Modal analysis — natural frequencies, mode shapes: Question bank

This is a rapid-fire self-test for Modal Analysis. Each line is a Question ::: Answer reveal — cover the answer, commit to a response with reasoning, then check. These target the exact misconceptions the topic breeds. If any term is unfamiliar, revisit Free vibration fundamentals and Multi-degree-of-freedom systems first.

True or false — justify

A natural frequency depends on how hard you pluck the structure.

False. Natural frequency is set by

ω_{i} = K_{i}^{*} / M_{i}^{*}

— only mass and stiffness. The amplitude of plucking changes how big the response is, never how fast the free vibration oscillates.

A mode shape tells you the actual displacement of each point in metres.

False. A mode shape

{ϕ_{i}}

is a relative pattern — it gives the ratio of motion between points, not absolute size. That is why we normalize it (e.g. set the first entry to 1); the real amplitude comes from the modal coordinate

q_{i} (t)

n

-degree-of-freedom structure has exactly

n

natural frequencies.

True. The characteristic polynomial

det ([K] - ω^{2} [M]) = 0

has degree

n

ω^{2}

, so it yields

n

eigenvalues

ω_{i}^{2}

and

n

corresponding mode shapes.

Two different mode shapes are always perpendicular in the ordinary dot-product sense.

False. They are orthogonal weighted by the matrices:

{ϕ_{j}}^{T} [M] {ϕ_{i}} = 0

and

{ϕ_{j}}^{T} [K] {ϕ_{i}} = 0

. The plain dot product

{ϕ_{j}}^{T} {ϕ_{i}}

is generally not zero unless

[M]

is a scalar multiple of the identity.

Resonance means the natural frequency itself grows over time.

False. The natural frequency is fixed. Resonance means you drive the structure at that fixed frequency, so the amplitude grows large (unbounded in the undamped ideal case); the frequency stays put.

Adding stiffness always raises every natural frequency.

False (as a blanket claim). Since

ω \propto K^{*} / M^{*}

, more stiffness tends to raise frequencies, but

K^{*}

is the generalized stiffness projected onto a specific mode. Stiffening a point that a given mode barely moves (near a node) does almost nothing to that mode's frequency.

The mass matrix

[M]

is positive definite.

True. Every real degree of freedom carries positive kinetic energy for any nonzero velocity, so

{v}^{T} [M] {v} > 0

for all

{v} \neq = {0}

— that is the definition of positive definite. This is why

[M]

can be inverted safely.

The stiffness matrix

[K]

is only positive semi-definite, not definite.

True for a free (unconstrained) structure. Rigid-body motion stores no strain energy, so

{r}^{T} [K] {r} = 0

for a rigid displacement

{r}

— giving a zero eigenvalue. A fully constrained structure has

[K]

positive definite.

Modal coordinates

{q}

decouple the equations because we chose clever coordinates by luck.

False. It is not luck: orthogonality of mode shapes forces

[Φ]^{T} [M] [Φ]

and

[Φ]^{T} [K] [Φ]

to be diagonal, which is what turns the coupled system into

n

independent single-DOF oscillators.

Spot the error

A student writes

[K] {ϕ_{i}} = ω_{i}^{2} [K] {ϕ_{i}}

as the eigenvalue problem.

The right side must use the mass matrix:

[K] {ϕ_{i}} = ω_{i}^{2} [M] {ϕ_{i}}

. Writing

[K]

on both sides would force

ω_{i}^{2} = 1

for all modes, which is nonsense — mass is what supplies the inertia term

- ω^{2} M

A student cancels

sin (ω t + θ)

from the equation of motion and worries it might be zero.

We are allowed to cancel because we seek a solution valid for all time;

sin

is nonzero at almost every instant, so the bracketed factor

([K] - ω^{2} [M]) {ϕ}

must itself vanish. Cancelling is legitimate.

A student concludes: "

det ([K] - ω^{2} [M]) = 0

has the trivial solution

{ϕ} = {0}

, so that's a mode."

{ϕ} = {0}

is a solution of any homogeneous system and describes no motion at all — it is discarded. The whole point of setting the determinant to zero is to find frequencies where a nontrivial

{ϕ}

exists.

A student normalizes a mode shape and then says the frequency changed.

Scaling a mode shape by any nonzero constant leaves it an eigenvector of the same eigenvalue. Normalization changes only the numbers you write for

{ϕ_{i}}

, never

ω_{i}

A student claims the modal force is

Q_{i} = [Φ]^{T} {F}

for a single mode.

For a single mode

i

it is

Q_{i} = {ϕ_{i}}^{T} {F}

(a scalar).

[Φ]^{T} {F}

is the full vector of all modal forces at once; picking mode

i

means taking one row.

A student says damping breaks orthogonality so modal analysis is useless with damping.

Orthogonality of mode shapes (from

[M]

and

[K]

) still holds. Modal analysis stays valid; you only get a diagonal damping matrix automatically if damping is proportional (Rayleigh). See Structural damping mechanisms.

Why questions

Why must the trial solution be harmonic,

{x} = {ϕ} sin (ω t + θ)

An undamped linear system conserves energy and cannot spiral in or out, so free motion is pure sinusoid. Assuming this shape lets us replace

\overset{x}{¨}

with

- ω^{2} x

and turn a differential equation into an algebraic eigenvalue problem.

Why do we get an eigenvalue problem instead of just solving for

{ϕ}

directly?

The equation

([K] - ω^{2} [M]) {ϕ} = {0}

is homogeneous — its right side is zero. A nonzero

{ϕ}

exists only when the matrix is singular, i.e.

det = 0

. That singularity condition is the eigenvalue problem.

Why does orthogonality let us treat each mode as an independent single-DOF oscillator?

Orthogonality diagonalizes

[Φ]^{T} [M] [Φ]

and

[Φ]^{T} [K] [Φ]

, so the transformed equations have no cross-terms coupling

q_{i}

q_{j}

. Each row reads

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

— a lone spring-mass system.

Why do spacecraft engineers care about raising the fundamental frequency above the launch environment?

Launch vibration energy is concentrated in a low-frequency band. Keeping the first natural frequency above that band means the launcher cannot "pluck" a resonance, avoiding runaway amplitude. See Launch vehicle load environments.

Why is the generalized stiffness

K_{i}^{*}

a scalar even though

[K]

is a matrix?

K_{i}^{*} = {ϕ_{i}}^{T} [K] {ϕ_{i}}

sandwiches the matrix between a row and column vector, collapsing it to one number — the effective stiffness the structure feels when moving purely in mode

i

Why does the off-diagonal term in a coupling spring appear in

[K]

but the masses stay diagonal in

[M]

A spring between two masses exerts a force on each that depends on the other's displacement — that cross-influence is the off-diagonal

- k

. Each mass's inertia depends only on its own acceleration, so

[M]

stays diagonal.

Edge cases

What happens to the mode shape at a node?

A node is a point that stays still during that mode, so its entry in

{ϕ_{i}}

is (near) zero. Forcing applied exactly at a node cannot excite that mode, and stiffening there barely shifts its frequency.

What if two natural frequencies are equal,

ω_{i} = ω_{j}

The orthogonality proof relied on

(ω_{i}^{2} - ω_{j}^{2}) \neq = 0

to force

{ϕ_{j}}^{T} [M] {ϕ_{i}} = 0

. With repeated frequencies that step fails, but the degenerate subspace can still be chosen orthogonal — any combination of the two shapes is also a valid mode.

What is the "mode" corresponding to a zero natural frequency in a free-floating spacecraft?

ω = 0

is a rigid-body mode — the whole structure translates or rotates with no internal deformation, storing zero strain energy. A free spacecraft in orbit has up to six such modes (3 translation, 3 rotation).

If you excite a structure exactly between two natural frequencies, what responds?

No single mode resonates, but the response is a weighted blend of nearby modes — each contributes according to its frequency response amplitude at the driving frequency. Amplitudes stay modest because you are off every peak.

What happens to modal decoupling if the structure is nonlinear (large deflection)?

The eigenvalue problem assumes linear

[M]

and

[K]

. Under large deflections stiffness depends on displacement, mode shapes change with amplitude, and clean decoupling is lost — modal analysis becomes a local approximation only.

In the limit of infinite stiffness for one spring, what happens to that connection?

The two masses it joins become effectively rigidly linked — one degree of freedom disappears (they must move together), lowering the DOF count and eliminating the high-frequency mode that "stretched" that spring.

What does an undamped resonance predict for amplitude, and why is that unphysical?

The idealized undamped model predicts amplitude growing without bound at

ω = ω_{n}

. Real structures always have some damping, which caps the peak at a large but finite value — that finite peak is what design margins are built around.

Recall Quick self-check

One natural frequency is fixed by ::: mass and stiffness only, via $ω = K^{*} / M^{*}$ — never by amplitude. Mode shapes are orthogonal with respect to ::: the mass and stiffness matrices, not the plain dot product. A zero eigenvalue of $[K]$ signals ::: a rigid-body mode with no strain energy (free/unconstrained structure).