Worked examples — Modal analysis — natural frequencies, mode shapes

3.6.10 · D3 · Physics › Spacecraft Structures & Systems Engineering › Modal analysis — natural frequencies, mode shapes

Yeh page parent modal analysis note ki "koi case choot na jaye" companion hai. Wahan humne eigenvalue problem $[K]\{{\varphi }_i\}={\omega }_i^2[M]\{{\varphi }_i\}$ derive ki thi. Yahan hum use run karte hain har tarah ke input par jo tum encounter kar sakte ho — including weird wale (zero frequency, repeated roots, unequal masses, ek real launch number).

Koi bhi symbol aane se pehle, yeh hai woh vocabulary jis par hum lean karte hain, plain words mein:

Neeche sab kuch parent ke boxed result se bana hai. Prerequisites: Multi-degree-of-freedom systems, Free vibration fundamentals, aur Orthogonality of mode shapes.

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Scenario matrix

Ek discrete system ki modal analysis exactly yeh situations throw kar sakti hai tumhare saamne. Har ek ka worked example neeche diya hai.

Cell	Case class	Kya special hai	Example
A	Baseline 2-DOF, grounded	Dono frequencies nonzero, clean roots	Ex 1
B	Unequal masses	$[M]$ diagonal hai par $mI$ nahi — normalization badlti hai	Ex 2
C	Free–free (no wall)	${\omega }_1=0$: ek rigid-body mode, degenerate	Ex 3
D	Repeated frequencies	Do modes ek $\omega$ share karte hain; shape unique nahi	Ex 4
E	Sign / phase reading	Kaun si masses saath chalti hain vs. opposite	Ex 5
F	Limiting behaviour	Ek stiffness ko $\to 0$ ya $\to \infty$ hone do; $\omega$ dekho	Ex 6
G	Real-world word problem	kg, N/m ko launch-relevant Hz mein convert karo	Ex 7
H	Exam twist: orthogonality check	Verify karo ki modes $[M]$-orthogonal hain, decoupling ke liye use karo	Ex 8

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Example 1 — Cell A: grounded baseline

Forecast: compute karne se pehle guess karo — kya do masses kabhi opposite directions mein chalenge? Haan/nahi likh lo.

Step 1 — matrices banao. $[M]=\left[\begin{array}{cc}m& 0\\ 0& m\end{array}\right],[K]=\left[\begin{array}{cc}2k& -k\\ -k& k\end{array}\right]$ Yeh step kyun? Mass 1 ko do springs feel hoti hain (uske diagonal par $2k$); mass 2 ko sirf coupling spring feel hoti hai ($k$). Shared $-k$ off-diagonal coupling hai — yahi cheez ise interesting banati hai.

Step 2 — characteristic equation. Parent se, $\det ([K]-{\omega }^2[M])=0$ deta hai $m^2{\omega }^4-3km{\omega }^2+k^2=0.$ Yeh step kyun? Hum woh frequencies chahte hain jahan ek nonzero shape survive kare; yeh tabhi hota hai jab determinant vanish ho (ek nonzero eigenvector ko singular matrix chahiye).

Step 3 — ${\omega }^2$ ke liye solve karo. ${\omega }^2$ ko unknown treat karo: ${\omega }^2=\frac{3k\pm \sqrt{5}k}{2m}\Rightarrow {\omega }_1=0.618\sqrt{k/m},{\omega }_2=1.618\sqrt{k/m}.$ $k/m=800/2=400$ ke saath, $\sqrt{k/m}=20\text{ rad/s}$: ${\omega }_1=12.36\text{ rad/s},{\omega }_2=32.36\text{ rad/s}.$

Step 4 — mode shapes. Har ${\omega }^2$ ko row 1 mein plug karo, $2k{\varphi }_1-k{\varphi }_2={\omega }^{2}m{\varphi }_1$, ${\varphi }_1=1$ set karo: $\{{\varphi }_1\}=\left[\begin{array}{c}1\\ 1.618\end{array}\right](\text{in-phase}),\{{\varphi }_2\}=\left[\begin{array}{c}1\\ -0.618\end{array}\right](\text{out-of-phase}).$

Figure — arrows padho. Neeche ka chalkboard sketch dono systems draw karta hai, ek per row. Har coloured arrow ek mass ki displacement hai ek instant par, uski length mode-shape number se set hoti hai jo uske neeche print hai. Upar (blue) row dekho: dono arrows same direction mein point karte hain aur doosra longer hai ($1.618$) — yeh mode 1 hai, in-phase, mass 2 lead kar rahi hai. Ab neeche (pink) row: do arrows opposite directions mein point karte hain ($+1$ aur $-0.618$) — yeh mode 2 hai, masses coupling spring ke across ek doosre se lad rahi hain. Picture hi poori baat hai: ek "mode shape" literally arrows ka yeh frozen pattern hai.

Forecast ka jawab: haan — mode 2 mein masses opposite direction mein chalti hain (note karo ${\varphi }_2$ par minus sign, reversed pink arrow ke roop mein draw kiya gaya hai).

Recall Example 1 Verify Karo

$m^2{\omega }^4-3km{\omega }^2+k^2$ ke roots ka product $(k/m{)}^2$ hona chahiye ::: ${\omega }_1^2{\omega }_2^2=k^2/m^2=400^2$, aur $12.36^2\cdot 32.36^2\approx 160000$. ✓ ${\omega }^2$ ka sum ::: $12.36^2+32.36^2\approx 1200=3k/m$. ✓

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Example 2 — Cell B: unequal masses

Forecast: kya low mode mein lighter mass heavy wali se zyada ya kam chalegi?

Step 1 — unequal $[M]$ ke saath matrices. $[M]=\left[\begin{array}{cc}4& 0\\ 0& 1\end{array}\right],[K]=\left[\begin{array}{cc}200& -100\\ -100& 100\end{array}\right].$ Yeh step kyun? $[M]$ ki diagonal ab ek single scalar nahi hai, isliye Ex 1 ka equal-mass shortcut fail ho jaata hai. Humein $m_1,m_2$ alag se carry karne honge.

Step 2 — determinant. $\det \left[\begin{array}{cc}200-4{\omega }^2& -100\\ -100& 100-{\omega }^2\end{array}\right]=0$ $(200-4{\omega }^2)(100-{\omega }^2)-10000=4{\omega }^4-600{\omega }^2+10000=0.$ Yeh step kyun? Same singular-matrix logic; coefficients ab $m_1,m_2$ mix karte hain.

Step 3 — solve karo. ${\omega }^2=\frac{600\pm \sqrt{600^2-4\cdot 4\cdot 10000}}{8}=\frac{600\pm \sqrt{200000}}{8}.$ $\sqrt{200000}\approx 447.21$, isliye ${\omega }_1^2=19.10$, ${\omega }_2^2=130.90$: ${\omega }_1=4.37\text{ rad/s},{\omega }_2=11.44\text{ rad/s}.$

Step 4 — shapes row 2 se, $-100{\varphi }_1+(100-{\omega }^2){\varphi }_2=0\Rightarrow {\varphi }_2={\varphi }_1\cdot 100/(100-{\omega }^2)$, ${\varphi }_1=1$:

• Mode 1: ${\varphi }_2=100/(100-19.10)=1.236$ → $\{1,1.236\}$ (light mass zyada swing karti hai, in-phase).
• Mode 2: ${\varphi }_2=100/(100-130.90)=-3.236$ → $\{1,-3.236\}$ (light mass bahut zyada swing karti hai, opposite).

Forecast ka jawab: light mass zyada chalti hai — kam inertia use whip around karne deti hai.

Recall Example 2 Verify Karo

${\omega }^2$ ka sum = $600/4=150$ ::: $19.10+130.90=150.0$. ✓ ${\omega }^2$ ka product = $10000/4=2500$ ::: $19.10\times 130.90\approx 2500$. ✓

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Example 3 — Cell C: free–free, zero-frequency mode

Forecast: orbit mein ek spacecraft ka koi wall nahi hota. Kya phir bhi uske drift karne ke liye ek "frequency" milti hai?

Step 1 — matrices. $[M]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],[K]=\left[\begin{array}{cc}k& -k\\ -k& k\end{array}\right]=\left[\begin{array}{cc}50& -50\\ -50& 50\end{array}\right].$ Yeh step kyun? Koi wall nahi hone se, har mass sirf ek single coupling spring feel karti hai, isliye dono diagonal entries sirf $k$ hain.

Step 2 — determinant. $\det \left[\begin{array}{cc}50-{\omega }^2& -50\\ -50& 50-{\omega }^2\end{array}\right]=(50-{\omega }^2{)}^2-2500={\omega }^4-100{\omega }^2=0.$ ${\omega }^2({\omega }^2-100)=0\Rightarrow {\omega }_1^2=0,{\omega }_2^2=100.$

Step 3 — ${\omega }_1=0$ padho. Yeh ek rigid-body mode hai. Uski shape $(50-0){\varphi }_1-50{\varphi }_2=0$ se hai: $\{{\varphi }_1\}=\left[\begin{array}{c}1\\ 1\end{array}\right](\text{dono saath chalte hain — pure translation, koi spring stretch nahi}).$ Yeh step kyun? $[K]$ sirf positive semi-definite hoti hai jab unrestrained ho: poora body zero restoring force ke saath drift kar sakta hai, isliye uski "frequency" genuinely $0$ hai.

Step 4 — real mode. ${\omega }_2=10\text{ rad/s}$, shape $(50-100){\varphi }_1-50{\varphi }_2=0\Rightarrow \{{\varphi }_2\}=\left[\begin{array}{c}1\\ -1\end{array}\right]$ (masses spring ko symmetrically squeeze/stretch karti hain).

Figure — do extremes. Neeche ka sketch do modes stack karta hai. Upar (yellow) row mein dono arrows equal aur parallel hain: poori assembly ek saath slide karti hai, spring ki length kabhi nahi badlti, isliye kuch bhi push back nahi karta aur $\omega =0$. Neeche (pink) row mein arrows equal hain par opposite: masses seedhe spring mein drive karti hain, yeh stretch aur squeeze hoti hai, aur woh restoring force finite $\omega =10\text{ rad/s}$ deti hai. Notice karo ki kisi bhi row mein koi wall nahi hai — woh missing wall exactly isliye hai ki yellow "drift" mode exist karne ki permission rakhta hai.

Forecast ka jawab: haan aur nahi — tumhe ek mode milta hai, par uski frequency $0$ hai: free drift, vibration nahi.

Recall Example 3 Verify Karo

Ek root exactly zero hai ::: ${\omega }^4-100{\omega }^2={\omega }^2({\omega }^2-100)$ factor karne par roots $0$ aur $10$ milte hain. ✓

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Example 4 — Cell D: repeated (degenerate) frequencies

Forecast: do identical, uncoupled springs ke saath, mode shapes ka koi ek jawab hota hai?

Step 1 — determinant. $\det \left[\begin{array}{cc}9-{\omega }^2& 0\\ 0& 9-{\omega }^2\end{array}\right]=(9-{\omega }^2{)}^2=0\Rightarrow {\omega }^2=9(\text{twice}).$ Toh ${\omega }_1={\omega }_2=3\text{ rad/s}$ — ek repeated eigenvalue.

Step 2 — shapes ke liye solve karo. ${\omega }^2=9$ plug karo: $[K]-9I=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$. Har vector ise satisfy karta hai. Yeh step kyun? Jab matrix poori tarah zero ho jaata hai, poora plane eigenspace ban jaata hai — mode shape unique nahi hota.

Step 3 — ek orthonormal pair choose karo. Hum koi bhi do independent, $[M]$-orthogonal directions choose karne ke liye free hain, jaise $\{{\varphi }_1\}=\left[\begin{array}{c}1\\ 0\end{array}\right],\{{\varphi }_2\}=\left[\begin{array}{c}0\\ 1\end{array}\right]\text{ya }45^{\circ }\text{ rotated:}\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 1\end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ -1\end{array}\right].$

Forecast ka jawab: koi single answer nahi — plane ko span karne wala koi bhi orthonormal pair kaam karta hai.

Recall Example 4 Verify Karo

Dono frequencies 3 equal hain ::: $(9-{\omega }^2{)}^2$ ka repeated root $\omega =3$ hai. ✓

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Example 5 — Cell E: sign / phase padhna

Forecast: kis mode mein masses ke beech ek stationary "node" point hoga?

Step 1 — sign pattern. Mode 1: $(+,+)$ — koi sign change nahi. Mode 2: $(+,-)$ — ek sign change. Yeh step kyun? Adjacent DOFs ke beech ek sign change ka matlab hai deflection zero se guzar rahi hai — ek node, ek point jo still rehta hai.

Step 2 — general rule. Ek chain ke liye, mode $i$ (frequency ke order mein) ke $i-1$ internal nodes hote hain. Mode 1 → 0 nodes (fundamental, softest). Mode 2 → 1 node. Yeh step kyun? Zyada nodes = zyada bending = zyada stored elastic energy = zyada $\omega$. Yeh shape ko directly frequency ranking se link karta hai.

Step 3 — physical picture. In-phase mode: dono masses same direction mein lunge karti hain, spring barely stretch hoti hai → low restoring force → low ${\omega }_1$. Out-of-phase: spring hard kaam karti hai → high ${\omega }_2$. Ex 1 se ${\omega }_2>{\omega }_1$ se match karta hai.

Forecast ka jawab: mode 2 mein node hota hai.

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Example 6 — Cell F: limiting behaviour

Forecast: jab $c\to 0$ to do masses uncouple ho jaati hain — ${\omega }_2$ ka kya hota hai?

Step 1 — small $c$. $\det =(1+c-{\omega }^2)(c-{\omega }^2)-c^2={\omega }^4-(1+2c){\omega }^2+c=0$. Jab $c\to 0$: ${\omega }^2\to 0\text{aur}{\omega }^2\to 1.$ Yeh step kyun? Koi coupling nahi hone par, mass 2 free float karti hai ($\omega \to 0$, uska apna rigid mode) jabki mass 1 apna wall spring rakhti hai ($\omega =1$).

Step 2 — large $c$. $c\to \infty$ ke liye, ${\omega }^4-2c{\omega }^2+c\approx 0$ ke roots: ek small root ${\omega }^2\approx 1/2$, ek large ${\omega }^2\approx 2c\to \infty$. Yeh step kyun? Bahut stiff coupling masses ko ek rigid block of mass $2$ mein wall spring $1$ par glue kar deta hai → low mode ${\omega }^2=1/2$; high mode (masses huge spring ke saath lad rahi hain) infinity tak shoot ho jaata hai.

Step 3 — sanity numbers. $c=10$ par: ${\omega }^4-21{\omega }^2+10=0\Rightarrow {\omega }^2=0.489$ ya $20.51$. Low mode $0.5$ ke paas, high mode $2c=20$ ke paas. ✓ Yeh step kyun? Steps 1–2 mein limits predictions hain; ek concrete moderate $c=10$ plug karna check karta hai ki real roots un predicted endpoints ki taraf slide ho rahe hain ($0.489\to 1/2$ aur $20.51\to 2c$). Ek number ko limit ki taraf creep karte dekhna hi hai jis se tum limiting argument par trust karte ho instead of sirf assert karne ke.

Forecast ka jawab: ${\omega }_2\to 1$ (mass 1 akeli), aur doosra root $\to 0$.

Recall Example 6 Verify Karo

c=10 par, low mode near 0.5 aur high near 20 ::: ${\omega }^4-21{\omega }^2+10$ ke roots $\approx 0.489$ aur $20.51$ hain. ✓

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Example 7 — Cell G: real-world launch number

Forecast: safer hone ke liye stiffer ya softer? Compute karne se pehle guess karo.

Step 1 — single-DOF frequency. $\omega =\sqrt{k/m}$ se (parent ka ${\omega }_i=\sqrt{K^{\ast }/M^{\ast }}$ ek mode ke saath): $\omega =\sqrt{9500/6}=\sqrt{1583.3}=39.79\text{ rad/s}.$ Yeh step kyun? Ek flexible DOF eigenvalue problem ko ek scalar mein collapse kar deta hai — classic spring–mass result.

Step 2 — Hz mein convert karo. $f=\omega /(2\pi )=39.79/6.283=6.33\text{ Hz}$. Yeh step kyun? Launch specs (Launch vehicle load environments se) Hz mein quote hoti hain; humein like with like compare karna hai.

Step 3 — margin check. Panel $6.33\text{ Hz}$ par, forcing $8\text{ Hz}$ par. Yeh sirf $\frac{8-6.33}{6.33}\approx 26\%$ apart hain — standard rule of thumb chahta hai ki pehla mode forcing ke upar ho with $\gtrsim 40\%$ margin. Yeh panel bahut soft hai aur dangerously close hai. Yeh step kyun? Resonance response ko sharply amplify karta hai jab frequencies coincide hoti hain; margin design safety buffer hai.

Step 4 — fix karo. $f$ ko upar push karne ke liye hum $k$ raise karte hain. Target $f=11.2\text{ Hz}$ (8 se 40% upar): $k=m(2\pi f{)}^2=6\cdot (70.4{)}^2=29730\text{ N/m}$ chahiye — roughly stiffness triple karni hogi. Yeh step kyun? Step 3 ne bataya ki panel forcing frequency ke galat side par hai, isliye humein uska mode move karna hai. Kyunki $f=\frac{1}{2\pi }\sqrt{k/m}$ sirf $k$ ke saath rise karta hai (mass hardware ke zariye fixed hai), us formula ko target $f$ ke liye invert karna humein exactly batata hai ki redesign mein kitni stiffness add karni hogi — ek vague "stiffen it" ko ek hard number mein badalta hai jis par structures team build kar sake.

Forecast ka jawab: stiffer — $k$ raise karne se $\omega$ raise hoti hai, mode ko launch band ke upar move karke.

Recall Example 7 Verify Karo

Panel frequency ::: $\sqrt{9500/6}/(2\pi )\approx 6.33$ Hz. ✓ 11.2 Hz tak pahunchne ke liye stiffness ::: $6\cdot (2\pi \cdot 11.2{)}^2\approx 29730$ N/m. ✓

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Example 8 — Cell H: exam twist, orthogonality decouples

Forecast: agar $[M]$ ke through dot product zero nahi hai, to humne kya galat kiya?

Step 1 — mass orthogonality. $\{{\varphi }_1{\}}^T[M]\{{\varphi }_2\}=2(1\cdot 1+1.618\cdot (-0.618))=2(1-1.0)=0.$ Yeh step kyun? Parent ne prove kiya tha ki distinct modes satisfy karte hain $\{{\varphi }_j{\}}^T[M]\{{\varphi }_i\}=0$; near-zero result confirm karta hai ki hamare shapes correct eigenvectors hain (dekho Orthogonality of mode shapes).

Step 2 — generalized masses $M_i^{\ast }=\{{\varphi }_i{\}}^T[M]\{{\varphi }_i\}$: $M_1^{\ast }=2(1^2+1.618^2)=2(3.618)=7.236\text{ kg},$ $M_2^{\ast }=2(1^2+0.618^2)=2(1.382)=2.764\text{ kg}.$ Yeh step kyun? $M_i^{\ast }$ har decoupled single-DOF oscillator $M_i^{\ast }{\ddot{q}}_i+K_i^{\ast }{q}_i=Q_i$ ki effective inertia hai.

Step 3 — frequency ke saath consistency. Generalized stiffness $K_i^{\ast }={\omega }_i^2{M}_i^{\ast }$. Mode 1 ke liye $k=800$ ke saath: ${\omega }_1^2=0.618^2\cdot 400=152.8$, toh $K_1^{\ast }=152.8\times 7.236=1105.6\text{ N/m}$, aur $\sqrt{K_1^{\ast }/M_1^{\ast }}=\sqrt{152.8}=12.36\text{ rad/s}$ — Ex 1 ke ${\omega }_1$ par wapas. Yeh step kyun? Agar orthogonal transform correct hai, to decoupled single-DOF oscillator ko same frequency reproduce karni chahiye jo humne full 2×2 problem se nikali thi. $12.36\text{ rad/s}$ recover karna loop ka closing hai — proof ki modal coordinates ne sach mein ek coupled system ko independent oscillators mein split kar diya.

Forecast ka jawab: nonzero cross term ka matlab hai shapes galat hain (ya ${\omega }_i={\omega }_j$, Ex 4 ka degenerate case).

Recall Example 8 Verify Karo

Mass orthogonality ::: $1\cdot 1+1.618\cdot (-0.618)=1-1.0=0$. ✓ $M_1^{\ast },M_2^{\ast }$ ::: $2(3.618)=7.236$ aur $2(1.382)=2.764$ kg. ✓ Round-trip frequency ::: $\sqrt{152.8}=12.36$ rad/s, Ex 1 ke ${\omega }_1$ se match karta hai. ✓

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Recall

Recall Har cell, ek line

Zero-frequency mode ka matlab hai ::: ek rigid-body (unrestrained) mode — free drift, koi spring stretch nahi. Repeated eigenvalue ka matlab hai ::: mode shape unique nahi hota; us eigenspace mein koi bhi orthonormal pair kaam karta hai. Ek mode ko launch band se door karne ke liye tum ::: structure ko stiffen karte ho (raise $k$ → raise $\omega =\sqrt{k/m}$). Ek chain mein mode $i$ ke internal nodes ki sankhya ::: $i-1$. Check karo ki do computed shapes correct modes hain ::: verify karo $\{{\varphi }_i{\}}^T[M]\{{\varphi }_j\}=0$ for $i\ne j$.

Related: Frequency response functions dikhata hai kya hota hai jab forcing apply ki jaati hai in frequencies ke paas; Structural damping mechanisms us resonant peaks ko soften karta hai jise is page ne locate kiya.