Modal analysis — natural frequencies, mode shapes: Visual walkthrough

3.6.10 · D2 · Physics › Spacecraft Structures & Systems Engineering › Modal analysis — natural frequencies, mode shapes

Yeh page parent topic ka central result rebuild karta hai — woh eigenvalue problem jo humein ek structure ki natural frequencies aur mode shapes deta hai — sirf do masses aur springs se shuru karke, aur kuch nahi. Har symbol pehle earn kiya jaata hai, phir use hota hai. Har figure mein red aur amber arrows follow karo; pictures hi argument carry karti hain.

Step 1 — Do masses, do springs: poora universe

KYA HAI. Hum sabse simple structure banate hain jo phir bhi interesting vibration dikha sake: ek wall, ek spring, ek mass, ek aur spring, ek aur mass.

KYUN. Ek mass ek spring par ek hi boring tarike se bounce karta hai. Ek doosra mass add karo aur dono ya toh saath move kar sakte hain ya ek dusre se lad sakte hain — yeh "saath vs. against" har rocket mein har mode shape ka beej hai. Hum yahan se shuru karte hain kyunki yeh sabse chhota system hai jiske paas ek se zyada answers hain.

PICTURE. Neeche, $m_{1}$ aur $m_{2}$ do masses hain. Symbols $x_{1}$ aur $x_{2}$ sirf arrows hain jo measure karte hain ki har mass apni resting spot se kitna right side khisak gaya hai. Positive matlab right, negative matlab left. Bas itna hi $x$ ka matlab hai.

Yeh kisi bhi structure ka discretised skeleton hai — dekho Multi-degree-of-freedom systems ki real spacecraft kaise Finite element analysis (FEA) ke zariye hazaaron aisa blocks ban jaata hai.

Step 2 — Har mass par Newton's law → matrix equation

KYA HAI. Hum har block ke liye $F = ma$ likhte hain, uspar pull karne wale har spring ko count karte hue, phir dono lines ko ek saath ek numbers ke tidy box mein stack karte hain.

KYUN. Spring ki force $- k \times (uski stretch)$ hoti hai — minus sign yeh kehta hai ki yeh hamesha rest ki taraf wapas kheenchti hai. Mass 1 ko do springs feel hoti hain (wall spring $x_{1}$ se stretch hua, aur middle spring $x_{1} - x_{2}$ se stretch hua). Mass 2 ko sirf middle spring feel hoti hai. Dono likhna aur unhe stack karna hamaari jodi ko ek object ki tarah treat karne deta hai.

PICTURE. Figure mein har spring ki stretch ek coloured segment ke roop mein hai aur woh force arrow jo woh har mass par produce karta hai.

Mass 1: wall spring $x_{1}$ se stretch hua hai; middle spring $(x_{1} - x_{2})$ se stretch hua hai.

m accel of m_{1} \overset{x}{¨}_{1} = wall spring pulls back - k x_{1} middle spring pulls back - k (x_{1} - x_{2}) = - 2 k x_{1} + k x_{2}

Mass 2: sirf middle spring act karti hai.

m \overset{x}{¨}_{2} = middle spring - k (x_{2} - x_{1}) = k x_{1} - k x_{2}

Ab inhe stack karo. Hum numbers ke do grids invent karte hain taaki do equations ek line mein fit ho jayein:

[M] mass matrix [m 0 0 m] {\overset{x}{¨}} [\overset{x}{¨}_{1} \overset{x}{¨}_{2}] + [K] stiffness matrix [2 k - k - k k] {x} [x_{1} x_{2}] = [00]

Step 3 — Motion guess karo: sab kuch ek hi rhythm par pulse karta hai

KYA HAI. Hum guess karte hain ki ek natural mode mein, har mass time mein same sine wave par ride karta hai, sirf kitna bada uska swing hai usme farq hota hai.

KYUN. Bina damping ke, ek spring–mass system sinusoidally bajta hai — woh ek aisi motion hai jiska acceleration seedha zero ki taraf point karta hai, jo exactly wahi hai jo springs demand karte hain. Toh hum aisi solution try karte hain jahan shape (kaun kitna move karta hai) frozen ho aur sirf amplitude andar-bahar saans leta ho.

PICTURE. Do masses, ek shared clock. Lamba amber curve aur chhota cyan curve same period share karte hain; sirf unki heights alag hain. Frozen height-ratio hi mode shape hai.

{x (t)} = frozen shape: kaun kitna move karta hai {ϕ} shared clock sin (ω t + θ)

${ϕ}$ (kaho "phi") — mode shape: relative amplitudes ki list, jaise $[1 1.6]$ ka matlab hai "jab mass 1 1 move kare, mass 2 1.6 move karta hai." Iske paas na units hain na time — pure pattern.
$ω$ (kaho "omega") — angular frequency rad/s mein: shared clock kitni tez tick karta hai. Yahi woh natural frequency hai jo hum dhundh rahe hain.
$θ$ — ek phase offset (sine kahan se start hoti hai); yeh cancel ho jaayega aur kabhi matter nahi karega.

Kyunki $sin$ curves apne aap par wapas curve karti hain, time mein do baar differentiate karne se same shape milti hai times $- ω^{2}$ :

{\overset{x}{¨}} = - ω^{2} {ϕ} sin (ω t + θ)

$- ω^{2}$ sinusoidal motion ka fingerprint hai: fast pulsing (bada $ω$ ) matlab fierce acceleration.

Step 4 — Substitute karo aur clock cancel karo → eigenvalue problem

KYA HAI. Guess ko Step 2 ki equation mein plug karo aur us sine ko cancel karo jo har term share karta hai.

KYUN. Sine wave $sin (ω t + θ)$ har term ko identically multiply karta hai, aur yeh almost har instant par nonzero hota hai, toh hum ise divide kar sakte hain. Jo bachta hai woh shapes aur frequencies ke baare mein ek pure statement hai — time bilkul squeeze out ho gaya.

PICTURE. Clock $sin (ω t + θ)$ dono sides se common factor ki tarah lift off ho jaata hai, ek timeless equation chhodh ke shape ${ϕ}$ ke baare mein.

${\overset{x}{¨}} = - ω^{2} {ϕ} sin (\cdot)$ aur ${x} = {ϕ} sin (\cdot)$ substitute karke:

[M] (- ω^{2} {ϕ}) sin (\cdot) + [K] {ϕ} sin (\cdot) = {0}

Shared $sin (\cdot)$ divide out karo:

([K] - ω^{2} [M]) {ϕ} = {0}

Determinant kyun vanish hona chahiye? Hum boring answer ${ϕ} = {0}$ nahi chahte (kuch bhi move nahi). Ek nonzero shape tab hi matrix $([K] - ω^{2} [M])$ se multiply hokar zero mein survive kar sakti hai jab woh matrix squashing ho — mathematically, jab uska determinant zero ho:

det ([K] - ω^{2} [M]) = 0

Step 5 — Determinant ko polynomial mein badlo aur solve karo

KYA HAI. Hamare 2×2 numbers ke liye woh determinant compute karo, $ω^{2}$ mein ek polynomial lo, aur use solve karo.

KYUN. Determinant condition woh single equation hai jo filter karti hai ki kaun se $ω$ allowed hain. Expand karo, toh yeh variable $ω^{2}$ mein ek ordinary quadratic hai — aur quadratics hum hamesha quadratic formula se crack kar sakte hain.

PICTURE. $ω^{2}$ mein parabola; uske zero ke do crossings hi do allowed $ω^{2}$ hain.

det [2 k - m ω^{2} - k - k k - m ω^{2}] = (2 k - m ω^{2}) (k - m ω^{2}) - (- k) (- k) = 0

Term by term expand karke:

m^{2} ω^{4} - 3 k m ω^{2} + k^{2} = 0

$ω^{2}$ ko ek unknown treat karo; quadratic formula deta hai

ω^{2} = \frac{3 k \pm 9 k ^{2} - 4 k ^{2}}{2 m} = \frac{( 3 \pm 5 ) k}{2 m}

ω_{1} \approx 0.618 k / m, ω_{2} \approx 1.618 k / m

Do masses → do natural frequencies. (Generally $n$ masses $n$ frequencies dete hain.)

Step 6 — Do shapes recover karo

KYA HAI. Har $ω_{i}^{2}$ ko wapas $([K] - ω_{i}^{2} [M]) {ϕ} = {0}$ mein daalo aur amplitude ratio padhho.

KYUN. Har allowed frequency matrix ko exactly ek direction mein squash karti hai — woh surviving direction hi mode shape hai. Hum $ϕ_{1} = 1$ fix karte hain aur equation ko $ϕ_{2}$ batane dete hain, kyunki sirf ratio physical hai (Step 3 ne shape freeze kiya, size nahi).

PICTURE. Mode 1 mein dono masses same taraf swing karte hain (amber, in-phase); Mode 2 mein woh opposite swing karte hain (cyan, out-of-phase) unke beech ek still node ke saath.

Low frequency, $ω_{1}$ : pehli row $2 k ϕ_{11} - k ϕ_{21} = \frac{( 3 - 5 ) k}{2} ϕ_{11}$ deta hai

{ϕ_{1}} = [1 1.618] (in-phase; mass 2 zyada swing karta hai)

High frequency, $ω_{2}$ :

{ϕ_{2}} = [1 - 0.618] (out-of-phase; unke beech ek stationary node)

Step 7 — Shapes orthogonal hain (payoff)

KYA HAI. Do mode shapes ${ϕ_{1}}^{T} [M] {ϕ_{2}} = 0$ aur ${ϕ_{1}}^{T} [K] {ϕ_{2}} = 0$ obey karte hain — ek matrix-weighted right-angle.

KYUN. Yahi cheez modes ko useful banati hai: yeh independent building blocks hain. Structure ki koi bhi wobble in dono ka sum hai, aur — kyunki yeh orthogonal hain — har ek apne alag single-mass oscillator ki tarah evolve karta hai. Isi tarah ek monstrous coupled system simple pieces mein pighal jaata hai jinhein hum ek ek karke solve kar sakte hain. Poora proof aur consequences: Orthogonality of mode shapes.

PICTURE. $M$ -weighted picture mein do shape vectors ek clean right angle par baithe hain; structure ki motion ko har axis par project karna padhtha hai "us mode ka kitna hissa present hai."

$[Φ] = [{ϕ_{1}} {ϕ_{2}}]$ aur ${x} = [Φ] {q}$ use karke, equation of motion ko $[Φ]^{T}$ se pre-multiply karna coupled matrices ko diagonal bana deta hai:

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

$q_{i}$ — modal coordinate: abhi mode $i$ kitna strongly switched on hai.
$M_{i}^{*}, K_{i}^{*}$ — mode ki apni effective mass aur stiffness, jisme $ω_{i} = K_{i}^{*} / M_{i}^{*}$ .
$Q_{i} = {ϕ_{i}}^{T} {F}$ — ek external force (jaise launch shake) us mode ko kitna hard push karta hai.

Khatra, precisely bataaya gaya: agar ek launch load mein $ω_{i}$ par energy ho, toh woh ek decoupled oscillator resonate karta hai. Modal analysis har $ω_{i}$ pin down karta hai taaki designers use dodge ya damp kar sakein — dekho Structural damping mechanisms aur Frequency response functions.

Ek-picture summary

Ek wall, do springs aur do masses se, humne Newton's law likha (arrows), ek shared sinusoidal rhythm guess kiya, clock cancel karke $[K] {ϕ} = ω^{2} [M] {ϕ}$ tak pahunche, do frequencies ke liye ek quadratic solve kiya, do orthogonal shapes padhhe, aur poori cheez ko independent oscillators mein decouple kar diya.

Recall Feynman retelling — ek story ki tarah batao

Hamare paas do blocks hain springs par. Har block sirf "force equals mass times acceleration" follow karta hai, aur ek spring ki force minus-stiffness-times-stretch hoti hai. Dono saath likhne par do matrices banti hain, $M$ aur $K$ . Hum guess karte hain ki jab cheez freely bajti hai, toh sab log ek hi sine clock par wiggle karte hain aur sirf unke sizes alag hote hain — yeh size-pattern hi "mode shape" hai. Guess plug karo; sine clock har term par hai toh hum use cancel karte hain, aur hum ek timeless demand par aa jaate hain: stiffness shape par act kare equals (frequency squared) times mass shape par act kare. Yeh sirf khaas shapes aur khaas frequencies ke liye hold kar sakta hai — eigen-pairs. Do blocks ke liye yeh quadratic hai, toh do frequencies nikalti hain: ek low wali jahan dono blocks saath swing karte hain (middle spring barely stretch, gentle rhythm) aur ek high wali jahan woh ek doosre ke against swing karte hain (middle spring hard kheencha, fast rhythm). Woh do shapes "orthogonal" hain, jo ek fancy tarika hai yeh kehne ka ki woh ek doosre ke raaste mein nahi aate — toh hum messy coupled structure ko do clean single-block oscillators treat kar sakte hain. Aur isliye humein parwah hai: agar ek rocket launch un frequencies mein se kisi par shake kare, toh woh single oscillator blow up karta hai — resonance — toh hum pehle frequencies dhundhte hain aur unke around design karte hain.

Question:

([K] - ω^{2} [M]) {ϕ} = {0}

mein, hum determinant zero hone ki demand kyun karte hain?

Kyunki hum ek nonzero shape

{ϕ}

chahte hain; ek nonzero vector tab hi

{0}

mein bheja ja sakta hai jab matrix singular ho, yaani jab uska determinant vanish ho jaaye.

Question: Physically, kyun

ω_{1}

(in-phase)

ω_{2}

(out-of-phase) se lower hai?

In-phase motion coupling spring ko barely stretch karti hai, jo weak restoring force aur slow rhythm deta hai; out-of-phase motion use hard stretch karti hai, jo strong restoring force aur fast rhythm deta hai.

Question: Mode shapes ki orthogonality humein kya faayda deti hai?

Yeh

[M]

aur

[K]

ko diagonalise karti hai, system ko

n

independent single-DOF oscillators mein decouple karti hai, ek per mode.