The Gold Standard of Vibrations | Following Energy Between Two Freely Coupled Masses

Use two equal masses m. Each mass connects to a fixed support through its own spring of stiffness k. A third spring, stiffness κ, connects the masses. Displacements x₁ and x₂ are measured from the equilibrium configuration in the same positive direction.

Assume linear massless springs, negligible dissipation and no external forcing after the initial state is prepared. The model has two independent position coordinates. Its energy includes both masses and all three springs.

Drawing the coupling spring matters. If it is omitted from the energy account while retained in the force equations, the calculation will appear to gain or lose energy mysteriously. The inconsistency belongs to the chosen boundary, not the physical model.

We will start with the first mass displaced and the second at equilibrium, both initially at rest. That state excites more than one normal mode. It should not be confused with an externally driven periodic response or a system already prepared in a single mode.

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The first ground spring gives force −kx₁. The coupling spring acts according to relative displacement, giving −κ(x₁ − x₂) on the first mass. Thus m x₁” = −kx₁ − κ(x₁ − x₂).

On the second mass, the coupling force is opposite: +κ(x₁ − x₂). Its equation is m x₂” = −kx₂ + κ(x₁ − x₂). Internal forces are equal and opposite, but their work on the separate moving masses need not cancel instantaneously.

Check a simple state. If x₁ = x₂, the coupling spring has no relative extension and supplies no coupling force. If x₁ is positive and x₂ is zero, it pulls the first back and the second forward.

Those sign checks are worth doing before solving the equations. A wrong sign can produce growing motion or an impossible energy account while still leaving a page of apparently orderly algebra. The force expressions explain what each term means physically.

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Define q = (x₁ + x₂)/2 and d = (x₁ − x₂)/2. Then x₁ = q + d and x₂ = q − d. These are invertible changes of coordinates, not new physical masses.

Adding the force equations gives q” + (k/m)q = 0. Subtracting them gives d” + [(k + 2κ)/m]d = 0. The new coordinates therefore oscillate independently within this ideal linear symmetric model.

The q motion has both masses moving together. The d motion has them moving oppositely. Their angular frequencies are ωs = √(k/m) and ωa = √[(k + 2κ)/m]. The subscripts identify the symmetric and antisymmetric patterns.

A normal mode is more than a frequency number. It includes the relative motion of the components. Seeing two peaks in a measurement is not by itself evidence that the assumed masses, symmetry or mode shapes describe the real apparatus. Here those properties are explicit parts of the teaching contract.

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Take m = 1 kg, k = 100 N/m and κ = 22 N/m. Then ωs = 10 rad/s and ωa = 12 rad/s. Their average angular frequency is 11 rad/s, and half their difference is 1 rad/s.

These values are chosen to make several later states exactly checkable. They are not claimed typical values or design targets. The arithmetic convenience lets the learner verify the energy transfer at specific times without relying only on a numerical plot.

The coupling changes the antisymmetric frequency because opposite motion stretches the connecting spring. In symmetric motion that spring remains unextended, so its stiffness does not enter ωs in this particular topology.

Keep the supports visible in that explanation. The system is not a pair of otherwise free masses joined by one spring, which would have a different rigid-motion mode and different frequencies. A model’s topology is part of the formula, even when two sketches both contain two masses.

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Let x₁(0) = A, x₂(0) = 0 and both initial velocities be zero. Use A = 0.020 m for numerical energies. The normal coordinates begin at q(0) = A/2 and d(0) = A/2, with zero initial derivatives.

Therefore q(t) = (A/2)cos(10t) and d(t) = (A/2)cos(12t). Transforming back gives x₁ = (A/2)[cos(10t) + cos(12t)] and x₂ = (A/2)[cos(10t) − cos(12t)].

Check t = 0 directly. Both cosines equal one, so x₁ = A and x₂ = zero. Their derivatives vanish because the sine factors are zero. A solution that gets the frequencies right but misses the initial state answers another physical preparation.

The initial condition also explains why two modes appear with equal coordinate amplitudes. That equality follows from this displacement preparation. It is not a universal property of coupled oscillators or a condition every measured signal must satisfy.

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Apply the sum-to-product identities. The first motion is x₁ = A cos(11t)cos(t). The second is x₂ = A sin(11t)sin(t).

The slowly varying magnitude factors are A|cos(t)| and A|sin(t)|. The first envelope shrinks as the second grows over the interval from zero to π/2 seconds. Later their roles reverse.

This is a statement about the exact signal representation. A zero displacement can come from the fast oscillation factor, from the envelope factor or from both. Those cases should not be treated as identical evidence about stored energy.

The parent Vibrations guide introduces superposition and beats at a measurement point. Here we go further by deriving the motions from a coupled force model and accounting for their mechanical energy. A visually similar envelope does not automatically supply that physical explanation unless the model and energy pathways have also been established.

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The two masses have kinetic energies m v₁²/2 and m v₂²/2. Their ground springs store kx₁²/2 and kx₂²/2. The coupling spring stores κ(x₁ − x₂)²/2.

Total energy is therefore E = (m/2)(v₁² + v₂²) + (k/2)(x₁² + x₂²) + (κ/2)(x₁ − x₂)². Every term has units of joules.

At the initial state, velocities vanish, x₁ = A and x₂ = zero. Thus E₀ = (k + κ)A²/2. With the chosen numbers, E₀ = 61 × 0.020² = 0.0244 J.

The squared displacement also provides a unit-independent ratio check. Halving A at unchanged parameters quarters each initial spring-energy contribution. A percentage change in amplitude should therefore not be copied directly into an energy percentage. The coupling spring initially stores 0.0044 J, while the first ground spring stores 0.0200 J. The second mass and its ground spring initially carry no energy in this account. Saying that all energy begins in the first oscillator would omit the connecting spring unless the boundary convention was defined differently.

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Differentiate the total energy. The kinetic derivatives are m v₁x₁” + m v₂x₂”. The ground-spring derivatives are kx₁v₁ + kx₂v₂, and the coupling derivative is κ(x₁ − x₂)(v₁ − v₂).

Substitute the two force balances. Terms involving each ground spring cancel. The remaining coupling-work terms cancel the derivative of coupling energy. Hence dE/dt = 0.

This proof establishes conservation at every time within the model, not merely at a few convenient snapshots. Numerical agreement at selected times is a useful independent check, but it should not replace the underlying balance when the equations permit a direct proof.

If damping or an external drive is later added, the derivative changes. A preserved-looking curve should not be used to declare conservation without examining those new terms. Conversely, local amplitude change does not contradict the proof, because the proof concerns the complete energy account rather than one component’s displacement.

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Define E₁ = m v₁²/2 + kx₁²/2 and E₂ = m v₂²/2 + kx₂²/2. Keep Ec = κ(x₁ − x₂)²/2 separately, so E = E₁ + E₂ + Ec.

The local derivatives are dE₁/dt = −κ(x₁ − x₂)v₁ and dE₂/dt = +κ(x₁ − x₂)v₂. Their sum is generally not zero because the coupling spring can store or release energy.

Adding dEc/dt restores the zero total derivative. This explains why E₁ + E₂ can change even though the complete system is conservative. A changing two-oscillator subtotal does not imply energy has left the three-spring system.

Other bookkeeping conventions can assign part of the coupling energy to each side. Such a convention must be stated and used consistently. The total remains the same; the numerical meaning of each local share can change. A claim of complete transfer therefore needs a declared boundary, not only an appealing picture of one mass becoming quiet.

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At t = π/4 seconds, the exact motions give x₁ = −A/2 and x₂ = +A/2. Differentiating the product forms gives v₁ = −5A and v₂ = −5A, in metres per second when A is in metres.

With A = 0.020 m, both speeds are 0.100 m/s. Each mass has kinetic energy 0.005 J and its ground spring stores 0.005 J. Thus E₁ = E₂ = 0.0100 J.

The relative displacement is −A, so Ec = 0.0044 J. The complete total is 0.0100 + 0.0100 + 0.0044 = 0.0244 J, matching the initial value.

The equal velocities do not imply equal displacements, nor does their common direction make the coupling spring unextended. Coupling extension depends on position difference. This state is useful because it forces the learner to keep displacement, velocity and spring extension as separate quantities while checking one consistent physical account.

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At t = π/2 seconds, cos(t) is zero and sin(t) is one. The fast factors give x₁ = zero and x₂ = −A. Both velocities are also zero for these chosen frequencies.

The first local energy E₁ is zero. The second is kA²/2 = 0.0200 J. The coupling spring still stores κA²/2 = 0.0044 J. Total energy remains 0.0244 J.

The local ground-spring energy that initially belonged on the first side now appears on the second side, while the coupling contribution has the same magnitude as at the start. This is an exact exchanged configuration for our deliberately commensurate model.

At t = π seconds, x₁ returns to +A, x₂ to zero and both velocities to zero. The initial state is restored. Recurrence is entirely compatible with the conservative equations. It would be inconsistent with interpreting the initial local envelope decrease as irreversible loss of the whole system’s mechanical energy.

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The values below use the same A, masses and spring constants throughout. Local energies exclude the coupling spring, which remains an explicit third term.

These snapshots do not imply Ec is constant at every intermediate time. Its expression depends on the changing relative displacement, and other times give different values. The conservation proof supplies the all-time statement.

A useful independent exercise is to choose a time absent from the table and calculate all five position, velocity and energy ingredients directly. A table is evidence of specific calculations; the equations explain the continuous relationships behind them.

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At t = π/22 seconds, the fast factor cos(11t) is zero, so x₁ = zero. But differentiating x₁ gives v₁ = −11A cos(π/22), which is nonzero and large relative to the chosen displacement scale.

Thus the first mass is passing through equilibrium with substantial kinetic energy. A displacement photograph showing it at equilibrium would not reveal that energy. It would look locally centred while moving rapidly.

This is different from the t = π/2 exchanged configuration, where both x₁ and v₁ vanish. To establish zero local mechanical energy under our definition, both terms in E₁ must vanish.

The distinction is elementary but consequential. A zero on a displacement trace does not mean zero speed; a small displacement amplitude in one frequency component does not automatically mean no other motion; and a quiet-looking point does not define the energy of an entire coupled system. State which quantity has actually become small.

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In q and d coordinates, total energy becomes E = m(q’² + d’²) + kq² + (k + 2κ)d². There are two separate conserved mode contributions: Es = mq’² + kq² and Ea = md’² + (k + 2κ)d².

For our initial q = d = A/2 and zero derivatives, Es = kA²/4 = 0.0100 J. Ea = (k + 2κ)A²/4 = 0.0144 J. They sum to 0.0244 J and each stays constant.

There is no contradiction between constant mode energies and changing local energies. Modes are patterns spanning both masses, while E₁ and E₂ refer to physical components. They partition the same total in different coordinates.

This is the deeper lesson of the change of variables. A coordinate system can make equations independent without turning the original parts into isolated physical objects. The mode description explains the superposition; the local description explains where motion and stored energy appear at a given time.

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Set x₁(0) = x₂(0) = A with both velocities zero. Then d starts at zero with zero derivative and remains zero. Both masses move as A cos(ωs t), and the coupling spring remains unextended.

There is only one mode in this preparation, so the particular slow exchange envelope derived from two modes does not appear. The system is still coupled in its equations, but its initial state selects a motion that does not stretch the connecting spring.

For the antisymmetric preparation x₁(0) = A and x₂(0) = −A, q remains zero and both masses oscillate oppositely at ωa. The coupling spring now participates throughout the cycle.

A useful transfer check asks what changed between these cases and the original one-mass displacement. The masses and springs did not change; the initial condition did. Observing a different trace does not always require a changed physical parameter. State preparation and parameter identification answer different questions.

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For general positive k, κ and m, the two frequencies remain ωs = √(k/m) and ωa = √[(k + 2κ)/m]. Their difference controls the slow envelope scale for the original one-mass displacement preparation.

When κ becomes small compared with k, the frequencies approach one another. The slow envelope variation takes longer. At κ = zero they coincide, and the second mass remains still in this initial condition because no physical interaction transfers motion to it.

That limiting case is an important check on the formulas. The difference-of-cosines expression for x₂ becomes zero when the two frequencies are identical. A proposed solution predicting exchange with no connecting force would fail this basic test.

A larger coupling changes both frequency separation and the coupling-energy share. It should not be described only as making energy transfer faster while leaving every other property fixed. The complete force and energy account changes with κ, even when the topology and initial displacement remain the same.

The simple q,d separation also relies on equal ground-spring stiffnesses. Let those stiffnesses change to k + ε and k − ε while masses remain equal. Adding and subtracting the new force equations gives m q” + kq + εd = 0 and m d” + (k + 2κ)d + εq = 0.

The old coordinates are now coupled to each other through ε. Their previous separate frequencies and conserved mode-energy expressions cannot be carried over unchanged. The physical total energy still has a valid conservative account when all stiffnesses are positive, but finding its normal coordinates is a different calculation.

This is a useful example of a limited symmetry argument. The method worked because the original system had a particular symmetry, not because sum and difference coordinates always decouple any pair of oscillators. When a condition changes, return to the force balances before trusting a familiar mode formula.

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The exact exchanged configuration at π/2 used our chosen frequencies 10 and 12. For general frequencies, let Ω = (ωa + ωs)/2 and δ = (ωa − ωs)/2. Then x₁ = A cos(Ωt)cos(δt).

At the first envelope node δt = π/2, displacement x₁ is zero. But its velocity is −Aδ cos(Ωt), generally nonzero. Hence the first local energy can retain a kinetic contribution at that node.

Our numerical choice also made cos(Ωt) zero at the node, giving exact local rest. That extra phase coincidence should not be assumed for every coupling strength. A near-complete envelope exchange can be a useful approximation without being an exact all-state transfer.

This example teaches how to check a visual description against the full state. An envelope explains the slowly varying displacement magnitude. Energy also depends on derivatives and on the chosen partition of the coupling term. Reconstruct both before turning a graphical node into a stronger physical claim.

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At that general first envelope node, E₁ = mA²δ² cos²(Ωt)/2 because x₁ = zero. Its initial complete-system energy is (k + κ)A²/2.

The leftover fraction E₁/E₀ is therefore mδ² cos²(Ωt)/(k + κ), bounded above by mδ²/(k + κ). The bound follows from cos² being at most one, without needing the precise fast phase.

This is a conditional mathematical bound, not a guarantee about a real sensor or a damped installation. It explains why weak coupling can make a visual exchange look almost complete: δ can be small, limiting the remaining local kinetic contribution at an envelope node.

The amplitude A cancels from the fraction. Doubling the initial displacement multiplies all model energies by four while preserving this relative bound. Keep that scaling separate from the absolute energy, which still depends strongly on A. A small percentage and a small number of joules are different statements.

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Now add viscous damping forces −cv₁ and −cv₂ to the two masses, with c nonnegative. The same complete energy expression has derivative dE/dt = −c(v₁² + v₂²).

The total mechanical energy is nonincreasing. Coupling can still redistribute motion, but the complete account now also loses energy to the excluded dissipative degrees of freedom. Both mechanisms can appear together in one record.

If a damper also connects the two masses, its dissipated power is proportional to (v₁ − v₂)², adding another nonpositive term. A common damping coefficient cannot be copied between these different placements without changing the force model.

The key observation is that local amplitude behaviour alone cannot separate exchange and dissipation. The energy derivative identifies what the specified damping terms do. An actual identification needs appropriate input, motion and parameter evidence rather than an envelope interpreted in isolation. The original undamped solution no longer supplies the exact damped trajectory after these terms are introduced.

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The existing guide on a driven primary mass remaining still while its attachment vibrates investigates a particular externally driven frequency response. Its cancellation condition concerns an input, an output and a periodic component.

Our present model has no continuing external drive. Its motion follows from an initial state containing two natural modes. The local quiet intervals and recurrence therefore answer a different dynamical question.

Both cases warn against treating one small output as evidence that the entire system is motionless. Their force balances and energy sources differ, so one formula cannot be substituted for the other merely because a mass appears quiet.

For a learner, label the source of energy in each diagram. In the free example it is the prepared initial state. In a driven example the external input can do work over time. That boundary decision should come before comparing amplitudes or interpreting a response zero.

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One displacement trace can reveal a local pattern, but it does not directly supply both velocities, all spring extensions and the complete mechanical energy. Those quantities need appropriate evidence or a justified model relationship.

For our paper system, velocities can be differentiated analytically from the supplied exact motions. In measured records, numerical differentiation can amplify noise and depends on acquisition and processing choices. A smoothed derivative is another processed quantity with its own interpretation.

A reproducible comparison should identify measurement location, direction, units, timing and the model used to infer any unmeasured state. Relative timing between the two mass records matters because the coupling extension uses their simultaneous difference.

The guide on unknown driving force and damping identification examines a different inverse ambiguity. Use it when an observed motion is being asked to establish a damping value without the required input information. Here the primary lesson is to preserve the complete coupled energy boundary.

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Consider the pure symmetric mode x₁ = x₂ = A cos(10t), whose true coupling extension is zero. Suppose the second record is accidentally shifted by a delay h before subtraction. The computed difference becomes A[cos(10t) − cos(10t − 10h)].

That difference is generally nonzero even though the actual coupling spring is unextended. A small time shift can therefore create an apparent coupling-energy signal from two correct but misaligned local traces.

For small h, the mismatch is approximately h times the local velocity, up to the sign convention used for the delay. Its size depends on both timing error and rate of motion. Merely matching displacement units does not ensure a valid simultaneous energy calculation.

This is an original measurement-model check, not a claim about any particular instrument. It shows why state reconstruction needs a common time basis. Before explaining an unexpected energy subtotal through new physics, verify the relationship between the supplied records and the simultaneous quantities in the equation.

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Keep m = 1 kg but multiply both spring stiffnesses by four: k = 400 N/m and κ = 88 N/m. The two frequencies become 20 and 24 rad/s. With the same initial displacement A, the entire motion follows the original trajectory at twice the time rate.

The exact exchanged configuration now occurs at t = π/4 seconds, half the original time. Initial complete energy is (400 + 88)A²/2, four times the original 0.0244 J when A remains 0.020 m.

Now halve A as well. All displacements halve, velocities return to the original magnitudes at corresponding scaled states because the frequency doubled, and the initial energy returns to 0.0244 J.

This changed problem tests several relationships together. Stiffness sets frequency through a square root, energy through stiffness times displacement squared, and velocity through amplitude times frequency. A learner who can reconcile those changes has moved beyond memorising one beat period or one energy table.

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In the scaled model with A = 0.010 m, the exchanged state has second ground-spring energy 400 × 0.010²/2 = 0.0200 J. Coupling energy is 88 × 0.010²/2 = 0.0044 J, while E₁ is zero.

The complete total again equals 0.0244 J. This matches the initial scaled state and confirms the combined amplitude-and-stiffness transformation through direct arithmetic.

At the corresponding quarter-interval state, displacements are −0.005 and +0.005 m, with both velocities −0.100 m/s. Each local energy is 0.0100 J, and coupling energy is 0.0044 J, reproducing the original energy partition at a faster physical time.

Explain why equal energy tables do not imply identical motion records. The displacement scale and elapsed times differ. Energy compresses information about the state, and matching one scalar summary does not identify every coordinate or parameter. The same caution applies when comparing two real systems through one reported energy value.

Another independent scaling check changes every mass and every spring stiffness by the same positive factor λ while retaining the original displacement initial condition. The ratios setting both frequencies remain unchanged, so the displacement and velocity trajectories are identical.

Every kinetic and spring-energy term, however, is multiplied by λ. A system with doubled masses and doubled stiffnesses can therefore have the same motion traces but twice the mechanical energy for the same initial displacement. Motion agreement alone does not establish an absolute energy inventory without parameter information.

Conversely, changing only A multiplies displacement and velocity by the amplitude factor while multiplying energy by its square. These two transformations produce different relationships between traces and energy. A learner should be able to identify which parameter transformation was stated instead of inferring it from one matching summary.

This provides a practical interpretation check for any reconstructed energy record: name the parameter evidence supporting the conversion from motion into joules. The mathematics can be exact while the dimensional scale remains unidentified from motion alone.

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The exact motion satisfies the chosen equations and initial conditions. That verifies the solution of this paper model. It does not establish that an unfamiliar real trace was generated by this particular two-mass system.

Other modal combinations, forcing histories or measurement processes can produce similar local envelopes. A useful investigation seeks evidence distinguishing the alternatives: mode shapes, phase relationships, independent parameter information and the relevant input or boundary conditions.

The Penn State coupled-oscillator demonstration, updated in August 2025, provides a primary educational illustration of motion exchanging between coupled oscillators. Our numerical model, exact configurations and energy partitions are original calculations rather than measurements taken from that demonstration.

Source checked on 25 October 2026. Use the demonstration to strengthen the physical picture, then return to the equations to establish what our specified model predicts. An animation and a conservation proof play complementary roles; neither should be claimed as an experiment that was not performed.

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For repair, begin with one spring and the distinction between displacement and velocity at equilibrium. If the learner interprets every zero displacement as no energy, that relationship needs attention before a second mass is added.

For stabilisation, calculate the initial energy and the π/4 snapshot using all three springs. Then ask why E₁ + E₂ alone is not the complete invariant. The learner should point to the connecting spring rather than invoke an unexplained correction factor.

For extension, change the coupling or initial condition and require a new mode decomposition. Ask whether an envelope node guarantees local rest and which extra phase condition would be needed.

Progress appears when the learner can choose the boundary, reconstruct the relevant state and distinguish local exchange from a dissipative total-energy term. Remembering that coupled systems can beat is useful vocabulary. Explaining where the energy is accounted for gives that vocabulary a dependable physical meaning.

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Does a shrinking local envelope prove damping?

No. This conservative example produces shrinking and growing local envelopes through two-mode superposition and energy redistribution. Damping requires its own physical model and evidence.

Is energy exchanged between the normal modes here?

The two normal-mode energies remain constant in this ideal linear symmetric model. Local component energies change because those physical components combine both modes.

Does an envelope node mean zero local energy?

Not automatically. Displacement can vanish while velocity remains nonzero. Our exact exchanged configuration uses a special phase coincidence as well as an envelope node.

Can coupling energy be ignored when it is small?

An approximation may sometimes neglect it for a stated purpose, but an exact complete energy account includes it. The approximation and its error need to remain visible.

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Build the motion foundation

Return to Vibrations for displacement, velocity, acceleration, modes, sampling and amplitude conventions. The driven-response and damping-identification branches linked above develop different input-output questions.

Follow the general interaction idea

What Is Coupling? explains the broader relationship between interacting systems. Use the specific force and energy model here when the task requires a reconstructible two-mass calculation rather than a general interaction description.

Transfer the learning

Continue through Gold Standard Education and How X Works. For a new changing trace, identify what is moving, what the measurement leaves unseen and which complete balance could distinguish redistribution from an actual loss of mechanical energy.

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