THE SCIENCE AND APPLICATIONS OF ACOUSTICS - H. H. Arnold ...

5.2 Derivation of the Longitudinal Wave Equation for a Bar 91In the situation in which static forces are applied to a uniform bar, the strain is thesame for each point and is time-independent. But we are considering a dynamiccase in which the strain in the bar varies with coordinate x and with time t. Thistype of variation generates a longitudinal wave motion in the bar in a manneranalogous to the transverse waves in a string. When a bar undergoes strain, elasticforces are generated inside the bar. These forces act across each cross-sectionalplane and essentially constitute reactions to longitudinally applied forces. Welet F x = F x (x, t) denote these longitudinal forces and adopt the convention thatcompressive forces are represented by positive values of F x and tensile forces bynegative values of F x . The stress σ in the bar is defined byσ = F x /ÂHere  is the cross-sectional area of the bar. We can now apply Hooke’s lawσ = F xÂ=−Eε =−E∂ξ∂x(5.2)where E is the elastic constant or Young’s modulus, a property characteristic ofthe material constituting the bar. Table A in Appendix lists the values of Young’smoduli for a number of commonly used materials. Because E must always have apositive value, a negative sign is introduced in Equation (5.2) to accommodate thefact that a positive stress (compression) results in a negative strain, and a negativestress (tension) in a positive strain. We rewrite Equation (5.2) to express force F xat point x as follows:F x =−E Âε =−E  ∂ξ(5.3)∂xUnlike the static case where the strain ε = ∂ξ/∂x and hence the force F x , remainsconstant throughout the bar, both the strain and F x vary in the dynamic case,and a net force acts on element dx. F x represents the internal force at x, and soF x + (∂ F x /∂x)dx constitutes the force at x + dx. The net force acting to theright becomes(dF x = F x − F x + ∂ F )x∂x dxCombining Equations (5.3) and (5.4) results in=− ∂ F xdx (5.4)∂xdF x = E  ∂2 ξdx (5.5)∂x2The volume of the element dx is given by Âdx, and therefore the mass is ρ Âdx,where ρ denotes the density (kg/m 3 ) of the bar material. Applying Newton’sequation of motion, with acceleration ∂ 2 ξ/∂t 2 , to Equation (5.5), we obtainρ Âdx ∂2 ξ∂t 2= E  ∂2 ξ∂x 2 dx