motion through honey resists harder the faster you push. It's the simplest and physically common case, so we start here.

HOW to derive the equation of motion

Newton's second law: $m\ddot{x} = \sum F$ . Forces are spring $-kx$ and damping $-b\dot{x}$ :

$m\ddot{x} = -kx - b\dot{x}$

Move everything to one side and divide by $m$ :

$\ddot{x} + \frac{b}{m}\dot{x} + \frac{k}{m}x = 0$

Why this step? Dividing by $m$ exposes the two physically meaningful frequencies. We define:

Why the factor 2 in $2\gamma$ ? Pure convenience — it makes later square roots come out clean ( $\sqrt{\gamma^2-\omega_0^2}$ with no extra factors).

Solve by guessing exponentials

A linear constant-coefficient ODE is solved by $x = e^{\lambda t}$ . Why this guess? Because differentiation just multiplies $e^{\lambda t}$ by $\lambda$ , turning calculus into algebra. Substitute:

$\lambda^2 + 2\gamma\lambda + \omega_0^2 = 0 \quad\text{(characteristic equation)}$

Quadratic formula:

$\lambda = -\gamma \pm \sqrt{\gamma^2 - \omega_0^2}$

Everything hinges on the sign of $\gamma^2 - \omega_0^2$ — i.e. on whether the discriminant is negative, zero, or positive. That single sign creates the three regimes.

The three regimes

WHY oscillation? The imaginary part of $\lambda$ is an oscillation (Euler: $e^{i\omega_d t}$ = sines/cosines); the real part $-\gamma$ is the decay. The spring still wins the "wobble" fight, but every swing is smaller.

WHY slow? The slower root $\lambda_1 = -\gamma + \sqrt{\gamma^2-\omega_0^2}$ is small in magnitude (close to 0), so that term lingers a long time. Heavy damping makes the system sluggish, like a door closer set too tight.

WHY the extra $t$ ? With a repeated root, $e^{\lambda t}$ alone gives only ONE solution, but a 2nd-order ODE needs TWO. Multiplying by $t$ produces a genuinely independent second solution (you can verify it satisfies the ODE).

WHY is critical damping special/useful? It's the borderline. Any less damping ⇒ overshoot/wobble; any more ⇒ sluggish. So it's the fastest non-oscillatory settling — exactly what you want for car suspensions, analog meter needles, and door closers.

Figure — Damped oscillations — underdamped, critically damped, overdamped

Energy & "how damped is it?" — the Quality factor

Energy in an underdamped oscillator decays because amplitude $\propto e^{-\gamma t}$ and energy $\propto$ amplitude²:

$E(t) = E_0\, e^{-2\gamma t}$

Why the 2? Energy $\sim \tfrac12 k A^2$ , and $A^2 \to (e^{-\gamma t})^2 = e^{-2\gamma t}$ .

Worked examples

Common mistakes (steel-manned)

Recall Feynman: explain to a 12-year-old

Imagine pushing a kid on a swing, then walking away. Normally they'd keep swinging a long time. Now imagine the swing is moving through thick syrup instead of air.

A little syrup: the kid still swings back and forth, but each swing is smaller and smaller until they stop. (underdamped)
The perfect amount of syrup: the swing glides smoothly back to the middle and stops — no back-and-forth, and as fast as possible. (critically damped)
Way too much syrup: the swing oozes back to the middle super slowly, like moving through cold honey. (overdamped) The syrup is the "damping." How thick it is decides which of the three things happens.

Flashcards

What force model defines a (linearly) damped oscillator?

A resistive force opposing velocity,

F=-b\dot x

Write the standard-form damped ODE.

\ddot x + 2\gamma\dot x + \omega_0^2 x = 0

with

\gamma=b/2m

\omega_0=\sqrt{k/m}

What is the characteristic equation and its roots?

\lambda^2+2\gamma\lambda+\omega_0^2=0

⇒

\lambda=-\gamma\pm\sqrt{\gamma^2-\omega_0^2}

Condition for underdamped, and its solution?

\gamma<\omega_0

;

x=Ae^{-\gamma t}\cos(\omega_d t+\phi)

with

\omega_d=\sqrt{\omega_0^2-\gamma^2}

Condition for overdamped, and its solution?

\gamma>\omega_0

;

x=Ae^{\lambda_1 t}+Be^{\lambda_2 t}

, both roots real & negative, no oscillation.

Condition for critically damped, and solution form?

\gamma=\omega_0

;

x=(A+Bt)e^{-\gamma t}

Why does critical damping have a

t e^{-\gamma t}

term?

Repeated root gives only one exponential; a 2nd independent solution is needed for a 2nd-order ODE.

What is the damped frequency

\omega_d

\omega_d=\sqrt{\omega_0^2-\gamma^2}

(always

<\omega_0

Which regime returns to equilibrium fastest without overshoot?

Critically damped.

How does energy decay in an underdamped oscillator?

E(t)=E_0 e^{-2\gamma t}

(since

E\propto A^2

, amplitude

\propto e^{-\gamma t}

Quality factor definition and critical value?

Q=\omega_0/(2\gamma)

;

Q=1/2

at critical damping.

Find

b

for critical damping given

m,k

b_{crit}=2\sqrt{km}=2m\omega_0

Connections

Simple Harmonic Motion — the $\gamma=0$ limit ( $\omega_d\to\omega_0$ , no decay).
Forced Oscillations & Resonance — add a driving force; damping sets resonance sharpness ( $Q$ ).
Second-order linear ODEs — the characteristic-root method used here.
Quality factor & bandwidth — how $Q$ links damping to resonance width.
RLC circuits — electrical twin: $L\ddot q + R\dot q + q/C=0$ , same three regimes.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Socho ek spring pe lagi mass jhool rahi hai. Agar koi friction na ho to woh hamesha oscillate karti rahegi. Lekin real life mein air drag ya viscous friction velocity ke against ek force lagati hai — ise hum $F=-b\dot x$ likhte hain. Newton's law lagao to equation banta hai $\ddot x + 2\gamma\dot x + \omega_0^2 x = 0$ , jahan $\gamma=b/2m$ (kitna damping) aur $\omega_0=\sqrt{k/m}$ (spring ki natural frequency).

Ab pura khel ek hi cheez pe depend karta hai: $\gamma$ bada hai ya $\omega_0$ bada hai. Iske teen cases hain. Underdamped ( $\gamma<\omega_0$ ): mass abhi bhi aage-peeche jhoolti hai par har swing chhoti hoti jaati hai, jaise ek shrinking envelope $e^{-\gamma t}$ ke andar. Critically damped ( $\gamma=\omega_0$ ): bina jhoole, seedha aur sabse jaldi equilibrium pe aa jaati hai. Overdamped ( $\gamma>\omega_0$ ): bohot zyada friction, isliye dheere-dheere, syrup mein chalne ki tarah, slowly wapas aati hai — koi oscillation nahi.

Yaad rakhne ka simple test: kam damping → wobble karta hai, zyada damping → crawl karta hai, exactly barabar → fastest clean stop. Critical damping isliye important hai kyunki car ke shock absorbers, door closers, aur meter ki needle sabko yahi chahiye — jaldi settle ho, par overshoot na kare.

Energy ke liye: amplitude $e^{-\gamma t}$ se girti hai, aur energy amplitude ka square hoti hai, isliye $E=E_0 e^{-2\gamma t}$ . Quality factor $Q=\omega_0/2\gamma$ batata hai system kitni baar "ring" karega — tuning fork ka $Q$ bohot high hota hai isliye lambi der tak bajta hai.

Damped oscillations — underdamped, critically damped, overdamped

WHAT are we describing?

HOW to derive the equation of motion

Solve by guessing exponentials

The three regimes

Energy & "how damped is it?" — the Quality factor

Worked examples

Common mistakes (steel-manned)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Oscillations & Waves

Connections