Normal modes — coupled oscillators, normal coordinates: Question bank

Picture the two modes before you test yourself

Everything on this page refers back to one image: two masses, three springs, walls on both sides. The left and right springs (stiffness $k$ ) tie each mass to a wall; the middle coupling spring (stiffness $k_{c}$ ) ties the two masses to each other. Displacements $x_{1}, x_{2}$ are measured from rest.

Figure — Normal modes — coupled oscillators, normal coordinates

Look at the sketch. Top row (symmetric mode): both arrows point the same way, equal length — the masses drift together, so the middle spring keeps its natural length. It never stretches, so it exerts no force: the masses feel only the wall springs, and the frequency is the lazy $ω_{1} = k / m$ . Bottom row (antisymmetric mode): the arrows point opposite ways — the middle spring is stretched and squeezed the hardest, adding an extra restoring pull, so the frequency climbs to $ω_{2} = (k + 2 k_{c}) / m$ . The blue dot marks the node at the centre: in the antisymmetric mode that point never moves.

The panel below shows how the frequencies drift as you change the coupling — read it before the edge-case questions.

True or false — justify

Two equal masses, outer springs $k$ , coupling spring $k_{c}$ , unless a line says otherwise.

In a normal mode, both masses reach their turning points at the same instant.

True — they share one frequency and (for real

A_{1}, A_{2}

) one phase, so they stop and reverse together; only the amplitudes differ. This is what "moving in fixed proportion" means.

The symmetric mode has a higher frequency than the antisymmetric mode.

False — symmetric is slower: the coupling spring never stretches, so its stiffness

k_{c}

is absent and

ω_{1}^{2} = k / m

. Antisymmetric fights that spring, giving the larger

ω_{2}^{2} = (k + 2 k_{c}) / m

If you release the system from an arbitrary configuration, the resulting motion is itself a normal mode.

False — any start

(x_{1}, x_{2})

splits into

η_{1}, η_{2}

, and both are generally non-zero, so two frequencies run at once (giving beats). It is a single mode only when the start lies along one eigenvector, killing the other

η

The number of normal modes equals the number of masses (degrees of freedom).

True — an

N

-coordinate small-oscillation system has exactly

N

normal modes, since the secular equation

det (K - ω^{2} M) = 0

is a polynomial of degree

N

ω^{2}

Making the coupling spring

k_{c}

stronger changes both normal-mode frequencies.

False — only

ω_{2}^{2} = (k + 2 k_{c}) / m

depends on

k_{c}

; the symmetric

ω_{1}^{2} = k / m

is untouched because that mode never stretches the coupling spring, no matter how stiff it is.

In the antisymmetric mode the centre point between the masses stays perfectly still.

True — with

x_{2} = - x_{1}

the midpoint of the coupling spring is a node; it neither moves nor accelerates, which is why you can imagine cutting the spring there.

If the two masses were unequal, you could still just add and subtract the equations to decouple them.

False — the neat

\frac{1}{2} (x_{1} \pm x_{2})

combinations only diagonalise the equal-mass case. For unequal masses the eigenvectors tilt, and you must solve the generalized problem

KA = ω^{2} MA

The total mechanical energy of the coupled system stays constant even while energy sloshes between the two masses.

True — no damping means total

E

is conserved; "sloshing" is a transfer between mass 1 and mass 2, not a loss. Each normal coordinate separately conserves its own energy too.

Spot the error

Each line contains a plausible-sounding but wrong statement. Reveal gives the correction.

"Since

k_{c}

appears once in

V

, the antisymmetric frequency is

ω_{2}^{2} = (k + k_{c}) / m

The antisymmetric stretch is

(x_{1} - x_{2}) = 2 x_{1}

, so the coupling spring's effective stiffness doubles:

ω_{2}^{2} = (k + 2 k_{c}) / m

. Always read it off the secular determinant, never eyeball

V

"To find the modes I solve

KA = λ A

, an ordinary eigenvalue problem."

It is the generalized eigenvalue problem

KA = ω^{2} MA

. It only looks ordinary here because

M = m 1

; with different masses the

M

cannot be dropped. See Eigenvalues and Eigenvectors.

"Each mass keeps its own natural frequency; mass 1 swings at its rate, mass 2 at its rate."

Coupling forces a shared frequency inside each mode. A single mass's actual motion is a sum of both shared-frequency modes — there is no one frequency you can pin to one mass.

"The eigenvector tells you the two masses' displacements at all times."

It tells you only their fixed ratio

A_{2} / A_{1}

; the overall size and phase are set by initial conditions. The eigenvector is a direction in coordinate space, not a full trajectory.

"Normal coordinates must have units of length, since

η = \frac{1}{2} (x_{1} + x_{2})

With the

\frac{1}{2}

chosen for

M

-orthonormality the

η_{a}

generally carry mixed units of

mass \times length

; only up to the (dimensionless here) normalisation do they look like lengths. What matters is that each obeys

\overset{η}{¨}_{a} = - ω_{a}^{2} η_{a}

"Because the system is coupled, its equations of motion can never be written as independent oscillators."

The whole point of normal coordinates is that they can — the coupling is a feature of the chosen coordinates

x_{1}, x_{2}

, not of the physics. Switch to

η_{a}

and the coupling vanishes.

"Beats happen because there are three or more masses interfering."

Beats arise from just two frequencies

ω_{1}, ω_{2}

that are close; two coupled masses already do it. Weak coupling makes

ω_{1} \approx ω_{2}

, giving a slow beat envelope.

Why questions

Why do we bother introducing normal coordinates instead of solving the coupled equations directly?

They turn one tangled system of coupled ODEs into

N

separate one-line SHM equations

\overset{η}{¨}_{a} = - ω_{a}^{2} η_{a}

, each trivially solvable. This is the payoff of the whole method — see Small Oscillations.

Why is the secular equation set to

det (K - ω^{2} M) = 0

rather than any other value?

A matrix times a nonzero vector gives zero only when the matrix is singular, i.e. its determinant vanishes. We want a non-trivial motion (

A \neq = 0

), so we demand singularity.

Why can symmetry tell you the symmetric mode is slower before doing any algebra?

In the symmetric motion the coupling spring never stretches, so it contributes no restoring force; less restoring force means a lower frequency. Symmetry pins the physics without touching the determinant.

Why do we guess

x_{j} = A_{j} cos (ω t + ϕ)

with a single

ω

for all masses?

It is a deliberate forecast that normal modes exist: we ask whether there are motions where everything shares one frequency. Substituting converts calculus (derivatives) into algebra (a matrix equation), which we can then verify has solutions.

Why does the Lagrangian route avoid the sign errors that force-balancing invites?

Energies

T

and

V

are scalars with no direction to get backwards; the Euler–Lagrange derivative then produces the correct signs automatically via the chain rule. See Lagrangian Mechanics.

Why does exciting only an eigenvector as the initial condition give a single-frequency motion forever?

An eigenvector sets one normal coordinate to zero and keeps it zero (no force feeds it), so only the other

η_{a}

oscillates — leaving a pure single-frequency mode.

Why are normal modes the bridge to crystal physics?

A crystal is just a huge chain of coupled oscillators; its normal modes become the collective vibrations we quantise into phonons. The two-mass problem is the seed of that whole subject.

Edge cases

What happens to the two frequencies as the coupling

k_{c} \to 0

Both approach

k / m

ω_{1}^{2} = k / m

and

ω_{2}^{2} \to k / m

. The masses become independent oscillators and any "beats" slow to an infinitely long, unobservable exchange — see the left edge of the frequency panel.

What happens as

k_{c} \to \infty

(a rigid rod between the masses)?

ω_{2}^{2} = (k + 2 k_{c}) / m \to \infty

: the antisymmetric mode costs infinite energy and is frozen out, so the pair moves only symmetrically as one rigid block at

ω_{1} = k / m

— the upper curve in the panel shoots off while the lower stays flat.

If you start with

x_{1} (0) = x_{2} (0) = a

and zero velocity, what motion results?

Pure symmetric mode:

η_{2} (0) = 0

stays zero forever, so both masses oscillate together at

ω_{1} = k / m

with the coupling spring untouched — no beats, no energy transfer.

If you release from

x_{1} (0) = a, x_{2} (0) = 0

, does energy ever fully reach mass 2?

Yes — with equal amplitudes in both modes the beat envelope drives

x_{1}

down to zero while

x_{2}

reaches

a

, then back. Full transfer requires equal mode amplitudes, which this symmetric-vs-antisymmetric mix provides.

What if a normal frequency comes out as

ω = 0

(a zero eigenvalue)?

That is a zero mode — not oscillation but free translation/rotation of the whole system with no restoring force. It appears when a symmetry (e.g. no wall springs) leaves a direction with no potential cost.

Can two different normal modes ever share the same frequency?

Yes — this is degeneracy, common in symmetric systems (e.g. a 2D lattice). Then any linear combination of the degenerate eigenvectors is also a valid mode, so the mode shapes are not uniquely fixed.

What if the masses are unequal (

m_{1} \neq = m_{2}

)? Does

η = \frac{1}{2} (x_{1} \pm x_{2})

still decouple them?

No — the eigenvectors are no longer simple sum/difference, and orthogonality holds with respect to

M

(

A_{a}^{T} M A_{b} = 0

), not the identity. You must solve the generalized problem honestly.

Recall One-line self-test

Name the single sentence that kills the "each mass has its own frequency" myth. ::: In any normal mode all parts share one common $ω$ ; a mass's real motion is a superposition of these shared-frequency modes, never one frequency of its own.