Watch energy trade between the spring and the block
You will be able to: Use energy bars and conservation to describe an ideal horizontal spring oscillator.
Where does the energy go as an oscillator crosses the middle?
At a spring block’s rightmost point, the block is momentarily at rest and the spring stores all the oscillator’s mechanical energy. As the block heads toward the middle, spring energy becomes kinetic energy. At the middle, speed is greatest.
A useful starting point: Choose a sine or cosine model that matches the start →
Words and symbols before equations
- Elastic potential energy U
- For an ideal horizontal spring, U=½kx² with zero at its natural-length equilibrium.
- Kinetic energy K
- Energy of motion: K=½mv².
- Mechanical energy E
- Sum K+U for the selected oscillator system.
- Isolated ideal model
- No net energy transfer from outside and no mechanical energy dissipated by friction or drag.
What this picture assumes
Horizontal ideal oscillator: k=8 N/m, m=2 kg, A=0.5 m. Block plus spring system, no dissipation or external work. U=0 at the natural-length equilibrium. Slider selects a position, not elapsed time or velocity direction.
Connect the picture to the physics
Choose block plus spring as the system. On a level frictionless surface, the normal force and weight do no work along the motion; with an ideal fixed support, no external work changes the oscillator’s energy. E=K+U stays constant.
At either turning point x=±A, v=0, so E=½kA². At any intermediate x, U=½kx² and K=E−U. The squared displacement means equal U at equal distances on opposite sides, and equal speed magnitude there. Direction of velocity is still a separate part of the state.
At x=0, U is minimum and K maximum. At ±A, U is maximum and K=0. This energy zero is chosen for the horizontal spring. A vertical oscillator also needs gravitational potential energy included; one cannot blindly treat its actual spring extension as displacement from loaded equilibrium in the same energy bookkeeping.
A worked example, step by step
A horizontal oscillator has k=8 N/m and A=0.50 m. Find total energy and the energy split at x=+0.25 m.
- At the turning point, E=½kA²=½(8)(0.50²)=1.00 J.
- At x=0.25 m, U=½(8)(0.25²)=0.25 J.
- K=E−U=1.00−0.25=0.75 J.
- Half the amplitude gives one-quarter of the total energy in the spring, not one-half, because U depends on x².
Potential energy depends on displacement squared. Equal energy bars do not generally occur at half the amplitude.
At x=A/2, what fraction of total energy is kinetic?
Compare with an explanation
Three quarters: U/E=x²/A²=1/4, so K/E=3/4.
Predict. Change one thing. Explain.
Move the position from −A to +A. Predict the spring and kinetic energy fractions at x=0, A/2 and A. Keep total energy fixed and explain why the two sides give the same energy bars.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
x/A=0.5, so x=0.25 m. U=0.25 J; K=0.75 J; E=1 J. Fractions: spring 25%, kinetic 75%. The sign of x does not change the split.
Horizontal ideal oscillator: k=8 N/m, m=2 kg, A=0.5 m. Block plus spring system, no dissipation or external work. U=0 at the natural-length equilibrium. Slider selects a position, not elapsed time or velocity direction.
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 horizontal ideal spring has k=50 N/m and amplitude 0.20 m. (a) Find E. (b) Find U at x=0.10 m. (c) Find K there. (d) Explain what changes in these energies at x=−0.10 m.
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Compare with the answer and four-point rubric
- 1 point: E=½(50)(0.20²)=1 J.
- 1 point: U=½(50)(0.10²)=0.25 J.
- 1 point: K=0.75 J.
- 1 point: Neither energy changes because U depends on x²; velocity direction is not fixed by energy alone.
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 1Where is spring potential energy greatest?
At x=±A.
RECALL 2Where is kinetic energy greatest?
At equilibrium.
RECALL 3Which quantity stays constant in the isolated ideal model?
Total mechanical energy E=K+U.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Watch energy trade between the spring and the block
- E=K+U=½kA² for the isolated horizontal ideal spring oscillator.
- U=½kx²; K=½k(A²−x²).
- At ±A: K=0. At x=0: K=E.
Remember: Potential energy depends on displacement squared. Equal energy bars do not generally occur at half the amplitude.
Conditions: Horizontal ideal oscillator: k=8 N/m, m=2 kg, A=0.5 m. Block plus spring system, no dissipation or external work. U=0 at the natural-length equilibrium. Slider selects a position, not elapsed time or velocity direction.
Refresh Kid · Unit 7 · Objectives 7.4.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 7.4, objectives 7.4.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.
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