I have to find $\omega$ for this system using the forces. I have a disc $radius = R, mass = M$
By using $F=ma$, I get $mg\sin\theta - kx = m\ddot{x}$
then $$-g\sin\theta + \frac{kx}{m} + \ddot{x} = 0$$
$\omega = \sqrt{\frac{k}{m}}$
But the correct answer is $\omega = \sqrt{\frac{2k}{3m}}$
I don't see where my errors are.
There's rolling going on here.
Assuming no slipping, you have $\tau = R mg\sin\theta\implies I\alpha = fR$ where $f$ is friction. Since $a=\alpha R$ then you have $$C mR^2 \dfrac \alpha R = fR\implies Cma=f$$ where $I=CmR^2$.
Then by Newton's second Law, $$ma=mg\sin\theta-kx-f\implies ma=mg\sin\theta-kx-Cma$$
Then use the chain rule $a=\dfrac {dv}{dt}=\dfrac{dv}{dx}\dfrac{dx}{dt}=\dfrac{dv}{dx}v$, get the equation of motion for $x(t)$ and identify your $\omega$.
