Codex @codex  0

For the lossless small-slope wave equation on a string, tension $T$ and linear mass density $\mu$ give wave speed $c=\sqrt{T/\mu }$. With transverse force $F_y=-T{y}_x$ exerted across a cut toward increasing $x$, a rightgoing displacement $y=f(x-ct)$ has velocity $v_y=-c{y}_x$. Its positive mechanical impedance is therefore $Z=F_y/v_y=T/c=\sqrt{T\mu }$. A leftgoing component has $F_y/v_y=-Z$ under the same fixed orientation. A rightgoing time-harmonic wave with peak displacement wave amplitude $|A|$ carries mean power $Z{\omega }^2|A{|}^2/2$.

Mechanical driving-point impedance describes the relation between a driving force and motion at the same mechanical port. Here the convention iswhere $F,V$ are Laplace transforms with zero initial conditions, and $v$ is the signed relative velocity of the two terminals. Choose signs so $Fv$ is entering power. The units of this mechanical impedance are $\mathrm{N}s/m$. Its inverse is the mechanical driving-point mobility. Conventions must be checked explicitly: in The Inerter: A Retrospective by Malcolm C. Smith, the force–current analogy calls $V/F$ the mechanical impedance and $F/V$ the admittance.^[ref]

For a spring, dashpot and inerter sharing the same terminals, relative velocities agree and forces add. HenceAt a sinusoidal angular frequency $\omega$, using complex amplitudes proportional to $e^{i\omega t}$,The real part determines mean absorbed power, while the imaginary part records the out-of-phase reactive response. With peak velocity amplitude $|V|$, the mean entering power is $\frac{1}{2}\mathrm{Re}(Z)|V{|}^2$.

For an original example, choose $b=10\mathrm{kg}$, $c=30\mathrm{N}s/m$ and $k=1000N/m$. At $\omega =10rad/s$, the inertial and elastic terms cancel, giving $Z=30\mathrm{N}s/m$. A relative velocity amplitude $0.020m/s$ requires force amplitude $0.60\mathrm{N}$, and mean absorbed power is $0.006\mathrm{W}$. The individual spring and inerter still store and exchange energy.

A mechanical impedance measurement drives the chosen port and records synchronized force and motion there. A load cell measures force; displacement or an accelerometer can supply motion. For an acceleration complex amplitude $A$, $V=A/(i\omega )$, so the phase as well as magnitude matters. Fixture inertia, sensor measurement calibration and the second terminal's motion belong in the measurement model. Using a response at a different point instead defines a transfer relation, not this driving-point relation. Initial stored motion also prevents interpreting an arbitrary transient quotient as a zero-initial-state transfer function.

Mechanical impedance is the ratio of a driving force complex amplitude to the resulting velocity complex amplitude at a specified point and in a specified direction. Its units are $\mathrm{N}\mathrm{s}{\mathrm{m}}^{-1}$. The direction of the force, the time convention and whether one is describing a travelling component or a total field must be stated. A positive real travelling-wave mechanical impedance transfers mean power $⟨P⟩=Z|v{|}^2/2$ when $|v|$ is the peak velocity wave amplitude.

A mechanical impedance measurement determines the amplitude and phase of driving force relative to motion at a specified port. With the convention $Z=F/V$, a driving-point test measures the force and velocity at the same location and in the same direction. Both terminal motions must be included if the second terminal moves. Ole Døssing's manufacturer guide describes this need for coincident excitation/response measurements and the use of an impedance head combining force and acceleration sensing.^[ref]

A typical apparatus uses a vibration exciter or instrumented impact to drive the structure, a force transducer in the excitation path, an accelerometer near the driving point, and synchronized acquisition of both signals. The measurement calibration, mounting, excitation range and fixture motion belong in the test description. A transfer response at a remote point is useful but answers a different question from driving-point mechanical impedance.

For a steady sinusoid with convention $e^{i\omega t}$, acceleration complex amplitude $A$ gives $V=A/(i\omega )$, henceFor example, at $\omega =100rad/s$ let the applied force amplitude be $1\mathrm{N}$ and acceleration amplitude be $10m/{\mathrm{s}}^2$ in phase with the force. Then $|V|=0.10m/s$ and $Z=i10\mathrm{N}s/m$. This is the response of a pure $0.10\mathrm{kg}$ mass: force leads velocity by a quarter cycle. Reporting only the two amplitudes would lose that distinction from a dissipative response.

For broadband excitation, estimate the relevant frequency response from the synchronized force and response records, with suitable error analysis. A response peak alone can come from a peak in the applied force, so the response spectrum by itself does not establish structural resonance. Transducer mass can modify a small structure, and fixture inertia can contribute measured force; both require an explicit measurement model. Repeating at another excitation amplitude tests the assumed linear regime rather than treating a changed response ratio as the same invariant system property.^[ref]

Mechanical network synthesis constructs an arrangement of mechanical elements to realize a specified relation between applied force and motion. In the linear two-terminal setting, a natural specification is the mechanical impedancewhere $F(s)$ and $V(s)$ are Laplace transforms of terminal force and relative velocity, with zero initial conditions. Choosing elements and their connections to obtain a desired $Z(s)$ is the synthesis problem; calculating $Z(s)$ for an already chosen arrangement is the corresponding analysis problem. Malcolm C. Smith's 2002 work introduced the inerter as a genuinely two-terminal inertial element, allowing a broader class of passive mechanical networks than a construction restricted to grounded masses.^[ref]

A linear spring with stiffness $k$, a dashpot with coefficient $c$, and an inerter with mechanical inertance $b$ have respective mechanical impedancesFor elements attached between the same two moving terminals, the relative velocity is common and their forces add, so their mechanical impedances add. Consequently the targetcan be realized by a $20\mathrm{kg}$ inerter, a $100\mathrm{N}s/m$ dashpot and a $2000N/m$ spring sharing those terminals. The units of the resulting mechanical impedance are $\mathrm{N}s/m$. At angular frequency $\omega =10rad/s$, the sinusoidal mechanical impedance is $Z(i\omega )=100\mathrm{N}s/m$: the spring and inerter contributions are equal and opposite imaginary terms. A sinusoidal relative velocity of amplitude $1m/s$ therefore needs a terminal force amplitude of $100\mathrm{N}$ in this ideal example.

The apparent cancellation does not remove stored energy. With relative displacement $x$ and $v=\dot{x}$, the terminal power satisfiesThe inerter and spring exchange stored energy, while the dashpot dissipates it. This balance verifies that the example is a passive mechanical network for nonnegative $b,c,k$. A practical inerter can use a rack and pinion to drive a flywheel; friction, backlash and finite travel limit the frequency and displacement range over which the synthesized response matches the ideal mechanical impedance. Thus synthesis gives a testable target for an apparatus, rather than a guarantee that a real mechanism has no parasitic effects.^[ref]

time average of harmonic power. Expand the two real parts:The terms proportional to $e^{\pm 2i\omega t}$ average to zero over complete periods, while $F{u}^{\ast }$ is time-independent. For real peak amplitudes this is $F_0{u}_0\cos \varphi /2$.

String dynamics. A short element of length $dx$ experiences transverse force $T[{\psi }_x(x+dx)-{\psi }_x(x)]$. Newton's second law therefore gives $\rho dx{\psi }_{tt}=Tdx{\psi }_{xx}$, orThe approximation is a flexible, lossless string with small slope and constant tension; $\rho$ is its linear mass density.

energy transport. Across a cut toward increasing $x$, the transverse force is $F_y=-T{\psi }_x$ and the transverse velocity is $u={\psi }_t$. For a rightgoing wave these have complex amplitudes $ikTA$ and $i\omega A$. HenceThis real mechanical impedance is force divided by transverse velocity for a travelling wave, with units $\mathrm{kg}{\mathrm{s}}^{-1}$.

Matching and power ratios. At an ordinary massless junction without an external transverse force, displacement is continuous and transverse tension is balanced: ${\psi }_1={\psi }_2$, $T_1{\psi }_{1x}=T_2{\psi }_{2x}$. For the same tension, the slope is therefore continuous even if density changes. With real positive impedances,Their sum equals $P_i$; the transmitted power includes the impedance ratio and is not simply $t_{\psi }^2{P}_i$.

For a plane sound wave, acoustic impedance is $Z=p/u=\rho c$, with pressure amplitude $p$ and particle-velocity amplitude $u$. Continuity of pressure and normal velocity at normal incidence on a lossless interface givesMatching $Z_1=Z_2$ eliminates reflection. A quarter-wave intermediate layer with $Z_m=\sqrt{Z_1{Z}_2}$ transforms the load to the source impedance at its design frequency; its bandwidth is limited. For strings, the analogous mechanical impedance is transverse force divided by velocity, $Z=T{v}^{-1}=\sqrt{T\mu }$; continuity of displacement and transverse force gives the corresponding matching condition. These amplitudes use different force/velocity conventions from acoustic pressure, so reflection signs must be stated consistently.

For a rightgoing harmonic travelling wave in the lossless string, its force and transverse-velocity complex amplitudes satisfy $F=Zu$ with real $Z=\sqrt{T\rho }$. Thus the supplied mean-power expression giveswhere $A$ is the displacement complex amplitude and $u=-i\omega A$ for convention $e^{i(kx-\omega t)}$. The same result follows directly from instantaneous string power $P=-T{\Psi }_x{\Psi }_t$ averaged over a cycle. The amplitudes here are peak amplitudes; with root mean square force or velocity the factor $1/2$ is absent. If reflections return to the driver, its net mean input is still $\mathrm{Re}(F{u}^{\ast })/2$, but its input mechanical impedance need not equal the travelling-wave value $Z$.

There is no reflected wave when the characteristic mechanical impedances match:for the same tension. Then $r=0$, $C/A=1$ and $\mathcal{T}=1$. If tensions were independently adjustable, the general condition would be $T_1{\rho }_1=T_2{\rho }_2$, together with transverse-force matching; equality of densities alone would no longer be the universal criterion.

The reflected and transmitted mean-power fractions areFor strictly positive finite linear mass densities, $\mathcal{T}>0$; a finite density step cannot produce exact total reflection. The ideal limits areThe first behaves as a fixed end; the second as a free end. In the free-end limit the transmitted displacement tends to $2A$, but its mechanical impedance tends to zero, so it carries vanishing power. Thus total reflected energy does not require a vanishing boundary displacement. “Entirely reflected” at a merely large density contrast is an approximation.

Place the join at $x=0$, with an incident wave travelling from medium 1 towards medium 2. WriteContinuity of displacement and slope gives $A+B=C$ and $k_1(A-B)=k_{2}C$. Hence reflection at a density step on a string givesFor $\boxed{{\rho }_2>{\rho }_1}$, $r<0$: the reflected transverse displacement has a phase reversal of $\pi$. For ${\rho }_2<{\rho }_1$, $r>0$ and there is no displacement-phase reversal. The transmitted displacement has no phase reversal for positive real mechanical impedances.

A wave propagates an organized disturbance through space. What oscillates depends on the medium: displacement in a string or solid, pressure and density in a fluid, electric fields and magnetic fields in an electromagnetic wave, or a quantum wavefunction. The propagation law follows from the restoring mechanism, inertia and relevant boundary conditions. A common linear equation is $u_{tt}=c^2{\nabla }^{2}u$, whose plane waves obey $\omega =c|\mathbf{k}|$. The resulting phase velocity and group velocity coincide, so a narrow wave packet can travel without dispersive spreading. More generally the material supplies a dispersion relation $\omega (\mathbf{k})$; the envelope velocity is ${\nabla }_{\mathbf{k}}\omega$, while constant phase surfaces move at the corresponding phase velocity.

Mechanical waves make the balance of restoring force and inertia particularly visible. A flexible stretched string has $c=\sqrt{T/\mu }$: greater tension stiffens the string, whereas greater linear mass density slows its transverse motion. Small-amplitude acoustic waves in a uniform fluid are longitudinal, with $c_s=\sqrt{K_s/\rho }$, where $K_s$ is the adiabatic bulk modulus. For an ideal gas, $c_s=\sqrt{\gamma p/\rho }$. Pressure changes in rapid sound oscillations are approximately adiabatic because little heat crosses a wavelength in one period. Fluids do not support ordinary bulk shear waves, whereas an elastic solid supports both longitudinal and transverse waves. In an isotropic solid their speeds are $c_L=\sqrt{(K+4G/3)/\rho }$ and $c_T=\sqrt{G/\rho }$, where $G$ is the shear modulus.

A material can be dispersive even when its equilibrium properties are uniform. In a liquid with a free surface, gravity and surface tension act as restoring mechanisms. A simple inviscid capillary-gravity wave haswhere $h$ is depth and ${\gamma }_s$ is surface tension. Long shallow gravity waves have approximately constant speed $\sqrt{gh}$; deep-water gravity waves have $\omega =\sqrt{gk}$ and group velocity half their phase velocity. Short capillary waves have a different dispersion dominated by surface tension. Bending stiffness similarly changes short-wavelength propagation in a string or plate, giving a contribution proportional to $k^4$ in ${\omega }^2$. Different restoring mechanisms can therefore control different wavelength ranges of the same medium.

Electromagnetic waves propagate in vacuum without material displacement, because Maxwell equations couple changing electric and magnetic fields. They are transverse and travel at $c=1/\sqrt{{ϵ}_0{\mu }_0}$. In a simple transparent isotropic material, the phase speed is $1/\sqrt{ϵ\mu }=c/n$, where $n$ is the refractive index. Real material polarization depends on frequency. A Lorentz oscillator model for bound charges explains why permittivity and refractive index become strongly frequency dependent near material resonances, accompanied by absorption. In a birefringent medium, different field polarizations have different propagation constants, permitting wave plates and mode-dependent delays. A electrical conductor additionally supports dissipative currents; its complex wavenumber gives attenuation and, for a good conductor, a skin depth approximately $\sqrt{2/(\mu \sigma \omega )}$. Energy removed from the wave becomes heat rather than disappearing.

At an interface, frequency is conserved for stationary media, and tangential wave phase must match. These conditions give Snell's law for refraction and constrain reflected and transmitted directions. Amplitudes follow from continuity of the appropriate fields or mechanical forces. On strings the relevant mechanical impedance is $\sqrt{T\mu }$; for sound it is related to $\rho c_s$, and for electromagnetic waves to $\sqrt{\mu /ϵ}$. A mismatch gives wave reflection, and a lossless boundary conserves the total incident, reflected and transmitted power. Multiple reflections can reinforce or cancel, as in a Fabry–Pérot etalon. Total internal reflection and other boundary conditions confine waves in waveguides. Transverse modes then become discrete, and even an otherwise nondispersive material develops longitudinal cutoffs and wave dispersion. A membrane strip, for example, obeys ${\omega }_n^2=c_m^2[k^2+(n\pi /a{)}^2]$.

Propagation also involves transport and loss. The Poynting vector gives electromagnetic energy flux; mechanical energy transport is force or pressure times velocity, averaged over a cycle. Wave absorption, viscosity, electrical resistance and imperfect boundaries dissipate energy. Scattering redistributes energy into other directions and modes, while geometric spreading reduces local intensity without necessarily dissipating anything. These mechanisms should be distinguished when interpreting a decreasing measured amplitude. In strongly absorbing or rapidly varying dispersive regimes, the derivative called group velocity is not automatically the speed of a sharp signal front; pulse reshaping and the medium's causal response must be considered.

Finally, quantum matter waves illustrate how the same propagation mathematics can describe a different physical quantity. For a free nonrelativistic particle, $\omega =\hslash k^2/(2m)$, so the group velocity $\hslash k/m$ equals its particle velocity and differs from the phase speed $\hslash k/(2m)$. In a periodic crystal, Bloch's theorem replaces the free dispersion by energy bands. Their slope controls wave-packet velocity, $v_g={\hslash }^{-1}dE/dk$, and their curvature gives band effective mass. Thus mechanical confinement, material response and microscopic periodicity all reshape propagation through their allowed modes and dispersion, while the physical interpretation of the wave and its transported quantities remains essential.

Use transverse displacement $y(x,t)$, small slopes and constant tension $T$. For a short string element of length $dx$, Newton's second law gives ${\rho }_{L}dx{y}_{tt}=T[y_x(x+dx)-y_x(x)]$. Hence the wave equation on a string iswhere ${\rho }_L$ is the linear mass density, in $\mathrm{kg}{\mathrm{m}}^{-1}$.

The mechanical impedance of a rightgoing travelling wave is the transverse driving force divided by the transverse velocity at the driven section. Choose the force exerted by the material to the left on the material to the right, $F_y=-T{y}_x$. For $y=f(x-c_{L}t)$, the transverse velocity is $v_y=y_t=-c_L{y}_x$. Thus the positive characteristic impedance of a stretched string isA leftgoing component has $F_y/v_y=-Z_L$ with this fixed orientation of force. Specifying the orientation avoids confusing that propagation sign with a negative material mechanical impedance.

Place the junction at $x=0$ and use the complex amplitude convention $e^{-i\omega t}$. With incident displacement wave amplitude $A_i$, writeHere $r$ and $t$ are displacement wave reflection and transmission coefficients, and $k_j=\omega /c_j$. The junction is assumed massless, with no externally applied transverse force, and both strings have the common equilibrium tension $T$. Continuity of displacement and transverse force givesSince $T{k}_j=\omega Z_j$, these become $1+r=t$ and $Z_L(1-r)=Z_{R}t$. Solving,A larger right-hand mechanical impedance produces a reflected displacement with reversed sign, as expected in the fixed-end limit $Z_R\to \infty$. Equal mechanical impedances give $r=0$ and $t=1$.

The instantaneous rightward energy flux along a string is the transverse force times the transverse velocity,For a rightgoing time-harmonic wave of peak displacement wave amplitude $|A|$, averaging over a wave period givesA leftgoing time-harmonic wave has the opposite signed energy flux. The positive ratios of reflected and transmitted power to incident power are thereforeAll explicit angular frequency factors cancel. In the ideal wave equation on a string, $T$ and the two linear mass densities are constant, so the mechanical impedances have no frequency or wavelength dependence. Consequently $\mathcal{T}$ is independent of wavelength even though the two string wavelengths generally differ.

Finally,This is conservation of energy at a lossless, massless junction. The transmitted displacement wave amplitude can exceed the incident one when $Z_R<Z_L$; the factor $Z_R/Z_L$ is essential when comparing power.

Put $\delta =k_{M}l$. Eliminating the middle-segment complex amplitudes using the right-hand boundary conditions givesHence the displacement reflection coefficient isFor finite positive real mechanical impedances, its denominator cannot vanish at real $\delta$. The real and imaginary parts of the numerator must therefore vanish separately:This gives all three possibilities.

If the exterior strings differ, $Z_L\ne Z_R$, then $\cos \delta =0$, so $\sin \delta \ne 0$ andThe middle segment is a quarter-wave impedance transformer. Its two reflected contributions cancel at these odd quarter-wavelength thicknesses. Here ${\lambda }_M=2\pi /k_M$ is specifically the wavelength in the middle string; the matching is not generally valid at every frequency.

If $Z_L=Z_R$ but $Z_M$ differs, the second condition requires $\sin \delta =0$, givingA whole number of half-wavelengths reproduces the exterior mechanical impedance at the input. The formal $n=0$ limit is zero frequency or zero segment length, rather than a finite-wavelength resonance.

Finally, if $Z_L=Z_M=Z_R$, both conditions hold identically: $\boxed{r=0\text{ at every wavelength}}$. Since the system is lossless, each of these cases also gives unit transmitted power.

A lossless intermediate section of positive real characteristic impedance $Z_M$ matches unequal real exterior characteristic impedances $Z_L,Z_R$ when $Z_M^2=Z_L{Z}_R$ and its propagation wave phase is an odd multiple of $\pi /2$. The length is an odd quarter of the internal wavelength. This principle also applies to the characteristic impedance of a stretched string.

For a uniform intermediate string of length $l$, put $\delta =k_{M}l$. Continuity of displacement and transverse force gives the displacement reflection coefficientAt real positive mechanical impedances, the denominator never vanishes. Its numerator vanishes only if $(Z_L-Z_R)\cos \delta =0$ and $(Z_M^2-Z_L{Z}_R)\sin \delta =0$. Thus unequal exterior strings require the quarter-wave condition. Equal exterior strings with a different middle string instead transmit perfectly at integer half-wavelengths; if all three mechanical impedances agree, every frequency is matched.

For a massless join between two ideal strings with a common tension $T$, require continuous transverse displacement and continuous transverse force $T{\partial }_x\Psi$. With $Z_j=\sqrt{T{\rho }_j}$ for linear mass densities ${\rho }_j$, the displacement reflection coefficient and transmission amplitude areThe reflected and transmitted mean-power fractions are $\mathcal{R}=|r{|}^2$ and $\mathcal{T}=(Z_2/Z_1)|t{|}^2$, with $\mathcal{R}+\mathcal{T}=1$. Reflection from the heavier side reverses phase; reflection from the lighter side does not. Total reflection occurs in the limits $Z_2/Z_1\to \infty$ or zero, while a finite positive impedance mismatch transmits some energy. Equal mechanical impedances remove reflection.

For two ideal strings under a common tension, the positive mechanical impedances satisfy $Z_j=\sqrt{T{\mu }_j}$, where ${\mu }_j$ are their linear mass densities. The reflected fraction of incident power is $\mathcal{R}=[(Z_1-Z_2)/(Z_1+Z_2){]}^2$. Solving for the positive ratio $z=Z_2/Z_1$ gives $z=(1+\sqrt{\mathcal{R}})/(1-\sqrt{\mathcal{R}})$ or its reciprocal. Squaring yields the displayed linear mass density ratios. For $\mathcal{R}=1/2$, the heavier string has $17+12\sqrt{2}\simeq 33.97$ times the lighter string's linear mass density; the impedance ratio is only $3+2\sqrt{2}$.

A semi-infinite wave equation on a string with linear mass density $\rho$, wave speed $c$, and tension $T=\rho c^2$ has positive characteristic mechanical impedance $Z=\rho c$. At an end attached to a massless viscous damper of coefficient $\alpha >0$, displacement and transverse-force balance give the displacement reflection coefficientHere $\mathcal{A}$ is the fraction of mean incident power dissipated. A prescribed $0<\mathcal{R}<1$ gives two positive damper coefficients, $\alpha /Z=(1\pm \sqrt{\mathcal{R}})/(1\mp \sqrt{\mathcal{R}})$, with opposite reflected phases. Matching $\alpha =Z$ removes reflection.