Modal analysis — natural frequencies, mode shapes: Question bank

3.6.10 · D5 · Physics › Spacecraft Structures & Systems Engineering › Modal analysis — natural frequencies, mode shapes

Yeh Modal Analysis ka ek rapid-fire self-test hai. Har line ek Question ::: Answer reveal hai — pehle answer ko cover karo, reasoning ke saath jawab do, phir check karo. Yeh un exact misconceptions ko target karta hai jo yeh topic paida karta hai. Agar koi term unfamiliar lage, toh pehle Free vibration fundamentals aur Multi-degree-of-freedom systems dobara dekho.

True ya false — justify karo

Natural frequency is par depend karti hai ki tune structure ko kitna zor se pluck kiya.

False. Natural frequency sirf

ω_{i} = K_{i}^{*} / M_{i}^{*}

se tay hoti hai — sirf mass aur stiffness. Plucking ki amplitude yeh change karti hai ki response kitna bada hoga, na ki free vibration kitni tezi se oscillate karegi.

Ek mode shape tumhe har point ki actual displacement metres mein batata hai.

False. Ek mode shape

{ϕ_{i}}

ek relative pattern hai — yeh points ke beech motion ka ratio deta hai, absolute size nahi. Isliye hum ise normalize karte hain (jaise pehli entry 1 set karna); real amplitude modal coordinate

q_{i} (t)

se aati hai.

n

-degree-of-freedom structure mein exactly

n

natural frequencies hoti hain.

True. Characteristic polynomial

det ([K] - ω^{2} [M]) = 0

ki degree

ω^{2}

mein

n

hai, isliye yeh

n

eigenvalues

ω_{i}^{2}

aur

n

corresponding mode shapes deta hai.

Do alag mode shapes hamesha ordinary dot-product sense mein perpendicular hote hain.

False. Yeh matrices ke saath weighted orthogonal hote hain:

{ϕ_{j}}^{T} [M] {ϕ_{i}} = 0

aur

{ϕ_{j}}^{T} [K] {ϕ_{i}} = 0

. Plain dot product

{ϕ_{j}}^{T} {ϕ_{i}}

generally zero nahi hota jab tak

[M]

identity ka scalar multiple na ho.

Resonance ka matlab hai ki natural frequency khud time ke saath badhti hai.

False. Natural frequency fixed hoti hai. Resonance ka matlab hai ki tum structure ko us fixed frequency par drive karte ho, isliye amplitude bada ho jaata hai (ideal undamped case mein unbounded); frequency wahi rehti hai.

Stiffness add karna hamesha har natural frequency ko badhata hai.

False (blanket claim ke taur par). Kyunki

ω \propto K^{*} / M^{*}

, zyada stiffness frequencies ko badhane ki tendency rakhti hai, lekin

K^{*}

ek specific mode par projected generalized stiffness hai. Kisi aisa point ko stiffen karna jahan ek given mode barely move karta hai (node ke paas) us mode ki frequency par almost kuch nahi karta.

Mass matrix

[M]

positive definite hoti hai.

True. Har real degree of freedom kisi bhi nonzero velocity ke liye positive kinetic energy carry karta hai, isliye

{v}^{T} [M] {v} > 0

sabhi

{v} \neq = {0}

ke liye — yahi positive definite ki definition hai. Isliye

[M]

ko safely invert kiya ja sakta hai.

Stiffness matrix

[K]

sirf positive semi-definite hai, definite nahi.

Free (unconstrained) structure ke liye True. Rigid-body motion mein koi strain energy store nahi hoti, isliye

{r}^{T} [K] {r} = 0

rigid displacement

{r}

ke liye — ek zero eigenvalue deta hai. Ek fully constrained structure mein

[K]

positive definite hoti hai.

Modal coordinates

{q}

equations ko decouple karte hain kyunki humne luck se clever coordinates choose kiye.

False. Yeh luck nahi hai: mode shapes ki orthogonality force karti hai ki

[Φ]^{T} [M] [Φ]

aur

[Φ]^{T} [K] [Φ]

diagonal hon, jo hi coupled system ko

n

independent single-DOF oscillators mein turn karta hai.

Error dhundho

Ek student eigenvalue problem ko

[K] {ϕ_{i}} = ω_{i}^{2} [K] {ϕ_{i}}

likhta hai.

Right side mein mass matrix honi chahiye:

[K] {ϕ_{i}} = ω_{i}^{2} [M] {ϕ_{i}}

. Dono sides par

[K]

likhne se sabhi modes ke liye

ω_{i}^{2} = 1

force ho jaata, jo bakwaas hai — mass hi inertia term

- ω^{2} M

supply karta hai.

Ek student equation of motion se

sin (ω t + θ)

cancel karta hai aur chinta karta hai ki yeh zero ho sakta hai.

Cancel karna allowed hai kyunki hum sabhi time ke liye valid solution dhundh rahe hain;

sin

almost har instant mein nonzero hota hai, isliye bracketed factor

([K] - ω^{2} [M]) {ϕ}

khud hi zero hona chahiye. Cancel karna legitimate hai.

Ek student conclude karta hai: "

det ([K] - ω^{2} [M]) = 0

ka trivial solution

{ϕ} = {0}

hai, toh yeh ek mode hai."

{ϕ} = {0}

kisi bhi homogeneous system ka solution hai aur koi motion describe nahi karta — ise discard kiya jaata hai. Determinant ko zero set karne ka pura point hi yahi hai ki aise frequencies dhundhe jaayein jahan ek nontrivial

{ϕ}

exist kare.

Ek student ek mode shape ko normalize karta hai aur phir kehta hai ki frequency change ho gayi.

Kisi mode shape ko kisi bhi nonzero constant se scale karna use usi eigenvalue ka eigenvector rehne deta hai. Normalization sirf woh numbers change karta hai jo tum

{ϕ_{i}}

ke liye likhte ho,

ω_{i}

kabhi nahi.

Ek student claim karta hai ki ek single mode ke liye modal force

Q_{i} = [Φ]^{T} {F}

hai.

Ek single mode

i

ke liye yeh

Q_{i} = {ϕ_{i}}^{T} {F}

hai (ek scalar).

[Φ]^{T} {F}

ek saath sabhi modal forces ka full vector hai; mode

i

pick karne ka matlab hai ek row lena.

Ek student kehta hai ki damping orthogonality tod deta hai isliye modal analysis damping ke saath useless hai.

Mode shapes ki orthogonality (

[M]

aur

[K]

se) abhi bhi hold karti hai. Modal analysis valid rehta hai; tumhe sirf automatic diagonal damping matrix tab milti hai jab damping proportional ho (Rayleigh). Dekho Structural damping mechanisms.

Why questions

Trial solution harmonic kyun hona chahiye,

{x} = {ϕ} sin (ω t + θ)

Ek undamped linear system energy conserve karta hai aur spiral in ya out nahi ho sakta, isliye free motion pure sinusoid hai. Is shape ko assume karne se hum

\overset{x}{¨}

- ω^{2} x

se replace kar sakte hain aur ek differential equation ko algebraic eigenvalue problem mein convert kar sakte hain.

Hum directly

{ϕ}

solve karne ki jagah eigenvalue problem kyun lete hain?

Equation

([K] - ω^{2} [M]) {ϕ} = {0}

homogeneous hai — iska right side zero hai. Nonzero

{ϕ}

tabhi exist karta hai jab matrix singular ho, yaani

det = 0

. Woh singularity condition hi eigenvalue problem hai.

Orthogonality har mode ko ek independent single-DOF oscillator ki tarah treat karne kyun deta hai?

Orthogonality

[Φ]^{T} [M] [Φ]

aur

[Φ]^{T} [K] [Φ]

ko diagonalize karti hai, isliye transformed equations mein

q_{i}

aur

q_{j}

ko couple karne wale koi cross-terms nahi hote. Har row padhta hai

M_{i}^{*} \overset{q}{¨}_{i} + K_{i}^{*} q_{i} = Q_{i}

— ek akela spring-mass system.

Spacecraft engineers fundamental frequency ko launch environment ke upar raise karne ki kyun parwah karte hain?

Launch vibration energy low-frequency band mein concentrated hoti hai. Pehli natural frequency ko us band ke upar rakhne ka matlab hai ki launcher koi resonance "pluck" nahi kar sakta, runaway amplitude se bachata hai. Dekho Launch vehicle load environments.

Generalized stiffness

K_{i}^{*}

ek scalar kyun hai jabki

[K]

ek matrix hai?

K_{i}^{*} = {ϕ_{i}}^{T} [K] {ϕ_{i}}

matrix ko ek row aur column vector ke beech sandwich karta hai, ise ek number mein collapse karta hai — woh effective stiffness jo structure feel karta hai jab purely mode

i

mein move karta hai.

Coupling spring ka off-diagonal term

[K]

mein kyun appear karta hai lekin masses

[M]

mein diagonal rehte hain?

Do masses ke beech ek spring dono par ek force exert karta hai jo doosre ki displacement par depend karta hai — yahi cross-influence off-diagonal

- k

hai. Har mass ki inertia sirf apni acceleration par depend karti hai, isliye

[M]

diagonal rehta hai.

Edge cases

Node par mode shape ka kya hota hai?

Node ek aisa point hai jo us mode ke dauran still rehta hai, isliye

{ϕ_{i}}

mein uski entry (near) zero hoti hai. Node par exactly apply kiya gaya forcing us mode ko excite nahi kar sakta, aur wahan stiffening karne se uski frequency barely shift hoti hai.

Agar do natural frequencies equal hon,

ω_{i} = ω_{j}

Orthogonality proof

(ω_{i}^{2} - ω_{j}^{2}) \neq = 0

par depend karta tha

{ϕ_{j}}^{T} [M] {ϕ_{i}} = 0

force karne ke liye. Repeated frequencies ke saath woh step fail hota hai, lekin degenerate subspace ko phir bhi orthogonal choose kiya ja sakta hai — do shapes ka koi bhi combination ek valid mode bhi hai.

Free-floating spacecraft mein zero natural frequency ke corresponding "mode" kya hai?

ω = 0

ek rigid-body mode hai — poora structure bina kisi internal deformation ke translate ya rotate karta hai, zero strain energy store karta hai. Orbit mein ek free spacecraft mein up to six aise modes ho sakte hain (3 translation, 3 rotation).

Agar tum structure ko exactly do natural frequencies ke beech excite karo toh kya respond karta hai?

Koi single mode resonate nahi karta, lekin response nearby modes ka weighted blend hota hai — har mode driving frequency par apne frequency response amplitude ke hisaab se contribute karta hai. Amplitudes modest rehte hain kyunki tum har peak se off ho.

Agar structure nonlinear ho (large deflection) toh modal decoupling ka kya hota hai?

Eigenvalue problem linear

[M]

aur

[K]

assume karta hai. Large deflections ke under stiffness displacement par depend karti hai, mode shapes amplitude ke saath change hote hain, aur clean decoupling kho jaati hai — modal analysis sirf ek local approximation ban jaata hai.

Ek spring ke liye infinite stiffness ki limit mein us connection ka kya hota hai?

Jo do masses ise join karti hain woh effectively rigidly linked ho jaati hain — ek degree of freedom disappear ho jaata hai (unhe saath move karna hoga), DOF count kam ho jaata hai aur high-frequency mode jo us spring ko "stretch" karta tha eliminate ho jaata hai.

Undamped resonance amplitude ke liye kya predict karta hai, aur yeh unphysical kyun hai?

Idealized undamped model

ω = ω_{n}

par amplitude ka bina bound ke badhna predict karta hai. Real structures mein hamesha kuch damping hoti hai, jo peak ko ek bade lekin finite value par cap karti hai — woh finite peak hi hai jiske around design margins banaye jaate hain.

Recall Quick self-check

Ek natural frequency kis cheez se fix hoti hai ::: sirf mass aur stiffness se, $ω = K^{*} / M^{*}$ ke zariye — kabhi bhi amplitude se nahi. Mode shapes kis cheez ke respect mein orthogonal hote hain ::: mass aur stiffness matrices ke, plain dot product ke nahi. $[K]$ ka zero eigenvalue kya signal karta hai ::: koi strain energy ke bina ek rigid-body mode (free/unconstrained structure).