Hooke's law

Hooke's law states that the restoring force or stress is proportional to a sufficiently small displacement or strain. For a one-dimensional spring, $f=kx$ in magnitude.

A taut spring of equilibrium length $L$ and initial tension $T_0$ whose ends acquire a small relative transverse displacement $y$ tilts through $y/L$. Its leading transverse force is therefore $-T_{0}y/L$. The length change is quadratic in $y$, so using only incremental axial extension misses this first-order force. A Hooke spring with natural length $l_0$ has $T_0=k(L-l_0)$ and $k_{\perp }=k(1-l_0/L)$. The unstretched limit has no linear transverse stiffness under this geometry.

An ideal spring with stiffness $k>0$ and zero natural length has extension equal to the distance between its endpoints. Its elastic potential energy is exactly $k|{\mathbf{r}}_2-{\mathbf{r}}_1{|}^2/2$, and its force on endpoint 2 is $-k({\mathbf{r}}_2-{\mathbf{r}}_1)$. Thus its Cartesian stiffness is isotropic and constant. Coupled systems of such springs have exactly linear displacement equations after a constant force, such as gravity, is absorbed into the equilibrium position. This special model differs from an ordinary finite-natural-length spring, whose transverse small-motion stiffness depends on its tension and equilibrium length.

A linear spring stores elastic energy $k{x}^2/2$ when displaced by $x$ from its relaxed position. The external force maintaining that displacement is $kx$; its restoring force on the moving attachment has the opposite sign. Its spring constant $k$ has units force per length.

A zero-rest-length spring has quadratic potential energy in its actual length $s$, with no subtracted natural length. It is a useful idealization, not an ordinary unstretched spring of arbitrary length. For an end-to-end spring on a two-link pendulum, changing the natural length changes the quadratic bending stiffness about the straight configuration.

Identical linear springs of stiffness $k$ have equivalent stiffness $2k$ in parallel and $k/2$ in series. Parallel springs have equal displacement and add forces; series springs carry equal force and add extensions. A mass $m$ attached to an equivalent massless spring has angular frequency $\sqrt{k_{\mathrm{eff}}/m}$.

For point masses joined by unstretched central springs obeying Hooke's law, let $n_{ij}$ be the equilibrium unit vector along an edge and $u_i$ the small displacement of its endpoint. The leading extension is $n_{ij}\cdot (u_j-u_i)$, so its quadratic elastic potential energy is $k_{ij}[n_{ij}\cdot (u_j-u_i){]}^2/2$. Consequently the block stiffness matrix has off-diagonal blocks $K_{ij}=-k_{ij}{n}_{ij}{n}_{ij}^T$ and diagonal blocks $K_{ii}={\sum }_j{k}_{ij}{n}_{ij}{n}_{ij}^T$, with absent edges contributing zero. It is symmetric and positive semidefinite. The normal modes solve $Kq={\omega }^{2}Mq$, where $M$ is the mass matrix. Rigid translations and infinitesimal rotations are rigid-motion zero modes of a spring framework. A prestressed spring can add transverse quadratic terms, so the unstretched assumption matters.

An ideal linear torsion spring exerts a restoring torque proportional to the relative angle between its two ends. Integrating the torque gives its elastic potential energy. A spring connecting two rotating bodies contributes $\kappa ({\theta }_2-{\theta }_1{)}^2/2$ to the potential energy; a spring connected to a fixed frame is obtained by setting one angle to zero. The coefficient $\kappa$ is the torsional spring constant.

The derivative of an axial torque with respect to relative end-rotation angle, evaluated at the operating configuration. In a linear torsion spring, $N=K_{\phi }\phi$.

For a uniform rod of length $L$ and twist per unit length $\tau$, the end-rotation angle is $\phi =L\tau$. The chain rule gives $K_{\tau }=L{K}_{\phi }$, where $K_{\phi }$ is the torsional stiffness. If a source defines a normalized torque $M=N/(2\pi )$, its derivative is $dM/d\tau =K_{\tau }/(2\pi )$.

A circular shaft twisted uniformly through small end angle $\varphi$ over length $L$ has local engineering shear strain $r\varphi /L$ at radius $r$. Its shear stress is $Gr\varphi /L$, so the integrated torque is $C={\int }_{A}r(Gr\varphi /L)dA=G{J}_p\varphi /L$, using the polar second moment of area. The restoring torque has the opposite sign to the imposed twist. The stored elastic energy is $C\varphi /2=G{J}_p{\varphi }^2/(2L)$. The circular cross-section is important: general noncircular sections can warp and have a different torsion constant.

The spring constant is the positive proportionality coefficient in the ideal spring restoring force $F=-kx$. Its SI unit is newtons per metre. With a constant attached mass $m$, it sets the angular frequency $\omega =\sqrt{k/m}$ of simple harmonic motion. Doubling $k$ doubles the restoring force at a fixed displacement, but increases that angular frequency only by a factor $\sqrt{2}$.