What sets a spring oscillator’s period?
You will be able to: Use mass and spring stiffness to predict period and frequency changes.
Does a heavier spring block oscillate faster or slower?
Attach a heavier block to the same spring. The spring has more inertia to accelerate, so oscillations take longer. Replace the spring with a stiffer one at the same mass and the stronger restoring response makes each cycle shorter.
A useful starting point: Count full cycles: period and frequency →
Words and symbols before equations
- Mass m
- Oscillating mass in kg; spring mass is neglected in this model.
- Spring constant k
- Stiffness in N/m.
- Natural period
- Period of the ideal undriven spring–mass oscillator.
- π (pi)
- Approximately 3.142; 2π appears in one full sinusoidal cycle.
What this picture assumes
Ideal spring, negligible spring mass and resistance. Release from rest at x=+0.2 m at time zero. Each setting is a separate trial, not a mass or stiffness change during motion.
Connect the picture to the physics
For an ideal mass–spring oscillator, T=2π√(m/k). The square root matters: four times the mass doubles T, while four times the stiffness halves T. Frequency is the reciprocal, f=(1/2π)√(k/m).
The restoring acceleration a=−(k/m)x shows why the ratio k/m controls the timing. More stiffness gives larger acceleration at a given displacement; more mass gives smaller acceleration. The period formula applies to both horizontal and vertical ideal spring motion about the appropriate equilibrium.
Amplitude does not appear in this formula. A larger amplitude gives a longer distance to travel, but also larger forces, accelerations and speeds. In ideal SHM these changes balance so the period stays the same. Real springs can depart from the model when stretched beyond their linear range.
A worked example, step by step
A 1 kg mass oscillates on an ideal spring with k=4 N/m. Calculate its period. Then replace it with a 4 kg mass on the same spring.
- T_1=2π√(1/4)=π≈3.14 s.
- T_2=2π√(4/4)=2π≈6.28 s.
- The mass increased by a factor of 4; the period increased by √4=2.
- The frequency therefore halves. Changing amplitude alone would not change this ideal period.
Period scales with the square root of m/k, not directly with mass or stiffness.
If both m and k double, does T change?
Compare with an explanation
No. The ratio m/k stays the same.
Predict. Change one thing. Explain.
Hold k fixed and quadruple m. Then reset and quadruple k while holding m fixed. Use the period readout and the time graph to compare the two changes.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
m=1 kg; k=4 N/m; T=3.142 s and f=0.318 Hz. Amplitude remains 0.2 m. These are separate ideal trials.
Ideal spring, negligible spring mass and resistance. Release from rest at x=+0.2 m at time zero. Each setting is a separate trial, not a mass or stiffness change during motion.
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 questionAn ideal spring system has m=0.5 kg and k=8 N/m. (a) Find T. (b) Find f. (c) Predict T if m becomes 2 kg. (d) Explain the effect of doubling amplitude within the ideal range.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: T=2π√(0.5/8)=π/2≈1.57 s.
- 1 point: f=2/π≈0.637 Hz.
- 1 point: Mass quadruples, so T doubles to π≈3.14 s.
- 1 point: Period remains unchanged; maximum speed and restoring acceleration increase.
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 1Spring period formula?
T=2π√(m/k).
RECALL 2What does larger k do at fixed m?
It reduces T and increases f.
RECALL 3Does gravity set the ideal vertical spring period?
No. It shifts equilibrium; T still depends on m/k.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
What sets a spring oscillator’s period?
- T=2π√(m/k); f=(1/2π)√(k/m).
- Ideal spring period is independent of amplitude and g.
Remember: Period scales with the square root of m/k, not directly with mass or stiffness.
Conditions: Ideal spring, negligible spring mass and resistance. Release from rest at x=+0.2 m at time zero. Each setting is a separate trial, not a mass or stiffness change during motion.
Refresh Kid · Unit 7 · Objectives 7.2.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 7.2, objectives 7.2.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 What sets a spring oscillator’s period?. Your explanation and answers remain free to access.