# Is disturbance propagation velocity equal to wave velocity in a solid body?

How do the disturbance propagation velocity and wave propagation velocity relate to each other?

To explain my question in details I will describe the following situation from the theory of acoustic waves in solid bodies.

A one-dimensional chain of $N$ identical atoms at distances $a$ from each other is considered, so that total chain length $L_c = Na$, allowed set of $N$ wave numbers $k_i=f(L_c,i)$, where i=1,2,..N, corresponding allowed set of $N$ angular frequencies $\omega_i = f(k_i)$ and $N$ wave lengths $L_i = f(k_i)$.

Maximal wave length is $L_{max} = Na = L_c$ in accordance with Born-Karman boundary conditions. Thus we can calculate corresponding to $L_{max}$ minimal possible natural frequency of the chain $\omega_{ min}$.

Phase velocity of the acoustic wave is ${v_p}_{(\omega = 0)}$ at $\omega = 0$. Group velocity is ${v_g}_{(\omega = 0)}$ at $\omega = 0$.

For any other allowed wave number $k_i$ and corresponding frequency $\omega_i$ we obtain corresponding smaller velocities ${v_g}_{(\omega>0)}<{v_g}_{(\omega = 0)}$ and ${v_p}_{(\omega>0)}<{v_p}_{(\omega = 0)}$.

That means that also for an oscillation at the smallest possible natural frequency $\omega_{min}$ (i.e. at wave length $L_{max}$) we will have ${v_g}_{(\omega = \omega_{min})}<{v_g}_{(\omega = 0)}$ and ${v_p}_{(\omega = \omega_{min})}<{v_p}_{(\omega = 0)}$.

Now let's consider the following situation. An initial disturbance propagates in the chain with some velocity $v_{dist}$.

While it propagates, we still do not know the length of the chain $L_c$. Thus we cannot calculate $L_{max}$ and $\omega_{ min}$, and also ${v_g}_{(\omega = \omega_{min})}$ and ${v_p}_{(\omega = \omega_{min})}$ are thus unknown.

Then the wave reaches the boundary. Now we know the length of the chain $L_c\Rightarrow$ can calculate $L_{max}$, $\omega_{min}$, ${v_g}_{(\omega = \omega_{min})}$, ${v_p}_{(\omega = \omega_{min})}$.

Now the wave is reflected from the boundary. The reflected wave propagates in opposite direction at velocity $v_{refl}$.

If the energy of the initial disturbance was enough to excite vibration at the minimal natural frequency $\omega_{ min}$, then vibration of the chain takes place with corresponding phase ${v_p}_{(\omega = \omega_{min})}$ and group ${v_g}_{(\omega = \omega_{min})}$ velocities.

The question is if the initial disturbance propagation velocity $v_{dist}$ is equal to phase velocity ${v_p}_{(\omega = 0)}$ corresponding to frequency $\omega = 0$ or to phase velocity ${v_p}_{(\omega = \omega_{min})}$ corresponding to frequency $\omega = \omega_{min}$, or is there another relation between $v_{dist}$, $v_{refl}$ and ${v_p}_{(\omega)}$ or ${v_g}_{(\omega)}$ ?

Thanks very-very much for all answers!