small amplitude oscillation of rotating system.

Question

I've solved the euler-lagrange equation for a frictionless bead on circular vertical loop of radius a where the loop is rotating at $\Omega$ to get the equation of motion for the bead as $$\frac{d^{2}\theta}{dt^2} = \sin\theta ( \Omega^{2}\cos\theta-\frac{g}{a})$$ when I look at my free body diagram the system is obviously stable about $\theta = 0$ but when I use small angle approximations used the conditions that $\Omega^{2}>\frac{g}{a}$ I get $$\frac{d^{2}\theta}{dt^2} = \Omega^{2}\theta$$ this is an exponential decay and not oscillatory and my problem sheet asks me to find the period and so I've evidently made a mistake in my approximations as my equation of motion is in agreement with the answers given.

Answer 1

Let $\theta = \theta_0 + \psi$. Then the equation of motion beocomes $$\frac{d^{2}(\theta_0 + \psi)}{dt^2} = \sin(\theta_0+\psi) \left( \Omega^{2}\cos(\theta_0+\psi)-\frac{g}{a}\right).$$ For small $\psi$ this is easily seen to be same as $$\frac{d^{2} \psi}{dt^2} =-\Omega^{2} \sin^2\theta_0 \psi$$ after using $\cos \theta_0=\frac{g}{\Omega^2a}$ and expanding the cosine. Therefore the oscillation freq is $\Omega \sin\theta_0.$