Codex @codex  0

Response and resonance. The terms $m\ddot{x}$, $b\dot{x}$ and $kx$ describe inertia, linear viscous resistance and the spring's restoring force; the right side is the applied harmonic force. Substituting the complex trial motion and taking real parts givesThe steady response excludes the decaying homogeneous transient. Its magnitude is $[(k-m{\omega }^2{)}^2+b^2{\omega }^2{]}^{-1/2}$. Differentiating the denominator gives $2\omega [-2m(k-m{\omega }^2)+b^2]=0$, soA nonzero-frequency amplitude maximum exists only for $b^2<2km$; otherwise the maximum is at zero frequency. Increasing damping lowers and broadens the peak and shifts it downward. Every curve starts at $|R(0)|=1/k$ and tends to $1/(m{\omega }^2)$ at high frequency.

Series circuit. The PDF circuit has the resistor, inductor and capacitor in one loop. With current $I=\dot{q}$, Kirchhoff's voltage law gives $V=L\dot{I}+RI+q/C$, henceThus $m,b,k$ correspond to $L,R,1/C$. The resistor's rms voltage isIt is maximized exactly at ${\omega }_{\max }={\omega }_0=1/\sqrt{LC}$ for $R>0$. The charge-amplitude resonance is lower. Expanding its square root for $R^{2}C/L\ll 1$ givesThe two maxima refer to different observables: resistor voltage follows current, whereas charge follows displacement.

Two parallel partially reflecting surfaces in a Fabry–Pérot interferometer produce a sequence of coherent transmitted beams. For equal lossless mirror intensity reflectance $\mathcal{R}$ and round-trip phase $\delta =4\pi nd\cos {\theta }_t/\lambda$, summing the geometric series of amplitudes givesresonances have $2nd\cos {\theta }_t=m\lambda$ and unit peak transmission in this symmetric lossless model. Varying angle gives concentric equal-inclination rings; changing spacing scans the resonance wavelengths.

At normal incidence with negligible dispersion, the free spectral range is $\Delta {\nu }_{\mathrm{FSR}}=c/(2nd)$. For high reflectance, the peak full width gives finesse and resolving powerThe coefficient $\mathcal{F}$ in the transmission formula and the finesse $\mathcal{F}$ are different quantities. Multiple round trips sharpen a peak, allowing high-resolution spectroscopy, frequency selection and cavity measurements. Absorption, mirror mismatch, finite angular spread, imperfect parallelism and surface errors reduce usable transmission or finesse. Adjacent orders can overlap in a broadband measurement, requiring order selection or prior spectral information.

For pendulum length $\ell$, small angular displacement gives $x\simeq \ell \theta$ and horizontal restoring force $-mgx/\ell$. Thus ${\omega }_0^2=g/\ell$ andFor $\omega \ne {\omega }_0$, the complementary solution and a sinusoidal particular solution giveInitial rest at the origin fixes $A=0$ and $B=-F_0\omega /[m{\omega }_0({\omega }_0^2-{\omega }^2)]$, givingAt exact resonance the corresponding limit is $x=F_0[\sin {\omega }_{0}t-{\omega }_{0}t\cos {\omega }_{0}t]/(2m{\omega }_0^2)$; unrestricted initial conditions add the complementary solution.

For $\omega =2{\omega }_0$, set $\tau ={\omega }_{0}t$. Then $m{\omega }_0^{2}x/F_0=(2\sin \tau -\sin 2\tau )/3$: the period is $2\pi$ in $\tau$, zeros are at integer multiples of $\pi$, and the first extrema are $(2\pi /3,\sqrt{3}/2)$ and $(4\pi /3,-\sqrt{3}/2)$. Both initial displacement and slope vanish.

Substitution of the damped complex ansatz gives $[m({\omega }_0^2-{\omega }^2)+ib\omega ]x_0=F_0$. HenceFor the near-resonant mechanical susceptibility, put $\gamma =b/(2m)\ll {\omega }_0$ and $\delta =\omega -{\omega }_0$, and neglect ${\delta }^2$ and $b\delta$ near resonance:Thus $|R{|}^2\simeq [4m^2{\omega }_0^2({\delta }^2+{\gamma }^2){]}^{-1}$, with angular-frequency full width $2\gamma =b/m$. The displacement lags the force by $\pi /2$ at ${\omega }_0$.

Take $m,k,b>0$, and define ${\omega }_0=\sqrt{k/m}$ and $\gamma =b/(2m)$. Substitution of $e^{rt}$ into the unforced equation gives $r^2+2\gamma r+{\omega }_0^2=0$. The roots determine whether the return is oscillatory.

For return after a displacement in this ideal model, critical damping, case (b), is the usual comfort choice: it removes repeated bouncing while avoiding the slow relaxation mode of strong overdamping. For release from rest it gives a monotone return with the fastest asymptotic decay available without oscillation. This criterion does not prove optimal comfort for every road profile or initial velocity; actual comfort also depends on acceleration, road forcing and suspension travel.

For the forced response, set $\Omega =2\pi u/d$ and use $x_{\mathrm{ss}}=\mathrm{Re}(\hat{x}{e}^{i\Omega t})$. ThenThe general forced solution adds the appropriate homogeneous transient from (a), (b) or (c). Initial data were not specified, so its constants cannot be fixed. Positive damping removes this transient at long times. This solution uses the particular driving force in the supplied equation; it does not silently replace it by a different moving-support model.

Maximising $X$ is minimising its squared denominator $D$. Differentiation givesA nonzero displacement resonance exists only if $b^2<2mk$. Here ${\Omega }_r^2=0.175{\mathrm{s}}^{-2}$, and thereforeThe low-frequency limit is $X(0)=A$, whereas $X\to 0$ at large $\Omega$. The parameters give damping ratio $\zeta =b/(2\sqrt{mk})=0.456$, consistent with an underdamped system and the existence of this amplitude maximum.

The Q factor measures stored energy relative to loss: $Q=2\pi E/(\text{energy lost per cycle})$, or $Q={\omega }_{0}E/P_{\mathrm{loss}}$ in the weak-loss limit. For viscous damping $m\ddot{x}+b\dot{x}+kx=0$, write $\gamma =b/(2m)$, ${\omega }_0=\sqrt{k/m}$ and ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}$. The conventional oscillator parameter isIn free ringdown, fit the amplitude envelope $A(t)=A_0{e}^{-\gamma t}$. Its decay time is ${\tau }_A=1/\gamma$, whereas the cycle-averaged energy envelope decays on ${\tau }_E=1/(2\gamma )$; thus $Q={\omega }_0{\tau }_A/2={\omega }_0{\tau }_E$. Equivalently the logarithmic decrement between same-sign peaks is $\delta =\ln (A_n/A_{n+1})=2\pi \gamma /{\omega }_d$. It gives $Q\simeq \pi /\delta$ for weak damping, or the exact parameter relation $Q=\sqrt{1/4+(\pi /\delta {)}^2}$. Fit multiple peaks above the noise floor, measure the damped period, and verify an exponential envelope rather than assuming all amplitude loss is viscous.

For a driven measurement, allow transients to settle and scan frequency at fixed force amplitude. Near resonance and at high $Q$, the response is Lorentzian in displacement power, with half-power points at displacement amplitudes $1/\sqrt{2}$ of the maximum. Their angular-frequency separation gives $Q\simeq {\omega }_0/\Delta \omega$. Halving the amplitude is not a half-power criterion.

The absorbed-power bandwidth of a viscously damped oscillator can be found exactly. Sinceits maximum is at $\Omega ={\omega }_0$, and the half-power frequencies satisfy $|k-m{\Omega }^2|=b\Omega$. HenceFor broad resonances, displacement and absorbed-power peaks differ, so the measured observable must be specified. Frequency-dependent drive, backgrounds or several overlapping modes can otherwise masquerade as a changed Q factor.

The capacitor-charge complex amplitude isFor fixed driving-voltage amplitude, maximizing $|q_0|$ minimizesDifferentiate:The positive-frequency minimum therefore hasThis is the resonance of charge, or capacitor voltage. If the stated inequality fails, their maximum is at zero frequency, and no positive displacement-resonance peak exists. The electric current has the additional factor $\omega$, and instead peaks exactly at ${\omega }_0=1/\sqrt{LC}$ for every $R>0$.

A mass $M$ on a spring of stiffness $K$, subject to linear viscous damping and a sinusoidal drive, obeys the forced harmonic oscillator equationHere $x$ is displacement, $2M\gamma$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, and $F_0$ is a force in newtons. A series RLC circuit is an electrical analogue, with $M\leftrightarrow L$, $2M\gamma \leftrightarrow R$, $K\leftrightarrow 1/C$ and $x\leftrightarrow q$.

For free motion set $F_0=0$ and solve the characteristic equation $s^2+2\gamma s+{\omega }_0^2=0$. The roots $s=-\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}$ distinguish three cases:

The constants are set by initial position and velocity. The mechanical energy and its rate of change in free motion areFor weak damping, the cycle-averaged energy decays approximately as $e^{-2\gamma t}$. The quality factor is $Q={\omega }_0/(2\gamma )$; for $Q\gg 1$ it equals $2\pi$ times stored energy divided by the loss per cycle. Logarithmic decrement of successive free-decay peaks determines $\gamma$ experimentally.

For driven motion the complete answer is the decaying free transient plus a steady response at the driving angular frequency:At low frequency, $X\to F_0/K$ and the displacement is nearly in phase with the drive. At $\Omega ={\omega }_0$, the lag is $\pi /2$ and $X=F_0/(2M\gamma {\omega }_0)$. At high frequency, $X\simeq F_0/(M{\Omega }^2)$ and the lag tends to $\pi$. The displacement resonance maximum lies atonly when $\gamma <{\omega }_0/\sqrt{2}$. Thus an underdamped oscillator need not have a nonzero-frequency displacement resonance peak.

In the steady state the mean input power equals the mean dissipated power:This absorption peaks exactly at $\Omega ={\omega }_0$, rather than at ${\Omega }_r$. Its half-power points have bandwidth $\Delta \Omega =2\gamma$, so $Q={\omega }_0/\Delta \Omega$ for this linear model. For weak damping, a high quality factor therefore means a narrow strong resonance and a long free-decay time. Exactly zero damping at resonant driving has no finite steady response: a secular amplitude grows with time.

A normal mode is a motion in which all coordinates have a fixed relative pattern and oscillate at one frequency. Near a stable equilibrium, let the quadratic kinetic and potential energies be $T={\dot{\mathbf{q}}}^{\mathsf{T}}\mathbf{M}\dot{\mathbf{q}}/2$ and $V={\mathbf{q}}^{\mathsf{T}}\mathbf{Kq}/2$. With positive-definite mass matrix $\mathbf{M}$, the equations and harmonic ansatz giveThe stiffness and mass matrices are symmetric. For distinct frequencies, their symmetry gives $({\omega }_s^2-{\omega }_r^2){\mathbf{a}}_r^{\mathsf{T}}\mathbf{M}{\mathbf{a}}_s=0$, so the mode vectors are orthogonal in the mass inner product. A basis can also be chosen in a degenerate eigenspace. Expanding $\mathbf{q}={\sum }_s{Q}_s{\mathbf{a}}_s$ decouples the equations into ${\ddot{Q}}_s+{\omega }_s^2{Q}_s=0$; initial displacements and velocities determine the coefficients. Positive stiffness gives real frequencies, a zero eigenvalue can describe a free symmetry motion, and a negative one signals instability rather than an oscillatory mode.^[ref]

For two equal masses between fixed walls with three equal springs, the stiffness matrix is $k\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$ and the mass matrix is $m1$. The patterns $(1,1)$ and $(1,-1)$ have frequencies $\sqrt{k/m}$ and $\sqrt{3k/m}$. An initial displacement of one mass excites both modes and can produce beats.

For a uniform taut string fixed at both ends, the wave equation ${\partial }_t^{2}y=c_s^2{\partial }_x^{2}y$, with $c_s=\sqrt{\mathcal{T}/{\mu }_{\ell }}$, has patterns and angular frequenciesThe boundary conditions select the spectrum. Fourier expansion of arbitrary admissible initial data provides the mode amplitudes, illustrating the extension from a finite set of coordinates to a field.

For the normal modes of inductively coupled LC circuits, take two identical circuits with self-inductance $L$, capacitance $C$ and mutual inductance $\mathcal{M}$, where $|\mathcal{M}|<L$. The capacitor charges obey $L{\ddot{q}}_1+\mathcal{M}{\ddot{q}}_2+q_1/C=0$ and the exchanged equation. In-phase and opposite-phase charge patterns haveThe positive inductance matrix for two identical coils is the physical counterpart of positive mass. Here inductance plays the role of the mass matrix. Mechanical, wave and electrical examples therefore share the same eigenvalue structure. Weak forcing excites modes according to its spatial overlap with each pattern; damping broadens the resonances, while strong nonlinearity can invalidate a fixed linear-mode decomposition.

Coupled oscillators have more than one collective pattern of motion. For two equal masses $m$, each attached to a fixed support by a spring of stiffness $k$ and linked to each other by stiffness $\kappa$, small displacements obeyHere equal linear damping $b$ is included to make finite steady-state responses at resonance. The normal mode coordinates $q_{+}=(x_1+x_2)/\sqrt{2}$ and $q_{-}=(x_1-x_2)/\sqrt{2}$ diagonalize these equations. Their natural angular frequencies areThe symmetric normal mode leaves the coupling spring unextended; the antisymmetric normal mode stretches it and therefore has the higher frequency. With $F_{\pm }=(F_1\pm F_2)/\sqrt{2}$, their complex steady-state responses areThus the drive's spatial pattern matters as well as its frequency. Equal in-phase forces drive only the symmetric normal mode; opposite forces drive only the antisymmetric one; driving a single mass generally excites both. Finite damping broadens the resonance peaks and produces a phase lag. In the ideal undamped system, a resonant drive with nonzero projection onto a normal mode produces secular growth rather than a bounded resonant steady state.

The displacement of a particular mass adds contributions from both normal modes. They can cancel at an antiresonance. For $F_2=0$ and no damping, $x_1$ vanishes at ${\Omega }^2=(k+\kappa )/m$, while the partner still moves. This differs from the two nearby resonances, where a mode responds strongly; a small measured response need not mean that no motion or stored energy exists elsewhere.

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 $\mathbf{u}$ describes the mode shape, while $\omega$ is its angular frequency. Linear combinations of modes describe arbitrary sufficiently small motions when the modes form a complete basis.

Let $q_i$ be displacements of independent generalized coordinates from equilibrium. Expand the Lagrangian to quadratic order:The mass matrix $M$ defines a positive-definite quadratic form on the independent coordinates, while the stiffness matrix $K$ 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 $M^{-1/2}K{M}^{-1/2}$. This guarantees real ${\omega }^2$ and a complete orthogonal basis in mass-weighted coordinates.

For distinct eigenvalues, the fact that $K$ and $M$ are real symmetric matrices impliesThus distinct modes are orthogonal in the inner product weighted by the mass matrix. Choosing ${\mathbf{u}}_i^{\mathsf{T}}M{\mathbf{u}}_j={\delta }_{ij}$ decouples the motion into ${\ddot{Q}}_i+{\omega }_i^2{Q}_i=0$, with total energy ${\sum }_i({\dot{Q}}_i^2+{\omega }_i^2{Q}_i^2)/2$. A positive-definite matrix $K$ 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 $m$, each attached to a fixed support by a spring of stiffness $k$ and coupled to the other by a spring of stiffness $k_c$,The symmetric mode ${\mathbf{u}}_{+}\propto (1,1)$ leaves the coupling spring unstretched and has ${\omega }_{+}^2=k/m$. The antisymmetric mode ${\mathbf{u}}_{-}\propto (1,-1)$ stretches it and has ${\omega }_{-}^2=(k+2k_c)/m$. 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 $L$, tension $\mathcal{T}$ and mass per unit length ${\mu }_{\ell }$, 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 $m$ joined by springs $k_s$, with spacing $a$, has travelling normal modes $q_j\propto e^{i(kja-\omega t)}$ 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 $\omega =c|\mathbf{k}|$ in vacuum. An elastic beam instead has a fourth-order spatial equation and flexural dispersion $\omega \propto k^2$. 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 $M$ or $K$ 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.

For the series-fed parallel LC voltage response, use phasors with time dependence $e^{i\omega t}$. The parallel branch has impedance $iX$, where $X=\omega L/(1-{\omega }^{2}LC)$. ThereforeAt the ideal parallel resonance ${\omega }_0=1/\sqrt{LC}$ its input impedance diverges. The input-current phase changes sign on crossing this pole; expressions there are interpreted by limits.

The capacitor voltage equals the parallel-branch voltage. Define $q=R(1-{\omega }^{2}LC)/(\omega L)$. ThenAt low frequency $H\sim i\omega L/R$; at high frequency $H\sim -i/(\omega RC)$; at resonance $H=1$. Thus it is a band-pass voltage filter or resonant frequency selector. Its phase decreases continuously from $+\pi /2$ through zero to $-\pi /2$. Half-power points satisfy $|q|=1$ and

For the supplied component values and $\omega =2\pi (50){\mathrm{s}}^{-1}$, $X=421.745\Omega$ and $V_{C,\mathrm{pk}}=99.3046\mathrm{V}$. The element peak currents areInductor and capacitor currents are opposite phasors; their magnitudes subtract here rather than add. The resistor carries the input current, withThese results assume ideal components and sinusoidal steady state.

In sinusoidal steady state, complex impedance is the voltage phasor divided by the current phasor. Its modulus is the amplitude ratio and its argument is the voltage lead. With the $e^{i\omega t}$ convention, $I=dq/dt=i\omega CV$, so $Z_C=1/(i\omega C)$.

The PDF labels $V_1$ across $R_1$, not across the complete driven network. If $I$ is the current through $R_1$, then $V_1=I{R}_1$ and $I=Y_p{V}_2$, whereThe resistor-referenced parallel-resonator response isWriting $b(\omega )=\omega C-1/(\omega L)$,The maximum occurs at ${\omega }_0=1/\sqrt{LC}$, where $|H|=R/R_1$ and the phase is zero. Below resonance the phase is positive, tending to $+\pi /2$; above it the phase is negative, tending to $-\pi /2$. The amplitude tends to zero at both extremes: the inductor shorts the parallel network at low frequency, and the capacitor does so at high frequency. At resonance their reactive currents cancel, leaving only the resistor current.

If instead $V_{\mathrm{in}}$ denotes the voltage across the complete circuit, its different transfer ratio is $V_2/V_{\mathrm{in}}=1/[1+R_1/R+i{R}_{1}b]$, with peak $R/(R+R_1)$. This distinction follows directly from the printed voltage endpoints.

For sinusoidal steady state with convention $e^{i\omega t}$, complex impedance is the voltage phasor divided by the current phasor. Its modulus is the amplitude ratio and its argument is the voltage-current phase difference. Since $V_L=LdI/dt$,The original circuit is a series RLC circuit with output across the inductor. Its transfer function isFor the original capacitance, the outputs are $10.15\mathrm{mV}$ at $9\mathrm{MHz}$ and $4.65\mathrm{mV}$ at $10\mathrm{MHz}$, a ratio $0.458$. The current resonance is $f_0=1/(2\pi \sqrt{LC})=8.967\mathrm{MHz}$, so the circuit favors the first transmitter.

The resistor voltage is in phase with the current; the inductor voltage leads it by $\pi /2$ and the capacitor voltage lags by $\pi /2$. Large reactive voltages therefore nearly cancel in their phasor sum. A large inductor voltage is consistent with a small supply voltage: it is inductor voltage magnification, not power gain in a passive circuit.

Reducing the capacitance raises $f_0$ to $9.452\mathrm{MHz}$, between the two transmitters. Direct evaluation gives $7.02\mathrm{mV}$ and $7.22\mathrm{mV}$, respectively. Their similar detunings give approximately equal amplitudes; they are not exactly equal, partly because the numerator of $H$ also depends on frequency.

The two-tone output of a linear resonant circuit, for coherent monochromatic sources of equal input amplitude and arbitrary fixed input phases, isWith $A\simeq B=V_o$, this is approximatelyThus a $9.5\mathrm{MHz}$ rapid oscillation has a beat envelope repeating in magnitude at $1\mathrm{MHz}$. The signed envelope cosine has frequency $0.5\mathrm{MHz}$. Unequal amplitudes leave a small nonzero envelope minimum; independently drifting transmitter phases make the beating drift. No new frequency is generated by this ideal linear circuit.

Measure the mass's displacement $x$ from its static equilibrium height. This removes the constant spring extension supporting its weight. If the spring's top moves with the box, its incremental extension is $y_2-x$, so its incremental upward force is $k(y_2-x)$. The specified damping force is against the mass's velocity in the static frame. Newton's second law therefore givesIn absolute height coordinates an additional constant on the right expresses the spring's rest length, its fixed attachment height and gravity; translating to static equilibrium removes it. Replacing the given damping by $-b(\dot{x}-{\dot{y}}_2)$ would describe a different base-excited oscillator.

After the decaying transient, write the steady-state response as $x=A(\omega )\cos (\omega t-\Phi )$. Using a complex sinusoid givessoFor positive frequency, the phase lag lies between zero and $\pi$, passing through $\pi /2$ at ${\omega }_0=\sqrt{k/m}$. For small $b$, the frequency response begins at $a_s$, develops a narrow resonance near ${\omega }_0$, and falls asThe precise amplitude-peak frequency is $\sqrt{k/m-b^2/(2m^2)}$ when that expression is positive.

The pen records the relative displacement $z=x-y_2$. Its complex transfer function isAt high frequency, $x$ is small while $z\simeq -y_2$. The inertial mass then stays nearly stationary and the recorded relative motion reproduces the ground displacement, with reversed sign. At low frequency the mass follows the box and their relative motion is small; near resonance the response is strongly frequency-dependent. Here “high” means compared with the suspension's natural frequency, rather than an absolute seismic-frequency threshold.

Comparing with the standard wave equation gives the shallow-water wave speedTwice differentiating the proposed standing wave multiplies its spatial factors by $-k_x^2$ and $-k_y^2$, and its temporal factor by $-{\omega }^2$. It solves the wave equation precisely whenTo obtain the local gravitational potential energy, integrate through a water column of height $d+h$ and unit horizontal area:The time averages are $⟨h⟩=0$ and $⟨h^2⟩=A^2{\cos }^2(k_{x}x){\cos }^2(k_{y}y)/2$. ThusThe constant is retained because the specified zero of gravitational potential is the tank bottom. This is local time-averaged potential energy, not the sum of potential energy and kinetic energy.

The no-normal-flow wall conditions are Neumann boundary conditions. Differentiating the cosine factors at the opposite walls givesNegative integers give the same cosine modes. The formal $m=n=0$ case has $\omega =0$: it is a uniform change of the mean level, not a finite-frequency oscillation. In a closed tank with fixed water volume even that mean-level change is excluded from the oscillatory disturbance.

Since $a<b$, the lowest finite frequency is the mode $(m,n)=(0,1)$:It is uniform across the shorter direction, has opposite-sign displacement antinodes at the two ends of the longer direction, and a displacement node midway along it.

For the open-ended Bay of Fundy model, approximate the ocean mouth as a displacement node and the closed head as an antinode. The fundamental is a quarter-wave basin resonance, so its wavelength is $4L$ andTaking a roughly $12$-hour semidiurnal tide gives $d\simeq 79\mathrm{m}$. Using the principal lunar period of approximately $12\mathrm{h}25\min$ gives $d\simeq 73\mathrm{m}$. The idealized estimate is therefore about $7\text{–}8\times 10^1\mathrm{m}$. The lunar period is half of a roughly $24$-hour $50$-minute lunar day.^[ref]

This estimate treats the mouth as an ideal open boundary and replaces the real basin by a uniform depth. The supplied large tidal height motivates resonance; it is not enough by itself to determine damping or an amplification factor in this model.

Use distinct symbols for the dimensions of mass, length, time, temperature and amount of substance. The molar gas constant has dimensions $\mathsf{M}{\mathsf{L}}^2{\mathsf{T}}^{-2}{\mathsf{\Theta }}^{-1}{\mathsf{N}}^{-1}$, while the given $M$ is a molar mass with dimensions $\mathsf{M}{\mathsf{N}}^{-1}$. Matching dimensions in the proposed power law giveswith the amount-of-substance equation repeating $\alpha +\gamma =0$. ThereforeDimensional analysis fixes these exponents but not a dimensionless multiplier. For example, an open-tube mode number and the gas's heat-capacity ratio enter the actual resonance frequency without changing the dimensions.

Expand the cloud's potential energy about a stable equilibrium: $U(z)=U(0)+U^{\prime }(0)z+U^{\prime \prime }(0)z^2/2+\cdots$. Equilibrium gives $U^{\prime }(0)=0$, stability gives $U^{\prime \prime }(0)=k>0$, and reflection symmetry also removes odd terms. For small displacements, $U-U(0)\simeq k{z}^2/2$ and $F=-dU/dz\simeq -kz$.

Including the electric and damping forces, Newton's second law is $m\ddot{z}=qE(t)-b\dot{z}-kz$. Dividing by $m$ gives the Lorentz oscillator modelThe specified convention uses $2\gamma$ as the damping coefficient per unit mass.

For the steady response, represent the field by $E_0{e}^{i\omega t}$ and the displacement by $\tilde{z}{e}^{i\omega t}$, taking real parts for physical quantities. Substitution givesDefine the nonnegative amplitude $z_0=|\tilde{z}|$ and $\delta =atan2(2\gamma \omega ,{\omega }_0^2-{\omega }^2)$. ThenFor positive $q$, $\varphi =-\delta$. For electrons $q<0$, the phase of a negatively charged driven oscillator is $\varphi =\pi -\delta$: the low-frequency displacement is opposite to the field, rather than in phase with it. A signed amplitude could absorb this extra $\pi$, but the phase convention must then be stated. The induced dipole $p=qz$ has phase $-\delta$ for either charge sign. Transient free oscillations have been omitted because the question asks for steady state.

For the electron cloud, $q/m=-e/m_e$ andThe visible range corresponds to $\omega =2\pi c/\lambda$ between $2.35\times 10^{15}$ and $4.71\times 10^{15}\mathrm{rad}{\mathrm{s}}^{-1}$, below this resonance. The resonant wavelength is about $180\mathrm{nm}$, in the ultraviolet.