Find equilibrium before measuring displacement
You will be able to: Distinguish spring extension from displacement about a loaded equilibrium position.
Why does a hanging spring oscillate around a stretched position?
Hang a 1 kg mass from a spring. It stretches until the upward spring force balances the 10 N weight. Pull the mass slightly farther down and release it. The oscillation is centered on that loaded balance point, not the spring’s unstretched length.
A useful starting point: What makes an oscillation simple harmonic? →
Words and symbols before equations
- Extension z
- Stretch measured from the spring’s natural length, in m; downward positive.
- Equilibrium extension z_eq
- Extension at static balance, where kz_eq=mg.
- Oscillation displacement y
- Signed displacement from loaded equilibrium: y=z−z_eq.
- g
- Gravitational acceleration; use 10 m/s² in this example.
What this picture assumes
Ideal vertical spring: m=1 kg, k=20 N/m, g=10 m/s². Downward positive. Natural-length extension z and equilibrium displacement y are distinct. All allowed states leave the spring stretched. Bars show signed forces at a snapshot, not energy.
Connect the picture to the physics
Draw the mass’s forces: weight mg downward and spring force kz upward for a stretched spring. Thus F_net=mg−kz. At equilibrium, z_eq=mg/k; the spring force is not zero there, but the two forces balance.
Write z=z_eq+y. Then F_net=mg−k(z_eq+y)=−ky because mg=kz_eq. The same linear restoring form appears when displacement is measured from the correct equilibrium. Above equilibrium y<0 and the net force is downward; below equilibrium y>0 and it is upward.
Gravity shifts the balance position. In this ideal model it does not change the restoring coefficient k about that position. A mass can pass through the balance point with nonzero speed; equilibrium describes a position and force balance, not necessarily a resting state.
A worked example, step by step
For m=1 kg, k=20 N/m and g=10 m/s², find equilibrium extension and the net force when the mass is 0.10 m below that equilibrium.
- Balance forces: z_eq=mg/k=10/20=0.50 m.
- Below equilibrium means y=+0.10 m, so actual extension z=0.60 m.
- Weight=10 N down; spring force=20(0.60)=12 N up.
- Net force=10−12=−2 N, upward. Equivalently, −ky=−20(0.10)=−2 N.
Spring extension and displacement from equilibrium are different quantities for a vertical oscillator.
At loaded equilibrium, is the vertical spring unstretched?
Compare with an explanation
No. Its extension mg/k produces the upward force needed to balance weight.
Predict. Change one thing. Explain.
Move the mass above and below its loaded equilibrium. Compare the individual weight and spring-force bars with the net-force bar. Notice that zero net force does not mean either individual force vanishes.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Loaded equilibrium extension=0.5 m. y=0.1 m gives actual extension z=0.6 m. Weight=+10 N; spring=-12 N; net=-2 N. Net force points toward the loaded equilibrium.
Ideal vertical spring: m=1 kg, k=20 N/m, g=10 m/s². Downward positive. Natural-length extension z and equilibrium displacement y are distinct. All allowed states leave the spring stretched. Bars show signed forces at a snapshot, not energy.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant force, motion or energy relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA mass hangs from a spring with k=50 N/m. Its weight is 10 N. (a) Find equilibrium extension. (b) Find extension when y=−0.04 m. (c) Find net force there. (d) Explain why this force points toward equilibrium.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: z_eq=10/50=0.20 m.
- 1 point: z=0.20−0.04=0.16 m.
- 1 point: F_net=10−50(0.16)=+2 N downward.
- 1 point: The mass is above equilibrium, so downward acceleration points back toward it.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1At vertical equilibrium, what balances?
Upward spring force and downward weight.
RECALL 2Which displacement belongs in F_net=−ky?
Displacement from loaded equilibrium.
RECALL 3Is the individual spring force zero at that equilibrium?
No; it equals the weight in magnitude.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Find equilibrium before measuring displacement
- z_eq=mg/k; y=z−z_eq.
- F_net=mg−kz=−ky about loaded equilibrium.
Remember: Spring extension and displacement from equilibrium are different quantities for a vertical oscillator.
Conditions: Ideal vertical spring: m=1 kg, k=20 N/m, g=10 m/s². Downward positive. Natural-length extension z and equilibrium displacement y are distinct. All allowed states leave the spring stretched. Bars show signed forces at a snapshot, not energy.
Refresh Kid · Unit 7 · Objectives 7.1.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 7.1, objectives 7.1.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
Want to work through this with a tutor?
Bring your question about Find equilibrium before measuring displacement. Your explanation and answers remain free to access.