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We know that sound waves are longitudinal waves and that the propagation of longitudinal waves depends on the bulk modulus $B$ (volumetric elasticity). Why, then, does the speed of sound in a solid depend on the Young’s modulus $Y$ of the medium and the density, $$v = \sqrt{(Y/\text{density of the medium})}?$$ The form should be the same as for liquids and gases: $$v = \sqrt{(B/\text{density of the medium})}.$$

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There is more than one speed of sound in a solid. In an isotropic solid the speed depend on the two Lame coeffcients $\lambda$ and $\mu$. The speed of longitudinal "p" waves in a bulk solid is $$ c_{p}= \sqrt{\frac{E(1-\nu)}{\rho(1+\nu)(1-2\nu)}} $$ where $$ E=\mu\left(\frac{3\lambda+2\mu}{\lambda+\mu}\right) $$ is Young's modulus and $$ \nu=\frac 12 \frac{\lambda}{\lambda+\mu} $$ is Poisson's ratio. The speed of transverse "s" waves is $$ c_s= \sqrt{\frac{\mu}{\rho}}. $$ The simple formula $$ c_{rod}= \sqrt{\frac {E}{\rho}} $$ applies only to longitudinal waves in a long thin rod.

The physical reason for the differences in the longitudinal speeds is that in a long thin rod, the material is free to contract laterally as the wave propagates, but in the infinite medium it is not.

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  • $\begingroup$ Is the short answer that the Poisson's ratio of a fluid $\nu = 0$ which makes the first expression $c_p = c_{rod}$. $\endgroup$ Jan 22, 2021 at 1:06
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    $\begingroup$ No. It is the shear modulus $\mu$ of a fluid that is zero. The fluid cannot sustain shear waves. Then Poisson's ratio is $\nu=1/2$ and Young's modulus is zero. This is because you can stretch a fluid with no effort as long as you allow it to shrink transversely so that its volume stays fixed. In a fluid it is the bulk modulus $\kappa$ that counts, and with $\mu=0$ we have $\kappa=\lambda$ and $c_{fluid}= \sqrt{\kappa/\rho}$. $\endgroup$
    – mike stone
    Jan 22, 2021 at 18:54
  • $\begingroup$ Nice. The above comment completes the answer as it bridges liquids to solids. $\endgroup$ Jan 23, 2021 at 0:54

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