 Solution

Write $A=\pi a^2$ for the cross-sectional area, and use $z$ consistently for the longitudinal coordinate: the paper's later $x$ is the same coordinate. The canonical momentum is a density per reference length,The Legendre transform gives the Hamiltonian density in its proper canonical variables:The Euler-Lagrange equation isFor a twice-differentiable wave packet $\psi =f(z-ct)$, the derivatives are $\ddot{\psi }=c^2{f}^{\prime \prime }$ and ${\psi }^{\prime \prime }=f^{\prime \prime }$. Thus the nontrivial packet solves the wave equation whenThe negative sign gives the opposite propagation direction. Here $\Pi$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, $\mathcal{H}$ has units $\mathrm{J}{\mathrm{m}}^{-1}$, and $c$ has units $\mathrm{m}{\mathrm{s}}^{-1}$. The result follows from the supplied axial stiffness in the Lagrangian. For a freely contracting thin elastic rod the relevant longitudinal modulus is normally Young's modulus, rather than the bulk modulus; the question's given Lagrangian fixes which coefficient is being used.