For most wire shapes, solving for the motion of the bead will require numerical methods. In any case, the specific time dependence of the bead's motion will depend on the shape of the wire. I wondered what wire shape would cause the bead's

*vertical*motion to be the same as if the bead were simply bobbing on a spring.

I did a minute or two of searching and didn't see the problem solved anywhere online, so I worked it out. Here's an animation of the solution:

At all times, the height of the "skateboarder" is the same as the height of the mass that bobs on the spring. (This is called "simple harmonic motion.")

The shape of the wire is given by the graph of \(x = \pm p(y)\), where \[p(y) = \sqrt{y_0\,y-y^2}+y_0\sin^{-1}\!\!\sqrt{y/y_0}\,.\] Here, the constant \(y_0 =\frac{2g}{\omega^2}\) is the height of the half-pipe, which is also double the amplitude of the spring motion; and \(\omega\) is the angular frequency of the simple harmonic motion you want to produce. (There are also more complicated solutions that don't turn vertical at the ends.)

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To find this solution, I worked in the first quadrant with the wire given by \(x = p(y)\) and \(p(y)\) unknown. Let the bead begin from rest at \((x,y) = (p(y_0), y_0)\). From energy conservation,

\[\frac{1}{2}mv^2 + mgy = mgy_0\,.\]

Express the speed as \(v^2 = ((p^\prime)^2 + 1)\dot{y}^2\). Now if we specify the vertical motion as \(y(t) = y_0\cos^2\left(\frac{\omega t}{2}\right)\), then we can express the time derivative in terms of the coordinate itself: \(\dot{y}^2 = \omega^2(y_0\,y - y^2)\). This eventually leads to a separable first-order nonlinear differential equation for \(p(y)\), \[\frac{dp}{dy} = \sqrt{\gamma\frac{y_0}{y}-1}\,\] where \(\gamma = \frac{2g}{\omega^2y_0}\). Note that \(\gamma \geq 1\), because the vertical acceleration can't possibly exceed \(g\), so that \(\frac{1}{2}\omega^2y_0 \leq g\) or in other words \(\gamma \geq 1\). Also note that \(\frac{dp}{dy}\rightarrow\infty\) as \(y\rightarrow 0\), which makes sense because the half-pipe must be horizontal where the vertical motion reverses direction.

Integrating with the boundary condition \(p(0) = 0\) so that the half-pipe goes through the origin, we get, with the help of Dwight 194.21 and 192.11, \[p = \int_0^y{\sqrt{\gamma \frac{y_0}{\xi}-1}\,d\xi} = \sqrt{\gamma\,y\,y_0 - y^2} + \gamma y_0 \left(\frac{\pi}{2}-\tan^{-1}\!\!\sqrt{\gamma\frac{y_0}{y}-1}\right)\,.\] If we take \(y_0 = \frac{2g}{\omega^2}\) then \(\gamma = 1\) and this simplifies to \[p = \sqrt{y\,y_0-y^2} + y_0\left(\frac{\pi}{2}-\tan^{-1}\!\!\sqrt{\frac{y_0}{y}-1}\right)\,.\] The arctangent can be expressed more simply as \(\cos^{-1}\!\!\sqrt{\frac{y}{y_0}}\) (to see this, draw a right triangle with legs \(y\) and \(\sqrt{y\,y_0-y^2}\)), whereupon we also notice that \(\frac{\pi}{2} - \cos^{-1}(\cdots) = \sin^{-1}(\cdots)\), so in the end \[p(y) = \sqrt{y_0\,y-y^2}+y_0\sin^{-1}\!\!\sqrt{y/y_0}\] as above.

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