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For the initial radius $r_0=l/2$ to be circular, the radial equation requiresConsider a small radial perturbation at fixed conserved angular momentum $J$. Using $J^2/(m{r}_0^3)=Mg$ and $r=r_0+ϵ$ in the radial equation givesConsequently the linearized radial motion is simple harmonic motion:The positive curvature of the effective potential establishes radial stability. This is the small-amplitude limit, not an exact finite-amplitude sinusoid. A perturbation that changes $J$ shifts the circular radius; its radial oscillations are about the new radius to the same order. A radial displacement or radial impulse preserving $J$ gives the stated oscillations about $l/2$.

Inertia is resistance to a change of motion. Translational inertia is measured by mass in Newton's second law; rotational inertia about a specified fixed axis is measured by the moment of inertiawhere $r_{\perp }$ is perpendicular distance from that axis. For a thin uniform rod, take axes perpendicular to the rod. With linear mass density $\lambda =m/l$,The second result uses the parallel axis theorem; direct integration over $0\le x\le l$ gives the same value. The axis specification is essential: an ideal thin rod has zero moment of inertia about its own longitudinal axis.

During the short elastic collision, neglect the impulse from gravity. The pivot can exert an external impulse, so total linear momentum of rod plus particle need not be conserved. Its torque about the pivot is zero, so conservation of angular momentum about that point applies. Let $w$ be the particle's signed outgoing horizontal velocity, and $\omega$ the rod's outgoing angular velocity, positive towards the incident motion. ThenSubstituting $\omega =3(v_0-w)/l$ into the kinetic energy equation givesDiscarding the zero-impulse branch $w=v_0$, the actual collision givesThe rod's lower end initially moves forward at $l\omega =3v_0/2$, so the bodies separate. Afterwards the free particle undergoes projectile motion. With the pivot as origin and $z$ downward, its trajectory is $x=(v_0/2)t$, $z=l+g{t}^2/2$, where $t=0$ at the collision. The rod instead swings about the fixed pivot in a vertical plane as a physical pendulum.

The rod's initial rotational kinetic energy is $I_{\mathrm{end}}{\omega }^2/2=3m{v}_0^2/8$. Reaching the horizontal raises its centre of mass by $l/2$, increasing gravitational potential energy by $mgl/2$. Consequently,For the threshold case, measure the rod's angle $\theta$ from the downward vertical and reset $t=0$ at its first horizontal turning point, so $\theta (0)=\pi /2$ and $\dot{\theta }(0)=0$. The restoring torque about the pivot gives the exact equation of motionAt the horizontal, $\ddot{\theta }=-3g/(2l)$, so the rod immediately swings back. Conservation of mechanical energy yieldsOn the initial descent its explicit time dependence is specified by the definite integralIt continues through the vertical to $-\pi /2$, reverses, and repeats without damping. An alternative expression valid throughout the oscillation uses the Jacobi elliptic sine. Put $k=1/\sqrt{2}$, ${\omega }_r=\sqrt{3g/(2l)}$, and let $K(k)$ denote the complete elliptic integral of the first kind. ThenThis is not simple harmonic motion at an amplitude of $\pi /2$. Its oscillation period exceeds its own small-amplitude value $2\pi \sqrt{2l/(3g)}$ by the factor $2K(1/\sqrt{2})/\pi \simeq 1.18034$.

A simple pendulum of length $l/2$ obeys $\ddot{\theta }+(2g/l)\sin \theta =0$. At the same amplitude $\pi /2$, its oscillation period is $T_p=4\sqrt{l/(2g)}K(1/\sqrt{2})$, soThe rod swings more slowly. Its equivalent simple pendulum length is $I_{\mathrm{end}}/[m(l/2)]=2l/3$, not the distance $l/2$ to its centre of mass. The same oscillation period ratio holds in the common small-amplitude limit, but replacing the actual 90-degree motion by that limit would underestimate both oscillation periods.

Axis theorems. The parallel axis theorem stateswhere $I_G$ is the moment of inertia about an axis through the centre of mass, $I_O$ is about a parallel axis, $m$ is the body's total mass, and $d$ is the perpendicular separation of the axes. Expanding $\int [(x-d{)}^2+y^2]dm$ proves the result because $\int xdm=0$ at the centre of mass. The perpendicular axis theorem applies to a planar lamina in the $xy$ plane:All three axes meet at the same point, $x,y$ lie in the lamina and $z$ is normal to it; the identity follows from $x^2+y^2$ being the squared distance to the $z$ axis. It is not a theorem for an arbitrary three-dimensional body.

Cylinder integrals. Put the cylinder's midpoint at $z=0$ and its symmetry axis along $z$. Its uniform mass density is $\rho =m/(\pi r^{2}l)$. The transverse moment of inertia of a uniform solid cylinder about the $x$ axis isEquivalently each thin circular slice has diametral moment of inertia $r^{2}dm/4$, by the perpendicular axis theorem and symmetry, and shifting it by $z$ adds $z^{2}dm$. The parallel axis theorem then gives the transverse end-axis moment of inertia:The axial moment of inertia of a uniform solid cylinder is $I_{\parallel }=m{r}^2/2$. Equating it to $I_{\perp ,G}$ givesHere the source's principal axis means the cylinder's symmetry axis; its transverse centre axes are also principal axes. This equality describes a short cylinder rather than the thin-rod limit.

Hammer dynamics and forces. Treat the compact head as a point mass and neglect its intrinsic moment of inertia. Let $\theta$ be the displacement from downward vertical, and let ${\mathbf{e}}_r$ point along the rod from pivot to head, with ${\mathbf{e}}_{\theta }$ in the direction of increasing $\theta$. DefineThe two external weights $Mg$ and $mg$ act at distances $l$ and $l/2$ from the pivot. The frictionless pivot supplies a reaction $\mathbf{R}$ but no torque about its axis. The total restoring torque and rotational equation of motion areFor completeness, Newton's second law for the centre of mass gives both components of the pivot force:The negative radial component points towards the pivot. These follow from the centre of mass acceleration $-d{\dot{\theta }}^2{\mathbf{e}}_r+d\ddot{\theta }{\mathbf{e}}_{\theta }$ and the total weight $m_{T}g(\cos \theta {\mathbf{e}}_r-\sin \theta {\mathbf{e}}_{\theta })$. They specify the external forces even when the oscillation is not small.

With the small-angle approximation, $\sin \theta \simeq \theta$, this is simple harmonic motion. The physical pendulum has small-amplitude oscillation periodFor release from a finite amplitude $0<{\theta }_0<\pi$, conservation of mechanical energy instead gives $\frac{1}{2}I{\dot{\theta }}^2=gA(\cos \theta -\cos {\theta }_0)$. Integrating a quarter-cycle gives the exact oscillation periodwhere $K$ is the complete elliptic integral of the first kind; $K(0)=\pi /2$ recovers $T_0$.

For $m=M/10$ and $r=l/20$, compare with the simple pendulum value $T_s=2\pi \sqrt{l/g}$:For small oscillations, the hammer's oscillation period is about $0.794\%$ shorter. The extra rod weight adds restoring torque proportionately faster than it adds pivot moment of inertia.

For a displacement $(0,y)$ of the middle wire, the horizontal attractions towards the two fixed outer wires cancel. Each attraction has magnitude ${\mu }_0{I}^2/(2\pi \sqrt{d^2+y^2})$ per unit length. Projecting them in the $y$ direction givesNeglecting dissipative effects and other transverse forces, this is simple harmonic motion. Release from rest at $y_0$ givesThe inverse-angular-frequency timescale is ${\Omega }^{-1}=0.500\mathrm{s}$; it differs from the oscillation period by $2\pi$. The assumed displacement plane matters: small displacements towards one outer wire along $x$ are unstable, rather than oscillatory.

This result concerns the specified magnetic motion with gravity absent from, or compensated in, the $y$ direction. The source diagram fixes coordinate directions, not their orientation relative to gravity. If $y$ is physically vertical with upward positive and the wire is unsupported, add $-{\mu }_{m}g$ to the force per unit length. The maximum restoring magnetic force is only ${\mu }_0{I}^2/(2\pi d)=2.0\times 10^{-5}\mathrm{N}{\mathrm{m}}^{-1}$, compared with weight $9.8\times 10^{-3}\mathrm{N}{\mathrm{m}}^{-1}$; that different setup would fall and could not oscillate about the original plane.

Let $m$ be the sliding point charge's mass. Expanding the axial field of a uniformly charged annulus for $|x|\ll a,b$ givesIts electric charge is $-Q$, so its electric force points toward the centre. Newton's second law becomesTo leading order this is simple harmonic motion, withThe mass is an additional parameter, since the source supplies no numerical value. The constraint is essential: stability along the axis does not imply unconstrained three-dimensional trapping by a static electric field.

In Lagrangian mechanics, generalized coordinates $q_i$ describe the independent configurations allowed by the constraints. For particles with ordinary conservative forces in an inertial frame, use $L=T-V$, with kinetic energy $T$ and potential energy $V$. Holonomic constraints can often be built into the coordinates, avoiding explicit constraint forces.

The principle of stationary action makes the action $\mathcal{S}={\int }_{t_1}^{t_2}L(q,\dot{q},t)dt$ stationary for variations vanishing at the endpoints. Integrating the velocity variation by parts givesSince each independent variation is arbitrary, the Euler-Lagrange equations areIf nonconservative generalized forces are present, their components appear on the right. The formalism is invariant under suitable coordinate changes and adding a total time derivative to $L$ leaves the fixed-endpoint equations unchanged.

The generalized momentum is $p_i=\partial L/\partial {\dot{q}}_i$. If $q_i$ is cyclic, $\partial L/\partial q_i=0$ and $p_i$ is conserved. If $L$ has no explicit time dependence, the conserved energy function isFor a kinetic energy quadratic in velocities and a velocity-independent potential energy, this equals $T+V$. This connects symmetry to conservation laws.

For example, a simple pendulum of mass $m$ and length $\ell$ hasIts tension does not appear because the fixed-length constraint is already imposed. Expanding for small $\theta$ gives simple harmonic motion with $\omega =\sqrt{g/\ell }$. More generally, expansion of $T$ and $V$ to quadratic order about a stable equilibrium leads directly to the normal-mode eigenvalue problem.

For the gyrocompass oscillation, use the same local basis and define $\theta$ about the stable orientation, with spin angular momentum directed north at $\theta =0$. Because the vertical is perpendicular to the horizontal axle, the extra vertical angular momentum from the slow azimuthal motion is $I_2\dot{\theta }$. To leading order in Earth's small angular velocity,A constant vertical contribution $I_2{\Omega }_E\sin \lambda$ from Earth's rotation has zero terrestrial time derivative; the transverse terrestrial contributions affect the following equation only at order ${\Omega }_E^2$. Taking the vertical component of the torque equation givesThe terrestrial derivative of $J_s\mathbf{u}$ is horizontal, as is ${\mathbf{\Omega }}_E\times (I_2\dot{\theta }\mathbf{z})$, so neither contributes to this component. Thus the small angle satisfies, using the small-angle approximation $\sin \theta \simeq \theta$,This is simple harmonic motion, $\theta =A\cos ({\omega }_{\mathrm{shm}}t+\varphi )$, with positive squared angular frequency away from the poles. The approximation requires rapid spin, small angular displacement and negligible bearing friction. Near the opposite orientation, putting $\theta =\pi +\delta$ instead yields $\ddot{\delta }-{\omega }_{\mathrm{shm}}^2\delta \simeq 0$, which is unstable. Consequently, the paper's oscillation claim applies to the stable equilibrium, not both north--south spin directions. Without damping the ideal gyrocompass continues to oscillate rather than settling.

For the same massless spring of spring constant $k$, Hooke's law gives the vertical simple harmonic motion about the displaced equilibrium position:Gravity shifts the equilibrium extension but does not change this frequency. Use the measured sample means in the ratio estimate. For small measurement errors, logarithmic differentiation givesAssuming the two frequency estimates have independent errors, their variances add. Writing $ϵ=\delta \nu /\overline{\nu }$ for their common fractional standard uncertainty, monomial uncertainty propagation yieldsThis is a first-order uncertainty formula, rather than an exact identity for an arbitrary noisy ratio. It assumes that errors in the two measured frequencies have zero covariance.

There is a statistical distinction in the wording: the standard deviation of individual repetitions is not the standard error of their sample mean. If the supplied $\sigma ({\nu }_i)$ describes individual trials, the frequency means have errors $\sigma ({\nu }_i)/\sqrt{n}$, and the ratio formed from those means instead satisfiesThus the printed formula describes either a single-pair ratio or a convention in which $\delta \nu$ already denotes the uncertainty of each averaged frequency. Under either consistent interpretation, independent repeated measurements reduce statistical error in proportion to $n^{-1/2}$. Halving it requires $4n$ trials in total for each mass, or $3n$ additional trials. Repetition alone does not reduce a fixed systematic error.

Axial moment of inertia. In the original diagram $R_1$ is the outer radius and $R_2$ the inner radius. For uniform mass density $\rho$, a coaxial annulus of thickness $dr$ has $dm=2\pi \rho Lrdr$. Therefore the axial moment of inertia of a hollow cylinder isSince $M=\pi \rho L(R_1^2-R_2^2)$, factorizing the fourth-power difference givesFor a solid cylinder $R_2\to 0$, giving $I=M{R}_1^2/2$. For a thin cylindrical shell with fixed mass and $R_1,R_2\to R$, $I=M{R}^2$. The latter limit agrees with every mass element being at radius $R$.

Undamped rolling oscillations. Let $x$ be axle displacement from the spring equilibrium. Rolling without slipping imposes $|{\omega }_{\mathrm{roll}}|=|\dot{x}|/R$. The sum of translational and rotational kinetic energy isThe potential energy is $U=k{x}^2/2$. Differentiating $T+U$ in the absence of damping, or applying the force and torque equations below, givesFor the thin shell $I=m{R}^2$, the effective inertia is $2m$, and the angular frequency of the simple harmonic motion isStatic contact friction supplies the required torque but does no work when the contact point is instantaneously stationary.

Damped force balance. Take rightward force as positive and counterclockwise rotation as positive. Write $f$ for the signed static friction at the bottom contact. ThenEliminate $f=-I\ddot{x}/R^2$ and use $I=m{R}^2$:Here $b$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, so $B$ has units ${\mathrm{s}}^{-1}$ and $C$ units ${\mathrm{s}}^{-2}$. The stated drag force is interpreted as acting through the axle, with no additional viscous torque. An oil bath could also exert such a torque physically, but its magnitude is not specified by this force-only model.

Light damping. Substitution of $z=A{e}^{-pt}$ into the differential equation gives $p^2-Bp+C=0$. In the underdamped regime $B^2<4C$,Taking real linear combinations gives the general damped harmonic oscillator displacementThe constants are fixed by the initial displacement and velocity. For example, release from rest at $x_0$ gives $A_1=x_0$ and $A_2=b{x}_0/(4m{\omega }_d)$. Light damping means $b^2\ll 8mk$, in which case ${\omega }_d\simeq {\omega }_0$ and the envelope decays slowly compared with an oscillation.

A conservative vector field has path-independent work: the integral ${\int }_A^B\mathbf{F}\cdot d\mathbf{r}$ depends only on its endpoints. Equivalently, the line integral around every closed path in its domain vanishes. One can then define a single-valued potential energy $U$ byThe arbitrary additive constant in $U$ has no effect on the force. A useful zero is often chosen at infinity for a localized attractive interaction.

A differentiable conservative force has $\nabla \times \mathbf{F}=0$ because the curl of a gradient vanishes. Conversely, a smooth curl-free field on a simply connected domain admits a global scalar potential. The domain condition matters: on the punctured plane, $\mathbf{F}=(C/r)\hat{\mathbf{\theta }}$ has zero curl away from the missing origin but closed-loop work $2\pi C$ around it. It is not globally conservative there. The existence of a local potential does not establish a single-valued potential throughout a domain with holes.

The work-energy theorem gives $\Delta T=W$ for kinetic energy $T$. If all forces are conservative and $U$ has no explicit time dependence, thenThis is conservation of mechanical energy. In one dimension, an equilibrium occurs at $dU/dx=0$; a local minimum with $U^{\prime \prime }>0$ is stable, while a maximum with $U^{\prime \prime }<0$ is unstable. Expanding about a minimum gives $U\simeq U_0+U^{\prime \prime }(x_0)(x-x_0{)}^2/2$, producing simple harmonic motion with ${\omega }^2=U^{\prime \prime }(x_0)/m$.

Examples include the spring force $F_x=-kx$ with $U=k{x}^2/2$, the near-surface gravitational force $-mg\hat{\mathbf{z}}$ with $U=mgz$, and the central Newtonian gravitational field with $U=-GMm/r$. An electrostatic force on a fixed electric charge $q$ has $U=q\Phi$, where $\Phi$ is the electric potential.

Sliding friction provides a contrasting physical test: traversing a closed path still dissipates mechanical energy, since the frictional force opposes the local motion on both outward and return legs. Its work depends on path length and cannot be represented by a position-only $U$. Total energy can remain conserved when the resulting internal energy is included. Also, even if $\mathbf{F}=-\nabla U(\mathbf{r},t)$ at each instant, explicit time dependence gives $d(T+U)/dt=\partial U/\partial t$; a static conservative-field argument alone does not guarantee constant mechanical energy in that situation.

Let the cylinder length be $\ell =0.020\mathrm{m}$, cross-sectional area $A$, and the fluid mass density ${\rho }_f$. At floating equilibrium, Archimedes' principle givesFor a small downward displacement $z$ from this equilibrium, the displaced fluid volume increases by $Az$. The additional upward buoyant force is ${\rho }_{f}gAz$, and Newton's second law becomesThis is simple harmonic motion, with angular frequency $\omega =\sqrt{2g/\ell }$ and oscillation periodThe result for vertical oscillations of a floating cylinder is independent of its cross-sectional area and fluid mass density because those quantities cancel using the known immersion fraction. It neglects drag, surface effects and the extra fluid inertia from the motion; the cylinder must remain partly immersed throughout the small motion.

Put $\delta ={\varphi }_2-{\varphi }_1$. If both coordinate amplitudes are nonzero, eliminate the common phase to obtainFor $\sin \delta \ne 0$, this is an ellipse centred at the origin. The following special cases exhaust the possible nontrivial shapes:

The degenerate zero-amplitude solution is the stationary origin. There are no general multi-loop patterns here because the two angular frequencies are equal. Another useful description is $\mathbf{r}=\mathbf{a}\cos \omega t+\mathbf{b}\sin \omega t$, the image of a circle under a linear map.

Let $x$ be signed displacement along the rails, measured from the centre. The diagram in the original paper shows a guide extending across the centre, so the signed coordinate can pass through zero. Hooke's law gives the radial force $-kx$. In the rotating reference frame, the radial centrifugal force is $+m{\Omega }^{2}x$. The Coriolis force is perpendicular to the rails and is balanced by their reaction; it has no radial component. Therefore Newton's second law givesEquivalently, the radial component of acceleration in polar coordinates in the inertial frame is $\ddot{x}-{\Omega }^{2}x$. The radial spring oscillator on a rotating guide has simple harmonic motion only when $k/m-{\Omega }^2>0$, withThe angular frequency is in $\mathrm{rad}{\mathrm{s}}^{-1}$ and the frequency in $\mathrm{Hz}$. The critical angular speed isStrictly this is an upper bound, not an attained maximum for oscillations: at equality $\ddot{x}=0$, so a displaced object released from rest stays at its displaced position relative to the rails; above it, the central equilibrium is unstable.

In the small-amplitude approximation, $\dot{x}$ is the speed along the track, and $y=B{x}^2$ supplies the gravitational potential. The conserved energy isDifferentiating and extending through turning points by continuity givesThis is simple harmonic motion with ${\omega }_0^2=4gB/\{(r_2/r_1{)}^2+2\}$. HenceThe frequency is lower than for a sliding point mass because energy is also stored in rotation.

Subtracting the two linearized equations eliminates their acceleration terms:For $g,k>0$, the assumed nontrivial simple harmonic motion therefore hasThe support displacement and pendulum angle oscillate in phase. Inserting this relation into either Euler-Lagrange equation yieldsso the one positive finite angular frequency isThe negative frequency describes the same real oscillation after a phase change. For a general phase, replace the sine in both coordinates by $A\cos (\omega t)+B\sin (\omega t)$.

There is no missing second finite branch. The quadratic Lagrangian iswhose kinetic matrix is $\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$ and has rank one. Its characteristic determinant is ${\omega }_0^2{\omega }_1^2-({\omega }_0^2+{\omega }_1^2){\omega }^2$, not a quadratic polynomial in ${\omega }^2$. This singular kinetic matrix in small oscillations results from taking the support mass to be exactly zero. Adding a small support mass restores a second mode that runs to infinite frequency as that mass vanishes.

For a stiff spring,The hinge becomes fixed, recovering the simple pendulum frequency, as expected for the pendulum with a massless spring-mounted support. At finite $k$, horizontal support motion makes the frequency smaller.

Constraint and example. A holonomic constraint can be written $f_{\alpha }(q_1,\dots ,q_n,t)=0$, with no irreducible dependence on velocities. A velocity relation is holonomic if it integrates to such an equation. For example, a particle constrained to a stationary circle satisfies $x^2+y^2={\ell }^2$, reducing its two Cartesian degrees of freedom to one angle. A nonholonomic constraint cannot be integrated in this way.

Kinematics and kinetic energy. Set $d=R-r>0$. In the plane of the original diagram, choose the orbital angle $\theta$ positive counterclockwise from vertically downward. The marked radius of the inner cylinder then has absolute orientation $\alpha =\theta -\varphi$ measured from the same downward direction. Its signed spin angular velocity is $\Omega =\dot{\theta }-\dot{\varphi }$, whileThe moment of inertia of a uniform solid cylinder is $I=m{r}^2/2$. Applying the kinetic energy decomposition about the center of mass gives

No slip. With tangential unit vector ${\mathbf{e}}_{\theta }=(\cos \theta ,\sin \theta )$, the inner contact point has velocity $(d\dot{\theta }+r\Omega ){\mathbf{e}}_{\theta }$. The fixed wall requires this to vanish. HenceBecause $r,R$ are constant, integration giveswhere the integration constant is zero with the marked-dot reference in the diagram and continuously unwrapped angles. This is a holonomic constraint, even though it was first obtained from a condition on velocities.

Oscillations. The gravitational potential energy above the lowest position is $V=mgd(1-\cos \theta )$. Eliminating $\varphi$ gives the reduced LagrangianThe Euler-Lagrange equation becomesThus the bottom is a stable equilibrium, and for $|\theta |\ll 1$ there is simple harmonic motion, withThe extra rotational kinetic energy reduces the frequency compared with a particle sliding without friction along the same circular path. The small oscillations inside a circular cylinder formula applies with inertia ratio $\gamma =1/2$.

Friction and energy. Let $F_t$ be the signed tangential contact force in the ${\mathbf{e}}_{\theta }$ direction. Applying Newton's second law to translation and the torque equation to rotation givesUsing the rolling without slipping condition yieldsExcept at special instants, static friction is needed to produce the spin acceleration. There is no sliding at the contact point, whose instantaneous velocity is zero; therefore this force does no total work. Its translational and rotational powers cancel: $F_{t}d\dot{\theta }+F_{t}r\Omega =0$. The stationary normal reaction likewise does no work. With a rigid wall, adequate static friction, and no rolling resistance, conservation of energy gives

Angular momentum. The cylinder's spin angular momentum $I\Omega$ is generally not conserved because the contact friction exerts a torque about its center of mass. About the fixed outer-cylinder center $O$, the angular momentum decomposition about the centre of mass givesGravity and friction exert the external torqueThus the rolling cylinder does not generally conserve angular momentum, although it conserves mechanical energy. There is a special geometric cancellation at $R=3r/2$: then $L_O$ and its torque vanish identically for this planar motion. Angular momentum of a larger isolated system including the wall and Earth is a separate conservation statement.

Scope of the holonomic reduction. The integrated constraint is exact at every angle for planar rolling without slipping with sustained contact; it is not restricted to small oscillations. Completing a full turn requires tracking angle winding rather than reducing each angle separately modulo $2\pi$. This global bookkeeping does not make the local velocity relation nonintegrable.

However, arbitrary initial conditions need not sustain this ideal rolling motion. The inward normal reaction and the required friction areContact requires $N\ge 0$, and a finite coefficient ${\mu }_s$ of static friction requires $|F_t|\le {\mu }_{s}N$. If either fails, the cylinder detaches or slips and the one-coordinate rolling Lagrangian no longer applies. Slipping contact allows $\theta$ and $\varphi$ to be independent; detachment removes the fixed center-distance constraint. These changes alone do not establish a nonholonomic constraint. Allowing general three-dimensional reorientation can introduce nonintegrable rolling constraints, but that motion is outside the diagram's two-angle planar model. Thus the stated planar rolling constraint is holonomic throughout its regime of validity; it cannot be imposed on every physically possible motion.

For fixed $\ell$ and $g$, the exact conical pendulum oscillation period satisfies $P/P_0=\sqrt{\cos \theta }$. In the small-angle approximation, with $\theta$ in radians,Thus the limiting revolution period equals the small-amplitude oscillation period of a planar simple pendulum. This equality has a geometric explanation: to first order, $x(t)=r\cos (\omega t)$ and $y(t)=r\sin (\omega t)$ are perpendicular simple harmonic motions of equal amplitude, a quarter period apart, with $\omega \to \sqrt{g/\ell }$. At finite angle, however, the conical pendulum's oscillation period decreases with $\theta$, whereas a planar simple pendulum's oscillation period increases with its angular amplitude.

The endpoint $\theta =0$ itself is a stationary bob with $r=0$ and no observable revolution; $P_0$ is a limit of nonzero circular motions, not a measured revolution period at rest. For fixed $\ell$, nonzero conical pendulum motion also requires $\omega >\sqrt{g/\ell }$, because $\cos \theta =g/(\ell {\omega }^2)<1$. At the other endpoint, $\theta \to \pi /2$ gives $h\to 0$, $P\to 0$, and unbounded tension and speed. These mathematical limits explain why the ideal string cannot sustain a horizontal orbit at finite tension; a real experiment ceases to satisfy the ideal assumptions before that limit.

A charge $q$ undergoing simple harmonic motion of angular frequency $\omega$ and slowly varying amplitude $A$ radiates the averaged powerIts mechanical energy $E=m{A}^2{\omega }^2/2$ therefore has exponential decay rate $\gamma ={\mu }_0{q}^2{\omega }^2/(6\pi mc)$ and half-life

A restoring force acts toward an equilibrium when a body is displaced from it. In one dimension, taking displacement $x=0$ at equilibrium, this means $xF(x)<0$ for sufficiently small nonzero $x$. An ideal linear spring has $F(x)=-kx$ with positive spring constant $k$, producing simple harmonic motion for a constant mass without damping. More general restoring forces need not be linear.

The Rüchardt experiment infers heat capacity ratio from the period of an oscillation of a piston supported by trapped gas. Use a vertical precision cylinder with a close-fitting movable piston, known moving mass $M$, face area $A$, trapped equilibrium volume $V_0$, and local atmospheric pressure $p_{\mathrm{atm}}$. With negligible static friction, force balance gives the absolute gas pressure $p_0=p_{\mathrm{atm}}+Mg/A$, where $g$ is gravitational acceleration. Displace the piston slightly and release it: compression raises pressure, and expansion lowers it, producing a restoring force. The oscillation is a mechanical probe of thermal response, without separately measuring two heat inputs. ^[ref]

Let $x$ be upward displacement from equilibrium, so $V=V_0+Ax$. For constant heat capacity ratio and a reversible adiabatic process, the net upward force, including the ambient load and weight, isThe linear approximation requires $|Ax|/V_0\ll 1$. Applying Newton's second law gives simple harmonic motion withHere ${\omega }_0$ is the undamped angular frequency and $\mathcal{T}$ the undamped oscillation period, rather than temperature. Record repeated cycles, measure $M,A,V_0,p_{\mathrm{atm}}$ independently, and include all connected trapped volume. An illustrative model with $M=0.020\mathrm{kg}$, $A=2.0\times 10^{-4}{\mathrm{m}}^2$, $V_0=0.010{\mathrm{m}}^3$, $p_{\mathrm{atm}}=100000\mathrm{Pa}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\gamma =1.4$ gives $p_0=100981\mathrm{Pa}$ and $\mathcal{T}\simeq 1.182\mathrm{s}$. These stipulated values are not measurements.