Sine Wave Sliding

See the energy-based solution from @Brian Moehring , in which he goes into detail about the constraints on the bead's motion, and about the computation of the final answer. The only thing I will add here is an alternate Lagrangian solution for the second-derivative of $x$ . The resulting expression can be numerically integrated to get the solution. This results in a list of $(t,x,y)$ coordinates which describes the particle trajectory.

Positions and velocities:
$\large{x = x \\ y = sin(x) \\ \dot{x} = \dot{x} \\ \dot{y} = cos(x) \, \dot{x}}$

Kinetic Energy:

$\large{E = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) = \frac{1}{2} m \dot{x}^2 [1 + cos^2 (x)] }$

Potential Energy:

$\large{U = m g \, y = m g \, sin(x) }$

Lagrangian:

$\large{L = E - U = \frac{1}{2} m \dot{x}^2 [1 + cos^2 (x)] - m g \, sin(x) }$

Equation of Motion:

$\large{\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{x}}} = \frac{\partial{L}}{\partial{x}}}$

Inner left side:

$\large{ \frac{\partial{L}}{\partial{\dot{x}}} = m \dot{x} [1 + cos^2 (x)] }$

Left side:

$\large{\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{x}}} = m \dot{x} [-2 \, sin(x) \, cos(x) \, \dot{x}] + m \ddot{x} [1 + cos^2 (x)] }$

Right side:

$\large{\frac{\partial{L}}{\partial{x}} = \frac{1}{2} m \dot{x}^2 [-2 \, sin(x) \, cos(x)] - m g \, cos(x) }$

Equating left and right sides:

$\large{ -2 \dot{x}^2 \, sin(x) \, cos(x) + \ddot{x} [1 + cos^2 (x)] = - \dot{x}^2 sin(x) \, cos(x) - g \, cos(x) \\ \ddot{x} [1 + cos^2 (x)] = \dot{x}^2 sin(x) \, cos(x) - g \, cos(x) }$

Final result:

$\large{\ddot{x} = \frac{\dot{x}^2 sin(x) \, cos(x) - g \, cos(x)}{1 + cos^2 (x) }}$

By the conservation of energy, $\dot{x}^2 + \dot{y}^2 + 2gy = \dot{x}_0^2 + \dot{y}_0^2 = \dot{x}_0^2 + (y'(x_0))^2\dot{x}_0^2 = 2\dot{x}_0^2 = 2(0.95^2 g)$ Here, we note that this gives $y=0.95^2 < 1$ when the velocity is zero, so the bead won't escape from between the two peaks at $x=-\frac{3\pi}{2},\frac{\pi}{2}$ .

Between these two peaks, $x = \sin^{-1}(y)$ or $x = -\frac{\pi}{2} - \sin^{-1}(y)$ , so we have $\dot{x} = \pm \frac{1}{\sqrt{1-y^2}}\dot{y}$ , and the above equation becomes $\frac{1}{1-y^2}\dot{y}^2 + \dot{y}^2 + 2gy = 2(0.95^2 g) \implies \left(\sqrt{\frac{2-y^2}{(1-y^2)(0.95^2-y)}}\right) \dot{y} = \pm \sqrt{2g}$

Now, if the position and velocity first match their initial values at time $t = T$ , then $T$ will be the time period for the movement of the bead. By the symmetry in the graph about the low point at $x=-\frac{\pi}{2}$ and the symmetry in the time axis, we conclude that it takes $\frac{T}{4}$ for the bead to move from the low point at $y=-1$ to its highest point at $y=0.95^2$ . If we suppose that the bead is at $y=-1$ at $t=t_{low}$ , then we can integrate the equation we found: $\int_{-1}^{0.95^2} \sqrt{\frac{2-y^2}{(1-y^2)(0.95^2-y)}} \, dy = \int_{t_{low}}^{t_{low} + \frac{T}{4}} \sqrt{2g} \,dt = \frac{T\sqrt{2g}}{4}$ and solving for $T$ gives $T = \frac{4}{\sqrt{2g}} \int_{-1}^{0.95^2} \sqrt{\frac{2-y^2}{(1-y^2)(0.95^2-y)}} \, dy \approx \boxed{4.5035661875494}$