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Take an ideal flexible mechanical string with constant tension $T>0$ and uniform linear mass density $\mu >0$. Let $x$ measure distance along its straight equilibrium position and let $y(x,t)$ be its small transverse displacement. Neglect damping, bending stiffness and external distributed forces, and assume $|y_x|\ll 1$.

A segment of reference length $\Delta x$ has mass $\mu \Delta x$. The endpoint tension forces have transverse resultantsince $\sin \theta \simeq \tan \theta =y_x$ for small slopes. Newton's second law, divided by $\Delta x$ and taken to the limit, givesHere $t$ is time, the subscripts denote partial derivatives, and $c$ is the wave speed. The transverse acceleration is $y_{tt}$; $y_{xx}$ describes the small-slope curvature responsible for the net transverse force.

Put $\xi =x+vt$ and $F(\xi )=A{e}^{-{\xi }^2/x_0^2}$, with $x_0\ne 0$. The chain rule givesFor a nonzero pulse the wave equation on a string therefore requiresIt is not a solution for arbitrary $v$; the zero pulse $A=0$ is the trivial exception. Here $\mu$ is linear mass density, and $T>0$ is tension. With $v>0$, the maximum lies at $x=-vt$, so the Gaussian function translates towards decreasing $x$ by $v(t_2-t_1)$ between the two times. Its amplitude and width stay constant.

The product-to-sum identity gives $V_1=V_0$, $V_2=V_0/2$ and $V_3=2V_0/3$, with all other modal coefficients zero. If $\mu$ is the string's linear mass density, the initial potential energy vanishes and orthogonality gives the modal energy of a stringConsequently $\boxed{E_1:E_2:E_3=36:9:16}$ for $V_0\ne 0$. The fundamental mode carries the most energy, with fraction $36/61$. Comparing displacement coefficients without their frequency factors would give the wrong energy test.

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 uniform rod has linear mass density $m/\ell$. Integrating squared distance from the pivot gives its moment of inertia about the axis perpendicular to the plane of motion:Measure $\theta$ from the downward vertical. The center of mass is at distance $\ell /2$ from the pivot, so gravity produces the restoring torque $-mg(\ell /2)\sin \theta$. The fixed-axis rotational equation of motion is $I_O\ddot{\theta }={\tau }_O$. Therefore the physical pendulum obeysThe pivot force has zero torque about the pivot.

Use the ideal transverse-motion model of the question: constant tension, small slopes, no bending stiffness, and no transverse body-force term. On a short segment away from $x=0$, the vertical components of the endpoint tensions are, to first order in $y_x$, $T{y}_x(x+\Delta x,t)$ and $-T{y}_x(x,t)$. Its mass is $\rho \Delta x$, since $\rho$ is linear mass density. Newton's second law givesDividing by $\Delta x$ and taking its limit proves the wave equation on a string on each side of the mass:The string on the right pulls the point mass with vertical component $T{y}_x(0^{+},t)$, while the string on the left contributes $-T{y}_x(0^{-},t)$. HenceThe fixed supports and attachment impose the boundary conditionsTogether with the mass equation, these are the endpoint and matching conditions. The slope can jump at $x=0$: the jump supplies the mass's vertical acceleration. Requiring the slope to be continuous would incorrectly remove its dynamics.

For a short string element of length $\Delta x$, the small-slope approximation makes the upward components of tension at its two ends $T{y}_x(x+\Delta x,t)$ and $-T{y}_x(x,t)$. Its mass is $\rho \Delta x$. Newton's second law givesDivide by $\Delta x$ and take the limit to obtain the wave equation on a string,Here $\rho$ is linear mass density, and the derivation assumes small transverse slopes, uniform tension, and a flexible string.

At the right endpoint the string pulls the ring toward the interior. Its vertical force is therefore $-T{y}_x(L,t)$ to first order; the frictionless wire supplies no vertical force. The ring obeysThis is the dynamic endpoint law for a string with an endpoint mass.

For a separated oscillatory solution, write $y=X(x)q(t)$ with $q^{\prime \prime }=-{\omega }^{2}q$ and $X^{\prime \prime }+(\omega /c{)}^{2}X=0$. The fixed left end removes the spatial cosine, so the nonzero normal modes haveThe endpoint law gives $M{\omega }_n^2\sin ({\omega }_{n}L/c)=T({\omega }_n/c)\cos ({\omega }_{n}L/c)$. Since $\sin ({\omega }_{n}L/c)=0$ would also force the cosine to vanish, it can be divided out for a nonzero mode. Using $T=\rho c^2$ givesFor physical $M\ge 0$, these give all positive modal frequencies. There is no nonzero static linear mode: the left value and right slope conditions would force it to vanish. Positivity also follows by integration by parts from

For $M=0$, the right endpoint is free, with the Neumann boundary condition $y_x(L,t)=0$. At $L=\pi$, the allowed wave numbers are $k_n=n+1/2$, $n\ge 0$, and ${\omega }_n=c{k}_n$. Zero initial displacement removes every cosine-in-time coefficient. The initial velocity contains only the modes $k=1/2$ and $k=5/2$, each with coefficient one. Dividing those velocity coefficients by their angular frequencies givesAt $x=0$ each term vanishes, at $x=\pi$ each spatial derivative vanishes, and at $t=0$ the displacement and its time derivative reproduce the two initial conditions.

For the uniform unloaded space-elevator cable, let the constant linear mass density be $\lambda$ and the outer endpoint be $b={\eta }^{\prime }R$. In the co-rotating frame the outward effective force per unit length is $\lambda ({\omega }^{2}r-G{M}_E/r^2)$. The cable is just clear of the surface, with neither a ground load nor an endpoint counterweight, so both ends have zero applied tension. The integrated effective force must vanish:Using $G{M}_E={\omega }^2{\eta }^3{R}^3$, this becomesThe root ${\eta }^{\prime }=1$ is a zero-length cable. The physical nonzero cable hasIt extends well beyond the geostationary orbit so its outer centrifugal force balances the inner segment's weight. If a cable is anchored with a nonzero endpoint load or has a counterweight, that load changes the integrated balance; its length is not fixed by the zero-end-load formula.

Measure position $x$ along the ideal thin rod from its centre of mass. Its perpendicular distance from the inclined axis is $|x|\sin \alpha$, and its linear mass density is $M/L$. The moment of inertia is thereforeThis tends to zero for an axis along an infinitely thin rod and to $M{L}^2/12$ for a perpendicular central axis. A finite rod thickness would add a contribution absent from the stated thin-rod model.

Inertia is resistance to a change of motion. Translational inertia is measured by mass in Newton's second law; rotational inertia about a specified fixed axis is measured by the moment of inertiawhere $r_{\perp }$ is perpendicular distance from that axis. For a thin uniform rod, take axes perpendicular to the rod. With linear mass density $\lambda =m/l$,The second result uses the parallel axis theorem; direct integration over $0\le x\le l$ gives the same value. The axis specification is essential: an ideal thin rod has zero moment of inertia about its own longitudinal axis.

During the short elastic collision, neglect the impulse from gravity. The pivot can exert an external impulse, so total linear momentum of rod plus particle need not be conserved. Its torque about the pivot is zero, so conservation of angular momentum about that point applies. Let $w$ be the particle's signed outgoing horizontal velocity, and $\omega$ the rod's outgoing angular velocity, positive towards the incident motion. ThenSubstituting $\omega =3(v_0-w)/l$ into the kinetic energy equation givesDiscarding the zero-impulse branch $w=v_0$, the actual collision givesThe rod's lower end initially moves forward at $l\omega =3v_0/2$, so the bodies separate. Afterwards the free particle undergoes projectile motion. With the pivot as origin and $z$ downward, its trajectory is $x=(v_0/2)t$, $z=l+g{t}^2/2$, where $t=0$ at the collision. The rod instead swings about the fixed pivot in a vertical plane as a physical pendulum.

The rod's initial rotational kinetic energy is $I_{\mathrm{end}}{\omega }^2/2=3m{v}_0^2/8$. Reaching the horizontal raises its centre of mass by $l/2$, increasing gravitational potential energy by $mgl/2$. Consequently,For the threshold case, measure the rod's angle $\theta$ from the downward vertical and reset $t=0$ at its first horizontal turning point, so $\theta (0)=\pi /2$ and $\dot{\theta }(0)=0$. The restoring torque about the pivot gives the exact equation of motionAt the horizontal, $\ddot{\theta }=-3g/(2l)$, so the rod immediately swings back. Conservation of mechanical energy yieldsOn the initial descent its explicit time dependence is specified by the definite integralIt continues through the vertical to $-\pi /2$, reverses, and repeats without damping. An alternative expression valid throughout the oscillation uses the Jacobi elliptic sine. Put $k=1/\sqrt{2}$, ${\omega }_r=\sqrt{3g/(2l)}$, and let $K(k)$ denote the complete elliptic integral of the first kind. ThenThis is not simple harmonic motion at an amplitude of $\pi /2$. Its oscillation period exceeds its own small-amplitude value $2\pi \sqrt{2l/(3g)}$ by the factor $2K(1/\sqrt{2})/\pi \simeq 1.18034$.

A simple pendulum of length $l/2$ obeys $\ddot{\theta }+(2g/l)\sin \theta =0$. At the same amplitude $\pi /2$, its oscillation period is $T_p=4\sqrt{l/(2g)}K(1/\sqrt{2})$, soThe rod swings more slowly. Its equivalent simple pendulum length is $I_{\mathrm{end}}/[m(l/2)]=2l/3$, not the distance $l/2$ to its centre of mass. The same oscillation period ratio holds in the common small-amplitude limit, but replacing the actual 90-degree motion by that limit would underestimate both oscillation periods.

time average of harmonic power. Expand the two real parts:The terms proportional to $e^{\pm 2i\omega t}$ average to zero over complete periods, while $F{u}^{\ast }$ is time-independent. For real peak amplitudes this is $F_0{u}_0\cos \varphi /2$.

String dynamics. A short element of length $dx$ experiences transverse force $T[{\psi }_x(x+dx)-{\psi }_x(x)]$. Newton's second law therefore gives $\rho dx{\psi }_{tt}=Tdx{\psi }_{xx}$, orThe approximation is a flexible, lossless string with small slope and constant tension; $\rho$ is its linear mass density.

energy transport. Across a cut toward increasing $x$, the transverse force is $F_y=-T{\psi }_x$ and the transverse velocity is $u={\psi }_t$. For a rightgoing wave these have complex amplitudes $ikTA$ and $i\omega A$. HenceThis real mechanical impedance is force divided by transverse velocity for a travelling wave, with units $\mathrm{kg}{\mathrm{s}}^{-1}$.

Matching and power ratios. At an ordinary massless junction without an external transverse force, displacement is continuous and transverse tension is balanced: ${\psi }_1={\psi }_2$, $T_1{\psi }_{1x}=T_2{\psi }_{2x}$. For the same tension, the slope is therefore continuous even if density changes. With real positive impedances,Their sum equals $P_i$; the transmitted power includes the impedance ratio and is not simply $t_{\psi }^2{P}_i$.

Let $T$ be the common tension and ${\mu }_j$ the linear mass densities. The characteristic impedance of a stretched string is $Z_j=T/c_j=\sqrt{T{\mu }_j}$. Continuity of displacement and transverse force at a massless junction gives $1+r=t$ and $Z_1(1-r)=Z_{2}t$. ConsequentlyThe reflected power is $7.18\%$. The minus sign of $r$ represents a phase reversal, not negative power. Both incident and reflected waves travel in the first string, so their power ratio is simply the squared displacement-amplitude ratio. As a check, $\mathcal{T}=(Z_2/Z_1)|t{|}^2=0.92820$ and $\mathcal{R}+\mathcal{T}=1$.

The symbol $\rho$ here is linear mass density, with units $\mathrm{kg}{\mathrm{m}}^{-1}$. A short element $[x,x+\delta x]$ has transverse forcebecause small slopes give $\sin \theta \simeq \tan \theta ={\Psi }_x$. Its mass is $\rho \delta x$. Applying Newton's second law, dividing by $\delta x$, and taking the limit gives the wave equation on a string,The derivation assumes constant positive tension, small transverse displacements and slopes, a flexible string with negligible bending stiffness, and negligible damping. Variations of tension due to the motion enter beyond the linear approximation.

At an ordinary point or a massless join, the string is continuous, soWith no concentrated transverse force or point mass, transverse force balance requiresFor the common tension in this problem, the slope $\partial \Psi /\partial x$ is therefore continuous as well. A jump in linear mass density changes wave speed, but introduces no finite inertial force at a zero-length join. A point mass, external support force, or unequal tensions would require modified boundary conditions.

The reflected and transmitted mean-power fractions areFor strictly positive finite linear mass densities, $\mathcal{T}>0$; a finite density step cannot produce exact total reflection. The ideal limits areThe first behaves as a fixed end; the second as a free end. In the free-end limit the transmitted displacement tends to $2A$, but its mechanical impedance tends to zero, so it carries vanishing power. Thus total reflected energy does not require a vanishing boundary displacement. “Entirely reflected” at a merely large density contrast is an approximation.

Assume a massless junction and common tension $T$. The characteristic impedance of a stretched string is $Z_j=\sqrt{T{\mu }_j}$, where ${\mu }_j$ is its linear mass density. The displacement reflection coefficient is $r=(Z_1-Z_2)/(Z_1+Z_2)$, so the reflected fraction of power isSet $\mathcal{R}=1/2$. The positive solutions are $z=3\pm 2\sqrt{2}$, which are reciprocals. Thus the string density ratio for prescribed reflected power isThe impedance ratio is about $5.83$, rather than the required density ratio. Reversing which string is on the incident side changes the reflected displacement sign but not the reflected power fraction.

A wave propagates an organized disturbance through space. What oscillates depends on the medium: displacement in a string or solid, pressure and density in a fluid, electric fields and magnetic fields in an electromagnetic wave, or a quantum wavefunction. The propagation law follows from the restoring mechanism, inertia and relevant boundary conditions. A common linear equation is $u_{tt}=c^2{\nabla }^{2}u$, whose plane waves obey $\omega =c|\mathbf{k}|$. The resulting phase velocity and group velocity coincide, so a narrow wave packet can travel without dispersive spreading. More generally the material supplies a dispersion relation $\omega (\mathbf{k})$; the envelope velocity is ${\nabla }_{\mathbf{k}}\omega$, while constant phase surfaces move at the corresponding phase velocity.

Mechanical waves make the balance of restoring force and inertia particularly visible. A flexible stretched string has $c=\sqrt{T/\mu }$: greater tension stiffens the string, whereas greater linear mass density slows its transverse motion. Small-amplitude acoustic waves in a uniform fluid are longitudinal, with $c_s=\sqrt{K_s/\rho }$, where $K_s$ is the adiabatic bulk modulus. For an ideal gas, $c_s=\sqrt{\gamma p/\rho }$. Pressure changes in rapid sound oscillations are approximately adiabatic because little heat crosses a wavelength in one period. Fluids do not support ordinary bulk shear waves, whereas an elastic solid supports both longitudinal and transverse waves. In an isotropic solid their speeds are $c_L=\sqrt{(K+4G/3)/\rho }$ and $c_T=\sqrt{G/\rho }$, where $G$ is the shear modulus.

A material can be dispersive even when its equilibrium properties are uniform. In a liquid with a free surface, gravity and surface tension act as restoring mechanisms. A simple inviscid capillary-gravity wave haswhere $h$ is depth and ${\gamma }_s$ is surface tension. Long shallow gravity waves have approximately constant speed $\sqrt{gh}$; deep-water gravity waves have $\omega =\sqrt{gk}$ and group velocity half their phase velocity. Short capillary waves have a different dispersion dominated by surface tension. Bending stiffness similarly changes short-wavelength propagation in a string or plate, giving a contribution proportional to $k^4$ in ${\omega }^2$. Different restoring mechanisms can therefore control different wavelength ranges of the same medium.

Electromagnetic waves propagate in vacuum without material displacement, because Maxwell equations couple changing electric and magnetic fields. They are transverse and travel at $c=1/\sqrt{{ϵ}_0{\mu }_0}$. In a simple transparent isotropic material, the phase speed is $1/\sqrt{ϵ\mu }=c/n$, where $n$ is the refractive index. Real material polarization depends on frequency. A Lorentz oscillator model for bound charges explains why permittivity and refractive index become strongly frequency dependent near material resonances, accompanied by absorption. In a birefringent medium, different field polarizations have different propagation constants, permitting wave plates and mode-dependent delays. A electrical conductor additionally supports dissipative currents; its complex wavenumber gives attenuation and, for a good conductor, a skin depth approximately $\sqrt{2/(\mu \sigma \omega )}$. Energy removed from the wave becomes heat rather than disappearing.

At an interface, frequency is conserved for stationary media, and tangential wave phase must match. These conditions give Snell's law for refraction and constrain reflected and transmitted directions. Amplitudes follow from continuity of the appropriate fields or mechanical forces. On strings the relevant mechanical impedance is $\sqrt{T\mu }$; for sound it is related to $\rho c_s$, and for electromagnetic waves to $\sqrt{\mu /ϵ}$. A mismatch gives wave reflection, and a lossless boundary conserves the total incident, reflected and transmitted power. Multiple reflections can reinforce or cancel, as in a Fabry–Pérot etalon. Total internal reflection and other boundary conditions confine waves in waveguides. Transverse modes then become discrete, and even an otherwise nondispersive material develops longitudinal cutoffs and wave dispersion. A membrane strip, for example, obeys ${\omega }_n^2=c_m^2[k^2+(n\pi /a{)}^2]$.

Propagation also involves transport and loss. The Poynting vector gives electromagnetic energy flux; mechanical energy transport is force or pressure times velocity, averaged over a cycle. Wave absorption, viscosity, electrical resistance and imperfect boundaries dissipate energy. Scattering redistributes energy into other directions and modes, while geometric spreading reduces local intensity without necessarily dissipating anything. These mechanisms should be distinguished when interpreting a decreasing measured amplitude. In strongly absorbing or rapidly varying dispersive regimes, the derivative called group velocity is not automatically the speed of a sharp signal front; pulse reshaping and the medium's causal response must be considered.

Finally, quantum matter waves illustrate how the same propagation mathematics can describe a different physical quantity. For a free nonrelativistic particle, $\omega =\hslash k^2/(2m)$, so the group velocity $\hslash k/m$ equals its particle velocity and differs from the phase speed $\hslash k/(2m)$. In a periodic crystal, Bloch's theorem replaces the free dispersion by energy bands. Their slope controls wave-packet velocity, $v_g={\hslash }^{-1}dE/dk$, and their curvature gives band effective mass. Thus mechanical confinement, material response and microscopic periodicity all reshape propagation through their allowed modes and dispersion, while the physical interpretation of the wave and its transported quantities remains essential.

Use transverse displacement $y(x,t)$, small slopes and constant tension $T$. For a short string element of length $dx$, Newton's second law gives ${\rho }_{L}dx{y}_{tt}=T[y_x(x+dx)-y_x(x)]$. Hence the wave equation on a string iswhere ${\rho }_L$ is the linear mass density, in $\mathrm{kg}{\mathrm{m}}^{-1}$.

The mechanical impedance of a rightgoing travelling wave is the transverse driving force divided by the transverse velocity at the driven section. Choose the force exerted by the material to the left on the material to the right, $F_y=-T{y}_x$. For $y=f(x-c_{L}t)$, the transverse velocity is $v_y=y_t=-c_L{y}_x$. Thus the positive characteristic impedance of a stretched string isA leftgoing component has $F_y/v_y=-Z_L$ with this fixed orientation of force. Specifying the orientation avoids confusing that propagation sign with a negative material mechanical impedance.

The instantaneous rightward energy flux along a string is the transverse force times the transverse velocity,For a rightgoing time-harmonic wave of peak displacement wave amplitude $|A|$, averaging over a wave period givesA leftgoing time-harmonic wave has the opposite signed energy flux. The positive ratios of reflected and transmitted power to incident power are thereforeAll explicit angular frequency factors cancel. In the ideal wave equation on a string, $T$ and the two linear mass densities are constant, so the mechanical impedances have no frequency or wavelength dependence. Consequently $\mathcal{T}$ is independent of wavelength even though the two string wavelengths generally differ.

Finally,This is conservation of energy at a lossless, massless junction. The transmitted displacement wave amplitude can exceed the incident one when $Z_R<Z_L$; the factor $Z_R/Z_L$ is essential when comparing power.

For point masses, Newton's force is attractive along their separation with magnitude $G{m}_1{m}_2/r^2$. Treat the narrow rod as uniform, with linear mass density $\mu =M/(2a)$; a mass profile would otherwise be needed beyond total mass and length.

For axial and transverse gravity of a uniform rod, on the axis at $b>a$, integrating the element force givesdirected toward the rod. On the perpendicular bisector, axial components cancel and transverse components add:At $r=b\gg a$,Both approach the point-mass force, but $F_1/F_2=1+3a^2/(2b^2)+O(a^4/b^4)$, so the axial attraction is slightly larger.

For a circular orbit perpendicular to a uniform rod, rotational symmetry about the rod makes a centred circular orbit possible in its perpendicular-bisector plane. With $m\ll M$, centrifugal balance givesFor $r\gg a$, it is Keplerian, $v\sim \sqrt{GM/r}$. For rod thickness $\ll r\ll a$, the force approximates that of an infinite line and $v\to \sqrt{GM/a}$, a flat speed limit rather than a divergence.