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Linear mass density is mass per unit length, measured in $\mathrm{kg}{\mathrm{m}}^{-1}$. A uniform string of mass $m$ and length $L$ has $\mu =m/L$; with constant tension $T$, its transverse wave speed is $\sqrt{T/\mu }$. Mass density instead measures mass per unit volume.

An ideal string of negligible mass and fixed length, transmitting tension when taut. In an ideal guide with no friction, the tension magnitude is the same on both sides: otherwise the force balance on an arbitrarily short massless segment would fail. Its fixed length gives a mechanical constraint relating the endpoint velocities. The two endpoint tension powers sum to zero when the guide is stationary; the string can transfer energy between connected particles without adding net energy.

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.

This coefficient multiplies the squared mean-curvature term $\frac{\kappa }{2}\int (2H-C_0{)}^{2}dA$ in the Helfrich membrane energy. It has units of energy, unlike a beam's flexural rigidity. For small height modes of a nearly flat membrane of area $A$, with Fourier normalization $h_q=A^{-1}\int h(\mathbf{x})e^{-i\mathbf{q}\cdot \mathbf{x}}{d}^{2}x$, quadratic bending and tension energies give $⟨|h_q{|}^2⟩=k_{B}T/[A(\kappa q^4+\sigma q^2)]$ by the equipartition theorem. Thermal-flicker measurements can therefore separate short-wavelength bending from longer-wavelength tension effects.

In a membrane-mechanics experiment, a narrow glass pipette applies a controlled pressure difference and draws a tongue of membrane into its opening. Imaging the aspirated length and vesicle geometry estimates the change in apparent membrane area. At low tension, much of that change can flatten thermal undulations; at higher tension it can measure actual area dilation. For ideal spherical caps with pipette radius $R_p$ and vesicle radius $R_v>R_p$, the Young–Laplace equation gives $\Delta p=2\sigma (1/R_p-1/R_v)$. Bending at the rim, adhesion and a membrane-associated cytoskeleton can require corrections to this simple geometric inference.

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.

For the undamped first normal mode of a fixed string, start at maximum displacement with $u(x,t)=A\sin (\pi x/L)\cos ({\omega }_{1}t)$. In the small-slope wave equation on a string, its kinetic energy and excess potential energy areThe potential energy comes from the extra length under tension: $\sqrt{1+u_x^2}-1\simeq u_x^2/2$. It depends on slope, so a node of a standing wave need not have zero local potential energy density. These are physical energies, with the mass and tension factors restored to the normalized wave energy. ^[ref]

Integrating ${\sin }^2(\pi x/L)$ and ${\cos }^2(\pi x/L)$ over the string gives $L/2$ in both cases. Using $\mu {\omega }_1^2=\tau {\pi }^2/L^2$ yieldsFor the model $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$, $\tau =1\mathrm{N}$, $L=1\mathrm{m}$, and $A=0.010\mathrm{m}$, the angular frequency is ${\omega }_1=\pi \mathrm{rad}{\mathrm{s}}^{-1}$, the frequency is $0.5\mathrm{Hz}$, and $E\simeq 2.4674\times 10^{-4}\mathrm{J}$. The maximum slope $A\pi /L\simeq 0.0314$ is small. At $t=0$ all the energy is potential energy; at $t=T_1/4=0.5\mathrm{s}$ the string is straight and all the energy is kinetic energy. At $t=T_1/2=1\mathrm{s}$ the displacement is reversed and all the energy is again potential energy. The energy curves repeat twice during one displacement cycle, while conservation of energy keeps their sum constant.

Take $N$ equal point masses $m>0$ on a massless string, with fixed ends, equal horizontal spacing $\ell >0$ and constant tension $\tau >0$. Small transverse displacements obey $m{\ddot{z}}_i=(\tau /\ell )(z_{i-1}-2z_i+z_{i+1})$, with $z_0=z_{N+1}=0$. The normal modes arewith the displayed positive frequencies. Substitution uses $\sin ((i-1)q)+\sin ((i+1)q)=2\cos q\sin (iq)$. The endpoint conditions quantize $q=j\pi /(N+1)$. These are the discrete sine transform modes and form an orthogonal basis, so each initial displacement and velocity decomposes into independent harmonic oscillations. For three masses, convenient relative shapes are $(1,\sqrt{2},1)$, $(1,0,-1)$ and $(1,-\sqrt{2},1)$, with $m\ell {\omega }^2/\tau =2-\sqrt{2},2,2+\sqrt{2}$ respectively.

For $N$ independent links pointing left or right with length $a$, fixed signed extension $L$ gives $N_R=(N+L/a)/2$ and $N_L=(N-L/a)/2$. The multiplicity is $\left(\genfrac{}{}{0ex}{}{N}{N_L}\right)$ when these counts are nonnegative integers. Leading Stirling's approximation gives $\mathcal{F}=(k_{B}T/(2a))\ln (N_R/N_L)$ as the externally applied tension conjugate to $L$. Its small-extension limit is $k_{B}TL/(N{a}^2)$. At fixed tension, the exact one-link partition function is $2\cosh (\beta \mathcal{F}a)$ and the mean extension is $Na\tanh (\beta \mathcal{F}a)$, confirming the same inverse relation without using Stirling at the full-extension endpoint.

The main generic costs of strongly confining a charged semiflexible polymer are bending it away from its preferred straight shape, crowding neighboring segments, and reducing its available configurations. Closely approaching charged strands also repel through electrostatic and hydration interactions; mobile ions screen or modify these interactions, while counterion organization contributes to the free-energy balance. The bending cost is especially large in regions of high curvature. A pressure on the confining wall and a force resisting further insertion are related thermodynamic responses, but are not identical quantities with the same units.

At the level of measurement principles, an optical tweezers apparatus acts as a calibrated compliant force sensor. A displaced trapped bead experiences approximately $F_{\mathrm{trap}}=k_{\mathrm{trap}}\Delta x$. A tether transmits tension between that sensor and the material under load; the opposing support fixes the other end. The conceptual force balance isThe extension and restoring force supply a mechanical record of the load. This is a force-sensing schematic, not a biological preparation or packaging protocol. The measured tension can include bending, interstrand interactions, dissipation and any other mechanically coupled contribution; it cannot be identified with only the bending term without additional assumptions. In a reversible generic model $F_{\mathrm{conf}}=\partial G/\partial L$ at specified geometry, while an internal pressure is $-\partial G/\partial \mathcal{V}$ at specified polymer content. These derivatives clarify what a force measurement can and cannot determine.

In spectrin unfolding by force spectroscopy, the cantilever measures tension through its calibrated deflection, while displacement of the pulling point increases the separation of the protein attachments. The protein extension is obtained after allowing for cantilever compliance and any linker extension; it is not simply the commanded displacement.

A folded spectrin repeat stores much of its backbone in a compact domain. Between unfolding events, the already released chain is stretched: increasingly straight conformations have fewer accessible configurations, and the resisting entropic force rises nonlinearly. A peak is reached when a still-folded domain crosses an unfolding barrier on the timescale of the imposed loading. Its released backbone increases the available contour length abruptly. At nearly the same end-to-end distance the fractional extension $x/L$ is then smaller, so tension drops. Pulling resumes on the next, longer-chain elastic branch. Repeated release of domains produces the saw-tooth sequence.

The record is therefore evidence for sequential mechanical unfolding of modular structure, not repeated stretching of a fixed-length rod. For the nominal eight-domain construct, successive branches correspond to increasing numbers of unfolded repeats. The order and rupture forces can fluctuate because barrier crossing is thermally activated and depends on loading rate, structural heterogeneity and pulling geometry; equivalent repeats need not rupture at identical force.

The horizontal separation between worm-like-chain fits at the same force estimates the added contour length, whereas the vertical force drop at one extension reflects redistribution of tension. These are distinct measurements. A repeat increment near the specified $31.7\mathrm{nm}$ is consistent with repeated similar structural units. The gain is a net release relative to the folded span and attachment geometry; $31.7/106\mathrm{nm}$ is not automatically the universal contour length of a peptide residue.

During unloading, relaxation and possible refolding can restore compact domains. Hysteresis would indicate dissipative or kinetically delayed processes. Neither the peak force alone nor an arbitrary saw-tooth area directly equals an equilibrium unfolding free energy: one must separate chain elasticity, cantilever storage, irreversible loss and pulling-rate effects. The worm-like-chain branches describe released-chain elasticity, not the molecular transition state responsible for a rupture peak.

After a domain unfolds, its backbone joins the same continuous chain as previously unfolded material. The new branch differs mainly by an added contour length, while the local backbone bending statistics remain similar. It is therefore consistent to fit successive branches with different $L_j$ and one common $L_p\simeq 0.8\mathrm{nm}$. Tension is transmitted through serial portions of the same chain; one should not add their tensile forces as though they were independent parallel springs.

The persistence length is short compared with the contour increment per repeat, so each released domain contributes many flexible bending segments. Stretching suppresses those thermal modes and raises free energy through loss of entropy, which is precisely the worm-like-chain mechanism. The common $L_p$ is a property of the unfolded backbone under the measurement conditions, not the folded domain's axial length or the number of domains already released. With $T=300\mathrm{K}$ only as an illustrative value, $k_{B}T{L}_p\simeq 3.3\mathrm{pN}\mathrm{n}{\mathrm{m}}^2$.

A common persistence-length fit is evidence for approximately repeat-independent chain flexibility over the observed branches; it is not proof that a continuum inextensible model describes every molecular scale. Residual folded structure, heterogeneous linkers, force-induced backbone stretching, and very short-chain effects can cause deviations. The formal force divergence at $x=L$ signals the model's inextensibility assumption, not a physical infinite tensile force.

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.

Take $z$ positive upwards, so the source's sign convention makes $p$ a downward applied load. The small-slope tension contribution on a short interval is $T{w}^{\prime \prime }$. Let $r\ge 0$ be the upward contact reaction. Force balance gives $T{w}^{\prime \prime }-p+r=0$. Nonpenetration is $w\ge f$, and reaction is absent where $w>f$. ThereforeThe contact boundary must have continuous $w$ to avoid a broken string, and continuous slope in this smooth-contact model with no concentrated point force. Integrating force balance over a vanishing interval around a contact endpoint gives $T[w^{\prime }]=0$ when the load and reaction are locally bounded. This is the one-dimensional obstacle problem, with its reaction acting as a nonnegative constraint force.

The small-slope excess length is $\int (w^{\prime }{)}^2/2$, so the first term is the work needed against uniform tension. The second is the potential energy of the downward distributed load. Their sum is total potential energy subject to the obstacle constraint.

Expand the difference exactly:The first two integrals together are nonnegative by part (b), and the square is nonnegative. ThusFor $T>0$ this also proves uniqueness: equality for two minimizers forces identical derivatives, and the fixed endpoints remove their additive constant. The convex admissible set and strictly convex energy make the variational inequality a global minimization condition.

Work to first order in the slope $w^{\prime }$ and take $B$ constant. The elastic beam is slender, its cross-sections remain plane and normal to its centreline, transverse shear deformation is neglected, and the loading is quasi-static. There are no distributed couples. In a non-contact interval the distributed force is $-p{\mathbf{e}}_z$ and there is no axial distributed force. These are the assumptions of the small-deformation Euler-Bernoulli beam equation, with an optional prescribed tension retained in transverse force balance.

Choose the internal section resultant on the right face of a segment aswhere $N$ is the transverse shear force, and take the section bending moment as $M{\mathbf{e}}_y$. Longitudinal and transverse force balance on a segment of length $dx$ give $T^{\prime }=0$ and $(T{w}^{\prime }+N{)}^{\prime }=p$. Balancing the torques about its left end gives $M^{\prime }=N$: the torque from the transverse section resultant is $-Ndx{\mathbf{e}}_y$. ThereforeWith this positive-$y$ bending moment convention, the linear relation between bending moment and curvature is $M=-B{w}^{\prime \prime }$. Thus $N=-B{w}^{\prime \prime \prime }$, and substitution givesThe tension need not be small compared with the transverse restoring terms, but products arising solely from finite deformation, such as the change of axial extension proportional to $(w^{\prime }{)}^2$, are outside this linear model. A separate axial constitutive constraint would be required to determine an unknown tension from fixed end positions.

In an open contact interval, the upward distributed normal contact force is $R$. At this order the smooth obstacle supplies no axial force; its vertical resultant changes the downward load to $p-R$. HenceFor nonadhesive unilateral beam contact, the obstacle can push but cannot pull, so $R\ge 0$. The accompanying constraints are $w\ge f$ and $(w-f)R=0$. A negative $R$ would require adhesion and would invalidate the proposed contact region. If $f$ is four times differentiable, the distributed normal contact force inside an open contact interval is $R=B{f}^{\prime \prime \prime \prime }-T{f}^{\prime \prime }+p$.

The jump conditions express a connected elastic beam with no concentrated horizontal force or applied couple. Continuity of tension, $[T]=0$, follows from horizontal force balance. Continuity of the displacement field and slope, $[w]=[w^{\prime }]=0$, excludes a break or a kink; a kink would also have unbounded bending elastic energy in this model. Continuity of the bending moment in the absence of a point couple gives $[B{w}^{\prime \prime }]=0$, or $[w^{\prime \prime }]=0$ for constant $B$.

The only allowed discontinuous section resultant is the transverse shear force. Integrating vertical force balance through an upward point force $R_i$ gives $[N]=-R_i$. Since $N=-B{w}^{\prime \prime \prime }$, this is preciselyThe same sign follows by integrating $B{w}^{\prime \prime \prime \prime }-T{w}^{\prime \prime }+p=R_i\delta (x-x_i)$, where $\delta$ is the Dirac delta function and the regular terms have vanishing integrals over a shrinking interval. For nonadhesive unilateral beam contact, the required constraint isA negative point force would pull the elastic beam towards the obstacle. At an endpoint where $R_i=0$, the third derivative is continuous as well.

With zero tension and zero distributed load, a free interval satisfies $w^{\prime \prime \prime \prime }=0$. Before contact the entire elastic beam is free, so $w$ is a cubic polynomial. The four end boundary conditions determine it uniquely. Reflection $x\mapsto -x$ preserves those boundary conditions, so the unique solution is an even function and its odd terms vanish. Direct substitution givesFor $\alpha \ge 0$, its minimum occurs at $x=0$ and is $-\alpha L/2$. The constraint $w\ge -H$ is therefore satisfied exactly when $\alpha \le 2H/L$. The first contact is at the midpoint:At equality the normal contact force is still zero. The uniqueness used here can also be checked by integrating $v^{\prime \prime \prime \prime }v$ for the difference of two solutions with homogeneous clamped boundary conditions, which gives ${\int }_{-L}^L(v^{\prime \prime }{)}^{2}dx=0$ and hence $v=0$.

For an element $dx$, the vertical components of the approximately constant tension are $T{y}_x$ at its two ends. Their difference is $T{y}_{xx}dx$ to first order in the small slope. Its inertial response is $\rho dx{y}_{tt}$. Thus Newton's second law givesHere $\rho$ is mass per unit length. Small slopes, negligible gravity, uniform tension and negligible bending stiffness or dissipation are the assumptions in this wave equation on a string.

For an equilibrium string segment of length $dx$, the mass is $\rho dx$. Under the small-slope approximation, the transverse tension component at an endpoint is $T{y}_x$. The difference between its two endpoint contributions is $T[y_x(x+dx,t)-y_x(x,t)]\simeq T{y}_{xx}dx$.

Newton's second law in the transverse direction gives $\rho y_{tt}dx=T{y}_{xx}dx$. Cancelling the segment length producesThis wave equation assumes a uniform string, small slopes, negligible gravity and negligible bending stiffness. Longitudinal changes in tension and the difference between actual and equilibrium segment length contribute only beyond the linear approximation used here.