Normal modes — coupled oscillators, normal coordinates: Worked examples

2.1.20 · D3 · Physics › Analytical Mechanics › Normal modes — coupled oscillators, normal coordinates

Yeh page parent note on normal modes ke liye practice arena hai. Wahan humne machinery derive ki thi. Yahan hum use har tarah ki situation ke khilaf run karte hain jo yeh topic dे sakta hai — clean modes, energy sloshing, weird masses, zero coupling, infinite coupling, ek real-world word problem, aur ek exam trap.

Numbers chhune se pehle, puri landscape lay out karte hain taaki koi bhi case surprise na kare.

Matrix mein do symbols aayenge, toh pehle inhe plain words mein name karte hain:

Scenario matrix

#	Case class	Kya special hai	Example
A	Pure symmetric kick	initial condition is eigenvector 1 → ek frequency	Ex 1
B	Pure antisymmetric kick	initial condition is eigenvector 2 → ek frequency	Ex 2
C	Mixed kick (beats)	general start → superposition, energy sloshes	Ex 3
D	Weak-coupling limit $k_{c} ≪ k$	modes nearly degenerate → bahut slow beats	Ex 4
E	Zero coupling $k_{c} = 0$	degenerate case: coupling vanish ho jaata hai, masses independent	Ex 5
F	Strong / stiff-link limit $k_{c} ≫ k$	ek mode race karta hai, link rigid act karta hai	Ex 6
G	Unequal masses $M \neq = m 1$	generalized eigenproblem, M-orthogonality bites	Ex 7
H	Real-world word problem (units!)	do atoms / do carts, Hz mein real frequency nikalo	Ex 8
I	Velocity kick + timing twist	struck (pull nahi) start, aur full-transfer time	Ex 9

Nau examples, nau cells. Note karo ki "kick" ek displacement ho sakta hai (mass ko pull karke release karo) ya ek velocity (rest mein mass ko strike karo) — Cell I velocity case cover karta hai taaki initial conditions ka poora space exhaust ho jaye. Chalo shuru karte hain.

Throughout, system parent ka two-mass chain hai (jab tak cell kuch aur na kahe), jinke results hum reuse karenge:

Case A — Pure symmetric kick (ek frequency)

Worked example Dono masses ek hi direction mein pull kiye

System: $m = 1 kg$ , $k = 4 N/m$ , $k_{c} = 3 N/m$ . Kick: $x_{1} (0) = x_{2} (0) = 0.05 m$ , dono rest mein. Find: motion, aur woh frequency jo tum measure karoge.

Forecast: Padhne se pehle guess karo — kya tum beats dekhoge, ya ek clean tone? (Ek clean tone: kick exactly eigenvector 1 hai.)

$t = 0$ par normal coordinates compute karo. $η_{1} (0) = \frac{1}{2} (0.05 + 0.05) = \frac{0.10}{2}$ , $η_{2} (0) = \frac{1}{2} (0.05 - 0.05) = 0$ . Yeh step kyun? Normal coordinates hi woh ek maatra variables hain jo independently move karte hain. $t = 0$ par inhe padhne se pata chalta hai ki kaun se modes switched on hain.
Note karo ki $η_{2} \equiv 0$ hamesha rahega. Uska equation $\overset{η}{¨}_{2} = - ω_{2}^{2} η_{2}$ hai jisme $η_{2} (0) = 0$ aur $\overset{η}{˙}_{2} (0) = 0$ → zero hi rahega. Yeh step kyun? Ek oscillator jo origin par rest se start ho woh kabhi wahan se nahi jaata. Mode 2 dead hai.
Sirf mode 1 alive hai: $η_{1} (t) = \frac{0.10}{2} cos ω_{1} t$ jisme $ω_{1} = k / m = 4 = 2 rad/s$ . Yeh step kyun? Ek surviving normal coordinate $\overset{η}{¨}_{1} = - ω_{1}^{2} η_{1}$ obey karta hai, jiska zero-velocity solution usi mode ki apni frequency par ek pure cosine hota hai.
Back-transform: $x_{1} = x_{2} = \frac{1}{2} η_{1} = 0.05 cos (2 t) m$ . Yeh step kyun? $x_{j} = \frac{1}{2} (η_{1} \pm η_{2})$ ; $η_{2} = 0$ ke saath dono masses ek motion share karte hain.

Verify: Dono masses identically move karte hain, toh middle spring kabhi stretch nahi hoti — usse $ω$ affect nahi honi chahiye. Indeed $ω = k / m$ mein koi $k_{c}$ nahi hai. ✓ Units: $(N/m) / kg = s^{- 2} = s^{- 1}$ . ✓ (Cell A)

Case B — Pure antisymmetric kick (ek frequency)

Worked example Masses opposite directions mein pull kiye

Same system ( $m = 1$ , $k = 4$ , $k_{c} = 3$ ). Kick: $x_{1} (0) = + 0.05$ , $x_{2} (0) = - 0.05$ , rest mein. Find: frequency.

Forecast: Case A se faster ya slower? (Faster — coupling spring ab fight kar rahi hai.)

Normal coordinates: $η_{1} (0) = \frac{1}{2} (0.05 - 0.05) = 0$ , $η_{2} (0) = \frac{1}{2} (0.05 + 0.05) = \frac{0.10}{2}$ . Yeh step kyun? Same reason as A: find karo kaun sa mode lit hai. Ab woh mode 2 hai.
Mode 1 dead, mode 2 alive: $ω_{2} = (k + 2 k_{c}) / m = (4 + 6) /1 = 10 \approx 3.162 rad/s$ . Yeh step kyun? Antisymmetric stretch coupling spring ka effect double kar deta hai: $k + 2 k_{c}$ , na ki $k + k_{c}$ .
Motion: $x_{1} = - x_{2} = 0.05 cos (10 t) m$ . Yeh step kyun? Back-transform $x_{j} = \frac{1}{2} (η_{1} \pm η_{2})$ jisme $η_{1} = 0$ single frequency $ω_{2}$ par equal-and-opposite motion deta hai.

Verify: $ω_{2} > ω_{1}$ ( $3.162 > 2$ ) ✓ — mnemonic SLOW=Same, FAST=Fight se match karta hai. Factor-2 trap $k + 2 k_{c}$ use karke dodge kiya. ✓ (Cell B)

Case C — Mixed kick → beats

Worked example Ek hi mass pull karo

Same system ( $m = 1$ , $k = 4$ , $k_{c} = 3$ ). Kick: $x_{1} (0) = a = 0.05$ , $x_{2} (0) = 0$ , rest mein. Find: $x_{1} (t)$ , aur motion describe karo.

Forecast: Kya mass 2, jo still start karta hai, still rahega? (Nahi — energy flood hoti hai; beats milte hain.)

Normal coords: $η_{1} (0) = \frac{a}{2}$ , $η_{2} (0) = \frac{a}{2}$ — dono modes lit, equally. Yeh step kyun? Ek one-sided pull symmetric aur antisymmetric ka 50/50 mix hai; wahi mix sloshing produce karta hai.
Har ek evolve karo: $η_{1} = \frac{a}{2} cos ω_{1} t$ , $η_{2} = \frac{a}{2} cos ω_{2} t$ , jisme $ω_{1} = 2$ , $ω_{2} = 10$ . Yeh step kyun? Har normal coordinate ek independent SHM hai; rest se start kiya toh woh bas apni frequency par cosines karta hai, aur ab dono frequencies alag hain.
Back-transform karo & sum-to-product identity use karo: $x_{1} = \frac{a}{2} (cos ω_{1} t + cos ω_{2} t) = a cos (\frac{ω _{2} - ω _{1}}{2} t) cos (\frac{ω _{1} + ω _{2}}{2} t)$ Yeh step kyun? Identity isse split karta hai ek fast carrier (average frequency) ke andar ek slow envelope (half the difference) mein — yahi exactly beats hote hain.
Envelope frequency: $\frac{ω _{2} - ω _{1}}{2} = \frac{3.162 - 2}{2} \approx 0.581 rad/s$ . Yeh step kyun? Yahi slow rate hai jo tumhari aankh ko actually energy "sloshing" back and forth dikhti hai; yeh set karta hai ki ek full transfer kitna time leta hai (Case I mein phir use hoga).

Neeche ka figure yeh exact motion plot karta hai. Horizontal axis par time $t$ (s) hai aur vertical axis par $x_{1}$ (units of $a$ mein) hai, legend mein har curve ka naam hai. Magenta curve dekho jisko "x1(t) full motion" label kiya hai: yeh ek fast wiggle hai jo violet dashed envelope ke andar trapped hai jisko "slow envelope" label kiya hai, equal to $\pm a cos (\frac{ω _{2} - ω _{1}}{2} t)$ . Jahan envelope zero par pinch hota hai (orange dotted line), mass 1 quiet ho jaata hai aur saari energy mass 2 par cross ho jaati hai — woh beat hai.

Verify: $t = 0$ par, $x_{1} = a \cdot 1 \cdot 1 = a$ ✓ aur $x_{2} = \frac{a}{2} (cos ω_{1} t - cos ω_{2} t) = 0$ ✓. Energy periodically mass 1 se bilkul leave ho jaati hai (envelope zero hit karta hai). ✓ (Cell C)

Case D — Weak coupling ( $k_{c} ≪ k$ ): slow beats

Worked example Loosely tied swings

System: $m = 1$ , $k = 100$ , $k_{c} = 1$ . Case C jaisa one-sided pull. Find: beat period ek single swing period se kaise compare karta hai.

Forecast: Energy exchange fast hai ya slow? (Bahut slow — weak link ka matlab hai modes almost same pitch par hain.)

Frequencies: $ω_{1} = 100 = 10$ , $ω_{2} = 102 \approx 10.0995$ . Yeh step kyun? Humein dono eigenfrequencies chahiye inki difference form karne ke liye; tiny $k_{c}$ ke saath woh almost coincide karte hain, yahi is limit ka poora point hai.
Envelope frequency: $\frac{ω _{2} - ω _{1}}{2} \approx \frac{0.0995}{2} \approx 0.0498 rad/s$ . Yeh step kyun? Beat rate frequency difference hai; tiny difference ⇒ tiny beat rate ⇒ lamba slosh time.
Beat period $T_{beat} = \frac{2 π}{∣ ω _{2} - ω _{1} ∣} = \frac{2 π}{0.0995} \approx 63.1 s$ , versus ek single swing period $\approx \frac{2 π}{10} \approx 0.628 s$ . Yeh step kyun? Ek full slosh over and back envelope ka ek period leta hai, aur envelope difference frequency par oscillate karta hai — toh $∣ ω_{2} - ω_{1} ∣$ se $2 π$ divide karna us rate ko ek actual time mein convert karta hai jo ek single swing se compare ho sake.

Verify: Beat period $≫$ swing period ( $63 ≫ 0.63$ , ratio $\approx 100$ ) — weak coupling ⇒ har slow slosh mein bahut saare fast swings, loosely coupled oscillators ki hallmark. ✓ (Cell D)

Case E — Zero coupling ( $k_{c} = 0$ ): degenerate limit

Worked example Middle spring kaat do

System: $m = 1$ , $k = 4$ , $k_{c} = 0$ . Kick: $x_{1} (0) = 0.05$ , $x_{2} (0) = 0$ , rest mein. Find: har mass kya karta hai.

Forecast: Koi link nahi, toh kya energy mass 2 par cross hogi? (Nahi — ab woh independent hain.)

Frequencies collapse together: $ω_{1}^{2} = k / m = 4$ , $ω_{2}^{2} = (k + 0) / m = 4$ . Dono $= 2 rad/s$ — degenerate. Yeh step kyun? $k_{c} = 0$ ke saath dono modes ki same frequency hai; system ab truly coupled nahi hai.
Beat envelope frequency $= \frac{ω _{2} - ω _{1}}{2} = 0$ → envelope kabhi nahi dipti. Yeh step kyun? Zero frequency difference ka matlab hai koi sloshing nahi; mass 2 ko kabhi energy nahi milti.
Motion: $x_{1} = 0.05 cos (2 t)$ , $x_{2} = 0$ har time ke liye. Yeh step kyun? Ek baar decouple hone par, mass 2 $m \overset{x}{¨}_{2} = - k x_{2}$ obey karta hai jisme $x_{2} (0) = \overset{x}{˙}_{2} (0) = 0$ , jiska ek hi solution $x_{2} \equiv 0$ hai; mass 1 ek lone cosine oscillator ki tarah $ω_{1}$ par reh jaata hai.

Verify: $x_{2}$ kabhi $0$ se depart nahi karta — equations instant mein decouple ho gaye jab $k_{c} \to 0$ : $m \overset{x}{¨}_{2} = - k x_{2}$ jisme $x_{2} (0) = \overset{x}{˙}_{2} (0) = 0$ . ✓ Yeh sanity anchor hai: zero coupling zaroori hai independent SHM dene ke liye. (Cell E)

Case F — Strong coupling ( $k_{c} ≫ k$ ): rigid-link limit

Worked example Masses ko weld kar do

System: $m = 1$ , $k = 4$ , $k_{c} = 1000$ . Find: do frequencies aur fast mode ka kya "matlab" hai.

Forecast: Jab $k_{c} \to \infty$ , $ω_{2}$ ka kya hoga? (Yeh infinity tak run karta hai — antisymmetric motion impossibly stiff ho jaata hai.)

Symmetric mode untouched rehta hai: $ω_{1} = k / m = 2 rad/s$ , chahe $k_{c}$ kitna bhi large ho. Yeh step kyun? Symmetric mode mein coupling spring kabhi stretch nahi hoti, toh uski stiffness irrelevant hai.
Antisymmetric mode: $ω_{2} = (k + 2 k_{c}) / m = 2004 \approx 44.77 rad/s$ . Yeh step kyun? Masses ki koi bhi relative motion bahut stiff link ko stretch karti hai, enormous energy cost karta hai — hence bahut high frequency.
Physical reading: slow motions ke liye sirf $ω_{1}$ excitable hai; pair ek rigid block ki tarah outer springs par $2 m$ mass ke saath move karta hai. Stiff mode "freeze out" ho jaata hai. Yeh step kyun? Kyunki $ω_{2}$ itna far above $ω_{1}$ hai, koi bhi gentle, slow driving sirf symmetric mode stir kar sakta hai — antisymmetric ek ke liye frequencies itni high chahiye jo reach nahi ho sakti, toh effectively dono masses ek single rigid object mein lock ho jaate hain aur sirf mode 1 observable rehta hai.

Verify: $ω_{1}$ $2$ par raha ( $k_{c}$ se independent) ✓; $ω_{2} ≫ ω_{1}$ ( $44.8 ≫ 2$ ) ✓. Yeh freeze-out exactly waise hai jaise high-frequency optical branches acoustic ones se alag hote hain Phonons and Lattice Vibrations mein. (Cell F)

Case G — Unequal masses: generalized eigenproblem

Worked example Heavy mass, light mass

System: $m_{1} = 1$ , $m_{2} = 2$ (kg), har ek wall se spring $k = 6$ se tied, coupled by $k_{c} = 6$ (N/m). Ab $M = (1002)$ nahi hai $m 1$ . Find: do frequencies.

Forecast: Kya eigenvectors phir bhi $(1, 1)$ aur $(1, - 1)$ honge? (Nahi — unequal masses modes ko tilt kar dete hain; simple symmetric/antisymmetric shapes break ho jaati hain.)

Matrices likho. Stiffness (potential energy $V$ se, $- \partial V / \partial x$ ke through): $K = (k + k_{c} - k_{c} - k_{c} k + k_{c}) = (12 - 6 - 6 12)$ Yeh step kyun? $K$ potential energy se aata hai aur masses par depend nahi karta.
Generalized secular equation solve karo $det (K - ω^{2} M) = 0$ : $det (12 - ω^{2} - 6 - 6 12 - 2 ω^{2}) = (12 - ω^{2}) (12 - 2 ω^{2}) - 36 = 0.$ Yeh step kyun? Yeh generalized hai — $ω^{2}$ $M$ se multiply hota hai, identity se nahi. Yeh woh trap hai jo parent ke mistake box mein flag kiya gaya tha.
Expand karo: $2 ω^{4} - 36 ω^{2} + 108 = 0 \Rightarrow ω^{4} - 18 ω^{2} + 54 = 0$ . $ω^{2}$ mein quadratic: $ω^{2} = \frac{18 \pm 324 - 216}{2} = 9 \pm 27 = 9 \pm 33 .$ Toh $ω^{2} \approx 14.196$ ya $3.804$ , yaani $ω \approx 3.768$ aur $ω \approx 1.950 rad/s$ . Yeh step kyun? Determinant multiply out karne se matrix condition ek ordinary polynomial ban jaati hai; $ω^{2}$ ko ek unknown treat karne se yeh ek plain quadratic ban jaata hai jo hum standard formula se solve kar sakte hain, aur har positive root ek squared mode frequency hai.

Verify: Dono $ω^{2} > 0$ (stable) ✓. Roots ka product $= (9)^{2} - (33)^{2} = 81 - 27 = 54 = det K / det M = 108/2$ ✓. Eigenvectors $(1, \pm 1)$ nahi hain aur orthogonal sirf $M$ ke saathe hain — dekho Eigenvalues and Eigenvectors. (Cell G)

Case H — Real-world word problem (units matter)

Worked example Do coupled pendulum-carts

Do carts, har ek $m = 0.50 kg$ , har ek wall se spring $k = 8.0 N/m$ se anchored aur $k_{c} = 2.0 N/m$ se coupled. Hertz mein do natural frequencies nikalo.

Forecast: Dono kuch Hz? (Haan — light carts, moderate springs.)

Symmetric: $ω_{1} = k / m = 8.0/0.50 = 16 = 4.0 rad/s$ . Yeh step kyun? Woh mode use karo jahan link so rahi ho.
Antisymmetric: $ω_{2} = (k + 2 k_{c}) / m = (8.0 + 4.0) /0.50 = 24 \approx 4.899 rad/s$ . Yeh step kyun? Opposite-way motion coupling spring ko work karta hai, effective stiffness mein $2 k_{c}$ add karta hai.
Hz mein convert karo $f = ω /2 π$ use karke: $f_{1} = 4.0/ (2 π) \approx 0.637 Hz$ , $f_{2} \approx 4.899/ (2 π) \approx 0.780 Hz$ . Yeh step kyun? Real instruments cycles per second padhte hain, rad/s nahi; angular frequency ko $2 π$ (radians per cycle) se divide karna hertz mein convert karta hai.

Verify: Units: $(N/m) / kg = s^{- 1}$ ✓; $2 π$ (rad/cycle) se divide karna Hz deta hai ✓. $f_{2} > f_{1}$ ✓. (Cell H)

Case I — Velocity kick + full-transfer timing

Worked example Ek mass strike karo, phir hand-off time karo

System ( $m = 1$ , $k = 4$ , $k_{c} = 3$ ). Ab kick ek velocity hai, pull nahi: dono masses equilibrium par baithe hain, aur mass 1 ko strike kiya jaata hai taaki $\overset{x}{˙}_{1} (0) = v_{0} = 0.30 m/s$ , jisme $x_{1} (0) = x_{2} (0) = 0$ aur $\overset{x}{˙}_{2} (0) = 0$ . Find: (a) $x_{1} (t)$ ; (b) kitna time lagega jab mass 1 momentarily quiet ho aur mass 2 full swing carry kare.

Forecast: Zero initial displacement ke saath, kya phir bhi beats milte hain? (Haan — velocity kick bhi dono modes excite karta hai; ab sine beats milte hain cosine ki jagah.)

Normal coordinates aur unke velocities $t = 0$ par. $η_{1} (0) = η_{2} (0) = 0$ ; $\overset{η}{˙}_{1} (0) = \frac{1}{2} (\overset{x}{˙}_{1} + \overset{x}{˙}_{2}) = \frac{v _{0}}{2}$ , $\overset{η}{˙}_{2} (0) = \frac{1}{2} (\overset{x}{˙}_{1} - \overset{x}{˙}_{2}) = \frac{v _{0}}{2}$ . Yeh step kyun? Ek SHM jo origin par speed ke saath start hota hai woh ek sine hai, cosine nahi — toh humein positions nahi, initial velocities padhni chahiye har mode ka amplitude set karne ke liye.
Har mode ko sine ke roop mein evolve karo. $\overset{η}{¨}_{a} = - ω_{a}^{2} η_{a}$ ke liye jisme $η_{a} (0) = 0$ , $\overset{η}{˙}_{a} (0) = u$ : $η_{a} (t) = \frac{u}{ω _{a}} sin ω_{a} t$ . Toh $η_{1} = \frac{v _{0}}{2 ω _{1}} sin ω_{1} t$ , $η_{2} = \frac{v _{0}}{2 ω _{2}} sin ω_{2} t$ , jisme $ω_{1} = 2$ , $ω_{2} = 10$ . Yeh step kyun? Sine solution woh hai jo "origin par, move karte hue" match karta hai; $ω_{a}$ se divide karna units correct rakhta hai (ek velocity ek frequency se divide = ek length).
Back-transform karo: $x_{1} = \frac{1}{2} (η_{1} + η_{2}) = \frac{v _{0}}{2} (\frac{s i n ω _{1} t}{ω _{1}} + \frac{s i n ω _{2} t}{ω _{2}})$ . Yeh step kyun? Same reconstruction rule jaise hamesha; do nearly-equal frequencies phir se ek slow beat banate hain, ab sine waves par.
Hand-off time. Is near-beat ka envelope pehli baar $t = \frac{π}{ω _{2} - ω _{1}} = \frac{π}{10 - 2} \approx 2.703 s$ par pinch karta hai. Yeh step kyun? Full energy transfer tab hota hai jab dono modes half a cycle out of step drift kar jaate hain — exactly wahi $\frac{π}{ω _{2} - ω _{1}}$ timing jaise displacement beat mein, kyunki timing sirf frequency difference par depend karti hai, iss par nahi ki tumne kick kaisa lagaya.

Verify: $t = 0$ par: $x_{1} = 0$ ✓ (struck-from-rest position match karta hai) aur $\overset{x}{˙}_{1} (0) = \frac{v _{0}}{2} (cos 0 + cos 0) = v_{0}$ ✓. Transfer time $= π / (10 - 2) \approx 2.703 s$ , aur yeh half beat period $T_{beat} /2$ ke equal hai jisme $T_{beat} = 2 π / (10 - 2) \approx 5.406 s$ ✓. Toh displacement kicks aur velocity kicks same slosh clock share karte hain. (Cell I)

Recall Matrix par quick self-test

Kaun se cell mein $ω_{1} = ω_{2}$ hai? ::: Cell E — zero coupling, modes degenerate hain. Kaun se cell mein $ω_{2} \to \infty$ hota hai jab link tighten karo? ::: Cell F — strong coupling, antisymmetric mode freeze out ho jaata hai. Cell G mein mode shapes $(1, \pm 1)$ kyun NAHI hain? ::: Kyunki unequal masses ise ek generalized eigenproblem banate hain; modes eigenvectors hain jo $M$ ke saathe orthogonal hain, identity ke nahi, toh tidy symmetric/antisymmetric shapes ab nahi chalti. Kya velocity kick phir bhi beats produce karta hai? ::: Haan — Cell I; yeh dono modes ko sines ki tarah excite karta hai, same slosh timing $π / (ω_{2} - ω_{1})$ ke saath sine beats deta hai.

Deep-dive built on Lagrangian Mechanics (the EOMs), Small Oscillations (kyun linearizing legal hai), aur Eigenvalues and Eigenvectors (the modes). Intuitive story ke liye dekho Hinglish companion.