Two-degree-of-freedom systems: natural frequencies and mode shapes
Equations of motion in matrix form, the frequency equation, mode shapes, orthogonality, semi-definite systems and the link to the bending–torsion typical section.
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Why it matters
Real structures have more than one natural frequency, and each comes with its own pattern of motion. The two-degree-of-freedom (2-DOF) system is the smallest model that shows this, and it is exactly the model used for the classic aeroelastic problem: a wing section that can plunge (bend) and pitch (twist). Flutter happens when two such modes interact, so solving a 2-DOF eigenvalue problem by hand is the core skill for the aeroelasticity topics that follow.
Key ideas
Equations of motion. Take the chain: wall, spring k₁, mass m₁, spring k₂, mass m₂, with displacements x₁ and x₂. Free-body diagrams give
m₁·ẍ₁ + (k₁ + k₂)·x₁ − k₂·x₂ = 0 m₂·ẍ₂ − k₂·x₁ + k₂·x₂ = 0
In matrix form [M]{ẍ} + [K]{x} = {0}, with [M] = diag(m₁, m₂) and [K] = [[k₁ + k₂, −k₂], [−k₂, k₂]]. The off-diagonal terms couple the coordinates: here the coupling is through stiffness (static coupling). If the mass matrix had off-diagonal terms the coupling would be dynamic (inertial), as in a wing section whose centre of mass is not on its elastic axis.
Eigenvalue problem. Assume synchronous harmonic motion {x} = {X}·sin(ω·t). Then ([K] − ω²·[M])·{X} = {0}. A non-trivial solution needs det([K] − ω²·[M]) = 0, the characteristic (frequency) equation. For the chain above, with λ = ω²:
m₁·m₂·λ² − (m₁·k₂ + m₂·(k₁ + k₂))·λ + k₁·k₂ = 0
Its two positive roots give the natural frequencies ω₁ < ω₂.
Mode shapes. Substituting each ω² back into either row gives the amplitude ratio, for example from row 1: X₂/X₁ = (k₁ + k₂ − m₁·ω²)/k₂. Only the ratio is fixed; the overall size is arbitrary (set X₁ = 1, or normalise to unit modal mass). In the first (lower) mode the masses move in phase; in the second they move in opposite directions, with a node (a point of zero motion) on the connecting spring.
Orthogonality. Distinct modes satisfy {X₁}ᵀ·[M]·{X₂} = 0 and {X₁}ᵀ·[K]·{X₂} = 0. This is what lets modal analysis decouple the equations (next topic). Note it is M- and K-orthogonality, not ordinary orthogonality {X₁}ᵀ·{X₂} = 0.
General free vibration. Any free motion is a superposition of the two modes, each at its own frequency, with amplitudes and phases set by the four initial conditions. If the initial displacement matches a mode shape exactly, only that mode is excited. A mix of modes is not periodic unless ω₂/ω₁ is rational.
Semi-definite systems. A free–free system (two masses joined by a spring, or an aircraft in free flight) has a zero frequency: the rigid-body mode. The elastic mode is ω² = k·(m₁ + m₂)/(m₁·m₂).
Close frequencies. If ω₁ and ω₂ are close and both modes are excited, the response beats, with energy swapping between the coordinates at the difference frequency.
Damping. With light damping proportional to [M] and [K] the mode shapes stay the same and the frequencies change very little.
Aeroelastic link. For a typical section with plunge h and pitch α, the mass matrix has the coupling term S_α (static unbalance = mass × distance of the centre of mass behind the elastic axis). In flight, aerodynamic forces add speed-dependent stiffness and damping; as speed rises the two frequencies move together, and flutter is the coalescence of the bending and torsion branches.
Formulas
[M]·{ẍ} + [K]·{x} = {0} — free vibration; [M] mass matrix (kg), [K] stiffness matrix (N/m).
det([K] − ω²·[M]) = 0 — frequency equation.
m₁·m₂·ω⁴ − (m₁·k₂ + m₂·(k₁ + k₂))·ω² + k₁·k₂ = 0 — wall–k₁–m₁–k₂–m₂ chain.
X₂/X₁ = (k₁ + k₂ − m₁·ω²)/k₂ — mode shape ratio for the same chain.
{X_i}ᵀ·[M]·{X_j} = 0, {X_i}ᵀ·[K]·{X_j} = 0 (i ≠ j) — orthogonality.
ω² = k·(m₁ + m₂)/(m₁·m₂) — elastic mode of a free–free two-mass system (the other ω = 0).
ω_i² = {X_i}ᵀ·[K]·{X_i} / {X_i}ᵀ·[M]·{X_i} — Rayleigh quotient for a known mode shape.
Worked examples
Example 1 (standard). Chain with m₁ = m₂ = 1 kg and k₁ = k₂ = 100 N/m. Find ω₁, ω₂ and the mode shapes.
- Frequency equation: 1·λ² − (100 + 200)·λ + 10 000 = 0, i.e. λ² − 300·λ + 10 000 = 0.
- λ = (300 ± √(90 000 − 40 000))/2 = (300 ± 223.6)/2 = 38.20 or 261.80 (rad/s)².
- ω₁ = √38.20 = 6.18 rad/s; ω₂ = √261.80 = 16.18 rad/s.
X₂/X₁ = (k₁ + k₂ − m₁·ω²)/k₂: mode 1, (200 − 38.20)/100 = 1.618; mode 2, (200 − 261.80)/100 = −0.618.
Answer: ω₁ = 6.18 rad/s with shape {1, 1.618}; ω₂ = 16.18 rad/s with shape {1, −0.618}.
Example 2 (GATE level). Chain with m₁ = 2 kg, m₂ = 1 kg, k₁ = 200 N/m, k₂ = 100 N/m. Find the natural frequencies in Hz, the mode shapes, and check M-orthogonality.
- Coefficients: m₁·m₂ = 2; m₁·k₂ + m₂·(k₁ + k₂) = 200 + 300 = 500; k₁·k₂ = 20 000.
- 2·λ² − 500·λ + 20 000 = 0, so λ² − 250·λ + 10 000 = 0, λ = (250 ± √(62 500 − 40 000))/2 = (250 ± 150)/2 = 50 or 200.
- ω₁ = √50 = 7.071 rad/s (1.125 Hz); ω₂ = √200 = 14.14 rad/s (2.251 Hz).
- Mode 1: X₂/X₁ = (300 − 2 × 50)/100 = 2, shape {1, 2}. Mode 2: (300 − 2 × 200)/100 = −1, shape {1, −1}.
- Orthogonality: {1, 2}·diag(2, 1)·{1, −1} = 2 × 1 × 1 + 1 × 2 × (−1) = 0. Stiffness check: [K]{1, −1} = {300 + 100, −100 − 100} = {400, −200}; {1, 2}·{400, −200} = 400 − 400 = 0.
Answer: f₁ = 1.13 Hz, mode {1, 2}; f₂ = 2.25 Hz, mode {1, −1}; both orthogonality checks give zero.
Common mistakes
- Writing k₂ instead of k₁ + k₂ in the first diagonal stiffness term: a mass feels every spring attached to it.
- Getting the off-diagonal sign wrong; for this chain it is −k₂ and the matrix is symmetric.
- Treating the mode shape as absolute amplitudes; only ratios are defined.
- Expecting {X₁}ᵀ·{X₂} = 0; orthogonality is with respect to [M] and [K].
- Forgetting the zero-frequency rigid-body mode of an unrestrained system.
- Taking the square root at the wrong step: the characteristic equation gives ω², not ω.
For GATE AE
Expect frequency equations for two-mass chains, free–free systems and simple bar or pendulum combinations; mode-shape ratios and node locations; orthogonality checks; and statements about static versus dynamic coupling. Practise writing [M] and [K] directly from a sketch and solving the quadratic in ω² quickly. The typical-section (plunge–pitch) 2-DOF model reappears in flutter problems.
Quick check
- How many natural frequencies does a 2-DOF system have?
- In the lower mode of a two-mass chain, do the masses move in or out of phase?
- Two masses of 2 kg and 1 kg joined by a 300 N/m spring float freely. Find the non-zero ω.
- What does an off-diagonal term in [M] indicate?
- Mode shapes {1, 2} and {1, −1} with M = diag(2, 1): are they M-orthogonal?
Answers: 1. two; 2. in phase; 3. ω = √(300 × 3/2) = 21.2 rad/s; 4. inertial (dynamic) coupling; 5. yes, 2 − 2 = 0.
Interview questions
All Structural Dynamics and Aeroelasticity interview questionsTry answering each one aloud before you open it.
1.What is a two-degree-of-freedom system in the context of structural dynamics?Concept
A two-degree-of-freedom system in structural dynamics refers to a system that can move in two independent ways. This typically involves two masses connected by springs and dampers, where each mass can move independently in response to external forces. Such systems are used to model and analyze the dynamic behavior of structures that have more than one mode of vibration.
2.Explain the concept of natural frequencies in a two-degree-of-freedom system.Concept
Natural frequencies are the frequencies at which a system tends to oscillate in the absence of any driving or damping force. In a two-degree-of-freedom system, there are typically two natural frequencies, corresponding to the two independent modes of vibration. These frequencies are determined by the system's mass and stiffness properties.
3.What are mode shapes in a two-degree-of-freedom system?Concept
Mode shapes describe the pattern of motion that a system undergoes at each of its natural frequencies. In a two-degree-of-freedom system, each mode shape corresponds to one of the natural frequencies and shows how the masses move relative to each other. Mode shapes are important for understanding how energy is distributed in the system during vibration.
4.Why is it important to determine the natural frequencies and mode shapes of a structure?Application
Determining the natural frequencies and mode shapes of a structure is crucial for predicting its dynamic response to external forces, such as wind or seismic activity. This information helps engineers design structures that can withstand these forces without resonating, which could lead to excessive vibrations and potential structural failure.
5.How does damping affect the natural frequencies and mode shapes of a two-degree-of-freedom system?Application
Damping generally reduces the amplitude of vibrations and can slightly alter the natural frequencies of a system. However, it does not significantly change the mode shapes. In a two-degree-of-freedom system, damping helps control vibrations and dissipate energy, but the fundamental characteristics of the mode shapes remain largely unchanged.
6.What happens if the natural frequencies of a two-degree-of-freedom system are very close to each other?Application
If both modes are excited, their responses add with slightly different frequencies and the motion beats: energy swaps back and forth between the two coordinates at the difference frequency, as in two weakly coupled pendulums. A harmonic force in that band excites both modes strongly at once, and in a test it becomes hard to separate the modes. In aeroelasticity close frequencies matter even more: bending and torsion frequencies that approach each other with airspeed and coalesce are the classic route to flutter, so a large torsion-to-bending frequency ratio is a design aim.
7.In what scenarios would you use a two-degree-of-freedom model instead of a single-degree-of-freedom model?Application
A two-degree-of-freedom model is used when a structure or system has two significant modes of vibration that need to be analyzed. This is common in systems where two components can move independently, such as in certain types of machinery, vehicles, or complex structural elements. It provides a more accurate representation of the dynamic behavior compared to a single-degree-of-freedom model.
8.Calculate the natural frequencies of a two-degree-of-freedom chain (wall – k₁ – m₁ – k₂ – m₂) with m₁ = 2 kg, m₂ = 3 kg, k₁ = 100 N/m and k₂ = 150 N/m.Numerical
[M] = diag(2, 3) and [K] = [[k₁ + k₂, −k₂], [−k₂, k₂]] = [[250, −150], [−150, 150]] N/m. det([K] − λ[M]) = 0 gives 6λ² − (2 × 150 + 3 × 250)·λ + 100 × 150 = 0, i.e. 6λ² − 1050λ + 15 000 = 0, so λ = 15.69 and 159.3 (rad/s)². The natural frequencies are ω₁ = 3.96 rad/s (0.63 Hz) and ω₂ = 12.62 rad/s (2.01 Hz), with mode shapes X₂/X₁ = (250 − 2λ)/150 = 1.457 (in phase) and −0.457 (out of phase).
9.Given a two-degree-of-freedom system with known natural frequencies, how would you determine the mode shapes?Numerical
To determine the mode shapes, solve the eigenvalue problem associated with the system's equations of motion. This involves substituting the natural frequencies into the system's characteristic equation and solving for the eigenvectors. These eigenvectors represent the mode shapes, showing the relative motion of the masses at each natural frequency.
10.Explain how aeroelasticity can influence the natural frequencies and mode shapes of an aircraft wing.Application
Aeroelasticity involves the interaction between aerodynamic forces and structural elasticity. In an aircraft wing, this interaction can alter the natural frequencies and mode shapes by changing the effective stiffness and mass distribution. Aeroelastic effects can lead to phenomena like flutter, where the wing experiences unstable oscillations. Engineers must account for these effects to ensure the wing's structural integrity and performance.
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