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A static lossless taper preserves frequency. If characteristic impedance varies slowly on the local wavelength scale, reflection is small and mean power $Z{\omega }^2{A}^2/2$ is conserved. At fixed tension this gives the displayed amplitude law, while wave speed and wavelength scale as ${\rho }^{-1/2}$. A long section containing an abrupt step is not automatically adiabatic.

For a plate under uniform tension per length $T$, with surface mass density $\sigma$ and flexural rigidity $D_b>0$, the small-displacement equation is $\sigma z_{tt}=T{\nabla }^{2}z-D_b{\nabla }^{4}z$. If the edge conditions admit a transverse sine mode of wavenumber $q_n$, its longitudinal dispersion relation isBending raises mode cutoff frequencies of a membrane waveguide and changes large-$k$ behaviour from $\omega \propto |k|$ to $\omega \propto k^2$. A genuinely clamped plate also fixes the edge slope, so its transverse mode shapes need not remain simple sine functions. The qualitative increase in restoring force survives that change of boundary conditions.

A flat, unsupported film's bending and tension energies give modal stiffness $\kappa q^4+\sigma q^2$. Negative tension represents compression, destabilizing sufficiently long waves. A finite sheet restricts allowed wave numbers, while a supporting foundation adds stiffness and can select a wrinkle wavelength. This is generic continuum mechanics, independent of biological membrane structure.

A catenary is the static shape of an ideal flexible, inextensible chain hanging under its own gravitational weight in a uniform gravitational field, with uniform linear mass density along its arc length. With the lowest point above $x=0$, one convenient vertical origin givesHere $a$ has units of length. The supports and the chain's arc length determine the scale and position; horizontal and vertical shifts give the general catenary. ^[ref]

For $a=1$, the three points at $x=-1,0,1$ have heights $\cosh 1\approx 1.543081$, $1$, and $\cosh 1\approx 1.543081$. If heights should instead be measured from the lowest point, subtract $a$:The same three heights are then approximately $0.543081$, $0$, and $0.543081$. The shift changes the vertical origin and preserves the catenary shape.

The scale also measures tension: if $H>0$ is its constant horizontal component, $\mu$ is the linear mass density, and $g$ is the magnitude of gravitational acceleration, then $a=H/(\mu g)$. A short force balance explains this. For the portion from the lowest point to $x\ge 0$, with arc length $s(x)$, vertical tension balances its gravitational weight and horizontal tension givesApplying differentiation yields $a{y}^{\prime \prime }=\sqrt{1+y^{\prime 2}}$. The proposed hyperbolic cosine solves this differential equation, because $y^{\prime }=\sinh (x/a)$ and $a{y}^{\prime \prime }=\cosh (x/a)=\sqrt{1+{\sinh }^2(x/a)}$. This is the mechanism behind the shape, under the stated ideal chain assumptions.

The same identity simplifies the arc length between symmetric supports at $x=\pm b$:For $a=b=1$, $L\approx 2.350402$, exceeding the horizontal span of two. A larger $a$ at fixed $b$ gives a shallower sag: the Taylor series gives $a(\cosh (x/a)-1)=x^2/(2a)+x^4/(24a^3)+\cdots$. Thus a parabola is a local approximation, while the exact ideal chain shape is a catenary. The useful connection is that the same hyperbolic Pythagorean identity controls its coordinates, force balance, and arc length.

A small arc of a rotating thin hoop needs inward force per length $\rho A{\omega }^{2}r$. Its circumferential tension supplies this through curvature, giving tension $\rho A{\omega }^2{r}^2$ and tensile stress $\rho {\omega }^2{r}^2$. Hooke's law converts that stress to circumferential strain, which equals fractional radius change. Small strain, a slender cross-section and uniaxial linear elasticity are required; the cross-sectional area cancels at leading order.

A centripetal force is the inward resultant force needed for circular motion. Newton's second law and centripetal acceleration give magnitude $m{v}^2/r$ for a particle of mass $m$, speed $v$, and path radius $r$. It is a role played by an actual interaction, such as tension or an electric force, rather than an additional interaction to add to a free-body diagram. A purely radial centripetal force is perpendicular to the velocity, so it does no work and leaves the speed unchanged.

For the lossless small-slope wave equation on a string, tension $T$ and linear mass density $\mu$ give wave speed $c=\sqrt{T/\mu }$. With transverse force $F_y=-T{y}_x$ exerted across a cut toward increasing $x$, a rightgoing displacement $y=f(x-ct)$ has velocity $v_y=-c{y}_x$. Its positive mechanical impedance is therefore $Z=F_y/v_y=T/c=\sqrt{T\mu }$. A leftgoing component has $F_y/v_y=-Z$ under the same fixed orientation. A rightgoing time-harmonic wave with peak displacement wave amplitude $|A|$ carries mean power $Z{\omega }^2|A{|}^2/2$.

A conical pendulum is a pendulum whose bob executes uniform horizontal circular motion beneath a fixed pivot, with its string at a constant angle $\theta$ to the downward vertical. The string sweeps a right circular conical surface; the bob follows its circular rim, without any physical cone supporting it. The ideal model has a point mass $m>0$, a taut, massless, inextensible string of length $\ell >0$, uniform gravitational acceleration of magnitude $g>0$, and negligible drag force and pivot friction. A nonzero circular motion requires $0<\theta <\pi /2$. This geometry and the corresponding forces are described in Richard Fitzpatrick, The conical pendulum, equations 262–267.

The right triangle formed by the pivot, the orbit centre, and the bob gives $r=\ell \sin \theta$ and $h=\ell \cos \theta$. Here $r$ is the orbit radius and $h$ is the vertical drop from the pivot to the bob's horizontal plane, not the bob's height above a table. Write $T$ for tension, $P$ for the oscillation period of one revolution, and $\omega =2\pi /P$ for the angular speed; then the bob's speed is $v=\omega r$. A side view compresses the horizontal circle to a line segment; the top view reveals the circle and its tangent velocity.

The two actual forces on the bob are tension toward the pivot and gravitational weight vertically downward. The vertical acceleration is zero, while the horizontal centripetal acceleration is $v^2/r={\omega }^{2}r$. Thus Newton's second law givesThe inward net force is the horizontal component of tension, rather than an extra interaction to add to these two forces. Constant speed still requires acceleration because the velocity changes direction. ^[ref]

Dividing the two component equations and using $r=\ell \sin \theta$ gives the useful collection of predictionsIf $\varphi$ is the azimuthal angle, then $\dot{\varphi }=\pm \omega$; either direction of rotation satisfies ${\dot{\varphi }}^2\cos \theta =g/\ell$. The mass cancels from the angular speed and oscillation period, but not from the tension. The motion is a relative equilibrium: in a rotating frame following the bob at angular speed $\omega$, its position is fixed. This is a concrete way to see how rotational symmetry can turn an apparently moving solution into a stationary one.

For a spool of outer radius $R$, axle radius $r$, mass $M$ and axial moment of inertia $I$, a lower-tangent pull of tension $T$ at upward angle $\theta$ produces horizontal acceleration $a=TR(R\cos \theta -r)/(I+M{R}^2)$ while rolling without slipping. This follows by combining $Ma=T\cos \theta +f$, $I\alpha =Tr+Rf$ and $a=-R\alpha$, with rightward translation and counterclockwise rotation positive. The acceleration reverses at ${\theta }_c$; zero acceleration means equilibrium only if initially at rest. At that angle the static friction limit gives $T_{\max }=\mu Mg/(\cos {\theta }_c+\mu \sin {\theta }_c)$.

For transverse waves on two strings at common tension, displacement continuity gives $1+r=t$, while transverse-force continuity gives $(1-r)/c_1=t/c_2$. Solving gives the displayed displacement-amplitude coefficients. A denser transmitted string has lower speed and a negative reflected coefficient. The energy fractions are $r^2$ and $(c_1/c_2)t^2$, summing to1. These are displacement coefficients; force or power-normalized wave amplitudes use different conventions.

Choose Cartesian coordinates with origin at the upper pivot, $x$ to the right and $y$ upward. The two position vectors areThese expressions satisfy the rod holonomic constraints $|{\mathbf{r}}_1|=\ell$ and $|{\mathbf{r}}_2-{\mathbf{r}}_1|=\ell$ automatically. This is the practical advantage of using the two angles as coordinates: the rod tensions need not be unknowns in the motion calculation.^[ref]

Differentiate these position vectors to obtain the velocities. The dot product between their two rod contributions uses $\cos ({\theta }_1-{\theta }_2)=\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2$, givingAdding the two kinetic energies and measuring gravitational potential energy from the downward equilibrium givesThe coefficient $2$ has a geometric meaning: the upper rod raises both masses, and its motion contributes to both kinetic energies. The coupling in $T$ is present because the lower mass has a moving pivot. The zero chosen for $V$ changes $L$ by a constant and leaves the Euler-Lagrange equations unchanged.^[ref]

The principle of stationary action applies to $S=\int Ldt$. Its fixed-endpoint first variation, after integration by parts, isIndependent allowed variations therefore give one Euler-Lagrange equation for each angle. Stationary action alone does not assert a minimum.^[ref]

The University of California, Santa Barbara's cord demonstration uses a function generator and mechanical driver loosely attached near a clamped cord end. The other end passes over a pulley to a mass hanger, setting tension. Its photograph shows operation with a strobe; the page reports measured and calculated lowest-mode frequencies. Varying drive frequency at fixed loading selects higher normal modes. This clamp arrangement differs from a string tied directly to a moving driver.^[ref]

A controlled frequency scan holds the cord, endpoint separation and hanging load fixed, then varies frequency slowly to locate large steady amplitudes. Read stationary nodes of a standing wave and count loops; compare the inferred $\lambda =2L/n$ with $c=f\lambda$. As additional checks, compare small drive settings, repeat the scan in reverse, and change only the hanging load to test the predicted square-root tension dependence. These checks test the ideal model rather than adding measurements to the demonstration.

An elastic cord's stretched linear mass density can differ from its relaxed value; the published calculation uses the relaxed value. A strobe can freeze motion at misleading settings, so identify nodes of a standing wave across several snapshots. A moving attachment need not be an exact node of a standing wave. These limits qualify any comparison with ideal fixed-end eigenfrequencies.^[ref]

A geometrically similar taut membrane of radius $r$, uniform isotropic tension $T$ per unit length and areal mass density $\sigma$ has normal mode frequencies $f=C{r}^{-1}\sqrt{T/\sigma }$. Matching mass, length and time dimensions fixes these exponents; the dimensionless constant depends on the normal mode and boundary shape. The result does not apply unchanged when bending stiffness adds an independent material parameter.

Place two equal masses $M$ on a massless mechanical string with spans $L,2L,L$ and uniform tension $T$. Fix one end and drive the other with displacement $h(t)$. Linear transverse force balance gives ${\ddot{y}}_1+{\omega }_0^2(3y_1-y_2)=0$ and ${\ddot{y}}_2+{\omega }_0^2(3y_2-y_1)=2{\omega }_0^{2}h$. The mass-normalized stiffness matrix has symmetric and antisymmetric normal modes with squared angular frequencies $2{\omega }_0^2$ and $4{\omega }_0^2$. The mode coordinates $(y_1+y_2)/2$ and $(y_1-y_2)/2$ separate the forced equations.

For a small-slope travelling wave $y=A\cos (kx-\omega t+\varphi )$ on a uniform string with linear mass density $\mu$ and tension $T$, the wave equation on a string requires ${\omega }^2=T{k}^2/\mu$. Its elastic energy of a small-slope string and kinetic energy density are equal, giving total energy density $u=\mu A^2{\omega }^2{\sin }^2(kx-\omega t+\varphi )$. Its spatial average is $\mu A^2{\omega }^2/2$, and integrating over a full wavelength $\lambda =2\pi /|k|$ gives the title formula. Omitting $\lambda$ produces an energy density, not an energy; units and a direct one-period integral detect the mistake independently.

An entropic force is the part of a generalized force arising from the dependence of entropy on a coordinate. At fixed temperature, the system's force is $-\partial A/\partial x=-\partial U/\partial x+T\partial S/\partial x$, where $A=U-TS$ is the Helmholtz free energy. If the internal energy is independent of the coordinate, the force is entirely entropic. For an ideal chain whose configurations become less numerous on extension, it acts toward shorter extension. The external tension holding the chain extended has the opposite sign. Thus the force depends on the statistical number of available configurations, even when the energy of each configuration is unchanged.

For a uniform fixed-end mechanical string of length $l$ and tension $T$, a single sine normal mode of amplitude $A_n$ has conserved energy $E_n=n^2{\pi }^{2}T{A}_n^2/(4l)$. Hence amplitudes proportional to $1/n$ give equal energies, even though their displacement magnitudes differ. In particular $A_2=A_1/2$ makes the first two modal energies equal. Orthogonality removes cross terms in the integrated kinetic energy and potential energy, so the modal energies add.

For the ideal uniform wave equation on a string, the wave speed is $c=\sqrt{\tau /\mu }$, where $\tau$ is constant tension and $\mu$ is linear mass density. Assume small transverse slopes, negligible bending stiffness and no damping. Fix both ends at zero displacement, giving the homogeneous Dirichlet boundary conditions $y(0,t)=y(L,t)=0$. Our standing wave already satisfies the first. For a nonzero amplitude, the second requires $\sin (kL)=0$, henceThe value $n=0$ gives zero displacement in this sine family, not a vibrating normal mode. Each positive $n$ fits $n$ half-wavelengths into the string: it has $n$ antinodes, $n-1$ interior nodes of a standing wave, and two endpoint nodes of a standing wave. For $L=1\mathrm{m}$ and $c=20\mathrm{m}{\mathrm{s}}^{-1}$, $f_n=10n\mathrm{Hz}$; the third normal mode has $f_3=30\mathrm{Hz}$ and wavelength $2/3\mathrm{m}$.

The equally spaced eigenfrequencies follow from this string's uniform wave speed and fixed-end boundary conditions. They are not a universal spectrum for every vibrating object. Changing an endpoint constraint changes the allowed mode shapes; changing tension changes wave speed. At fixed $L$ and $\mu$, quadrupling $\tau$ doubles every $f_n$. The more general construction of arbitrary initial motion belongs to normal mode and wave equation on a string.

Use local east, north, and up unit vectors, with $x$ eastward, $y$ northward, and positive orientation $\psi$ anticlockwise as viewed from above. Let $\phi$ be signed geographical latitude, positive north, and $\Omega >0$ the magnitude of Earth's angular velocity. In the spherical-Earth approximation,Here $g$ is the local effective gravity in a rotating frame, including the steady terrestrial centrifugal acceleration. For small horizontal displacements compared with $\ell$, the horizontal restoring acceleration from tension is $-{\omega }_0^2(x,y)$. Vertical displacement is second order in the swing amplitude, so vertical velocity can be omitted at this order. The horizontal Coriolis acceleration in local Earth coordinates then gives the coupled linear differential equationsThese equations neglect damping, finite-amplitude corrections, support anisotropy, and spatial changes of the centrifugal force of order ${\Omega }^2$; they retain the leading terrestrial Coriolis force. A related derivation uses south/east axes instead.^[ref]

Introduce the complex number $z=x+iy$. Multiplying the second equation by $i$ and adding givesSet $z=e^{-i\alpha t}u$. Direct differentiation cancels the first derivative of $u$ and yieldsThe ${\alpha }^2$ in $\beta$ comes from solving these reduced equations exactly; it is not a complete prediction of terrestrial corrections of order ${\Omega }^2$. Thus the horizontal trajectory is an ellipse in coordinates rotating at angular velocity $-\alpha$ relative to the building. When $|\alpha |\ll {\omega }_0$, each rapid swing samples an almost fixed long axis; its slow precession isThe sidereal day $T_{\mathrm{sid}}=2\pi /\Omega$ is a rotation relative to distant directions. $T_{\mathrm{p}}$ denotes a full $2\pi$ turn of a continuously followed orientation; an unlabelled swing line repeats after $T_{\mathrm{p}}/2$. At the equator there is no finite precession period in this model. The Coriolis parameter is $f=2\alpha$, so the plane's rate is $-f/2$, not $-f$.

The limiting cases check the sign and scale. At the north pole $\alpha =\Omega$: the swing direction turns clockwise once per sidereal day. At the equator $\alpha =0$: both horizontal equations reduce to independent simple harmonic motions and a line launch stays on its original line. At the south pole $\alpha =-\Omega$: the direction turns anticlockwise once per sidereal day. Vanishing horizontal precession at the equator does not mean that every component of Coriolis acceleration vanishes there.

For a stipulated $\Omega =7.292115\times 10^{-5}{\mathrm{s}}^{-1}$, the sidereal day is approximately $23.9345\mathrm{h}$. At $45^{\circ }$ north the predicted rate is $-10.6356^{\circ }/\mathrm{h}$ and $T_{\mathrm{p}}\simeq 33.8485\mathrm{h}$; at $45^{\circ }$ south the rate changes sign. At the illustrative geographical latitude $34^{\circ }$ south, the prediction is $+8.4109^{\circ }/\mathrm{h}$. These calculations are model predictions, not new observations of the UNSW apparatus.

A swing plane fixed among the stars is generally only a polar picture. A north–south swing at the equator turns with Earth in an inertial frame, while having no precession against the building.^[ref]

Let a massless bucket approach a wall under constant tension $T>0$, with remaining distance $x$ and remaining sand mass $m=Mx/L$. Assume zero-relative-velocity mass shedding horizontally. Then $m\ddot{x}=-T$ and, from rest at $x=L$, the speed satisfies $u^2=(2TL/M)\ln (L/x)$. Its kinetic energy is $K=Tx\ln (L/x)$ and its momentum magnitude is $p=x\sqrt{(2MT/L)\ln (L/x)}$. Differentiating gives $x_K=L/e$ and $x_p=L/\sqrt{e}$: the kinetic energy peaks later, closer to the wall, because its dependence on the increasing speed is stronger. Both tend to zero as the remaining mass vanishes, despite divergent ideal speed.