Creepy Oscillations (Part 3)

This is a very nice problem. At any general time $t$ , let the coordinates of point $B$ be $(R\cos{\phi},R\sin{\phi})$ . The total potential energy of the mass is:

$\mathcal{V} = \mathcal{V}_{\mathrm{spring}} + \mathcal{V}_{\mathrm{gravity}}$ $\mathcal{V} = \frac{K}{2} \left( 2R-\sqrt{(R + R\cos{\phi})^2 + R^2\sin^2{\phi}} \right)^2 + mgR\sin{\phi}$ $\implies \mathcal{V} = \sin{\phi} + 2\left(1 - \cos\left(\frac{\phi}{2}\right)\right)^2$

Now, the particle attains equilibrium at that angle where its potential energy is minimum. In other words, the equilibrium configuration satisfies:

$\frac{d \mathcal{V} }{d\phi} = \cos{\phi} - \sin{\phi} + 2\sin\left(\frac{\phi}{2}\right) = 0$

Let the equilibrium angle be $\phi_{\mathrm{eq}} = \alpha$ . So: $\cos{\alpha} - \sin{\alpha} + 2\sin\left(\frac{\alpha}{2}\right) = 0$

The above relation will be used later in the solution. Now, since the energy of the system is conserved the sum of its kinetic and potential energy is a constant. This means:

$\frac{\dot{\phi}^2}{2} + \sin{\phi} + 2\left(1 - \cos\left(\frac{\phi}{2}\right)\right)^2 = E$

Differentiating the above equation with respect to time and recognising that $\dot{\phi} \ne 0$ leads to the equation of motion which is:

$\ddot{\phi} + \cos{\phi} - \sin{\phi} + 2\sin\left(\frac{\phi}{2}\right) = 0$

Now, to analyse small oscillations about the equilibrium, we consider the mass to have displaced by a small angle $\delta$ from the equilibrium point. In other words, we analyse the system when:

$\phi = \alpha + \delta$

Where $\delta$ is small. Plugging this into the equation of motion and knowing that $\alpha$ is a constant leads to:

$\ddot{\delta} + \cos(\alpha + \delta) - \sin(\alpha + \delta) + 2\sin\left(\frac{\alpha + \delta}{2}\right) = 0$

Expanding all the terms using trigonometric identities and applying the small angle approximation $\sin{\delta} \approx \delta$ and $\cos{\delta}\approx 1$ leads to the following. The steps involved here are left out in this solution. The reader may attempt this him/her self.

$\ddot{\delta} + \cos{\alpha} - \sin{\alpha} + 2\sin\left(\frac{\alpha}{2}\right) + \delta\left( -\cos{\alpha} - \sin{\alpha} + \cos\left(\frac{\alpha}{2}\right) \right)=0$

Now, recognising that: $\cos{\alpha} - \sin{\alpha} + 2\sin\left(\frac{\alpha}{2}\right) = 0$

Transforms the linearised equation of motion to:

$\ddot{\delta} + \left( -\cos{\alpha} - \sin{\alpha} + \cos\left(\frac{\alpha}{2}\right) \right)\delta=0$

$\implies \omega^2 = -\cos{\alpha} - \sin{\alpha} + \cos\left(\frac{\alpha}{2}\right)$

The final step is to calculate $\alpha$ . I did this using a computer. I plotted the potential energy as a function of $\phi$ in the interval $-\frac{\pi}{2} \le \phi \le \frac{\pi}{2}$ . The minimum occurs at $\phi_{\mathrm{eq}} = \alpha = -1.3136 \ \mathrm{rad}$ .

Finally:

$\omega = \sqrt{-\cos{\alpha} - \sin{\alpha} + \cos\left(\frac{\alpha}{2}\right)}$ $\boxed{f = \frac{1}{2\pi} \sqrt{-\cos{\alpha} - \sin{\alpha} + \cos\left(\frac{\alpha}{2}\right)}}$