Potential-Energy Diagrams & Equilibrium

The energy landscape is modern science's favourite picture. Chemists draw a bond as a valley in $U(r)$ — the Morse curve — whose depth is the bond energy and whose turning points are the vibrating molecule's stretch limits; infrared spectroscopy measures exactly those vibrations. Biophysicists describe protein folding as a marble descending a rugged funnel-shaped landscape with countless local valleys (misfolds) en route to the global minimum (the working protein); drug designers hunt molecules that reshape the funnel. Space agencies park telescopes at Lagrange points — literal equilibria of a gravitational landscape, with station-keeping budgets sized by whether the point is a valley or a saddle. And materials scientists model glasses, batteries and catalysts by mapping $U$ over millions of atomic configurations. One curve, one line, read the gap: the protocol you drilled here runs from NCERT Exercise to Nature papers unchanged.

The honest way to build intuition is the marble in a smooth bowl. Its potential energy $mgh(x)$ literally is a landscape — the bowl's profile. Release the marble on the rim: it accelerates down the wall (steep slope, big force), crosses the bottom at maximum speed (lowest $U$, biggest gap below the energy line), climbs the far side losing speed, and stops exactly at the height it started from — the turning point — before running the film backward, forever. Every feature of the motion was drawn in the bowl's shape before the marble ever moved. The power move — the one this module exists to teach — is that the same reading works when the curve is not a physical hill: the $U(x)$ of a spring, of two atoms in a molecule, of a planet in a gravity well. The landscape is abstract; the protocol is identical.

Hold on to two disciplines while reading landscapes. First, the particle does not live on the curve — the graph's vertical axis is energy, not altitude; the particle moves only along the $x$-axis. (For the marble-in-a-bowl the two happen to coincide, which is why it is the perfect trainer and the perfect trap.) Second, the energy line is sacred: it is horizontal because mechanical energy is conserved — this whole instrument works only for conservative systems. Add friction and $E$ becomes a slowly sagging line, the oscillation shrinks, and the particle eventually beaches at the valley bottom.

The master identity is conservation, rearranged to read the graph:

where: $E$ — the total mechanical energy, fixed by the initial conditions (launch position and speed) and drawn as a horizontal line; $U(x)$ — the curve. Read it as an inequality too: motion requires $KE \ge 0$, so the particle is confined to $U(x) \le E$. Speed follows directly: $v(x) = \sqrt{2\left(E - U(x)\right)/m}$ — the gap, converted to a speed at any position you like, no dynamics solved. Turning points are the equality case:

where: the gap closes, $v = 0$ for an instant, and the force (see below) throws the particle back the way it came. Between two turning points the motion is bound — a shuttle run repeated forever in the frictionless ideal. If, on one side, the curve never again climbs to $E$, the particle escapes that way and never returns: unbound. The force is the landscape's slope, with a minus sign:

where: "slope" is the graph-reading (tangent) slope — rise of $U$ per run of $x$. (In calculus dress, $F = -dU/dx$; the graphical reading is all this module needs, and it is exact.) The sign does the physics: where $U$ climbs to the right, the force points left — always downhill. Steep cliff, strong force; flat ground, no force. Equation (3) is why equilibria are the flat spots, and it is also the inverse of a fact you own already: the spring's $U = \tfrac12 kx^2$ has slope $kx$, so $F = -kx$ — Hooke's law, recovered from the parabola.

1 · Board level — the NCERT graph audit (CBSE pattern, 3 marks)

A particle of total energy $E = 4.0$ J moves in a potential $U(x)$ whose graph is a single smooth valley: $U = 8$ J at $x = 0$, falling to a minimum $U = 1.0$ J at $x = 3$ m, rising back through $4.0$ J at $x = 5$ m and to $9$ J by $x = 7$ m. The energy line crosses the curve at $x = 1$ m and $x = 5$ m. Find (a) the region(s) where the particle can be found, (b) its kinetic energy and speed at $x = 3$ m if $m = 0.5$ kg, (c) what happens at $x = 5$ m.

Step 1. (a) Motion requires $U \le E$: allowed only where the curve sits below the 4.0 J line — between the crossings: $\boxed{1\ \mathrm{m} \le x \le 5\ \mathrm{m}}$. Outside is forbidden (the marker wants the word).

Step 2. (b) At the bottom: $KE = E - U = 4.0 - 1.0 = \boxed{3.0\ \mathrm{J}}$; $v = \sqrt{2 \times 3.0 / 0.5} = \boxed{3.5\ \mathrm{m/s}}$ — the fastest point of the whole motion, being the deepest.

Step 3. (c) At $x = 5$ m, $U = E$: the gap closes, $KE = 0$ — a turning point. The particle halts for an instant and the downhill force (curve climbs to the right there, so force points left) sends it back toward the valley. It shuttles between 1 m and 5 m indefinitely.

2 · Board level — the spring's landscape, quantified (CBSE/JEE-Main bridge)

A 0.80 kg block on a spring ($k = 100$ N/m) is set moving so that its total energy is $2.0$ J (zero of $U$ at natural length). Using the energy diagram, find (a) the amplitude, (b) the maximum speed, (c) the block's speed at $x = 10$ cm.

Step 1. (a) Turning points: $\tfrac12 kA^2 = E \Rightarrow A = \sqrt{2E/k} = \sqrt{0.04} = \boxed{0.20\ \mathrm{m}}$.

Step 2. (b) Valley bottom: $v_{\max} = \sqrt{2E/m} = \sqrt{5.0} = \boxed{2.2\ \mathrm{m/s}}$.

Step 3. (c) At $x = 0.10$ m: $U = \tfrac12 \times 100 \times 0.01 = 0.50$ J, so $KE = 1.5$ J and $v = \sqrt{2 \times 1.5/0.8} = \boxed{1.9\ \mathrm{m/s}}$. Note the non-linearity: at half the amplitude the block has lost only a quarter of its speed budget — the parabola's gap shrinks slowly at first.

3 · Competitive level — algebra finds the valley (JEE-Main pattern, ~3 minutes)

A particle of mass $2.0$ kg moves along $x$ with potential energy $U(x) = \left(x^2 - 4x\right)$ J ($x$ in metres). Its total energy is $E = 1.0$ J. Find (a) the equilibrium position and its type, (b) the turning points, (c) the maximum speed.

Step 1. (a) Complete the square — the algebra-only route to the vertex: $U = (x-2)^2 - 4$. A upward parabola with minimum at $\boxed{x = 2\ \mathrm{m}}$ (where $U = -4$ J): a valley → stable equilibrium. (Slope reading agrees: $U$ falls before 2 and rises after, so the force points toward 2 from both sides.)

Step 2. (b) Turning points where $U = E$: $(x-2)^2 - 4 = 1 \Rightarrow (x-2)^2 = 5 \Rightarrow x = 2 \pm \sqrt 5$, i.e. $\boxed{x \approx -0.24\ \mathrm{m}\ \text{and}\ x \approx 4.24\ \mathrm{m}}$.

Step 3. (c) Maximum speed at the bottom: $KE_{\max} = E - U_{\min} = 1.0 - (-4.0) = 5.0$ J, so $v_{\max} = \sqrt{2 \times 5.0/2.0} = \boxed{2.2\ \mathrm{m/s}}$.

Step 4. Sanity: negative $U$ alarmed no one — the zero of potential energy is a convention, and the gap $E - U_{\min} = 5$ J is what physics spends. Timing note: ~3 min. Completing the square replaces calculus everywhere a quadratic $U$ appears — vertex = equilibrium, and the turning-point quadratic is already solved for you. The designed error in (c): using $E$ alone ($v = 1$ m/s) instead of the full gap above the minimum.

4 · Competitive level — the double well, read at three energies (JEE-Advanced flavour, ~4 minutes)

A landscape has two valleys with bottoms at $U = 2$ J ($x_1$) and $U = 5$ J ($x_2$), separated by a hill whose crest is $U = 8$ J; far away on both sides the curve climbs above $12$ J. A particle starts in the LEFT valley. Describe the motion when its total energy is (a) $E = 4$ J, (b) $E = 10$ J, and (c) exactly $8$ J having started left of the crest; (d) state where the particle moves fastest in case (b).

Step 1. (a) $E = 4$ J clears neither the crest (8 J) nor the outer walls: the particle is bound in the left valley, shuttling between the two points where the left valley's walls reach 4 J. The right valley, though it has room below 4 J, is unreachable — a forbidden barrier stands between. (Quantum mechanics will later let particles tunnel through such barriers; classical particles never do.)

Step 2. (b) $E = 10$ J clears the crest (8 J) but not the outer walls (12 J): the particle roams both valleys, turning around on the far outer walls where $U = 10$ J. It slows to a crawl passing the crest (gap only 2 J) and speeds up in each valley.

Step 3. (c) $E = 8$ J exactly: the turning "point" coincides with the unstable equilibrium at the crest. The particle climbs ever more slowly, approaching the crest without quite stopping short of it — the marginal, knife-edge case (in the ideal, it takes unlimited time to arrive). Any energy sliver below 8 J and it turns back; any sliver above and it crosses. Setters love this edge for assertion–reason items.

Step 4. (d) Fastest at the deepest accessible point — the left valley's bottom, $U = 2$ J: $KE = 8$ J there, versus 5 J over the right valley's bottom and 2 J at the crest. Timing note: ~4 min, almost all reading. Answer with the checklist — clears which barriers? turning points where $U = E$? fastest at deepest? — and each sub-case is one sentence. The classic slip in (b): putting the fastest point in the newly opened right valley; depth, not novelty, decides.

Turning point confused with equilibrium. At a turning point the particle is momentarily at rest but the force is nonzero (the slope isn't flat) — it must reverse. At an equilibrium the force is zero but the particle may be moving through. Rest + zero force happens only when the energy line touches a flat spot.

Forbidden-region amnesia. Any region with $U > E$ is off-limits, including pockets beyond a barrier that the particle's energy cannot clear — reachability requires a connected allowed path from the starting point, not just $U \le E$ locally.

Force read as the curve's value. The force is minus the slope, not minus $U$. A deep flat valley floor has large negative $U$ and exactly zero force; a shallow but steep shoulder can carry a huge force.

Downhill mis-signed. Where $U$ increases to the right, $F$ points left. If your force arrows point uphill, the minus sign in (3) got dropped — and every stability verdict that follows will invert.

"Slow at the top" upgraded to "stops at the top". Crossing a barrier with $E$ above the crest, the particle slows but does NOT stop; stopping needs $U = E$ exactly. The knife-edge case $E =$ crest is its own species (worked example 4c).

Energy below the minimum. No motion exists with $E < U_{\min}$ — an option offering it is unphysical. Similarly, at $E = U_{\min}$ the "motion" is rest at the bottom.

Friction imported silently. The whole instrument assumes conservation — a horizontal $E$-line. If the problem adds friction, the line sags between snapshots and only the energy-audit method (loss $= f s$) survives; the geometric readings (fixed turning points, perpetual oscillation) do not.

Landscape mistaken for a track. The pitfall box's error, in its exam costume: assigning the particle speed for "descending" the curve's vertical drop. The vertical axis is joules.

Texts. NCERT Physics Class 11, Work, Energy and Power chapter (potential energy of a spring; the potential-energy-function exercise) · HC Verma, Concepts of Physics Vol. 1, Ch. 8 · Halliday–Resnick–Walker Ch. 8 (potential-energy curves — the treatment this module downshifts).

Provenance. All results are standard. A verifier should recompute: derivation A's numbers ($A = 0.1$ m, $v_{\max} = 2.0$ m/s, $v(5\,\mathrm{cm}) = 1.73$ m/s); example 1 (3.0 J, 3.5 m/s); example 2 (0.20 m, 2.2 m/s, 1.9 m/s); example 3 (vertex at 2 m, turning points $2 \pm \sqrt5$, $v_{\max} = 2.2$ m/s); example 4's three readings and the fastest-point verdict (KE = 8 J over the left bottom).

Prerequisites. Potential Energy: Gravitational & Elastic · Conservative & Non-Conservative Forces · Work by a Variable Force & F–x Graphs. Leads to. Simple Harmonic Motion (oscillations chapter) · TECHNIQUE: Force Method or Energy Method? · Potential-Energy Diagrams (UG deep dive).