The relation

$$c=\sqrt{\frac{E}{\rho}}$$

gives the speed of small longitudinal disturbances in an ideal slender elastic rod. A continuum model explains where that result comes from and shows that the disturbance obeys a wave equation.

Let $u(x,t)$ be the small longitudinal displacement of material that was initially at coordinate $x$. The rod has uniform cross-sectional area $A$, density $\rho$, and Young's modulus $E$.

Local strain and stress

Two nearby material points separated initially by $dx$ have a displacement difference approximately

$$\frac{\partial u}{\partial x}dx.$$

Their fractional change in separation is therefore the longitudinal strain

$$\varepsilon=\frac{\partial u}{\partial x}.$$

In the linear elastic regime,

$$\sigma=E\varepsilon,$$

so

$$\boxed{\sigma=E\frac{\partial u}{\partial x}}.$$

The axial force transmitted across a section is

$$F=A\sigma.$$

Force balance on a differential element

Consider the short rod element between $x$ and $x+dx$. The force at its right face differs from the force at its left face by

$$dF\approx A\frac{\partial\sigma}{\partial x}dx.$$

The element has mass

$$dm=\rho A,dx.$$

Its longitudinal acceleration is

$$\frac{\partial^2u}{\partial t^2}.$$

Newton's second law therefore gives

$$\rho A,dx\frac{\partial^2u}{\partial t^2} =A\frac{\partial\sigma}{\partial x}dx.$$

Canceling $A,dx$,

$$\rho\frac{\partial^2u}{\partial t^2} =\frac{\partial\sigma}{\partial x}.$$

For uniform $E$,

$$\frac{\partial\sigma}{\partial x} =E\frac{\partial^2u}{\partial x^2}.$$

Thus

$$\boxed{\frac{\partial^2u}{\partial t^2} =\frac{E}{\rho}\frac{\partial^2u}{\partial x^2}}.$$

Comparing with the one-dimensional wave equation

$$\frac{\partial^2u}{\partial t^2} =c^2\frac{\partial^2u}{\partial x^2}$$

gives

$$\boxed{c=\sqrt{\frac{E}{\rho}}}.$$

The continuum derivation therefore makes the stiffness-versus-inertia interpretation precise: a strain gradient creates a net elastic force, while density determines the mass that must accelerate.

Example: checking the propagation speed

For

$$E=200\times10^9,\mathrm{Pa},\qquad \rho=7850,\mathrm{kg/m^3},$$

the coefficient of the spatial derivative is

$$\frac{E}{\rho}\approx2.55\times10^7,\mathrm{m^2/s^2},$$

so

$$c\approx5.05\times10^3,\mathrm{m/s}.$$

This agrees with the elastic-wave speed obtained directly from the material properties.

Scope of the derivation

The model assumes small longitudinal deformation, a slender rod, uniform material properties and one-dimensional motion. In bulk three-dimensional solids, longitudinal and shear waves involve the full elastic response and their speeds can depend on additional elastic constants and, for anisotropic materials, propagation direction.

The important result of the derivation is not only the formula for $c$: it shows how local constitutive behavior plus Newton's law produces a field equation that propagates disturbances through a continuous medium.