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The second-order central difference approximation to the fixed-end wave equation on $n$ interior points has normal modesHere $h=L/(n+1)$ and the endpoint displacements are zero. The eigenvectors form an orthonormal basis in the ordinary Euclidean norm, without an extra grid-spacing weight. They sample the continuous standing wave shapes exactly, while their angular frequencies differ from $k\pi c/L$ because of spatial discretization. For fixed $k$, the difference tends to zero as $h\to 0$.

Identical linear springs of stiffness $k$ have equivalent stiffness $2k$ in parallel and $k/2$ in series. Parallel springs have equal displacement and add forces; series springs carry equal force and add extensions. A mass $m$ attached to an equivalent massless spring has angular frequency $\sqrt{k_{\mathrm{eff}}/m}$.

Consider a stipulated series circuit containing an ideal resistor of electrical resistance $R=100\Omega$ and an ideal capacitor of capacitance $C=1\mu \mathrm{F}$. The same electric current passes through both. Their terminal electric potential differences add by Kirchhoff's voltage law, soA quick derivation uses $i=Cd{v}_C/dt$: taking the derivative of the complex exponential multiplies the capacitor voltage phasor by $\mathrm{i}\omega$, hence ${\hat{V}}_C=\hat{I}/(\mathrm{i}\omega C)$. Dimensionally, $1/(\omega C)$ has units $s/F=V/A=\Omega$, matching $R$.

At frequency $f=1000\mathrm{Hz}$, the angular frequency is $\omega =2\pi f$. The signed capacitive electrical reactance, modulus and complex argument of the total electrical impedance areApply $V_{\mathrm{rms}}=1.00\mathrm{V}$ across the complete series circuit. Its root mean square electric current isWith voltage as the zero-phase reference, the current phasor has complex argument $+57.858092^{\circ }$: electric current leads. The phasor drops have moduli $V_{R,\mathrm{rms}}=0.532018\mathrm{V}$ and $V_{C,\mathrm{rms}}=0.846733\mathrm{V}$; these add as complex numbers, not as positive magnitudes. Their squared magnitudes sum to $(1.00\mathrm{V}{)}^2$.

The mean absorbed power uses the root mean square values:The ideal capacitor has zero cycle-averaged absorbed power; all the mean absorbed power becomes Joule heating in the resistor.^[ref] To see why the phase factor appears, average $p(t)=v(t)i(t)$ over a full cycle with $v(t)=\sqrt{2}{V}_{\mathrm{rms}}\cos \omega t$ and $i(t)=\sqrt{2}{I}_{\mathrm{rms}}\cos (\omega t-\varphi )$. The cosine addition formula reduces the average to $V_{\mathrm{rms}}{I}_{\mathrm{rms}}\cos \varphi$; the term at $2\omega$ averages to zero. Independently, $p_C=d(C{v}_C^2/2)/dt$, and the capacitor energy returns to its starting value after each cycle.

At fixed applied $V_{\mathrm{rms}}$, the limit $\omega \to 0^{+}$ gives $|Z|\to \infty$, $\varphi \to -\pi /2$ and $I_{\mathrm{rms}}\to 0$: the ideal capacitor blocks steady electric current. The limit $\omega \to \infty$ gives $Z\to R$, $\varphi \to 0$ and $I_{\mathrm{rms}}\to V_{\mathrm{rms}}/R$. These are limits of the ideal lumped electrical circuit; unwanted lead inductance and component losses can invalidate that model at high frequency. At $\omega RC=1$, the equal legs of the impedance right triangle give $|Z|=\sqrt{2}R$ and $\varphi =-\pi /4$ without numerical evaluation. Measuring both the magnitude and the sign of the phase difference distinguishes this capacitor response from an equally large positive electrical reactance.

In a series RLC circuit, the same electric current $i=q^{\prime }$ flows through the resistor with electrical resistance $R$, the inductor, and the capacitor. Choose $q$ as the signed electric charge on the capacitor plate into which positive $i$ flows. With terminal drops measured in that direction, Kirchhoff's voltage law givesHere $V(t)$ is the applied electric potential difference; assume constant $L>0$, $C>0$, and $R>0$. With the applied source set to zero and the loop kept closed, the free response satisfiesThe angular frequency ${\omega }_0$ is the value for lossless free oscillation; $\alpha$ is the decay parameter. The three possibilities are underdamped RLC response for $\alpha <{\omega }_0$, critically damped RLC response for $\alpha ={\omega }_0$, and overdamped RLC response for $\alpha >{\omega }_0$. ^[ref].

For a fair comparison, start each circuit with the same electric charge $q(0)=q_0$ and zero electric current $q^{\prime }(0)=0$. Set $\tau ={\omega }_{0}t$ and $a=\alpha /{\omega }_0$; then $x=q/q_0$ obeys $d^{2}x/d{\tau }^2+2adx/d\tau +x=0$. The plot compares $a=0.5,1,2$. These dimensionless curves apply to any $L,C$ after rescaling time and electric charge.

The stored energy is the sum of magnetic energy and capacitor energy:Differentiating and using the circuit equation gives the complete power balance,In the free response, $E^{\prime }=-R{i}^2\le 0$: Joule heating accounts exactly for the lost energy. For the plotted initial conditions, $E(0)=q_0^2/(2C)$ and $E/E(0)=x^2+(dx/d\tau {)}^2$. Since every free solution tends to zero for $R>0$,In the lossless limit $R=0$, energy stays constant and shuttles between the capacitor and inductor. Thus an oscillation of electric charge and a decrease of total energy are compatible statements.

The analogy with a damped harmonic oscillator is exact within these ideal models: electric charge corresponds to displacement, electric current to velocity, inductance to mass, $1/C$ to spring constant, and $R$ to the coefficient of a resisting force proportional to velocity. The two stored energy terms play the roles of kinetic energy and elastic potential energy. ^[ref].

Differentiating the driven equation also gives $L{i}^{\prime \prime }+R{i}^{\prime }+i/C=V^{\prime }$. When using this equation to solve for electric current, retain the original constraint $L{i}^{\prime }(0)+Ri(0)+q(0)/C=V(0)$: the initial electric charge fixes the initial rate of change of electric current.

Write an ideal point-transfer model using the Dirac delta function:For this arrival stream with independent increments, distinct arrivals contribute the product of the means, which cancels when centering. The same-arrival term leaves the autocovarianceIts Fourier transform gives the two-sided power spectral density, while folding negative frequencies onto positive ones gives the one-sided convention:These spectra describe $\delta i$, excluding the spectral line from the mean direct current. The amplitude spectral density is $\sqrt{2qI}$ in $\mathrm{A}/\sqrt{\mathrm{Hz}}$; it is not a root mean square electric current until a measurement signal bandwidth is specified. Approximate white noise requires frequencies low enough that the single-transfer pulse shape and any correlations can be neglected. Neither the ideal impulses nor a flat spectrum extending to infinity describe a real detector at every timescale.

All spectra above are per hertz, with $f$ in cycles per second. One possible angular frequency convention defines a density ${\mathcal{S}}_I^{(2)}(\omega )$ integrated using $d\omega$ alone. Since $\omega =2\pi f$, it hasAnother common convention keeps the numerical density $qI$ but integrates it with $d\omega /(2\pi )$. State the integration measure as well as whether a spectrum is one-sided: a change of variable and a folding of signs are separate operations.

Simple harmonic motion is motion in which the acceleration is proportional to the displacement from equilibrium and points back towards it:Here $x$ is measured from equilibrium, and $\omega$ is the constant angular frequency. For a point mass attached to an ideal spring, Hooke's law gives the restoring force $F=-kx$, with stiffness $k>0$, and Newton's second law yields $\omega =\sqrt{k/m}$. The minus sign matters: for $x>0$ the force is negative, and for $x<0$ it is positive. A restoring force need not oppose velocity; on the return towards equilibrium it increases speed.

The general solution isThe amplitude $A\ge 0$ and oscillation phase $\phi$ encode the initial conditions. We use $P$ for the oscillation period to distinguish it from kinetic energy $T$. For $A>0$, the oscillation period is independent of amplitude, the property of isochronous harmonic motion. At $A=0$ the body remains at equilibrium. ^[ref]

Choose $m=0.5\mathrm{kg}$, $k=2N/m$, and release the mass from rest at displacement $A=0.3\mathrm{m}$. Then $\omega =2{\mathrm{s}}^{-1}$, $P=\pi \mathrm{s}$, and $\phi =0$. Writing time $t$ in seconds,At $t=0$ the velocity vanishes, but the acceleration is $-1.2m/{\mathrm{s}}^2$: a turning point is not a state of rest that persists. At $t=\pi /4\mathrm{s}$, the body crosses equilibrium with velocity $-0.6m/s$ and zero acceleration. At $t=\pi /2\mathrm{s}$ it reaches the opposite turning point, $x=-0.3\mathrm{m}$.

Taking potential energy zero at equilibrium, the same solution givesAt each turning point all mechanical energy is potential energy; at each equilibrium crossing it is kinetic energy. Both repeat twice during one complete position cycle because squaring removes the sign. At $t=\pi /8\mathrm{s}$, for example, each contributes $0.045\mathrm{J}$.

The physical position trace covers a line segment, $-0.3\le x\le 0.3$ in metres. In the phase plane, the paired position and velocity instead trace the ellipseThis is precisely $m{v}^2/2+k{x}^2/2=E$, rewritten: the single conserved mechanical energy fixes the amplitude and the whole ellipse. It does not fix the starting oscillation phase. At $x=0$ both $v=-0.6m/s$ and $v=+0.6m/s$ lie on it; position alone cannot distinguish them. Replacing $v$ by momentum $p=mv$ gives the equivalent phase-space trajectory. The arrows show increasing time, clockwise in $(x,v)$ coordinates.

This closed ellipse assumes a time-independent spring and no damping or applied drive. Adding a resisting force $-bv$, $b>0$, gives $m\ddot{x}+b\dot{x}+kx=0$ and $dE/dt=-b{v}^2\le 0$. An underdamped damped harmonic oscillator spirals inward in the phase plane, crossing successively smaller energy contours. The ideal harmonic oscillator is also useful beyond springs: near a smooth potential energy minimum with $V^{\prime \prime }(0)>0$, its Taylor expansion begins with $V(0)+V^{\prime \prime }(0)x^2/2$. For sufficiently small displacements, this gives the same restoring force to leading order, explaining why small oscillations share one solvable pattern.

For an overdamped harmonic optical trap, impose a known sinusoidal local fluid velocity of amplitude $u_0$ and angular frequency $\Omega$. The coherent bead response has amplitude $A_x=u_0/\sqrt{(k/\zeta {)}^2+{\Omega }^2}$. Measuring the thermal corner frequency $f_c=k/(2\pi \zeta )$ and detector-voltage amplitude $A_s=A_x/\beta$ therefore determines the position sensitivity in the displayed formula without a drag coefficient as input. Use a thermal record with the drive line removed to obtain stiffness from the equipartition theorem. Fluid-following, detector response and hydrodynamic corrections must remain valid. Combined thermal and stage measurements have experimentally calibrated an optical trap without drag as an input.^[ref]

The slew rate bounds the large-signal output-voltage change per unit time. It is distinct from a small-signal gain-bandwidth product. A sinusoidal output of amplitude $V_p$ and angular frequency $\omega$ needs $\omega V_p$ below the available slew rate to avoid this source of distortion.

Slowness is the reciprocal of phase velocity, measured in time per distance. A plane wave with wavevector $\mathbf{k}$ and angular frequency $\omega$ has vector slowness $\mathbf{p}=\mathbf{k}/\omega$. Its direction is the wavefront normal, which need not be the group velocity direction in an anisotropic medium.

For a spherical pendulum with nonzero conserved vertical angular momentum $h$, eliminate the azimuthal angular velocity using $h=m{l}^2{\sin }^2\theta \dot{\varphi }$. The polar effective potential is $V_{\mathrm{eff}}=h^2/(2m{l}^2{\sin }^2\theta )-mgl\cos \theta$. A conical pendulum at $0<{\theta }_0<\pi /2$ is a minimum satisfying $h^2=m^{2}g{l}^3{\sin }^4{\theta }_0/\cos {\theta }_0$. Linearization at fixed $h$ yields polar angular frequency ${\omega }_{\theta }^2=V_{\mathrm{eff}}^{\prime \prime }({\theta }_0)/(m{l}^2)={\Omega }^2(1+3{\cos }^2{\theta }_0)$, where ${\Omega }^2=g/(l\cos {\theta }_0)$ is the conical revolution frequency squared. Holding $\dot{\varphi }$ fixed during the perturbation would describe a different, externally driven constraint. The polar perturbation is a nutation; the exactly planar $h=0$ limit requires separate treatment.

A uniform rigid body disk of mass $m$ and radius $a$ moves in a frictionless plane. Let $N\ge 3$ identical springs attach equally spaced rim points to fixed anchors in their radial directions. The equilibrium spring length is $l$, natural length $l_0<l$, and spring constant $k$. Write the small centre displacement as $\mathbf{q}$ and the rotation as $\theta$. For a spring with radial and tangential unit vectors $\mathbf{n},\mathbf{t}$, its length isThis follows by a second-order Taylor series for the distance between $\mathbf{q}+aR(\theta )\mathbf{n}$ and $(a+l)\mathbf{n}$. In the elastic potential energy $k(s-l_0{)}^2/2$, the equilibrium tension $k\tau$ multiplies the second-order change of length. Thus it contributes to the linearized stiffness even though the displacement is small.

The equally spaced directions satisfy $\sum \mathbf{n}=\sum \mathbf{t}=0$ and $\sum \mathbf{n}{\mathbf{n}}^T=\sum \mathbf{t}{\mathbf{t}}^T=(N/2)I$. Summing the spring energies givesThe kinetic energy is $m|\dot{\mathbf{q}}{|}^2/2+m{a}^2{\dot{\theta }}^2/4$, using the moment of inertia of a uniform disc. The Euler-Lagrange equations give the displayed squared angular frequencies. The translational normal modes are degenerate; rotation is independent. At zero prestretch the rotational frequency vanishes at quadratic order, and finite-angle spring stretching first contributes at higher order.

For fixed real positive angular frequency $\omega$, a transverse Alfvén wave proportional to $e^{i(kz-\omega t)}$ in a uniform fluid has $k^2={\omega }^2/(v_A^2-i{\eta }_m\omega )$ when magnetic diffusivity ${\eta }_m$ is retained and viscosity is neglected. Choose the branch with positive real and imaginary parts for propagation and attenuation along $+z$. For $ϵ\ll 1$, $k=(\omega /v_A)(1+iϵ/2+O({ϵ}^2))$. The amplitude attenuation length is $L_A=1/\mathrm{Im}k$, whereas the energy attenuation length is $L_E=1/(2\mathrm{Im}k)$. With wavelength $\lambda =2\pi /\mathrm{Re}k$, $L_A/\lambda \sim 1/(\pi ϵ)$ and $L_E/\lambda \sim 1/(2\pi ϵ)$. Fixed-real-wavenumber temporal Resistive damping of an Alfvén wave is a different boundary-value prescription.

Use a complex envelope $A(t)$ normalized so that $|A(t){|}^2=P(t)$; its units are ${\mathrm{W}}^{1/2}$. The optical carrier convention is $e^{-i{\omega }_{0}t}$: the physical electric field is proportional to the real part of $A(t)e^{-i{\omega }_{0}t}$ for a fixed transverse mode. With angular frequency offset $\Omega =\omega -{\omega }_0$, fix the Fourier transform conventionWrite $\tilde{A}=B{e}^{i\varphi }$ with real $B\ge 0$. A dispersive element adds spectral phaseConstant phase rotates the envelope; the linear term delays it by ${\varphi }_1$. The group delay is $d\varphi /d\Omega$. The quadratic coefficient ${\varphi }_2$ is group delay dispersion, abbreviated GDD, with units of time squared. Positive GDD makes higher frequencies arrive later with this convention. It is accumulated dispersion of a component; material group velocity dispersion is its corresponding coefficient per unit length.

To undo a quadratic stretcher spectral phase, the compressor must supply opposite GDD. Include the amplifier and all intervening optics in the sum:Setting this sum to zero does not cancel third-order dispersion, ${\varphi }_3$, or still higher orders. Residual spectral phase can produce asymmetry and satellite pulses even when the quadratic term vanishes. A diffraction grating system can make path length depend on frequency; matching the complete spectral phase over the occupied spectrum is more demanding than merely choosing opposite quadratic coefficients.

There is a useful proof of what phase correction can accomplish. By Parseval's identity,For a smooth decaying spectrum, the Fourier transform of $tA$ is $-id\tilde{A}/d\Omega$. Consequently the power-weighted mean arrival time and variance obeyAt fixed spectral amplitude, constant group delay minimizes this root mean square duration. Such a pulse is a transform-limited pulse, up to a time shift and constant phase. The nonnegative second term measures temporal broadening from unequal arrival times; the first term remains even after perfect phase correction.

The spring constant is the positive proportionality coefficient in the ideal spring restoring force $F=-kx$. Its SI unit is newtons per metre. With a constant attached mass $m$, it sets the angular frequency $\omega =\sqrt{k/m}$ of simple harmonic motion. Doubling $k$ doubles the restoring force at a fixed displacement, but increases that angular frequency only by a factor $\sqrt{2}$.

Attach a spring of spring constant $k$ to the axle of a round body of mass $m$, radius $R$ and axial moment of inertia $I$. Assume rolling without slipping, a horizontal drag force $-b\dot{x}$ applied through the axle, and no additional resisting torque. The translational force equation is $m\ddot{x}=-kx-b\dot{x}+f$, where $f$ is signed static friction. The rotational equation and rolling constraint give $f=-I\ddot{x}/R^2$, yielding the displayed equation. Equivalently, the kinetic energy is $(m+I/R^2){\dot{x}}^2/2$. This added effective inertia lowers the undamped angular frequency to $\sqrt{k/(m+I/R^2)}$. A distributed viscous torque would add a separate damping term and cannot be inferred from the drag force alone.

A pendulum bob and a horizontally moving support of equal mass $m$, with support spring spring constant $k$, have the quadratic potential energy $k{x}_1^2/2+mg(x_2-x_1{)}^2/(2l)$. With $\kappa =k/m$, $q=g/l$, the normal mode eigenvectors can be normalized to $(1,r_{\pm })$, where $r_{\pm }=(\kappa /2\mp \sqrt{{\kappa }^2/4+q^2})/q$. The lower mode is in phase and the upper mode in opposite phase. The positive eigenvalues of the symmetric stiffness-per-mass matrix give the displayed squared angular frequencies.

A standing wave results from the superposition principle applied to equal-frequency, equal-amplitude waves travelling in opposite directions. Its spatial nodes of a standing wave remain fixed while its displacement oscillates in time.

For a pure standing wave, the two travelling waves also have equal amplitudes: their interference leaves a fixed mode shape, rather than a profile moving along the string. Choose the position origin and time origin so that their transverse displacements areHere $a>0$ is each constituent's amplitude, $k=2\pi /\lambda$ is the wavenumber, $\lambda$ is the wavelength, and $\omega =2\pi f$ is the angular frequency corresponding to frequency $f$. Their wave speed is $c=\omega /k$. The superposition principle and the sine addition formula giveThe spatial factor stays put; the temporal factor changes its size and sign. Other constant wave phases shift the position or time origin. Unequal constituent amplitudes generally leave moving displacement at the would-be nodes of a standing wave, rather than this exact factorization.^[ref]

For an original ideal-string example, set $L=1\mathrm{m}$, $c=20\mathrm{m}{\mathrm{s}}^{-1}$, and choose the third normal mode. Then $k=3\pi {\mathrm{m}}^{-1}$, $\omega =60\pi {\mathrm{s}}^{-1}$, and $f=30\mathrm{Hz}$ by fixed-end standing wave frequencies. With each constituent amplitude $a=5\mathrm{mm}$,Its nodes of a standing wave are at $x=0,1/3,2/3,1\mathrm{m}$ and its antinodes at $x=1/6,1/2,5/6\mathrm{m}$. These are constructed model parameters, not experimental readings.

At $t=1/120\mathrm{s}$, every point has zero displacement, but most have nonzero transverse velocity: the string is crossing its straight configuration. The black dots alone mark permanent nodes of a standing wave. At an antinode the greatest transverse speed is $2a\omega =0.6\pi \mathrm{m}{\mathrm{s}}^{-1}$, much smaller than the chosen $20\mathrm{m}{\mathrm{s}}^{-1}$ wave speed. One is motion of string material; the other is propagation of the constituent profiles.

The beauty of a standing wave is that motion reveals geometry: fixed nodes of a standing wave turn a moving disturbance into a spatial ruler. Their separation measures wavelength, and boundary conditions turn the continuous choice of wavenumber into a discrete set of normal modes. This is useful both for interpreting string vibrations and for understanding how a bounded wave equation selects its eigenfrequencies.

Consider a straight chain with a fixed wall, spring constants $k_1,k_2,k_3>0$, and consecutive point masses $m_1,m_2,m_3>0$, with the third mass freely ending the chain. Displacements are measured from equilibrium. If the middle mass stays at equilibrium, the two adjacent spring forces must cancel: $k_2{q}_1+k_3{q}_3=0$. The outer equations of motion then reduce to $m_1{\ddot{q}}_1=-(k_1+k_2)q_1$ and $m_3{\ddot{q}}_3=-k_3{q}_3$.

A nonzero oscillatory normal mode with $q_2=0$ therefore exists exactly when the two outer squared angular frequencies agree, as displayed. Its mode shape is proportional to $(1,0,-k_2/k_3)$ and its frequency does not depend on $m_2$. Appropriate initial displacements and velocities confined to this shape keep the middle mass still. For $k_1=2k$, $k_2=k_3=k$, $m_1=3m$, $m_3=m$, the common frequency is $\sqrt{k/m}$ and the outer displacements are equal and opposite.

A steady periodic thermal state repeats each drive period $P$ after the startup transient has decayed. It is time dependent, unlike a time-independent steady state. For angular frequency $\omega =2\pi /P$, its single-frequency part has the phasor representation $\theta (x,t)=\mathrm{Re}\{\hat{\theta }(x)e^{i\omega t}\}$. Here $\theta$ is the oscillating temperature about the mean profile. Repeating the fitted amplitudes and wave phases over successive cycles checks that startup drift is small; merely completing one drive cycle is insufficient.

A nonlinear optics process producing a field at the sum of two input angular frequencies. The driven electric polarization contains a term proportional to the product of the input complex amplitudes. With undepleted plane-wave inputs, integration along a crystal gives an output amplitude proportional to $L{e}^{i\Delta kL/2}\mathrm{sinc}(\Delta kL/2)$; the sinc function therefore determines the ideal length and phase mismatch dependence. A nonzero initial output field also interferes with this generated field.