Question about coupled pendula

Consider two pendula, as pictured below, consisting of point masses $m_1, m_2$ and massless rods of length $L_1, L_2$, with $L_2 = 2 L_1$ and $m_1 = 2 m_2$. The two pendula are connected by a spring of constant $s$ as pictured, such that its natural length coincides with the distance between the two rods when they are parallel. For small pertrubations from the equilibrium point (such that $sin theta _1 = theta _1$, $sin theta _2 = theta _2$), determine the frequencies $omega _1 , omega_2$ of the normal modes of oscillation as a function of $omega _0 = frac{g}{L_1} = frac{s}{m_1}$.

I have been asked this question as part of a course assignment. I came up with a straightforward solution, and ended up finding that $omega _{1,2}^2 = frac{3 pm sqrt{3}}{2} omega _0 ^2$, but a number of other students have said that the answer is $omega _1 = omega_2 = omega_0$. Is there something fundamentally wrong with my reasoning? I suppose if there is a mistake, it lies in the differential equations I end up with, because the rest is just standard procedures.

Specifically, I simply started by using the fact that $sum tau = I ddot{θ}$ for each of the systems $m_1 – L_1$, $m_2 – L_2$.

The only forces generating torque on each of the systems are the weights of the masses and the force of the spring.

Thus, for the first system we get (letting $x_1 = theta _1 L_1$ and similarly for $x_2$):
$-m_1 g x_1 + s(x_2 frac{L_1}{L_2} – x_1)L_1 cos theta _1 = m_1 L_1 ^2 frac{ddot{x}_1}{L_1} Leftrightarrow$
$- frac{g}{L_1} x_1 + frac{s}{m_1}(x_2 frac{L_1}{L_2} – x_1) = ddot{x}_1 Leftrightarrow – omega _0 ^2 x_1 + omega _0 ^2 (frac{x_2}{2} – x_1) = ddot{x} _1$

Similarly, for the second mass we get:
$-m_2 g x_2 + s(x_1 – x_2 frac{L_1}{L_2})L_1 cos theta _2 = m_2 L_2 ^2 frac{ddot{x} _2}{L_2} Leftrightarrow$
$- frac{g}{2L_1} x_2 + frac{s}{2 m_2} (x_1 – x_2 frac{L_1}{L_2}) = ddot{x} _2 Leftrightarrow$
$- frac{omega _0 ^2}{2} x_2 + omega _0 ^2 (x_1 – frac{x_2}{2}) = ddot{x} _2$

After that, the problem boils down to substituting $x_1 = A sin( omega t)$, $x_2 = B sin(omega t)$ and making sure the determinant of the system is zero, so I doubt there are any mistakes there.