Modal analysis — natural frequencies, mode shapes: Exercises

3.6.10 · D4 · Physics › Spacecraft Structures & Systems Engineering › Modal analysis — natural frequencies, mode shapes

Yeh page ek self-test ladder hai. Har problem parent note ke upar build karti hai, simple recognition se lekar full synthesis tak. Har solution ek collapsible callout ke andar chhupa hua hai — pehle problem try karo, phir reveal karo.

Shuru karne se pehle, ek reminder un tools ka jo hum baar baar use karte rahenge, taaki koi bhi symbol bina samjhe na aaye:

Definition Symbols jo hum neeche har jagah use karte hain

$x_{i}$ — mass $i$ ki displacement uski rest position se (metres mein). $\overset{x}{¨}_{i}$ uska acceleration hai (doosri time-derivative). Ye raw physical coordinates hain.
$[M]$ — mass matrix. Numbers ka ek grid jahan row $i$ , column $j$ batata hai ki mass $j$ ka acceleration, mass $i$ ki equation mein kaisa "feel" hota hai. Simple lumped masses ke liye ye diagonal hoti hai (har mass sirf apna acceleration feel karta hai).
$[K]$ — stiffness matrix. Row $i$ , column $j$ woh force hai jo mass $i$ par lagti hai jab mass $j$ ko unit displacement diya jaaye. Off-diagonal entries non-zero hoti hain jab koi spring do masses ko connect karti hai.
${ϕ_{i}}$ — ek mode shape: numbers ka ek column jo har mass ki relative displacement batata hai jab structure ek pure frequency par ring karta hai. Sirf ratios matter karte hain, isliye hum ise scale karne ke liye free hain.
$ω_{i}$ — mode $i$ ki natural frequency (rad/s). $f_{i} = ω_{i} / (2 π)$ Hz mein.
$λ$ — eigenvalue ka ek shorthand, defined as $λ \equiv ω^{2}$ . Hum ise sirf determinant algebra ko tidy rakhne ke liye use karte hain; end mein hum $ω = λ$ lete hain.
$det (\cdot)$ — determinant: ek square matrix se compute kiya gaya ek single number; ye exactly tab zero hota hai jab matrix kisi non-zero vector ko zero par squash kar de. Woh "koi non-zero vector" hi hai jo ek mode ko exist karne deta hai.

Woh ek master equation jisse sab kuch flow karta hai (dekho Multi-degree-of-freedom systems): $([K] - ω^{2} [M]) {ϕ} = {0}$

Yahan us equation ke peeche ki picture hai — actual mass–spring chains jo ye exercises use karti hain, taaki neeche har matrix ke liye ek physical drawing ho jis par point kar sako:

Figure — Modal analysis — natural frequencies, mode shapes

Prerequisites jinhe dekhna worth hai: Free vibration fundamentals, Orthogonality of mode shapes, Frequency response functions.

Level 1 — Recognition

Exercise 1.1

Ek single mass $m$ ek single spring $k$ par hang karti hai (one degree of freedom). Iski natural frequency likho aur batao ki sirf ek hi mode kyun hai.

Recall Solution

Hum kya karte hain: Ek DOF ke liye "matrices" sirf scalars hain: $[M] = m$ , $[K] = k$ . Eigenvalue problem $([K] - ω^{2} [M]) ϕ = 0$ ban jaata hai $(k - ω^{2} m) ϕ = 0$ . Kyun: Ek non-zero $ϕ$ ke liye bracket ka vanish hona zaroori hai, isliye $ω^{2} = k / m$ . $ω_{n} = \frac{k}{m}$ Sirf ek mode kyunki ek hi coordinate hai — ek $n$ -DOF system mein exactly $n$ modes hote hain. Yahan $n = 1$ .

Exercise 1.2

Parent note ke two-DOF system ke liye do natural frequencies thi $ω_{1} \approx 0.618 k / m$ aur $ω_{2} \approx 1.618 k / m$ . Kaun sa fundamental mode hai, aur "fundamental" ka matlab kya hai?

Recall Solution

Fundamental mode woh hai jisme sabse kam natural frequency ho: $ω_{1} \approx 0.618 k / m$ . Kyun important hai: launch energy low frequencies par concentrated hoti hai (dekho Launch vehicle load environments), isliye sabse low mode usually sabse dangerous hota hai — ise "pluck" karna sabse aasaan hai.

Exercise 1.3

Mode shapes ${ϕ_{1}} = [1 1.618]$ aur ${ϕ_{2}} = [1 - 0.618]$ diye hain, kaun se mode mein do masses in phase move karti hain aur kaun se mein out of phase?

Recall Solution

Har vector ke andar signs dekho.

${ϕ_{1}}$ : dono entries positive hain → dono masses ek hi direction mein ek hi instant par move karti hain → in phase. Neeche figure dekho, mode 1: dono arrows right ki taraf point kar rahe hain.
${ϕ_{2}}$ : entries ke opposite signs hain → masses opposite directions mein move karti hain → out of phase, unke beech ek stationary point (a node) hai.

Level 2 — Application

Exercise 2.1

Do equal masses $m$ ek line mein: wall— $k$ — $m_{1}$ — $k$ — $m_{2}$ — $k$ —wall (dono ends fixed), upar ke schematic mein lower chain ki tarah dikhaya gaya hai. Yahan $x_{1}$ aur $x_{2}$ do masses ke displacements hain (global symbol list dekho). $[M]$ aur $[K]$ banao.

Recall Solution

Mass 1 (displacement $x_{1}$ ): wall se aur mass 2 se springs. $m \overset{x}{¨}_{1} = - k x_{1} - k (x_{1} - x_{2}) = - 2 k x_{1} + k x_{2}$ Mass 2 (displacement $x_{2}$ ): mass 1 se aur door wali wall se springs. $m \overset{x}{¨}_{2} = - k (x_{2} - x_{1}) - k x_{2} = k x_{1} - 2 k x_{2}$ Right-hand sides par $x_{1}, x_{2}$ ke coefficients padhkar: $[M] = [m 0 0 m], [K] = [2 k - k - k 2 k]$ Note karo ki $[K]$ symmetric hai — ek general property jo real frequencies guarantee karti hai (dekho Orthogonality of mode shapes).

Exercise 2.2

2.1 ke system ke liye, dono natural frequencies nikalo.

Recall Solution

Hum kya karte hain: $det ([K] - ω^{2} [M]) = 0$ set karo. Shorthand $λ \equiv ω^{2}$ substitute karo (symbol list se) taaki algebra ek letter mein polynomial ho. $det [2 k - mλ - k - k 2 k - mλ] = (2 k - mλ)^{2} - k^{2} = 0$ Determinant kyun: ye exactly tab zero hota hai jab ek non-zero mode shape survive kar sake. Difference of squares ki tarah factor kyun karte hain: left side ka exact shape $A^{2} - B^{2}$ hai jahan $A = 2 k - mλ$ aur $B = k$ hai. Hum woh pattern recognize karte hain kyunki ye instantly $(A - B) (A + B)$ mein factor ho jaata hai — full quadratic expand karke re-solve karne se kahin aasaan. Yahi reason hai ise spot karne ka. $(2 k - mλ - k) (2 k - mλ + k) = 0 \Rightarrow (k - mλ) (3 k - mλ) = 0$ Har factor ek root $λ = ω^{2}$ deta hai, aur $ω = λ$ : $ω_{1} = \frac{k}{m}, ω_{2} = \frac{3 k}{m}$

Exercise 2.3

2.1 ke mode shapes nikalo.

Recall Solution

Mode 1 ( $λ_{1} = k / m$ ): $([K] - λ_{1} [M]) {ϕ} = 0$ ki pehli row deti hai $(2 k - k) ϕ_{11} - k ϕ_{21} = 0 \Rightarrow k ϕ_{11} = k ϕ_{21} \Rightarrow ϕ_{21} = ϕ_{11} .$ Kyun hum ab "simply $ϕ_{11} = 1$ set karte hain": eigenvalue equation sirf ratio $ϕ_{21} / ϕ_{11}$ fix karta hai, absolute size kabhi nahi — ek mode shape motion ka ek pattern describe karta hai, aur har entry ko double karna wahi pattern hai do guna bada. Isliye hum ek entry ke liye koi bhi non-zero value choose kar sakte hain; $ϕ_{11} = 1$ choose karna sabse tidy hai. ${ϕ_{1}} = [11] — symmetric, masses saath move karti hain.$

Mode 2 ( $λ_{2} = 3 k / m$ ): $(2 k - 3 k) ϕ_{12} - k ϕ_{22} = 0 \Rightarrow - k ϕ_{12} = k ϕ_{22} \Rightarrow ϕ_{22} = - ϕ_{12} .$ Phir scale $ϕ_{12} = 1$ se fix karke: ${ϕ_{2}} = [1 - 1] — antisymmetric, masses opposite direction mein move karti hain, beech mein node hai.$

Level 3 — Analysis

Exercise 3.1

2.3 ke do mode shapes ki mass-orthogonality verify karo: dikhao ki ${ϕ_{1}}^{T} [M] {ϕ_{2}} = 0$ .

Recall Solution

$[M] = m I$ ke saath, ${ϕ_{1}}^{T} [M] {ϕ_{2}} = m [11] [1 - 1] = m (1 \cdot 1 + 1 \cdot (- 1)) = 0.$ Ye zero kyun hona tha: dono frequencies alag hain ( $k / m \neq = 3 k / m$ ), aur orthogonality theorem guarantee karta hai ki distinct modes mass-orthogonal hote hain. Yahi woh cheez hai jo equations ko decouple karne deti hai (dekho Orthogonality of mode shapes).

Exercise 3.2

2.1 ke mode 2 ke liye generalized (modal) mass $M_{i}^{*} = {ϕ_{i}}^{T} [M] {ϕ_{i}}$ aur generalized stiffness $K_{i}^{*} = {ϕ_{i}}^{T} [K] {ϕ_{i}}$ compute karo, phir confirm karo ki $ω_{2} = K_{2}^{*} / M_{2}^{*}$ .

Recall Solution

${ϕ_{2}} = [1 - 1]$ . $M_{2}^{*} = m [1 - 1] [1 - 1] = m (1 + 1) = 2 m .$ $[K] {ϕ_{2}} = [2 k - k - k 2 k] [1 - 1] = [3 k - 3 k] .$ $K_{2}^{*} = [1 - 1] [3 k - 3 k] = 3 k + 3 k = 6 k .$ $\frac{K _{2}^{*}}{M _{2}^{*}} = \frac{6 k}{2 m} = \frac{3 k}{m} = ω_{2} . ✓$ Ye kyun kaam karta hai: full matrices ko ek mode par project karna poore system ko ek single scalar oscillator $M_{i}^{*} \overset{q}{¨}_{i} + K_{i}^{*} q_{i} = Q_{i}$ mein collapse kar deta hai.

Exercise 3.3

Ek cantilever ka first bending mode $f_{1} = 25 Hz$ par measure kiya gaya hai. Uske launch vehicle ka design rule demand karta hai ki $f_{1} \geq 35 Hz$ (lateral). Mass unchanged rakhte hue, stiffness $k$ ko kis factor se increase karna hoga taaki rule just meet ho?

Recall Solution

Stiffness kyun, mass kyun nahi: $f \propto k / m$ . $k$ badhane se $f$ badhta hai; mass add karne se ghatta hai. Fixed mass ke saath frequency badhane ke liye hum $k$ increase karte hain. $\frac{f _{new}}{f _{old}} = \frac{k _{new}}{k _{old}} \Rightarrow \frac{k _{new}}{k _{old}} = (\frac{35}{25})^{2} = 1.96.$ Answer: stiffness ko apni current value ke $1.96 \times$ tak badhna hoga. Is scaling ke liye frequency-vs-stiffness curve neeche draw ki gayi hai (marked point dekho).

Level 4 — Synthesis

Exercise 4.1

Ek spacecraft ka first mode $f_{1} = 30 Hz$ hai jisme modal mass $M_{1}^{*} = 40 kg$ aur modal damping ratio $ζ_{1} = 0.02$ hai. Ek sine sweep ke dauran base exactly resonance par ek modal force amplitude $Q_{0} = 50 N$ ke saath is mode ko excite karta hai. Peak modal displacement amplitude $q_{m a x}$ estimate karo.

Recall Solution

Tool: ek single-DOF oscillator ki resonant amplitude (har mode ek hi hoti hai). Resonance par $M^{*} \overset{q}{¨} + C^{*} \overset{q}{˙} + K^{*} q = Q_{0} sin ω t$ ka response hai $q_{m a x} = \frac{Q _{0}}{K _{1}^{*}} \cdot \frac{1}{2 ζ _{1}} = \frac{Q _{0}}{M _{1}^{*} ω _{1}^{2}} \cdot \frac{1}{2 ζ _{1}} .$ $1/ (2 ζ)$ kyun: resonance par stiffness aur inertia forces cancel ho jaati hain; sirf damping resist karti hai, isliye amplitude static deflection $Q_{0} / K^{*}$ hai jo quality factor $1/ (2 ζ)$ se magnified hoti hai (dekho Frequency response functions). Numbers: $ω_{1} = 2 π (30) = 188.50 rad/s$ , $ω_{1}^{2} = 35530 (rad/s)^{2}$ , $K_{1}^{*} = M_{1}^{*} ω_{1}^{2} = 1.4212 \times 1 0^{6} N/m$ . $q_{m a x} = \frac{50}{1.4212 \times 1 0 ^{6}} \cdot \frac{1}{2 ( 0.02 )} = 3.52 \times 1 0^{- 5} \cdot 25 = 8.80 \times 1 0^{- 4} m \approx 0.88 mm .$

Exercise 4.2

Do candidate designs ke first modes $28 Hz$ (design A) aur $45 Hz$ (design B) par hain. Launch sine environment mein $5$ – $40 Hz$ se strong energy hai. Kaun sa design choose karoge aur kyun? Phir ek quantitative margin bhi batao.

Recall Solution

B choose karo. Iski $45 Hz$ fundamental environment ki $40 Hz$ upper edge ke upar hai, isliye koi bhi launch energy directly ise hit nahi karti — frequency-separation (ya "clearance") design philosophy. Design A $28 Hz$ par $5$ – $40 Hz$ ke andar baithti hai; ye resonance par drive hogi aur $1/ (2 ζ)$ se amplify karengi. B ke liye quantitative margin: separation $= (45 - 40) /40 = 0.125 = 12.5%$ . Typical primary-structure rules $≳$ itna maangti hain, isliye B (barely) acceptable hai jabki A disqualified hai.

Level 5 — Mastery

Exercise 5.1

Design task. Tumhare paas parent note se two-DOF chain wall— $k$ — $m$ — $k$ — $m$ (free right end) hai — is page ke upar schematic mein upper chain — jisme $m = 2 kg$ , $k = 8000 N/m$ hai; iski fundamental $ω_{1} \approx 0.618 k / m$ hai. (a) $f_{1}$ Hz mein compute karo. (b) Requirement hai $f_{1} \geq 12 Hz$ . Agar mass fixed hai, toh minimum $k$ nikalo jo ise satisfy kare. (c) Apna naya $k$ confirm karo ki kaam karta hai.

Recall Solution

(a) $k / m = 8000/2 = 4000 = 63.25 rad/s$ . $ω_{1} = 0.618 \times 63.25 = 39.09 rad/s \Rightarrow f_{1} = ω_{1} / (2 π) = 6.22 Hz$ . $12 Hz$ rule fail karta hai.

(b) Kyunki $ω_{1} = 0.618 k / m$ aur $f_{1} \propto k$ , frequency ratio squared se scale karo: $\frac{k _{new}}{k _{old}} = (\frac{12}{6.22})^{2} = 3.723 \Rightarrow k_{new} = 3.723 \times 8000 = 29784 N/m .$ Round up karke $k_{new} = 29800 N/m$ .

(c) Check: $k_{new} / m = 29800/2 = 14900 = 122.07 rad/s$ . $ω_{1} = 0.618 \times 122.07 = 75.44 rad/s \Rightarrow f_{1} = 12.01 Hz \geq 12. ✓$

Exercise 5.2

Usi free-end system ke liye, decouple karo. ${ϕ_{1}} = [1 1.618]$ , $[M] = m I$ diye hain, generalized mass $M_{1}^{*}$ compute karo, aur $ω_{1}^{2} = (3 - 5) k / (2 m)$ use karke generalized stiffness $K_{1}^{*}$ ; verify karo ki $ω_{1} = K_{1}^{*} / M_{1}^{*}$ .

Recall Solution

$M_{1}^{*} = m [1 1.618] [1 1.618] = m (1 + 1.61 8^{2}) = m (1 + 2.618) = 3.618 m .$ Kyunki kisi mode ke liye $K_{1}^{*} = ω_{1}^{2} M_{1}^{*}$ , $K_{1}^{*} = \frac{( 3 - 5 ) k}{2 m} \cdot 3.618 m = \frac{0.7639 k}{2} \cdot 3.618 = 1.3820 k .$ $\frac{K _{1}^{*}}{M _{1}^{*}} = \frac{1.3820 k}{3.618 m} = 0.3820 k / m = 0.618 k / m = ω_{1} . ✓$ Matlab: poora coupled system, is mode ke liye, ek mass $3.618 m$ ek spring $1.382 k$ par ki tarah behave karta hai.