Modal analysis — natural frequencies, mode shapes: Exercises

This page is a self-test ladder. Each problem builds on the parent note, climbing from simple recognition to full synthesis. Every solution is hidden inside a collapsible callout — try the problem first, then reveal.

Before we start, one reminder of the tools we will keep re-using, so no symbol appears un-earned:

Definition Symbols we use everywhere below

$x_{i}$ — the displacement of mass $i$ from its rest position (metres). $\overset{x}{¨}_{i}$ is its acceleration (second time-derivative). These are the raw physical coordinates.
$[M]$ — the mass matrix. A grid of numbers where row $i$ , column $j$ tells you how the acceleration of mass $j$ feels in the equation for mass $i$ . For simple lumped masses it is diagonal (each mass only feels its own acceleration).
$[K]$ — the stiffness matrix. Row $i$ , column $j$ is the force on mass $i$ caused by a unit displacement of mass $j$ . Off-diagonal entries are non-zero when a spring connects two masses.
${ϕ_{i}}$ — a mode shape: a column of numbers giving the relative displacement of every mass while the structure rings at one pure frequency. Only ratios matter, so we are free to scale it.
$ω_{i}$ — the natural frequency (rad/s) of mode $i$ . $f_{i} = ω_{i} / (2 π)$ in Hz.
$λ$ — a shorthand eigenvalue, defined as $λ \equiv ω^{2}$ . We use it only to keep the determinant algebra tidy; at the end we take $ω = λ$ .
$det (\cdot)$ — the determinant: a single number computed from a square matrix; it is zero exactly when the matrix squashes some non-zero vector to zero. That "some non-zero vector" is what lets a mode exist.

The one master equation everything flows from (see Multi-degree-of-freedom systems): $([K] - ω^{2} [M]) {ϕ} = {0}$

Here is the picture behind that equation — the actual mass–spring chains these exercises use, so every matrix below has a physical drawing to point at:

Figure — Modal analysis — natural frequencies, mode shapes

Prerequisites worth a look: Free vibration fundamentals, Orthogonality of mode shapes, Frequency response functions.

Level 1 — Recognition

Exercise 1.1

A single mass $m$ hangs on a single spring $k$ (one degree of freedom). Write its natural frequency and say why there is only one mode.

Recall Solution

What we do: For one DOF the "matrices" are just scalars: $[M] = m$ , $[K] = k$ . The eigenvalue problem $([K] - ω^{2} [M]) ϕ = 0$ becomes $(k - ω^{2} m) ϕ = 0$ . Why: A non-zero $ϕ$ needs the bracket to vanish, so $ω^{2} = k / m$ . $ω_{n} = \frac{k}{m}$ One mode only because there is one coordinate — an $n$ -DOF system has exactly $n$ modes. Here $n = 1$ .

Exercise 1.2

For the parent note's two-DOF system the two natural frequencies were $ω_{1} \approx 0.618 k / m$ and $ω_{2} \approx 1.618 k / m$ . Which is the fundamental mode, and what does "fundamental" mean?

Recall Solution

The fundamental mode is the one with the lowest natural frequency: $ω_{1} \approx 0.618 k / m$ . Why it matters: launch energy is concentrated at low frequencies (see Launch vehicle load environments), so the lowest mode is usually the most dangerous — it is the easiest to "pluck".

Exercise 1.3

Given mode shapes ${ϕ_{1}} = [1 1.618]$ and ${ϕ_{2}} = [1 - 0.618]$ , which mode has the two masses moving in phase and which out of phase?

Recall Solution

Look at the signs inside each vector.

${ϕ_{1}}$ : both entries positive → both masses move the same direction at the same instant → in phase. Look at figure below, mode 1: both arrows point right.
${ϕ_{2}}$ : entries have opposite signs → masses move opposite directions → out of phase, with a stationary point (a node) between them.

Level 2 — Application

Exercise 2.1

Two equal masses $m$ in a line: wall— $k$ — $m_{1}$ — $k$ — $m_{2}$ — $k$ —wall (fixed at both ends), shown as the lower chain in the schematic above. Here $x_{1}$ and $x_{2}$ are the displacements of the two masses (see the global symbol list). Build $[M]$ and $[K]$ .

Recall Solution

Mass 1 (displacement $x_{1}$ ): springs to the wall and to mass 2. $m \overset{x}{¨}_{1} = - k x_{1} - k (x_{1} - x_{2}) = - 2 k x_{1} + k x_{2}$ Mass 2 (displacement $x_{2}$ ): springs to mass 1 and to the far wall. $m \overset{x}{¨}_{2} = - k (x_{2} - x_{1}) - k x_{2} = k x_{1} - 2 k x_{2}$ Reading off the coefficients of $x_{1}, x_{2}$ on the right-hand sides: $[M] = [m 0 0 m], [K] = [2 k - k - k 2 k]$ Note the symmetric $[K]$ — a general property that guarantees real frequencies (see Orthogonality of mode shapes).

Exercise 2.2

For the system of 2.1, find both natural frequencies.

Recall Solution

What we do: set $det ([K] - ω^{2} [M]) = 0$ . Substitute the shorthand $λ \equiv ω^{2}$ (from the symbol list) so the algebra is a polynomial in one letter. $det [2 k - mλ - k - k 2 k - mλ] = (2 k - mλ)^{2} - k^{2} = 0$ Why a determinant: it is zero exactly when a non-zero mode shape can survive. Why factor as a difference of squares: the left side has the exact shape $A^{2} - B^{2}$ with $A = 2 k - mλ$ and $B = k$ . We recognise that pattern because it factors instantly into $(A - B) (A + B)$ — far easier than expanding to a full quadratic and re-solving. That is the whole reason to spot it. $(2 k - mλ - k) (2 k - mλ + k) = 0 \Rightarrow (k - mλ) (3 k - mλ) = 0$ Each factor gives one root $λ = ω^{2}$ , and $ω = λ$ : $ω_{1} = \frac{k}{m}, ω_{2} = \frac{3 k}{m}$

Exercise 2.3

Find the mode shapes for 2.1.

Recall Solution

Mode 1 ( $λ_{1} = k / m$ ): first row of $([K] - λ_{1} [M]) {ϕ} = 0$ gives $(2 k - k) ϕ_{11} - k ϕ_{21} = 0 \Rightarrow k ϕ_{11} = k ϕ_{21} \Rightarrow ϕ_{21} = ϕ_{11} .$ Why we now just "set $ϕ_{11} = 1$ ": the eigenvalue equation only fixes the ratio $ϕ_{21} / ϕ_{11}$ , never the absolute size — a mode shape describes a pattern of motion, and doubling every entry is the same pattern twice as big. So we are free to pick any non-zero value for one entry; choosing $ϕ_{11} = 1$ is the tidiest. ${ϕ_{1}} = [11] — symmetric, masses move together.$

Mode 2 ( $λ_{2} = 3 k / m$ ): $(2 k - 3 k) ϕ_{12} - k ϕ_{22} = 0 \Rightarrow - k ϕ_{12} = k ϕ_{22} \Rightarrow ϕ_{22} = - ϕ_{12} .$ Again fixing the scale with $ϕ_{12} = 1$ : ${ϕ_{2}} = [1 - 1] — antisymmetric, masses move opposite, node in the middle.$

Level 3 — Analysis

Exercise 3.1

Verify mass-orthogonality of the two mode shapes from 2.3: show ${ϕ_{1}}^{T} [M] {ϕ_{2}} = 0$ .

Recall Solution

With $[M] = m I$ , ${ϕ_{1}}^{T} [M] {ϕ_{2}} = m [11] [1 - 1] = m (1 \cdot 1 + 1 \cdot (- 1)) = 0.$ Why this had to be zero: the two frequencies differ ( $k / m \neq = 3 k / m$ ), and the orthogonality theorem guarantees distinct modes are mass-orthogonal. This is what lets us decouple the equations (see Orthogonality of mode shapes).

Exercise 3.2

Compute the generalized (modal) mass $M_{i}^{*} = {ϕ_{i}}^{T} [M] {ϕ_{i}}$ and generalized stiffness $K_{i}^{*} = {ϕ_{i}}^{T} [K] {ϕ_{i}}$ for mode 2 of 2.1, then confirm $ω_{2} = K_{2}^{*} / M_{2}^{*}$ .

Recall Solution

${ϕ_{2}} = [1 - 1]$ . $M_{2}^{*} = m [1 - 1] [1 - 1] = m (1 + 1) = 2 m .$ $[K] {ϕ_{2}} = [2 k - k - k 2 k] [1 - 1] = [3 k - 3 k] .$ $K_{2}^{*} = [1 - 1] [3 k - 3 k] = 3 k + 3 k = 6 k .$ $\frac{K _{2}^{*}}{M _{2}^{*}} = \frac{6 k}{2 m} = \frac{3 k}{m} = ω_{2} . ✓$ Why this works: projecting the full matrices onto one mode collapses the system to a single scalar oscillator $M_{i}^{*} \overset{q}{¨}_{i} + K_{i}^{*} q_{i} = Q_{i}$ .

Exercise 3.3

A cantilever's first bending mode is measured at $f_{1} = 25 Hz$ . The design rule for its launch vehicle demands $f_{1} \geq 35 Hz$ (lateral). By what factor must the stiffness $k$ increase (mass unchanged) to just meet the rule?

Recall Solution

Why stiffness, not mass: $f \propto k / m$ . Raising $k$ raises $f$ ; adding mass lowers it. To raise frequency with fixed mass we increase $k$ . $\frac{f _{new}}{f _{old}} = \frac{k _{new}}{k _{old}} \Rightarrow \frac{k _{new}}{k _{old}} = (\frac{35}{25})^{2} = 1.96.$ Answer: stiffness must rise to $1.96 \times$ its current value. The frequency-vs-stiffness curve for this scaling is drawn below (see the marked point).

Level 4 — Synthesis

Exercise 4.1

A spacecraft's first mode is $f_{1} = 30 Hz$ with modal mass $M_{1}^{*} = 40 kg$ and modal damping ratio $ζ_{1} = 0.02$ . During a sine sweep the base excites this mode at exactly resonance with a modal force amplitude $Q_{0} = 50 N$ . Estimate the peak modal displacement amplitude $q_{m a x}$ .

Recall Solution

Tool: the resonant amplitude of a single-DOF oscillator (each mode is one). At resonance the response of $M^{*} \overset{q}{¨} + C^{*} \overset{q}{˙} + K^{*} q = Q_{0} sin ω t$ is $q_{m a x} = \frac{Q _{0}}{K _{1}^{*}} \cdot \frac{1}{2 ζ _{1}} = \frac{Q _{0}}{M _{1}^{*} ω _{1}^{2}} \cdot \frac{1}{2 ζ _{1}} .$ Why $1/ (2 ζ)$ : at resonance the stiffness and inertia forces cancel; only damping resists, so the amplitude is the static deflection $Q_{0} / K^{*}$ magnified by the quality factor $1/ (2 ζ)$ (see 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

Two candidate designs have first modes at $28 Hz$ (design A) and $45 Hz$ (design B). The launch sine environment has strong energy from $5$ – $40 Hz$ . Which design do you pick and why? Then state one quantitative margin.

Recall Solution

Pick B. Its $45 Hz$ fundamental lies above the $40 Hz$ upper edge of the environment, so no launch energy directly hits it — the frequency-separation (or "clearance") design philosophy. Design A at $28 Hz$ sits inside $5$ – $40 Hz$ ; it will be driven at resonance and amplify by $1/ (2 ζ)$ . Quantitative margin for B: separation $= (45 - 40) /40 = 0.125 = 12.5%$ . Typical primary-structure rules ask for $≳$ this, so B is (barely) acceptable while A is disqualified.

Level 5 — Mastery

Exercise 5.1

Design task. You have the two-DOF chain wall— $k$ — $m$ — $k$ — $m$ (free right end) from the parent note — the upper chain in the schematic near the top of this page — with $m = 2 kg$ , $k = 8000 N/m$ ; its fundamental is $ω_{1} \approx 0.618 k / m$ . (a) Compute $f_{1}$ in Hz. (b) The requirement is $f_{1} \geq 12 Hz$ . If mass is fixed, find the minimum $k$ that satisfies it. (c) Confirm your new $k$ works.

Recall Solution

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

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

(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

For that same free-end system, decouple. Given ${ϕ_{1}} = [1 1.618]$ , $[M] = m I$ , compute the generalized mass $M_{1}^{*}$ and, using $ω_{1}^{2} = (3 - 5) k / (2 m)$ , the generalized stiffness $K_{1}^{*}$ ; verify $ω_{1} = K_{1}^{*} / M_{1}^{*}$ .

Recall Solution

$M_{1}^{*} = m [1 1.618] [1 1.618] = m (1 + 1.61 8^{2}) = m (1 + 2.618) = 3.618 m .$ Because $K_{1}^{*} = ω_{1}^{2} M_{1}^{*}$ for a mode, $K_{1}^{*} = \frac{( 3 - 5 ) k}{2 m} \cdot 3.618 m = \frac{0.7639 k}{2} \cdot 3.618 = 1.3820 k .$ $\frac{K _{1}^{*}}{M _{1}^{*}} = \frac{1.3820 k}{3.618 m} = 0.3820 k / m = 0.618 k / m = ω_{1} . ✓$ Meaning: the whole coupled system, for this mode, behaves like one mass $3.618 m$ on one spring $1.382 k$ .