Coupled oscillators and normal modes

Release three connected masses

Three labelled bodies A, B and C, each of mass 2 kg, move along one fixed inertial line. Positive position X points right. Their relaxed reference positions are −2 m, 0 m and 2 m. Two massless ideal springs connect A to B and B to C; each has natural length 2 m and stiffness k = 2 N/m. Neither endpoint is attached to a wall.

Let q_A, q_B and q_C be each body’s signed displacement from its own reference position. At t = 0 s, supply q = (0.1, −0.2, 0.1) m and velocities v = (0.05, 0.05, 0.05) m/s. For example, the actual initial positions are (−1.9, −0.2, 2.1) m. Equal displacements of A and C do not put those bodies at the same position.

The guide contributes no longitudinal force. There is no damping, friction, drive or collision in this supplied model. The linear spring force depends on extension relative to natural length, including signed compression. These exact equations teach a conditional model; no useful extension range for a material spring is supplied.

Newton’s laws and free-body diagrams and Friction, tension, drag, springs, and contact forces are published preparation for force signs. The definitions and force calculation are repeated here. On a first reading, follow the release, the two histories and the selected-pattern examples. Matrices, linear algebra, and eigenvalue problems and Derivatives, Taylor series, and local approximation are published optional return paths for the compact matrix and derivative checks. Small oscillations and normal modes offers a separate wall-bound comparison; it is not necessary to solve this release.

Use relative spring extensions

The A–B spring length is 2 m+q_B−q_A, so its signed extension is q_B−q_A. Its force on A is k(q_B−q_A); its force on B is the negative of that. The B–C extension is q_C−q_B. Thus F_A = k(q_B−q_A), F_B = −k(q_B−q_A)+k(q_C−q_B), and F_C = −k(q_C−q_B). Each force is in newtons.

At release the extensions are −0.3 m and +0.3 m. The first spring pushes its endpoints apart and the second pulls its endpoints together. The three forces are (−0.6, +1.2, −0.6) N. Dividing by 2 kg gives the initial accelerations (−0.3, +0.6, −0.3) m/s². The common initial velocity does not change these ideal spring forces.

A dot means a derivative with respect to actual time: q̇ is velocity and q̈ is acceleration. Newton’s law gives q̈_A = (1 s⁻²)(−q_A+q_B), q̈_B = (1 s⁻²)(q_A−2q_B+q_C), and q̈_C = (1 s⁻²)(q_B−q_C). Multiplying a metre displacement by s⁻² gives m/s².

Adding the three forces gives zero at every time under these equations. With fixed masses, total momentum P = 2 kg(v_A+v_B+v_C) therefore stays at its initial value 0.3 kg m/s. Internal springs can accelerate individual bodies while the center continues at constant velocity.

Test fixed displacement patterns

Try a collective motion q = a(t)e, where e = (e_A,e_B,e_C) is a nonzero, dimensionless fixed pattern and the common coordinate a has units of metres. A positive-frequency normal mode has a sinusoidal a(t); its nonzero components have fixed signed amplitude ratios. A zero component stays at its reference position for that mode.

The pattern (1, 0, −1) gives accelerations proportional to (−1, 0, +1). Hence ä = −(1 s⁻²)a. The pattern (1, −2, 1) gives accelerations proportional to (−3, +6, −3), so ä = −(3 s⁻²)a. Their angular frequencies are 1 rad/s and √3 rad/s. The middle body has twice the outer amplitude and the opposite sign in the second pattern.

The pattern (1, 1, 1) changes neither spring extension and gives ä = 0 m/s². Its coordinate is a(t) = a(0)+ȧ(0)t. This is the zero-frequency mode of this free chain, including uniform translation or a constant shift. It is not a positive-frequency periodic oscillation.

For a compact return, define D = (1 s⁻²)[[1, −1, 0], [−1, 2, −1], [0, −1, 1]], so q̈ = −Dq. A tested pattern satisfies De = λe, where λ is an eigenvalue with units s⁻²; positive λ gives ω = √λ as an inverse-time value. Here the three patterns form a basis: the reconstruction formulas below recover any three displacements. This finite check avoids assuming that every arbitrary matrix has a complete set of oscillatory modes.

Select motion with six initial values

Write q = a₀(1,1,1)+a₁(1,0,−1)+a₂(1,−2,1). This means q_A = a₀+a₁+a₂, q_B = a₀−2a₂ and q_C = a₀−a₁+a₂. Adding or subtracting these equations gives a₀ = (q_A+q_B+q_C)/3, a₁ = (q_A−q_C)/2 and a₂ = (q_A−2q_B+q_C)/6. These formulas also apply to velocities after differentiating.

The supplied positions give (a₀,a₁,a₂) = (0,0,0.1) m initially. The velocities give (ȧ₀,ȧ₁,ȧ₂) = (0.05,0,0) m/s. Thus the antisymmetric coordinate a₁ stays zero, the translation is a₀ = (0.05 m/s)t, and a₂ = (0.1 m)cos(√3τ), where τ = t/(1 s) is dimensionless.

The resulting motion is q_A = q_C = [τ/20+cos(√3τ)/10] m and q_B = [τ/20−cos(√3τ)/5] m. The cosine argument is the numerical phase in radians. Substitution into the force equations checks the accelerations, and substitution at t = 0 checks all six initial values.

The model’s patterns and frequencies depend on its masses, springs and boundary conditions. Initial data select their coordinates and phases. An individual body in a superposition need not oscillate sinusoidally about its fixed reference; here every body also carries the steadily growing translation.

Compare fixed and internal histories

For equal masses the center displacement is Q = (q_A+q_B+q_C)/3 = a₀. The relaxed reference center is X = 0 m, so Q also equals the actual center position in this example. Define internal displacements r_i = q_i−Q. Here r_A = r_C = (0.1 m)cos(√3τ) and r_B = −(0.2 m)cos(√3τ). The lower panel reveals their fixed ratio after translation has been subtracted.

Blue circles represent the shared A/C displacement value, violet squares represent B, and green triangles represent Q in the upper history. One A/C glyph has two body memberships; there are still three distinct bodies. The schematic shows only their relaxed reference arrangement. It is not the initial state or a selected-time snapshot.

Both histories use the same time axis and 300 pixels per metre. Their horizontal zero-displacement lines are physical zeros inside the plots; the lower time rulers are separate. The reference schematic instead uses 145 pixels per metre of actual position. Compare numerical coordinates and units rather than lengths across those differently scaled views.

The mathematical solution is defined for t ≥ 0 s; the drawing shows 0 ≤ t ≤ 4 s. Short line segments approximate its smooth curves. Selected hollow markers are exact model states, not measurements. Near some sampled times their outlines overlap because the displacements are close; the shape keys and exact table retain each quantity’s meaning.

In the table, c_n means cos(√3n), with n = t/(1 s) at the listed integer times. Every entry is exact; no decimal precision is inferred from the drawing.

Separate center motion from vibration

The center velocity is Q̇ = P/(6 kg) = 0.05 m/s, and its acceleration is zero. The internal displacements instead have zero mass-weighted sum. A moving center does not imply that the springs are unchanging: their extensions are −(0.3 m)cos(√3τ) and +(0.3 m)cos(√3τ).

The actual spring lengths are therefore 2 m−(0.3 m)cos(√3τ) and 2 m+(0.3 m)cos(√3τ). Both stay between 1.7 m and 2.3 m for all t ≥ 0 s, so A, B and C remain ordered. Also X_C−X_A = 4 m throughout. These checks justify the geometry of this solution; they do not verify an actual spring’s material law over that range.

If the only initial displacement were (0.1,0.1,0.1) m with the same common velocity 0.05 m/s, every q_i would be 0.1 m+(0.05 m/s)t. Neither spring would change length. That pure translation differs from the main release, which excites translation and one vibrating pattern.

In contrast, the release q = (0.1,0,−0.1) m with all velocities zero selects only the pattern (1,0,−1). The middle body stays at its reference position and the other two move with angular frequency 1 rad/s. The chain has not changed; its initial data have selected a different motion.

Check energy and pattern normalization

For this undamped model the kinetic energy is K = Σ(2 kg/2)v_i². The spring energy is U = (k/2)(q_B−q_A)²+(k/2)(q_C−q_B)², with zero energy at relaxed spring lengths. Multiplying each motion equation by its body velocity and adding gives d(K+U)/dt = 0: each spring’s work cancels the derivative of its stored energy.

With s = sin(√3τ) and c = cos(√3τ), the main solution has K = 3/400 J+(9/50 J)s² and U = (9/50 J)c². The constant 3/400 J is the center’s kinetic energy. Internal kinetic and spring energy exchange while their sum stays 9/50 J; the total is 3/16 J. Individual body kinetic energies need not be constant.

A pattern’s normalization is a convention. If e₂ = (1,−2,1) is replaced by e₂/√6, multiply its coordinate by √6 to keep the same q. Its initial coordinate becomes √6/10 m, and its frequency does not change. A pattern alone cannot determine a physical amplitude.

For an optional matrix return, the mass-weighted product of two patterns is Σm_i e_i f_i. The three listed patterns have pairwise zero products. Their effective masses Σm_i e_i² are 6 kg, 4 kg and 12 kg, respectively. For q = a_j e_j alone, kinetic energy is (M_j/2)ȧ_j² and spring energy is (M_jω_j²/2)a_j² for a positive-frequency mode. These formulas and orthogonality are checked for this symmetric conservative chain with fixed positive masses, not asserted for arbitrary damped or nonsymmetric equations.

Keep the system and boundaries visible

Changing a mass changes the pattern test. If only B’s mass becomes 4 kg, the old pattern (1,−2,1) produces acceleration coefficients (−3,+3,−3) s⁻² times its metre coordinate. The outer equations require frequency squared 3 s⁻², while the middle requires 3/2 s⁻². One shared frequency cannot satisfy both. The new pattern (1,−1,1) instead has frequency squared 2 s⁻².

Changing boundaries also changes the problem. Add a 1 N/m wall spring to each endpoint, relaxed at its reference position. A common displacement Q then gives endpoint forces −(1 N/m)Q and middle force zero. The old uniform pattern is not preserved, and the total external force is −(2 N/m)Q. Its center acceleration at that common-displacement state is −(1/3 s⁻²)Q; constant center velocity is no longer a general conclusion.

In the original free chain the linear spring law was supplied exactly, so the modal solution is exact within that model. In a nonlinear apparatus, replacing forces by their local linear terms would require a declared equilibrium, an appropriate small-displacement range and an error assessment. Damping, driving, different constraints or collisions likewise require a new equation and mode analysis.

A useful check follows a short sequence: define the system and reference positions, derive signed relative forces, test a collective pattern, then use both initial positions and velocities. Center motion, frequency, amplitude and stored energy are separate questions. The two histories let you see which one a plotted coordinate answers.

Worked examples

Try it yourself

1. In the main release, q_A = q_C at every time. Does this make A and C coincide? Find their actual separation and the range of each spring length.

   Show the answer and reasoning

   They do not coincide. X_C−X_A = 4 m, and each spring length remains between 1.7 m and 2.3 m.

   The bodies have different reference positions. Their equal q values cancel in X_C−X_A, leaving 2−(−2) = 4 m. Each spring length is 2 m plus or minus (0.3 m)cos(√3τ), whose cosine lies between −1 and 1.

2. Supply q(0) = (0.1,0.1,0.1) m and v(0) = (0.05,0.05,0.05) m/s in the original free chain. What happens to the springs, and what are the total momentum and energy?

   Show the answer and reasoning

   q_i = 0.1 m+(0.05 m/s)t for all three bodies. The springs stay relaxed, P = 0.3 kg m/s and E = 3/400 J.

   All extensions are zero, so all accelerations are zero. The three kinetic energies add to 3×(2 kg/2)×(0.05 m/s)². This is uniform translation, not a positive-frequency oscillation.

3. Replace e₂ = (1,−2,1) by e₂/√6 in the main release. What initial coordinate must multiply it, what is its effective mass, and does its frequency change?

   Show the answer and reasoning

   The initial coordinate is √6/10 m, the effective mass is 2 kg, and the angular frequency stays √3 rad/s.

   Multiplying the coordinate by √6 compensates for dividing the pattern by √6. Its effective mass is (12 kg)/6. Consequently (2 kg/2)(3 s⁻²)(√6/10 m)² = 9/50 J, preserving the initial internal energy as well as the displacement.

4. Change only B’s mass to 4 kg. Can (1,−2,1) still be one fixed normal-mode pattern? Test (1,−1,1) as an alternative.

   Show the answer and reasoning

   The old pattern fails the common-frequency test. The alternative (1,−1,1) satisfies ω² = 2 s⁻², giving ω = √2 rad/s.

   For the old pattern, the force coefficients are (−6,+12,−6) N/m. Dividing by masses (2,4,2) kg gives (−3,+3,−3) s⁻², inconsistent with one multiple of (1,−2,1). For the alternative, forces per metre are (−4,+8,−4) N/m and accelerations (−2,+2,−2) s⁻², equal to −2 times its pattern.

5. Add the two 1 N/m endpoint wall springs described above. At a state where every q_i = Q, find the total longitudinal force and center acceleration. Is uniform translation still a zero-frequency pattern?

   Show the answer and reasoning

   F_total = −(2 N/m)Q and Q̈ = −(1/3 s⁻²)Q. Uniform translation is no longer a zero-frequency pattern of this modified system.

   The internal extensions vanish at that state, but the two wall springs exert external endpoint forces. Divide their total force by 6 kg for the center acceleration. The middle acceleration is zero while the endpoint accelerations are −(1/2 s⁻²)Q, so a nonzero common displacement does not keep one common acceleration.

6. In the main release, when cos(√3τ) = 0, is all kinetic energy internal vibration? Give the center, internal and total kinetic energies at such an instant.

   Show the answer and reasoning

   Center kinetic energy is 3/400 J, internal kinetic energy is 9/50 J and total kinetic energy is 3/16 J. It is not all internal vibration.

   The spring extensions vanish, so U = 0 J. The internal sine-squared factor equals 1, but the constant center velocity retains its own kinetic energy. Adding those non-overlapping contributions preserves the total energy.

What to carry forward

• A normal-mode pattern describes collective signed displacement ratios; initial positions and velocities select its coordinate and amplitude.
• A free-chain zero-frequency coordinate can translate steadily. Subtracting center motion reveals the internal vibration.
• Changing masses, supports or force laws changes the pattern test; keep those conditions with the frequency and energy conclusions.