Normal Coordinates for N Coupled Oscillators

Derivation

Normal Coordinates for N Coupled Oscillators

D-063 Home PU-103 Threads symmetry · energy Depends on Coupled Oscillators and Normal Modes, simultaneous-diagonalization

Statement

For a conservative system of \(N\) degrees of freedom executing small oscillations about a stable equilibrium, with quadratic kinetic energy \(T=\tfrac12\,\dot{\mathbf q}^{\mathsf T}\mathbf M\,\dot{\mathbf q}\) and potential \(V=\tfrac12\,\mathbf q^{\mathsf T}\mathbf K\,\mathbf q\) (\(\mathbf M\) symmetric positive-definite, \(\mathbf K\) symmetric positive-semidefinite), there exists a single real linear transformation \(\mathbf q=\mathbf A\,\boldsymbol\eta\) that simultaneously diagonalizes \(\mathbf M\) and \(\mathbf K\), giving \(\mathbf A^{\mathsf T}\mathbf M\mathbf A=\mathbf I\) and \(\mathbf A^{\mathsf T}\mathbf K\mathbf A=\boldsymbol\Omega^2=\mathrm{diag}(\omega_1^2,\dots,\omega_N^2)\). In the normal coordinates \(\eta_r\) the Lagrangian separates into \(N\) uncoupled harmonic oscillators \(\ddot\eta_r+\omega_r^2\eta_r=0\), whose frequencies \(\omega_r^2\) are the roots of \(\det(\mathbf K-\omega^2\mathbf M)=0\).

Why it matters

Coupling is a statement about coordinates, not about physics. Normal coordinates are the choice in which every mode oscillates independently, so an intractable set of \(N\) coupled second-order equations collapses into \(N\) one-line problems already solved in first year. This is the structural backbone of lattice dynamics, molecular vibrations, coupled circuits, and the field-theoretic passage to phonons and photons.

The construction also explains why the naive eigenvalue problem \(\mathbf K\mathbf x=\lambda\mathbf x\) is the wrong one when masses differ: the correct object is the generalized eigenproblem \(\mathbf K\mathbf x=\omega^2\mathbf M\mathbf x\), and the theorem guaranteeing a full real spectrum is simultaneous diagonalization by congruence, not by orthogonal similarity.

Assumptions

Motion is small about a stable equilibrium.If displacements are not small, cubic and higher terms in \(V\) survive; the equations become nonlinear, modes couple through anharmonicity, and the normal-coordinate separation fails.

\(\mathbf M\) is symmetric positive-definite.If \(\mathbf M\) is only positive-semidefinite (a massless or constrained coordinate), \(T\) does not define an inner product, \(\mathbf M^{-1/2}\) does not exist, and the reduction to a standard symmetric eigenproblem breaks down.

\(\mathbf K\) is symmetric positive-semidefinite.If \(\mathbf K\) has a negative eigenvalue the equilibrium is unstable; some \(\omega_r^2<0\) gives an exponentially growing normal coordinate rather than an oscillation.

The system is conservative and time-independent (no damping, no gyroscopic forces).Velocity-dependent terms \(\dot{\mathbf q}^{\mathsf T}\mathbf G\,\dot{\mathbf q}\) or a Rayleigh dissipation function add an antisymmetric or symmetric first-order matrix; \(\mathbf M\) and \(\mathbf K\) can no longer in general be diagonalized simultaneously and modes become complex or coupled.

Degeneracy is handled within each eigenspace.If two roots \(\omega_r^2\) coincide, the corresponding eigenvectors are not fixed uniquely; one must Gram-Schmidt within the degenerate subspace under the \(\mathbf M\)-inner product to keep the diagonalization valid.

Derivation

\[ L=T-V=\tfrac12\,\dot{\mathbf q}^{\mathsf T}\mathbf M\,\dot{\mathbf q}-\tfrac12\,\mathbf q^{\mathsf T}\mathbf K\,\mathbf q,\qquad \mathbf M=\mathbf M^{\mathsf T},\ \mathbf K=\mathbf K^{\mathsf T}. \]

Expand \(T\) and \(V\) to second order about equilibrium \(\mathbf q=\mathbf 0\); linear terms in \(V\) vanish at equilibrium and the constant is dropped. Only the symmetric parts of \(\mathbf M,\mathbf K\) contribute to the quadratic forms, so they may be taken symmetric, with \(K_{ij}=\partial^2 V/\partial q_i\partial q_j\). A

\[ \frac{d}{dt}\frac{\partial L}{\partial \dot q_i}-\frac{\partial L}{\partial q_i}=0 \;\Longrightarrow\; \mathbf M\,\ddot{\mathbf q}+\mathbf K\,\mathbf q=\mathbf 0. \]

Apply the Euler-Lagrange equations to the quadratic \(L\); symmetry of \(\mathbf M,\mathbf K\) makes \(\partial(\tfrac12\mathbf q^{\mathsf T}\mathbf K\mathbf q)/\partial\mathbf q=\mathbf K\mathbf q\). A

\[ \mathbf q(t)=\mathbf a\,e^{i\omega t}\;\Longrightarrow\;\big(\mathbf K-\omega^2\mathbf M\big)\mathbf a=\mathbf 0. \]

Try a synchronous (single-frequency) solution in which every coordinate shares the same time dependence; the ansatz turns the ODE into an algebraic generalized eigenproblem. A

\[ \det\!\big(\mathbf K-\omega^2\mathbf M\big)=0. \]

A nontrivial amplitude \(\mathbf a\neq\mathbf 0\) requires the coefficient matrix to be singular; this characteristic (secular) equation is a polynomial of degree \(N\) in \(\omega^2\). A

\[ \mathbf M=\mathbf M^{1/2}\mathbf M^{1/2},\qquad \mathbf y=\mathbf M^{1/2}\mathbf a,\qquad \big(\mathbf M^{-1/2}\mathbf K\,\mathbf M^{-1/2}-\omega^2\mathbf I\big)\mathbf y=\mathbf 0. \]

Because \(\mathbf M\succ0\) it has a unique symmetric positive-definite square root \(\mathbf M^{1/2}\) (spectral theorem applied to \(\mathbf M\)). Left-multiply step 3 by \(\mathbf M^{-1/2}\) and insert \(\mathbf M^{-1/2}\mathbf M^{1/2}=\mathbf I\); this whitening converts the generalized problem into a standard eigenproblem for the symmetric matrix \(\tilde{\mathbf K}=\mathbf M^{-1/2}\mathbf K\,\mathbf M^{-1/2}\). B

\[ \tilde{\mathbf K}=\tilde{\mathbf K}^{\mathsf T}\ \Rightarrow\ \tilde{\mathbf K}=\mathbf R\,\boldsymbol\Omega^2\,\mathbf R^{\mathsf T},\quad \mathbf R^{\mathsf T}\mathbf R=\mathbf I,\quad \boldsymbol\Omega^2=\mathrm{diag}(\omega_1^2,\dots,\omega_N^2),\ \omega_r^2\in\mathbb R. \]

The spectral theorem for real symmetric matrices: \(\tilde{\mathbf K}\) has an orthonormal eigenbasis and real eigenvalues. Positive-semidefiniteness of \(\mathbf K\), hence of the congruent \(\tilde{\mathbf K}\), gives \(\omega_r^2\ge0\), so all frequencies are real. B

\[ \mathbf A\equiv\mathbf M^{-1/2}\mathbf R,\qquad \mathbf A^{\mathsf T}\mathbf M\,\mathbf A=\mathbf R^{\mathsf T}\mathbf R=\mathbf I,\qquad \mathbf A^{\mathsf T}\mathbf K\,\mathbf A=\mathbf R^{\mathsf T}\tilde{\mathbf K}\,\mathbf R=\boldsymbol\Omega^2. \]

Undo the whitening: the columns of \(\mathbf A\) are the physical mode shapes \(\mathbf a_r=\mathbf M^{-1/2}\mathbf r_r\). The same congruence \(\mathbf A\) reduces \(\mathbf M\) to the identity and \(\mathbf K\) to a diagonal, so the columns solve \(\mathbf K\mathbf a_r=\omega_r^2\mathbf M\mathbf a_r\) with \(\mathbf M\)-orthonormality \(\mathbf a_r^{\mathsf T}\mathbf M\,\mathbf a_s=\delta_{rs}\). C

\[ \mathbf q=\mathbf A\,\boldsymbol\eta\ \Rightarrow\ L=\tfrac12\,\dot{\boldsymbol\eta}^{\mathsf T}(\mathbf A^{\mathsf T}\mathbf M\mathbf A)\dot{\boldsymbol\eta}-\tfrac12\,\boldsymbol\eta^{\mathsf T}(\mathbf A^{\mathsf T}\mathbf K\mathbf A)\boldsymbol\eta=\tfrac12\sum_{r=1}^{N}\big(\dot\eta_r^2-\omega_r^2\eta_r^2\big). \]

Change variables to the normal coordinates \(\boldsymbol\eta=\mathbf A^{-1}\mathbf q\). Substituting the two congruence results collapses both quadratic forms to diagonal sums; the Lagrangian becomes a sum of decoupled oscillators. B

\[ \frac{d}{dt}\frac{\partial L}{\partial\dot\eta_r}-\frac{\partial L}{\partial\eta_r}=0\ \Longrightarrow\ \ddot\eta_r+\omega_r^2\,\eta_r=0,\qquad r=1,\dots,N. \]

Euler-Lagrange in the new coordinates; each \(\eta_r\) appears in one term only, so the equations are independent. This is the target form. A

Result

\[ \boxed{\;\mathbf q(t)=\sum_{r=1}^{N}\mathbf a_r\,\eta_r(t),\quad \eta_r(t)=C_r\cos(\omega_r t+\phi_r),\quad (\mathbf K-\omega_r^2\mathbf M)\mathbf a_r=\mathbf 0,\quad \mathbf a_r^{\mathsf T}\mathbf M\,\mathbf a_s=\delta_{rs}\;}\]

Reading. Any small motion of the system is a superposition of \(N\) independent normal modes. Mode \(r\) has a fixed spatial pattern \(\mathbf a_r\) (all coordinates in phase or antiphase) oscillating at its own frequency \(\omega_r\); the amplitude \(C_r\) and phase \(\phi_r\) are set by the initial \(\mathbf q\) and \(\dot{\mathbf q}\). The mode shapes are orthonormal not in the ordinary sense but under the mass matrix as metric: the physically meaningful inner product is \(\langle\mathbf x,\mathbf y\rangle_{\mathbf M}=\mathbf x^{\mathsf T}\mathbf M\mathbf y\).

Units check. \([\mathbf M]=\mathrm{kg}\), \([\mathbf K]=\mathrm{N\,m^{-1}}=\mathrm{kg\,s^{-2}}\), so \(\det(\mathbf K-\omega^2\mathbf M)=0\) requires \([\omega^2]=[\mathbf K]/[\mathbf M]=\mathrm{s^{-2}}\), giving \([\omega]=\mathrm{s^{-1}}\). The \(\mathbf M\)-normalization \(\mathbf a_r^{\mathsf T}\mathbf M\mathbf a_s=\delta_{rs}\) forces \([\mathbf a_r]=\mathrm{kg^{-1/2}}\), so each \(\eta_r=\mathbf a_r^{\mathsf T}\mathbf M\mathbf q\) has units \(\mathrm{kg^{1/2}\,m}\); then \(\tfrac12\dot\eta_r^2\) has units \(\mathrm{kg\,m^2\,s^{-2}}=\mathrm{J}\), correctly an energy.

Limiting cases

Equal masses, \(\mathbf M=m\mathbf I\). The generalized problem reduces to the ordinary one \(\mathbf K\mathbf a=m\omega^2\mathbf a\); mode shapes become ordinary orthonormal eigenvectors of \(\mathbf K\) and \(\omega_r^2=\lambda_r/m\).
Decoupled system, \(\mathbf M\) and \(\mathbf K\) diagonal. \(\mathbf A=\mathbf M^{-1/2}\) already diagonalizes both; normal coordinates coincide with the original coordinates rescaled by \(\sqrt{M_{ii}}\) and no rotation is needed.
Zero-frequency (rigid) mode. If \(\mathbf K\) is singular, some \(\omega_r=0\): a translation or rotation of the whole system that costs no potential energy, giving \(\eta_r(t)=C_r+D_r t\) (free drift) rather than oscillation.
Weak coupling. If off-diagonal stiffness \(\kappa\ll k\), the frequencies split symmetrically, \(\omega_\pm^2\approx\omega_0^2(1\pm\kappa/k)\), and energy beats slowly between the near-degenerate coordinates.
Continuum limit \(N\to\infty\). The discrete mode index \(r\) becomes a wavevector \(k\); \(\det(\mathbf K-\omega^2\mathbf M)=0\) passes into a dispersion relation \(\omega(k)\) and normal coordinates become Fourier amplitudes, the seed of phonons.

Breaks when

Anharmonicity or large amplitude. Once cubic terms in \(V\) matter, the equations \(\ddot\eta_r+\omega_r^2\eta_r=-\partial V_3/\partial\eta_r\) couple the normal coordinates; energy flows between modes (Fermi-Pasta-Ulam-Tsingou), and no single \(\mathbf A\) decouples the motion for all time.
Non-conservative or gyroscopic forces. Damping \(\mathbf C\dot{\mathbf q}\) or Coriolis/magnetic terms add a third matrix. Unless \(\mathbf C\) is a linear combination of \(\mathbf M\) and \(\mathbf K\) (Rayleigh/proportional damping), \(\mathbf M,\mathbf K,\mathbf C\) cannot be simultaneously diagonalized; modes become complex and generally coupled.
Indefinite stiffness (unstable equilibrium). If any \(\omega_r^2<0\), the mode is \(\eta_r=C_r e^{|\omega_r|t}+D_r e^{-|\omega_r|t}\), an instability; the harmonic-superposition result no longer describes bounded oscillation.
Singular or time-dependent mass matrix. If \(\mathbf M\) is not positive-definite (massless coordinate, ideal constraint) or depends on \(\mathbf q\)/\(t\), \(\mathbf M^{-1/2}\) does not exist and the whitening step fails; the DOF count must first be reduced by eliminating the constrained coordinate.

Failure modes

Diagonalizing \(\mathbf K\) alone. Using orthogonal eigenvectors of \(\mathbf K\) when \(\mathbf M\neq m\mathbf I\); these do not decouple \(T\), so the modes still exchange kinetic energy. One must solve the generalized problem \(\mathbf K\mathbf a=\omega^2\mathbf M\mathbf a\).
Solving \(\det(\mathbf K-\omega^2\mathbf I)=0\). Forgetting the \(\mathbf M\): the correct secular equation is \(\det(\mathbf K-\omega^2\mathbf M)=0\) whenever \(\mathbf M\neq m\mathbf I\).
Using ordinary orthonormality \(\mathbf a_r^{\mathsf T}\mathbf a_s=\delta_{rs}\). The correct condition is \(\mathbf a_r^{\mathsf T}\mathbf M\mathbf a_s=\delta_{rs}\); normalizing without the mass metric puts the wrong scale on \(\eta_r\) and corrupts the energy bookkeeping.
Treating \(\mathbf M^{-1/2}\) entrywise. \(\mathbf M^{-1/2}\) is a matrix square root, not the reciprocal of each entry; the two coincide only for diagonal \(\mathbf M\).
Diagonalizing \(\mathbf M^{-1}\mathbf K\) and expecting orthogonal eigenvectors. \(\mathbf M^{-1}\mathbf K\) is generally not symmetric, so its eigenvectors are not orthogonal even though its eigenvalues are still the correct \(\omega_r^2\).
Dropping the zero mode. Treating a free (unclamped) system as if all \(\omega_r>0\) misses the rigid-body drift and miscounts the degrees of freedom.
Mishandling degeneracy. When two \(\omega_r^2\) coincide, blindly picking eigenvectors gives a non-\(\mathbf M\)-orthogonal pair; the fix is Gram-Schmidt within the degenerate subspace using the \(\mathbf M\)-inner product.

Discussion

The essential content is that two quadratic forms, kinetic and potential energy, can always be brought to standard form together by one real change of basis, provided one of them is positive-definite. This is stronger than the spectral theorem, which diagonalizes a single symmetric matrix by an orthogonal transformation. Here the transformation \(\mathbf A\) is generally not orthogonal; it is orthogonal with respect to the metric \(\mathbf M\). The kinetic energy defines that metric, and the whitening \(\mathbf q\to\mathbf M^{1/2}\mathbf q\) is exactly the change to coordinates in which the metric is Euclidean, after which ordinary orthogonal diagonalization finishes the job. That two-step view, whiten then rotate, is why a single congruence handles both matrices. The same geometry reappears wherever a system carries its own natural inner product: generalized eigenproblems in structural engineering, the metric lowering indices in general relativity, and the overlap matrix in quantum chemistry all play the role of \(\mathbf M\).

The threads of this unit meet here. Symmetry: if the system is invariant under a group of coordinate permutations or rotations, that group commutes with both \(\mathbf M\) and \(\mathbf K\), so the modes organize into irreducible representations and much of the diagonalization can be done by symmetry alone before any determinant is expanded; this is how one predicts which molecular vibrations are infrared or Raman active without solving a single secular equation. Energy: because \(\mathbf A\) diagonalizes \(T\) and \(V\) at once, the total energy is a sum \(E=\sum_r\tfrac12(\dot\eta_r^2+\omega_r^2\eta_r^2)\) of independently conserved mode energies. Nothing couples them at quadratic order, so energy placed in one mode stays there forever, which is precisely the paradox FPUT set out to test.

Physically, a normal mode answers the question "how must I start the system so that it rings at a single pure frequency?" The mode shape \(\mathbf a_r\) is that special initial displacement pattern; released from it, every particle executes simple harmonic motion in lockstep, passing through equilibrium together. A generic initial condition is a weighted sum of these patterns, and because the modes evolve independently, the future is obtained by advancing each phase \(\omega_r t\) separately and resumming, the discrete analogue of Fourier's method for the wave equation.

At the deepest level this construction is the classical skeleton of second quantization. Promoting each \(\eta_r\) and its conjugate momentum \(p_r=\dot\eta_r\) to operators and defining \(a_r=\sqrt{\omega_r/2\hbar}\,(\eta_r+ip_r/\omega_r)\) turns the decoupled Hamiltonian \(H=\sum_r\tfrac12(p_r^2+\omega_r^2\eta_r^2)\) into \(H=\sum_r\hbar\omega_r(a_r^\dagger a_r+\tfrac12)\): a gas of non-interacting quanta, one oscillator per mode. Phonons in a crystal and photons in a cavity are literally the normal coordinates of coupled classical oscillators, quantized. Promoted to infinite dimensions the same structure is Sturm-Liouville theory, with \(\mathbf M\) becoming a positive weight function and \(\mathbf M\)-orthogonality becoming weighted \(L^2\) orthogonality of eigenfunctions. The simultaneous diagonalization performed here is therefore the step that makes a many-body problem soluble; interactions reappear only when the neglected cubic and quartic terms are restored, and those become phonon-phonon scattering.

Common misconceptions. A normal frequency is a property of the whole system, not of any one particle; "the frequency of mass 2" is meaningless because mass 2 participates in every mode at every \(\omega_r\). Decoupling is a change of coordinates, not a change of physics: the particles are still connected by springs, and it is only the abstract normal coordinates \(\eta_r\) that evolve independently. And \(\omega_r^2\) are not the diagonal entries of \(\mathbf K\) divided by those of \(\mathbf M\) unless both are already diagonal; they are the generalized eigenvalues, which mix all entries.

Worked examples

Two unequal masses \(m_1=1\ \mathrm{kg}\), \(m_2=4\ \mathrm{kg}\), joined by a single spring \(k=12\ \mathrm{N\,m^{-1}}\) on a frictionless line (no walls). \[ \mathbf M=\begin{pmatrix}1&0\\ 0&4\end{pmatrix},\quad \mathbf K=k\begin{pmatrix}1&-1\\ -1&1\end{pmatrix}=\begin{pmatrix}12&-12\\ -12&12\end{pmatrix}. \]

Set up the mass and stiffness matrices for a two-body spring; \(\mathbf K\) is the standard graph Laplacian of one bond and \(\mathbf M\) is genuinely non-scalar. A

\[ \det(\mathbf K-\omega^2\mathbf M)=\det\!\begin{pmatrix}12-\omega^2 & -12\\ -12 & 12-4\omega^2\end{pmatrix}=(12-\omega^2)(12-4\omega^2)-144=0. \]

Impose the secular equation for a nontrivial amplitude. A

\[ 4\omega^4-60\omega^2=0\ \Rightarrow\ \omega^2(4\omega^2-60)=0\ \Rightarrow\ \omega_1^2=0,\quad \omega_2^2=15\ \mathrm{s^{-2}}. \]

Expand and factor. The zero root is the free-translation (rigid) mode; the nonzero root is the true vibration, \(\omega_2=\sqrt{15}\approx3.87\ \mathrm{s^{-1}}\). A

\[ \omega_2^2=15:\ (12-15)a_1-12a_2=0\Rightarrow a_2=-\tfrac14 a_1\ \Rightarrow\ \mathbf a_2\propto\begin{pmatrix}1\\ -\tfrac14\end{pmatrix},\quad\text{check }m_1a_1+m_2a_2=1-1=0.\]

Solve for the mode shape; the momentum sum vanishes, confirming the masses move in antiphase about a fixed centre of mass. This is the reduced-mass oscillation \(\mu=m_1m_2/(m_1+m_2)=0.8\ \mathrm{kg}\), and \(k/\mu=12/0.8=15\ \mathrm{s^{-2}}\). B

\[ \mathbf a_2^{\mathsf T}\mathbf M\mathbf a_2=c^2\!\left(1\cdot1+4\cdot\tfrac1{16}\right)=\tfrac54 c^2=1\Rightarrow c=\sqrt{4/5}\ \Rightarrow\ \mathbf a_2=\begin{pmatrix}0.894\\ -0.224\end{pmatrix}\mathrm{kg^{-1/2}}. \]

Normalize under the mass metric so that \(\eta_2\) carries a clean unit energy. B

\[ \boxed{\;\omega_1=0,\quad \omega_2=\sqrt{15}\approx 3.87\ \mathrm{s^{-1}},\quad \mathbf a_2\propto(1,\,-\tfrac14)\;}\]

Reading. One rigid-body drift plus one vibration at the reduced-mass frequency \(\sqrt{k/\mu}\); the light mass swings four times as far as the heavy one, in antiphase.

Units check. \(k/\mu=(12\ \mathrm{kg\,s^{-2}})/(0.8\ \mathrm{kg})=15\ \mathrm{s^{-2}}\), so \(\omega_2\) is in \(\mathrm{s^{-1}}\). Good.

Three equal masses \(m=0.5\ \mathrm{kg}\) on a line, nearest-neighbour springs \(k=2\ \mathrm{N\,m^{-1}}\), fixed walls at both ends (so each end mass feels two springs). \[ \mathbf M=m\mathbf I_3,\quad \mathbf K=k\begin{pmatrix}2&-1&0\\ -1&2&-1\\ 0&-1&2\end{pmatrix}. \]

Fixed-fixed linear chain of three beads; the stiffness matrix is the standard tridiagonal \((2,-1)\) form. Equal masses, so the problem reduces to an ordinary eigenproblem. A

\[ \omega^2=\frac{k}{m}\,\lambda,\qquad \lambda_r=2-2\cos\theta_r,\quad \theta_r=\frac{r\pi}{N+1}=\frac{r\pi}{4},\quad r=1,2,3, \]

The eigenvalues of the fixed-fixed \((2,-1)\) tridiagonal of size \(N=3\) are known in closed form; the mode shapes are \(a^{(r)}_j=\sin(jr\pi/4)\). B

\[ \lambda_1=2-\sqrt2\approx0.586,\quad \lambda_2=2,\quad \lambda_3=2+\sqrt2\approx3.414. \]

Evaluate \(\cos(\pi/4)=\tfrac{\sqrt2}{2}\), \(\cos(\pi/2)=0\), \(\cos(3\pi/4)=-\tfrac{\sqrt2}{2}\). A

\[ \frac{k}{m}=\frac{2}{0.5}=4\ \mathrm{s^{-2}}\ \Rightarrow\ \omega_r=2\sqrt{\lambda_r}:\quad \omega_1=1.53,\ \ \omega_2=2.83,\ \ \omega_3=3.70\ \mathrm{s^{-1}}. \]

Scale by \(\sqrt{k/m}=2\); numerically \(\omega_1=2\sqrt{0.586}\), \(\omega_2=2\sqrt2\), \(\omega_3=2\sqrt{3.414}\). A

\[ \mathbf a^{(1)}\propto(1,\sqrt2,1),\quad \mathbf a^{(2)}\propto(1,0,-1),\quad \mathbf a^{(3)}\propto(1,-\sqrt2,1). \]

Insert \(\sin(jr\pi/4)\). Mode 1 has all masses in phase (lowest frequency, no internal nodes); mode 2 has a node at the centre mass; mode 3 alternates in sign (highest frequency, two nodes). B

\[ \boxed{\;\omega_1=1.53,\ \omega_2=2.83,\ \omega_3=3.70\ \mathrm{s^{-1}};\quad \mathbf a^{(1)}\!\propto(1,\sqrt2,1),\ \mathbf a^{(2)}\!\propto(1,0,-1),\ \mathbf a^{(3)}\!\propto(1,-\sqrt2,1)\;}\]

Reading. Three modes ordered by number of nodes; frequency rises with spatial roughness, the discrete precursor of the dispersion relation \(\omega(k)=2\sqrt{k/m}\,\lvert\sin(ka/2)\rvert\).

Units check. \(\sqrt{k/m}=\sqrt{4\ \mathrm{s^{-2}}}=2\ \mathrm{s^{-1}}\), and \(\lambda_r\) is dimensionless, so every \(\omega_r\) is in \(\mathrm{s^{-1}}\). Good.

Problems

Two equal masses \(m\) in a line joined by three identical springs \(k\) to two fixed walls (wall-\(m\)-\(k\)-\(m\)-wall). Find the two normal frequencies and mode shapes.
Solution
\(\mathbf M=m\mathbf I\), \(\mathbf K=k\begin{pmatrix}2&-1\\ -1&2\end{pmatrix}\). Eigenvalues of the matrix: \(2\mp1\), i.e. \(\lambda=1,3\). So \(\omega_1=\sqrt{k/m}\) with symmetric shape \(\mathbf a_1\propto(1,1)\) (masses move together, middle spring unstretched), and \(\omega_2=\sqrt{3k/m}\) with antisymmetric shape \(\mathbf a_2\propto(1,-1)\) (masses move oppositely, middle spring doubly active). Numerically for \(m=1,k=1\): \(\omega_1=1\ \mathrm{s^{-1}}\), \(\omega_2=\sqrt3\approx1.73\ \mathrm{s^{-1}}\).
For the unequal-mass pair of Worked Example 1 (\(m_1=1,m_2=4,k=12\)), the system starts with \(m_1\) displaced by \(x_1(0)=0.05\ \mathrm{m}\), \(m_2\) at rest at the origin, both velocities zero. Find \(x_1(t)\) and \(x_2(t)\).
Solution
Modes: \(\omega_1=0\) (translation \(\mathbf a_1\propto(1,1)\)) and \(\omega_2=\sqrt{15}\) (\(\mathbf a_2\propto(1,-\tfrac14)\)). Write \(\mathbf x(0)=\alpha(1,1)+\beta(1,-\tfrac14)\). From \(x_1(0)=\alpha+\beta=0.05\), \(x_2(0)=\alpha-\tfrac14\beta=0\Rightarrow\alpha=\tfrac14\beta\). Then \(\tfrac14\beta+\beta=0.05\Rightarrow\beta=0.04,\ \alpha=0.01\). Zero initial velocity keeps the translation mode static (centre of mass fixed, total momentum zero) and the vibration purely cosine: \(x_1(t)=0.01+0.04\cos(\sqrt{15}\,t)\ \mathrm{m}\), \(x_2(t)=0.01-0.01\cos(\sqrt{15}\,t)\ \mathrm{m}\). Check \(m_1x_1+m_2x_2=0.05\) constant (centre of mass at \(0.01\ \mathrm{m}\)).
Show that for any two mode shapes with distinct frequencies, \(\mathbf a_r^{\mathsf T}\mathbf M\mathbf a_s=0\) (mass-orthogonality).
Solution
From \(\mathbf K\mathbf a_r=\omega_r^2\mathbf M\mathbf a_r\) and \(\mathbf K\mathbf a_s=\omega_s^2\mathbf M\mathbf a_s\), left-multiply the first by \(\mathbf a_s^{\mathsf T}\) and the second by \(\mathbf a_r^{\mathsf T}\): \(\mathbf a_s^{\mathsf T}\mathbf K\mathbf a_r=\omega_r^2\,\mathbf a_s^{\mathsf T}\mathbf M\mathbf a_r\) and \(\mathbf a_r^{\mathsf T}\mathbf K\mathbf a_s=\omega_s^2\,\mathbf a_r^{\mathsf T}\mathbf M\mathbf a_s\). Because \(\mathbf K,\mathbf M\) are symmetric the left sides are equal and \(\mathbf a_s^{\mathsf T}\mathbf M\mathbf a_r=\mathbf a_r^{\mathsf T}\mathbf M\mathbf a_s\). Subtracting: \((\omega_r^2-\omega_s^2)\,\mathbf a_r^{\mathsf T}\mathbf M\mathbf a_s=0\). Since \(\omega_r^2\neq\omega_s^2\), the mass inner product vanishes. This is the generalized-eigenproblem analogue of orthogonality of eigenvectors of a symmetric matrix.
Three equal masses \(m\) on a ring joined by three identical springs \(k\) (each mass to its two neighbours). Argue on symmetry grounds how many distinct nonzero frequencies there are, identify the zero mode, then find the nonzero frequency.
Solution
\(\mathbf M=m\mathbf I\), \(\mathbf K=k\begin{pmatrix}2&-1&-1\\ -1&2&-1\\ -1&-1&2\end{pmatrix}\). The uniform vector \((1,1,1)\) gives \(\mathbf K(1,1,1)^{\mathsf T}=\mathbf 0\), the rigid rotation around the ring, so \(\omega=0\). The remaining two eigenvalues are equal by the three-fold symmetry: \(\mathbf K\) has trace \(6k\) and one zero eigenvalue, so the other two sum to \(6k\) and, being equal, are each \(3k\). Hence one doubly-degenerate nonzero frequency \(\omega=\sqrt{3k/m}\). The degenerate pair spans the plane orthogonal to \((1,1,1)\); any orthonormal basis of it (e.g. running-wave modes \(e^{\pm 2\pi i/3}\)) is a valid mode set.
A CO\(_2\)-like linear triatomic: masses \(m,M,m\) with \(m=16\ \mathrm{u}\), \(M=12\ \mathrm{u}\) (use \(1\ \mathrm{u}=1.66\times10^{-27}\ \mathrm{kg}\)), two equal bonds \(k=1400\ \mathrm{N\,m^{-1}}\), longitudinal motion, no walls. Find the two nonzero longitudinal frequencies.
Solution
\(\mathbf K=k\begin{pmatrix}1&-1&0\\ -1&2&-1\\ 0&-1&1\end{pmatrix}\), \(\mathbf M=\mathrm{diag}(m,M,m)\). The three roots of \(\det(\mathbf K-\omega^2\mathbf M)=0\) are: (i) \(\omega=0\) (translation, \(\mathbf a\propto(1,1,1)\)); (ii) symmetric stretch with the central atom fixed, \(\mathbf a\propto(1,0,-1)\), giving \(\omega_s=\sqrt{k/m}\); (iii) antisymmetric stretch, \(\mathbf a\propto(1,-2m/M,1)\), giving \(\omega_a=\sqrt{k(1/m+2/M)}=\sqrt{(k/m)(1+2m/M)}\). Numbers: \(m=2.656\times10^{-26}\ \mathrm{kg}\), \(k/m=1400/2.656\times10^{-26}=5.27\times10^{28}\ \mathrm{s^{-2}}\Rightarrow\omega_s=2.30\times10^{14}\ \mathrm{rad\,s^{-1}}\). Factor \(1+2m/M=1+2(16/12)=3.667\Rightarrow\omega_a^2=1.93\times10^{29}\), \(\omega_a=4.40\times10^{14}\ \mathrm{rad\,s^{-1}}\). Both lie in the infrared, the correct order of magnitude for CO\(_2\) stretches; the antisymmetric mode is higher because the light atoms move against the recoiling central mass.