Small oscillations and normal modes

Release two coupled masses

Two 2 kg point masses lie on a fixed straight line, with right positive. Their equilibrium positions are X₁ = 1 m and X₂ = 3 m. Each mass connects to its nearest fixed wall, at 0 m or 4 m, through a spring of stiffness k = 8 N/m. A third spring of stiffness k_c = 12 N/m joins the masses. The three natural lengths are 1 m, 2 m and 1 m, so every spring is relaxed at equilibrium.

Let q₁ = X₁ − 1 m and q₂ = X₂ − 3 m be signed displacements from those fixed equilibrium positions. Release the masses with q₁ = +0.06 m, q₂ = +0.02 m and both velocities zero at t = 0. The joining spring is compressed by 0.04 m. Its push on the second mass can outweigh that mass’s wall spring; displacement alone does not determine each acceleration’s direction.

These are exact stipulated Newtonian model values, not measurements. The frame is inertial, the masses are constant, motion stays on the declared line, and all longitudinal forces come from ideal linear springs. We exclude damping and driving. We study 0 ≤ t ≤ π s; physical spring accuracy outside a tested range is a separate question.

Write forces before choosing patterns

A linear spring pulls or pushes against its signed extension. The joining spring’s extension is q₂ − q₁. Thus the longitudinal forces are F₁ = −k q₁ + k_c(q₂ − q₁) and F₂ = −k q₂ − k_c(q₂ − q₁). The joining forces on the two masses are opposite; their wall forces are separate interactions.

Velocity is the instantaneous rate of change of displacement, v_i = dq_i/dt. Acceleration is its rate of change, a_i = d²q_i/dt². A dot is shorthand for one time derivative, and two dots mean acceleration. Newton’s equation in this frame is m q̈_i = F_i. Force has units N, while acceleration has units m/s².

For m = 2 kg the equations become q̈₁ = −(10 s⁻²)q₁ + (6 s⁻²)q₂ and q̈₂ = (6 s⁻²)q₁ − (10 s⁻²)q₂. They are coupled: each mass’s acceleration depends on both displacements. Two masses need two initial displacements and two initial velocities to select their motion.

Separate together and opposite patterns

Define Q₊ = (q₁ + q₂)/2 and Q₋ = (q₁ − q₂)/2. Adding the two force equations and dividing by two gives Q̈₊ = −(4 s⁻²)Q₊. Subtracting gives Q̈₋ = −(16 s⁻²)Q₋. The normal coordinates Q₊ and Q₋ each obey their own restoring equation; neither equation contains the other coordinate.

Recover the physical displacements with q₁ = Q₊ + Q₋ and q₂ = Q₊ − Q₋. A pure Q₊ motion has q₁ = q₂: the masses move together and the joining spring keeps its natural length. A pure Q₋ motion has q₁ = −q₂: the masses move oppositely and both ends contribute to the joining spring’s extension.

A normal mode here is a motion with one frequency and a fixed displacement pattern, (1, 1) or (1, −1). The pattern is dimensionless and can be multiplied by a displacement amplitude. Opposite signs mean opposite motion, not unequal frequencies. The normal coordinates still have units m; they are combinations of the same two masses’ displacements, not two additional particles.

Supply phases and initial data

The cosine function repeats when its numerical radian argument increases by 2π. Write the phase angles as θ₊ = φ₊ rad and θ₋ = φ₋ rad, with dimensionless φ₊ = (2 s⁻¹)t and φ₋ = (4 s⁻¹)t. The physical angular frequencies are 2 rad/s and 4 rad/s. Their periods are T₊ = π s and T₋ = (π/2) s; a period is the time for a complete cycle.

Our initial data give Q₊(0) = 0.04 m and Q₋(0) = 0.02 m, with both modal velocities zero. The matching solutions are Q₊(t) = (0.04 m) cos φ₊ and Q₋(t) = (0.02 m) cos φ₋. Each starts at its positive amplitude, the largest magnitude of that normal coordinate.

Therefore q₁(t) = (0.04 m) cos φ₊ + (0.02 m) cos φ₋ and q₂(t) = (0.04 m) cos φ₊ − (0.02 m) cos φ₋. Their velocities are v₁ = −(0.08 m/s) sin φ₊ − (0.08 m/s) sin φ₋ and v₂ = −(0.08 m/s) sin φ₊ + (0.08 m/s) sin φ₋. Differentiating again checks the coupled equations for the whole interval, not only at the table’s sample times.

Read motion and normal coordinates together

The upper plot shows q₁ and q₂. The lower plot shows their half-sum Q₊ and half-difference Q₋. Both use the same actual time and metre scales. One lower curve completes a cycle while the other completes two; their sums and differences produce the more complicated upper motion.

The symbols identify exact table times, not observations. The joining spring’s length is X₂ − X₁ = 2 m − 2Q₋, between 1.96 m and 2.04 m. Also |q₁| and |q₂| are at most 0.06 m by the sum of the modal amplitudes. These bounds keep the supplied point masses separated.

At t = (π/4) s and t = (3π/4) s the displacements are identical, but the velocities have opposite signs. A repeated configuration is not necessarily a repeated state. At t = π s both displacements and both velocities return to their initial values. The selected mixed motion has period π s.

Track energy in the normal coordinates

For the chosen reference, V = k(q₁² + q₂²)/2 + k_c(q₁ − q₂)²/2 is the energy stored in all three springs. Kinetic energy is K = m(v₁² + v₂²)/2. With no damping or driving and fixed spring parameters, K + V is constant along this motion.

Substitute q₁ = Q₊ + Q₋ and q₂ = Q₊ − Q₋. The cross terms cancel: K = m(Q̇₊² + Q̇₋²) and V = k Q₊² + (k + 2k_c)Q₋². Thus E₊ = m Q̇₊² + k Q₊² and E₋ = m Q̇₋² + (k + 2k_c)Q₋² are separately constant in this undamped linear model.

Because our normal coordinates use a half-sum and half-difference, each kinetic term is (4 kg)Q̇²/2. Treating a normal coordinate as an extra 2 kg physical mass would miss this factor. Initially E₊ = 0.0128 J and E₋ = 0.0128 J, so the total is 0.0256 J. Their equality follows these particular amplitudes; normal modes do not generally have equal energies.

Check what “small” means

Our supplied quadratic spring potential makes its linear equations exact. More generally, a small-oscillation model keeps the quadratic part of a sufficiently smooth potential near equilibrium and the leading kinetic terms. At equilibrium the first-order potential change vanishes. For regular independent local coordinates, positive inertia and strictly positive restoring quadratic energy give the stable oscillatory case described here. These conditions are assumptions to check, not consequences of calling a displacement small.

A separate one-coordinate example makes the accuracy issue concrete. Let V(q) = kq²/2 + λq⁴/4, with k = 8 N/m and λ = 800 N/m³. Its exact force is F = −kq − λq³. For nonzero q the magnitude of the neglected cubic force relative to the retained linear force is λq²/k. At q = 0.01 m it is 1%; at q = 0.05 m it is 25%. The denominator is the retained force magnitude.

That ratio compares instantaneous forces at a supplied displacement. It does not by itself bound the difference between full trajectories over a time interval. This quartic example is a different model, not another force silently added to our two-mass system. Damping, driving, zero restoring directions and nonlinear coupling need their own analysis; undamped oscillations do not settle into rest.

Connect a pattern to an eigenvalue

A matrix is a table of coefficients. Multiplying each row by the displacement column gives the corresponding component of S q; the physical force column is −S q. Here the stiffness matrix S has rows (20, −12) and (−12, 20), with entries in N/m, and m q̈ = −S q. It sends the nonzero patterns (1, 1) and (1, −1) to 8(1, 1) N/m and 32(1, −1) N/m respectively. A pattern preserved up to a multiplier is an eigenvector; the multiplier is its eigenvalue.

For equal mass m, a pattern with stiffness eigenvalue s has numerical phase-rate square ν² = s/m. Thus (8 N/m)/(2 kg) = 4 s⁻² and (32 N/m)/(2 kg) = 16 s⁻². Taking positive square roots gives 2 s⁻¹ and 4 s⁻¹. A stiffness eigenvalue is not itself a frequency. Different positive masses require a mass matrix and a corresponding generalized eigenvalue problem.

Begin with the release, force equations, half-sum and half-difference, then read the exact table and plots. You can return to the derivative and matrix checks after the relevant mathematical-tool lessons. The next subject, Hamiltonian mechanics and phase space, will revisit a state through coordinates and momenta; none of its machinery is required for this supplied model.

Worked examples

Try it yourself

1. At t = (π/2) s, q₁ = −0.02 m and q₂ = −0.06 m. Recover Q₊ and Q₋ without reading the lower graph.

   Show the answer and reasoning

   Q₊ = −0.04 m and Q₋ = +0.02 m.

   The half-sum is (−0.02 − 0.06)m/2 = −0.04 m. The half-difference is (−0.02 + 0.06)m/2 = +0.02 m. Recombining them returns the two physical displacements.

2. Instead release both masses at q₁ = q₂ = +0.04 m with both velocities zero. Which mode is excited, and what are the initial accelerations and period?

   Show the answer and reasoning

   Only Q₊ is excited. Both accelerations are −0.16 m/s² and the period is π s.

   Q₋ and Q̇₋ are zero, so their homogeneous equation keeps Q₋ zero. The joining spring has zero extension. Each wall supplies −0.32 N, which gives −0.16 m/s² for 2 kg. The together-mode phase rate is 2 s⁻¹, giving T = 2π/(2 s⁻¹).

3. The table gives the same q₁ and q₂ at (π/4) s and (3π/4) s. Is the full state repeated?

   Show the answer and reasoning

   No. Both velocities change from −0.08 m/s to +0.08 m/s.

   Displacements specify configuration; this Newtonian state also needs velocities. The two instants have the same normal-coordinate values but different rates. The selected full state returns after π s, not after π/2 s.

4. The opposite pattern has stiffness eigenvalue 32 N/m. A reader calls this “32 rad/s”. Calculate the correct angular frequency and period for mass 2 kg.

   Show the answer and reasoning

   The angular frequency is 4 rad/s and the period is (π/2) s.

   First divide stiffness by mass: 32/2 = 16 s⁻². Its positive square root is the numerical phase rate 4 s⁻¹, corresponding to 4 rad/s. A full numerical phase change of 2π takes 2π/(4 s⁻¹).

5. At t = (π/4) s use q₁ = −0.02 m, q₂ = +0.02 m and v₁ = v₂ = −0.08 m/s. Find K, V and K + V from the original two-mass formulas.

   Show the answer and reasoning

   K = 0.0128 J, V = 0.0128 J and K + V = 0.0256 J.

   K = (2 kg)[2(0.08 m/s)²]/2 = 0.0128 J. The two wall terms total 0.0032 J; the joining term is (12 N/m)(0.04 m)²/2 = 0.0096 J. Their sum is V = 0.0128 J. Equal K and V at this instant does not make them equal at every instant.

6. For the separate quartic model at q = 0.05 m, the cubic force is 25% of the linear force. Does that prove the trajectory’s position error stays below 25% throughout 0 ≤ t ≤ π s?

   Show the answer and reasoning

   No. It is an instantaneous force comparison with a specified denominator.

   Position differences depend on the equations, initial data and elapsed time. The local ratio supplies no such trajectory bound. It also does not alter the exact linear model used for the two-mass plots.

What to carry forward

• Normal modes separate this linear coupled motion into fixed together and opposite patterns; the physical masses remain coupled.
• Initial displacements and velocities determine the modal amplitudes and phases. A repeated configuration need not be a repeated state.
• Stiffness divided by mass gives a squared phase rate; frequency, displacement, acceleration and energy retain different units.
• The chosen normal coordinates give two constant modal energies, whose equality here follows the chosen release.
• A quadratic ideal model is exact by stipulation. A small-oscillation approximation needs conditions and checks on neglected terms; a local force ratio is not a trajectory-error bound.