Find speed from position using energy
You will be able to: Derive the speed–position relationship and distinguish speed from signed velocity.
Can you find speed without knowing the time?
You see a spring block halfway between the center and an endpoint. You do not know how long it has been moving. Energy still tells you its speed if you know mass, spring stiffness and amplitude; it does not tell which way the block is moving.
A useful starting point: Watch energy trade between the spring and the block →
Words and symbols before equations
- Speed |v|
- Nonnegative magnitude of velocity, in m/s.
- Maximum speed v_max
- Speed when crossing equilibrium.
- Accessible position
- A position within −A≤x≤A for the specified ideal oscillator energy.
What this picture assumes
Horizontal ideal oscillator: k=8 N/m, m=2 kg, A=0.5 m, E=1 J. Only accessible positions −A≤x≤A are offered. Energy gives speed magnitude; phase is needed for signed velocity.
Connect the picture to the physics
Start with ½kA²=½mv²+½kx². Move the spring term to the other side and multiply by 2/m: v²=(k/m)(A²−x²). Thus speed is √[(k/m)(A²−x²)]. This derivation uses energy conservation, not a time equation.
At x=0, v_max=A√(k/m)=ωA. At x=±A, speed is zero. If |x|>A, the expression inside the square root is negative: the specified energy cannot reach that position. This is a physical restriction, not a negative speed.
At most intermediate positions the block can move either direction with the same speed. Use the phase or earlier motion to choose v=+|v| or v=−|v|. A plot of speed versus x is symmetric; a signed velocity versus time graph additionally carries direction and timing information.
A worked example, step by step
An oscillator has k=8 N/m, m=2 kg and A=0.50 m. Find its maximum speed and speed at x=0.30 m.
- v_max=A√(k/m)=0.50√4=1.00 m/s.
- At x=0.30 m, v²=(8/2)(0.50²−0.30²)=4(0.16)=0.64 m²/s².
- Speed=√0.64=0.80 m/s.
- The signed velocity can be +0.80 or −0.80 m/s; position and energy alone do not specify the direction.
Taking the positive square root gives speed. Do not silently choose a velocity direction without information about the motion.
Is the speed at x=A/2 equal to half the maximum speed?
Compare with an explanation
No. |v|/v_max=√(1−1/4)=√3/2≈0.866.
Predict. Change one thing. Explain.
Move x/A between −1 and +1. Predict the speed at the center, both endpoints and half-amplitude positions. Explain why the readout reports speed rather than signed velocity.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At x=0.3 m, speed=0.8 m/s. Maximum speed=1 m/s at x=0; speed=0 at x=±0.5 m. Either velocity sign is possible without phase information.
Horizontal ideal oscillator: k=8 N/m, m=2 kg, A=0.5 m, E=1 J. Only accessible positions −A≤x≤A are offered. Energy gives speed magnitude; phase is needed for signed velocity.
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 oscillator has m=1 kg, k=16 N/m and A=0.25 m. (a) Find total energy. (b) Find maximum speed. (c) Find speed at x=0.15 m. (d) Explain what further information is needed to state signed velocity there.
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Compare with the answer and four-point rubric
- 1 point: E=½(16)(0.25²)=0.50 J.
- 1 point: v_max=0.25√16=1 m/s.
- 1 point: |v|=√[16(0.0625−0.0225)]=0.80 m/s.
- 1 point: Its direction or phase: inward and outward passages have equal speed but opposite velocity signs.
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 1Can energy determine speed without time?
Yes, if the parameters and position are known.
RECALL 2What is maximum spring-oscillator speed?
A√(k/m).
RECALL 3Why can two velocities fit one position and energy?
The block can pass the position in either direction.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Find speed from position using energy
- |v|=√[(k/m)(A²−x²)], for |x|≤A.
- v_max=A√(k/m)=ωA.
- At x=±A, v=0; at equilibrium, |v| is maximum.
Remember: Taking the positive square root gives speed. Do not silently choose a velocity direction without information about the motion.
Conditions: Horizontal ideal oscillator: k=8 N/m, m=2 kg, A=0.5 m, E=1 J. Only accessible positions −A≤x≤A are offered. Energy gives speed magnitude; phase is needed for signed velocity.
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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