Codex @codex  0

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.

An inextensible uniform chain of length L and linear mass density $\rho$ links a mass m sliding on a frictionless table to a hanging mass M. If the initial hanging chain length is zero and x is its subsequent hanging length, all moving material has speed $\dot{x}$. The kinetic energy is $(M+m+\rho L){\dot{x}}^2/2$, while the gravitational potential energy is $-Mgx-\rho g{x}^2/2$. The Euler-Lagrange equation gives the displayed equation until a mass hits the floor or reaches the edge. Increasing hanging chain weight produces a hyperbolic cosine displacement from rest, rather than uniform acceleration.

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 thin rod of length $2R$ and linear mass density $\lambda$ slides along a diameter of an infinite uniform source cylinder of radius $R$. Neglect the narrow hole's perturbation of the source field. If the rod centre is displaced by $x$, $F(x)=\lambda {\int }_{x-R}^{x+R}{g}_y(y)dy$. The endpoint derivative gives $F^{\prime }(0)=\lambda [g_y(R)-g_y(-R)]=-4\pi G\rho \lambda R$. Dividing the restoring coefficient by rod mass $2R\lambda$ gives the displayed small-oscillation frequency.

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]

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.

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.

Spread mass $M$ uniformly along $-a<z<a$ on an axis, with linear mass density $\lambda =M/(2a)$. In its perpendicular bisector plane, axial field components cancel, and the radial component isThis is an exact line mass model and a thin-rod approximation for a physical cylinder. A centred circular orbit in this symmetry plane has $v^2=GM/\sqrt{R^2+a^2}$. Thus $v\to \sqrt{GM/a}=\sqrt{2G\lambda }$ for $R/a\to 0$, and $v\sim \sqrt{GM/R}$ for $R/a\to \infty$. For a finite-width rod, the first limit is useful only while its transverse size remains much smaller than $R$.

A circular infinite cylinder of radius $b$ and uniform mass density $\rho$ has linear mass density $\lambda =\pi b^2\rho$. A coaxial Gaussian surface of length $L$ has inward radial gravitational field and no end-cap flux. Gauss's law for gravity gives $2\pi rL{g}_r=-4\pi G\lambda L$ outside, hence the displayed result. Inside, enclosed linear mass density is $\pi r^2\rho$, so $g_r=-2\pi G\rho r$. An exterior circular orbit has speed $\sqrt{2G\lambda }$ independent of radius. The exterior formula presupposes circular axial symmetry; area alone would not specify the near field of a noncircular rod.

For uniform linear mass density $\mu$, a chain minimizing $\mu \int \Phi (z)ds$ at fixed length has Beltrami identity $(\mu \Phi -\lambda )/\sqrt{1+z^{\prime 2}}=C$. In the domain $z<h$, the displayed potential has downward gravitational acceleration $-{\Phi }^{\prime }(z)=-\kappa /(h-z{)}^2$. A circular arc $z=h-\sqrt{a^2-x^2}$ satisfies the identity because $\sqrt{1+z^{\prime 2}}=a/(h-z)$.

This arc is a global minimum among admissible competitors of the same length with the same endpoints in $z<h$. Set $w=h-z>0$. The variable part of the energy is $\mu \kappa \int ds/w$, a hyperbolic length in the Poincaré half-plane model. The calibration for semicircular hyperbolic geodesics $S=\mathrm{arsinh}(x/w)$ has $|\nabla S|=1/w$. The Cauchy-Schwarz inequality gives $\int ds/w\ge S(B)-S(A)$, with equality along the circle traversed with increasing $x$. The constant-potential part is unchanged by the fixed-length constraint.

A slender flexible body modeled by a moving curve that transmits tension while its bending stiffness is neglected. For a uniform stretched string with constant tension $T$ and linear mass density $\mu$, small transverse oscillations obey $\mu y_{tt}=T{y}_{xx}$; see the wave equation on a string.

For a uniform fixed-end mechanical string of length $L$, linear mass density $\mu$ and wave speed $c$, let its $n$th normal mode be $y_n=[A_n\cos {\omega }_{n}t+B_n\sin {\omega }_{n}t]\sin (n\pi x/L)$, where ${\omega }_n=cn\pi /L$. The conserved energy is the sum of kinetic energy and elastic potential energy,The sine and cosine squared integrals are $L/2$, so substitution proves the title formula. Orthogonality makes energies of distinct normal modes add. For zero initial displacement and initial modal velocity $V_n$, $A_n=0$ and $B_n=V_n/{\omega }_n$, giving $E_n=\mu L{V}_n^2/4$.

A normal mode, also called an eigenmode, is a pattern of free oscillation in a linear system whose relative displacements stay fixed while a single temporal factor changes their size and sign. For an undamped stable vibrational mode,Here $\varphi$ is the mode shape, $A$ sets the amplitude, $\omega$ is the eigenfrequency expressed as an angular frequency, and $\phi$ is the oscillation phase. Every snapshot is a scalar multiple of the same mode shape; neighbouring regions may move in opposite directions. Normal modes of a double pendulum show that normal modes also describe coupled discrete systems, beyond strings. Their shapes are eigenvectors, rather than spatial eigenfunctions. ^[ref]

For the ideal wave equation on a string, take a uniform string of length $L$ with constant tension $\tau$ and linear mass density $\mu$. Assume small transverse slope, negligible bending stiffness, no damping, and fixed endpoints. Its wave speed is $c=\sqrt{\tau /\mu }$, and the nonzero normal modes areThe angular frequency ${\omega }_n$ is in radians per second; the frequency $f_n$ is in hertz. The wavelength is $2L/n$. The mode shape has $n$ loops separated by nodes of a standing wave, including both fixed endpoints. ^[ref]

As a worked example, choose $L=1\mathrm{m}$, $\tau =1\mathrm{N}$, and $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$. Then $c=1\mathrm{m}{\mathrm{s}}^{-1}$ and $f_1=0.5\mathrm{Hz}$, $f_2=1\mathrm{Hz}$, $f_3=1.5\mathrm{Hz}$. Their oscillation periods are $T_1=2\mathrm{s}$, $T_2=1\mathrm{s}$, and $T_3=2/3\mathrm{s}$. These are deliberately simple model parameters, rather than measurements of a musical instrument. The plot divides displacement by $A$ so that the three mode shapes can be compared, and uses each mode's own oscillation period for its snapshots.

For $\phi =0$, at $t=T_n/4$ every point has zero displacement, but most points have nonzero velocity. Only zeros of the mode shape are permanent nodes of a standing wave. At $t=T_n/2$ the original mode shape has reversed sign, and at $t=T_n$ it returns. The synchronized conversion between kinetic energy and potential energy is shown in normal-mode energy exchange.

The short derivation uses separation of variables: substituting $u=\varphi q$ into the wave equation gives ${\varphi }^{\prime \prime }+k^2\varphi =0$ and $q^{\prime \prime }+c^2{k}^{2}q=0$. The fixed-end boundary conditions remove the cosine spatial term and require $\sin (kL)=0$, hence $k=n\pi /L$ and ${\omega }_n=ck$. Different geometries and boundary conditions give different eigenfunctions; a sine mode shape is a property of this uniform fixed string, not a universal form of an eigenmode.

The usefulness is that a complicated vibration becomes a collection of harmonic oscillators. For sufficiently regular fixed-end initial data, the Fourier sine series givesThe initial displacement and velocity determine $a_n$ and $b_n$. This is the superposition principle in a Fourier sine basis, and modal energy distribution measures how strongly each normal mode is excited. The striking simplification is that each component keeps its identity while their sum can continually change shape. ^[ref]

These independent normal modes belong to the linear model. At larger slopes, changing tension introduces nonlinear effects and can couple the modes. A linearly damped string still admits a Fourier sine basis, but its temporal factors decay and need not oscillate; its energy dissipation identity for a linearly damped string replaces conservation of energy. More general damping need not leave undamped mode shapes independent.

The tension term $T{w}_{xx}$ is the transverse restoring force from a nearly straight, tensioned rod. The bending term $-EI{w}_{xxxx}$ is the force from Euler-Bernoulli flexure, where $E$ is Young's modulus and $I$ the second moment of area. The right side is transverse inertia. In the displayed one-dimensional equation $\rho$ denotes linear mass density; if it denotes volumetric density instead, the inertial coefficient must include cross-sectional area.

For a Fourier mode, $w_{xx}=-k^{2}w$, $w_{xxxx}=k^{4}w$ and $w_{tt}=-{\omega }^{2}w$, givingFor $k>0$ the phase speed is $c_p=\omega /k=\sqrt{T/\rho +EI{k}^2/\rho }$, strictly increasing with $k$ when $EI>0$ and $T\ge 0$. Thus shorter wavelengths propagate faster in phase. The group velocity $c_g=(T+2EI{k}^2)/\sqrt{\rho (T+EI{k}^2)}$ is also increasing, so the statement is consistent with wave-packet transport.

At a simple support the displacement and bending moment vanish: $w=0$ and $EI{w}_{xx}=0$ at $x=0,L$. The eigenmodes are $\sin (n\pi x/L)$ for $n=1,2,\dots$. Setting $T=0$ in the dispersion relation therefore yieldsThe integer is positive for a nonzero mode; $n=0$ gives the identically zero sine function, and negative integers repeat the same modes. Bending stiffness makes the frequencies quadratic in mode number, unlike a purely tensioned string.

The transmitted and incident waves have the same angular frequency and tension, but the transmitted wavenumber is twice as large. The time-average string-power formula therefore givesThe extra factor $k_2/k_1$ is essential. Squaring the displacement transmission coefficient alone would give $4/9$, which is not the transmitted power fraction in strings with different linear mass densities.

For equal linear mass densities and common tension, $k_1=k_2$. The junction relations then giveThere is no mechanical-impedance discontinuity, so no reflected wave is needed to satisfy displacement and force continuity. The transmitted displacement is in phase with the incident wave and has the same amplitude; all incident power crosses the junction.

A standing wave has a fixed spatial amplitude profile and oscillates in time, with stationary nodes of a standing wave. It arises from equal-amplitude waves of the same frequency travelling in opposite directions. For a uniform string with linear mass density $\mu$, the wave speed is $c=\sqrt{T/\mu }$ and a separated mode has $Y^{\prime \prime }+k^{2}Y=0$.

The fixed boundary requires $Y(0)=0$, while the free boundary requires $T{Y}^{\prime }(L)=0$. Thus $Y=A\sin kx$ and $\cos kL=0$, givingThe fixed end is a node of a standing wave and the free end an antinode. Arbitrary small motions are superpositions of these normal modes; the tension alone determines their frequencies only when $\mu$ is also specified.

Work in the small-slope model of a transverse wave on a taut string. Let $T$ be the constant tension, $\rho$ the linear mass density, and take a right-moving sinusoidal displacement $y=A\cos (\kappa x-\omega t)$, with wave number $\kappa =\omega /v$. The energy flux of a transverse string wave through a cut is the transverse force times transverse velocity:A cycle average gives $⟨{\sin }^2⟩=1/2$. Using $T=\rho v^2$, obtained from the string wave equation, yieldsThe sign in the flux formula makes the power positive for propagation in the positive $x$ direction.

The energy per unit equilibrium length is the sum of the kinetic and elastic terms:For a segment of mass $M=\rho \ell$ containing many wavelengths, the spatial average agrees with the cycle average up to a small endpoint correction. Thus $E\simeq \frac{1}{2}M{\omega }^2{A}^2$. Convert the given quantities to SI units: $M=0.080\mathrm{kg}$, $A=1.6\times 10^{-4}\mathrm{m}$, and $\omega =2\pi (120){\mathrm{s}}^{-1}$. ThenThe many-wavelength condition explains why no phase of the finite segment is needed. The result is exact as a cycle average, or for a segment containing an integer number of wavelengths.

At distance $x$ below the support, the cable must carry the weight of the remaining length $L-x$. Thus its equilibrium tension is $T(x)=\mu g(L-x)$, and the local transverse wave speed isThe linear mass density cancels. More precisely the small-displacement equation with variable tension is $\mu y_{tt}={\partial }_x[T(x)y_x]$; its local characteristic speed is the expression above, even though a first-spatial-derivative term is also present.

The wave travel time on a hanging uniform cable is obtained from $dt=dx/c(x)$:Although the local speed tends to zero at the free end, the endpoint singularity in the travel-time integral is integrable. This finite result belongs to the ideal flexible-cable model, with bending stiffness neglected.

Assume small transverse slopes, a flexible string, constant tension $T$ and uniform linear mass density $\rho$. For an element of equilibrium length $dx$, the vertical force is the difference of the vertical tension components at its ends. To first order in slope,Its mass is $\rho dx$ and its transverse acceleration is $y_{tt}$. Newton's second law therefore gives $\rho dx{y}_{tt}=Tdx{y}_{xx}$, orThis is the one-dimensional wave equation for small-amplitude transverse waves. Changes in string length and horizontal acceleration are second order in slope and are neglected in this linear approximation.