Worked examples — Modal analysis — natural frequencies, mode shapes

This page is the "no case left behind" companion to the parent modal analysis note. There we derived the eigenvalue problem $[K]\{{\varphi }_i\}={\omega }_i^2[M]\{{\varphi }_i\}$. Here we run it on every kind of input you can meet — including the weird ones (zero frequency, repeated roots, unequal masses, a real launch number).

Before any symbol appears, here is the vocabulary we lean on, in plain words:

Everything below is built from the parent's boxed result. Prerequisites: Multi-degree-of-freedom systems, Free vibration fundamentals, and Orthogonality of mode shapes.

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The scenario matrix

Modal analysis of a discrete system can throw exactly these situations at you. Every one gets a worked example below.

Cell	Case class	What's special	Example
A	Baseline 2-DOF, grounded	Both frequencies nonzero, clean roots	Ex 1
B	Unequal masses	$[M]$ diagonal but not $mI$ — normalization changes	Ex 2
C	Free–free (no wall)	${\omega }_1=0$: a rigid-body mode, degenerate	Ex 3
D	Repeated frequencies	Two modes share one $\omega$; shape not unique	Ex 4
E	Sign / phase reading	Which masses move together vs. opposite	Ex 5
F	Limiting behaviour	Let a stiffness $\to 0$ or $\to \infty$; watch $\omega$	Ex 6
G	Real-world word problem	Convert kg, N/m to a launch-relevant Hz	Ex 7
H	Exam twist: orthogonality check	Verify modes are $[M]$-orthogonal, use for decoupling	Ex 8

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Example 1 — Cell A: the grounded baseline

Forecast: guess before computing — will the two masses ever move in opposite directions? Jot yes/no.

Step 1 — build the matrices. $[M]=\left[\begin{array}{cc}m& 0\\ 0& m\end{array}\right],[K]=\left[\begin{array}{cc}2k& -k\\ -k& k\end{array}\right]$ Why this step? Mass 1 feels two springs ($2k$ on its diagonal); mass 2 feels only the coupling spring ($k$). The shared $-k$ off-diagonal is the coupling — that's what makes this interesting.

Step 2 — characteristic equation. From the parent, $\det ([K]-{\omega }^2[M])=0$ gives $m^2{\omega }^4-3km{\omega }^2+k^2=0.$ Why this step? We want the frequencies where a nonzero shape survives; that only happens where the determinant vanishes (a nonzero eigenvector needs a singular matrix).

Step 3 — solve for ${\omega }^2$. Treat ${\omega }^2$ as the unknown: ${\omega }^2=\frac{3k\pm \sqrt{5}k}{2m}\Rightarrow {\omega }_1=0.618\sqrt{k/m},{\omega }_2=1.618\sqrt{k/m}.$ With $k/m=800/2=400$, $\sqrt{k/m}=20\text{ rad/s}$: ${\omega }_1=12.36\text{ rad/s},{\omega }_2=32.36\text{ rad/s}.$

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

Figure — read the arrows. The chalkboard sketch below draws both systems, one per row. Each coloured arrow is a mass's displacement at one instant, its length set by the mode-shape number printed beneath it. Look at the top (blue) row: both arrows point the same way and the second is longer ($1.618$) — that's mode 1, in-phase, mass 2 leading. Now the bottom (pink) row: the two arrows point opposite ways ($+1$ and $-0.618$) — that's mode 2, the masses fighting each other across the coupling spring. The picture is the whole point: a "mode shape" is literally this frozen pattern of arrows.

Answer to forecast: yes — mode 2 has the masses moving oppositely (note the minus sign on ${\varphi }_2$, drawn as the reversed pink arrow).

Recall Verify Example 1

Product of roots of $m^2{\omega }^4-3km{\omega }^2+k^2$ should be $(k/m{)}^2$ ::: ${\omega }_1^2{\omega }_2^2=k^2/m^2=400^2$, and $12.36^2\cdot 32.36^2\approx 160000$. ✓ Sum of ${\omega }^2$ ::: $12.36^2+32.36^2\approx 1200=3k/m$. ✓

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Example 2 — Cell B: unequal masses

Forecast: will the lighter mass move more or less than the heavy one in the low mode?

Step 1 — matrices with unequal $[M]$. $[M]=\left[\begin{array}{cc}4& 0\\ 0& 1\end{array}\right],[K]=\left[\begin{array}{cc}200& -100\\ -100& 100\end{array}\right].$ Why this step? The diagonal of $[M]$ is no longer a single scalar, so the equal-mass shortcut from Ex 1 fails. We must carry $m_1,m_2$ separately.

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

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

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

• Mode 1: ${\varphi }_2=100/(100-19.10)=1.236$ → $\{1,1.236\}$ (light mass swings more, in-phase).
• Mode 2: ${\varphi }_2=100/(100-130.90)=-3.236$ → $\{1,-3.236\}$ (light mass swings far, opposite).

Answer to forecast: the light mass moves more — small inertia lets it whip around.

Recall Verify Example 2

Sum of ${\omega }^2$ = $600/4=150$ ::: $19.10+130.90=150.0$. ✓ Product of ${\omega }^2$ = $10000/4=2500$ ::: $19.10\times 130.90\approx 2500$. ✓

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

Forecast: a spacecraft in orbit has no wall. Do you still get a "frequency" for it drifting as a whole?

Step 1 — matrices. $[M]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],[K]=\left[\begin{array}{cc}k& -k\\ -k& k\end{array}\right]=\left[\begin{array}{cc}50& -50\\ -50& 50\end{array}\right].$ Why this step? With no wall, each mass feels only the single coupling spring, so both diagonal entries are just $k$.

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 — read ${\omega }_1=0$. This is a rigid-body mode. Its shape from $(50-0){\varphi }_1-50{\varphi }_2=0$ is $\{{\varphi }_1\}=\left[\begin{array}{c}1\\ 1\end{array}\right](\text{both move together — pure translation, no spring stretch}).$ Why this step? $[K]$ is only positive semi-definite when unrestrained: the whole body can drift with zero restoring force, so its "frequency" is genuinely $0$.

Step 4 — the 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 squeeze/stretch the spring symmetrically).

Figure — the two extremes. The sketch below stacks the two modes. The top (yellow) row has both arrows equal and parallel: the whole assembly slides as one, the spring never changes length, so nothing pushes back and $\omega =0$. The bottom (pink) row has the arrows equal but opposite: the masses drive straight into the spring, it stretches and squeezes hard, and that restoring force gives the finite $\omega =10\text{ rad/s}$. Notice there is no wall in either row — that missing wall is exactly why the yellow "drift" mode is allowed to exist.

Answer to forecast: yes and no — you get a mode, but its frequency is $0$: free drift, not vibration.

Recall Verify Example 3

One root is exactly zero ::: factoring ${\omega }^4-100{\omega }^2={\omega }^2({\omega }^2-100)$ gives roots $0$ and $10$. ✓

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Example 4 — Cell D: repeated (degenerate) frequencies

Forecast: with two identical, uncoupled springs, is there one answer for the mode shapes?

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

Step 2 — solve for shapes. Plug ${\omega }^2=9$: $[K]-9I=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$. Every vector satisfies it. Why this step? When the matrix becomes fully zero, the whole plane is the eigenspace — the mode shape is not unique.

Step 3 — choose an orthonormal pair. We are free to pick any two independent, $[M]$-orthogonal directions, e.g. $\{{\varphi }_1\}=\left[\begin{array}{c}1\\ 0\end{array}\right],\{{\varphi }_2\}=\left[\begin{array}{c}0\\ 1\end{array}\right]\text{or rotated }45^{\circ }:\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 1\end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ -1\end{array}\right].$

Answer to forecast: no single answer — any orthonormal pair spanning the plane works.

Recall Verify Example 4

Both frequencies equal 3 ::: repeated root of $(9-{\omega }^2{)}^2$ is $\omega =3$. ✓

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Example 5 — Cell E: reading sign / phase

Forecast: which mode has a stationary "node" point between the masses?

Step 1 — sign pattern. Mode 1: $(+,+)$ — no sign change. Mode 2: $(+,-)$ — one sign change. Why this step? A sign change between adjacent DOFs means the deflection passes through zero — a node, a point that stays still.

Step 2 — general rule. For a chain, mode $i$ (ordered by frequency) has $i-1$ internal nodes. Mode 1 → 0 nodes (fundamental, softest). Mode 2 → 1 node. Why this step? More nodes = more bending = more stored elastic energy = higher $\omega$. This links shape directly to frequency ranking.

Step 3 — physical picture. In-phase mode: both masses lunge the same way, spring barely stretches → low restoring force → low ${\omega }_1$. Out-of-phase: spring is worked hard → high ${\omega }_2$. Matches ${\omega }_2>{\omega }_1$ from Ex 1.

Answer to forecast: mode 2 has the node.

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Example 6 — Cell F: limiting behaviour

Forecast: as $c\to 0$ the two masses uncouple — what happens to ${\omega }_2$?

Step 1 — small $c$. $\det =(1+c-{\omega }^2)(c-{\omega }^2)-c^2={\omega }^4-(1+2c){\omega }^2+c=0$. As $c\to 0$: ${\omega }^2\to 0\text{and}{\omega }^2\to 1.$ Why this step? With no coupling, mass 2 floats free ($\omega \to 0$, its own rigid mode) while mass 1 keeps its wall spring ($\omega =1$).

Step 2 — large $c$. For $c\to \infty$, roots of ${\omega }^4-2c{\omega }^2+c\approx 0$: one small root ${\omega }^2\approx 1/2$, one large ${\omega }^2\approx 2c\to \infty$. Why this step? A very stiff coupling glues the masses into one rigid block of mass $2$ on a wall spring $1$ → low mode ${\omega }^2=1/2$; the high mode (masses fighting the huge spring) shoots to infinity.

Step 3 — sanity numbers. At $c=10$: ${\omega }^4-21{\omega }^2+10=0\Rightarrow {\omega }^2=0.489$ or $20.51$. Low mode near $0.5$, high mode near $2c=20$. ✓ Why this step? The limits in Steps 1–2 are predictions; plugging a concrete moderate $c=10$ checks that the real roots are already sliding toward those predicted endpoints ($0.489\to 1/2$ and $20.51\to 2c$). Watching a number crawl toward the limit is how you trust the limiting argument instead of just asserting it.

Answer to forecast: ${\omega }_2\to 1$ (mass 1 alone), and the other root $\to 0$.

Recall Verify Example 6

At c=10, low mode near 0.5 and high near 20 ::: roots of ${\omega }^4-21{\omega }^2+10$ are $\approx 0.489$ and $20.51$. ✓

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Example 7 — Cell G: real-world launch number

Forecast: stiffer or softer to be safer? Guess before computing.

Step 1 — single-DOF frequency. From $\omega =\sqrt{k/m}$ (parent's ${\omega }_i=\sqrt{K^{\ast }/M^{\ast }}$ with one mode): $\omega =\sqrt{9500/6}=\sqrt{1583.3}=39.79\text{ rad/s}.$ Why this step? One flexible DOF collapses the eigenvalue problem to a scalar — the classic spring–mass result.

Step 2 — convert to Hz. $f=\omega /(2\pi )=39.79/6.283=6.33\text{ Hz}$. Why this step? Launch specs (from Launch vehicle load environments) are quoted in Hz; we must compare like with like.

Step 3 — margin check. Panel at $6.33\text{ Hz}$, forcing at $8\text{ Hz}$. They are only $\frac{8-6.33}{6.33}\approx 26\%$ apart — the standard rule of thumb wants the first mode above forcing with $\gtrsim 40\%$ margin. This panel is too soft and sits dangerously close. Why this step? Resonance amplifies response sharply when frequencies coincide; margin is the design safety buffer.

Step 4 — fix. To push $f$ up we raise $k$. Target $f=11.2\text{ Hz}$ (40% above 8): need $k=m(2\pi f{)}^2=6\cdot (70.4{)}^2=29730\text{ N/m}$ — roughly triple the stiffness. Why this step? Step 3 told us the panel is on the wrong side of the forcing frequency, so we must move its mode. Since $f=\frac{1}{2\pi }\sqrt{k/m}$ rises only with $k$ (mass is fixed by the hardware), inverting that formula for the target $f$ tells us exactly how much stiffness the redesign must add — turning a vague "stiffen it" into a hard number the structures team can build to.

Answer to forecast: stiffer — raising $k$ raises $\omega$, moving the mode above the launch band.

Recall Verify Example 7

Panel frequency ::: $\sqrt{9500/6}/(2\pi )\approx 6.33$ Hz. ✓ Stiffness to reach 11.2 Hz ::: $6\cdot (2\pi \cdot 11.2{)}^2\approx 29730$ N/m. ✓

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Example 8 — Cell H: exam twist, orthogonality decouples

Forecast: if the dot product through $[M]$ isn't zero, what did we do wrong?

Step 1 — mass orthogonality. $\{{\varphi }_1{\}}^T[M]\{{\varphi }_2\}=2(1\cdot 1+1.618\cdot (-0.618))=2(1-1.0)=0.$ Why this step? The parent proved distinct modes satisfy $\{{\varphi }_j{\}}^T[M]\{{\varphi }_i\}=0$; a near-zero result confirms our shapes are correct eigenvectors (see Orthogonality of mode shapes).

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

Step 3 — consistency with frequency. Generalized stiffness $K_i^{\ast }={\omega }_i^2{M}_i^{\ast }$. For mode 1 with $k=800$: ${\omega }_1^2=0.618^2\cdot 400=152.8$, so $K_1^{\ast }=152.8\times 7.236=1105.6\text{ N/m}$, and $\sqrt{K_1^{\ast }/M_1^{\ast }}=\sqrt{152.8}=12.36\text{ rad/s}$ — back to Ex 1's ${\omega }_1$. Why this step? If the orthogonal transform is correct, the decoupled single-DOF oscillator must reproduce the same frequency we found from the full 2×2 problem. Recovering $12.36\text{ rad/s}$ is the closing of the loop — proof that modal coordinates really did split one coupled system into independent oscillators.

Answer to forecast: a nonzero cross term means the shapes are wrong (or ${\omega }_i={\omega }_j$, the degenerate case of Ex 4).

Recall Verify Example 8

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

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Recall

Recall Every cell, one line

Zero-frequency mode means ::: a rigid-body (unrestrained) mode — free drift, no spring stretch. Repeated eigenvalue means ::: the mode shape is not unique; any orthonormal pair in that eigenspace works. To move a mode off the launch band you ::: stiffen the structure (raise $k$ → raise $\omega =\sqrt{k/m}$). Number of internal nodes of mode $i$ in a chain ::: $i-1$. Test that two computed shapes are correct modes ::: check $\{{\varphi }_i{\}}^T[M]\{{\varphi }_j\}=0$ for $i\ne j$.

Related: Frequency response functions shows what happens near these frequencies when forcing is applied; Structural damping mechanisms softens the resonant peaks this page located.