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Coupled normal-contact penalty modes are small oscillations generated by several simultaneously compressed penalty contacts acting through shared body inertia. A displacement at one contact can accelerate another because both load the same translating or rotating bodies. Separate scalar effective mass in a normal-contact penalty mode estimates do not generally give the coupled frequencies.

Freeze a contact configuration, use independent generalized displacement perturbations $q$, positive-definite mass matrix $M$, and gap perturbations $\xi =Jq$. Let $K$ be a positive diagonal matrix of normal penalty stiffnesses. After subtracting equilibrium loads, the restoring generalized force is $-J^{\mathsf{T}}KJq$, soThe contact-Jacobian force/velocity pairing and stiff compliant-contact model come from the primary multibody formulation; the local modal reduction here follows by substitution.^[ref]

Although $AK$ need not be symmetric, it is similar to the symmetric positive-semidefinite matrix $K^{1/2}A{K}^{1/2}$. Set $z=K^{1/2}\xi$ to obtainIts positive eigenvalues are squared angular frequencies. A zero eigenvalue may describe redundant gap coordinates outside the realizable displacement subspace; it should not automatically be interpreted as an extra physical oscillation. Free generalized motions satisfying $Jq=0$ are a separate possibility.

For an original planar example, a rigid bar has vertical translation $y$, small rotation $\theta$, mass $m$, center-of-mass moment of inertia $I$, and two vertical contacts at horizontal offsets $-a,+a$. ThenWith equal contact stiffness $k$, the equal-gap mode is vertical translation and the opposite-gap mode is rocking. Their angular frequencies areTake $m=2\mathrm{kg}$, $I=0.005\mathrm{kg}{\mathrm{m}}^2$, $a=0.10\mathrm{m}$ and $k=10^{5}N/m$. Then $A$ has diagonal entries $2.5\mathrm{k}{\mathrm{g}}^{-1}$ and off-diagonal entries $-1.5\mathrm{k}{\mathrm{g}}^{-1}$. Its eigenvalues are $1$ and $4\mathrm{k}{\mathrm{g}}^{-1}$, giving $316.2$ and $632.5rad/s$. Treating the contacts independently would assign each $500rad/s$ and miss the faster rocking mode.

Contact-mode timestep stability depends on the largest active frequency and on the integrator. For the undamped central-difference recurrence, one mode has amplification equationThe distinct oscillatory roots lie on the unit circle for $0<h\omega <2$; the endpoints need separate treatment. The example therefore requires $h<2/632.5\approx 3.162\mathrm{ms}$ for this linear stability condition. Resolving the waveform accurately normally requires a smaller step. This is not a stability rule for every integration method.

The derivation applies while both contacts remain compressed and geometry is locally constant. A sufficiently large rocking amplitude can unload one contact, changing the active model. A proposed check preloads both supports, applies small symmetric and antisymmetric perturbations, and compares measured or simulated frequencies with the two predictions. Repeat with smaller amplitudes and smaller timesteps to separate linearization error, contact opening and numerical error.

The effective mass in a normal-contact penalty mode is the inertia seen by one specified relative normal displacement, rather than automatically the total mass of either body. A contact away from a center of mass can excite rotation as well as translation, changing that inertia and the normal contact penalty time scale.

Use independent generalized velocities $v$, a positive-definite mass matrix $M$, and a normal gap $g$ whose local velocity is $\dot{g}=Jv$. A scalar normal force $\lambda$ produces generalized force $J^{\mathsf{T}}\lambda$. The primary multibody formulation gives this contact-Jacobian velocity/force pairing explicitly.^[ref] At a frozen configuration, the part of gap acceleration caused by that force isIf the scalar $a=J{M}^{-1}{J}^{\mathsf{T}}$ is positive, defineThe definition uses a physical normal gap measured in metres and a matching normal force measured in newtons. Arbitrarily rescaling the gap coordinate changes the associated inertia. Constraints must first be accounted for by independent coordinates or the appropriate constrained response.

A short normal impulse $P$ gives $\Delta v=M^{-1}{J}^{\mathsf{T}}P$, hence $\Delta \dot{g}=P/m_{\mathrm{eff}}$. This supplies a direct interpretation: less effective mass means a larger relative normal-velocity change for the same impulse. At a fixed local configuration, a compressed linear penalty spring of stiffness $k$ gives the small-motion estimateIt omits changing contact geometry, other driving forces and damping; the source separately explains why stiff compliant contact challenges numerical integration.^[ref]

For two freely translating particles, the relative gap sees $a=1/m_1+1/m_2$, so $m_{\mathrm{eff}}=m_1{m}_2/(m_1+m_2)$. This is the reduced mass. Equal $2\mathrm{kg}$ particles give $m_{\mathrm{eff}}=1\mathrm{kg}$; replacing one by a fixed wall gives $2\mathrm{kg}$ instead.

For a planar rigid body above a fixed plane, let its mass be $m$, rotational moment of inertia about its center be $I$, and the contact normal be vertical. A contact a horizontal distance $r$ from the center hasFor $m=2\mathrm{kg}$, $I=0.020\mathrm{kg}{\mathrm{m}}^2$ and $r=0.10\mathrm{m}$, $m_{\mathrm{eff}}=1\mathrm{kg}$. At the center it is $2\mathrm{kg}$. With $k=10^{5}N/m$, the off-center angular frequency is $316.2rad/s$, compared with $223.6rad/s$ at the center. Using the full mass in both cases misses this rotational response.

Several simultaneous contacts require more than separate scalar replacements. Their matrix $A=J{M}^{-1}{J}^{\mathsf{T}}$ generally has off-diagonal terms: a force at one contact changes another contact’s acceleration. With frozen geometry and positive diagonal stiffness matrix $K$, compressed gap perturbations satisfy $\ddot{\xi }+AK\xi =0$; its nonzero modal squared frequencies are those of the symmetric matrix $K^{1/2}A{K}^{1/2}$. Coupled normal-contact penalty modes can therefore be faster than estimates treating each contact independently. A proposed verification applies a small known impulse or displacement perturbation at a stated configuration and checks relative contact motion, followed by a timestep convergence study.

The normal contact penalty time scale describes the fast dynamics introduced by a stiff normal-contact approximation. For a compressed linear penalty spring with stiffness $k$ acting on a point mass $m$, small oscillations about equilibrium satisfy $m\ddot{x}+kx=0$. Their angular frequency, reference time and period areThe reference time $t_c$ is not the full oscillation period. Rescaling time by $s=t/t_c$ gives $d^{2}x/d{s}^2+x=0$, so a numerical step is controlled by its dimensionless size $h/t_c$, not just its duration in seconds. The primary compliant-contact treatment highlights the numerical stiffness caused by such springs.^[ref]

A modeling and numerical budget interact. Limiting equilibrium overlap under a specified maximum supported load $F_{\max }$ to $d_{\max }$ requires $k\ge F_{\max }/d_{\max }$ in this linear model. For an original choice $F_{\max }=10\mathrm{N}$ and $d_{\max }=10\mu \mathrm{m}$, choose $k=10^6\mathrm{N}{\mathrm{m}}^{-1}$. At $m=0.5\mathrm{kg}$,A chosen resolution target of twenty steps per contact oscillation would set $h\le 0.2221\mathrm{ms}$. This is a resolution target, not a universal accuracy or stability theorem for every integrator.

Increasing the stiffness by a factor of 100 reduces the same-load overlap by 100 but reduces these times by only ten. Keeping the old step then gives roughly two samples per new oscillation instead of twenty. The parent penalty approximation of unilateral contact also shows why the Forward Euler method's undamped-oscillator instability cannot be cured merely by calling the step small.

For a multibody contact, the effective mass in a normal-contact penalty mode replaces an arbitrary total body mass. Additional damping and nonlinear force laws can introduce other time scales. A convergence study should therefore check force histories and contact events as well as maximum overlap; smaller geometric penetration alone does not establish a more accurate simulation.