Potential-Energy Diagrams

Everything starts with E, and E comes from one moment where you know both position and speed. Take a 2.0 kg slider in a well where U(0.30 m) = −1.18 J, moving at 0.62 m s⁻¹ there: K = ½(2.0 kg)(0.62 m s⁻¹)² = 0.38 J, so E = 0.38 J + (−1.18 J) = −0.80 J. A negative total energy is not a problem; it only says the slider sits below the zero chosen for U. With only conservative forces doing work that value never changes, so it is drawn as a horizontal line across the whole diagram. The zero of U is a free choice. Add 2.0 J to U everywhere and U(0.30 m) becomes 0.82 J while E becomes 1.20 J, yet the gap between them is still 0.38 J. Every prediction the diagram makes depends on gaps and slopes, never on where the zero sits, provided E is computed with the same zero as the curve.

Since K = ½mv² ≥ 0, the particle can only be where E ≥ U(x). Shade the stretches of x where the curve pokes above the line: those are classically forbidden, and the rest is allowed. The allowed set can come in separate pieces. In the double well U(x) = U₀[(x/a)⁴ − 2(x/a)²] with U₀ = 2.0 J and a = 0.50 m, a line at E = −0.80 J lies above the curve for 0.24 m ≤ |x| ≤ 0.67 m: two disjoint intervals, one per well. A slider that starts in the right-hand well stays there for ever, because reaching the other one would mean crossing x = 0, where U = 0 > E. Raise the line above the barrier top and the two intervals merge into one. So an energy alone does not settle the motion; you need E and the interval the particle starts in.

Where the energy line meets the curve, U = E and the speed is zero. The force there is set by the slope, F = −dU/dx, and at a crossing the slope is almost never zero. On the right edge of an allowed interval the curve is rising, so dU/dx > 0 and F points toward −x, back inside; on the left edge the reverse holds. In the double well at E = −0.80 J, the force at x = 0.67 m is −16.5 N and at x = 0.24 m it is +5.9 N, both pointing into the allowed interval. That is why the particle reverses instead of stopping. Two turning points mean the particle shuttles between them for ever: bound, periodic motion. One turning point, with the curve staying below the line all the way out on the other side, means it arrives, reverses once and leaves: unbound motion.

Flat points of the curve, dU/dx = 0, are equilibria, and the shape around them decides what a small push does. At a minimum the curve rises on both sides, so the force points back toward it: stable. At a maximum the curve falls away on both sides and the force points away: unstable. A flat stretch is neutral. The energy line turns these into motion. With E equal to a minimum value the allowed region shrinks to a single point, and the particle sits there at rest; a little above, it rocks inside a narrow band. To leave a well, E must clear the lower rim: a 0.10 kg bead at the floor of a well whose lower rim is 0.45 J higher needs more than √(2 × 0.45 J ÷ 0.10 kg) = 3.0 m s⁻¹. With E exactly at a maximum, a turning point and an equilibrium coincide, and the particle creeps toward the top without ever arriving.

Expand U about a minimum x₀. The first-derivative term vanishes because dU/dx = 0 there, so U(x) ≈ U(x₀) + ½U″(x₀)(x − x₀)², where U″ is the second derivative. Differentiating gives F ≈ −U″(x₀)(x − x₀): a restoring force proportional to displacement, the signature of simple harmonic motion, with U″(x₀) playing the role of the spring constant. Hence ω = √(U″(x₀)/m). For the double well, U″ = 8U₀/a² at x₀ = ±a, which is 64 N m⁻¹ for U₀ = 2.0 J and a = 0.50 m; a 2.0 kg slider then oscillates with ω = 5.66 rad s⁻¹ and T = 1.11 s. Curvature, not depth, is what counts: a well with U₀ = 8.0 J and a = 1.0 m is four times deeper, yet its floor has the same 64 N m⁻¹ curvature and the same small-oscillation period.

The diagram gives |v| at each x, not the direction of travel, which comes from the initial conditions. It gives time only through T = 2∫dx/|v|: a parabola returns the same period at every energy, but in the V-shaped well U = F₀|x| the period grows as √E, so quadrupling E doubles it. The harmonic result is a small-amplitude statement. In the double well with U₀ = 2.0 J, a slider 0.2 J above the floor turns at 0.413 m and 0.574 m, close to the parabola's 0.421 m and 0.579 m; at 1.8 J above the floor it turns at 0.113 m and 0.698 m, where the parabola predicts 0.263 m and 0.737 m. The real well is softer toward the barrier and stiffer outside. Finally, the level line assumes no friction: a dissipative force lowers E as the particle moves, and the turning points creep inward cycle by cycle.