Coupled Oscillators and Normal Modes

Derivation

Coupled Oscillators and Normal Modes

D-033 Home PU-101 Threads waves · energy · symmetry Depends on Small Oscillations and Simple Harmonic Motion, symmetric-matrix-orthogonal-diagonalization

Statement

For a mechanical system near a stable equilibrium, the linearized equations of motion \( \mathbf{M}\ddot{\mathbf{q}} = -\mathbf{K}\mathbf{q} \), with symmetric positive-definite mass matrix \( \mathbf{M} \) and symmetric positive-semidefinite stiffness matrix \( \mathbf{K} \), are decoupled by the generalized eigenvalue problem \( \mathbf{K}\mathbf{a}^{(r)} = \omega_r^2\,\mathbf{M}\mathbf{a}^{(r)} \): the eigenvectors \( \mathbf{a}^{(r)} \) are the normal modes (rigid spatial patterns that oscillate in unison) and the eigenvalues \( \omega_r^2 \) are the squared normal frequencies, so every free motion is a superposition \( \mathbf{q}(t)=\sum_r \mathbf{a}^{(r)}\,(A_r\cos\omega_r t + B_r\sin\omega_r t) \).

Why it matters

A system of \( n \) coupled oscillators looks intractable: every coordinate pushes on every other, and no single coordinate executes simple harmonic motion. The normal-mode theorem states that the coupling is an artifact of the coordinates. There exists a linear change of variables in which the \( n \)-body problem falls apart into \( n \) independent one-dimensional harmonic oscillators, each with its own frequency. Solving \( n \) coupled second-order ODEs is replaced by one algebraic eigenvalue problem.

This is the archetype for the whole of linear physics near equilibrium. Phonons in crystals, molecular vibrational spectra, the modes of a loaded string or a bridge, LC ladder filters, and — after the same diagonalization is quantized — the treatment of a field as independent mode oscillators all rest on this construction. The eigenvectors also expose symmetry: modes organize into patterns dictated by the symmetry group of the equilibrium.

Assumptions

Motion stays near a stable equilibrium so the potential may be truncated at quadratic order.If displacements are large the neglected cubic and higher terms matter, the equations become nonlinear, modes couple and exchange energy, and no fixed frequencies exist.

The kinetic energy is a positive-definite quadratic form in the velocities, giving a symmetric positive-definite mass matrix \( \mathbf{M} \).If \( \mathbf{M} \) is singular some coordinate carries no inertia, the generalized eigenvalue problem is ill-posed, and that direction is not a genuine dynamical degree of freedom.

The equilibrium is stable, so the Hessian \( \mathbf{K} \) of the potential is positive-semidefinite.If \( \mathbf{K} \) has a negative eigenvalue the corresponding \( \omega_r^2<0 \), \( \omega_r \) is imaginary, and that mode grows exponentially rather than oscillating — the equilibrium is unstable.

The system is conservative: forces derive from a potential and there is no dissipation.With damping the equations gain a term \( \mathbf{C}\dot{\mathbf{q}} \); \( \mathbf{M} \) and \( \mathbf{K} \) alone no longer diagonalize simultaneously, modes acquire complex frequencies, and in general no single orthogonal transformation decouples the motion.

The coefficients \( \mathbf{M} \) and \( \mathbf{K} \) are constant in time.If they vary the problem is non-autonomous (parametric), Floquet theory replaces the eigenvalue analysis, and phenomena such as parametric resonance appear.

Derivation

\[ T=\tfrac{1}{2}\sum_{i,j} M_{ij}\,\dot q_i\,\dot q_j=\tfrac{1}{2}\dot{\mathbf{q}}^{\mathsf T}\mathbf{M}\dot{\mathbf{q}},\qquad V=V_0+\tfrac{1}{2}\sum_{i,j}\left.\frac{\partial^2 V}{\partial q_i\,\partial q_j}\right|_0 q_i q_j=\tfrac{1}{2}\mathbf{q}^{\mathsf T}\mathbf{K}\mathbf{q} \]

Choose generalized coordinates \( q_i \) measuring displacement from equilibrium (\( \mathbf{q}=0 \)). Taylor-expand \( V \); the linear term vanishes at equilibrium and \( V_0 \) is an irrelevant constant, leaving a quadratic form. \( K_{ij}=\partial^2 V/\partial q_i\partial q_j|_0 \) is the Hessian, symmetric by equality of mixed partials. A

\[ \frac{d}{dt}\frac{\partial L}{\partial \dot q_k}-\frac{\partial L}{\partial q_k}=0,\qquad L=T-V \;\Longrightarrow\; \mathbf{M}\ddot{\mathbf{q}}+\mathbf{K}\mathbf{q}=\mathbf{0} \]

Apply the Euler–Lagrange equations. With \( T \) quadratic in \( \dot{\mathbf q} \) and \( V \) quadratic in \( \mathbf q \), \( \partial L/\partial\dot q_k=\sum_j M_{kj}\dot q_j \) and \( \partial L/\partial q_k=-\sum_j K_{kj}q_j \); symmetry of \( \mathbf M,\mathbf K \) collapses the double sums. B

\[ \mathbf{q}(t)=\mathbf{a}\,e^{i\omega t},\qquad \mathbf{a}\in\mathbb{C}^n\ \text{constant} \]

Since the coefficients are constant and the system linear, seek solutions in which every coordinate oscillates at one common frequency with a fixed spatial profile \( \mathbf a \). The physical displacement is the real part; the complex form only bookkeeps amplitude and phase. A

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

Substitute the trial form into step 2 and cancel the never-zero scalar \( e^{i\omega t} \). This is a generalized eigenvalue problem: \( \omega^2 \) is the eigenvalue, \( \mathbf a \) the eigenvector, with respect to the pair \( (\mathbf K,\mathbf M) \). B

\[ \det\!\left(\mathbf{K}-\omega^2\mathbf{M}\right)=0 \]

A nonzero \( \mathbf a \) exists only if the matrix is singular. This characteristic equation is a polynomial of degree \( n \) in \( \omega^2 \); its \( n \) roots \( \omega_1^2,\dots,\omega_n^2 \) are the squared normal frequencies. B

\[ \mathbf{M}=\mathbf{M}^{1/2}\mathbf{M}^{1/2},\qquad \mathbf{u}=\mathbf{M}^{1/2}\mathbf{a},\qquad \widetilde{\mathbf{K}}=\mathbf{M}^{-1/2}\,\mathbf{K}\,\mathbf{M}^{-1/2} \;\Longrightarrow\; \widetilde{\mathbf{K}}\,\mathbf{u}=\omega^2\mathbf{u} \]

Because \( \mathbf M \) is symmetric positive-definite it has a unique symmetric positive-definite square root \( \mathbf M^{1/2} \). Mass-weighting converts the generalized problem into an ordinary one. Crucially \( \widetilde{\mathbf K}=\mathbf M^{-1/2}\mathbf K\mathbf M^{-1/2} \) is symmetric: \( \widetilde{\mathbf K}^{\mathsf T}=\mathbf M^{-1/2}\mathbf K^{\mathsf T}\mathbf M^{-1/2}=\widetilde{\mathbf K} \). C

\[ \widetilde{\mathbf{K}}=\mathbf{O}\,\mathrm{diag}(\omega_1^2,\dots,\omega_n^2)\,\mathbf{O}^{\mathsf T},\qquad \mathbf{O}^{\mathsf T}\mathbf{O}=\mathbf{I},\qquad \omega_r^2\in\mathbb{R},\ \omega_r^2\ge 0 \]

By the spectral theorem (prior result: symmetric-matrix-orthogonal-diagonalization) a real symmetric matrix is diagonalized by an orthogonal matrix of orthonormal eigenvectors, and its eigenvalues are real. Positive-semidefiniteness of \( \mathbf K \) (stable equilibrium) forces \( \omega_r^2\ge 0 \), so every \( \omega_r \) is real. C

\[ \mathbf{a}^{(r)}=\mathbf{M}^{-1/2}\mathbf{u}^{(r)}\;\Longrightarrow\;\big(\mathbf{a}^{(r)}\big)^{\mathsf T}\mathbf{M}\,\mathbf{a}^{(s)}=\big(\mathbf{u}^{(r)}\big)^{\mathsf T}\mathbf{u}^{(s)}=\delta_{rs} \]

Transform the orthonormal \( \mathbf u^{(r)} \) back to physical amplitudes. The modes are not orthogonal in the ordinary sense but are \( \mathbf M \)-orthonormal: orthogonal with the mass matrix as metric. This is the inner product under which the modes decouple. C

\[ Q_r(t)\equiv\big(\mathbf{a}^{(r)}\big)^{\mathsf T}\mathbf{M}\,\mathbf{q}(t)\;\Longrightarrow\;\ddot{Q}_r+\omega_r^2\,Q_r=0 \]

Define the normal coordinates by projecting the motion onto each mode with the \( \mathbf M \)-metric. Left-multiplying step 2 by \( (\mathbf a^{(r)})^{\mathsf T} \) and using \( \mathbf K\mathbf a^{(r)}=\omega_r^2\mathbf M\mathbf a^{(r)} \) plus \( \mathbf M \)-orthonormality yields one uncoupled SHM equation per mode (prior result: simple-harmonic-motion-linearization). B

\[ \mathbf{q}(t)=\sum_{r=1}^{n}\mathbf{a}^{(r)}\,Q_r(t)=\sum_{r=1}^{n}\mathbf{a}^{(r)}\big(A_r\cos\omega_r t+B_r\sin\omega_r t\big) \]

Each \( Q_r \) is a free harmonic oscillator with the general solution shown; \( 2n \) constants \( A_r,B_r \) are fixed by the \( 2n \) initial data \( \mathbf q(0),\dot{\mathbf q}(0) \). Reassembling recovers the physical motion as a superposition of modes. A

Result

\[ \mathbf{K}\,\mathbf{a}^{(r)}=\omega_r^2\,\mathbf{M}\,\mathbf{a}^{(r)},\qquad \mathbf{q}(t)=\sum_{r=1}^{n}\mathbf{a}^{(r)}\big(A_r\cos\omega_r t+B_r\sin\omega_r t\big),\qquad \big(\mathbf{a}^{(r)}\big)^{\mathsf T}\mathbf{M}\,\mathbf{a}^{(s)}=\delta_{rs} \]

Reading. The eigenvalues of the pair \( (\mathbf K,\mathbf M) \) are the squared normal frequencies and the eigenvectors are the normal modes — spatial patterns in which every coordinate moves at the same frequency, passing through equilibrium together. Any free motion is a weighted sum of these modes; the weights (normal coordinates) each obey independent simple harmonic motion, so energy sloshes within a mode but never leaks between modes. The modes form an \( \mathbf M \)-orthonormal basis, guaranteeing the decomposition is unique.

Units check. \( \mathbf K \) carries units of stiffness (\( \mathrm{N\,m^{-1}} = \mathrm{kg\,s^{-2}} \)) and \( \mathbf M \) units of mass (\( \mathrm{kg} \)); the eigenvalue \( \omega_r^2 = (\mathbf a^{\mathsf T}\mathbf K\mathbf a)/(\mathbf a^{\mathsf T}\mathbf M\mathbf a) \) has units \( \mathrm{kg\,s^{-2}}/\mathrm{kg}=\mathrm{s^{-2}} \), so \( \omega_r \) is a frequency in \( \mathrm{s^{-1}} \). The relation \( \mathbf K\mathbf a=\omega^2\mathbf M\mathbf a \) balances \( \mathrm{kg\,s^{-2}}\cdot[\mathbf a] = \mathrm{s^{-2}}\cdot\mathrm{kg}\cdot[\mathbf a] \), consistent for any amplitude units \( [\mathbf a] \).

Limiting cases

Zero coupling. If \( \mathbf K \) is diagonal in the chosen coordinates, the modes are the coordinate axes themselves and \( \omega_r^2=K_{rr}/M_{rr} \): the oscillators are already independent.
Identical symmetric pair. Two equal masses \( m \), outer springs \( k \), coupling spring \( k_c \): modes are symmetric \( (1,1) \) with \( \omega_-^2=k/m \) and antisymmetric \( (1,-1) \) with \( \omega_+^2=(k+2k_c)/m \).
Weak coupling \( k_c\ll k \). The two frequencies nearly coincide; energy beats slowly between the masses at the beat frequency \( \omega_+-\omega_-\approx k_c/\sqrt{mk} \).
Degenerate spectrum. If two \( \omega_r^2 \) coincide (often by symmetry), any linear combination within that eigenspace is also a mode; the mode directions are fixed only up to rotation in the degenerate subspace.
Zero mode. A vanishing eigenvalue \( \omega_r^2=0 \) corresponds to a rigid translation or rotation — a free drift \( Q_r=A_r+B_r t \), not an oscillation.

Breaks when

Amplitudes leave the harmonic regime. Once cubic terms in \( V \) are comparable to the quadratic term, the neglected nonlinearity couples the modes: frequencies become amplitude-dependent and energy flows between modes (Fermi–Pasta–Ulam–Tsingou). The fixed-frequency, non-mixing picture fails.
Dissipation or gyroscopic forces are present. A damping matrix \( \mathbf C \) (or a velocity-dependent Coriolis/magnetic term) is generally not simultaneously diagonalizable with \( \mathbf M \) and \( \mathbf K \). Frequencies go complex, modes are no longer real standing patterns, and the clean real-orthogonal decomposition is lost unless \( \mathbf C \) happens to lie in the span of \( \mathbf M \) and \( \mathbf K \) (Rayleigh damping).
The equilibrium is unstable. If \( \mathbf K \) has a negative eigenvalue, the corresponding \( \omega_r^2<0 \) gives a real growth rate: \( Q_r\propto e^{\pm|\omega_r|t} \) runs away and the linear theory only describes the initial departure from equilibrium.
Parameters vary in time. If \( \mathbf M(t) \) or \( \mathbf K(t) \) change (a pumped pendulum, a modulated spring), the autonomous eigenproblem does not apply; Floquet/parametric analysis is required and resonances can pump energy in without bound.

Failure modes

Using ordinary orthogonality instead of \( \mathbf M \)-orthogonality. Requiring \( (\mathbf a^{(r)})^{\mathsf T}\mathbf a^{(s)}=0 \) is wrong when masses differ; modes are orthogonal only in the \( \mathbf M \)-metric, and normal coordinates built the naive way fail to decouple.
Solving \( \mathbf K\mathbf a=\omega^2\mathbf a \) instead of \( \mathbf K\mathbf a=\omega^2\mathbf M\mathbf a \). Dropping \( \mathbf M \) gives wrong frequencies whenever the masses are unequal; the eigenvalues of \( \mathbf M^{-1}\mathbf K \), not of \( \mathbf K \), are the \( \omega_r^2 \).
Assuming \( \mathbf M^{-1}\mathbf K \) is symmetric. The product of two symmetric matrices is generally not symmetric, so one cannot invoke the spectral theorem on it directly; the mass-weighted \( \widetilde{\mathbf K} \) is the symmetric object to diagonalize.
Reading the eigenvector as a displacement rather than an amplitude ratio. A mode fixes only the relative amplitudes and phases; its overall scale is arbitrary and set by initial conditions, not by the eigenvalue problem.
Forgetting the sine term for a mode started with velocity. Writing \( \mathbf q=\sum_r\mathbf a^{(r)}A_r\cos\omega_r t \) alone cannot match nonzero \( \dot{\mathbf q}(0) \); both \( A_r \) and \( B_r \) are needed.
Taking \( \omega_r=\sqrt{\omega_r^2} \) as necessarily real without checking stability. A negative eigenvalue is a physical instability, not an arithmetic slip to be discarded.

Discussion

The deep content is that coupling is coordinate-dependent, not physical. The same motion that looks like an intractable tangle in the position coordinates \( q_i \) is, in the normal coordinates \( Q_r \), simply \( n \) independent pendulums. The eigenvalue problem is precisely the search for the coordinates in which the Lagrangian's two quadratic forms — kinetic and potential energy — are simultaneously diagonal. That two symmetric forms can be diagonalized together, when one of them (\( \mathbf M \)) is positive-definite, is the linear-algebra fact underwriting the whole construction, and it is why the mass matrix acts as the natural metric on configuration space.

Energy is the invariant that makes the modes physically meaningful. In normal coordinates the total energy separates additively, \( E=\sum_r\tfrac12(\dot Q_r^2+\omega_r^2 Q_r^2) \), so each mode carries a conserved energy of its own and there is no cross term to shuttle energy between modes. This is the classical seed of the quantum result that a field is a collection of independent oscillators, each mode later becoming a set of quanta (phonons, photons) with energy \( \hbar\omega_r \). The connection to the waves thread is direct: for a chain of many identical masses the mode index becomes a wavevector and the discrete spectrum \( \omega_r \) becomes a dispersion relation \( \omega(k) \).

Symmetry organizes the spectrum before any calculation. If the equilibrium is invariant under a group of transformations, that group commutes with \( \mathbf M^{-1}\mathbf K \), so eigenvectors fall into the irreducible representations of the group and degeneracies are forced by representation dimension. In a symmetric triatomic molecule, reflection symmetry immediately splits the modes into symmetric and antisymmetric stretches without solving a determinant. This is the entry point from the symmetry thread: group theory predicts how many modes of each type exist, their degeneracies, and their infrared/Raman activity, reducing the eigenvalue problem to labeling.

More abstractly, the generalized eigenvalue problem is a statement about a pencil of quadratic forms. When \( \mathbf M \) is positive-definite, the pencil \( \mathbf K-\lambda\mathbf M \) is equivalent under congruence to a diagonal pencil, and the \( \omega_r^2 \) are congruence invariants — they do not depend on the coordinate choice, unlike the individual matrix entries. If \( \mathbf M \) loses definiteness (constraints, gauge systems), the clean simultaneous diagonalization can fail: Jordan blocks appear, the modes no longer span, and one enters the theory of defective pencils, which is exactly the mathematics behind exceptional points in non-Hermitian and \( \mathcal{PT} \)-symmetric physics.

Common misconceptions. A normal mode is not a single moving object but a synchronized pattern of the whole system; "the first mode" names a shape and a frequency, not one of the masses. Nor does a general motion "have a frequency" — only a pure mode does; a superposition of two modes is genuinely two-tone and, if the frequencies are incommensurate, never exactly repeats. And the modes are a property of the equilibrium configuration, computed once; the initial conditions merely choose how much of each is present.

Worked examples

Two equal masses \( m=0.50\ \mathrm{kg} \), three springs in a line (wall–\(k\)–mass–\(k_c\)–mass–\(k\)–wall), \( k=200\ \mathrm{N\,m^{-1}} \), \( k_c=50\ \mathrm{N\,m^{-1}} \). Find the normal frequencies and modes.

Set up \( \mathbf M,\mathbf K \) from \( m\ddot x_1=-kx_1-k_c(x_1-x_2) \), \( m\ddot x_2=-kx_2-k_c(x_2-x_1) \). A

\[ \mathbf M=m\mathbf I,\quad \mathbf K=\begin{pmatrix}k+k_c & -k_c\\ -k_c & k+k_c\end{pmatrix},\quad \det(\mathbf K-\omega^2 m\mathbf I)=0 \]

Equal masses make \( \mathbf M \) a scalar; the generalized problem reduces to the ordinary eigenproblem of \( \mathbf K \). A

\[ (k+k_c-m\omega^2)^2-k_c^2=0\;\Rightarrow\; m\omega^2=k+k_c\mp k_c=\{\,k,\ k+2k_c\,\} \]

The \( 2\times2 \) determinant factors as a difference of squares. Symmetric mode \( (1,1) \): the coupling spring never stretches, so \( \omega_-^2=k/m \). Antisymmetric mode \( (1,-1) \): coupling spring doubly loaded, \( \omega_+^2=(k+2k_c)/m \). B

\[ \omega_-=\sqrt{\frac{200}{0.50}}=20.0\ \mathrm{s^{-1}},\qquad \omega_+=\sqrt{\frac{200+100}{0.50}}=\sqrt{600}=24.5\ \mathrm{s^{-1}} \]

Insert numbers. Units: \( \sqrt{\mathrm{N\,m^{-1}}/\mathrm{kg}}=\sqrt{\mathrm{s^{-2}}}=\mathrm{s^{-1}} \). A

\[ \omega_-=20.0\ \mathrm{s^{-1}}\ \text{(in-phase, }(1,1)\text{)},\qquad \omega_+=24.5\ \mathrm{s^{-1}}\ \text{(out-of-phase, }(1,-1)\text{)} \]

Reading. The lower frequency is the sway where both masses move together and the middle spring is inert; the higher frequency is the breathing mode that stretches the coupling spring hardest. Units check. Both frequencies in \( \mathrm{s^{-1}} \), and \( \omega_+>\omega_- \) as expected since the antisymmetric mode feels extra restoring force.

Unequal masses on a common spring set: \( m_1=1.0\ \mathrm{kg} \), \( m_2=2.0\ \mathrm{kg} \), outer springs each \( k=300\ \mathrm{N\,m^{-1}} \), coupling \( k_c=100\ \mathrm{N\,m^{-1}} \). Find the normal frequencies.

Now \( \mathbf M \) is not scalar, so the mass matrix must be kept explicitly. A

\[ \mathbf M=\begin{pmatrix}1.0&0\\0&2.0\end{pmatrix},\quad \mathbf K=\begin{pmatrix}400&-100\\-100&400\end{pmatrix},\quad \det(\mathbf K-\omega^2\mathbf M)=0 \]

Diagonal stiffness entries are \( k+k_c=400\ \mathrm{N\,m^{-1}} \); off-diagonals are \( -k_c \). A

\[ (400-\omega^2)(400-2\omega^2)-(-100)^2=0\;\Rightarrow\;2\omega^4-1200\,\omega^2+150000=0 \]

Expand the determinant with the mass-weighted diagonal. Let \( \lambda=\omega^2 \): \( 2\lambda^2-1200\lambda+150000=0 \), i.e. \( \lambda^2-600\lambda+75000=0 \). B

\[ \lambda=\frac{600\pm\sqrt{600^2-4\cdot75000}}{2}=\frac{600\pm\sqrt{60000}}{2}=300\pm122.5 \]

Quadratic formula; \( \sqrt{60000}=244.9 \), half of it is \( 122.5 \). A

\[ \lambda_-=177.5\ \mathrm{s^{-2}},\ \ \omega_-=13.3\ \mathrm{s^{-1}};\qquad \lambda_+=422.5\ \mathrm{s^{-2}},\ \ \omega_+=20.6\ \mathrm{s^{-1}} \]

Take square roots. Amplitude ratio from the top row \( (400-\lambda)a_1=100\,a_2 \): for \( \lambda_- \), \( a_2/a_1=(400-177.5)/100=2.22 \) (in-phase); for \( \lambda_+ \), \( a_2/a_1=(400-422.5)/100=-0.225 \) (out-of-phase, heavy mass barely moves). B

\[ \omega_-=13.3\ \mathrm{s^{-1}}\ (a_2/a_1=+2.22),\qquad \omega_+=20.6\ \mathrm{s^{-1}}\ (a_2/a_1=-0.23) \]

Reading. With unequal masses the modes are no longer clean \( (1,\pm1) \) patterns: the low mode has both masses swinging the same way with the lighter mass leading, the high mode is dominated by the light mass with the heavy one nearly stationary. Units check. \( \lambda \) came out in \( \mathrm{s^{-2}} \) (as \( \mathrm{N\,m^{-1}/kg} \)), roots in \( \mathrm{s^{-1}} \); amplitude ratios dimensionless.

Problems

For two equal masses \( m=0.20\ \mathrm{kg} \) joined only to each other by one spring \( k_c=80\ \mathrm{N\,m^{-1}} \) (no walls), find both normal frequencies and identify the zero mode.
Solution
\( \mathbf M=m\mathbf I \), \( \mathbf K=\begin{pmatrix}k_c&-k_c\\-k_c&k_c\end{pmatrix} \). Eigenvalues of \( \mathbf K \): \( 0 \) and \( 2k_c \). So \( \omega_1=0 \) with mode \( (1,1) \) — rigid translation of the pair (center of mass drifts, \( Q_1=A_1+B_1t \)). \( \omega_2^2=2k_c/m=2(80)/0.20=800\ \mathrm{s^{-2}} \Rightarrow \omega_2=\sqrt{800}=28.3\ \mathrm{s^{-1}} \), mode \( (1,-1) \), the relative stretch. Physically this is the reduced-mass oscillation: \( \omega_2=\sqrt{k_c/\mu} \) with \( \mu=m/2=0.10\ \mathrm{kg} \), giving \( \sqrt{80/0.10}=28.3\ \mathrm{s^{-1}} \).
A single mass \( m=0.40\ \mathrm{kg} \) is fixed between two walls by springs \( k_1=100 \) and \( k_2=300\ \mathrm{N\,m^{-1}} \). Show this is a one-mode system and give its frequency; explain why "normal modes" still applies with \( n=1 \).
Solution
One degree of freedom: \( \mathbf M=(0.40) \), \( \mathbf K=(k_1+k_2)=(400) \). The generalized eigenproblem \( 400=\omega^2(0.40) \) gives \( \omega^2=1000\ \mathrm{s^{-2}} \), \( \omega=\sqrt{1000}=31.6\ \mathrm{s^{-1}} \). With \( n=1 \) the "matrix" is a scalar; the single eigenvector is the coordinate itself and there is nothing to decouple. Normal-mode theory contains ordinary SHM as its \( n=1 \) case: springs in parallel add stiffnesses.
For the symmetric pair of Worked Example 1 (\( m=0.50 \), \( k=200 \), \( k_c=50 \)), the system starts with mass 1 displaced by \( x_1(0)=A_0=0.030\ \mathrm{m} \), mass 2 at rest at origin, both velocities zero. Find \( x_1(t) \) and the beat period.
Solution
Decompose into modes: \( (A_0,0)=\tfrac{A_0}{2}(1,1)+\tfrac{A_0}{2}(1,-1) \). With zero initial velocity only cosines survive: \( x_1(t)=\tfrac{A_0}{2}\cos\omega_-t+\tfrac{A_0}{2}\cos\omega_+t \), \( x_2(t)=\tfrac{A_0}{2}\cos\omega_-t-\tfrac{A_0}{2}\cos\omega_+t \). Using \( \omega_-=20.0 \), \( \omega_+=24.5\ \mathrm{s^{-1}} \): \( x_1(t)=0.015\big(\cos 20.0t+\cos 24.5t\big)\ \mathrm{m}=0.030\cos(2.24 t)\cos(22.2 t)\ \mathrm{m} \) using \( \cos a+\cos b=2\cos\frac{a-b}{2}\cos\frac{a+b}{2} \). Energy transfers fully to mass 2 when the slow envelope \( \cos(2.24t) \) reaches zero: full energy-return period \( 2\pi/(\omega_+-\omega_-)=2\pi/4.49=1.40\ \mathrm{s} \). The envelope first vanishes at \( t=\pi/(2\cdot2.24)=0.70\ \mathrm{s} \), where mass 1 is momentarily still and mass 2 carries all the energy.
Three equal masses \( m \) on a ring joined by three identical springs \( k \) (each mass to its two neighbours). Set up \( \mathbf K \) and find the three normal frequencies, using the ring's symmetry.
Solution
\( \mathbf M=m\mathbf I \); \( \mathbf K=k\begin{pmatrix}2&-1&-1\\-1&2&-1\\-1&-1&2\end{pmatrix} \). The circulant is diagonalized by \( a_j^{(p)}=e^{2\pi i pj/3} \), \( p=0,1,2 \), with eigenvalues \( k(2-2\cos\frac{2\pi p}{3}) \). \( p=0 \): \( \lambda=0 \Rightarrow \omega_0=0 \) (rigid rotation of the ring). \( p=1,2 \): \( 2-2\cos120^\circ=2-2(-\tfrac12)=3 \), so \( \omega^2=3k/m \), \( \omega=\sqrt{3k/m} \), a doubly degenerate pair (the two counter-rotating standing patterns). Thus the spectrum is \( \{0,\ \sqrt{3k/m},\ \sqrt{3k/m}\} \); the degeneracy is forced by the 3-fold rotational symmetry.
Two pendulums of length \( \ell=1.0\ \mathrm{m} \), bob mass \( m \), are coupled by a weak spring giving \( k_c/m=0.20\ \mathrm{s^{-2}} \); take \( g=9.8\ \mathrm{m\,s^{-2}} \). Find both frequencies and the fractional splitting.
Solution
Small-angle EOM: \( \ddot\theta_1=-\frac{g}{\ell}\theta_1-\frac{k_c}{m}(\theta_1-\theta_2) \) and symmetrically. In-phase mode \( (1,1) \): spring relaxed, \( \omega_-^2=g/\ell=9.8\ \mathrm{s^{-2}} \Rightarrow \omega_-=3.13\ \mathrm{s^{-1}} \). Out-of-phase mode \( (1,-1) \): \( \omega_+^2=g/\ell+2k_c/m=9.8+0.40=10.2\ \mathrm{s^{-2}} \Rightarrow \omega_+=3.19\ \mathrm{s^{-1}} \). Fractional splitting \( (\omega_+-\omega_-)/\omega_-\approx \tfrac12\,\frac{2k_c/m}{g/\ell}=0.20/9.8=0.020 \), about 2%. Beat period for energy exchange \( 2\pi/(\omega_+-\omega_-)=2\pi/0.063=100\ \mathrm{s} \) — slow sloshing characteristic of weak coupling.