First you decided on the independent degrees of freedom. I choose the center of the plank and I track the translation $x$ and rotation $\theta$ from the equilibrium conditions.
The displacements of each spring are:
$$\begin{align}
x_1 & = x - \frac{\ell}{2} \theta \\
x_2 & = x + \frac{\ell}{2} \theta
\end{align} $$
Each spring force is:
$$\left. \begin{align}
F_1 & = a m g - k x_1 \\
F_2 & = b m g - k x_2
\end{align} \right\}
\begin{aligned}
F_1 & = a m g + \frac{\ell}{2} k \theta - k x \\
F_2 & = b m g - \frac{\ell}{2} k \theta - k x
\end{aligned}$$
The displacement of the center of mass is $x_C = x + \frac{\ell}{2} (b-a) \theta$ and hence the center of mass acceleration (needed for the equations of motion) is $ \ddot{x}_C = \ddot{x} + \frac{\ell}{2} (b-a) \ddot{ \theta}$.
The EOM are:
$$\begin{align}
m \ddot{x}_C & = F_1 + F_2 - m g \\
I_C \ddot{\theta} & = a \ell F_1 - b \ell F_2
\end{align} $$
where $I_C$ is the mass moment of inertia about the center of mass. The above is solved by $x(t) = X \sin \omega t$ and $\theta(t) = \Theta \sin \omega t$. This produces the system of equations of
$$ \begin{align}
2 k X & = m \omega^2 \left( X + \Theta \frac{\ell}{2} (b-a) \right) \\
k X \ell (a-b) + k \frac{\ell^2}{2} \Theta (a+b) & = I_C \Theta \omega^2
\end{align} $$
This has two solutions for frequency $\omega_T$ and $\omega_R$ for translational and rotational modes of vibration.
The two degrees of freedom are coupled with
$$ -\frac{X}{\Theta} = \frac{\frac{\ell}{2} m \omega^2 (a-b)}{ (2k-m \omega^2)}$$
The left hand side of this equation is the center of rotation position (distance) from the center of the plank. Pure translation occurs when $\omega^2 = 2 \frac{k}{m}$ and pure rotation when $a=b$.
$$\begin{align}
\omega^2_T & = \frac{k}{m} \left(1+ \frac{m \ell^2 (1-2 a b)}{2 I_C} + \sqrt{ 1 + \left( \frac{m \ell^2 (1-2 a b)}{2 I_C} \right)^2 - \frac{2 a b m \ell^2}{I_C} } \right) \\
\omega^2_R & = \frac{k}{m} \left(1+ \frac{m \ell^2 (1-2 a b)}{2 I_C} - \sqrt{ 1 + \left( \frac{m \ell^2 (1-2 a b)}{2 I_C} \right)^2 - \frac{2 a b m \ell^2}{I_C} } \right)
\end{align} $$
Edit 1
To estimate $a$ $b$ from the resulting motion, maybe you can solve the equations of motion using the normalized frequency $n^2 = \frac{\omega^2}{2 k/m}$ and center of rotation location $r=-\frac{X}{\Theta}$.
$$ \begin{align}
a &= \frac{2 I_C n^2}{m \ell^2} + \frac{ r (1-n^2) (2 r+\ell)}{n^2 \ell^2} \\
b &= \frac{2 I_C n^2}{m \ell^2} + \frac{ r (1-n^2) (2 r-\ell)}{n^2 \ell^2}
\end{align} $$
