The frequency is a function of the dimensions of the bar and its Young's modulus.
You need to know what mode of oscillation you are exciting in your bar - there is a hug difference between the flexural and longitudinal modes.
If the rod is bending, you can find the formulas here. The derivation goes on and on... but you should be able to use the formula on the first page (for free-free):
$$f = \frac{1}{2\pi}\left(\frac{22.373}{L^2}\right)\sqrt{\frac{EI}{\rho}}$$
In this formula, $I$ is the second moment of area of the rod - see the wiki article for an explanation and to find the appropriate value for the shape of your rod.
If you have a higher mode, you can find the position of two fixed nodes and use the fixed-fixed equation instead.
And if you have longitudinal vibration, the answer is much simpler - you just have to look at the transit time of the sound wave from end to end. One round trip corresponds to the fundamental frequency, so
$$f = \frac{v}{2L} = \sqrt{\frac{E}{4L^2\rho}}\\
E = \rho\; \left(\;2\;L\;f\;\right)^2$$
$A \approx \frac{W_0*L_0}{L}*\frac{T_0*L_0}{L}=\frac{W_0*T_0*L_0^2}{L^2}$
If my interpretation is correct, you are assuming that $W*L=W_0*L_0$ and $T*L=T_0*L_0$
That would make the volume: $W_0*T_0*L_0^2/L$, which decreases when you stretch the material. For small strain, the Poisson ratio would approach 1. Poisson ratio should be between -1 and 0.5 for a stable, isotropic, linear elastic material. I don't know what the material is you're testing, but..
$A \approx \frac{A_0*L_0}{L}$ seems more suitable imo.
Best Answer
There's no general mathematical relation between stiffness (as parameterized by an elastic modulus such as Young's modulus) and density for all materials, but a relationship can be defined for an ideal gas, and a general trend exists for condensed matter.
For fluids such as gases and liquids, Young's modulus is zero; you won't encounter any resistance if you slowly pull on these materials uniaxially. However, the bulk modulus (i.e., the resistance one encounters when trying to compress these materials using pressure) is not zero; for an ideal gas, it is exactly equal to the pressure. One can relate this relation to the density $m/V$ using the ideal gas law $PV=nRT$ and the molecular weight of the gas.
For condensed matter, the stiffness generally scales up with the density:
The reason is that both properties depend on the pair potential between atoms; if the atoms are strongly bonded, then their spacing is generally small; thus, the density is high. In addition, the curvature of the pair potential at the equilibrium position is large, which is exactly equivalent to saying that the material is stiff:
(A third implication to strong bonding is that the pair potential dips deep and that the material is refractory, i.e., it has a high melting temperature):
An active area of research is to precisely define the pair potential in condensed matter to theoretically predict both stiffness and density; this is something you might pursue in an academic or industry environment.