There is a relation which holds in solids: $$v_{t}=\sqrt{\frac{E}{\rho}}$$ where $v_t$ is the the velocity of sound (transversal elastic waves), and $E$ and $\rho$ are the Young's modulus and density of the medium respectively.
How would I derive this or can someone give me a reference where the derivation is obtained? (I would guess somehow connecting the wave equation and Hooke's law maybe?)
All you need is the Navier's equation of motion (you can consult it in any book of elasticity)
$$ \rho \frac{\partial^2 w}{ \partial t^2 } = \mu \nabla^2w + (\lambda + \mu)\nabla\nabla.w $$
Naturally, you can decompose a wave $ w$ in a transversal and a longitudinal part:
$$ w = w_L + w_T $$
with the following properties :
$$ \nabla \times w_L = 0$$ $$ \nabla .w_T = 0 $$
If we focused only in the transversal part, retaking the Navier equation we have:
$$ \rho \frac{\partial^2 w_T}{ \partial t^2 } = \mu \nabla^2 w_T $$
Now you can remember that the general wave equation is
$$ \frac{\partial^2 u}{ \partial t^2 } = c^2 \nabla^2 u $$
where c is the phase velocity of the wave $ u $
So, that's it
$$ v_T = \sqrt \frac \mu\rho$$
I hope you help
J.
To add to J0KerSpin's good answer, $\mu$ is not the elastic modulus, but rather the Lamé parameter and is equivalent to the shear modulus, $G$, rather than the Young's modulus for linear isotropic solids.
This question may be in regards to using ultrasound to measure the elastic modulus. Usually, ultrasound measurements apply longitudinal waves, $w_L$ rather than transverse waves, $w_T$, in order to measure the longitudinal velocity through the sample. Going back to the Navier-Cauchy equation, we have:
\begin{equation} \rho \frac{\partial^2w}{\partial t^2} = \mu \nabla^2 w + (\lambda + \mu) \nabla (\nabla \cdot w) \end{equation}
We use the vector identity:
\begin{equation} \nabla(\nabla \cdot w) = \nabla^2w + \nabla \times w \end{equation}
With longitudinal waves, we have the property: $$ \nabla \times w_L = 0 $$ Therefore, applying these equations to the Navier-Cauchy equation, we get:
$$ \frac{\rho}{2\mu+\lambda} \frac{\partial^2w_L}{\partial t^2} = \nabla^2 w_L $$
This is the wave equation:
$$ \frac{\partial^2w_L}{\partial t^2} = v_L \nabla^2 w_L $$
Where $v_L$ is the longitudinal velocity. Thus we get that the longitudinal velocity is:
$$ v_L = \sqrt{\frac{2\mu + \lambda}{\rho}} $$
[1] shows that the Lamé parameters can be written in terms of the Poisson's ratio $\nu$ and the Young's modulus, $E$ for linear isotropic systems:
$$ \lambda = \frac{E\nu}{(1+\nu) (1-2\nu)} \\ $$
$$ \nu = \frac{E}{2(1+\nu)} $$
Thus we end up with the relationship between the longitudinal velocity, the Young's modulus, and the Poisson's ratio:
$$ v_L = \sqrt{\frac{E (1-\nu)}{\rho(1+\nu)(1-2\nu)}} $$
This result does NOT agree with the commonly cited result that was presented in the question of this post, $v_L = \sqrt{E/\rho}$. However, I have seen this equation being used in the literature, see [2] for example, and some have argued it is the more correct equation to use when trying to extract the Young's modulus from ultrasound velocity measurements, see [3].
Best, -Jacob
References: [1] https://web.mit.edu/16.20/homepage/3_Constitutive/Constitutive_files/module_3_with_solutions.pdf
[2] https://pubs.acs.org/doi/10.1021/acsami.6b06612
[3] http://koski.ucdavis.edu/BRILLOUIN/CRYSTALS/LongitudinalModulus.html
