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Let the uniform string have constant tension $\tau >0$ and linear mass density $\mu >0$. Assume small transverse motion with $|u_x|\ll 1$, negligible longitudinal motion, and no external transverse force. If the string makes angle $\beta$ with the horizontal, then $\tan \beta =u_x$, and the small-angle approximation gives $\sin \beta \simeq u_x$ and $ds\simeq dx$ to first order.

The net upward force on the segment from $x$ to $x+\Delta x$ is thereforeBy Newton's second law, its mass $\mu \Delta x$ times its transverse acceleration equals this force, to the order retained in the small displacement model. Divide by $\Delta x$ and let $\Delta x\to 0$ to obtainThus the wave equation on a string iswhere $c$ is the wave speed. The horizontal force balance makes the horizontal component of tension constant; to first order it equals $\tau$.

The small-angle approximation $\sin \theta =\theta +O({\theta }^3)$ reduces ii to $\ddot{\theta }+{\omega }^2\theta =0$, whereThe coefficient is positive, so the solution is simple harmonic motion, $\theta (t)=C\cos \omega t+D\sin \omega t$. Here $\omega$ is the angular frequency; the ordinary frequency and period are

For a fixed attachment ratio $\delta >0$, making the disc much heavier than the rod givesIf the disc is also small relative to the attachment distance, $\alpha /\delta =a/d\to 0$, its moment of inertia about its own center is negligible compared with $M{d}^2$. ThusThis is a simple pendulum with a point mass at the disc's center and effective length $d$; it is length $\ell$ only if $d=\ell$. Even before the small-angle approximation, its equation tends to $\ddot{\theta }=-(g/d)\sin \theta$.

If $\delta$ is allowed to tend to zero together with the other parameters, the bare conditions $\alpha \ll 1$ and $\mu \gg 1$ do not alone justify that reduction. The needed dominance conditions are $\mu \delta \gg 1$, $\mu {\delta }^2\gg 1$, and $\alpha /\delta \ll 1$, so that the rod contributes negligible gravitational torque and moment of inertia, and the disc behaves as a point mass. They follow automatically in the combined limit with fixed $\delta >0$.

The magnitude of the gravitational force remains approximately $mg$ during a laboratory swing. Its tangential component changes with position, however. For simple pendulum length $\ell$ and angle $\theta$ measured from the downward vertical,This restoring component is zero at the lowest point and largest in magnitude at the turning points. It always accelerates the bob towards the lowest point. The radial gravitational component is $mg\cos \theta$; the string tension $S$ must additionally supply the centripetal acceleration, so $S-mg\cos \theta =m{v}^2/\ell$. Thus it is the radial resultant, not gravity alone, that bends the trajectory. The bob has maximum speed at the bottom even though its tangential acceleration is then zero.

For a small angular amplitude, the small-angle approximation gives simple harmonic motion with oscillation period $T=2\pi \sqrt{\ell /g}$. Measure $\ell$ from pivot to the bob's centre of mass, time many complete swings, divide the elapsed time by their number, and calculateRepeat with several lengths and plot $T^2$ against $\ell$; the ideal slope is $4{\pi }^2/g$. The first-order relative uncertainty obeys $\delta g/g\simeq \delta \ell /\ell -2\delta T/T$, with absolute uncertainty contributions combined according to their dependence. Keep the amplitude small: the finite-amplitude period is $T=T_0(1+{\theta }_0^2/16+\cdots )$, with ${\theta }_0$ in radians. Using the longer finite-amplitude period in the small-amplitude formula would underestimate $g$. Pivot friction, a finite bob and support motion are further experimental corrections.^[ref]

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.

Take the positive coordinate from $O$ toward $S$. The centre of mass of the remaining material has coordinateThus $SX=a+a/6=7a/6$. If $\varphi$ is the small angular displacement from the stable hanging orientation, gravity gives torque $-(3m/4)g(7a/6)\sin \varphi$ about the suspension point. The physical pendulum equation isUsing the small-angle approximation $\sin \varphi \simeq \varphi$ gives simple harmonic motion withBoth the reduced mass and displaced centre of mass matter; changing only the moment of inertia would give an incorrect frequency.

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 the small-angle approximation the physical pendulum is a harmonic oscillator with $\omega =2\pi /T$. Its classical energy is $E=m{\omega }^2{A}^2/2=3.55\times 10^{-3}\mathrm{J}$. Equating this to $(n+1/2)\hslash \omega$ givesAn oscillating bob is not a single stationary energy eigenstate. An ideal coherent state has a Poisson distribution of oscillator occupations centred on this mean, with relative spread $1/\sqrt{\bar{n}}$, exceptionally small here. This quantum occupation scale of a classical oscillator fixes the excitation scale without requiring the real, damped clock to be a pure coherent state.

For time dependence $e^{-i\omega t}$, a scalar Fresnel-Kirchhoff aperture formula for a point-source amplitude $A$ isHere $s=|S-Q|$, $r=|P-Q|$, and $K$ is the obliquity factor, approximately the mean of the incident and outgoing direction cosines measured with the forward aperture normal. The factors describe incident spherical spreading, outgoing spreading and accumulated phase. This is the scalar Kirchhoff aperture approximation; it does not supply an exact electromagnetic edge boundary solution.

In the Fresnel regime, expand path lengths to quadratic order in aperture coordinates while replacing slowly varying amplitude and obliquity factors by their axial values. In the Fraunhofer regime the quadratic aperture phase is negligible, leaving the aperture Fourier transform. For characteristic aperture size $D$, the latter requires $D^2/(\lambda R)\ll 1$ in the relevant finite-source geometry; a lens can supply the Fourier-plane configuration.

For the one-dimensional Cornu construction, put $S,O,P$ on the unobstructed straight ray, with the aperture plane perpendicular at $O$. An element is displaced transversely by $x$, with $|x|\ll a,b$, where $a=SO$ and $b=OP$. A long straight edge or slit makes the other transverse direction separable. Then

For knife-edge diffraction at the fence, the edge is $h=0.5\mathrm{m}$ above the direct ray, with $a=5\mathrm{m}$, $b=45\mathrm{m}$, $\lambda \simeq 0.100\mathrm{m}$, and $R=4.5\mathrm{m}$. The unblocked aperture extends upward from $h$, so its Cornu chord runs from $u_h=h\sqrt{2/(\lambda R)}=1.054$ to infinity. Using $C(\infty )=S(\infty )=1/2$ and normalizing by the full unobstructed chord givesThe approximate chord coordinates are $C=0.775$, $S=0.492$. This is the standard single-knife-edge Fresnel normalization.^[ref]

For geometric spreading in a moving-receiver diffraction measurement, set $b=d$ while retaining $a=5$ and $h=0.5$:Normalized transmission decreases toward the fence. Absolute received intensity initially increases, because the free-space spherical-wave intensity rises as source–receiver distance decreases. In this model it reaches a maximum near $d\simeq 2.3\mathrm{m}$, then falls. The formal paraxial small-$d$ behavior is $I\propto d/(5+d{)}^3$; extrapolation to $d\lesssim h$ is outside the small-angle approximation. The sketch distinguishes the absolute signal from its ratio to the unobstructed signal at each position.

At distance $L$, the small-angle approximation gives $\ell =L\theta$. Thus the transverse resolution is approximately $2.5\times 10^{10}\mathrm{m}$, or $3.0\times 10^{10}\mathrm{m}$ with the circular-aperture Rayleigh criterion. This is a transverse separation at the source, rather than a determination of its distance along the line of sight.

For a photon, $E=hc/\lambda$. Using Planck constant $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, speed of light $c=2.99792458\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$ and $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$ givesThe Rayleigh criterion for a circular aperture of diameter $D$ is ${\theta }_R=1.22\lambda /D$, set by the first zero of its Airy disk. Thus the ideal diffraction-limited resolution isThe conversion to transverse separation uses the small-angle approximation. An order-of-magnitude estimate $\lambda /D$ instead gives $2.07\times 10^{-7}\mathrm{rad}$ and $4.13\times 10^{10}\mathrm{m}$; the distinction is the chosen resolution criterion. These are aperture limits, before instrumental aberrations and detector sampling.

Write $C_R=4R/(1-R{)}^2$, keeping this coefficient distinct from the finesse $\mathcal{F}$. The transmission has maxima at $\delta =2\pi m$. At a displacement $ϵ$ from a maximum, its ratio to peak transmission is $[1+C_R{\sin }^2(ϵ/2){]}^{-1}$. Half maximum therefore requiresThis is the half-width in wave phase; the full width at half maximum is $2{\delta }_{1/2}$. Successive resonances are $2\pi$ apart, so the exact ideal-model result isFor narrow peaks, $(1-R)/(2\sqrt{R})\ll 1$, the small-angle approximation to the inverse sine givesThus the formula requested in the paper is the high-reflectivity approximation, not an exact identity at arbitrary $R$. Ordinary half-maximum peaks exist only when $(1-R)/(2\sqrt{R})\le 1$, or $R\ge 3-2\sqrt{2}$; below that threshold the transmission minimum never reaches half the peak. The prefactor $T^2/(1-R{)}^2$ cancels from this relative-width calculation.

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.

Newton's second law takes its familiar form $m\mathbf{a}=\mathbf{F}$ in an inertial frame. Axes fixed to a turning apparatus or to the Earth form a rotating reference frame. Extra inertial forces account for the acceleration of these coordinates. Their values depend on the chosen reference frame; physical interactions such as gravity, tension and contact forces must still be included separately.

Let the rotating origin have inertial acceleration $\mathbf{A}$, and let its axes have angular velocity $\mathbf{\Omega }(t)$. For any vector $\mathbf{q}$, the rotating-frame derivative formula isEach rotating basis vector changes by $\mathbf{\Omega }\times {\mathbf{e}}_i$, which proves this rule by expanding $\mathbf{q}$ in that basis. Apply it twice to the position relative to the moving origin. With ${\mathbf{v}}^{\prime }$ and ${\mathbf{a}}^{\prime }$ measured in the rotating axes,Consequently the equation in the rotating reference frame isThe additional terms are the translational inertial force, Coriolis force, Euler force and centrifugal force, respectively. The last two simplify when the angular velocity is constant: the Euler force vanishes, while the other rotation terms remain.

The centrifugal force points away from the rotation axis. If ${\mathbf{r}}_{\perp }$ is the perpendicular displacement from that axis, thenFor uniform rotation it has potential energyA mass at rest relative to a turntable can therefore be described as having an inward tension or friction balancing its outward centrifugal force. In an inertial frame, the inward physical force instead produces the required centripetal acceleration. These are two descriptions of the same motion; one must not insert the rotating-frame outward term into the inertial equation as an additional physical interaction.

A rotating liquid gives a useful quantitative example. For a liquid in hydrostatic equilibrium relative to a vessel rotating uniformly about the vertical axis,Along the free surface the pressure is constant, so ${\Omega }^{2}sds-gdz=0$ andThe surface is a elliptic paraboloid. This is the balance of gravity and the centrifugal force in the rotating description. On the Earth, the same outward acceleration reduces effective gravity, most strongly at the equator, and contributes to the planet's equatorial bulge.

The Coriolis force depends on relative motion:It is perpendicular to both $\mathbf{\Omega }$ and ${\mathbf{v}}^{\prime }$, vanishes for motion along the rotation axis, and reverses when the relative velocity reverses. The dot product givesThus the Coriolis force does no work in the rotating reference frame: it changes the direction of motion rather than directly changing the rotating-frame kinetic energy. For example, on a horizontal turntable rotating anticlockwise as viewed from above, an outward-moving slider has a clockwise Coriolis acceleration, whereas an inward-moving slider has an anticlockwise one. The deflection can be observed by comparing a laboratory camera with a camera rotating with the table.

For slow horizontal motion on the Earth at latitude $\lambda$, keep the vertical component ${\Omega }_E\sin \lambda$ of its angular velocity. In east-north coordinates $(u,v)$ the horizontal Coriolis acceleration isIt deflects motion to the right in the northern hemisphere and to the left in the southern hemisphere. In large-scale atmospheric and oceanic motion it can balance a horizontal pressure gradient, giving geostrophic balance. With $\hat{\mathbf{z}}$ upward, the steady horizontal equation without friction isThe resulting velocity runs along lines of equal pressure. This approximation requires rotation to dominate local inertial changes; it is inappropriate close to the equator where $f$ tends to zero.

The Foucault pendulum gives another terrestrial example. In the small-angle approximation, its horizontal coordinates satisfyTo first order in the small ratio ${\Omega }_E/{\omega }_0$, the oscillation plane precesses relative to the ground at angular rate $-{\Omega }_E\sin \lambda$: clockwise in the northern hemisphere, zero at the equator, and one revolution per sidereal day at a pole. The Coriolis force explains this changing orientation without requiring energy input. Together these examples show why inertial forces are useful whenever the apparatus or environment supplies naturally rotating coordinates.

The phase step between adjacent slits isFor equal amplitudes $a_0$, the six unmodified contributions form a closed regular hexagon in the complex plane:Reversing one slit at an array minimum replaces slit $j$'s contribution by its negative, so $S=S_0-2a_0{e}^{ij\pi /3}$. ConsequentlyThis is independent of which slit is reversed. The result uses the far-field, narrow-slit and small-angle approximations; retaining $\sin \theta$ changes the phase step only by about $7\times 10^{-7}$ radians.

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.

Use the original diagram's positive distances: $u$ from the object to the vertex, $d$ from the vertex to the image, and $R$ from the vertex toward the centre of curvature on the outgoing side. Let a paraxial light ray strike at height $Y$. Its incoming inclination is $\alpha \simeq Y/u$ above the optical axis, its outgoing inclination is $\beta \simeq Y/d$ below it, and the inward surface normal is inclined by $\theta \simeq Y/R$ below it. Thus the incidence and refraction angles relative to that normal are respectively $\alpha +\theta$ and $\theta -\beta$.

The local tangent-plane form of Snell's law, using the small-angle approximation, givesSubstitute the three height-to-distance ratios and cancel $Y$:This refraction at a spherical surface result keeps only first-order ray angles. The surface sag shifts the height's axial reference only at higher order. Distances for other object, image or curvature orientations must be signed consistently; the displayed positive-distance form follows the particular orientation in the paper.

Interpret the stated detector width $d=20\mu \mathrm{m}$ as one square spatial sample. At the point directly below the spacecraft, the small-angle approximation gives pixel angular width $d/f\simeq \Delta x/h$. ThusThere is a definition error in the source: the usual focal ratio, or f-number, divides focal length by the entrance-aperture diameter $D$, whereas the printed parenthesis divides by detector width $d$. Reporting both avoids silently changing the question:The actual optical design is therefore approximately f/1.36, a fast optical telescope, not an optically slow f/34000 instrument. Such a focal ratio is demanding but possible in principle: optical aberrations, obscuration, field coverage, stray light and alignment need careful design. The single-sample angular width is $2.94\times 10^{-5}\mathrm{rad}$. Spatial sampling alone does not guarantee true resolving power; diffraction, pointing and signal-to-noise ratio also matter. At large viewing angles, the same angular pixel covers a larger, distorted surface footprint. If the detector is literally one element, scanning is additionally needed to form a map.

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.