Damped oscillations — underdamped, critically damped, overdamped: Exercises

1.6.9 · D4 · Physics › Oscillations & Waves › Damped oscillations — underdamped, critically damped, overda

Kuch symbols har jagah aate hain, toh kisi bhi problem se pehle unhe plain words mein pin down karte hain:

Definition Woh symbols jo hum baar baar use karte rehte hain

$ω_{0} = k / m$ — natural frequency: spring akela (bina drag ke) kitni tezi se oscillate karta. Units: radians per second.
$γ = b /2 m$ — damping coefficient: drag motion ko kitni strongly drain karta hai. Units: per second ( $s^{- 1}$ ).
$ω_{d} = ω_{0}^{2} - γ^{2}$ — damped frequency: jab system oscillate karta hai toh jo actual rhythm observe hoti hai. Yeh tabhi exist karta hai (real hota hai) jab $γ < ω_{0}$ ; drag ise $ω_{0}$ se thoda slower bana deta hai. Units: radians per second.
$A, B$ — integration constants: har general solution mein do free numbers, jo baad mein initial conditions (starting position $x (0)$ aur starting velocity $\overset{x}{˙} (0)$ ) se fix hote hain. Yeh physics nahi hain — yeh "motion kahaan se aur kaise shuru hui" ka bookkeeping hai.

Figure — Damped oscillations — underdamped, critically damped, overdamped

Is figure ko kaise padhein (pure page ka map): horizontal axis damping $γ$ hai (jahan $ω_{0}$ fixed hai $1$ par); vertical axis discriminant $Δ = γ^{2} - ω_{0}^{2}$ hai. Slate curve $Δ$ khud hai. Jahan curve dashed zero line ke neeche hai (lavender band, $γ < ω_{0}$ ) tum underdamped ho; jahan yeh zero cross karta hai (mint line, $γ = ω_{0}$ ) tum critical ho; jahan yeh zero ke upar uthta hai (coral band, $γ > ω_{0}$ ) tum overdamped ho. Apna $γ$ ko $ω_{0}$ ke relative dekho, $Δ$ ka sign padho, aur kuch aur karne se pehle hi regime pata chal jaati hai.

Level 1 — Recognition

Exercise 1.1

Ek mass–spring–damper mein $m = 4 kg$ , $k = 36 N/m$ , $b = 6 N\cdots/m$ hai. Kaun sa regime hai?

Recall Solution

KYA compute karna hai: do frequencies, phir discriminant ke through unhe compare karna.

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

Answer: underdamped. Damping spring se kaafi weak hai, toh mass ek shrinking envelope ke andar aage-peeche swing karta hai.

Exercise 1.2

$m = 1 kg$ , $k = 9 N/m$ ke liye, woh value of $b$ batao jo system ko wobbling aur crawling ke beech exactly boundary par rakhti hai.

Recall Solution

KYA: boundary critical damping hai, $γ = ω_{0}$ , yaani discriminant $Δ = 0$ .

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

Answer: $b_{cr i t} = 6 N\cdots/m$ . Isse neeche underdamp hoga; isse upar overdamp. (KYU: $γ = ω_{0}$ discriminant $Δ = γ^{2} - ω_{0}^{2}$ ko exactly zero par laata hai — borderline ki definition yahi hai.)

Level 2 — Application

Exercise 2.1

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

Recall Solution

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

KYU $ω_{d} < ω_{0}$ : drag har swing ko thoda slow karta hai, isliye observed rhythm ideal undamped wali se thodi slower hoti hai.

Exercise 2.2

Wahi system $x (0) = 0.05 m$ se rest se release hoti hai. Amplitude envelope $A e^{- γ t}$ hai jahan $A = 0.05 m$ (approximately, light damping ke liye). $t = 1 s$ ke baad initial amplitude ka kitna fraction bachta hai? Initial energy ka kitna fraction?

Recall Solution

KYA: decaying envelope evaluate karo, phir energy ke liye square karo.

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

KYU energy ke liye factor 2: $E \sim \frac{1}{2} k A^{2}$ , aur $e^{- γ t}$ ko square karne par $e^{- 2 γ t}$ milta hai. Energy hamesha amplitude se do guna tezi se (exponent mein) leakti hai.

Level 3 — Analysis

Exercise 3.1

Underdamped oscillator jahan $ω_{0} = 5 rad/s$ , $γ = 0.5 s^{- 1}$ hai. Quality factor $Q = ω_{0} /2 γ$ compute karo, aur estimate karo ki amplitude apni start ki $1/ e$ tak girne se pehle kitne full oscillation cycles hote hain.

Recall Solution

$Q = \frac{ω _{0}}{2 γ} = \frac{5}{2 \cdot 0.5} = \frac{5}{1} = 5$ .
Amplitude $1/ e$ par pahunchti hai jab $e^{- γ t} = e^{- 1}$ , yaani $t = 1/ γ = 1/0.5 = 2 s$ par.
Us time mein cycles ki number: period $T = 2 π / ω_{d}$ . Yahan $ω_{d} = 25 - 0.25 = 24.75 \approx 4.975 rad/s$ , toh $T \approx 2 π /4.975 \approx 1.263 s$ .
Cycles $= t / T = 2/1.263 \approx 1.58$ cycles. (KYU $t$ ko $T$ se divide karte hain: ek period $T$ exactly ek complete cycle ka time hota hai, isliye elapsed time $t$ mein $t / T$ aisi periods hain — woh count hi cycles ki number hai.)

$Q$ se sanity check — " $Q / π$ " kahaan se aata hai: $1/ e$ tak pahunchne mein cycles ki number hai $N = \frac{t _{1/ e}}{T} = \frac{1/ γ}{2 π / ω _{d}} = \frac{ω _{d}}{2 π γ} .$ Light damping ke liye $ω_{d} \approx ω_{0}$ , isliye $N \approx \frac{ω _{0}}{2 π γ} = \frac{1}{π} \cdot \frac{ω _{0}}{2 γ} = \frac{Q}{π}$ . Yahan $Q / π = 5/ π \approx 1.59$ — upar ke $1.58$ se match karta hai. (KYU yeh kaam karta hai: $Q = ω_{0} /2 γ$ pehle se hi "oscillation speed over decay speed" package karta hai; $π$ se divide karna sirf us ratio ko whole cycles mein convert karta hai.)

Exercise 3.2

Overdamped system: $ω_{0} = 3 rad/s$ , $γ = 5 s^{- 1}$ . Dono real roots $λ_{1, 2}$ find karo, aur batao ki kaun sa term long-time behaviour mein dominate karta hai aur kyun.

Recall Solution

KYU roots at all: equation of motion $m \overset{x}{¨} + b \overset{x}{˙} + k x = 0$ ( $m$ se divide karne par, $\overset{x}{¨} + 2 γ \overset{x}{˙} + ω_{0}^{2} x = 0$ milta hai) $x = e^{λ t}$ guess karke solve hoti hai. Har derivative sirf $λ$ se multiply karta hai, isliye ODE characteristic equation $λ^{2} + 2 γ λ + ω_{0}^{2} = 0$ mein collapse ho jaata hai. Us quadratic ko quadratic formula se solve karne par neeche ke roots milte hain — yahi poora reason hai ki $λ_{1, 2}$ kyun aate hain.

$λ = - γ \pm γ^{2} - ω_{0}^{2} = - 5 \pm 25 - 9 = - 5 \pm 16 = - 5 \pm 4$ .
Toh $λ_{1} = - 5 + 4 = - 1 s^{- 1}$ aur $λ_{2} = - 5 - 4 = - 9 s^{- 1}$ .
Solution shape: $x (t) = A e^{- t} + B e^{- 9 t}$ , jahan $A$ aur $B$ integration constants hain jo initial conditions $x (0)$ aur $\overset{x}{˙} (0)$ se fix hote hain — yeh set karte hain ki har exponential ka kitna hissa present hai, lekin decay rates nahi.
Long times par dominant: $e^{- t}$ term. Iska decay rate magnitude mein sabse chhota hai ( $∣ λ_{1} ∣ = 1$ ), isliye $e^{- 9 t}$ almost instantly gayab ho jaata hai jabki $e^{- t}$ linger karta hai.

KYU overdamped sluggish lagta hai: slow root $λ_{1}$ zero ke paas baithta hai, isliye rest par return ki slow tail dominate karti hai. Zyaada damping → aur slower crawl, faster nahi.

Do exponentials aur unka sum neeche plot kiya gaya hai — notice karo ki coral fast root ( $e^{- 9 t}$ ) almost turant chala jaata hai jabki lavender slow root ( $e^{- t}$ ) poori dashed curve ki pace set karta hai:

Dashed slate curve (sum $x (t)$ ) dekho: $t \approx 0.5 s$ ke baad yeh almost exactly lavender slow root ko track karta hai — visual proof ki $λ_{1} = - 1$ tail govern karta hai.

Level 4 — Synthesis

Exercise 4.1

Ek car suspension mein effective $m = 250 kg$ (per wheel) aur spring $k = 40000 N/m$ hai. Engineers critical damping chahte hain (sabse tezi settle, koi bounce nahi). Required damping constant $b_{cr i t}$ find karo, aur corresponding $Q$ confirm karo.

Recall Solution

KYA: critical damping ke liye $γ = ω_{0}$ chahiye, yaani $b_{cr i t} = 2 k m$ .

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

KYU $2 k m$ : $γ = ω_{0}$ set karo → $b /2 m = k / m$ → $b = 2 m k / m = 2 k m$ . $m$ s cleanly combine ho jaate hain.

Exercise 4.2

Parent-note system $m = 2$ , $k = 50$ lo aur teen damping values par settling behaviour compare karo: $b = 8$ (underdamped), $b = 20$ (critical), $b = 40$ (overdamped). Har ek ke liye $γ$ do, regime naam do, aur woh slowest decay rate identify karo jo tail govern karta hai.

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 $γ = 2$ par decay karta hai
20	$5$	critical ( $5 = 5$ )	$e^{- 5 t}$ (ek $t$ factor ke saath)
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: $γ = 5 s^{- 1}$ — overall sabse tezi envelope.
Overdamped slow root: $λ_{1} \approx - 1.34 s^{- 1}$ — dono doosron se slower, crawl confirm karta hai.

" $t$ factor" KAHAAN se aata hai (critical case): jab $γ = ω_{0}$ hota hai toh discriminant $Δ = 0$ hota hai, isliye do roots merge ho jaate hain ek single repeated root $λ = - γ$ mein. Ek 2nd-order ODE ko hamesha do independent solutions chahiye, lekin merge sirf ek chhodta hai, $e^{- γ t}$ . Repeated root ke liye standard remedy doosra missing solution $t e^{- γ t}$ ke roop mein supply karta hai (tum verify kar sakte ho ki yeh $\overset{x}{¨} + 2 γ \overset{x}{˙} + ω_{0}^{2} x = 0$ satisfy karta hai jab $γ = ω_{0}$ ho). Isliye general form hai $x (t) = (A + B t) e^{- γ t},$ jahan $A, B$ phir se initial conditions se fix hote hain. Envelope par effect: $t$ ka akela factor sirf linearly badhta hai jabki $e^{- γ t}$ exponentially shrink karta hai, isliye product zero par decay karta hi hai — bas ek brief rise-then-fall bump ke saath instead of pure exponential ke. Isliye critical damping itni cleanly aur quickly settle ho sakti hai.

Punchline: critical ( $γ = 5$ ) ka sabse fastest-decaying dominant behaviour hai. Isse aage $b = 40$ push karo aur ek $- 1.34 s^{- 1}$ tail appear hoti hai — underdamped case se bhi slower. Yahi hai kyun critical damping engineering sweet spot hai.

Neeche teenon responses ek saath plot kiye hain (sab normalized hain $x = 1$ se start karne ke liye, rest se release), taaki tum sweet spot dekh sako na ki sirf padho:

Har curve trace karo: lavender underdamped waali zero ke neeche jaati hai (wobble karti hai), mint critical curve bina cross kiye sabse tezi se zero slide karti hai, aur coral overdamped curve poori tarah mint wali ke upar crawl karti hai — mint curve ka sabse pehle zero reach karna "critical sabse teza clean stop hai" ka geometric meaning hai.

Level 5 — Mastery

Exercise 5.1

Ek underdamped oscillator ki amplitude $4$ full oscillations mein $12 cm$ se $3 cm$ tak girti hai. Observed damped period $T = 0.5 s$ hai. (a) damping coefficient $γ$ , aur (b) quality factor $Q$ find karo (light-damping estimate ke liye $ω_{d} \approx ω_{0}$ use karo).

Recall Solution

KYA: ek known time ke across envelope ratio use karo.

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

KYU logs: exponential decay log-space mein linear ho jaata hai, "ratio over time" ko $γ$ ke liye clean solve mein badal deta hai. Yeh logarithmic decrement trick damping measure karne ka standard experimental tarika hai.

Exercise 5.2

Wahi physical damper ek RLC circuit analogy mein daala gaya hai: $L = 0.5 H$ , $C = 200 μ F$ , $R = 20 Ω$ . Mechanical–electrical dictionary $m \to L$ , $b \to R$ , $1/ k \to C$ use karke, circuit ki transient response classify karo.

Recall Solution

KYA: same $γ$ vs $ω_{0}$ test par map karo. RLC ke liye, $ω_{0} = 1/ L C$ aur $γ = R /2 L$ .

$ω_{0} = \frac{1}{L C} = \frac{1}{0.5 \cdot 200 \times 1 0 ^{- 6}} = \frac{1}{1 0 ^{- 4}} = \frac{1}{0.01} = 100 rad/s$ .
$γ = \frac{R}{2 L} = \frac{20}{2 \cdot 0.5} = \frac{20}{1} = 20 s^{- 1}$ .
Compare: $γ = 20 < ω_{0} = 100$ → underdamped — current/charge ring karta hai aur decay karta hai, bilkul mechanical spring ki tarah.
Bonus $Q = ω_{0} /2 γ = 100/40 = 2.5$ : ek modestly ringing circuit.

KYU same test kaam karta hai: governing ODE $L \overset{q}{¨} + R \overset{q}{˙} + q / C = 0$ form mein $m \overset{x}{¨} + b \overset{x}{˙} + k x = 0$ se identical hai. Same maths → same teen regimes → same $γ$ -vs- $ω_{0}$ decision. Second-order linear ODEs dekho.

Recall Quick self-check (answers reveal karo)

Ek line mein regime test? ::: Discriminant $Δ = γ^{2} - ω_{0}^{2}$ ka sign check karo; negative → under, zero → critical, positive → over. Amplitude vs energy exponents? ::: $A \propto e^{- γ t}$ , $E \propto e^{- 2 γ t}$ . $b$ ke terms mein critical damping? ::: $b_{cr i t} = 2 k m$ , equivalently $Q = \frac{1}{2}$ . Overdamped slow kyun hota hai? ::: Iska slow root $λ_{1} = - γ + γ^{2} - ω_{0}^{2}$ zero ke paas baithta hai. Critical damping par extra $t$ kyun? ::: Repeated root sirf ek solution $e^{- γ t}$ deta hai; doosra independent wala $t e^{- γ t}$ hota hai.