Past exam of the physics B course of the University of Cambridge 2017 Paper 1 Section C C12 Solution 2026-10-18
Normal-mode idea. Near a stable equilibrium, expand the potential energy to quadratic order and the kinetic energy as a quadratic form in generalized velocities. The Euler-Lagrange equations then take the linear form . A normal mode has fixed displacement ratios, , with . Real systems admit this description for sufficiently small motion, before higher-order nonlinearities or strong damping matter. Superposition permits a general motion to be assembled from the modes. A free symmetry direction may instead be a zero mode, with constant or linearly changing coordinate.
Lagrangian and bead equations. Take each as an arc-length displacement positive in the same sense around the fixed circular wire. Spring length changes are , and . Equal equilibrium spring tensions give no net tangential force; their linear energy terms sum to zero. Dropping an irrelevant constant, the Lagrangian isFor example, differentiating the potential with respect to gives . The three Euler-Lagrange equations therefore areThe wire supplies normal constraint forces but does no tangential work. These are the equal-mass case of the normal modes of three beads connected by springs on a circular hoop.
Frequencies. The stiffness matrix divided by isIt annihilates . On the two-dimensional subspace , its action is multiplication by three. Thus the angular frequencies of the normal modes areThe repeated frequency is a degeneracy of normal modes, while the zero frequency expresses free rigid rotation of all beads along the wire.
General solution. Two independent vectors spanning the vibrating subspace are and . Each amplitude satisfies , with . The uniform coordinate satisfies . HenceThe six freely specifiable real constants give the six initial data; is fixed by and is not an additional independent integration constant. The linear zero-mode term must be included to allow nonzero total tangential momentum.
Shapes and uniqueness. In the first term beads 1 and 2 have opposite equal displacements and bead 3 remains at its equilibrium position. In the second term beads 1 and 3 do this while bead 2 stays fixed. The last term moves all three identically, leaving spring extensions unchanged: is a rigid phase shift and a uniform drift around the wire. The sketches show positive modal amplitudes; all arrows reverse half a vibration period later.
Original displacement sketches for two independent equal-frequency bead modes and the rigid-rotation zero mode; arrows indicate displacement along the ring rather than its radius
. Details. The two nonzero-frequency eigenvectors are not unique, because any independent linear combinations in their degenerate two-dimensional eigenspace form an equally valid basis. In particular the displayed pair is independent but not orthogonal; one could use and instead. The zero eigenspace is one-dimensional, so its vector is fixed up to scale and sign.
Specified release. Zero initial velocities permit and require . At , the three displacement conditions become , , . Therefore one convenient set of constants isEquivalently,Their sum remains , as required by zero initial total tangential momentum. Alternative amplitude signs with compensating phase shifts represent the same motion.
Past exam of the physics B course of the University of Cambridge 2018 Paper 2 Section B B9 Solution 2026-10-18
Definition and equations. A normal mode is a motion in which every component of a linear system has the same time dependence with fixed amplitude ratios and phases. For a conservative system near equilibrium, the most useful form isThe vector describes the mode shape, while is its angular frequency. Linear combinations of modes describe arbitrary sufficiently small motions when the modes form a complete basis.
Let be displacements of independent generalized coordinates from equilibrium. Expand the Lagrangian to quadratic order:The mass matrix defines a positive-definite quadratic form on the independent coordinates, while the stiffness matrix is the Hessian of the potential energy. Substituting a mode gives the generalized eigenvalue problem for small oscillationsA practical calculation identifies the equilibrium, chooses independent coordinates, computes the quadratic kinetic and potential energies, solves for the eigenvalues and then obtains their eigenvectors. One can equivalently diagonalize the real symmetric matrix . This guarantees real and a complete orthogonal basis in mass-weighted coordinates.
For distinct eigenvalues, the fact that and are real symmetric matrices impliesThus distinct modes are orthogonal in the inner product weighted by the mass matrix. Choosing decouples the motion into , with total energy . A positive-definite matrix gives positive eigenvalues and oscillatory stability. Zero modes can describe rigid translations or rotations without restoring force; a negative eigenvalue gives exponential instability rather than an oscillatory mode.
Symmetry. A symmetry preserving both quadratic energies maps modes into modes of the same frequency. Coordinates adapted to symmetry divide the eigenvalue problem into smaller independent blocks. For two identical subsystems, exchanging them gives symmetric and antisymmetric modes. The degeneracy of normal modes means that independent modes share one frequency. Higher-dimensional group representations of a symmetry can enforce such degeneracies; symmetry does not require every pair of modes to be degenerate, and accidental degeneracies can also occur. Within a degenerate eigenspace, orthogonal linear combinations are equally valid mode shapes.
For two equal masses , each attached to a fixed support by a spring of stiffness and coupled to the other by a spring of stiffness ,The symmetric mode leaves the coupling spring unstretched and has . The antisymmetric mode stretches it and has . Superposing the two gives beats when their frequencies are close.
Continuous systems. The same idea replaces matrices by differential operators and mode vectors by functions satisfying boundary conditions. For a taut string of length , tension and mass per unit length , the wave equation and fixed endpoints giveOrthogonality of the sine functions allows an arbitrary initial displacement to be decomposed into modes. Changing the endpoints changes the spectrum: boundary conditions are part of the physical mode problem.
A uniform chain of equal masses joined by springs , with spacing , has travelling normal modes withTheir quantization gives phonons; the long-wavelength limit is a sound wave. Rectangular electromagnetic cavity modes satisfy Maxwell's equations and conductor boundary conditions, with in vacuum. An elastic beam instead has a fourth-order spatial equation and flexural dispersion . These examples show how restoring forces and geometry determine the frequency spectrum.
Real systems. Small-amplitude motion is often approximately linear even when the full equations are nonlinear. Large amplitudes produce frequency shifts, mode coupling and transfer of energy between modes. Friction broadens resonances and makes free oscillations decay. A damping matrix proportional to or can retain the undamped mode shapes; general damping need not, and the appropriate eigenfrequencies and modes can be complex. External periodic forcing predominantly excites modes near resonance, with amplitude set by coupling to the forcing and dissipation. Imperfections split ideal degeneracies, while radiation or coupling to a surrounding fluid can remove energy from the chosen subsystem. Normal modes remain useful when linearization, the boundary conditions and the treatment of damping accurately describe the motion being measured.
