Classical-Mechanics · Unit 17 · Video 2 · Interactive Practice

The Spring Energy Diagram: Turning Points and Forbidden Regions

IKey Formulas

Formula	Name	What you need
$U^s(x) = \tfrac{1}{2}kx^2$	Spring potential energy	The reference choice $U^s(0) \equiv 0$
$F_x^s = -\dfrac{dU^s(x)}{dx} = -kx$	Force is the negative slope	The slope of the curve at $x$
$K(x) = E_m - U^s(x)$	Kinetic energy on the diagram	The gap between the energy line and the curve
$x_{\max} = \sqrt{\dfrac{2E_m}{k}}$	Turning points	The conserved energy $E_m$ and the spring constant $k$

Key Insight: The parabola alone fixes the force everywhere. Add one horizontal line at height $E_m$ and the same picture also fixes the turning points, the kinetic energy at every position, and the regions the object can never enter — all before a single equation of motion is solved.

IIVisualization 1 — Force Is the Negative Slope

At every position the spring force is minus the slope of the potential energy curve.

IIIVisualization 2 — Where the Motion Stops

A horizontal energy line cuts the parabola at the two turning points that bound the motion.

Energy E (J) 2.00

💡 A quantum object on the same diagram has a small but nonzero probability of being found inside the shaded bands — those regions are forbidden only classically.

IVVisualization 3 — The Gap Is the Kinetic Energy

The vertical gap between the energy line and the curve is $K$ : widest at equilibrium, zero at the turning points.

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VQuiz Questions

Problem 1 · Force From the Curve

Given: a spring with $k = 40\ \text{N/m}$ and $U^s(x) = \tfrac{1}{2}kx^2$ , with $U^s(0) \equiv 0$ — find the force component $F_x^s$ when the spring is stretched to $x = 0.15\ \text{m}$ .

✅ Correct!

F_x^s = -\dfrac{d}{dx}\left(\tfrac{1}{2}kx^2\right) = -kx = -(40)(0.15) = -6.0\ \text{N}

— negative, so the stretched spring pulls back toward

x = 0

❌ Check the sign. The magnitude is right. A stretched spring sits on the rising arm of the parabola, where the slope

dU^s/dx

is positive — and the force is minus that slope.

❌ Not quite. The factor

\tfrac{1}{2}

belongs to the potential energy, not to the force: differentiating

\tfrac{1}{2}kx^2

brings the exponent

2

down, and it cancels the

\tfrac{1}{2}

❌ That is an energy, not a force.

0.45\ \text{J}

is the value of

U^s(0.15)

itself; the force is the negative derivative of that curve, not its height.

❌ Not quite. Differentiate first, then attach the minus sign:

F_x^s = -dU^s/dx = -kx

Show solution

The force component is the negative slope of the potential energy curve:

F_x^s = -\frac{dU^s(x)}{dx} = -\frac{d}{dx}\left(\frac{1}{2}kx^2\right) = -kx

Substituting $k = 40\ \text{N/m}$ and $x = 0.15\ \text{m}$ :

F_x^s = -(40\ \text{N/m})(0.15\ \text{m}) = -6.0\ \text{N}

The minus sign is the restoring property: with the spring stretched ( $x > 0$ ) the force points in the $-x$ direction, back toward equilibrium. For comparison, the stored energy there is $U^s(0.15) = \tfrac{1}{2}(40)(0.15)^2 = 0.45\ \text{J}$ — a different quantity with different units.

Problem 2 · Locating the Turning Points

Given: a frictionless spring–object system with $k = 50\ \text{N/m}$ and conserved mechanical energy $E_m = 4.0\ \text{J}$ , with $U^s(0) \equiv 0$ — find $x_{\max}$ , the maximum extension of the spring.

✅ Correct! At a turning point

K = 0

, so

E_m = \tfrac{1}{2}kx_{\max}^2

and

x_{\max} = \sqrt{2E_m/k} = \sqrt{0.16} = 0.40\ \text{m}

❌ One step short.

2E_m/k = 0.16

x_{\max}^2

, not

x_{\max}

— and its units are

\text{m}^2

. Take the square root.

❌ The factor of 2 went missing.

\tfrac{1}{2}kx_{\max}^2 = E_m

gives

x_{\max}^2 = 2E_m/k

, not

E_m/k

;

\sqrt{E_m/k} = 0.28\ \text{m}

is the answer to a different equation.

❌ Two steps short.

E_m/k = 0.08

drops the factor of

2

and never takes the square root — and

\text{J}/(\text{N/m}) = \text{m}^2

, so it is not even a length. From

E_m = \tfrac{1}{2}kx_{\max}^2

: multiply by

2

, divide by

k

, then take the root.

❌ Not quite. Set the energy line equal to the curve,

E_m = \tfrac{1}{2}kx_{\max}^2

, and solve for

x_{\max}

Show solution

A turning point is where the horizontal energy line meets the parabola, so all the energy is potential and $K = 0$ :

E_m = U^s(x_{\max}) = \frac{1}{2}kx_{\max}^2

Solving for $x_{\max}$ :

x_{\max} = \sqrt{\frac{2E_m}{k}} = \sqrt{\frac{2(4.0\ \text{J})}{50\ \text{N/m}}} = \sqrt{0.16\ \text{m}^2} = 0.40\ \text{m}

The line meets the curve symmetrically, so the motion is confined to $-0.40\ \text{m} \le x \le 0.40\ \text{m}$ : $-x_{\max}$ is the maximum compression and $+x_{\max}$ the maximum extension.

Problem 3 · Reading the Gap

Given: an object of mass $m = 0.50\ \text{kg}$ on a frictionless surface, attached to a spring with $k = 200\ \text{N/m}$ , is pulled to $x = 0.10\ \text{m}$ and released from rest — find its kinetic energy and its speed as it passes $x = 0.060\ \text{m}$ .

Kinetic energy there?

Speed there?

✅ Correct! Released from rest at

x = 0.10\ \text{m}

fixes

E_m = 1.00\ \text{J}

; the gap at

x = 0.060\ \text{m}

K = 1.00 - 0.36 = 0.64\ \text{J}

, and

v = \sqrt{2K/m} = 1.60\ \text{m/s}

❌ Check the gap. Released from rest, the object starts at a turning point, so

E_m = U^s(0.10) = 1.00\ \text{J}

. The kinetic energy is the difference

E_m - U^s(0.060)

, not either height on its own.

❌ Check the speed step. Invert

K = \tfrac{1}{2}mv^2

to get

v = \sqrt{2K/m}

— using

K

, not

E_m

, and keeping the factor of

2

inside the root.

Show solution

Step 1 — the energy line. Released from rest, $K = 0$ there, so $x = 0.10\ \text{m}$ is a turning point and its height sets $E_m$ :

E_m = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.10)^2 = 1.00\ \text{J}

Step 2 — the curve at the new position.

U^s(0.060) = \frac{1}{2}(200)(0.060)^2 = 0.36\ \text{J}

Step 3 — the gap is the kinetic energy.

K = E_m - U^s(x) = 1.00\ \text{J} - 0.36\ \text{J} = 0.64\ \text{J}

Step 4 — turn energy into speed.

v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2(0.64\ \text{J})}{0.50\ \text{kg}}} = \sqrt{2.56\ \text{m}^2/\text{s}^2} = 1.60\ \text{m/s}

Moving inward from $0.060\ \text{m}$ the gap keeps widening, and the speed peaks at $x = 0$ with $v = \sqrt{2(1.00)/0.50} = 2.00\ \text{m/s}$ .

Problem 4 · Beyond the Turning Point

Given: $U^s(x) = \tfrac{1}{2}kx^2$ with $k = 100\ \text{N/m}$ and an object whose conserved mechanical energy is $E_m = 2.0\ \text{J}$ — decide what the diagram says about the position $x = 0.25\ \text{m}$ .

✅ Correct!

U^s(0.25) = 3.125\ \text{J}

rises above the energy line at

2.0\ \text{J}

, so

K

would be negative there. The turning point is

x_{\max} = \sqrt{2(2.0)/100} = 0.20\ \text{m}

❌ That number is the curve, not the gap.

3.125\ \text{J}

U^s(0.25)

. The kinetic energy is

E_m - U^s

, and here that difference is negative.

❌ Check where the line actually meets the curve. Momentarily at rest means

E_m = U^s(x)

, which gives

x = \sqrt{2(2.0)/100} = 0.20\ \text{m}

— the object has already turned before reaching

0.25\ \text{m}

❌ The arithmetic is right, but read what it means.

E_m - U^s = -1.125\ \text{J}

is exactly the contradiction:

K = \tfrac{1}{2}mv^2

can never be negative, so the object is never found there at all.

❌ Not quite. Compare the height of the curve at

x = 0.25\ \text{m}

with the height of the energy line, then ask whether

K = E_m - U^s

could be negative.

Show solution

Step 1 — height of the curve at that position.

U^s(0.25) = \frac{1}{2}(100)(0.25)^2 = 3.125\ \text{J}

Step 2 — compare with the energy line. Since $U^s(0.25) = 3.125\ \text{J} > E_m = 2.0\ \text{J}$ , the parabola is above the line there, so

K = E_m - U^s(x) = 2.0\ \text{J} - 3.125\ \text{J} = -1.125\ \text{J}

Step 3 — reject it. Kinetic energy is $K = \tfrac{1}{2}mv^2 \ge 0$ for any real speed, so no object with this energy can ever be found at $x = 0.25\ \text{m}$ : it lies in the classically forbidden region.

Step 4 — where it does turn.

x_{\max} = \sqrt{\frac{2E_m}{k}} = \sqrt{\frac{2(2.0)}{100}} = \sqrt{0.04} = 0.20\ \text{m}

The motion is the bounded oscillation $-0.20\ \text{m} \le x \le 0.20\ \text{m}$ ; everything outside is forbidden — classically.

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