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Use the angle orientation in the PDF: $M$ has coordinates $(R\sin \theta ,-R\cos \theta )$ relative to the hoop centre. The wrapped string segment has length $R(\pi /2-\theta )$. As $\theta$ increases, it releases length $R\theta$, so the suspended mass moves down by $R\theta$. Both speeds have magnitude $R|\dot{\theta }|$.

Thus the hoop with a suspended counterweight hasand $L=T-V$. The Euler-Lagrange equation givesAt equilibrium,The positive curvature $V^{\prime \prime }({\theta }_0)=MgR\cos {\theta }_0$ establishes stability. Put $\theta ={\theta }_0+\delta$ and linearize. The equilibrium force cancels, leavingHenceFor $M\gg m$, ${\theta }_0\simeq m/M$ and $\omega \to \sqrt{g/R}$, the small-oscillation frequency of a simple pendulum of length $R$. The linearization assumes an amplitude small enough to remain inside the stated angular range and a taut string throughout.

The covering-law model explains an event by showing how it falls under general laws in specified circumstances. It captures a major form of scientific understanding, but entailment and predictability do not by themselves capture explanatory relevance or direction.

In the deductive-nomological model, let $L$ be laws, $C$ relevant circumstances, and $E$ the event description. The explanans is $L,C$; the explanandum is $E$. The argument has the formThe deduction must be valid, at least one law must be essential to it, the explanans must have empirical content, and its premises must be true for a sound explanation. Merely writing a valid argument from arbitrary premises is insufficient. Carl Hempel and Paul Oppenheim also identify a structural affinity between explanation and prediction: supplied beforehand, the same premises could in principle support the prediction.^[ref]

For an ideal simple pendulum at small amplitude, Newtonian mechanics and the specified constraints yieldHere $ell$ is the pendulum's length, $g$ the local gravitational acceleration, and $T$ its oscillation period. The deduction connects an observed regularity to more general laws and explains why changing length changes the oscillation period. The approximation must be stated: a finite-amplitude, dissipative simple pendulum does not exactly satisfy these premises. A useful model-based explanation can be approximately correct even when the ideal premises are not literally true.

The model's appeal is that it requires more than an evocative story. It displays assumptions and shows why they constrain the phenomenon. It can also explain laws through wider laws: geometrical optics can recover a refraction law from suitable descriptions of wave propagation. A principle theory can explain by constraints without supplying a detailed constructive theory of every constituent process. This is a strength of the account's generality.

The first difficulty is asymmetry. For a vertical pole of vertical length $h$ on level ground and the Sun's elevation angle $\alpha$, geometrical optics gives the shadow's lengthWith suitable premises, either deduction is valid. Yet the pole and illumination explain the shadow, whereas measuring the shadow ordinarily allows inference to the pole's vertical length without explaining why the pole has that vertical length. The same equation supports both calculations. Its logical form does not encode the relevant direction of dependence. Context can change the question, but the distinction between explanation and evidential reconstruction remains.

The second difficulty is irrelevance. Adding the premise that a laboratory wall is blue to a correct simple pendulum derivation preserves validity and the essential use of physical laws. The extra premise contributes nothing to explaining the oscillation period. More elaborate examples show the same problem when irrelevant factors are hidden inside statistical regularities. The explanatory account needs a way to identify what matters under relevant changes, not just what accompanies a valid derivation.

The inductive-statistical model extends the programme to explanations using statistical laws and high conditional probability, relative to appropriate available information. This is an extension beyond the deductive focus of the 1948 article. High probability is nevertheless not necessary for every explanation. Radioactive decay can explain an infrequent detector event through a physical process and its nuclear decay constant, although that particular event in a short interval was unlikely. Nor does a high conditional probability automatically establish explanatory relevance: a reliable indicator may be an effect of the outcome or share another cause with it.

A better assessment therefore retains law-based derivations as one explanatory resource while adding dependence, relevance and the question being asked. Mechanistic explanations identify productive processes; accounts based on interventions ask which changes would alter the outcome. Neither proposal needs to force every mathematical or constraint-based explanation into a causal story. The covering-law model illuminates how laws make phenomena intelligible, but it is too broad as a sufficient test and too restrictive as a universal necessary form.

Put $\theta ={\theta }_e+\delta$ at the stable equilibrium. Since $M\sin {\theta }_e=m$, linearization of d givesThusAs $m\to 0$, the result approaches $\sqrt{g/R}$, the simple pendulum value. As $m\uparrow M$, the restoring curvature tends to zero and this small-amplitude frequency tends to zero; the endpoint itself is not a stable quadratic minimum.

For the shorter uniform rod, the parallel axis theorem gives $I=m(a^2/12+x^2)$. Its equivalent simple pendulum length is $I/(mx)=x+a^2/(12x)$. The rod of length $2a$ pivoted at an end has equivalent length $2(2a)/3=4a/3$. Matching the periods therefore requiresA pivot on the shorter rod requires $0<x\le a/2$, leavingEither side of the rod's centre of mass gives this distance. The other algebraic root is beyond the rod and would require a support extension of negligible mass.

For the general Lagrangian just obtained, the Euler-Lagrange equation givesUsing the small-angle approximation $\sin \theta \simeq \theta$, this is simple harmonic motion with ${\omega }_0^2=A/I$. ThusThe endpoint mass cancels in the light-rod limit. A concrete comparison shows why the finite rod cannot silently be included: the same specified endpoint $m$ with $M_r=m$ gives $T=2\pi \sqrt{8l/(9g)}$, distinct from the negligible-rod result. This is a physical pendulum when the rod contributes, and a simple pendulum in the light-rod limit. Both expressions require small oscillation amplitude; the full sine equation remains nonlinear.

Constant azimuth. If $\varphi$ is constant, $\dot{\varphi }=0$ and the motion is a planar simple pendulum:For the small-angle approximation, $\sin \theta \simeq \theta$, giving simple harmonic motion withThe harmonic angular frequency requires small amplitude. At finite libration amplitude ${\theta }_{\max }<\pi$, the exact oscillation period is $4\sqrt{l/g}K(\sin ({\theta }_{\max }/2))$, where $K$ is the complete elliptic integral of the first kind; the nonlinear pendulum does not have the same amplitude-independent period.

Constant polar angle. If $\theta ={\theta }_0$ is nonzero and constant, the bob traces a horizontal circle: this is a conical pendulum. Setting $\ddot{\theta }=0$ and dividing by $\sin {\theta }_0$ givesThe sign distinguishes the two senses of rotation. The exactly downward configuration ${\theta }_0=0$ is stationary; its azimuth is undefined and it must not be obtained by dividing by $\sin {\theta }_0$. A nontrivial constant-angle orbit of this gravity-only system needs $\cos {\theta }_0>0$ even though the supporting rod can also push.

For the elastic pendulum, set $r=l+x>0$ and measure $\theta$ from the downward vertical. In the vertical plane, Cartesian coordinates are $X=r\sin \theta$ and $Z=-r\cos \theta$, with $Z$ upward. Differentiating gives ${\dot{X}}^2+{\dot{Z}}^2={\dot{x}}^2+r^2{\dot{\theta }}^2$. Thus the Lagrangian isFor $x$, the Euler-Lagrange equation givesFor $\theta$, the canonical momentum is $p_{\theta }=m(l+x{)}^2\dot{\theta }$. Its Euler-Lagrange equation isor, for positive length,At an equilibrium, all time derivatives vanish, so $kx=mg\cos \theta$ and $\sin \theta =0$. Modulo full rotations, the two formal possibilities areThe lower configuration always has positive length. The upper one is a physical straight-spring configuration only if $l>mg/k$. For example, $l=mg/(2k)$ would give the purported upper state a negative length $-mg/(2k)$; at $l=mg/k$ its zero length makes the angular coordinate singular. Also, the straight compressed spring is an idealization; a real unsupported compressed spring can buckle.

The potential energy is $V=k{x}^2/2-mg(l+x)\cos \theta$. At either formal stationary point its mixed second derivative vanishes, whileBoth curvatures are positive at the lower state, making it a stable equilibrium; the upper state, when it has positive length, has negative angular curvature and is unstable.

To linearize about the stable state, define the small displacement $\xi =x-mg/k$, not $x$ itself, and $r_0=l+mg/k$. Products such as ${\dot{\theta }}^2$, $\dot{\xi }\dot{\theta }$ and $\xi \ddot{\theta }$ are second order, while $\cos \theta =1+O({\theta }^2)$ and $\sin \theta =\theta +O({\theta }^3)$. The first-order equations of motion therefore decouple:There are two independent normal modes. The radial one is vertical simple harmonic motion with angular frequency ${\omega }_r=\sqrt{k/m}$. The transverse one is a simple pendulum oscillation at fixed, statically stretched length $r_0$, with angular frequency ${\omega }_{\theta }=\sqrt{g/r_0}$. Their decoupling is a first-order small-oscillation result, not an assertion about the nonlinear motion of the elastic pendulum.

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.

Interpret $h$ as the vertical drop of the cylinder's centre of mass, and initially take the suspended rod to be vertical. The moment of inertia of a uniform solid cylinder is $I=m{r}^2/2$. Before impact, rolling without slipping gives $\omega =u/r$, so conservation of mechanical energy givesUse the usual instantaneous smooth elastic collision model: the horizontal impact impulse passes through the cylinder's centre of mass, and the floor supplies negligible tangential impulse during impact. The cylinder's angular velocity is then unchanged during impact; its rotational kinetic energy does not transfer to the bob. The rod supplies no horizontal impulse when vertical. If $v$ and $V$ are the signed cylinder and bob velocities immediately afterwards, conservation of momentum and the unit coefficient of restitution giveDuring the subsequent simple pendulum swing, conservation of mechanical energy givesThe cylinder generally slips immediately after this impact, which is consistent with imposing rolling without slipping only on the descent.

Impact convention matters. The floor-impulse dependence of a rolling-cylinder collision gives a different answer if a different constraint is imposed. If instead the floor enforces rolling without slipping throughout the impulsive collision, it supplies a nonzero horizontal impulse. The cylinder's effective translational mass is then $m+I/r^2=3m/2$. Writing $J$ for the impulse delivered to the bob gives $(3m/2)(v-u)=-J$, $2mV=J$ and $V-v=u$. Hence $V=6u/7$ and $\cos {\theta }_{\max }=25/49$, giving $59.32^{\circ }$. The original wording specifies the descent constraint but does not specify an impulsive floor-friction law; the first boxed value uses the smooth-impact convention.

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.

In irrotational flow, the velocity has a velocity potential locally, and globally in a simply connected flow domain. incompressibility then gives $\nabla \cdot \mathbf{v}={\nabla }^2\Phi =0$. For a sphere moving at speed $U$ along $z$ through fluid at rest far away, the potential of a translating sphere is a decaying axisymmetric dipole. Its solution and its impermeability condition giveThese are laboratory fluid velocities expressed relative to a moving origin. A velocity field measured in the sphere-rest frame instead requires adding $-Ur\cos \theta$ to the potential.

The vector identity printed in the PDF has its signs reversed. The product rule gives the consistent identity $v^2=\nabla \cdot (\Phi \mathbf{v})-\Phi \nabla \cdot \mathbf{v}$. With outward normal of the exterior fluid equal to $-\hat{r}$ at the sphere, the divergence theorem yieldsThe integral at infinity vanishes. HenceThis added mass of a sphere is half the displaced fluid mass. For a spring of unchanged stiffness, buoyancy only shifts the static equilibrium. Its oscillation frequency with fluid added mass isFor a small-angle simple pendulum, buoyancy also reduces restoring torque from $mg\ell \theta$ to $(m-{\rho }_{0}V)g\ell \theta$, while the inertia is $(m+m_a){\ell }^2$. ThusThe downward equilibrium is stable only for $\rho >{\rho }_0$. The ideal results assume unbounded fluid, negligible viscosity, small excursions, and no important free-surface or container corrections.

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.

Lagrangian mechanics expresses dynamics through generalized coordinates $q_i$ that respect the constraints. For a system with conservative, velocity-independent forces and standard kinetic energy, the Lagrangian is $L=T-V$. Its equations follow from the principle of stationary action:for variations that vanish at the endpoints. Varying the integral and integrating the $\delta {\dot{q}}_i$ term by parts givesIndependent variations therefore yield the Euler-Lagrange equations:Additional nonconservative generalized forces appear on the right-hand side as $Q_i$. The action is stationary, not necessarily a minimum. For ideal holonomic constraints, using suitable generalized coordinates removes the need to solve explicitly for constraint forces.

The conjugate momentum is $p_i=\partial L/\partial {\dot{q}}_i$. If a coordinate is cyclic, $\partial L/\partial q_i=0$ and its conjugate momentum is conserved. If $L$ has no explicit time dependence, the energy functionis conserved. It equals $T+V$ for the usual quadratic kinetic energy with time-independent geometry, but that identification is not universal for every Lagrangian or constraint.

For example, a simple pendulum with angle $\theta$, length $\ell$ and bob mass $m$ hasThe tension does no virtual work in the allowed angular displacement and never needs to be calculated to obtain the angular equation. The method scales naturally to coupled systems and coordinates with complicated geometry.

For uniform Earth density, the shell theorem says that only mass within radius $r$ contributes to the gravitational field there. Since enclosed mass is proportional to $r^3$,At depth $h$, $g_h/g_0=1-h/R$. A simple pendulum has small-amplitude period $P=2\pi \sqrt{l/g}$, so the pendulum clock below a uniform-density surface runs at the fraction $P_0/P_h=\sqrt{1-h/R}$ of its surface-calibrated rate. After a surface-clock hour, its lag isThe mine clock is slower because gravity is weaker under the stated uniform-density assumption. The expansion $\Delta t\simeq 3600h/(2R)$ gives the same result. This is a change in the mechanical pendulum period, not a relativistic clock effect; a realistic Earth density profile would require a different interior-gravity calculation.

For a simple pendulum in the small-angle approximation, the oscillation period is $T=2\pi \sqrt{L/g}$, soInterpret the quoted random timing error and length error as standard uncertainties, with independent errors and no additional systematic error. For $N$ independent measurements, the standard error of the sample mean of the period is ${\sigma }_{\overline{T}}/T=0.02/\sqrt{N}$. The length has relative standard uncertainty ${\sigma }_L/L=0.1/25.0=0.004$. By propagation of uncertainty,Requiring a relative standard uncertainty no greater than $0.01$ givesFor $N=20$, the combined relative standard uncertainty is $0.009798$, or $0.980\%$; $N=19$ gives $1.001\%$ and just fails. Repeating the timing does not reduce the fixed length measurement uncertainty. The stated “accuracy” is consequently a statistical uncertainty target, not a guarantee for every experimental outcome. This is an example of repeated-timing uncertainty in a pendulum gravity measurement.

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.

Inside a spherical body of uniform density, the shell theorem gives gravitational acceleration $g(r)=g(R)r/R$. A simple pendulum at depth $h$ therefore has period $P_h=P_0/\sqrt{1-h/R}$. If its clock is calibrated at the surface, its indicated elapsed time is $t\sqrt{1-h/R}$ and it loses $t[1-\sqrt{1-h/R}]\simeq th/(2R)$. This is a mechanical clock-rate change caused by weaker local gravity, not gravitational time dilation. The uniform-density assumption is essential; Earth's actual density profile gives a different interior field.

For the nondimensional simple pendulum $\ddot{x}=-\sin x$, the separatrix has energy ${\dot{x}}^2/2+1-\cos x=2$. Starting at $x=0$ with velocity $2$, its time to angle $a<\pi$ is $\log (\sec (a/2)+\tan (a/2))$. The saddle at angle $\pi$ is approached only as time tends to infinity.

// Target: classical-mechanics.bigb

For a bob of mass $m$ on a massless rod of length $\ell$, let a massless support slide horizontally against a spring of stiffness $k$. With support displacement $s=\ell \eta$ and angle $\theta$ from downward vertical, the quadratic Lagrangian, divided by $m{\ell }^2$, is $(\dot{\eta }+\dot{\theta }{)}^2/2-(k/m){\eta }^2/2-(g/\ell ){\theta }^2/2$. Subtracting its two Euler-Lagrange equations gives $\eta =mg\theta /(k\ell )$. There is one finite normal mode, with the displayed frequency. A stiff spring fixes the support and recovers the simple pendulum frequency. The missing second finite frequency reflects a singular kinetic matrix in small oscillations, not an omitted oscillation branch.

A uniform rigid rod of length $l$ and mass $M_r$, pivoted at one end with a point mass $m$ at the other, has pivot moment of inertia $I=l^2(m+M_r/3)$. Its gravitational potential energy is $-gl(m+M_r/2)\cos \theta$. The planar Euler-Lagrange equation is $I\ddot{\theta }+gl(m+M_r/2)\sin \theta =0$, giving the displayed small-amplitude oscillation period. The limit $M_r=0$ gives a simple pendulum, whereas $m=0$ gives the uniform-rod physical pendulum. Thinness does not specify a rod's mass, so both contributions must be stated or explicitly neglected.