Energy Diagrams and Equilibrium

In short

An energy diagram plots potential energy U(x) against position x. The total mechanical energy E appears as a horizontal line. Where E > U(x), the particle moves; where E = U(x), it turns around (turning points); where E < U(x), it cannot go. The force at any point is F = -dU/dx — always pointing downhill on the curve. A minimum of U(x) is stable equilibrium; a maximum is unstable; a flat region is neutral.

Place a marble in a steel katori — the small bowl you eat dal-rice from every night. Push it to one side and release it. The marble rolls down, climbs up the opposite wall, slows, stops for an instant, and rolls back. It oscillates back and forth, trapped inside the bowl, never escaping over the rim.

Now flip the katori upside down and try to balance the marble on its curved top. It sits there — for a moment. The slightest breath of air sends it rolling off the edge. Same bowl, same marble, completely different behaviour. One position traps the marble; the other rejects it.

The difference between these two situations — one where a small push brings you back, one where a small push sends you away forever — is the entire subject of this article. You will learn to read it all from a single curve: the potential energy diagram.

The energy landscape

You already know from conservation of mechanical energy that when a particle moves under a conservative force, the total energy stays constant:

E = K + U(x) = \tfrac{1}{2}mv^2 + U(x) = \text{constant}

Rearrange this to isolate the kinetic energy:

K = E - U(x)

Why: since E is fixed for a given situation, this equation tells you the kinetic energy at every position. If you know the potential energy function U(x) and the total energy E, you know everything about how fast the particle moves at every point — without solving F = ma directly.

Here is the insight that changes how you think about mechanics: plot U(x) against x, and draw the total energy E as a horizontal line. The vertical gap between E and the U(x) curve at any position equals the kinetic energy there. Where the gap is large, the particle moves fast. Where the gap shrinks to zero, the particle stops momentarily. Where U(x) rises above E, the particle cannot exist at all — because kinetic energy would have to be negative, which is physically impossible.

A potential energy curve $U(x) = \frac{x^4}{4} - 2x^2$ (solid line). The dashed line shows one possible total energy $E = -2$ J. The valleys at $x = \pm 2$ are stable equilibrium points; the hilltop at $x = 0$ is unstable. A particle with this energy is trapped in one valley, oscillating between the two turning points where the dashed line intersects the curve.

Think of this curve as a physical landscape — a chain of hills and valleys viewed from the side. The particle is a ball rolling on this terrain. It speeds up going downhill (kinetic energy increases as potential energy decreases) and slows down going uphill (the reverse). The ball can never rise above the elevation set by its total energy, because that would require negative kinetic energy.

Imagine you are in an autorickshaw on a hilly road outside Munnar. The rickshaw rolls fast through the valleys and slows down on the uphill stretches. If the driver cuts the engine, your speed at any point depends entirely on how high you are relative to where you started — the total energy is fixed, and the height profile of the road is the potential energy landscape. This is exactly what the U(x) curve encodes.

Force from the curve: F = -dU/dx

The link between the energy diagram and Newton's second law comes from the definition of potential energy. For a conservative force acting along x, the change in potential energy over a small displacement is:

dU = -F\,dx

Why: this is the definition of potential energy — the work done by the conservative force (F\,dx) equals the decrease in potential energy. When the force does positive work (pushes in the direction of motion), potential energy decreases.

Dividing both sides by dx:

\boxed{F = -\frac{dU}{dx}}

Why: the force at any position is the negative of the slope of the U(x) curve at that position. This single equation is the bridge between the energy diagram and Newton's second law — every feature of the curve tells you something about the force.

Read the implications carefully:

Where U(x) slopes upward to the right (dU/dx > 0), the force is negative — it pushes left, back down the hill.
Where U(x) slopes downward to the right (dU/dx < 0), the force is positive — it pushes right, down the hill.
Where U(x) is flat (dU/dx = 0), the force is zero — the particle is in equilibrium.
Where the slope is steep, the magnitude of the force is large. A gentle slope means a weak force.

The rule is beautifully simple: the force always points downhill on the potential energy curve. A marble released on a slope rolls toward lower U(x). The steeper the slope, the harder the push.

The force always pushes the particle downhill on the $U(x)$ curve. On the left slope, the force points right (toward the minimum). On the right slope, it points left. At the bottom of the valley, the slope is zero and the force vanishes.

Three types of equilibrium

At any point where dU/dx = 0, the force is zero and the particle is in equilibrium. But the outcome of a small disturbance depends entirely on the curvature of the curve around that point.

Stable equilibrium — the valley

At a minimum of U(x), the curve opens upward like a bowl. Nudge the particle to the right: the slope becomes positive, so F = -dU/dx is negative — the force pushes left, back toward the minimum. Nudge it to the left: the slope becomes negative, so the force pushes right, again toward the minimum.

No matter which way you push, the force pulls you back. This is stable equilibrium — the marble sitting at the bottom of the katori, a pendulum hanging straight down, a cricket ball resting in a small dip on the pitch.

Mathematical test: \frac{dU}{dx} = 0 and \frac{d^2U}{dx^2} > 0 → the equilibrium is stable.

Unstable equilibrium — the hilltop

At a maximum of U(x), the curve opens downward. Nudge the particle to the right: the slope becomes negative, so the force is positive — it pushes further to the right. Nudge it to the left: the force pushes further left. The displacement grows. The particle accelerates away from the equilibrium position.

This is unstable equilibrium — the marble balanced on the inverted katori, a pencil standing on its tip, a ball perched on top of a hill. Theoretically at rest, but practically impossible to maintain.

Mathematical test: \frac{dU}{dx} = 0 and \frac{d^2U}{dx^2} < 0 → the equilibrium is unstable.

Neutral equilibrium — the flat plain

If U(x) is constant over a region, the slope is zero everywhere in that interval. Displace the particle and the force remains zero — it simply stays at its new position, with no tendency to return or to flee.

This is neutral equilibrium — a ball on a flat floor, a hockey puck on a smooth ice rink, a train on level tracks.

Mathematical test: \frac{dU}{dx} = 0 and \frac{d^2U}{dx^2} = 0 (with higher derivatives also zero in the region) → the equilibrium is neutral.

**Stable:** the displaced ball (faded) experiences a force back toward the minimum — a restoring force. **Unstable:** the displaced ball is pushed further away from the maximum. **Neutral:** the displaced ball feels no force and stays wherever you put it.

Turning points and the forbidden region

Return to the fundamental equation: K = E - U(x). Kinetic energy is \frac{1}{2}mv^2, which can never be negative — you cannot have a negative squared speed. This gives a hard constraint on where the particle can be:

E - U(x) \geq 0 \quad \Longrightarrow \quad U(x) \leq E

Why: if U(x) > E at some position, then K = E - U(x) < 0, which would mean \frac{1}{2}mv^2 < 0. Since mass and velocity-squared are both non-negative, this is impossible. The particle simply cannot be at that position.

The particle can only exist where U(x) \leq E. Any region where U(x) > E is a classically forbidden region — the particle cannot enter it.

The boundaries of the allowed region — the positions where U(x) = E exactly — are called turning points. At a turning point, K = 0 and v = 0: the particle momentarily stops and reverses direction. This is the marble at the highest point of its swing inside the katori, pausing for an instant before rolling back down.

Bounded motion

If the particle is trapped between two turning points with U(x) > E on both sides, the motion is bounded. The particle oscillates back and forth between the turning points indefinitely (assuming no friction). A mass on a spring is the simplest example: the turning points are the maximum extension and maximum compression.

Think of a Diwali rocket that does not have enough thrust. It shoots upward, slows, stops, and falls back to the ground. Its total energy was not enough to escape — the turning point is the maximum height, and the motion is bounded (it returns to where it started).

Unbounded motion

If the total energy is large enough that U(x) \leq E extends to infinity in at least one direction, the particle is free to escape. The motion is unbounded. The kinetic energy never drops to zero in that direction, so the particle never turns around.

Think of an ISRO rocket during a Mangalyaan-type mission. If the rocket's total energy exceeds the gravitational potential energy at infinite distance (conventionally set to zero), the spacecraft escapes Earth's gravity entirely. It will never fall back — the motion is unbounded. If the total energy is below zero, the spacecraft remains in orbit, oscillating between a closest approach (perigee) and farthest point (apogee) — bounded motion.

How to read any energy diagram — four steps

Draw the horizontal line at E across the U(x) plot.
Mark the intersections with the U(x) curve — these are the turning points.
Identify the allowed region where U(x) \leq E — the particle can move here.
Check if the allowed region is bounded (enclosed between two turning points) or unbounded (extends to infinity in at least one direction).

The vertical gap E - U(x) at any position is the kinetic energy there. Where this gap is largest, the particle moves fastest. At the turning points, the gap is zero and the particle is momentarily at rest. At the equilibrium position (the bottom of a well), the gap is widest and the speed is maximum.

Explore the energy diagram yourself

The potential below is a spring: U(x) = x^2 (in joules, with x in metres — corresponding to a spring constant k = 2 N/m). Drag the red dot up or down to change the total energy E, and watch the turning points shift.

Drag the red dot on the left axis to change the total energy $E$ (dashed line). The red dots on the parabola mark the turning points where $U(x) = E$. Between them is the allowed region. As you raise $E$, the turning points slide outward and the particle explores a wider range. The maximum kinetic energy (at $x = 0$, the bottom of the well) always equals $E$ itself.

Raise the energy and the turning points slide outward — the particle can explore a wider range. Lower the energy toward zero and the turning points close in toward x = 0 — the particle barely oscillates. This is the energy diagram doing all the work that a force equation would otherwise require.

Worked examples

Example 1: Puck on a spring

A 500 g puck on a smooth carrom board is tethered to the centre by a rubber band that behaves like a spring with k = 200 N/m. You pull the puck 0.20 m from the centre and release it from rest. Using the energy diagram, find (a) the turning points, (b) the maximum speed of the puck, and (c) the force on the puck at x = 0.10 m.

Step 1. Write the potential energy and the total energy.

U(x) = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(200)x^2 = 100x^2 \text{ J}

At release, v = 0 (all energy is potential):

E = U(0.20) = 100(0.20)^2 = 100 \times 0.04 = 4 \text{ J}

Why: the puck is released from rest, so K = 0 and E = U at that instant. This total energy is conserved throughout the motion.

Step 2. Find the turning points.

At a turning point, K = 0 and U(x) = E:

100x^2 = 4 \quad \Longrightarrow \quad x^2 = 0.04 \quad \Longrightarrow \quad x = \pm 0.20 \text{ m}

Why: the puck oscillates between x = -0.20 m and x = +0.20 m — the two positions where all energy is potential and the puck momentarily stops.

Step 3. Find the maximum speed.

Maximum K occurs where U(x) is smallest — at x = 0 (the equilibrium point):

K_{\max} = E - U(0) = 4 - 0 = 4 \text{ J}

\tfrac{1}{2}mv_{\max}^2 = 4 \quad \Longrightarrow \quad v_{\max}^2 = \frac{2 \times 4}{0.50} = 16 \quad \Longrightarrow \quad v_{\max} = 4 \text{ m/s}

Why: at the bottom of the potential well, all the energy is kinetic. The full 4 J drives the puck, giving its highest speed.

Step 4. Find the force at x = 0.10 m.

F = -\frac{dU}{dx} = -\frac{d}{dx}(100x^2) = -200x

At x = 0.10 m:

F = -200 \times 0.10 = -20 \text{ N}

Why: the negative sign means the force points toward the origin — a restoring force pulling the puck back to centre. This is exactly the spring force F = -kx.

Result: The puck oscillates between \pm 0.20 m. Its maximum speed is 4 m/s at the centre. At x = 0.10 m, the restoring force is 20 N toward the centre.

$U(x) = 100x^2$ (solid parabola) with $E = 4$ J (dashed line). The turning points at $\pm 0.20$ m are where the energy line meets the curve. At the centre ($x = 0$), the full 4 J gap is kinetic energy — the puck moves fastest there. At $x = 0.10$ m, the gap of 3 J is kinetic energy and 1 J is potential.

The graph confirms the algebra: the puck's allowed region is the stretch between the two red turning-point dots. At the centre, the gap between the dashed line and the parabola is widest — that is where the kinetic energy and speed are greatest.

Example 2: The double-well track

At your school's science exhibition, a track is shaped so that a marble rolling on it has potential energy U(x) = \frac{x^4}{4} - 2x^2 (in joules, with x in metres). The marble has total mechanical energy E = -3 J. Find the equilibrium positions, classify each one, identify the turning points, and determine whether the motion is bounded or unbounded.

Step 1. Find the equilibrium positions.

At equilibrium, F = 0, so dU/dx = 0:

\frac{dU}{dx} = x^3 - 4x = x(x^2 - 4) = x(x - 2)(x + 2) = 0

x = 0, \quad x = +2 \text{ m}, \quad x = -2 \text{ m}

Why: factoring reveals three equilibrium points — one at the origin and a symmetric pair at \pm 2 m. At each of these positions the slope of U(x) is zero and the net force vanishes.

Step 2. Classify each equilibrium.

Compute the second derivative: \frac{d^2U}{dx^2} = 3x^2 - 4.

At x = 0: \frac{d^2U}{dx^2} = -4 < 0 → unstable (local maximum). U(0) = 0 J.

At x = \pm 2: \frac{d^2U}{dx^2} = 3(4) - 4 = 8 > 0 → stable (local minima). U(\pm 2) = 4 - 8 = -4 J.

Why: positive second derivative means the curve is concave up (a valley) — stable. Negative second derivative means concave down (a hilltop) — unstable.

Step 3. Find the turning points for E = -3 J.

Set U(x) = E:

\frac{x^4}{4} - 2x^2 = -3

Multiply both sides by 4:

x^4 - 8x^2 + 12 = 0

Substitute u = x^2 to get a quadratic:

u^2 - 8u + 12 = 0 \quad \Longrightarrow \quad u = \frac{8 \pm \sqrt{64 - 48}}{2} = \frac{8 \pm 4}{2}

u = 6 \quad \text{or} \quad u = 2

So x^2 = 6 \Rightarrow x = \pm\sqrt{6} \approx \pm 2.45 m, and x^2 = 2 \Rightarrow x = \pm\sqrt{2} \approx \pm 1.41 m.

Why: the quartic in x becomes a clean quadratic in u = x^2, which factors to give four turning points: \pm\sqrt{2} and \pm\sqrt{6}.

Step 4. Determine the type of motion.

The energy E = -3 J is less than U(0) = 0 J. The particle cannot cross x = 0 — the hilltop at the origin acts as an energy barrier. It is confined to one of the two symmetric valleys.

If the marble starts in the right well (near x = 2), it oscillates between x = \sqrt{2} \approx 1.41 m and x = \sqrt{6} \approx 2.45 m. The motion is bounded.

Why: the total energy is below the potential energy barrier at x = 0, so the particle has insufficient energy to cross from one well to the other. It is trapped, oscillating between turning points within a single valley.

Result: Three equilibria: x = \pm 2 m (stable, U = -4 J) and x = 0 (unstable, U = 0). For E = -3 J, the marble oscillates between \sqrt{2} \approx 1.41 m and \sqrt{6} \approx 2.45 m in a single well. The motion is bounded. Raise E above 0 J and the marble could cross the hilltop, merging both wells into one large oscillation region.

The double-well potential $U(x) = \frac{x^4}{4} - 2x^2$ (solid) with $E = -3$ J (dashed). The marble in the right well oscillates between the turning points near $x = 1.41$ and $x = 2.45$ (red dots). The hilltop at $x = 0$ blocks passage to the left well.

The graph makes the physics vivid: the marble rolls back and forth in its valley, climbing the walls on each side until it runs out of kinetic energy at the turning points. If you raised E above 0 J — the height of the central barrier — the two separate oscillation regions would merge into one, and the marble would swing freely across the entire track.

Common confusions

"A turning point is where the particle bounces off a wall." There is no physical wall. The particle simply slows down as U(x) rises, stops when U(x) = E, and the conservative force (-dU/dx, pointing downhill) smoothly accelerates it back. The reversal is gradual, not a sudden impact.
"Unstable equilibrium means the particle cannot be there." The particle can sit at an unstable equilibrium — the force is zero there. The problem is that any infinitesimal disturbance grows. In practice, thermal vibrations and tiny perturbations always exist, so unstable equilibria are impossible to maintain for any real duration.
"Higher potential energy means higher total energy." Total energy E is constant throughout the motion. Potential energy U(x) varies with position. The particle trades U for K and back as it moves, but E = K + U never changes (in a conservative system).
"The force is strongest at the equilibrium point." The force is zero at equilibrium (dU/dx = 0 there). The force is strongest where the slope of U(x) is steepest — typically on the flanks of the potential well, not at the bottom.
"Negative total energy means the particle has no energy." A negative E simply reflects the choice of reference point for U(x). What matters is the difference E - U(x), which gives the kinetic energy — and that is always non-negative at allowed positions.

If you are comfortable reading energy diagrams for individual potentials, you can stop here — you have the tools to analyse any U(x) curve. What follows connects energy diagrams to oscillations and to a powerful approximation that appears regularly in JEE Advanced problems.

Every stable equilibrium looks like a spring

Take any smooth potential energy function U(x) with a stable minimum at x = x_0. Expand in a Taylor series:

U(x) = U(x_0) + \underbrace{\frac{dU}{dx}\bigg|_{x_0}}_{= \, 0}(x - x_0) + \frac{1}{2}\frac{d^2U}{dx^2}\bigg|_{x_0}(x - x_0)^2 + \cdots

Why: at equilibrium dU/dx = 0, so the linear term vanishes. The leading correction beyond the constant is the quadratic term — the same shape as a spring potential.

Define the effective spring constant k_{\text{eff}} = \frac{d^2U}{dx^2}\big|_{x_0} and the displacement \xi = x - x_0. For small displacements (neglecting cubic and higher terms):

U \approx U(x_0) + \tfrac{1}{2}k_{\text{eff}}\,\xi^2

The force near the equilibrium is:

F = -\frac{dU}{d\xi} \approx -k_{\text{eff}}\,\xi

This is a linear restoring force — exactly a spring. The particle executes simple harmonic motion with angular frequency:

\omega = \sqrt{\frac{k_{\text{eff}}}{m}} = \sqrt{\frac{1}{m}\frac{d^2U}{dx^2}\bigg|_{x_0}}

Why: near any smooth minimum, the potential is approximately parabolic. A parabolic potential gives a linear restoring force, which produces SHM. This is why small oscillations about any stable equilibrium are always simple harmonic — regardless of the actual shape of U(x).

Application: frequency of small oscillations in the double-well

For the double-well potential U(x) = \frac{x^4}{4} - 2x^2 from Example 2, the stable minimum is at x_0 = 2 m. The effective spring constant is:

k_{\text{eff}} = \frac{d^2U}{dx^2}\bigg|_{x=2} = 3(2)^2 - 4 = 8 \text{ N/m}

For a marble of mass 50 g (m = 0.050 kg):

\omega = \sqrt{\frac{8}{0.050}} = \sqrt{160} \approx 12.6 \text{ rad/s}

T = \frac{2\pi}{\omega} \approx 0.50 \text{ s}

The marble oscillates roughly twice per second near the bottom of the well. As the amplitude grows and the marble climbs higher on the walls, the cubic and quartic terms in U start mattering and the motion is no longer exactly sinusoidal — but for small oscillations, this SHM approximation is excellent.

The Lennard-Jones potential — a real-world energy diagram

The interaction between two atoms in a molecule is well-described by the Lennard-Jones potential:

U(r) = 4\varepsilon\left[\left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^{6}\right]

where \varepsilon is the depth of the potential well and \sigma is a characteristic distance. This curve has all the features you have studied: a steep repulsive wall at small r (atoms resist being squeezed together), a stable minimum at r_{\min} = 2^{1/6}\sigma (the equilibrium bond length), and a gradual approach to zero at large r (atoms barely interact when far apart).

If the total energy is below zero, the two atoms are bound — they oscillate about the equilibrium, forming a molecule. If the total energy is above zero, the atoms fly apart — the molecule dissociates. The energy diagram for two nitrogen atoms, two oxygen atoms, or two hydrogen atoms looks exactly like this: a valley with a steep left wall and a gently sloping right side. Bounded motion is a chemical bond; unbounded motion is a free pair of atoms. The same physics you learned from a marble in a katori governs the forces holding molecules together.

Where this leads next

Simple Harmonic Motion — the special case where U(x) = \frac{1}{2}kx^2 exactly. The energy diagram is a parabola, and the oscillation is solved completely.
Conservative Forces and Potential Energy Functions — how to build U(x) from a given force, and why only conservative forces have potential energy functions.
Non-Conservative Forces and Energy Dissipation — what happens when friction is present and total mechanical energy is not conserved.
Spring Force and Hooke's Law — the restoring force behind the parabolic potential well, and why the going-deeper section shows every stable equilibrium behaves like a spring for small oscillations.
Potential Energy — the foundational concept of stored energy that makes these diagrams possible.