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Stress pulse in a bar

In the case of a stress pulse propagating longitudinally through a bar, the bar acts much like an infinite number of springs in series and can be taken as an extension of the equation derived for Hooke's law. A uniform bar, i.e. of constant cross-section, made from a linear elastic material has a stiffness K given by K = E A L, {\displaystyle K={\frac {EA}{L}},} where A is the cross-sectional area, and E is the Young's modulus of the material. The wave equation becomes ∂ 2 u (x, t) ∂ t 2 = E A L M ∂ 2 u (x, t) ∂ x 2. {\displaystyle {\frac {\partial ^{2}u(x,t)}{\partial t^{2}}}={\frac {EAL}{M}}{\frac {\partial ^{2}u(x,t)}{\partial x^{2}}}.}

AL is equal to the volume of the bar, and therefore A L M = 1 ρ, {\displaystyle {\frac {AL}{M}}={\frac {1}{\rho }},} where ρ is the density of the material. The wave equation reduces to ∂ 2 u (x, t) ∂ t 2 = E ρ ∂ 2 u (x, t) ∂ x 2. {\displaystyle {\frac {\partial ^{2}u(x,t)}{\partial t^{2}}}={\frac {E}{\rho }}{\frac {\partial ^{2}u(x,t)}{\partial x^{2}}}.}

The speed of a stress wave in a bar is therefore E / ρ {\displaystyle {\sqrt {E/\rho }}}.