GRE Physics Companion: Small Oscillations and the Normal-Mode Eigenvalue Problem
For small conservative oscillations,
The normal modes satisfy
1 High-value GRE facts
- At equilibrium, the first derivatives of the potential vanish.
- The stiffness matrix is the potential-energy Hessian evaluated at equilibrium.
- The mass matrix need not be diagonal in generalized coordinates.
- small oscillations obey Mη + Kη = 0.
- The generalized eigenvalue is ω2, not ω.
- Distinct modes satisfy arT Ma
s = 0.
- mass-normalized modes satisfy ΦT MΦ = 1.
- The Rayleigh quotient estimates squared frequencies.
- A zero eigenvalue gives a neutral mode.
- A negative eigenvalue indicates linear instability.
Part I: Original GRE-style problems
Problem 1: equilibrium expansion
At a mechanical equilibrium,
- all second derivatives of U vanish
- all first derivatives of U vanish
- the kinetic energy vanishes for all time
- the mass matrix is the identity
- every Normal frequency is zero
Problem 2: stiffness matrix
The small-oscillation stiffness matrix is
- Kij = ∂U∕∂qi
- Kij = ∂2U∕∂q
i∂qj at equilibrium
- Kij = Mij−1
- Kij = qiqj
- always diagonal
Problem 3: generalized eigenproblem
The mode equation is
- Ka = ωMa
- Ka = ω2Ma
- Ma = ω2Ka only
- Ka = 0 for every mode
- MKa = 0
Problem 4: mass orthogonality
For two distinct-frequency modes of a conservative symmetric system,
- arT a
s = 1
- arT Ma
s = 0
- arT Ka
s = 1
- ar = as
- ωr = ωs
Problem 5: Rayleigh quotient
The Rayleigh quotient is
- (yT My)∕(yT Ky)
- (yT Ky)∕(yT My)
- det M∕ det K
- yT y
- tr K
Problem 6: zero mode
If
the modal equation of motion is
- Q + ω02Q = 0
- Q = 0
- Q = 0 necessarily
- Q = 0 necessarily
- Q − σ2Q = 0
Problem 7: unstable eigenvalue
If a generalized eigenvalue is negative,
- the mode has a real stable oscillation frequency
- the mode is a zero mode
- the linearized motion contains exponential growth and decay
- the mass matrix is singular
- the mode disappears
Problem 8: double-pendulum frequencies
For the equal-mass equal-length small-angle double pendulum,
are
- (g∕ℓ)(1 ± 1)
- (g∕ℓ)(2 ±
)
- (g∕ℓ)(2 ± 2)
- (g∕ℓ)(1 ±
)
- both g∕ℓ
Problem 9: lower double-pendulum mode shape
A mode vector for the lower-frequency equal-mass double-pendulum mode is proportional
to
![[ ]
0
1](https://images.physicslibrary.org/cache/objects/1497/make4ht/GREPhysicsCompanionSmallOscillationsAndTheNormalModeEigenvalueProblem10x.png)
Problem 10: numerical low frequency
For
the lower double-pendulum normal frequency is closest to
- 1.00 rad/s
- 2.40 rad/s
- 3.13 rad/s
- 5.79 rad/s
- 9.81 rad/s
Problem 11: mass normalization
If the columns of Φ are mass-normalized modes, then
- ΦT MΦ = 1
- ΦT KΦ = 1 always
- M = 1
- K = 0
- det Φ = 0
Problem 12: modal projection
For mass-normalized modes, the initial modal displacement is
- Q0 = ΦT η
0
- Q0 = ΦT Mη
0
- Q0 = Kη0
- Q0 = M−1η
0
- Q0 = 0
Part II: Complete worked solutions
Solution 1
At equilibrium,
This eliminates the linear term in the potential expansion. Answer: (B).
Solution 2
The local quadratic potential is determined by the Hessian:
Answer: (B).
Solution 3
Substituting
into the small-oscillation equations gives
Answer: (B).
Solution 4
For distinct eigenfrequencies and symmetric
the modes satisfy
Answer: (B).
Solution 5
The Rayleigh quotient is
For an exact mode it equals
Answer: (B).
Solution 6
With
the modal equation becomes
Answer: (B).
Solution 7
Write
Then
whose solutions contain
and
Answer: (C).
Solution 8
The equal-mass equal-length double pendulum gives
Answer: (B).
Solution 9
The lower-frequency eigenvector has same-sign angular components:
Answer: (C).
Solution 10
| ω− | =  | (22)
|
| ≈ 2.40 rad∕s. | (23) |
Answer: (B).
Solution 11
Mass normalization gives
Thus
Answer: (A).
Solution 12
From
multiply by
Therefore
Answer: (B).
2 GRE checklist
- Confirm that the expansion point is actually an equilibrium.
- Form both M and K; do not assume either is diagonal.
- Remember that the eigenvalue is ω2.
- Use the mass inner product for modal orthogonality.
- Mass normalize before applying the simplest projection formulas.
- Use the Rayleigh quotient as a frequency-squared estimate.
- Treat zero modes separately from oscillatory modes.
- Interpret negative eigenvalues as local instability.
References
References
[1] PhysicsLibrary, M06-04, Coupled Oscillators and Normal Modes.
[2] PhysicsLibrary, M06-03, Driven Harmonic Motion and Resonance.
[3] PhysicsLibrary, M06-01, Simple Harmonic Motion.
[4] A. P. French, Vibrations and Waves, W. W. Norton, 1971.
[5] J. R. Taylor, Classical Mechanics, University Science Books, 2005.
[6] H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley,
2002.