Visual walkthrough — Normal modes — coupled oscillators, normal coordinates

2.1.20 · D2 · Physics › Analytical Mechanics › Normal modes — coupled oscillators, normal coordinates

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Step 1 — Machine draw karo aur har arrow ko naam do

KYA. Do identical blocks, har ek ka mass $m$, ek frictionless line par hain. Ek wall spring baayein, ek wall spring daayein, aur ek middle spring jo dono blocks ko jodata hai. Left aur right springs ki stiffness $k$ hai; middle ki stiffness $k_c$ hai (c matlab "coupling").

KYUN. Kisi bhi algebra se pehle hume fix karna hai ki har symbol ka ek distance ke roop mein kya matlab hai. Agar yeh skip kiya, toh baad ka har sign ek coin-flip hoga. Toh hum mark karte hain $x_1$ = block 1 apni resting spot se kitna khisaka, $x_2$ = block 2 kitna khisaka, dono rightward-positive measure kiye hue.

PICTURE. Figure mein blue block, block 1 hai, orange block, block 2 hai. Green arrows displacements $x_1,x_2$ hain dashed equilibrium lines se.

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Step 2 — Har spring kitna stretch hua?

KYA. Spring ki stored energy uske stretch par depend karti hai — woh apni relaxed length se kitni zyada (ya kam) ho gayi. Hum picture se teen stretches padhte hain:

• Left spring stretch $=x_1$ (block 1 wall se door kheench raha hai).
• Right spring stretch $=x_2$.
• Middle spring stretch $=x_2-x_1$.

Middle wala $x_2-x_1$ kyun? Middle spring ke dono sir dono blocks hain. Uski nayi length badlti hai (right end kitna khisaka) minus (left end kitna khisaka) $=x_2-x_1$ se. Agar dono blocks same direction mein same amount se khisken, toh $x_2-x_1=0$: middle spring relaxed hai. Yeh baat yaad rakho — yahi slow mode ka poora raaz hai.

PICTURE. Red bracket $x_2-x_1$ highlight karta hai. Dekho: jab arrows same direction mein hain aur equal length ke, bracket zero ho jaata hai.

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Step 3 — Total energy likho, phir machine ko apna motion choose karne do

KYA. Moving energy (kinetic) aur stored spring energy (potential) add karo.

$T=\underset{\text{block 1 moving}}{\underbrace{\frac{1}{2}m{\dot{x}}_1^2}}+\underset{\text{block 2 moving}}{\underbrace{\frac{1}{2}m{\dot{x}}_2^2}}$

Yahan ${\dot{x}}_1$ matlab hai "$x_1$ ki speed" — woh distance per second kitni tezi se change ho rahi hai. (Dot Newton ka shorthand hai rate of change ke liye; ek dot = speed, do dots = acceleration.)

$V=\underset{\text{left spring}}{\underbrace{\frac{1}{2}k{x}_1^2}}+\underset{\text{right spring}}{\underbrace{\frac{1}{2}k{x}_2^2}}+\underset{\text{middle spring}}{\underbrace{\frac{1}{2}{k}_c(x_2-x_1{)}^2}}$

Forces ki jagah energy kyun? Forces ke liye tumhe track karna padta hai ki har spring kis direction mein push karta hai — pressure mein sign flip karna easy ho jaata hai. Energy sirf ek positive scalar hai; Lagrangian Mechanics ki machinery phir sahi forces automatically nikaalti hai, signs ke saath. Hum yeh rule use karte hain: $\frac{d}{dt}\frac{\partial L}{\partial {\dot{x}}_i}=\frac{\partial L}{\partial x_i},L=T-V.$ Ise padhte hain: "momentum ka rate of change $=$ force," jahan force hai $-\partial V/\partial x_i$.

Crank ghumaane par (V differentiate karo, chain rule $(x_2-x_1{)}^2$ term ko kaata hai):

$m{\ddot{x}}_1=\underset{\text{wall wapas kheenchti hai}}{\underbrace{-k{x}_1}}+\underset{\text{middle spring ka khinchaav}}{\underbrace{k_c(x_2-x_1)}}$ $m{\ddot{x}}_2=\underset{\text{wall wapas kheenchti hai}}{\underbrace{-k{x}_2}}-\underset{\text{wahi spring, doosra end}}{\underbrace{k_c(x_2-x_1)}}$

PICTURE. Middle-spring term dono lines mein opposite signs ke saath aata hai — yahi coupling hai jo ek single red spring ke roop mein dono blocks ko opposite directions mein kheechthi hai.

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Step 4 — Bold guess: sab log ek hi frequency par vibrate karte hain

KYA. Hum guess karte hain ki ek solution hai jahan dono blocks simple harmonic motion karte hain same frequency $\omega$ par, sirf amplitude mein alag: $x_j(t)=A_j\cos (\omega t+\varphi ).$

• $A_j$ = block $j$ ka amplitude (max swing) — ek number, possibly negative.
• $\omega$ = angular frequency, dono blocks mein shared.
• $\varphi$ = ek common phase (cycle mein hum kahan se shuru karte hain).

Yeh shape kyun? Pure cosine Simple Harmonic Motion ki pehchaan hai: acceleration displacement ke proportional. Do baar differentiate karne par ek sundar fact milta hai: ${\ddot{x}}_j=-{\omega }^2{x}_j,$ toh acceleration sirf $-{\omega }^2$ times position hai. Yeh messy time-differential equations ko numbers $A_1,A_2$ mein seedha algebra mein badal deta hai. Hum bet laga rahe hain ki aisi special synchronized motions exist karti hain; hum unhe dhundh kar verify karenge.

PICTURE. Do cosine waves same frequency ke; sirf unki heights (aur possibly unka upar/neeche flip) alag hain.

${\ddot{x}}_j=-{\omega }^2{x}_j$ substitute karne aur common cosine cancel karne par: $(k+k_c-m{\omega }^2)A_1-k_c{A}_2=0$ $-k_c{A}_1+(k+k_c-m{\omega }^2)A_2=0$

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Step 5 — Nonzero motion kab survive karta hai? Secular equation

KYA. Dono equations ko ek matrix ke roop mein likho jo ek vector ko zero kar deta hai: $\underset{\mathbf{K}-{\omega }^2\mathbf{M}}{\underbrace{\left(\begin{array}{cc}k+k_c-m{\omega }^2& -k_c\\ -k_c& k+k_c-m{\omega }^2\end{array}\right)}}\left(\begin{array}{c}{A}_1\\ A_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right).$

Determinant kyun? Ek matrix space ko ek factor se flatten karta hai jise uska determinant kehte hain. Agar woh factor zero nahi hai, toh woh sirf zero vector ko hi origin par bhej sakta hai — matlab $A_1=A_2=0$, koi motion nahi, boring. Ek real vibration paane ke liye hume matrix ko space flat crush karna hoga (determinant $=0$) taaki nonzero vectors ki ek poori line zero ho jaye. Woh surviving line hamaara mode hai. Yahi eigenvalue idea hai: ${\omega }^2$ ek eigenvalue hai, $(A_1,A_2)$ uska eigenvector.

PICTURE. Ek unit square ek line par squash ho raha hai — area zero — jis waqt ${\omega }^2$ ek special value par pahunchta hai.

Determinant zero set karo: $(k+k_c-m{\omega }^2{)}^2-k_c^2=0⟹k+k_c-m{\omega }^2=\pm k_c.$

Do sign choices se do frequencies milti hain: $\boxed{{\omega }_1^2=\frac{k}{m}}\boxed{{\omega }_2^2=\frac{k+2k_c}{m}}.$

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Step 6 — Har frequency actually kaisi dikhti hai (do mode shapes)

KYA. Har ${\omega }^2$ wapas daalo taaki ratio $A_2/A_1$ — us vibration ki shape — pata chale.

Mode 1, ${\omega }_1^2=k/m$: bracket $+k_c$ ke barabar hai, toh $k_c{A}_1-k_c{A}_2=0\Rightarrow A_2=A_1$. Blocks saath move karte hain.

Mode 2, ${\omega }_2^2=(k+2k_c)/m$: bracket $-k_c$ ke barabar hai, toh $A_2=-A_1$. Blocks opposite move karte hain.

Slow/fast split kyun? Mode 1 mein middle spring ka stretch $x_2-x_1=0$ hai — woh sota hai, koi extra force contribute nahi karta, toh frequency sirf lone-block value $\sqrt{k/m}$ hai. Mode 2 mein middle spring sabse zyada kaam karta hai (stretch $=2x_1$), extra stiffness $2k_c$ jaisi acting karta hai, frequency ko upar push karta hai.

PICTURE. Upar: symmetric, arrows parallel, middle spring relaxed (gray/limp drawn). Neeche: antisymmetric, arrows opposed, middle spring taut stretched (red).

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Step 7 — Naye coordinates jo equations ko ausaana kar dete hain (acche se)

KYA. Do combined variables invent karo: ${\eta }_1=\frac{1}{\sqrt{2}}(x_1+x_2),{\eta }_2=\frac{1}{\sqrt{2}}(x_1-x_2).$ ${\eta }_1$ = "dono kitna saath move karte hain," ${\eta }_2$ = "woh kitna alag move karte hain." ($\frac{1}{\sqrt{2}}$ sirf neat bookkeeping hai taaki energy clean dikhe.)

Yeh do kyun? Yeh exactly Step 6 ke mode shapes hain. Dono equations of motion add karo aur middle-spring terms cancel ho jaati hain; subtract karo aur middle-spring terms double ho jaati hain: ${\ddot{\eta }}_1=-{\omega }_1^2{\eta }_1,{\ddot{\eta }}_2=-{\omega }_2^2{\eta }_2.$ Har ek single simple harmonic oscillator hai — doosre se completely independent. Tangled 2-body problem do alag 1D springs ban gaya hai.

PICTURE. $x_1$–$x_2$ plane 45° rotate hoke ${\eta }_1$–${\eta }_2$ axes par; har naye axis ke along motion ek akela clean oscillation hai.

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Step 8 — Edge aur degenerate cases (reader ko kabhi stranded mat chodo)

KYA & KYUN — teen limits, har ek ek sanity check.

1. Koi coupling nahi, $k_c\to 0$. Tab ${\omega }_2^2\to k/m={\omega }_1^2$: dono frequencies merge ho jaati hain. Physically middle spring gayab ho jaata hai aur blocks do identical independent oscillators ban jaate hain — koi bhi motion allowed hai, mode split khatam ho jaata hai. Yeh degenerate case hi wajah hai ki weak coupling near-equal frequencies deta hai (Step 9 ke beats).

2. Rigid coupling, $k_c\to \infty$. ${\omega }_2^2\to \infty$: blocks ko alag move karne mein infinite energy lagti hai, toh Mode 2 freeze out ho jaata hai. Blocks ek saath lock ho jaate hain aur sirf slow ${\omega }_1$ survive karta hai — do wall springs par ek glued mass.

3. Exactly ek eigenvector se shuru karo. Dono ko $x_1(0)=x_2(0)=a$ se zero speed ke saath release karo: tab ${\eta }_2(0)=0$ aur hamesha ke liye zero rehta hai. ${\omega }_1$ par pure single-frequency motion, coupling spring kabhi nahi chua. Eigenvector ko apna starting point banana sirf ek mode excite karta hai.

PICTURE. Do frequencies ${\omega }_1,{\omega }_2$ coupling $k_c$ ke against plot ki gayi hain: $k_c=0$ par touch karti hain, aur $k_c$ badhne ke saath ${\omega }_2$ upar chadh jaati hai jabki ${\omega }_1$ flat rehti hai.

Recall Limits par khud check karo

$k_c\to 0$ hone par dono normal-mode frequencies ka kya hota hai? ::: Woh equal ho jaati hain, ${\omega }_2\to {\omega }_1=\sqrt{k/m}$ — mode split gayab ho jaata hai (degenerate). $k_c\to \infty$ hone par Mode 2 ka kya hota hai? ::: Uski frequency $\to \infty$ ho jaati hai; woh motion freeze ho jaata hai, blocks lock ho jaate hain aur sirf slow mode rehta hai.

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Step 9 — Koi bhi real motion dono ka sum hai: beats

KYA. Block 1 ko aside kheecho, block 2 rest par ($x_1(0)=a$, $x_2(0)=0$, koi initial speed nahi). Normal coordinates mein ${\eta }_1(0)={\eta }_2(0)=a/\sqrt{2}$, phases zero. Wapas transform karne par: $x_1(t)=\underset{\text{slow envelope}}{\underbrace{a\cos \left(\frac{{\omega }_2-{\omega }_1}{2}t\right)}}\underset{\text{fast wiggle}}{\underbrace{\cos \left(\frac{{\omega }_1+{\omega }_2}{2}t\right)}}.$

Yeh breathing jaisa kyun lagta hai? Do nearly-equal cosines add hone par ek fast oscillation ban jaati hai ek slowly pulsing envelope ke andar — beats (dekho Beats and Superposition). Energy poori tarah block 2 mein jaati hai, phir block 1 mein wapas, baar baar. Weak coupling ⇒ closer frequencies ⇒ slower sloshing.

PICTURE. Fast blue oscillation ek slow orange envelope mein hai; block 2 (green) bharta hai jab block 1 khaali hota hai.

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Ek-picture summary

Ek nazar, poori kahani: do springs + do blocks → energy → single-frequency guess → determinant $=0$ → do frequencies → do mode shapes (slow same / fast fight) → normal coordinates sab kuch decouple karte hain → real motion = unka superposition (beats).

Recall Feynman retelling — poora walkthrough seedhe shabdon mein

Do carts jo springs se thami hain, ek rubber band unhe baandh raha hai. Pehle maine sirf label kiya ki har cart kitna khiski aur ek cheez notice ki: rubber band sirf khiskaav ke difference se stretch hoti hai. Maine saari bounce-energy add ki aur energy rule ko anumati di ki woh bataye ki carts ek doosre ko kaise push karti hain — aur wahan tha, rubber band unhe opposite tarike se dhakelta hua. Phir maine ek wild bet lagai: shayad kuch khaas tarike hain jismein dono carts ek shared speed par wiggle karti hain. Woh bet plug karne se calculus arithmetic ban gaya, aur ek "kab kuch survive karta hai?" sawaal — determinant — exactly do speeds ugal deta hai. Aur gehrayi se dekhne par, ek speed carts ka saath chalana hai (rubber band sota hai, toh slow) aur ek unka alag chalana hai (rubber band fight karta hai, toh fast). Aakhir mein maine apne variables ko "saath" aur "alag" mein rename kiya, aur equations jaadui tarike se do alag simple springs mein split ho gayi. Toh koi bhi messy motion jo main shuru kar sakta hoon woh sirf do simple wiggles add ki gayi hai — aur agar frequencies close hain, tum literally dekhte ho energy ek cart se doosri mein aur wapas jaati hai: beats.

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See also: Eigenvalues and Eigenvectors · Small Oscillations · Phonons and Lattice Vibrations · Hinglish version