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Adding a spatially constant function of time to a velocity potential leaves its velocity $\mathbf{u}=\nabla \varphi$ unchanged. When integrating the Euler equations in space, this freedom can absorb the time-only integration term in the potential form of momentum balance. A spatially constant zero-frequency potential represents no acoustic velocity and must not be counted as a propagating normal mode merely because a frequency determinant vanishes at zero. Pressure conventions and time dependence must still be fixed consistently when the potential is used to recover physical fields.

An antinode is a position of maximum oscillation amplitude in a standing wave. For $u_n=A\sin (n\pi x/L)\cos ({\omega }_{n}t+\phi )$, the antinodes occur atwhere the amplitude is $|A|$. They sit halfway between consecutive nodes of a standing wave. For the first normal mode, the sole antinode is $x=L/2$. It reaches zero displacement at a quarter oscillation period while moving fastest, which illustrates why an antinode is defined by oscillation amplitude rather than instantaneous height.

An antiresonance is a suppression of a specified measured response through cancellation of contributions, often between normal modes. It does not generally mean that the entire system is motionless. In an ideal undamped two-oscillator system, the directly driven oscillator can have zero displacement while its partner still moves.

For a cosine drive of amplitude $h_0$, the periodic response of an end-driven two-bead string has signed amplitudes $A_1=2h_0/[(2-s)(4-s)]$ and $A_2=(3-s)A_1$, where $s={\omega }^2/{\omega }_0^2$. At $s=3$, $A_2=0$ while $A_1=-2h_0$. Equal and opposite transverse forces from the driven end and moving first bead cancel at the second bead. This antiresonance differs from a natural resonance: it is a zero of one measured response between the two natural frequencies. Free normal modes remain unless initial conditions exclude them.

In a constant-shear Eady model with slopes $\alpha$ and $\gamma$, set $A_0=\Lambda -N^2\alpha /f_0$ and $A_D=\Lambda -N^2\gamma /f_0$. The interior potential vorticity obeys $({\partial }_t+\Lambda z{\partial }_x)q^{\prime }=0$. In the zero-interior-PV subspace, integrating ${\psi }_x^{\prime }{q}^{\prime }$ by parts and using the two buoyancy boundary equations proves conservation of the displayed boundary quadratic form. Opposite signs of $A_0,A_D$ make it definite and exclude growing normal modes. For arbitrary nonzero interior PV, its derivative is $(N^2/f_0^3)\int {\psi }_x^{\prime }{q}^{\prime }dV$ and need not vanish. The restriction to zero interior PV is automatic for a growing smooth mode with complex phase speed.

A breathing mode is a collective normal mode in which a structure expands and contracts radially while preserving its overall symmetry. For a uniform scaling pattern, the displacement of point $i$ is proportional to its position relative to the center of mass, as displayed above. In a central spring framework this pattern is a normal mode only when applying the stiffness matrix returns a multiple of the same pattern with the appropriate mass matrix; radial motion alone does not establish an eigenvector.

A normal mode involving substantial stretching of a carbon–carbon double bond. conjugated systems alter its force constant and infrared intensity.

For a series of comparable metal carbonyls, greater metal-to-CO pi backbonding populates CO antibonding orbitals, weakens its internal bond and lowers its infrared stretching frequency. Coupled normal modes and changing geometry must be considered when comparing observed bands.

A normal mode with substantial variation of a carbonyl bond length. Its wavenumber depends on the effective force constant and coupled motions; substitution, conjugation and measurement conditions matter.

Releasing the central bob of three identical spring-coupled simple pendulums from displacement $d$ excites only the symmetric normal modes. In displacement ratios, $(0,d,0)=(d/3)(1,1,1)-(d/3)(1,-2,1)$. Thus the outer coordinates are $(d/3)(\cos {\omega }_{1}t-\cos {\omega }_{3}t)$ and the central coordinate is $(d/3)(\cos {\omega }_{1}t+2\cos {\omega }_{3}t)$, where ${\omega }_1^2=g/l$ and ${\omega }_3^2=g/l+3k/m$. The antisymmetric mode is absent by reflection symmetry. The central modulation envelope is $|d|\sqrt{5+4\cos [({\omega }_3-{\omega }_1)t]}/3$, so it never vanishes completely. Slow beats require a small frequency difference; the decomposition itself is exact in the linear model.

For point masses joined by unstretched central springs obeying Hooke's law, let $n_{ij}$ be the equilibrium unit vector along an edge and $u_i$ the small displacement of its endpoint. The leading extension is $n_{ij}\cdot (u_j-u_i)$, so its quadratic elastic potential energy is $k_{ij}[n_{ij}\cdot (u_j-u_i){]}^2/2$. Consequently the block stiffness matrix has off-diagonal blocks $K_{ij}=-k_{ij}{n}_{ij}{n}_{ij}^T$ and diagonal blocks $K_{ii}={\sum }_j{k}_{ij}{n}_{ij}{n}_{ij}^T$, with absent edges contributing zero. It is symmetric and positive semidefinite. The normal modes solve $Kq={\omega }^{2}Mq$, where $M$ is the mass matrix. Rigid translations and infinitesimal rotations are rigid-motion zero modes of a spring framework. A prestressed spring can add transverse quadratic terms, so the unstretched assumption matters.

An isotropic planar harmonic oscillator of positive charge in a perpendicular magnetic field has two circular normal modes of opposite handedness. With ${\omega }_c=qB/m>0$ and ${\omega }_0^2=k/m$, the signed angular frequencies have the displayed form for trajectories $x+iy\propto e^{i\omega t}$. A static electric field shifts the equilibrium without changing these frequencies. The clockwise mode has the larger frequency magnitude.

A normal mode of a uniformly bending elastic filament fixed in position and slope at both ends satisfies $W^{\prime \prime \prime \prime }=k^{4}W$. With $\beta =kL>0$, the allowed wavenumbers solve $\cos \beta \cosh \beta =1$. The first root is $4.730040745$ and ${\beta }_n=(n+\frac{1}{2})\pi +O(e^{-(n+1/2)\pi })$ for $n\ge 1$. The apparent root at zero is not an eigenfunction: the zero-eigenvalue cubic polynomial satisfying all four clamped boundary conditions is identically zero. This spectrum differs from that of a filament clamped only at one end.

A symmetric confocal optical resonator has two concave mirrors of equal curvature $R$, separated by $R$. Its symmetric fundamental Gaussian beam has a central beam waist and Rayleigh range $R/2$. Its ideal transverse spectrum has degeneracies, so transmission at one resonance need not certify exclusive excitation of the fundamental normal mode.

The elastic energy $U=A\int (W^{\prime \prime }{)}^{2}dx/2$ minimized without forcing gives $W^{\prime \prime \prime \prime }=0$. With fixed $\int W^{2}dx$, stationarity of $U-A{k}^4\int W^{2}dx/2$ gives $W^{\prime \prime \prime \prime }=k^{4}W$: a bending eigenmode. A general admissible shape expands in these normal modes. The natural boundary conditions for a free endpoint are $W^{\prime \prime }(L)=W^{\prime \prime \prime }(L)=0$, while the clamp imposes $W(0)=W^{\prime }(0)=0$.

Coupled oscillators exchange energy through interactions. Linear equations can be transformed into normal mode coordinates; the spatial pattern of an applied force determines which normal modes are driven.

Let two series loops have inductances $L_1,L_2>0$, capacitances $C_1,C_2>0$, total electrical resistances $R_1,R_2>0$, and reciprocal mutual inductance $M=k\sqrt{L_1{L}_2}$ with $0\le k<1$. For a steady sinusoid of angular frequency $\omega >0$, choose complex peak amplitudes with time convention $e^{\mathrm{i}\omega t}$. An ideal source of peak electric potential difference $V$ drives loop 1; loop 2 has no independent source. ThenEliminating $I_1$ proves the response formula. Reversing one coil changes the sign of $M$ and the relative wave phase, but leaves this amplitude response unchanged. The measured sense ratio is $|R_s{I}_2/V|$, with $R_s$ included in $R_2$, not added a second time.

For identical loops at their uncoupled resonant frequency ${\omega }_0=1/\sqrt{LC}$,Writing $x={\omega }_{0}M/R$ makes this $x/[R(1+x^2)]$. Its derivative has the sign of $1-x^2$, so response at this fixed frequency is largest when $x=1$, equivalently $k=1/Q$. Too little coupling transfers little excitation; excessive coupling reduces the response at the original centre frequency. This is a fixed-frequency optimum, not a universal optimum over every possible receiver and drive frequency.

For the chosen values, $k=1/Q=0.063246$ gives $M=0.31623\mathrm{mH}$. At source peak $V=0.100\mathrm{V}$, the secondary peak electric current is $0.500\mathrm{mA}$ and the sense peak electric potential difference is $25.0\mathrm{mV}$. The primary peak electric current is also $0.500\mathrm{mA}$, so source mean power is $25.0\mu \mathrm{W}$: each loop dissipates $12.5\mu \mathrm{W}$, checking conservation of energy. Here $V$ is the equivalent ideal source behind $R_1$; a generator's terminal reading must be converted using its measured Thévenin resistance.

In the lossless identical-loop limit, the normal modes have equal or opposite electric currents, making their effective inductances $L+M$ or $L-M$. ThusAt $k=0.20$ these are $45.944$ and $56.270\mathrm{kHz}$. Finite electrical resistance broadens and shifts the driven peaks, so the exact response formula, rather than these lossless values, generates the plot. Changing a capacitor moves the uncoupled tuning through $f_0\propto C^{-1/2}$: increasing it by $21\%$ lowers $f_0$ by the factor $1/1.1$. The model explains why tuning and coupling must be adjusted together; it supplies no transatlantic propagation or detector-threshold prediction.

Two identical torsional loads of moment of inertia $I$ hang in series from massless fibres of lengths ${\ell }_1,{\ell }_2$ and common filament torsional stiffness $K_t$. Put $R={\ell }_2/{\ell }_1$ and ${\omega }_0^2=K_t/(I{\ell }_1)$. The elastic energy is $K_t{\varphi }_1^2/(2{\ell }_1)+K_t({\varphi }_2-{\varphi }_1{)}^2/(2{\ell }_2)$ and the kinetic energy is $I({\dot{\varphi }}_1^2+{\dot{\varphi }}_2^2)/2$. Their normal modes have the displayed squared angular frequencies. Writing $\lambda ={\omega }^2/{\omega }_0^2$, their amplitude ratio is $A_2/A_1=1/(1-R\lambda )$. The lower mode twists in phase and approaches a combined inertia $2I$ as $R\to 0$, giving ${\omega }_{-}^2\to {\omega }_0^2/2$. The upper mode approaches the original single-load oscillation as $R\to \infty$, giving ${\omega }_{+}^2\to {\omega }_0^2$.

For an axisymmetric normal mode in a rigid rotating sphere, substitute $W=z(A{R}^2+B{z}^2+C)$ into the pressure equation and its impermeability condition. With $\beta =1-4{\Omega }^2/{\sigma }^2$, the interior equation gives $2A+3\beta B=0$. The boundary polynomial gives $3\beta B=(2+\beta )A$ and $\beta C=-(2+\beta )A{r}_0^2$. A nontrivial oscillatory solution has $\beta =-4$, hence the displayed frequency and $B=A/6$, $C=-A{r}_0^2/2$. It satisfies both equations identically.

The cutoff frequency of a membrane waveguide is the lowest frequency at which a specified transverse normal mode can propagate longitudinally. For fixed edges separated by $b$, ${\omega }_{c,n}=n\pi v/b$ and $f_{c,n}={\omega }_{c,n}/(2\pi )$. Below cutoff the longitudinal wavenumber is imaginary and a bounded excitation is an evanescent wave. Above it, $v_g=v\sqrt{1-({\omega }_{c,n}/\omega {)}^2}$ and $v_p{v}_g=v^2$.