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For a flexible uniform string with linear mass density $\rho$ and tension $T$, dimensional analysis gives $[T/\rho ]=L^2{T}^{-2}$ and hence $c\propto \sqrt{T/\rho }$. The small-slope force balance $\rho {\psi }_{tt}=T{\psi }_{xx}$ fixes the factor to one:Here $\rho$ has units kg/m, not the volume-density units used for a bulk fluid. The phase moves along the string at $c$, while a material point moves transversely with ${\psi }_t=A\omega \sin (kx-\omega t)$. Its maximum speed is $A\omega$, which is not the phase velocity. Small slope $Ak\ll 1$ makes $A\omega \ll c$ in this model.

The kinetic energy and extra elastic energy of a segment areThe second expression follows from extra length $\sqrt{1+{\psi }_x^2}dx-dx\simeq {\psi }_x^{2}dx/2$ against the pre-existing constant tension. Time averaging a harmonic wave gives $⟨{\mathcal{E}}_K⟩=\rho A^2{\omega }^2/4$ and $⟨{\mathcal{E}}_U⟩=T{A}^2{k}^2/4=\rho A^2{\omega }^2/4$. Therefore

At the driven endpoint the string pulls along its local tangent into $x>0$, with vertical component $T{\psi }_x$ for small slope. The point exerts the opposite force on the string, so the driven-string endpoint power isFor the outgoing wave $\psi =A\cos (kx-\omega t)$, at $x=0$ the derivatives are ${\psi }_x=Ak\sin \omega t$ and ${\psi }_t=-A\omega \sin \omega t$. Thus $P=T{A}^{2}k\omega {\sin }^2\omega t$ andThe energy flux of a transverse string wave follows by combining the energy density $\mathcal{E}=(\rho {\psi }_t^2+T{\psi }_x^2)/2$ and flux $P=-T{\psi }_x{\psi }_t$, which satisfy ${\partial }_t\mathcal{E}+{\partial }_{x}P=0$. For a pure right-going nondispersive wave, $P=c\mathcal{E}$ even instantaneously. energy is transported with the wave; a material string element merely oscillates about its unperturbed position.

Field of discrete masses. Newton's law of universal gravitation and superposition give the gravitational field away from the source points:It is the force per unit test mass, with units $\mathrm{N}\mathrm{k}{\mathrm{g}}^{-1}=\mathrm{m}{\mathrm{s}}^{-2}$, and points toward each positive source mass.

Infinite circular cylindrical rod. Let $b$ be the radius of the circular cross-section, so $A=\pi b^2$, and write $\lambda =\rho A$ for the linear mass density. Choose a coaxial cylindrical Gaussian surface with radius $r>b$ and length $L$. Translation and rotation symmetry make $\mathbf{g}=g_r(r)\hat{\mathbf{r}}$. The end caps have zero field flux because their normals are axial; on the curved surface the outward-normal component is the constant negative value $g_r$. Gauss's law for gravity therefore givesThe gravitational field of an infinite uniform cylinder inside the material instead has enclosed mass $\rho \pi r^{2}L$, giving $g_r=-2\pi G\rho r$. The exterior result assumes a circular cylindrical rod; area by itself would not specify the exact near field of a noncircular cross-section.

In the left panel the dashed circle is a transverse section of the coaxial Gaussian surface; the field points inward and the surface normals outward. Its side area is $2\pi rL$.

For an exterior circular orbit, Newton's second law equates the force required for centripetal acceleration to gravity:The orbital speed is independent of radius and of the orbiting object's mass, for orbits outside the infinite rod.

Finite rod. Taking the supplied field as the gravitational field of a finite uniform line segment, the corresponding balance at radius $R$ isSince $M=2a\rho A=2a\lambda$, its two limits areThe first reproduces the infinite-rod result: a nearby observer sees a very long source. The second is the familiar circular orbit speed about a point mass, since a distant observer cannot resolve the rod's length. The dimensionless curve isIt starts at a constant plateau and decreases smoothly, approaching an $R^{-1/2}$ dependence far away. For $R>0$, $dv/dR=-\sqrt{GM}R/[2(R^2+a^2{)}^{5/4}]<0$, consistent with the sketch.

Derivation of the finite-rod field. Put the source along $-a<z<a$ on the $z$ axis. In its central perpendicular plane, an element $dm=\lambda dz$ is at distance $s=\sqrt{r^2+z^2}$. Its field magnitude is $G\lambda dz/s^2$, and its inward radial component has the extra factor $r/s$. The axial components cancel between $z$ and $-z$. ThusReplacing $2a\lambda$ by $M$ givesThe integral treats the source as a line mass. For a finite cylinder of nonzero cross-section, the displayed finite-rod formula is a thin-rod approximation, not an exact field for all $r$. Its near-rod limit should therefore be read as $b\ll R\ll a$, keeping the orbit outside the material; the apparent line-source singularity at $r=0$ does not describe the interior of a real cylinder.

Since tension is a force and ${\lambda }_0$ is a linear mass density,Thus $c$ has units $\mathrm{m}{\mathrm{s}}^{-1}$. It is the transverse wave speed of the ideal uniform string. The wave equation on a string assumes small transverse slopes, constant positive tension, positive uniform linear mass density, and negligible damping and bending stiffness.

For a massless join between two ideal strings with a common tension $T$, require continuous transverse displacement and continuous transverse force $T{\partial }_x\Psi$. With $Z_j=\sqrt{T{\rho }_j}$ for linear mass densities ${\rho }_j$, the displacement reflection coefficient and transmission amplitude areThe reflected and transmitted mean-power fractions are $\mathcal{R}=|r{|}^2$ and $\mathcal{T}=(Z_2/Z_1)|t{|}^2$, with $\mathcal{R}+\mathcal{T}=1$. Reflection from the heavier side reverses phase; reflection from the lighter side does not. Total reflection occurs in the limits $Z_2/Z_1\to \infty$ or zero, while a finite positive impedance mismatch transmits some energy. Equal mechanical impedances remove reflection.

For a semi-infinite ideal string on $x\le 0$ with tension $T$ and linear mass density $\mu$, let a free transverse endpoint mass be $m$. Use a time-harmonic wave $y=\mathrm{Re}\{A(e^{ikx}+R{e}^{-ikx})e^{-i\omega t}\}$, where $k=\omega \sqrt{\mu /T}>0$. The endpoint boundary condition from Newton's second law is $m{y}_{tt}=-T{y}_x$. It gives the title displacement reflection coefficient, with characteristic impedance of a stretched string $Z=\sqrt{T\mu }$. Since numerator and denominator have equal moduli, all mean incident power is reflected. The endpoint kinetic energy varies within a cycle, but its mean rate of change is zero. A free mass changes the reflected phase, and does not by itself provide dissipation. The limits $m\to 0$ and $m\to \infty$ give $R=+1$ and $R=-1$ respectively at fixed positive angular frequency.

For two ideal strings under a common tension, the positive mechanical impedances satisfy $Z_j=\sqrt{T{\mu }_j}$, where ${\mu }_j$ are their linear mass densities. The reflected fraction of incident power is $\mathcal{R}=[(Z_1-Z_2)/(Z_1+Z_2){]}^2$. Solving for the positive ratio $z=Z_2/Z_1$ gives $z=(1+\sqrt{\mathcal{R}})/(1-\sqrt{\mathcal{R}})$ or its reciprocal. Squaring yields the displayed linear mass density ratios. For $\mathcal{R}=1/2$, the heavier string has $17+12\sqrt{2}\simeq 33.97$ times the lighter string's linear mass density; the impedance ratio is only $3+2\sqrt{2}$.

A uniform string with fixed ends $x=\pm \ell$, tension $T$, linear mass density $\rho$, and attached central point mass $m>0$ obeys the wave equation away from zero, continuous displacement at zero, and $m{y}_{tt}(0,t)=T[y_x(0^{+},t)-y_x(0^{-},t)]$. The normal modes split into antisymmetric modes that leave the mass at rest and symmetric modes that move it.

To prove the spectral equations, use $X=A\sin (k(x+\ell ))$ on the left and $X=B\sin (k(\ell -x))$ on the right, with ${\omega }^2=(T/\rho )k^2$. Continuity gives $(A-B)\sin (k\ell )=0$ and the force law gives $Tk(A+B)\cos (k\ell )=m{\omega }^{2}A\sin (k\ell )$. If $\sin (k\ell )=0$, then $k\ell =n\pi$, $n\ge 1$, and $B=-A$. Otherwise $A=B$ and cancellation gives $\cot (k\ell )=mk/(2\rho )$. In dimensionless variables $z=k\ell$, $\alpha =m/(2\rho \ell )$, strict monotonicity gives one symmetric root in each $(n\pi ,n\pi +\pi /2)$, $n\ge 0$. The apparent zero root of the sine equation has only the zero profile and is excluded.

For $m/(\rho \ell )\gg 1$, the lowest root tends to zero. The Taylor series $\cot z=1/z-z/3+O(z^3)$ gives $z\sim {\alpha }^{-1/2}$ and hence $\omega \sim \sqrt{2T/(m\ell )}$. The two nearly straight half-strings supply a restoring force $-2TY/\ell$, explaining this harmonic oscillator limit.

For a uniform string with tension $T>0$, linear mass density $\rho >0$, left end fixed, and a freely transverse-moving point mass $M\ge 0$ at $x=L$, the wave equation and endpoint law areThe last equation is Newton's second law: the string pulls the mass back toward the interior, giving the minus sign. With wave speed $c=\sqrt{T/\rho }$, a nonzero normal mode has spatial part $X(x)=\sin (\omega x/c)$ and satisfiesFor $M>0$, its dimensionless roots $q=\omega L/c$ occur once in each $(n\pi ,n\pi +\pi /2)$, $n\ge 0$, because $\cot q$ decreases from infinity to zero while $Mq/(\rho L)$ increases. For $M=0$, $q=(n+1/2)\pi$: the right end has a homogeneous Neumann boundary condition. There is no nonzero zero-frequency mode, since a linear spatial profile vanishing at the left end and with zero right slope must vanish everywhere.

Multiplying the spatial equation by $X$ and using integration by parts givesFor two different modes, the same calculation gives the weighted inner product orthogonality $\rho {\int }_0^L{X}_m{X}_{n}dx+M{X}_m(L)X_n(L)=0$. The endpoint mass therefore contributes to both inertia and mode orthogonality.

Tension is a pulling force transmitted along a taut string, rope, or similar body. For an ideal uniform string with linear mass density $\mu$, constant tension $T$ sets the small transverse wave speed $c=\sqrt{T/\mu }$.

A radial co-rotating cable of constant linear mass density $\lambda$ extends from planetary radius $R$ to $b={\eta }^{\prime }R$, with zero applied endpoint tension. If the geostationary orbit has radius $\eta R$, integration of the effective force gives ${\eta }^{\prime }({\eta }^{\prime }+1)=2{\eta }^3$ for a nonzero cable. At position $r$ its tension is $\lambda [GM(1/R-1/r)-{\omega }^2(r^2-R^2)/2]$. Differentiation shows the maximum at the geostationary orbit. An anchor load or counterweight alters the integrated condition.

A semi-infinite wave equation on a string with linear mass density $\rho$, wave speed $c$, and tension $T=\rho c^2$ has positive characteristic mechanical impedance $Z=\rho c$. At an end attached to a massless viscous damper of coefficient $\alpha >0$, displacement and transverse-force balance give the displacement reflection coefficientHere $\mathcal{A}$ is the fraction of mean incident power dissipated. A prescribed $0<\mathcal{R}<1$ gives two positive damper coefficients, $\alpha /Z=(1\pm \sqrt{\mathcal{R}})/(1\mp \sqrt{\mathcal{R}})$, with opposite reflected phases. Matching $\alpha =Z$ removes reflection.

An ideal uniform, flexible string under constant tension $T>0$ has transverse displacement $y(x,t)$ satisfyingwhere $\mu >0$ is linear mass density, meaning mass per unit length of the straight reference string. The approximation assumes small slope $|y_x|\ll 1$, transverse motion, negligible bending stiffness, negligible damping, and no distributed external force. Constant tension and uniform linear mass density make $c$ constant. Some texts use $\rho$ for this same linear mass density, giving $c=\sqrt{T/\rho }$; here $\rho$ must have units $\mathrm{kg}{\mathrm{m}}^{-1}$, rather than units of volume mass density. ^[ref]

The units provide a useful check: $T/\mu$ has units $(\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-2})/(\mathrm{kg}{\mathrm{m}}^{-1})={\mathrm{m}}^2{\mathrm{s}}^{-2}$. For $T=100\mathrm{N}$ and $\mu =0.010\mathrm{kg}{\mathrm{m}}^{-1}$, the wave speed is $100\mathrm{m}{\mathrm{s}}^{-1}$, so a pulse crosses a one-metre string in $0.010\mathrm{s}$. Quadrupling tension doubles wave speed; quadrupling linear mass density halves it. These square-root laws explain how material and loading set the time scale of the vibration.

For a length-$L$ string with fixed endpoints, the homogeneous Dirichlet boundary conditions are $y(0,t)=y(L,t)=0$. Its nonzero normal modes haveThus $cn\pi /L$ is an angular frequency, in radians per second, whereas $cn/(2L)$ is the frequency in hertz. The wavelength is $2L/n$, and each node of a standing wave lies at $x=jL/n$ for some $j=0,\dots ,n$. For the numerical string above with $L=1\mathrm{m}$, the first three frequencies are $50$, $100$ and $150\mathrm{Hz}$, with angular frequencies $100\pi$, $200\pi$ and $300\pi {\mathrm{s}}^{-1}$. ^[ref]

Each normal mode is a standing wave, but it also has a travelling wave description:At a fixed endpoint an incident displacement pulse reflects with reversed sign, allowing incident and reflected displacements to cancel at the endpoint. Periodic reflection for a fixed-end string implements both endpoints through odd, $2L$-periodic extensions of the initial data. This is the bridge between moving pulses and the stationary Fourier sine basis. ^[ref]

A short derivation comes from Newton's second law. The transverse force on a small segment is approximately $T[y_x(x+\delta x,t)-y_x(x,t)]$, and its mass is $\mu \delta x$. Dividing the force balance by $\delta x$ and taking the limit gives $\mu y_{tt}=T{y}_{xx}$. Separation of variables then gives ${\varphi }^{\prime \prime }+k^2\varphi =0$; the fixed-end boundary conditions select $k=n\pi /L$, while the temporal equation is a harmonic oscillator equation with $\omega =ck$.

For the homogeneous one-dimensional wave equation $u_{tt}=c^2{u}_{xx}$ with constant $c>0$, wave speed means the speed at which a travelling wave profile translates. In $F(x-ct)$, a feature initially at $x=s$ follows $x=s+ct$, giving signed velocity $+c$; in $G(x+ct)$ it follows $x=s-ct$, giving signed velocity $-c$. The characteristic lines for the one-dimensional wave equation therefore have slopes $dx/dt=\pm c$. The speed of either profile is $c$, even though the displacement at a fixed point can change sign or pass through zero.^[ref]

The same $c$ bounds the causal cone for the constant-speed wave equation. For a wave equation on a string, $c=\sqrt{T/\mu }$ depends on tension and linear mass density; for small acoustic waves in stationary uniform air it is the speed of sound. This interpretation assumes the stated ideal wave equation. In media with wave dispersion, phase velocity, group velocity and the earliest signal arrival have different meanings: a measured wave crest speed need not be the speed of a wave envelope, and neither alone establishes a causal cone for the constant-speed wave equation.

At distance $x$ below the support of a flexible uniform cable of length $L$, the equilibrium tension is $\mu g(L-x)$, with linear mass density $\mu$. The local transverse speed is $\sqrt{g(L-x)}$. Integrating its reciprocal from zero to $L$ gives the displayed travel time. The speed vanishes at the free end, but the travel-time singularity is integrable. With variable tension, the small-amplitude wave equation is $\mu y_{tt}={\partial }_x(T{y}_x)$; its characteristic speed is still $\sqrt{T/\mu }$.