Two Gravitational Potentials

The equation of motion of the particle is $\ddot{\mathbf{r}} \; = \; \mathbf{R} - mg\binom{1}{1}$ where $\mathbf{r}$ is the position vector of the particle and $\mathbf{R}$ is the normal reaction from the wire. Now $\mathbf{r} = \binom{x}{x^2}$ , and hence $\ddot{\mathbf{r}} \; = \; \binom{1}{2x}\ddot{x} + \binom{0}{2}(\dot{x})^2$ Since the wire is smooth, $\mathbf{R}$ is normal to the tangent $\binom{1}{2x}$ of the curve. Thus, taking the scalar product with $\binom{1}{2x}$ , we see that $\begin{aligned} m(1 + 4x^2)\ddot{x} + 4mx(\dot{x})^2 & = \; -mg(1 + 2x) \\ \frac{d}{dx}\left[\tfrac12(1 + 4x^2)(\dot{x})^2 + g(x + x^2)\right] & = \; 0 \\ \tfrac12(1 + 4x^2)(\dot{x})^2 & = \; g(2 - x - x^2) \; = \; g(1-x)(2+x) \\ (\dot{x})^2 & = \; \frac{2g(1 - x)(2+x)}{1 +4x^2} \end{aligned}$ and so the particle oscillates between $x=1$ and $x=-2$ , and the period of oscillation is $T \; = \; 2\int_{-2}^1 \sqrt{\frac{1 + 4x^2}{2g(1 - x)(2 + x)}}\,dx \; = \; \int_{-2}^1 \sqrt{\frac{1 + 4x^2}{5(1-x)(2+x)}}\,dx \; = \; \boxed{3.28252}$

Similar to the solution posted by @Mark Hennings . The only difference here is that I avoid deriving the equation of motion. Since there is no dissipative element present in the scenario, energy is conserved.

At ant instant of time $t$ :

$PE_{initial} + KE_{initial} = PE_{present} + KE_{present}$

$2mg + 0 = mg(x+x^2) + \frac{1}{2}m\dot{x}^2(1+4x^2)$ $20 = 10(x+x^2) + \frac{1}{2}\dot{x}^2(1+4x^2)$

Rearranging and observing that as $t$ increases, $x$ reduces, gives:

$\dot{x} = -\sqrt{\frac{2(20 - 10(x+x^2))}{1+4x^2}} = f(x)$

Observing that the derivative of $x$ becomes zero at $x = 1$ and $x=-2$ , we can say that the bead moves between these two points at half a time period.

Therefore:

$\frac{T}{2} = \int_{-2}^{1}\frac{dx}{f(x)} \implies \boxed{T = 3.2825}$