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

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

A free mass $M$ driven by $F_0\cos \omega t$ can be coupled by a single spring of stiffness $k$ to an auxiliary free mass $m$. With no wall attachment or damping, the periodic displacement amplitude of $M$ is $A=F_0(k-m{\omega }^2)/\{{\omega }^2[Mm{\omega }^2-k(M+m)]\}$. Tuning $k=m{\omega }^2$ gives antiresonance: $A=0$ while the auxiliary mass has signed amplitude $-F_0/k$, supplying an exactly cancelling spring force. A nearby pole occurs at $k=Mm{\omega }^2/(M+m)$. For relative tuning error $ϵ=(k-m{\omega }^2)/(m{\omega }^2)$ sufficiently small, the primary amplitude relative to its free response grows as $(M/m)|ϵ|$. A light absorber is therefore narrowband and may require large auxiliary travel. This cancellation dissipates no energy, and undamped free modes persist unless excluded by initial conditions.

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.

Both masses in the source's absorber diagram are free: there is no additional spring attaching $M$ to a wall. Let their displacements be $X$ and $x$. The equations areFor a periodic particular response $X=A\cos \omega t$, $x=B\cos \omega t$, the amplitude equations are $(k-M{\omega }^2)A-kB=F_0$ and $-kA+(k-m{\omega }^2)B=0$. Their solution isAt $\boxed{k=m{\omega }^2}$, $A=0$ and $B=-F_0/k$: the spring force on the larger mass cancels the applied force at every instant. This is antiresonance, a frequency-selective cancellation, rather than dissipation of energy.

There is a nearby undamped resonance at $k_{\mathrm{res}}=Mm{\omega }^2/(M+m)$. For $m\ll M$ it lies close to the cancellation setting. More quantitatively, write $k=m{\omega }^2(1+ϵ)$. ThenFor $|ϵ|\ll m/(M+m)$, the amplitude relative to the unconnected mass's amplitude $F_0/(M{\omega }^2)$ is approximately $(M/m)|ϵ|$. A light absorber therefore needs very accurate tuning and loses suppression when the driving frequency changes. Its own travel $|B|=F_0/(m{\omega }^2)$ can also become large, limiting linearity and available stroke.

Without damping, resonance growth and persistent free oscillations remain; special initial conditions select the displayed periodic particular response. The free pair also has rigid-translation motion. These limitations prevent interpreting the cancellation as broadband damping or as a unique attracting late-time state.

At $\omega =\sqrt{3}{\omega }_0$, $s=3$, so the periodic amplitudes becomeThe second bead is stationary in this particular motion, although the endpoint oscillates. The first bead moves at twice the endpoint amplitude and in opposite phase. This is an antiresonance of the second-bead response, not a natural resonance of the complete two-bead system.

The force balance makes the cancellation explicit. At $y_2=0$, the middle segment supplies transverse force $T{y}_1/(2L)=-Th/L$, while the end segment supplies $Th/L$; these cancel on the second bead. The first bead then oscillates between two effectively fixed neighbours with restoring coefficient $3T/(2L)$, whose single-coordinate natural angular frequency is $\sqrt{3}{\omega }_0$.

As elsewhere in the undamped problem, this exact nodal motion presumes initial conditions selecting the periodic particular response. Additional free normal modes would move the second bead as well; they do not decay automatically.

A suitable electrical circuit has $R$ in series with the input and a shunt branch consisting of $C$ in parallel with the lossy inductor $R_L+i\omega L$; take the output voltage across that shunt branch. This is the middle arrangement in the circuit figure. In terms of its complex impedance,At direct current the capacitor is an open circuit and the inductor contributes only $R_L$, so $H(0)=R_L/(R+R_L)=0.01$. ThereforeWhen $R_L$ is neglected, $Z_p=i\omega L/(1-{\omega }^{2}LC)$. At the antiresonance ${\omega }_0^{2}LC=1$ its electrical admittance vanishes, so $|H|$ reaches its maximum value of one. ConsequentlyThis value uses precisely the requested lossless approximation; keeping $R_L$ changes both the peak position and its height. A transfer function peak and one DC measurement do not uniquely identify every possible two-port network, but the displayed arrangement meets both observations to the stipulated approximation. Also, the angle brackets in the source must denote AC wave amplitudes or root mean square values: the literal time averages of sinusoidal voltages are zero, whose ratio would be undefined.

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.