Questions: Coupled Oscillator Systems and Equations of Motion

Question 1

Two identical masses on a symmetric spring system are displaced equally in the same direction and released simultaneously. A student predicts the masses will slowly exchange energy back and forth (beats). Why is this prediction wrong?

• A. The masses will exchange energy because the coupling spring always transfers energy between them regardless of initial conditions
• B. Displacing both masses equally in the same direction excites only the symmetric normal mode, in which the coupling spring never stretches — both masses oscillate in unison at a single frequency indefinitely, with no energy exchange
• C. Energy exchange cannot occur because the masses are identical — only unequal masses produce beats
• D. Beats require the system to be driven by an external force; free oscillations never produce energy exchange

Answer: (B)

Beats (energy exchange) occur only when *multiple* normal modes are excited simultaneously. In the symmetric initial condition, both masses move identically — this is exactly the pattern of the symmetric normal mode, in which the coupling spring is neither compressed nor stretched. Since only one mode is excited, the system oscillates at that single frequency forever, with both masses moving in lockstep. Energy exchange requires a superposition of two modes with different frequencies; the interference between them produces the beat pattern.

Question 2

For a two-mass coupled oscillator, substituting x = v·cos(ωt) into the matrix equation Mẍ = −Kx reduces the problem to (K − ω²M)v = 0. What does finding non-trivial solutions to this equation tell you?

• A. That the masses must be equal for any oscillatory solution to exist
• B. That the values of ω² satisfying det(K − ω²M) = 0 give the system's natural frequencies, and the corresponding vectors v give the normal mode shapes — the configurations in which all masses oscillate at a single frequency
• C. That x = 0 is the only equilibrium, confirming the masses always return to rest
• D. That ω must be purely imaginary, indicating the motion is exponentially growing rather than oscillatory

Answer: (B)

For (K − ω²M)v = 0 to have a non-trivial solution v ≠ 0, the matrix (K − ω²M) must be singular — its determinant must be zero. This eigenvalue condition produces N values of ω² for an N-mass system, giving N natural frequencies. Each corresponding eigenvector v describes the *ratio of displacements* in that mode — how much each mass moves relative to the others. These normal mode frequencies and shapes are the complete characterization of the system's oscillatory behavior.

Question 3

In a two-mass coupled spring system, both normal mode frequencies equal the natural frequency of a single uncoupled mass-spring system.

• T. True
• F. False

Answer: False

Only one mode frequency equals the uncoupled frequency. For the symmetric two-mass system, the symmetric mode has frequency ω₁ = √(k/m) — equal to the single-mass frequency because the coupling spring carries no force (it never stretches). But the antisymmetric mode, where masses move in opposite directions, has a *higher* frequency ω₂ = √((k + 2kc)/m) because the coupling spring is alternately compressed and stretched, adding to the restoring force. Coupling splits the degenerate single-mass frequency into two distinct frequencies: one unchanged, one raised.

Question 4

The general motion of a two-mass coupled oscillator can always be expressed as a superposition of its two normal modes, with amplitudes and phases set by the initial conditions.

• T. True
• F. False

Answer: True

Because the equations of motion are linear, any solution can be written as a linear combination of the fundamental solutions — the two normal modes. The general solution is x(t) = A₁v₁cos(ω₁t + φ₁) + A₂v₂cos(ω₂t + φ₂), where v₁ and v₂ are the mode shapes, and the four constants (A₁, A₂, φ₁, φ₂) are determined by the four initial conditions (two initial positions, two initial velocities). This superposition principle is what makes mode decomposition powerful: even complicated correlated motion reduces to independent oscillations in the mode basis.

Question 5

Explain what a 'normal mode' is in a coupled oscillator system, and why finding the normal modes simplifies the analysis of arbitrary initial conditions.

Model answer: A normal mode is a special pattern of motion in which all masses oscillate at the same single frequency and maintain fixed ratios of displacement to each other. In a normal mode, the system behaves like a single harmonic oscillator — simple, periodic, non-transferring. For arbitrary initial conditions, the motion is a superposition of all normal modes with amplitudes and phases determined by those conditions. This simplifies analysis because instead of solving a coupled system of differential equations directly, you decompose the motion into independent oscillators (one per mode). The complicated correlated motion is just the sum of simple motions that happen at different frequencies.

The energy-exchange phenomenon (beats) illustrates why the decomposition is powerful: when you excite both modes with different frequencies, their superposition produces amplitude modulation — energy appears to slosh between the masses. But in the mode basis, nothing is being exchanged — each mode simply oscillates independently. The 'exchange' is a feature of the original mass coordinate description, not of the mode description.