Classical-Mechanics · Unit 17 · Video 3 · Interactive Practice

Stable vs Unstable Equilibrium on a Cubic Energy Diagram

IKey Formulas

Formula	Name	What you need
$U(x) = -U_1\left[\left(\dfrac{x}{x_1}\right)^3 - \left(\dfrac{x}{x_1}\right)^2\right]$	The cubic potential (Example 14.1)	Positive constants $U_1, x_1$ , with $U(0) = 0$
$F_x = -\dfrac{dU}{dx} = U_1\left(\dfrac{3x^2}{x_1^3} - \dfrac{2x}{x_1^2}\right)$	Force from the potential	Vanishes at $x = 0$ and $x = \tfrac{2}{3}x_1$
$\left.\dfrac{d^2U}{dx^2}\right\|_{x=0} = \dfrac{2U_1}{x_1^2} > 0, \qquad \left.\dfrac{d^2U}{dx^2}\right\|_{x=2x_1/3} = -\dfrac{2U_1}{x_1^2} < 0$	Curvature test for stability	The sign of the second derivative
$K(x) = E - U(x) \ge 0$	Allowed regions and turning points	The horizontal total-energy line $E$

Key Insight: zero force is not enough — curvature decides whether a resting particle stays put. And the crest height $U\!\left(\tfrac{2}{3}x_1\right) = \tfrac{4}{27}U_1$ splits every possible motion in two: below it the particle is trapped between turning points, above it the particle escapes to infinity.

IIVisualization 1 — Force Is the Negative Slope

The force is minus the slope of the curve, so wherever the curve levels off the particle can sit still.

Particle position x [m] 0.42

💡 The curve is not a hill in space: the particle only ever moves along the $x$ -axis, and the height of the curve is stored energy, not altitude.

IIIVisualization 2 — Where the Particle Is Allowed to Be

Kinetic energy is the gap between the energy line and the curve — and it can never be negative.

Total energy E [J] 0.20

IVVisualization 3 — The Threshold Launch Speed

How fast must the particle leave the origin to just clear the crest at $x = \tfrac{2}{3}x_1$ ?

Launch speed v₀ [m/s] 1.000

💡 Launch it toward $-x$ instead and no speed is enough: $U$ climbs without bound on the left, so the particle always turns back.

VQuiz Questions

Problem 1 · Locate the Equilibria

Given: $U(x) = -U_1\left[\left(\dfrac{x}{x_1}\right)^3 - \left(\dfrac{x}{x_1}\right)^2\right]$ with $U_1, x_1 > 0$ — find every position where the force on the particle vanishes.

✅ Correct! Factoring

x\left(3x - 2x_1\right) = 0

leaves exactly these two roots — the valley and the crest.

❌ Not quite. Those are the two places where

U = 0

(the curve crosses the axis), but equilibrium needs the slope to vanish, not the value.

❌ Close, but one is missing. The factored condition is

x\left(3x - 2x_1\right) = 0

, and the factor

x

contributes the root

x = 0

❌ Not quite. Differentiate first, then set

3x^2/x_1^3 = 2x/x_1^2

and clear the constants.

Show solution

Differentiate and flip the sign to get the force:

F_x = -\frac{dU}{dx} = U_1\left(\frac{3x^2}{x_1^3} - \frac{2x}{x_1^2}\right)

Set $F_x = 0$ and multiply through by $x_1^3/U_1$ :

3x^2 - 2x_1 x = 0 \quad\Longrightarrow\quad x\left(3x - 2x_1\right) = 0

So the force vanishes at

x = 0 \qquad\text{and}\qquad x = \frac{2}{3}x_1

Watch the trap: writing $U(x) = U_1\left(\dfrac{x}{x_1}\right)^2\left(1 - \dfrac{x}{x_1}\right)$ shows $U = 0$ at $x = 0$ and $x = x_1$ . Those are the zeros of $U$ , not the zeros of its slope — only $x = 0$ belongs to both lists.

Problem 2 · Stability from Curvature

Given: at the equilibrium $x = \tfrac{2}{3}x_1$ the second derivative evaluates to $\dfrac{d^2U}{dx^2} = -\dfrac{2U_1}{x_1^2}$ — what does that number tell you about the particle there?

✅ Correct! Negative curvature is a crest: just to the right of it

F_x > 0

and just to the left

F_x < 0

, so the force points away from equilibrium on both sides.

❌ Check the sign. A local minimum needs

d^2U/dx^2 > 0

; this value is negative, since

U_1, x_1^2 > 0

❌ That is the wrong derivative. The force is

F_x = -dU/dx

, which is zero here. The second derivative measures curvature, not force.

❌ Not quite. Equilibrium is fixed by the first derivative; the second derivative only sorts the equilibria into stable and unstable.

Show solution

The curvature of the potential is

\frac{d^2U}{dx^2} = -U_1\left(\frac{6x}{x_1^3} - \frac{2}{x_1^2}\right)

At the crest $x = \tfrac{2}{3}x_1$ :

-U_1\left(\frac{4}{x_1^2} - \frac{2}{x_1^2}\right) = -\frac{2U_1}{x_1^2} < 0

Negative curvature means a local maximum: displace the particle and the force $F_x = -dU/dx$ points away from the equilibrium, so the displacement grows. The point is unstable.

At the origin the same formula gives

-U_1\left(0 - \frac{2}{x_1^2}\right) = +\frac{2U_1}{x_1^2} > 0

a local minimum, hence a stable equilibrium: the force there is restoring. Both points have $F_x = 0$ ; only the curvature tells them apart.

Problem 3 · Barrier Height and Threshold Speed

Given: the particle of mass $m$ starts at $x = 0$ (where $U = 0$ ) with speed $v_0$ and must just barely reach the unstable equilibrium at $x = \tfrac{2}{3}x_1$ — find the barrier height and the required launch speed.

How high is the barrier?

What launch speed does that require?

✅ Correct! With the sample values

U_1 = \tfrac{27}{4}\ \text{J}

and

m = 1\ \text{kg}

this is

v_0 = \sqrt{2}\ \text{m/s} \approx 1.414\ \text{m/s}

❌ Check the barrier height. Both bracket terms survive:

\left(\tfrac{2}{3}\right)^3 = \tfrac{8}{27}

and

\left(\tfrac{2}{3}\right)^2 = \tfrac{12}{27}

, and the overall minus sign flips the difference positive.

❌ Check the speed. Start from

E = \tfrac{1}{2}mv_0^2

and solve for

v_0

— the factor

2

from the kinetic energy must survive, and

v_0

is a square root.

Show solution

Step 1 — barrier height. Substitute $x = \tfrac{2}{3}x_1$ , so $x/x_1 = \tfrac{2}{3}$ :

U\!\left(\tfrac{2}{3}x_1\right) = -U_1\left[\left(\tfrac{2}{3}\right)^3 - \left(\tfrac{2}{3}\right)^2\right] = -U_1\left[\frac{8}{27} - \frac{12}{27}\right] = \frac{4}{27}U_1

Step 2 — energy at the start. At $x = 0$ the potential energy is zero, so all of the energy is kinetic:

E = K(0) = \tfrac{1}{2}mv_0^2

Step 3 — set them equal. "Just barely reaching" the crest means arriving with $K \to 0$ , i.e. $E = \tfrac{4}{27}U_1$ :

\frac{4}{27}U_1 = \frac{1}{2}mv_0^2 \quad\Longrightarrow\quad v_0^2 = \frac{8U_1}{27m} \quad\Longrightarrow\quad v_0 = \sqrt{\frac{8U_1}{27m}}

Check with the sample values. $U_1 = \tfrac{27}{4}\ \text{J}$ , $m = 1\ \text{kg}$ give $v_0^2 = \dfrac{8}{27}\cdot\dfrac{27}{4} = 2$ , so $v_0 = \sqrt{2}\ \text{m/s}$ and $E = \tfrac{1}{2}(1)(2) = 1\ \text{J}$ — exactly the $1\ \text{J}$ crest.

Problem 4 · Reading a Fate off the Diagram

Given: the sample values $x_1 = 1.5\ \text{m}$ , $U_1 = \tfrac{27}{4}\ \text{J}$ , $m = 1\ \text{kg}$ — so the crest sits at $x = 1\ \text{m}$ with $U = 1\ \text{J}$ , and $U(1.6\ \text{m}) = -0.51\ \text{J}$ . A particle with total energy $E = 1.2\ \text{J}$ is at $x = 1.6\ \text{m}$ moving in the $-x$ direction. What happens to it?

✅ Correct! Above the crest height the cubic

E = U(x)

keeps only one root, so there is exactly one turning point and it lies on the far left.

❌ Negative $U$ is perfectly allowed. Only

K = E - U

must stay non-negative, and here

K = 1.2 - (-0.51) = 1.71\ \text{J}

❌ Trapping needs a barrier taller than $E$ . Here

E = 1.2\ \text{J} > 1\ \text{J}

, so the crest cannot stop the particle.

❌ Not quite. Compare

E

with the crest height first, then count the solutions of

E = U(x)

Show solution

With these sample values $U(x) = 3x^2 - 2x^3$ (in joules, $x$ in metres) and the crest height is

\tfrac{4}{27}U_1 = \tfrac{4}{27}\cdot\tfrac{27}{4}\ \text{J} = 1\ \text{J} < E = 1.2\ \text{J}

Kinetic energy at the start: $K = E - U(1.6) = 1.2 - (-0.51) = 1.71\ \text{J} > 0$ , so the particle is certainly allowed to be there.

At the crest: $K = 1.2 - 1 = 0.2\ \text{J} > 0$ , so moving left it sails over the maximum instead of stopping on it.

The one turning point: solving $E = U(x)$ ,

3x^2 - 2x^3 = 1.2 \quad\Longrightarrow\quad x_d \approx -0.54\ \text{m}

is the only root, on the steeply climbing left branch. The particle crosses the valley, stops there, reverses, runs back over the crest, and from then on $U$ falls without limit, so $K$ grows and it escapes to $+\infty$ . It can never enter $x < x_d$ .

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