<s>[INST] Oscillation of non-uniform plank on parallel springs A plank of length $l$ and mass $m$ is placed on two parallel springs, each with spring constant $k$ and equidistant from the plank's horizontal center of gravity. When the plank is displaced from it's equilibrium position, it's expected to oscillate at this frequency:
$$ f = \frac{1}{2\pi}\sqrt{\frac{2k}{m}} $$ 
Now, if the plank isn't of uniform density, such that at equilibrium, the force acting on spring 1 is $amg$ and that on spring 2 is $bmg$ where $a + b = 1$. At what frequency will the plank oscillate at when it's displaced from its equilibrium position?
Is it possible to obtain an estimate of its horizontal center of gravity at equilibrium (or rather, constants $a$, $b$) by observing the oscillations, say by taking a fourier transform of the plank's oscillation waveform?
 [/INST] 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 $r=-\frac{X}{\Theta}$.
$$ \begin{align}
  a &= \frac{2 I_C n^2}{m \ell^2} + \frac{ r (1-n^2) (2 r+\ell)}{n^2 \ell^2} \\
  b &= \frac{2 I_C n^2}{m \ell^2} + \frac{ r (1-n^2) (2 r-\ell)}{n^2 \ell^2} 
\end{align} $$
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