Normal modes — coupled oscillators, normal coordinates: Question bank

2.1.20 · D5 · Physics › Analytical Mechanics › Normal modes — coupled oscillators, normal coordinates

Test karne se pehle dono modes ko picture karo

Is page par sab kuch ek image se connect hota hai: do masses, teen springs, dono taraf walls. Left aur right springs (stiffness $k$ ) har mass ko wall se jodte hain; beech wali coupling spring (stiffness $k_{c}$ ) dono masses ko ek doosre se jodti hai. Displacements $x_{1}, x_{2}$ rest se measure kiye jaate hain.

Figure — Normal modes — coupled oscillators, normal coordinates

Sketch dekho. Top row (symmetric mode): dono arrows ek hi direction mein point karte hain, equal length — masses saath drift karte hain, isliye beech wali spring apni natural length mein rehti hai. Woh kabhi stretch nahi hoti, toh koi force exert nahi karti: masses sirf wall springs feel karte hain, aur frequency hai aaram se $ω_{1} = k / m$ . Bottom row (antisymmetric mode): arrows opposite directions mein hain — beech wali spring sabse zyada stretch aur squeeze hoti hai, ek extra restoring pull add hota hai, toh frequency $ω_{2} = (k + 2 k_{c}) / m$ tak badh jaati hai. Blue dot node mark karta hai center par: antisymmetric mode mein woh point kabhi nahi hilta.

Neeche wala panel dikhata hai ki jab tum coupling change karte ho toh frequencies kaise shift hoti hain — edge-case questions se pehle ise padho.

True ya false — justify karo

Do equal masses, outer springs $k$ , coupling spring $k_{c}$ , jab tak koi line kuch aur na kahe.

Normal mode mein, dono masses apne turning points par ek saath pahunchte hain.

True — woh ek frequency share karte hain aur (real

A_{1}, A_{2}

ke liye) ek phase, toh woh ek saath ruk kar reverse karte hain; sirf amplitudes alag hoti hain. Iska matlab yahi hai "fixed proportion mein move karna".

Symmetric mode ki frequency antisymmetric mode se zyada hoti hai.

False — symmetric slower hai: coupling spring kabhi stretch nahi hoti, toh uski stiffness

k_{c}

absent hai aur

ω_{1}^{2} = k / m

. Antisymmetric us spring se ladhta hai, isliye bada

ω_{2}^{2} = (k + 2 k_{c}) / m

deta hai.

Agar tum system ko kisi arbitrary configuration se release karo, toh resulting motion khud ek normal mode hoti hai.

False — koi bhi start

(x_{1}, x_{2})

η_{1}, η_{2}

mein split hota hai, aur dono generally non-zero hote hain, toh do frequencies ek saath chalti hain (beats deti hain). Yeh tab ek single mode hota hai jab start ek eigenvector ke along ho, doosre

η

ko khatam karte hue.

Normal modes ki sankhya masses (degrees of freedom) ki sankhya ke barabar hoti hai.

True — ek

N

-coordinate small-oscillation system mein exactly

N

normal modes hote hain, kyunki secular equation

det (K - ω^{2} M) = 0

ω^{2}

mein degree

N

ka polynomial hai.

Coupling spring

k_{c}

ko strong banana dono normal-mode frequencies ko change karta hai.

False — sirf

ω_{2}^{2} = (k + 2 k_{c}) / m

k_{c}

par depend karta hai; symmetric

ω_{1}^{2} = k / m

untouched rehta hai kyunki woh mode coupling spring ko kabhi stretch nahi karta, chahe woh kitni bhi stiff ho.

Antisymmetric mode mein masses ke beech ka center point bilkul still rehta hai.

True —

x_{2} = - x_{1}

ke saath coupling spring ka midpoint ek node hai; woh na hilta hai na accelerate karta hai, isliye tum imagine kar sakte ho ki spring ko wahan se kaat rahe ho.

Agar do masses unequal hote, tum phir bhi equations ko decouple karne ke liye sirf add aur subtract kar sakte.

False — neat

\frac{1}{2} (x_{1} \pm x_{2})

combinations sirf equal-mass case mein diagonalise karte hain. Unequal masses ke liye eigenvectors tilt ho jaate hain, aur tumhe generalized problem

KA = ω^{2} MA

solve karna hoga.

Coupled system ki total mechanical energy constant rehti hai chahe energy dono masses ke beech slosh kare.

True — no damping ka matlab total

E

conserved hai; "sloshing" mass 1 aur mass 2 ke beech ek transfer hai, koi loss nahi. Har normal coordinate apni energy alag se conserve karti hai bhi.

Error dhundho

Har line mein ek plausible-sounding lekin galat statement hai. Reveal correction deta hai.

"Kyunki

k_{c}

V

mein ek baar appear hota hai, antisymmetric frequency

ω_{2}^{2} = (k + k_{c}) / m

hai."

Antisymmetric stretch

(x_{1} - x_{2}) = 2 x_{1}

hai, toh coupling spring ki effective stiffness double ho jaati hai:

ω_{2}^{2} = (k + 2 k_{c}) / m

. Hamesha secular determinant se read karo, kabhi

V

ko eyeball mat karo.

"Modes find karne ke liye main

KA = λ A

solve karta hoon, ek ordinary eigenvalue problem."

Yeh generalized eigenvalue problem

KA = ω^{2} MA

hai. Yahan sirf isliye ordinary lagta hai kyunki

M = m 1

; alag masses ke saath

M

ko drop nahi kar sakte. Dekho Eigenvalues and Eigenvectors.

"Har mass apni natural frequency rakhta hai; mass 1 apni rate par swing karta hai, mass 2 apni rate par."

Coupling har mode ke andar ek shared frequency force karta hai. Ek mass ki actual motion dono shared-frequency modes ka sum hai — koi ek frequency nahi hai jo tum ek mass se pin kar sako.

"Eigenvector tumhe dono masses ke displacements har time par batata hai."

Woh sirf unka fixed ratio

A_{2} / A_{1}

batata hai; overall size aur phase initial conditions se set hoti hai. Eigenvector coordinate space mein ek direction hai, puri trajectory nahi.

"Normal coordinates ke units length hone chahiye, kyunki

η = \frac{1}{2} (x_{1} + x_{2})

M

-orthonormality ke liye choose kiye gaye

\frac{1}{2}

ke saath

η_{a}

generally

mass \times length

ke mixed units carry karte hain; sirf (yahan dimensionless) normalization tak woh lengths jaise lagte hain. Jo matter karta hai woh yeh hai ki har ek

\overset{η}{¨}_{a} = - ω_{a}^{2} η_{a}

obey karta hai.

"Kyunki system coupled hai, iske equations of motion kabhi independent oscillators ke roop mein nahi likhe ja sakte."

Normal coordinates ka poora point yahi hai ki ja sakte hain — coupling chosen coordinates

x_{1}, x_{2}

ki feature hai, physics ki nahi.

η_{a}

par switch karo aur coupling gaayab ho jaati hai.

"Beats tab hote hain jab teen ya zyada masses interfere karte hain."

Beats sirf do frequencies

ω_{1}, ω_{2}

se arise hote hain jo close hain; do coupled masses already yeh karte hain. Weak coupling

ω_{1} \approx ω_{2}

banata hai, ek slow beat envelope deta hai.

Why questions

Hum coupled equations directly solve karne ki jagah normal coordinates introduce karne ki taklif kyun lete hain?

Woh coupled ODEs ke ek ulajhe system ko

N

alag one-line SHM equations

\overset{η}{¨}_{a} = - ω_{a}^{2} η_{a}

mein badal dete hain, har ek trivially solvable. Poore method ka payoff yahi hai — dekho Small Oscillations.

Secular equation

det (K - ω^{2} M) = 0

par kisi aur value par nahi kyun set hoti hai?

Ek matrix ek nonzero vector ko zero tab hi deta hai jab matrix singular ho, yani uska determinant vanish ho. Hum non-trivial motion (

A \neq = 0

) chahte hain, toh hum singularity maangte hain.

Symmetry koi algebra kiye bina pehle kyun bata sakti hai ki symmetric mode slower hai?

Symmetric motion mein coupling spring kabhi stretch nahi hoti, toh woh koi restoring force contribute nahi karti; kam restoring force ka matlab kam frequency. Symmetry determinant chhue bina physics pin kar deti hai.

Hum sabhi masses ke liye ek single

ω

ke saath

x_{j} = A_{j} cos (ω t + ϕ)

guess kyun karte hain?

Yeh ek deliberate forecast hai ki normal modes exist karte hain: hum poochhte hain ki kya aisi motions hain jahan sab kuch ek frequency share kare. Substitute karna calculus (derivatives) ko algebra (ek matrix equation) mein convert karta hai, jise hum phir verify kar sakte hain ke solutions hain.

Lagrangian route force-balancing ke sign errors kyun avoid karta hai?

Energies

T

aur

V

scalars hain jinmein ulta hone ke liye koi direction nahi; Euler-Lagrange derivative phir chain rule ke zariye automatically correct signs produce karta hai. Dekho Lagrangian Mechanics.

Sirf ek eigenvector ko initial condition ke roop mein excite karna hamesha ke liye single-frequency motion kyun deta hai?

Ek eigenvector ek normal coordinate ko zero set karta hai aur ise zero rakhta hai (koi force ise feed nahi karta), toh sirf doosra

η_{a}

oscillate karta hai — ek pure single-frequency mode chhodta hai.

Normal modes crystal physics ka bridge kyun hain?

Ek crystal sirf ek bada coupled oscillators ka chain hai; iske normal modes woh collective vibrations ban jaate hain jinhe hum phonons mein quantise karte hain. Do-mass problem us poore subject ka seed hai.

Edge cases

Do frequencies ka kya hota hai jab coupling

k_{c} \to 0

Dono

k / m

approach karte hain:

ω_{1}^{2} = k / m

aur

ω_{2}^{2} \to k / m

. Masses independent oscillators ban jaate hain aur koi bhi "beats" infinitely lambe, unobservable exchange tak slow ho jaate hain — frequency panel ke left edge dekho.

Kya hota hai jab

k_{c} \to \infty

(masses ke beech ek rigid rod)?

ω_{2}^{2} = (k + 2 k_{c}) / m \to \infty

: antisymmetric mode infinite energy cost karta hai aur frozen out ho jaata hai, toh pair sirf symmetrically ek rigid block ke roop mein

ω_{1} = k / m

par move karta hai — panel mein upper curve shoot off karta hai jabki lower flat rehta hai.

Agar tum

x_{1} (0) = x_{2} (0) = a

aur zero velocity se start karo, toh kya motion hogi?

Pure symmetric mode:

η_{2} (0) = 0

hamesha zero rehta hai, toh dono masses

ω_{1} = k / m

par saath oscillate karte hain coupling spring untouched ke saath — no beats, no energy transfer.

Agar tum

x_{1} (0) = a, x_{2} (0) = 0

se release karo, toh kya energy kabhi poori tarah mass 2 tak pahunchi hai?

Haan — dono modes mein equal amplitudes ke saath beat envelope

x_{1}

ko zero tak drive karta hai jabki

x_{2}

a

reach karta hai, phir wapas. Full transfer ke liye equal mode amplitudes chahiye, jo yeh symmetric-vs-antisymmetric mix provide karta hai.

Kya hoga agar ek normal frequency

ω = 0

aaye (ek zero eigenvalue)?

Woh ek zero mode hai — oscillation nahi balki puri system ka free translation/rotation bina kisi restoring force ke. Yeh tab appear hota hai jab ek symmetry (jaise no wall springs) ek direction chhodti hai bina kisi potential cost ke.

Kya do alag normal modes kabhi ek hi frequency share kar sakte hain?

Haan — yeh degeneracy hai, symmetric systems mein common (jaise ek 2D lattice). Tab degenerate eigenvectors ka koi bhi linear combination bhi ek valid mode hai, toh mode shapes uniquely fixed nahi hote.

Kya hoga agar masses unequal hों (

m_{1} \neq = m_{2}

)? Kya

η = \frac{1}{2} (x_{1} \pm x_{2})

phir bhi unhe decouple karta hai?

Nahi — eigenvectors ab simple sum/difference nahi hote, aur orthogonality

M

ke respect se hold karta hai (

A_{a}^{T} M A_{b} = 0

), identity ke nahi. Tumhe generalized problem honestly solve karna hoga.

Recall One-line self-test

Woh single sentence naam batao jo "har mass ki apni frequency hoti hai" myth ko khatam karta hai. ::: Kisi bhi normal mode mein sab parts ek common $ω$ share karte hain; ek mass ki real motion in shared-frequency modes ka ek superposition hai, kabhi khud ki ek frequency nahi.