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Assume a slender homogeneous elastic beam obeying linear elasticity, with plane sections remaining plane and no resultant longitudinal force. A length element of its neutral axis subtending a small angle $d\varphi$ has length $Rd\varphi$. A neighbouring fibre at distance $z$ from that axis has a length change of magnitude $zd\varphi$, giving longitudinal strain $\epsilon =z/R$ with signs chosen consistently for tensile and compressed fibres. By Hooke's law, its longitudinal stress is $\sigma =E\epsilon =Ez/R$.

The torque of these longitudinal forces about the neutral axis is thereforeHere $E$ is Young's modulus and $I$ the second moment of area. The displayed relation uses a signed radius and bending moment; an unsigned geometrical radius is $|R|=EI/|B|$. For consistency with the negative bending moments in the later parts, take the deflection $y$ positive downward and the signed curvature as $1/R\simeq y^{\prime \prime }$. Then the small-slope Euler-Bernoulli beam equation reads $EI{y}^{\prime \prime }=B$. Reversing the vertical coordinate reverses the corresponding sign relation but leaves the physical deflection unchanged.

Interpret the applied pressure as a uniform increase of external hydrostatic pressure on all six faces, and approximate the rock by an isotropic material obeying linear elasticity. Taking tensile stress as positive, ${\sigma }_x={\sigma }_y={\sigma }_z=-p$. By Hooke's law, the strain along any one axis isTo first order in the small strain, the fractional volume change is the sum of the three axial strains:Thus the bulk modulus is $K=40\mathrm{GPa}$. Using the specified Young's modulus, Poisson's ratio, $V=1{\mathrm{m}}^3$ and $p=101325\mathrm{Pa}$ givesThe sign indicates contraction. This is the change relative to the unloaded reference state; simply maintaining a specimen already equilibrated at this same pressure would produce no further change.

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.

Equilibrium and undamped motion. Resolve forces along the incline, taking positive displacement down the slope. At equilibrium, Hooke's law balances the downhill component of weight:If $x$ is measured from that equilibrium position, Newton's second law becomesThus the motion is a harmonic oscillator withGravity shifts the equilibrium extension but drops out of this oscillation frequency because the downhill force component is constant.

Viscous damping. For a damped harmonic oscillator, put $\beta =b/(2m)$. Substitution of $z=A{e}^{-pt}$ in the supplied linear ordinary differential equation givesIn the underdamped regime $\beta <{\omega }_0$, define ${\omega }_d=\sqrt{{\omega }_0^2-{\beta }^2}$. The two characteristic roots are $p=\beta \pm i{\omega }_d$, and the general real displacement isHere $C,D$, or equivalently $A_{\mathrm{env}}$ and $\varphi$, are fixed by the initial conditions. For release from rest at $x_0$, the condition $\dot{x}(0)=0$ requires $D=\beta x_0/{\omega }_d$, not $D=0$:The curve oscillates about zero between exponentially shrinking envelopes $\pm A_{\mathrm{env}}{e}^{-\beta t}$, where $A_{\mathrm{env}}=|x_0|\sqrt{1+(\beta /{\omega }_d{)}^2}$ for this release. The upper panel illustrates that shape; its damping strength is chosen to make the envelope visible.

Frequency change. Viscous damping reduces the angular frequency and lengthens the oscillation period:For light damping, ${\omega }_d\simeq {\omega }_0[1-b^2/(8mk)]$, so the frequency shift is second order in $b$, even though the envelope decay rate is first order in $b$.

Will the block reach the stop? The given parameters giveFor $x_0=-0.260\mathrm{m}$, differentiating the release-from-rest solution givesIt is positive throughout the first half-cycle. The first positive turning point therefore occurs at $t_{+}=\pi /{\omega }_d=0.993460\mathrm{s}$. The turning points of an underdamped release from rest give its unobstructed height:Since $x_{+}>0.250\mathrm{m}$ and the displacement increases continuously up to that point,Solving $x(t)=0.250\mathrm{m}$ gives first contact at approximately $0.91117\mathrm{s}$. The plotted dashed continuation is the hypothetical free motion after that time; the post-impact motion is not determined by the supplied damping model.

The original diagram shows two equal balls, each of mass $m$, attached to the fixed upper pivot and light sliding collar by four equal rods. Each ball describes a horizontal circle of radius $r=L\sin \theta$. The pivot-to-collar distance is $2L\cos \theta$, so the upward collar displacement compresses the spring byThus Hooke's law gives a downward spring force $F_s=k\delta$ on the collar. Let $T_u,T_l$ be the tensions in the upper and lower rods attached to one ball. Vertical force balance on the light collar and then on either ball givesThe radial force on one ball supplies its centripetal acceleration:For $\theta \ne 0$, adding the vertical-balance results and eliminating the rod tensions givesFor the given centrifugal governor, $\cos 30^{\circ }=\sqrt{3}/2$ andHenceThe factor of two in the collar balance matters: both lower rods support the same light collar. The spring force is not an extra downward force of this full magnitude on each ball; its effect on each ball is transmitted through its own lower rod. The formula is the steady spring-loaded centrifugal governor balance.

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.

Let $X_1,X_2$ be longitudinal displacements. Hooke's law givesThe symmetric and antisymmetric eigenvectors are respectively $(1,1)$ and $(1,-1)$, soOrdinary frequencies are these values divided by $2\pi$.

For symmetric quadratic coefficient matrices, the Euler-Lagrange equations give $T\ddot{q}+Vq=0$, where $T$ is the positive-definite mass matrix. Substitute a normal-mode oscillation $q=v\cos (\omega t+\varphi )$ to getThe positive roots ${\omega }^2$ determine the squared normal frequencies and their nonzero null vectors give mode shapes. Repeated eigenvalues give multidimensional mode spaces.

The particle's kinetic energy is $m|\dot{q}{|}^2/2$. For a taut elastic string at length $r=\sqrt{x^2+y^2+z^2}$, Hooke's law gives tension $\lambda (r-a)/a$ and elastic energy $\lambda (r-a{)}^2/(2a)$. With $\lambda =mg$ and gravitational potential $mgz$, this gives exactlyThis is the taut-string regime containing the equilibrium. Differentiating the potential givesAt $(0,0,-2a)$, $r=2a$ and all three accelerations vanish.

With $z=-2a+\zeta$, expansion to quadratic order gives $r=2a-\zeta +(x^2+y^2)/(4a)+O(3)$. Substitution cancels the linear terms, leavingThe mass matrix remains $mI$, and the stiffness matrix is $\mathrm{diag}(mg/(2a),mg/(2a),mg/a)$. ThereforeThe horizontal frequency is doubly degenerate: a linearly polarized oscillation in any horizontal direction is a normal mode. The remaining mode is vertical stretching and contraction of the string, with horizontal coordinates zero.

Let the nearest-neighbour spacing be $d$, so the primitive unit cell has length $\ell =2d$. Denote the two longitudinal displacements in cell $n$ by $u_n,v_n$, with spring $K$ within each cell and spring $K^{\prime }$ between cells. Hooke's law and Newton's second law giveFor normal modes $u_n=u{e}^{i(qn\ell -\omega t)}$ and $v_n=v{e}^{i(qn\ell -\omega t)}$, the two amplitudes satisfyIts vanishing determinant gives the alternating-spring chain dispersion relationThe minus sign is the acoustic phonon branch, with ${\omega }_{-}(0)=0$ and $u=v$ at the origin. The plus sign is the optical phonon branch, with $u=-v$ at the origin. For positive springs the acoustic phonon branch rises from zero to $\sqrt{2\min (K,K^{\prime })/m}$ at the Brillouin zone edge; the optical phonon branch falls from $\sqrt{2(K+K^{\prime })/m}$ to $\sqrt{2\max (K,K^{\prime })/m}$. Their long-wavelength limits areThus the leading frequency in the strong-bond limit of the alternating-spring chain is $\sqrt{2K/m}$, the internal stretching frequency of a pair with reduced mass $m/2$.

The dynamic modulus, or complex viscoelastic modulus, is the ratio of complex stress amplitude to complex strain amplitude in the periodic steady response:This defines the sign convention; for passive response the loss modulus is nonnegative at positive frequency. A Newtonian fluid obeys $\sigma =\eta \dot{ϵ}$, while a Hooke's law solid obeys $\sigma =G_0ϵ$. HenceThe first is purely dissipative and the second purely elastic. Here $ϵ$ denotes the scalar deformation used in the problem, with any shear-versus-extension geometric factors already included in its modulus and viscosity.

Write the bead position in Cartesian coordinates as ${\mathbf{x}}_B=(a\cos {\theta }_B,a\sin {\theta }_B,b{\theta }_B)$. Differentiation gives $|{\dot{\mathbf{x}}}_B{|}^2=(a^2+b^2){\dot{\theta }}_B^2$, while the free particle has $|\dot{\mathbf{x}}{|}^2={\dot{r}}^2+r^2{\dot{\theta }}^2+{\dot{z}}^2$. Thus the kinetic energy is the sum of these squared speeds weighted by half the respective mass. The squared spring length isA spring obeying Hooke's law with zero natural length has potential energy $k{\ell }^2/2$. Consequently $L=T-k{\ell }^2/2$, and deleting the additive constant $-k{a}^2/2$ givesThe spring length is the straight separation of its endpoints, not distance measured along the helix. The fixed frictionless constraint is already incorporated by using the generalized coordinate ${\theta }_B$; it supplies no extra potential energy.

Let $\theta$ and $\phi$ be the lower and upper azimuthal angles, measured from a common fixed direction. The squared spring length isBy Hooke's law, a spring with zero natural length has potential energy $k{d}^2/2$. ThereforeConstant gravitational energies can be omitted. It is useful to define the moments of inertia $I_1=m{a}^2$, $I_2=M{A}^2$ and the angular coupling $\kappa =kaA$. The part of the potential that affects motion is simply $-\kappa \cos (\theta -\phi )$. In particular, $h$ changes the constant spring energy but not the angular equations, because the spring has zero natural length.

In small-strain elastoplasticity, total infinitesimal strain tensor splits into an elastic part and a generally irreversible plastic part. The elastic constitutive equation determines stress from $e^{\mathrm{e}}$, while a yield condition and a plastic flow law determine the evolution of $e^{\mathrm{p}}$. Removing external load need not remove plastic strain, so using Hooke's law on total strain after yielding is insufficient.

Potential energy assigns energy to a system's arrangement, so a change of arrangement accounts for the work done by a conservative force. For a time-independent potential energy $V(\mathbf{r})$,The gradient points toward increasing potential energy; the force points toward decreasing potential energy. Only differences matter: adding a constant to $V$ changes neither the force nor its work. This endpoint rule follows by substituting $-\nabla V$ into the line integral for work. ^[ref]

Near Earth's surface, stipulate constant gravitational acceleration $g=9.81m/{\mathrm{s}}^2$ and measure height $h$ upward from a chosen zero. Then gravitational potential energy is $V=mgh$. Raising a $2\mathrm{kg}$ mass by $0.5\mathrm{m}$ increases its gravitational potential energy by $9.81\mathrm{J}$, while gravity does $-9.81\mathrm{J}$ of work. These are computed model values, not measurements of a particular lift. An ideal spring obeying Hooke's law instead has $V=k{x}^2/2$, with displacement $x$ from its relaxed state and spring constant $k$: $k=100N/m$ and $x=0.2\mathrm{m}$ give $V=2\mathrm{J}$ relative to $x=0$. ^[ref]

In a homogeneous isotropic solid without body force, a radial displacement field satisfies $(\lambda +2\mu )(u^{\prime }+2u/r{)}^{\prime }=0$. For nonzero $\lambda +2\mu$, its volumetric strain is constant. Since Hooke's law gives $\mathrm{tr}\tau =(3\lambda +2\mu )(u^{\prime }+2u/r)$, its stress tensor has constant trace. After plasticity in continuum mechanics has occurred, this conclusion applies to purely elastic increments during elastic unloading, and need not apply to the total residual stress.

An axisymmetric small radial displacement $R(r)$ gives radial strain $R^{\prime }$ and hoop strain $R/r$. Radial force balance on an annular wedge gives ${\tau }_r^{\prime }+({\tau }_r-{\tau }_{\varphi })/r=0$. Under plane stress, isotropic Hooke's law gives ${\tau }_r=E(R^{\prime }+\nu R/r)/(1-{\nu }^2)$ and ${\tau }_{\varphi }=E(R/r+\nu R^{\prime })/(1-{\nu }^2)$. Substitution into equilibrium gives the displayed displacement equation. Its power-law exponents are $+1$ and $-1$, so $R=c_{1}r+c_2/r$. The hoop-stress solution is not based on treating a thick wall as a membrane.

A point mass constrained to a straight guide through the axis of a uniformly rotating reference frame, with signed central displacement $x$ and Hooke's law force $-kx$, satisfies $m\ddot{x}=-kx+m{\Omega }^{2}x$. The transverse Coriolis force is balanced by the guide. The central equilibrium is stable for ${\Omega }^2<k/m$, neutral at equality and unstable above it. The angular frequency is the positive square root in the title.

Simple harmonic motion is motion in which the acceleration is proportional to the displacement from equilibrium and points back towards it:Here $x$ is measured from equilibrium, and $\omega$ is the constant angular frequency. For a point mass attached to an ideal spring, Hooke's law gives the restoring force $F=-kx$, with stiffness $k>0$, and Newton's second law yields $\omega =\sqrt{k/m}$. The minus sign matters: for $x>0$ the force is negative, and for $x<0$ it is positive. A restoring force need not oppose velocity; on the return towards equilibrium it increases speed.

The general solution isThe amplitude $A\ge 0$ and oscillation phase $\phi$ encode the initial conditions. We use $P$ for the oscillation period to distinguish it from kinetic energy $T$. For $A>0$, the oscillation period is independent of amplitude, the property of isochronous harmonic motion. At $A=0$ the body remains at equilibrium. ^[ref]

Choose $m=0.5\mathrm{kg}$, $k=2N/m$, and release the mass from rest at displacement $A=0.3\mathrm{m}$. Then $\omega =2{\mathrm{s}}^{-1}$, $P=\pi \mathrm{s}$, and $\phi =0$. Writing time $t$ in seconds,At $t=0$ the velocity vanishes, but the acceleration is $-1.2m/{\mathrm{s}}^2$: a turning point is not a state of rest that persists. At $t=\pi /4\mathrm{s}$, the body crosses equilibrium with velocity $-0.6m/s$ and zero acceleration. At $t=\pi /2\mathrm{s}$ it reaches the opposite turning point, $x=-0.3\mathrm{m}$.

Taking potential energy zero at equilibrium, the same solution givesAt each turning point all mechanical energy is potential energy; at each equilibrium crossing it is kinetic energy. Both repeat twice during one complete position cycle because squaring removes the sign. At $t=\pi /8\mathrm{s}$, for example, each contributes $0.045\mathrm{J}$.

The physical position trace covers a line segment, $-0.3\le x\le 0.3$ in metres. In the phase plane, the paired position and velocity instead trace the ellipseThis is precisely $m{v}^2/2+k{x}^2/2=E$, rewritten: the single conserved mechanical energy fixes the amplitude and the whole ellipse. It does not fix the starting oscillation phase. At $x=0$ both $v=-0.6m/s$ and $v=+0.6m/s$ lie on it; position alone cannot distinguish them. Replacing $v$ by momentum $p=mv$ gives the equivalent phase-space trajectory. The arrows show increasing time, clockwise in $(x,v)$ coordinates.

This closed ellipse assumes a time-independent spring and no damping or applied drive. Adding a resisting force $-bv$, $b>0$, gives $m\ddot{x}+b\dot{x}+kx=0$ and $dE/dt=-b{v}^2\le 0$. An underdamped damped harmonic oscillator spirals inward in the phase plane, crossing successively smaller energy contours. The ideal harmonic oscillator is also useful beyond springs: near a smooth potential energy minimum with $V^{\prime \prime }(0)>0$, its Taylor expansion begins with $V(0)+V^{\prime \prime }(0)x^2/2$. For sufficiently small displacements, this gives the same restoring force to leading order, explaining why small oscillations share one solvable pattern.

For a uniform-load bending of a clamped filament, total load $F=wL$ and tip displacement $\delta =w{L}^4/(8B)$ obey Hooke's law in the form $F=k_{\mathrm{eff}}\delta$, with $k_{\mathrm{eff}}=8B/L^3$. This stiffness depends on the specified distributed loading. A concentrated tip force instead gives $3B/L^3$, so equal total forces with different distributions are not interchangeable.