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

1.6.9 · D3 · Physics › Oscillations & Waves › Damped oscillations — underdamped, critically damped, overda

Shuru karne se pehle, physical law ko dobara state karte hain taaki kuch bhi "out of nowhere" na lage. Mass $m$ ek spring of stiffness $k$ par, resistive force $- b \overset{x}{˙}$ jo velocity ke proportional hai, Newton's second law follow karta hai:

$m \overset{x}{¨} + b \overset{x}{˙} + k x = 0.$

Yahan $x$ displacement hai, $\overset{x}{˙}$ uski velocity, $\overset{x}{¨}$ uski acceleration; $b > 0$ damping constant hai (N·s/m). $m$ se divide karo standard form tak pahunchne ke liye:

$\overset{x}{¨} + 2 γ \overset{x}{˙} + ω_{0}^{2} x = 0,$

jiske characteristic roots hain: $λ = - γ \pm γ^{2} - ω_{0}^{2} .$ Root ke andar wali quantity, == $γ^{2} - ω_{0}^{2}$ == (ise discriminant $Δ$ kaho), yahi poori kahani hai.

$γ$ = damping coefficient $= \frac{b}{2 m}$ (kitni zor se syrup kheenchta hai) — yahan $b$ aur $m$ enter karte hain.
$ω_{0}$ = natural frequency $= \frac{k}{m}$ (spring kitni uthlaas se jhoolna chahta hai) — yahan $k$ aur $m$ enter karte hain.

Toh jab bhi neeche $γ$ ya $ω_{0}$ dekho, yaad rakho yeh sirf us original ODE ke physical $b$ , $m$ , $k$ ke bundled versions hain.

Har regime mein general solution — ek baar, shuru mein hi state karo

$Δ$ ka har sign solution ki ek alag shape produce karta hai. Yeh teen forms abhi yaad kar lo; neeche ke har example mein inhi mein se ek mein numbers plug hote hain.

Discriminant ko picture karo: complex plane mein roots

Yeh roots kahaan rehte hain? Har root $λ$ ko ek plane par dot ki tarah plot karo jiska horizontal axis real part hai (decay rate, hamesha $\leq 0$ yahan) aur vertical axis imaginary part (oscillation rate). Yeh plane "quadrant" word ke peeche ka asli picture hai: yeh complex plane hai, aur $Δ$ ka sign decide karta hai ki root dots real axis se door hain (oscillation) ya uspe hain (pure decay).

Figure — Damped oscillations — underdamped, critically damped, overdamped

Figure ko aise padho: jab damping $γ$ $0$ se badhti hai, do red root-dots conjugate pair ban ke imaginary axis par upar/neeche se shuru hote hain (underdamped — unki height hai, toh oscillate karte hain), decay badhne par left slide karte hain, phir $γ = ω_{0}$ par real axis par milte hain (critical — do dots merge ho jaate hain ek mein), aur finally real axis ke saath alag ho jaate hain (overdamped — dono real, koi height nahi, toh koi oscillation nahi). "Axis se door ⇒ jhoolna; axis par ⇒ rengna" — ek moving picture mein poora topic. Jab bhi koi example pooche kaun sa regime, tum actually pooch rahe ho red dots kahaan hain.

Scenario matrix

Is topic ka har question exactly inhi cells mein se ek mein aata hai. Neeche ka har example ek ya zyada cells ke saath tagged hai.

Cell	Case class	Sign of $Δ = γ^{2} - ω_{0}^{2}$	Kya hota hai	Example
A	Underdamped	$Δ < 0$	shrinking envelope mein oscillate karta hai	Ex 1, 6
B	Critically damped	$Δ = 0$	sabse tez clean stop	Ex 2
C	Overdamped	$Δ > 0$	do decaying exponentials, sluggish	Ex 3
D	Degenerate: $γ = 0$	$Δ < 0$ , koi decay nahi	pure SHM, hamesha bajta rahta hai	Ex 4
E	Limiting: $γ \to \infty$	$Δ > 0$ , ek root $\to 0$	ek term lingta hai, bahut slow	Ex 5
F	Boundary crossing ( $b$ dhundo)	set $Δ = 0$	upar/neeche classify karo	Ex 2, 7
G	Energy / $Q$ decay	underdamped	energy kitni tez drain hoti hai	Ex 8
H	Word problem (real device)	koi bhi	words ko $m, k, b$ mein translate karo	Ex 9
I	Exam twist ( $ω_{d}$ measure karo, $γ$ infer karo)	underdamped	formulas ko invert karo	Ex 10

Har row upar ki figure mein red root-dots ki ek position se correspond karta hai: rows A/D unhe real axis se door rakhte hain, row B uspar (merged), rows C/E uspe split.

Worked examples

Worked example Ex 1 — System classify karo · Cell A (underdamped)

$m = 0.5$ kg, $k = 200$ N/m, $b = 4$ N·s/m. Kaun sa regime hai, aur $ω_{d}$ kya hai?

Forecast: Compute karne se pehle guess karo — kya $b = 4$ yahan "chhota" hai? Underdamped / critical / overdamped likh lo.

$ω_{0} = k / m = 200/0.5 = 400 = 20$ rad/s. Yeh step kyun? $ω_{0}$ spring ki undamped eagerness hai — hume ise damping se compare karne ke liye chahiye.
$γ = b /2 m = 4/ (2 \cdot 0.5) = 4/1 = 4$ s⁻¹. Yeh step kyun? $γ$ standard-form damping hai; sirf $γ$ aur $ω_{0}$ hi $Δ$ mein appear karte hain.
$Δ = γ^{2} - ω_{0}^{2} = 16 - 400 = - 384 < 0$ . Yeh step kyun? $Δ$ ka sign cell pick karta hai. Negative ⇒ underdamped (Cell A) — root-dots axis se door.
$ω_{d} = ω_{0}^{2} - γ^{2} = 400 - 16 = 384 \approx 19.60$ rad/s, toh $x (t) = A e^{- 4 t} cos (19.60 t + ϕ)$ . Yeh step kyun? Jab $Δ < 0$ toh $λ$ ka imaginary part real oscillation rate hai; ise underdamped shape mein daalo.

Verify: $ω_{d} < ω_{0}$ (19.60 < 20) ✓ — damping hamesha frequency ghatata hai. Units: $(s^{- 2}) = s^{- 1}$ ✓.

Worked example Ex 2 — Critical damping tune karo · Cells B, F (boundary)

Same $m = 0.5$ kg, $k = 200$ N/m. Kaunsa $b$ critical damping deta hai? Repeated root kya hai?

Forecast: Kya critical $b$ , Ex 1 ke $b = 4$ se bada hoga ya chhota? Guess karo.

Critical matlab $Δ = 0$ , yani $γ = ω_{0}$ . Yeh step kyun? $Δ = 0$ wobble aur crawl ke beech ki boundary ki definition hai — woh moment jab red dots milte hain.
$γ = ω_{0} \Rightarrow \frac{b}{2 m} = \frac{k}{m} \Rightarrow b_{cr i t} = 2 k m$ . Yeh step kyun? $b$ ke liye algebraically solve karo; mass clean form $2 k m$ mein cancel ho jaata hai.
$b_{cr i t} = 2 200 \cdot 0.5 = 2100 = 2 \cdot 10 = 20$ N·s/m. Yeh step kyun? Numbers ko boundary formula mein plug karo.
Repeated root: $λ = - γ = - ω_{0} = - 20$ s⁻¹, solution $x = (A + B t) e^{- 20 t}$ . Yeh step kyun? $Δ = 0$ par do roots merge ho jaate hain; doosra independent solution $t$ ka factor carry karta hai — critically-damped shape.

Verify: $b_{cr i t} = 20 > 4$ (Ex 1 ka $b$ ) ✓ — Ex 1 mein less damping thi, correctly underdamped. Units: $N/m \cdot kg = N\cdotkg/m$ ; since N = kg·m/s², yeh hai $kg^{2} / s^{2} = kg/s = N\cdots/m$ ✓.

Worked example Ex 3 — Overdamped roots · Cell C (overdamped)

$m = 1$ kg, $k = 9$ N/m, $b = 10$ N·s/m. Dono decay rates dhundo.

Forecast: Do real roots — magnitude mein kaun sa bada hai? Long-time motion mein kaun sa dominate karta hai?

$ω_{0} = 9/1 = 3$ rad/s; $γ = b / (2 m) = 10/ (2 \cdot 1) = 10/2 = 5$ s⁻¹. Yeh step kyun? Standard-form parameters pehle, hamesha. Note karo $γ = b / (2 m)$ ke denominator mein mass chahiye — yahan $m = 1$ hai toh $10/2$ dikhta hai, lekin general formula mein $m$ ka factor carry hota hai.
$Δ = γ^{2} - ω_{0}^{2} = 25 - 9 = 16 > 0 \Rightarrow$ overdamped (Cell C), $Δ = 4$ — dots real axis ke saath split. Yeh step kyun? Positive $Δ$ ⇒ do real roots, koi oscillation nahi.
$λ_{1} = - γ + Δ = - 5 + 4 = - 1$ s⁻¹; $λ_{2} = - γ - Δ = - 5 - 4 = - 9$ s⁻¹, toh $x = C e^{- t} + D e^{- 9 t}$ . Yeh step kyun? Root ke dono signs overdamped shape ke do exponentials dete hain.
Long-time motion $\approx C e^{- t}$ — slow root $λ_{1} = - 1$ , $λ_{2} = - 9$ se 9× zyada time tak rehta hai. Yeh step kyun? Woh root jo zero ke paas hai sabse dheere decay karta hai aur isliye motion ki tail mein rule karta hai.

Verify: Dono roots negative ✓ (system ko decay karna hi chahiye). Product $λ_{1} λ_{2} = (- 1) (- 9) = 9 = ω_{0}^{2}$ ✓ aur sum $λ_{1} + λ_{2} = - 10 = - 2 γ$ ✓ — Vieta's formulas confirm karte hain ki yeh $λ^{2} + 2 γ λ + ω_{0}^{2} = 0$ solve karte hain.

Worked example Ex 4 — Degenerate case

$γ = 0$ · Cell D (no damping) $b = 0$ set karo, $m = 0.5$ , $k = 200$ ke saath. Motion kya hai?

Forecast: Zero syrup ke saath, kya yeh kabhi rukta hai?

$γ = b /2 m = 0$ . Yeh step kyun? Zero damping usi equation ka degenerate limit hai.
$λ = - 0 \pm 0 - ω_{0}^{2} = \pm i ω_{0}$ , purely imaginary — root-dots imaginary axis par baithe hain, zero decay. Yeh step kyun? Real part nahi ⇒ koi decay envelope nahi; sirf oscillation bachta hai.
$ω_{d} = ω_{0}^{2} - 0 = ω_{0} = 20$ rad/s, aur $x (t) = A cos (20 t + ϕ)$ — pure Simple Harmonic Motion. Yeh step kyun? Jab $γ = 0$ , $ω_{d}$ collapse hokar $ω_{0}$ ban jaata hai; underdamped shape apna envelope kho deta hai aur SHM ban jaata hai.

Verify: Envelope $e^{- γ t} = e^{0} = 1$ — constant amplitude, hamesha bajta rahta hai ✓. Yeh sanity anchor hai: damped theory ko SHM reduce karna chahiye jab damping vanish ho.

Worked example Ex 5 — Limiting behaviour

$γ \to \infty$ · Cell E (heavy overdamp) $ω_{0} = 3$ rad/s fix karo. Jab $γ$ bahut bada ho jaata hai, slow root $λ_{1}$ ka kya hota hai?

Forecast: Kya slow root $0$ ki taraf jaata hai ya $- \infty$ ki taraf? Physically iska kya matlab hai?

$λ_{1} = - γ + γ^{2} - ω_{0}^{2}$ . Yeh step kyun? $λ_{1}$ ( $+$ root) tail ke liye slow, dominant term hai.
Large $γ$ ke liye: $γ^{2} - ω_{0}^{2} = γ 1 - ω_{0}^{2} / γ^{2} \approx γ (1 - \frac{ω _{0}^{2}}{2 γ ^{2}}) = γ - \frac{ω _{0}^{2}}{2 γ}$ . Yeh step kyun? $γ ≫ ω_{0}$ ke liye $1 - ϵ \approx 1 - ϵ /2$ use karke root expand karo — woh tool jo "bahut bada" ko clean estimate mein badalta hai.
Toh $λ_{1} \approx - γ + γ - \frac{ω _{0}^{2}}{2 γ} = - \frac{ω _{0}^{2}}{2 γ} \to 0^{-}$ jab $γ \to \infty$ . Yeh step kyun? Dominant $- γ$ aur $+ γ$ cancel ho jaate hain, ek tiny negative number bachta hai — ek red dot origin ki taraf creep karta hai.
$γ = 100$ , $ω_{0} = 3$ par check karo: exact $λ_{1} = - 100 + 10000 - 9 = - 100 + 99.955 \approx - 0.04500$ ; approx $- 9/200 = - 0.04500$ . Yeh step kyun? Numerically confirm karo ki approximation trustworthy hai.

Verify: $λ_{1} \to 0^{-}$ matlab decay bina bound ke slow hoti jaati hai — zyada damping ise aur slow kar deta hai, classic overdamped paradox. Approx $- 0.04500$ exact $- 0.04500$ se match karta hai (4 dp tak) ✓.

Worked example Ex 6 — Initial conditions se amplitude & phase · Cell A (underdamped, full solve)

Underdamped with $γ = 1$ s⁻¹, $ω_{d} = 4$ rad/s. $x (0) = 2$ m se rest se release kiya. $x (t) = A e^{- γ t} cos (ω_{d} t + ϕ)$ mein $A$ aur $ϕ$ dhundo.

Forecast: "Rest se" matlab $\overset{x}{˙} (0) = 0$ . Kya $ϕ$ exactly $0$ hoga? Guess karo.

$x (0) = A cos ϕ = 2$ . Yeh step kyun? $t = 0$ par position condition apply karo; $e^{0} = 1$ .
$\overset{x}{˙} (t) = A e^{- γ t} [- γ cos (ω_{d} t + ϕ) - ω_{d} sin (ω_{d} t + ϕ)]$ ; $t = 0$ par: $- γ cos ϕ - ω_{d} sin ϕ = 0$ . Yeh step kyun? Differentiate karo (product rule) aur zero initial velocity impose karo.
$tan ϕ = - γ / ω_{d} = - 1/4 \Rightarrow ϕ = arctan (- 1/4) \approx - 0.2450$ rad. Yeh step kyun? Velocity equation ko phase ke liye solve karo. Branch choice: $tan ϕ = - 1/4$ ke ek turn mein do solutions hain, $ϕ \approx - 0.2450$ aur $ϕ \approx π - 0.2450 \approx 2.897$ . Humein $x (0) = A cos ϕ = 2 > 0$ chahiye with $A > 0$ , toh $cos ϕ$ positive hona chahiye — yeh second-quadrant value $2.897$ ko rule out karta hai (jahan $cos < 0$ ) aur principal branch $ϕ \approx - 0.2450$ rad force karta hai. Yahi $arctan$ ke peeche sign-based quadrant reasoning hai.
$A = 2/ cos ϕ = 2/ cos (- 0.2450) = 2/0.97014 \approx 2.0616$ m. Yeh step kyun? Amplitude paane ke liye $ϕ$ ko position condition mein back-substitute karo; kyunki $cos ϕ > 0$ hai hum correctly $A > 0$ paate hain.

Verify: $\overset{x}{˙} (0)$ recompute karo: $A (- γ cos ϕ - ω_{d} sin ϕ) = 2.0616 (- 1 \cdot 0.97014 - 4 \cdot (- 0.24254)) = 2.0616 (- 0.97014 + 0.97016) \approx 0$ ✓. Aur $A > x (0)$ (2.06 > 2) ✓ — envelope starting displacement se kam se kam utni hi upar se shuru honi chahiye.

Worked example Ex 7 — Boundary test (above/below critical) · Cell F (classify)

Ek system mein $k = 100$ N/m, $m = 1$ kg hai. Teen candidates: $b = 15, 20, 25$ N·s/m. Har ek ko classify karo.

Forecast: Kaun sa akela $b$ knife-edge hai?

$b_{cr i t} = 2 k m = 2 100 \cdot 1 = 2 \cdot 10 = 20$ N·s/m. Yeh step kyun? Ex 2 ke formula se boundary value; sab kuch isse compare hota hai.
$b = 15 < 20$ ⇒ underdamped (Cell A). Yeh step kyun? Critical se less damping ⇒ wobble bachta hai — dots axis se door.
$b = 20 = 20$ ⇒ critically damped (Cell B). Yeh step kyun? Exactly boundary par ⇒ repeated root — dots collide.
$b = 25 > 20$ ⇒ overdamped (Cell C). Yeh step kyun? Critical se zyada damping ⇒ do real roots, sluggish — dots split.

Verify: $γ$ values hain $7.5, 10, 12.5$ ; $ω_{0} = 100 = 10$ . Toh $γ < ω_{0}$ , $γ = ω_{0}$ , $γ > ω_{0}$ respectively ✓ — $b$ -comparison se exactly match karta hai.

Worked example Ex 8 — Energy aur Quality factor · Cell G (energy decay)

Underdamped, $ω_{0} = 50$ rad/s, $γ = 0.5$ s⁻¹. $Q$ dhundo, 2 s baad bachi energy fraction, aur tab tak oscillation ne kitne radians of phase ring kiye.

Forecast: Itne chhote $γ$ ke saath, kya 2 s mein zyaadatar energy bachegi ya chali jaayegi?

$Q = \frac{ω _{0}}{2 γ} = \frac{50}{2 \cdot 0.5} = \frac{50}{1} = 50$ . Yeh step kyun? $Q$ measure karta hai ki system kitna lightly damped hai — high $Q$ lambe time tak bajta hai. Dekho Quality factor & bandwidth.
Amplitude $e^{- γ t}$ ki tarah decay karta hai, toh $t = 2$ par: $e^{- 0.5 \cdot 2} = e^{- 1} \approx 0.3679$ original amplitude bachti hai. Yeh step kyun? Envelope $A e^{- γ t}$ hai; $t = 2$ set karo.
Energy $\propto$ amplitude², toh $E (t) = E_{0} e^{- 2 γ t}$ , fraction deta hai $e^{- 2 \cdot 0.5 \cdot 2} = e^{- 2} \approx 0.1353$ , yani $\approx 13.53%$ energy bachti hai. Yeh step kyun? Amplitude ratio ko square karo $(e^{- 1})^{2} = e^{- 2}$ — energy factor $2 γ$ produce hota hai.
Phase rung: exact $ω_{d} = 5 0^{2} - 0. 5^{2} = 2499.75 \approx 49.9975$ , relative error sirf $γ^{2} / (2 ω_{0}^{2}) = 0.25/5000 = 5 \times 1 0^{- 5}$ hai, toh $ω_{d} \approx ω_{0}$ safe hai. Phase $\approx ω_{0} t = 50 \cdot 2 = 100$ rad $\approx 100/ (2 π) \approx 15.9$ full cycles. Yeh step kyun? $ω_{d} \approx ω_{0}$ tabhi use karo jab $γ ≪ ω_{0}$ ; yahan hum error quantify karte hain aur confirm karte hain ki negligible hai.

Verify: Per-cycle rule se cross-check: amplitude har cycle mein $e^{- 2 π / Q} = e^{- 2 π /50} = e^{- 0.12566} \approx 0.8820$ factor se girti hai. $n = ω_{0} t /2 π = 100/2 π \approx 15.915$ cycles mein amplitude ratio hai $(e^{- 2 π / Q})^{n} = e^{- 2 π \cdot 15.915/50} = e^{- 1.0}$ , step 2 ke $e^{- 1}$ se exactly match ✓. Energy ratio $= (e^{- 1})^{2} = e^{- 2} \approx 0.1353$ ✓. $Q$ ki units: dimensionless ✓.

Worked example Ex 9 — Real device: ek car suspension · Cell H (word problem)

Ek car quarter-body of $m = 300$ kg ek spring $k = 27000$ N/m par rakhi hai. Engineers critical damping chahte hain (bump ke baad koi bounce nahi). Required shock-absorber constant $b$ dhundo aur confirm karo ki settle non-oscillatory hai.

Forecast: Kya critical $b$ , lab spring ke $20$ N·s/m se bada hoga ya chhota? Order of magnitude guess karo.

$ω_{0} = k / m = 27000/300 = 90 \approx 9.487$ rad/s. Yeh step kyun? Car body ki spring par bare bounce frequency.
Critical: $b_{cr i t} = 2 k m = 2 27000 \cdot 300 = 2 8.1 \times 1 0^{6} = 2 \cdot 2846.05 \approx 5692$ N·s/m. Yeh step kyun? Same boundary formula; real shocks waqai hazaaron N·s/m mein hote hain.
$b_{cr i t}$ par, $γ = b_{cr i t} / (2 m) = 5692/600 \approx 9.487 = ω_{0}$ toh $Δ = 0$ : single root $λ = - 9.487$ s⁻¹, $x = (A + B t) e^{- 9.487 t}$ — koi cosine nahi, koi bounce nahi. Yeh step kyun? $Δ$ se motion class confirm karo, design goal se match karo.

Verify: $γ \approx 9.487 = ω_{0}$ ✓, toh $Δ = 0$ exactly critical par. Isliye car shocks (near-)critical hote hain: koi oscillation nahi ke saath sabse tez settle — yeh bhi dekho Forced Oscillations & Resonance jisse pata chale kyun tum kabhi nahi chahte suspension underdamped ho road-bump frequencies ke paas.

Worked example Ex 10 — Exam twist:

$ω_{d}$ measure karo, $γ$ infer karo · Cell I (invert formulas) Tumne ek damped oscillation measure kiya jisme damped frequency $ω_{d} = 19.60$ rad/s hai, aur undamped frequency jaani hui hai $ω_{0} = 20$ rad/s. $γ$ aur damping constant $b$ dhundo agar $m = 0.5$ kg ho. (Yeh Ex 1 ko invert karta hai.)

Forecast: Kya recovered $b$ , Ex 1 ke $b = 4$ se match karega? Predict karo haan/nahi.

$ω_{d}^{2} = ω_{0}^{2} - γ^{2}$ se, solve karo $γ = ω_{0}^{2} - ω_{d}^{2}$ . Yeh step kyun? Hume outputs $ω_{d}, ω_{0}$ pata hain aur input $γ$ chahiye — defining relation ko rearrange karo.
$γ = 2 0^{2} - 19.6 0^{2} = 400 - 384.16 = 15.84 \approx 3.980$ s⁻¹. Yeh step kyun? Numbers plug karo; note karo ki do bade numbers ka near-cancellation hota hai, dhyan chahiye.
$b = 2 mγ = 2 \cdot 0.5 \cdot 3.980 = 3.980$ N·s/m. Yeh step kyun? Physical damping constant recover karne ke liye $γ = b /2 m$ invert karo.

Verify: Recovered $b \approx 3.98 \approx 4$ N·s/m — Ex 1 ke input se match karta hai ✓ (chhota gap isliye hai kyunki Ex 1 ka $ω_{d} = 384 = 19.5959$ round hokar $19.60$ hua tha). Round-trip close ho jaata hai.

Recall Kaun sa cell kaun sa hai — quick self-test

Sign of $Δ < 0$ matlab ::: underdamped (oscillate karta hai); root-dots real axis se door. Sign of $Δ = 0$ matlab ::: critically damped (sabse tez clean stop); dots real axis par merge. Sign of $Δ > 0$ matlab ::: overdamped (do decaying exponentials); dots real axis ke saath split. $γ = 0$ solution ko reduce karta hai ::: pure SHM, $x = A cos (ω_{0} t + ϕ)$ , koi decay nahi; dots imaginary axis par. Jab $γ \to \infty$ slow root $λ_{1}$ tend karta hai ::: $0^{-}$ ki taraf (motion kabhi bhi slow hoti jaati hai). Critical damping constant ka formula ::: $b_{cr i t} = 2 k m$ . $Q$ ke terms mein per-cycle amplitude decay factor ::: $e^{- 2 π / Q}$ .

General theory ke liye Second-order linear ODEs bhi dekho characteristic equation ke peeche, aur RLC circuits jahan $L, R, C$ exactly $m, b, k$ ki roles play karte hain.