Mechanical network synthesis

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]

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 driving-point mobility is the frequency response relating motion to excitation at the same mechanical port. Using relative velocity $v$ and signed entering force $F$, its zero-initial-state Laplace transform convention iswhere $Z$ is the force/velocity mechanical driving-point impedance. Mobility has units $m/(\mathrm{N}\mathrm{s})$. The measured direction, port and boundary conditions are part of its specification.^[ref]

For an inertial element $b$, dashpot coefficient $c$ and spring stiffness $k$ sharing a port,For $e^{i\omega t}$ complex amplitudes, mobility tends to $i\omega /k$ at low frequency when $k>0$ and to $1/(i\omega b)$ at high frequency when $b>0$. A spring therefore prevents a nonzero steady drift response to constant force. This differs from the steady mechanical mobility of a body under pure linear drag, which has no elastic restoring term.

Take $b=2\mathrm{kg}$, $c=4\mathrm{N}s/m$ and $k=200N/m$. At $\omega =10rad/s$ the reactive terms cancel and $Y=0.25m/(\mathrm{N}\mathrm{s})$. A peak force amplitude $0.8\mathrm{N}$ gives peak velocity $0.20m/s$ and mean entering power $0.08\mathrm{W}$. These statements concern the same port; a response measured elsewhere is a transfer mobility.

A response/force ratio using acceleration instead is mechanical accelerance, $A/F=sY$; using displacement gives mechanical receptance, $X/F=Y/s$. Confusing these ratios introduces a frequency-dependent magnitude and phase error. With several coupled ports, matrix inversion rather than elementwise reciprocation is needed to exchange complete impedance and mobility descriptions.

Mechanical receptance is a displacement-to-force frequency response. With zero initial conditions, the Laplace transform convention is $H(s)=X(s)/F(s)$, with units $m/N$. Mechanical driving-point mobility is $Y=sH$ and mechanical accelerance is $s^{2}H$. The port, direction and boundary conditions must accompany the ratio.^[ref]

For a single-coordinate model $m\ddot{x}+c\dot{x}+kx=F$,Its inverse is mechanical dynamic stiffness. If $k>0$, the zero-frequency limit $H(0)=1/k$ agrees with static mechanical compliance. At finite frequency, inertia and damping change both magnitude and phase; replacing the denominator by $k$ would discard them.

For an original example, use $m=2\mathrm{kg}$, $c=4\mathrm{N}s/m$ and $k=200N/m$. At $\omega =10rad/s$, stiffness and inertia cancel, giving $H=-0.025im/N$. A peak force amplitude $0.8\mathrm{N}$ therefore gives displacement amplitude $0.020\mathrm{m}$, lagging force by a quarter cycle. Static displacement under the same force magnitude would be only $0.004\mathrm{m}$.

For sufficiently high frequency, this model has $H(i\omega )\sim -1/(m{\omega }^2)$, revealing an inertia-dominated response with a phase reversal. A mechanical impedance measurement can infer receptance from acceleration by dividing the acceleration/force response by $-{\omega }^2$ for nonzero $\omega$. This conversion becomes sensitive to low-frequency measurement errors and is not a substitute for specifying the actual displacement measurement or integration model. With multiple coupled coordinates, the receptance matrix is the inverse of the complete dynamic-stiffness matrix, rather than a collection of elementwise reciprocal entries.

Mechanical dynamic stiffness is the complex force required per unit displacement in a specified zero-initial-state frequency response. If mechanical receptance is $H(s)=X(s)/F(s)$, its scalar inverse is $K_{\mathrm{dyn}}(s)=1/H(s)$, wherever that inverse exists. Units are $N/m$, but this dynamic quantity includes inertia and damping as well as the static spring stiffness. The measured coordinate, applied-force direction and boundary conditions are part of the definition.^[ref]

For $m\ddot{x}+c\dot{x}+kx=F$, zero initial conditions giveThus displacement and force generally differ in phase. With $m=2\mathrm{kg}$, $c=4\mathrm{N}s/m$ and $k=200N/m$, at angular frequency $\omega =10rad/s$ the result is $40iN/m$. A real displacement complex amplitude $X=0.020\mathrm{m}$ requires $F=0.80i\mathrm{N}$, rather than the static estimate $kX=4\mathrm{N}$. Elastic and inertial contributions cancel at this frequency; damping still requires an applied force.

For several coupled coordinates, a mechanical dynamic-stiffness matrix maps displacement to force:A matrix inverse cannot be replaced by reciprocating individual entries. For example, two masses connected by a spring of stiffness $k_c$, each also grounded by stiffness $k_0$, have $D_{11}=D_{22}=a=m{s}^2+k_0+k_c$ and $D_{12}=D_{21}=-k_c$. If coordinate2 is free of external force, $H_{11}=a/(a^2-k_c^2)$, so the driving-point dynamic stiffness is $a-k_c^2/a$ where defined. If coordinate2 is clamped, the required ratio is instead $a$. The measurement constraints determine which inverse is physically relevant.

A mechanical dynamic-stiffness matrix maps displacement coordinates to applied forces in a linear structural dynamics model at a specified transform variable or frequency. Starting with constant mass matrix $\mathbf{M}$, damping matrix $\mathbf{C}$ and mechanical stiffness matrix $\mathbf{K}$,zero initial conditions give $\mathbf{F}(s)=\mathbf{D}(s)\mathbf{X}(s)$ with $\mathbf{D}=s^2\mathbf{M}+s\mathbf{C}+\mathbf{K}$. For the $e^{i\omega t}$ convention, $\mathbf{D}(i\omega )=\mathbf{K}-{\omega }^2\mathbf{M}+i\omega \mathbf{C}$. Its matrix inverse, when it exists, is the complete mechanical receptance matrix. Models and measured responses must use matching coordinates and boundary conditions.^[ref]

Consider two $1\mathrm{kg}$ masses, each grounded by a $100N/m$ spring and joined by another $100N/m$ spring, with no damping. At angular frequency $5rad/s$,A force amplitude $F_1=1\mathrm{N}$ with $F_2=0$ gives $X_1=175/20625\mathrm{m}\approx 8.485\mathrm{mm}$ and $X_2=100/20625\mathrm{m}\approx 4.848\mathrm{mm}$. Clamping coordinate2 instead gives $X_1=1/175\mathrm{m}\approx 5.714\mathrm{mm}$, because the clamp supplies a reaction force.

Partitioning retained and internal coordinates gives mechanical dynamic condensation. With ${\mathbf{F}}_i=0$ and invertible ${\mathbf{D}}_{ii}$,The effective matrix is a Schur complement, generally frequency dependent even when the original mass matrix, damping and mechanical stiffness matrix are constant. In the example its scalar value is $175-100^2/175\approx 117.857N/m$, rather than175. A reduced force-free internal coordinate and a clamped coordinate describe different experiments. At a singular block, that particular elimination is unavailable; it does not establish that the full mechanical system lacks a valid solution.

Mechanical dynamic condensation eliminates selected coordinates from a frequency-dependent mechanical response while retaining their effect on the remaining coordinates. It is exact algebraic elimination at the chosen transform variable when the eliminated block is invertible. It is not equivalent to clamping those coordinates. Starting from the partitioned mechanical dynamic-stiffness matrix,solve the internal equation as ${\mathbf{X}}_i={\mathbf{D}}_{ii}^{-1}({\mathbf{F}}_i-{\mathbf{D}}_{ir}{\mathbf{X}}_r)$. Substitution givesThe effective matrix is a Schur complement; the right side is the condensed mechanical load vector. Both matter when the internal coordinate is externally forced. The coupled dynamic equations and measurement boundary conditions are the starting model, rather than independent scalar stiffness reciprocals.^[ref]

For the two-mass example at $\omega =5rad/s$, take $D_{11}=D_{22}=175N/m$ and $D_{12}=D_{21}=-100N/m$. With $F_1=0$ and $F_2=1\mathrm{N}$, eliminating coordinate2 gives effective stiffness $117.8571N/m$ and effective load $100/175\mathrm{N}$. Hence $X_1=100/20625\mathrm{m}\approx 4.848\mathrm{mm}$. Omitting the load correction would incorrectly predict no response at coordinate1.

For constant original mass matrix, damping and mechanical stiffness matrix, the condensed matrix is generally a rational function of frequency, not simply a smaller constant mass–damping–stiffness triple. A low-frequency approximation requires its own justification. The elimination can also become singular at an internal-block frequency even when the complete system matrix remains invertible. For example, $\left(\begin{array}{cc}0& -k_c\\ -k_c& 0\end{array}\right)$ has invertible full matrix for $k_c\ne 0$, but its lower-right scalar block cannot be inverted. One must then retain that coordinate or choose a different elimination.

The condensed mechanical load vector is the effective forcing on retained coordinates after mechanical dynamic condensation eliminates internal coordinates. It includes the effect of forces applied to the eliminated coordinates; it is not simply the retained subset of the original load vector. With partitioned mechanical dynamic-stiffness matrix $\mathbf{D}$ and invertible internal block ${\mathbf{D}}_{ii}$,and the retained equation is ${\hat{\mathbf{D}}}_{rr}{\mathbf{X}}_r={\hat{\mathbf{F}}}_r$. This follows by solving the internal block equation and substituting it into the retained equation, using the same Schur complement that constructs the reduced dynamic stiffness. Coupled force/response models require their boundary conditions to be stated.^[ref]

For an illustrative two-coordinate system at one frequency, let diagonal entries of $\mathbf{D}$ be175N/m and off-diagonal entries be $-100N/m$. Eliminating coordinate2 givesIf $F_1=0$ and $F_2=1\mathrm{N}$, then ${\hat{F}}_1=4/7\mathrm{N}$ and $X_1=100/20625\mathrm{m}$. Discarding the eliminated-coordinate force would incorrectly remove this response.

Conversely, choose simultaneous real complex amplitudes $F_1=1\mathrm{N}$ and $F_2=-1.75\mathrm{N}$. The condensed load vanishes and $X_1=0$, even though neither applied force is zero. The recovered internal displacement is $X_2=-0.010\mathrm{m}$, which satisfies both original equations. This is a frequency-specific response cancellation, rather than a vanishing total physical force: $F_1+F_2=-0.75\mathrm{N}$.

Because the elimination uses ${\mathbf{D}}_{ii}^{-1}(s)$, the effective load can depend on frequency and phase even when the original force amplitudes are simple. An arbitrary change of load or frequency therefore requires recomputing that transformation; a reduced matrix alone does not preserve every forced-response problem.

Mechanical accelerance is the acceleration-to-force frequency response, $G(s)=A(s)/F(s)$ under zero initial conditions. It has units $\mathrm{k}{\mathrm{g}}^{-1}$. It is related to mechanical receptance and mechanical driving-point mobility byThus an acceleration response cannot be labeled mobility without a frequency-dependent conversion. Its inverse is mechanical dynamic mass, which is generally complex and frequency dependent.^[ref]

For $m\ddot{x}+c\dot{x}+kx=F$,For a pure mass, $G=1/m$, independent of frequency in the ideal rigid model. A supported mass is different: spring and damping forces also enter its response. If $k>0$, accelerance tends to zero as $\omega$ tends to zero, while its high-frequency limit is $1/m$ within the model's useful range.

With $m=2\mathrm{kg}$, $c=4\mathrm{N}s/m$, $k=200N/m$ and $\omega =10rad/s$, the result is $G=2.5i\mathrm{k}{\mathrm{g}}^{-1}$. A peak force $0.8\mathrm{N}$ gives acceleration amplitude $2m/{\mathrm{s}}^2$, leading force by a quarter cycle. The reciprocal response is $-0.4i\mathrm{kg}$: it describes a dynamic ratio, rather than a negative physical amount of inertial material.

An impedance head measures force and acceleration together, making this ratio directly accessible after measurement calibration and signal processing. Driving-point accelerance uses coincident force and response directions. A response at another point is a transfer measurement; a small or vanishing transfer response can arise from the mode shapes and does not establish that the driven structure has infinite physical mass.

Mechanical dynamic mass is the complex force-to-acceleration ratio $M_{\mathrm{dyn}}(s)=F(s)/A(s)$ for a stated mechanical excitation and response. It is the scalar reciprocal of mechanical accelerance where that response is nonzero. Its units are kilograms, but a supported structure need not have a real, positive or constant dynamic ratio. An impedance head supplies separate force and acceleration channels from which this response can be estimated.^[ref]

For $m\ddot{x}+c\dot{x}+kx=F$ with zero initial conditions,Equivalently, mechanical dynamic stiffness equals $s^2{M}_{\mathrm{dyn}}$. A freely translating ideal mass gives $M_{\mathrm{dyn}}=m$; a spring-supported mass includes the elastic and dissipative terms. With $m=2\mathrm{kg}$, $k=200N/m$ and $c=4\mathrm{N}s/m$, at $\omega =5rad/s$ the result is $(-6-0.8i)\mathrm{kg}$. At $10rad/s$ it is $-0.4i\mathrm{kg}$. The negative real part in the first case describes a restoring-force contribution relative to acceleration; it is not a negative quantity of material.

An impedance-head mass correction subtracts the effective moving mass below the force-sensing plane from the measured scalar dynamic ratio, provided that material shares the measured acceleration. This is a force-plane correction, not removal of the sensor's physical influence on the structure. At an antiresonance, a small response can produce a very large reciprocal ratio; dividing by a channel near its noise floor makes that estimate unreliable. With coupled measurement coordinates, exchanging complete force/acceleration descriptions requires a matrix inverse, rather than the reciprocal of each transfer entry.

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]

An impedance head combines a force transducer and an accelerometer in one housing for a driving-point vibration test. It measures applied force and acceleration at nearly the same location and along the intended excitation direction. The raw channels yield mechanical accelerance or its reciprocal mechanical dynamic mass; deriving mechanical driving-point mobility or mechanical driving-point impedance additionally requires the acceleration-to-velocity conversion.^[ref]

In the manufacturer's Types8000/8001 example, a piezoelectric acceleration sensor sits above a force gauge in a stiff housing, with separate electrical outputs. The arrangement aims to preserve the phase relation needed for a complex response ratio. This solves a practical apparatus problem: a separate exciter and response sensor can otherwise occupy different points, particularly on a small structure.

An impedance-head mass correction accounts for moving material between the force-sensing plane and the specimen. In a rigid lumped approximation, choose a common positive axis and let that material have effective mass $m_h$ and acceleration complex amplitude $A$. Then Newton's second law givesFor an original in-phase example, let $m_h=0.020\mathrm{kg}$, $A=10m/{\mathrm{s}}^2$ and gauge force $1\mathrm{N}$. Specimen force is $0.80\mathrm{N}$, so specimen accelerance is $12.5\mathrm{k}{\mathrm{g}}^{-1}$ rather than the uncorrected value10. With general signals the correction is a complex subtraction, not subtraction of unsigned amplitudes.

This force-plane correction does not undo every effect of attaching the instrument: added mass and mounting stiffness can change the specimen's actual vibration. The test must also state mounting, calibration, usable frequency range and the motion of any second terminal. Manufacturer guidance distinguishes measurements requiring mass compensation and cases where separate force/acceleration sensors are preferable.^[ref]

An impedance-head mass correction removes the inertial force required to accelerate material between an impedance head's force-sensing plane and the specimen. The manufacturer's construction places the acceleration sensor above the force gauge and makes the connection to the driving point stiff. Its guidance recommends compensation when the specimen's mechanical dynamic mass is comparable to the mass below the gauge.^[ref]

Choose a common positive axis and assume a rigid intervening mass $m_h$ shares specimen acceleration $A$. Newton's second law givesHere $F_g,F_s,A$ are synchronized complex amplitudes at the same angular frequency. Hence the corrected mechanical dynamic mass is $F_s/A=F_g/A-m_h$, and corrected mechanical accelerance is $A/(F_g-m_{h}A)$. The effective intervening mass must include the applicable mounting hardware; it is not automatically the entire sensor mass.

For an original example, take $m_h=0.020\mathrm{kg}$, $A=10m/{\mathrm{s}}^2$ real, and $F_g=(0.60+0.80i)\mathrm{N}$. Then $F_s=(0.40+0.80i)\mathrm{N}$, with magnitude $\sqrt{0.80}\approx 0.8944\mathrm{N}$. Subtracting unsigned amplitudes would instead give $1-0.20=0.80\mathrm{N}$ and lose the correct phase. The corrected mechanical dynamic mass is $(0.040+0.080i)\mathrm{kg}$.

The model requires synchronized channels, appropriate measurement calibration and sufficiently rigid motion below the force gauge. If that assembly flexes, its motion cannot be represented by one frequency-independent $m_h$. Subtracting inertia also does not recover the unattached specimen's response: the attached head can still change its natural frequencies and mode shapes. A useful apparatus check is a rigid test mass, for which the force/acceleration ratio should agree with the total moving mass in the chosen force plane within the tested range.

A passive mechanical network can store and dissipate energy, but cannot supply more net work than was previously stored or delivered through its terminals. For a two-terminal mechanical element, choose signed force $F$ and relative velocity $v$ so that $p=Fv$ is power entering the network. With a nonnegative mechanical energy storage function $E$, passivity requiresIn particular, the net extracted energy over that interval cannot exceed $E(t_0)$. This permits temporary outward power: a stretched spring can release stored elastic potential energy. A requirement that $Fv$ be nonnegative at every instant would incorrectly exclude such a spring.^[ref]

For a spring, dashpot and inerter connected between the same two terminals, let $x$ be relative displacement and $v=\dot{x}$. Their constitutive equations giveFor $k,b,c\ge 0$, integration proves the passivity inequality: the extra term $\int c{v}^{2}dt$ is dissipated energy. The inerter stores kinetic energy associated with relative motion, so a passive mechanical network need not be a device that merely resists displacement or converts all input work into heat. This is central to mechanical network synthesis using springs, dashpots and inerters.^[ref]

For example, a spring with $k=1000N/m$ initially extended by $0.10\mathrm{m}$ stores $5\mathrm{J}$. It can deliver up to $5\mathrm{J}$ while relaxing, despite being passive. A purely viscous element with $c=-10\mathrm{N}s/m$ would instead have input power $-10v^2$: maintaining $v=1m/s$ for one second would extract $10\mathrm{J}$ even from an initially unenergized model. An apparatus producing that sustained behavior needs another energy source.

An experimental check records terminal force and relative velocity simultaneously and integrates their product, accounting for initial energy and every power-carrying connection. A brief interval of negative measured power is compatible with stored-energy release. For a linear zero-initial-state network, the mechanical driving-point impedance provides a complementary frequency description; a nonnegative real part at measured sinusoidal frequencies checks mean absorbed power there, but a finite frequency sweep alone does not establish the full time-domain passivity inequality.

A mechanical energy storage function assigns a nonnegative stored-energy quantity $E$ to the state of a mechanical model so that its increase never exceeds entering work. For a one-port passive mechanical network with entering power $Fv$, the mechanical dissipation inequality isThis state-based accounting includes energy present before the measurement begins. A temporary outward power is compatible with release of stored energy.^[ref]

For an ideal linear spring, $E_k=\frac{1}{2}k{x}^2$ is elastic potential energy. For an ideal inerter, $E_b=\frac{1}{2}b{v}^2$ is kinetic energy associated with terminal relative motion. An ideal dashpot need not store mechanical energy: its entering power $c{v}^2$ is dissipated. If these elements share terminals, their state variables obey $v=\dot{x}$ andThis equality proves the inequality for $c\ge 0$. In a larger network, summing element balances cancels internal transfers, leaving power through the external terminals and total dissipation.

For an original numerical check, let $k=800N/m$, $b=5\mathrm{kg}$ and $c=10\mathrm{N}s/m$. Start at $x=0.050\mathrm{m}$ and $v=0.20m/s$, so $E=1.10\mathrm{J}$. Maintain that relative velocity for $0.10\mathrm{s}$: displacement reaches $0.070\mathrm{m}$ and stored energy becomes $2.06\mathrm{J}$. Dissipation is $c{v}^2\Delta t=0.040\mathrm{J}$, so input work is $0.96+0.04=1.00\mathrm{J}$. Counting all entering work as newly stored energy would overestimate storage.

The mechanical available storage is the maximum net work that could be recovered under allowed future terminal motions; it is a related operational quantity, rather than automatically every form of internal physical energy. A model's storage function is a certificate that must respect its accessible states and port variables. With zero external power, a positive-definite storage function can supply a Lyapunov function, but nonincreasing energy alone does not establish that every motion decays to rest: a lossless spring–inerter oscillation keeps a constant nonzero stored energy.

Mechanical available storage is the supremum of net work that can be extracted from a system's current state through its allowed mechanical terminals. If $Fv$ is entering power and the initial state is $x_0$, defineThe zero-duration choice gives $S_a\ge 0$. Any nonnegative mechanical energy storage function satisfying the mechanical dissipation inequality bounds the extracted work, so $S_a(x_0)\le E(x_0)$. Available storage is thus an operational recoverability quantity; it need not equal every form of internal stored energy. The general supply/storage construction was developed by Jan C. Willems.^[ref]

Consider a spring and dashpot in parallel, with $F=kx+c\dot{x}$ and initially $x=x_0$. Unload to $x=0$ at constant speed over duration $T$, so $\dot{x}=-x_0/T$. The outgoing work isThe first term is released elastic potential energy; the second is lost in the dashpot. For $k=1000N/m$, $c=10\mathrm{N}s/m$ and $x_0=0.10\mathrm{m}$, the initial spring energy is $5\mathrm{J}$. A one-second unloading recovers $4.90\mathrm{J}$ and a ten-second unloading recovers $4.99\mathrm{J}$. With unrestricted duration and suitable terminal control, these paths approach the energy upper bound, proving $S_a=5\mathrm{J}$ for this ideal example. A supremum need not be attained in finite time.

The allowed operations are essential. If the same example must finish at zero extension within one second, the Cauchy-Schwarz inequality gives ${\int }_0^T{\dot{x}}^{2}dt\ge x_0^2/T$, so the best recoverable work under that extra requirement is $4.90\mathrm{J}$. A physically inaccessible internal motion may store energy without allowing its recovery at the chosen terminals. Thus stating the initial state, usable ports, time and actuator limits is part of an available-storage claim, not an optional qualification.

The mechanical dissipation inequality states that a mechanical energy storage function can increase no faster than energy supplied through the mechanical terminals. With one port and entering power $w=Fv$, its differentiable form and interval form areThe second form remains meaningful when a pointwise time derivative is unavailable. The mechanical supply rate $w$ records the external input; the difference $Fv-\dot{E}\ge 0$ records irreversible loss in the model. This is the mechanical specialization of dissipativity, not a demand that entering power be positive at every instant.^[ref]

For a spring, inerter and dashpot sharing a port,and multiplication by $v$ gives the exact balanceA spring may return energy with $Fv<0$ while $E$ decreases; that does not violate the inequality. Conversely, a negative viscous coefficient fails this storage balance by supplying energy to motion.

For an original free-decay example, omit the spring and take grounded inertial coefficient $b=2\mathrm{kg}$, $c=1\mathrm{N}s/m$ and initial velocity $v_0=1m/s$. With zero applied port force, $b\dot{v}+cv=0$, soAt $t=(b/(2c))\ln 2\approx 0.6931\mathrm{s}$, stored energy has fallen from $1\mathrm{J}$ to $0.5\mathrm{J}$; the missing $0.5\mathrm{J}$ equals ${\int }_0^{t}c{v}^{2}dt$. No external work entered during that interval.

For several ports, use ${\sum }_j{F}_j{v}_j$ as the total supply. When passive elements are connected with power-conserving internal constraints, their internal supply terms cancel, so the sum of their storage functions satisfies the same external inequality. This explains why passive interconnections can be checked element by element, provided every power-carrying terminal is accounted for.

A mechanical supply rate specifies the rate at which energy enters a mechanical system through its stated ports. For a translational mechanical power port, entering force $F$ and its conjugate velocity $v$ give $w=Fv$; for rotation, torque $\tau$ and angular velocity $\omega$ give $w=\tau \omega$. The supply rate has units of power, while its time integral is supplied work.^[ref]

A two-terminal element makes the sign convention explicit. If forces on the element are $F$ and $-F$ at terminal velocities $v_1,v_2$, total entering power isOnly when terminal2 is stationary may one replace relative velocity by $v_1$. For an original example, $F=100\mathrm{N}$, $v_1=0.030m/s$ and $v_2=0.010m/s$ give $w=2\mathrm{W}$. Over two seconds the first terminal supplies $6\mathrm{J}$ and the second returns $2\mathrm{J}$, leaving net input $4\mathrm{J}$. Ignoring terminal2 would overstate the net supply.

For several translational and rotational ports, sum all entering power terms:The mechanical dissipation inequality then requires $\dot{E}\le w$ for an appropriate nonnegative mechanical energy storage function. Negative instantaneous $w$ is permitted when stored energy returns through the ports; it does not automatically contradict a passive mechanical network.

The chosen supply rate also defines the question being certified. A mathematical quadratic supply used in dissipativity can encode another input/output property, but it represents physical mechanical power only when its variables, units and sign convention match the actual energy exchange. This is why port bookkeeping must precede interpreting a fitted response model as passive.

A mechanical power port specifies the variables and sign convention through which a mechanical system exchanges energy. For translational motion these variables are an entering force $F$ and the corresponding signed velocity $v$; entering power is $p=Fv$. A rotational port uses entering torque $\tau$ and angular velocity $\omega$, giving $p=\tau \omega$. The pairing matters: multiplying a force by a displacement gives work, not its rate. A mechanical supply rate sums the powers of all specified ports.^[ref]

A two-terminal element makes mechanical power-port orientation explicit. Equal and opposite terminal forces $F,-F$ acting at velocities $v_1,v_2$ giveThus the port's flow variable is relative velocity, unless the second terminal is fixed. Reversing both chosen directions leaves the physical power unchanged. For $F=100\mathrm{N}$, $v_1=0.030m/s$ and $v_2=0.010m/s$, net entering power is $2\mathrm{W}$; the separate terminal contributions are $3\mathrm{W}$ and $-1\mathrm{W}$.

At an ideal rigid interconnection, connected terminals have the same velocity, while their entering forces sum to zero by Newton's third law. Their port powers therefore cancel in the combined system. This cancellation explains why internal connections disappear from an overall mechanical energy balance. A spring at the remaining external port obeys $p=kx\dot{x}=d(k{x}^2/2)/dt$; a dashpot obeys $p=c{v}^2$. One stores energy, while the other dissipates it.

For sinusoidal peak complex amplitudes, mean entering power is $\overline{p}=\frac{1}{2}\mathrm{Re}(F{V}^{\ast })$. A pure mass can have nonzero force and velocity amplitudes yet zero mean power, because their phases differ by a quarter cycle. Measuring magnitudes alone therefore does not determine energy transmission or establish the passivity of a mechanical network synthesis realization.

Mechanical power-port orientation specifies the signs of the conjugate variables used to describe entering power at a mechanical power port. For a translational port with signed force $F$ and velocity $v$, use $p=Fv>0$ for net supply into the chosen system. Reversing the coordinate axis reverses both variables, so $(-F)(-v)=Fv$. Reversing just one variable instead changes the entering-versus-leaving convention and must be stated explicitly. This bookkeeping makes a mechanical supply rate meaningful in a mechanical dissipation inequality.^[ref]

For an element acted on by terminal forces $F,-F$ and terminal velocities $v_1,v_2$,Exchanging terminal labels gives $F^{\prime }=-F$ and $v_1^{\prime }-v_2^{\prime }=-(v_1-v_2)$, preserving power. For $F=100\mathrm{N}$, $v_1=0.030m/s$ and $v_2=0.010m/s$, both descriptions give $2\mathrm{W}$. Taking force from the exchanged convention while retaining the original relative velocity would incorrectly report $-2\mathrm{W}$.

More general coordinate changes require the corresponding conjugate-force change. If physical displacement coordinates satisfy $\mathbf{x}=\mathbf{Tq}$, then $\dot{\mathbf{x}}=\mathbf{T}\dot{\mathbf{q}}$. Equality of virtual work, or instantaneous power, requiresFor the scalar coordinate $q=2x$, the generalized force is $F_q=F_x/2$. Thus $F_x=10\mathrm{N}$ and $\dot{x}=0.10m/s$ correspond to $F_q=5\mathrm{N}$ and $\dot{q}=0.20m/s$, both giving1W. Merely rescaling the motion channel without its conjugate force changes the apparent power. At an ideal connection, equal terminal velocities and opposite entering forces cancel internal port powers, enabling the combined mechanical energy balance.