Normal modes — coupled oscillators, normal coordinates: Worked examples

This page is the practice arena for the parent note on normal modes. There we derived the machinery. Here we run it against every kind of situation the topic can hand you — clean modes, energy sloshing, weird masses, zero coupling, infinite coupling, a real-world word problem, and an exam trap.

Before touching numbers, let us lay out the full landscape so no case surprises you.

Two symbols will appear in the matrix, so let us name them in plain words first:

The scenario matrix

#	Case class	What is special about it	Example that hits it
A	Pure symmetric kick	initial condition is eigenvector 1 → one frequency	Ex 1
B	Pure antisymmetric kick	initial condition is eigenvector 2 → one frequency	Ex 2
C	Mixed kick (beats)	general start → superposition, energy sloshes	Ex 3
D	Weak-coupling limit $k_{c} ≪ k$	modes nearly degenerate → very slow beats	Ex 4
E	Zero coupling $k_{c} = 0$	degenerate case: coupling vanishes, masses independent	Ex 5
F	Strong / stiff-link limit $k_{c} ≫ k$	one mode races off, the link acts rigid	Ex 6
G	Unequal masses $M \neq = m 1$	generalized eigenproblem, M-orthogonality bites	Ex 7
H	Real-world word problem (units!)	two atoms / two carts, get a real frequency in Hz	Ex 8
I	Velocity kick + timing twist	struck (not pulled) start, and full-transfer time	Ex 9

Nine examples, nine cells. Note that a "kick" can be a displacement (pull a mass and release) or a velocity (strike a mass at rest) — Cell I covers the velocity case so the whole space of initial conditions is exhausted. Let's go.

Throughout, the system is the parent's two-mass chain (unless a cell says otherwise), with results we will reuse:

Case A — Pure symmetric kick (one frequency)

Worked example Both masses pulled the same way

System: $m = 1 kg$ , $k = 4 N/m$ , $k_{c} = 3 N/m$ . Kick: $x_{1} (0) = x_{2} (0) = 0.05 m$ , both at rest. Find: the motion, and the frequency you'd measure.

Forecast: Guess before reading — will you see beats, or one clean tone? (One clean tone: the kick is exactly eigenvector 1.)

Compute the normal coordinates at $t = 0$ . $η_{1} (0) = \frac{1}{2} (0.05 + 0.05) = \frac{0.10}{2}$ , $η_{2} (0) = \frac{1}{2} (0.05 - 0.05) = 0$ . Why this step? Normal coordinates are the only variables that move independently. Reading them at $t = 0$ tells us which modes are switched on.
Note $η_{2} \equiv 0$ forever. Its equation is $\overset{η}{¨}_{2} = - ω_{2}^{2} η_{2}$ with $η_{2} (0) = 0$ and $\overset{η}{˙}_{2} (0) = 0$ → stays zero. Why this step? An oscillator started at rest at the origin never leaves it. Mode 2 is dead.
Only mode 1 lives: $η_{1} (t) = \frac{0.10}{2} cos ω_{1} t$ with $ω_{1} = k / m = 4 = 2 rad/s$ . Why this step? A surviving normal coordinate obeys $\overset{η}{¨}_{1} = - ω_{1}^{2} η_{1}$ , whose zero-velocity solution is a pure cosine at that mode's own frequency.
Back-transform: $x_{1} = x_{2} = \frac{1}{2} η_{1} = 0.05 cos (2 t) m$ . Why this step? $x_{j} = \frac{1}{2} (η_{1} \pm η_{2})$ ; with $η_{2} = 0$ both masses share one motion.

Verify: Both masses move identically, so the middle spring never stretches — it should not affect $ω$ . Indeed $ω = k / m$ contains no $k_{c}$ . ✓ Units: $(N/m) / kg = s^{- 2} = s^{- 1}$ . ✓ (Cell A)

Case B — Pure antisymmetric kick (one frequency)

Worked example Masses pulled opposite ways

Same system ( $m = 1$ , $k = 4$ , $k_{c} = 3$ ). Kick: $x_{1} (0) = + 0.05$ , $x_{2} (0) = - 0.05$ , at rest. Find: the frequency.

Forecast: Faster or slower than Case A? (Faster — the coupling spring is now fighting.)

Normal coordinates: $η_{1} (0) = \frac{1}{2} (0.05 - 0.05) = 0$ , $η_{2} (0) = \frac{1}{2} (0.05 + 0.05) = \frac{0.10}{2}$ . Why this step? Same reason as A: find which mode is lit. Now it's mode 2.
Mode 1 dead, mode 2 lives: $ω_{2} = (k + 2 k_{c}) / m = (4 + 6) /1 = 10 \approx 3.162 rad/s$ . Why this step? The antisymmetric stretch doubles the coupling spring's effect: $k + 2 k_{c}$ , not $k + k_{c}$ .
Motion: $x_{1} = - x_{2} = 0.05 cos (10 t) m$ . Why this step? Back-transform $x_{j} = \frac{1}{2} (η_{1} \pm η_{2})$ with $η_{1} = 0$ gives equal-and-opposite motion at the single frequency $ω_{2}$ .

Verify: $ω_{2} > ω_{1}$ ( $3.162 > 2$ ) ✓ — matches the mnemonic SLOW=Same, FAST=Fight. The factor-2 trap is dodged by using $k + 2 k_{c}$ . ✓ (Cell B)

Case C — Mixed kick → beats

Worked example Pull one mass only

Same system ( $m = 1$ , $k = 4$ , $k_{c} = 3$ ). Kick: $x_{1} (0) = a = 0.05$ , $x_{2} (0) = 0$ , at rest. Find: $x_{1} (t)$ , and describe the motion.

Forecast: Will mass 2, starting still, stay still? (No — energy floods across; you get beats.)

Normal coords: $η_{1} (0) = \frac{a}{2}$ , $η_{2} (0) = \frac{a}{2}$ — both modes lit, equally. Why this step? A one-sided pull is a 50/50 mix of symmetric and antisymmetric; that mix is what produces sloshing.
Evolve each: $η_{1} = \frac{a}{2} cos ω_{1} t$ , $η_{2} = \frac{a}{2} cos ω_{2} t$ , with $ω_{1} = 2$ , $ω_{2} = 10$ . Why this step? Each normal coordinate is an independent SHM; started at rest it just cosines at its own frequency, and the two frequencies now differ.
Back-transform & use the sum-to-product identity: $x_{1} = \frac{a}{2} (cos ω_{1} t + cos ω_{2} t) = a cos (\frac{ω _{2} - ω _{1}}{2} t) cos (\frac{ω _{1} + ω _{2}}{2} t)$ Why this step? The identity splits it into a fast carrier (average frequency) inside a slow envelope (half the difference) — that is exactly what beats are.
Envelope frequency: $\frac{ω _{2} - ω _{1}}{2} = \frac{3.162 - 2}{2} \approx 0.581 rad/s$ . Why this step? This slow rate is what your eye actually sees as the energy "sloshing" back and forth; it sets how long a full transfer takes (used again in Case I).

The figure below plots this exact motion. It has time $t$ (s) on the horizontal axis and $x_{1}$ (in units of $a$ ) on the vertical axis, with a legend naming each curve. Look at the magenta curve labelled "x1(t) full motion": it is a fast wiggle trapped inside the violet dashed envelope labelled "slow envelope", equal to $\pm a cos (\frac{ω _{2} - ω _{1}}{2} t)$ . Where the envelope pinches to zero (the orange dotted line), mass 1 falls quiet and all the energy has crossed to mass 2 — that is the beat.

Verify: At $t = 0$ , $x_{1} = a \cdot 1 \cdot 1 = a$ ✓ and $x_{2} = \frac{a}{2} (cos ω_{1} t - cos ω_{2} t) = 0$ ✓. Energy periodically leaves mass 1 entirely (envelope hits zero). ✓ (Cell C)

Case D — Weak coupling ( $k_{c} ≪ k$ ): slow beats

Worked example Loosely tied swings

System: $m = 1$ , $k = 100$ , $k_{c} = 1$ . One-sided pull as in Case C. Find: how the beat period compares to a single swing period.

Forecast: Fast energy exchange or slow? (Very slow — weak link means the modes are nearly the same pitch.)

Frequencies: $ω_{1} = 100 = 10$ , $ω_{2} = 102 \approx 10.0995$ . Why this step? We need both eigenfrequencies to form their difference; with tiny $k_{c}$ they almost coincide, which is the whole point of this limit.
Envelope frequency: $\frac{ω _{2} - ω _{1}}{2} \approx \frac{0.0995}{2} \approx 0.0498 rad/s$ . Why this step? The beat rate is the frequency difference; tiny difference ⇒ tiny beat rate ⇒ long slosh time.
Beat period $T_{beat} = \frac{2 π}{∣ ω _{2} - ω _{1} ∣} = \frac{2 π}{0.0995} \approx 63.1 s$ , versus a single swing period $\approx \frac{2 π}{10} \approx 0.628 s$ . Why this step? A full slosh over and back takes one period of the envelope, and the envelope oscillates at the difference frequency — so dividing $2 π$ by $∣ ω_{2} - ω_{1} ∣$ turns that rate into an actual time to compare against a single swing.

Verify: Beat period $≫$ swing period ( $63 ≫ 0.63$ , ratio $\approx 100$ ) — weak coupling ⇒ many fast swings per slow slosh, the hallmark of loosely coupled oscillators. ✓ (Cell D)

Case E — Zero coupling ( $k_{c} = 0$ ): degenerate limit

Worked example Cut the middle spring

System: $m = 1$ , $k = 4$ , $k_{c} = 0$ . Kick: $x_{1} (0) = 0.05$ , $x_{2} (0) = 0$ , at rest. Find: what each mass does.

Forecast: With no link, does energy still cross to mass 2? (No — they're independent now.)

Frequencies collapse together: $ω_{1}^{2} = k / m = 4$ , $ω_{2}^{2} = (k + 0) / m = 4$ . Both $= 2 rad/s$ — degenerate. Why this step? With $k_{c} = 0$ the two modes have the same frequency; the system is no longer truly coupled.
Beat envelope frequency $= \frac{ω _{2} - ω _{1}}{2} = 0$ → the envelope never dips. Why this step? Zero frequency difference means no sloshing; mass 2 never gets energy.
Motion: $x_{1} = 0.05 cos (2 t)$ , $x_{2} = 0$ for all time. Why this step? Once decoupled, mass 2 obeys $m \overset{x}{¨}_{2} = - k x_{2}$ with $x_{2} (0) = \overset{x}{˙}_{2} (0) = 0$ , whose only solution is $x_{2} \equiv 0$ ; mass 1 is left as a lone cosine oscillator at $ω_{1}$ .

Verify: $x_{2}$ never departs from $0$ — the equations decoupled the instant $k_{c} \to 0$ : $m \overset{x}{¨}_{2} = - k x_{2}$ with $x_{2} (0) = \overset{x}{˙}_{2} (0) = 0$ . ✓ This is the sanity anchor: zero coupling must give independent SHM. (Cell E)

Case F — Strong coupling ( $k_{c} ≫ k$ ): rigid-link limit

Worked example Weld the masses together

System: $m = 1$ , $k = 4$ , $k_{c} = 1000$ . Find: the two frequencies and what the fast mode "means".

Forecast: As $k_{c} \to \infty$ , what happens to $ω_{2}$ ? (It runs to infinity — the antisymmetric motion becomes impossibly stiff.)

Symmetric mode is untouched: $ω_{1} = k / m = 2 rad/s$ , no matter how large $k_{c}$ is. Why this step? In the symmetric mode the coupling spring never stretches, so its stiffness is irrelevant.
Antisymmetric mode: $ω_{2} = (k + 2 k_{c}) / m = 2004 \approx 44.77 rad/s$ . Why this step? Any relative motion of the masses must stretch the very stiff link, costing enormous energy — hence a very high frequency.
Physical reading: for slow motions only $ω_{1}$ is excitable; the pair moves as one rigid block of mass $2 m$ on the outer springs. The stiff mode "freezes out". Why this step? Because $ω_{2}$ is so far above $ω_{1}$ , any gentle, slow driving can only stir the symmetric mode — the antisymmetric one needs frequencies too high to reach, so effectively the two masses lock into a single rigid object and only mode 1 remains observable.

Verify: $ω_{1}$ stayed at $2$ (independent of $k_{c}$ ) ✓; $ω_{2} ≫ ω_{1}$ ( $44.8 ≫ 2$ ) ✓. This freeze-out is exactly how high-frequency optical branches separate from acoustic ones in Phonons and Lattice Vibrations. (Cell F)

Case G — Unequal masses: the generalized eigenproblem

Worked example Heavy mass, light mass

System: $m_{1} = 1$ , $m_{2} = 2$ (kg), each tied to a wall by spring $k = 6$ , coupled by $k_{c} = 6$ (N/m). Now $M = (1002)$ is not $m 1$ . Find: the two frequencies.

Forecast: Will the eigenvectors still be $(1, 1)$ and $(1, - 1)$ ? (No — unequal masses tilt the modes; the simple symmetric/antisymmetric shapes break.)

Write the matrices. Stiffness (from the potential energy $V$ , via $- \partial V / \partial x$ ): $K = (k + k_{c} - k_{c} - k_{c} k + k_{c}) = (12 - 6 - 6 12)$ Why this step? $K$ comes from potential energy and does not depend on the masses.
Solve the generalized secular equation $det (K - ω^{2} M) = 0$ : $det (12 - ω^{2} - 6 - 6 12 - 2 ω^{2}) = (12 - ω^{2}) (12 - 2 ω^{2}) - 36 = 0.$ Why this step? It's generalized — the $ω^{2}$ multiplies $M$ , not the identity. This is the trap flagged in the parent's mistake box.
Expand: $2 ω^{4} - 36 ω^{2} + 108 = 0 \Rightarrow ω^{4} - 18 ω^{2} + 54 = 0$ . Quadratic in $ω^{2}$ : $ω^{2} = \frac{18 \pm 324 - 216}{2} = 9 \pm 27 = 9 \pm 33 .$ So $ω^{2} \approx 14.196$ or $3.804$ , i.e. $ω \approx 3.768$ and $ω \approx 1.950 rad/s$ . Why this step? Multiplying out the determinant turns the matrix condition into an ordinary polynomial; treating $ω^{2}$ as one unknown makes it a plain quadratic we can solve with the standard formula, and each positive root is a squared mode frequency.

Verify: Both $ω^{2} > 0$ (stable) ✓. Product of roots $= (9)^{2} - (33)^{2} = 81 - 27 = 54 = det K / det M = 108/2$ ✓. The eigenvectors are not $(1, \pm 1)$ and are orthogonal only with respect to $M$ — see Eigenvalues and Eigenvectors. (Cell G)

Case H — Real-world word problem (units matter)

Worked example Two coupled pendulum-carts

Two carts, each $m = 0.50 kg$ , each anchored to a wall by a spring $k = 8.0 N/m$ and coupled together by $k_{c} = 2.0 N/m$ . Find the two natural frequencies in hertz.

Forecast: Both a few Hz? (Yes — light carts, moderate springs.)

Symmetric: $ω_{1} = k / m = 8.0/0.50 = 16 = 4.0 rad/s$ . Why this step? Use the mode where the link sleeps.
Antisymmetric: $ω_{2} = (k + 2 k_{c}) / m = (8.0 + 4.0) /0.50 = 24 \approx 4.899 rad/s$ . Why this step? The opposite-way motion works the coupling spring, adding $2 k_{c}$ to the effective stiffness.
Convert to Hz using $f = ω /2 π$ : $f_{1} = 4.0/ (2 π) \approx 0.637 Hz$ , $f_{2} \approx 4.899/ (2 π) \approx 0.780 Hz$ . Why this step? Real instruments read cycles per second, not rad/s; dividing angular frequency by $2 π$ (radians per cycle) converts to hertz.

Verify: Units: $(N/m) / kg = s^{- 1}$ ✓; dividing by $2 π$ (rad/cycle) gives Hz ✓. $f_{2} > f_{1}$ ✓. (Cell H)

Case I — Velocity kick + full-transfer timing

Worked example Strike one mass, then time the hand-off

System ( $m = 1$ , $k = 4$ , $k_{c} = 3$ ). Now the kick is a velocity, not a pull: both masses sit at equilibrium, and mass 1 is struck so that $\overset{x}{˙}_{1} (0) = v_{0} = 0.30 m/s$ , with $x_{1} (0) = x_{2} (0) = 0$ and $\overset{x}{˙}_{2} (0) = 0$ . Find: (a) $x_{1} (t)$ ; (b) how long until mass 1 is momentarily quiet and mass 2 carries the full swing.

Forecast: With zero initial displacement, do we still get beats? (Yes — a velocity kick also excites both modes; you now get sine beats instead of cosine ones.)

Normal coordinates and their velocities at $t = 0$ . $η_{1} (0) = η_{2} (0) = 0$ ; $\overset{η}{˙}_{1} (0) = \frac{1}{2} (\overset{x}{˙}_{1} + \overset{x}{˙}_{2}) = \frac{v _{0}}{2}$ , $\overset{η}{˙}_{2} (0) = \frac{1}{2} (\overset{x}{˙}_{1} - \overset{x}{˙}_{2}) = \frac{v _{0}}{2}$ . Why this step? An SHM started at the origin with speed is a sine, not a cosine — so we must read the initial velocities, not positions, to set each mode's amplitude.
Evolve each mode as a sine. For $\overset{η}{¨}_{a} = - ω_{a}^{2} η_{a}$ with $η_{a} (0) = 0$ , $\overset{η}{˙}_{a} (0) = u$ : $η_{a} (t) = \frac{u}{ω _{a}} sin ω_{a} t$ . So $η_{1} = \frac{v _{0}}{2 ω _{1}} sin ω_{1} t$ , $η_{2} = \frac{v _{0}}{2 ω _{2}} sin ω_{2} t$ , with $ω_{1} = 2$ , $ω_{2} = 10$ . Why this step? The sine solution is the one matching "at the origin, moving"; dividing by $ω_{a}$ keeps the units (a velocity over a frequency is a length).
Back-transform: $x_{1} = \frac{1}{2} (η_{1} + η_{2}) = \frac{v _{0}}{2} (\frac{s i n ω _{1} t}{ω _{1}} + \frac{s i n ω _{2} t}{ω _{2}})$ . Why this step? Same reconstruction rule as always; the two nearly-equal frequencies again make a slow beat, now on sine waves.
Timing the hand-off. The envelope of this near-beat first pinches at $t = \frac{π}{ω _{2} - ω _{1}} = \frac{π}{10 - 2} \approx 2.703 s$ . Why this step? Full energy transfer happens when the two modes drift half a cycle out of step — exactly the same $\frac{π}{ω _{2} - ω _{1}}$ timing as a displacement beat, since the timing depends only on the frequency difference, not on how you kicked.

Verify: At $t = 0$ : $x_{1} = 0$ ✓ (matches struck-from-rest position) and $\overset{x}{˙}_{1} (0) = \frac{v _{0}}{2} (cos 0 + cos 0) = v_{0}$ ✓. Transfer time $= π / (10 - 2) \approx 2.703 s$ , and this equals half the beat period $T_{beat} /2$ with $T_{beat} = 2 π / (10 - 2) \approx 5.406 s$ ✓. So displacement kicks and velocity kicks share the same slosh clock. (Cell I)

Recall Quick self-test across the matrix

Which cell has $ω_{1} = ω_{2}$ ? ::: Cell E — zero coupling, the modes are degenerate. Which cell has $ω_{2} \to \infty$ as you tighten the link? ::: Cell F — strong coupling, antisymmetric mode freezes out. Why are the mode shapes NOT $(1, \pm 1)$ in Cell G? ::: Because unequal masses make it a generalized eigenproblem; the modes are eigenvectors that are orthogonal with respect to $M$ , not the identity, so the tidy symmetric/antisymmetric shapes no longer hold. Does a velocity kick still produce beats? ::: Yes — Cell I; it excites both modes as sines, giving sine beats with the same slosh timing $π / (ω_{2} - ω_{1})$ .

Deep-dive built on Lagrangian Mechanics (the EOMs), Small Oscillations (why linearizing is legal), and Eigenvalues and Eigenvectors (the modes). For the intuitive story see the Hinglish companion.