How to calculate the period of the movement from a potential? I have an assignment, where I have an object moving in 1-D with a given mass  and energy, and the potential V(x), and I'm supposed to calculate the period of the movement as a function of the energy 
$$
V(x)=\begin{cases}\infty &x < -a \\
                  0      &-a < x < 0\\
                  \alpha x^2 & x>0
\end{cases}
$$
Should I find 3 Lagrangians for the 3 separate parts of the potential? And then how would I come to the period of the movement? Thanks in advance!!
 A: It seems, based on the comments above, that you have figured it out. Just for closure, I am writing the steps out.
If you had just a parabolic potential well, $V(x) = \alpha x^2$, you could get the period quite easily - for a given mass $m$, the frequency would be
$$\omega = \sqrt{\frac{2\alpha}{m}}\\
T = \frac{2\pi}{\omega} = \pi \sqrt{\frac{2m}{\alpha}}$$
For half the parabola, the time from the bottom to the side and back will be half this:
$$T_1 = \frac12 T = \pi \sqrt{\frac{m}{2\alpha}}$$
The mass will slide along the bottom of the well with a constant velocity given by the energy (which is all kinetic at this point):
$$v = \sqrt{\frac{2E}{m}}$$
So the time taken to cover the distance $a$ (there and back) is
$$T_2 = \frac{2a}{v} = a\sqrt{\frac{2m}{E}}$$ 
When the mass gets to the infinite potential wall it will just turn around.
Summing these two times give you the period of the oscillation:
$$T = \pi \sqrt{\frac{m}{2\alpha}} + a\sqrt{\frac{2m}{E}}$$
As expected, one term depends on how steep the potential well is on one side, and the other term decreases as the energy goes up. And when the width of the well increases, the period increases as well.
A: You can calculate the action integral as a function of energy 
$$
J(E) =  \oint p_x dx =  2\sqrt {2m} \int_{x_0(E)}^{x_1(E)} \sqrt {E-V(x)}dx
$$
where $x_0$ and $x_1$ are the turning points. (In your case: $x_0 = -a$ and $x_1 = \sqrt{E/\alpha}$). The Period is then  given by the derivative 
$$
T(E) =  dJ/dE
$$
