to Eqs.(\[Shear-1\]),(\[Rotations23-1\]) and (\[VolumeVariation-1\]) are respectively: $$\omega(k)=\sqrt{\omega_{s}^{2}+k^{2}c_{g}^{2}},\qquad\omega(k)=\sqrt{\omega_{r}^{2}+k^{2}c_{g}^{2}},\qquad\omega(k)=\sqrt{\omega_{s}^{2}+k^{2}c_{g}^{2}}.\label{Eigen1-1}$$

\[NumericalSim\]The relaxed micromorphic model: numerical results 
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In this section, following Mindlin [@Mindlin; @EringenBook], we will show the dispersion relations $\omega=\omega(k)$ associated to the considered relaxed micromorphic model. The analysis of dispersion relations in [@Mindlin; @EringenBook] was of qualitative nature, due to the huge number of constitutive parameters (18 elastic coefficients) which are assumed to be non-vanishing and to computational difficulties. On the other hand, thanks to the relaxed constitutive assumption (\[KinPot\]), we are able to clearly show precise dispersion relations for the considered cases and to associate to few constitutive parameters the main mechanisms associated to wave propagation in micromorphic media. As principal obtained result, we will clearly point out that band-gaps can be forecast only when considering a non-vanishing Cosserat couple modulus $\mu_{c}$ in our relaxed micromorphic model. This parameter is associated to the relative rotation of the microstructure with respect to the continuum matrix.

             Parameter                   Value              Unit
  ------------------------------- -------------------- ---------------
             $\mu_{e}$                   $200$              $MPa$
      $\lambda_{e}=2\mu_{e}$              400               $MPa$
       $\mu_{c}=2.2\mu_{e}$              $ $440             $MPa$
             $\mu_{h}$                    100               $MPa$
           $\lambda_{h}$                 $100$              $MPa$
             $L_{c}\ $                    $3$               $mm$
             $L_{g}\ $                    $3$               $mm$
   $\alpha_{c}=\mu_{e}L_{c}^{2}$   $1.8\times10^{-3}$   $MPa\: m^{2}$
   $\alpha_{g}=\mu_{e}L_{g}^{2}$   $1.8\times10^{-3}$   $MPa\: m^{2}$
              $\rho$                     $2000$          $Kg/m^{3}$
              $\rho$’                    $2500$          $Kg/m^{3}$
                $d$                       $2$               