= 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,
$$
\boxed{\Pi(z,t)=\frac{\partial\mathcal L}{\partial\dot\psi}=A\rho\dot\psi}.
$$
The <Legendre transform> gives the <Hamiltonian density> in its proper canonical variables:
$$
\boxed{\mathcal H(\psi,\Pi)=\Pi\dot\psi-\mathcal L=\frac{\Pi^2}{2A\rho}+\frac{AK}{2}(\psi')^2}.
$$
The <Euler-Lagrange equation> is
$$
\partial_t(A\rho\dot\psi)+\partial_z(-AK\psi')=0,
\qquad \boxed{\ddot\psi=\frac K\rho\psi''}.
$$
For a twice-differentiable <wave packet> $\psi=f(z-ct)$, the derivatives are $\ddot\psi=c^2f''$ and $\psi''=f''$. Thus the nontrivial packet solves the <wave equation> when
$$
\boxed{c=\sqrt{K/\rho}}.
$$
The negative sign gives the opposite propagation direction. Here $\Pi$ has units $\mathrm{kg\,s^{-1}}$, $\mathcal H$ has units $\mathrm{J\,m^{-1}}$, and $c$ has units $\mathrm{m\,s^{-1}}$. 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.
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