A graph of $U(x)$ with a horizontal line at the total energy $E$ tells you nearly everything about one-dimensional motion without solving any equation. It combines Force from Potential Energy (force is minus the slope) with Concavity and the Second Derivative Test (the shape of a graph near a flat point).
Key idea
With total energy $E$, the kinetic energy is $K = E - U(x)$, the gap between the line and the curve.
- Allowed regions: where $U(x) \le E$, since $K$ can't be negative.
- Turning points: where $U(x) = E$; the object stops and reverses.
- Equilibria: where $dU/dx = 0$. A minimum ($d^2U/dx^2 > 0$) is stable; a maximum ($d^2U/dx^2 < 0$) is unstable.
Why it works
Energy conservation says $\tfrac{1}{2}mv^2 + U(x) = E$, so $v = \pm\sqrt{2(E - U)/m}$: fastest where $U$ is lowest, zero where $U = E$. At a minimum of $U$, a small nudge either way puts the object uphill, and the force $-dU/dx$ points back toward the minimum: it rocks back and forth there (stable). At a maximum, a nudge puts it downhill, and the force pushes it farther away (unstable).
Worked example: Equilibria of a double well
Find and classify the equilibria of $U(x) = x^4 - 8x^2$ (J, m).
- $\dfrac{dU}{dx} = 4x^3 - 16x = 4x(x^2 - 4) = 0$ at $x = 0, \pm 2$.
- $\dfrac{d^2U}{dx^2} = 12x^2 - 16$: at $x = 0$ it is $-16$ (a maximum, unstable); at $x = \pm 2$ it is $32$ (minima, stable).
Answer: Unstable at $x = 0$; stable at $x = \pm 2\ \text{m}$, where $U = -16\ \text{J}$.
Worked example: Turning points
A particle in the same well has $E = -7\ \text{J}$ and starts in the right-hand well. Where does it turn around?
- Set $U = E$: $x^4 - 8x^2 + 7 = 0$, a quadratic in $x^2$: $(x^2 - 1)(x^2 - 7) = 0$.
- On the right: $x = 1\ \text{m}$ and $x = \sqrt{7} = 2.65\ \text{m}$.
- Between them $U < E$; outside them $U > E$, which is forbidden. Since $E < U(0) = 0$, it can't cross to the left well.
Answer: At $x = 1\ \text{m}$ and $x = 2.65\ \text{m}$.
Common mistake
Thinking the object can pass through a region where $E < U$, perhaps "with enough momentum." Its momentum is already counted in $E$. Where $U > E$, the kinetic energy $E - U$ would be negative, and $\tfrac{1}{2}mv^2$ can't be. Check by reading the graph: if the energy line is below the curve somewhere, the object can't get there from where it is (in classical physics; quantum tunneling is a different story, met in later physics).
Near a stable minimum, any smooth $U$ looks like a parabola, $U \approx U_0 + \tfrac{1}{2}U''(x_0)(x - x_0)^2$: a spring with $k = U''(x_0)$. That is why small oscillations everywhere in nature are close to simple harmonic, as you'll see in The Equation of Simple Harmonic Motion.