Learning
LESSON 12 / 12 · TOPIC 7.4

More amplitude: more energy, same ideal period

You will be able to: Compare amplitude, maximum speed, maximum acceleration and total energy under controlled conditions.

Free study resourceReview editionTeacher review pending

Why does pulling a spring twice as far not double its period?

Release the same ideal spring block from 0.10 m and then from 0.20 m. The second trial travels twice as far each cycle, but it also moves faster. Both have the same period, while the second trial starts with four times the mechanical energy.

A useful starting point: Find speed from position using energy →

Words and symbols before equations

Controlled comparison
Change one input while keeping the stated other quantities fixed.
Maximum acceleration a_max
Largest acceleration magnitude, (k/m)A.
External work W_ext
Energy transferred to the selected system by an external interaction.
Dissipation
Conversion of mechanical energy into other forms, such as thermal energy, through resistance.
Energy changes with amplitude squared0+Reference A=0.2 m0.4Selected trial energy0.4Energy (J) · same scale for every bar · full half-axis 1.6
Read this model snapshot. A=0.2 m (1× reference); E=0.4 J (1×). v_max=0.894 m/s; a_max=4 m/s². T=1.405 s stays fixed. Both maxima scale by 1×.
What this picture assumes

Separate ideal horizontal spring trials with m=1 kg and k=20 N/m held fixed. Reference amplitude is 0.2 m. Each trial receives its own starting energy; the slider does not inject work during an ongoing trajectory.

Connect the picture to the physics

At fixed m and k, T=2π√(m/k) does not depend on A. Maximum speed is A√(k/m), and maximum acceleration magnitude is (k/m)A. Both scale directly with amplitude, whereas E=½kA² scales with its square.

An amplitude change between trials requires a different energy input or initial condition. To increase amplitude from A_1 to A_2 without dissipation, the net added mechanical energy is ½k(A_2²−A_1²). Amplitude does not grow spontaneously in an isolated ideal oscillator.

Real oscillations often decay as resistance converts mechanical energy to thermal energy. Then the isolated constant-energy SHM model no longer describes the complete process. For this lesson, compare ideal trials separately and keep m and k fixed; changing mass or stiffness at the same time changes the timing too.

Double A; keep m and k fixed
QuantityDependence on AChange
PeriodIndependent of A in ideal SHMUnchanged
Maximum speedProportional to ADoubles
Maximum accelerationProportional to ADoubles
Total mechanical energyProportional to A²Quadruples

A worked example, step by step

A 1 kg block on a 20 N/m spring is released first from A=0.10 m and then from A=0.20 m. Compare energy, maximum speed and period.

  1. E_1=½(20)(0.10²)=0.10 J; E_2=0.40 J: four times larger.
  2. v_max,1=0.10√20≈0.447 m/s; v_max,2≈0.894 m/s: twice as large.
  3. Both periods are 2π√(1/20)≈1.405 s.
  4. The second release needs 0.30 J more starting mechanical energy. The longer path is traveled with proportionally larger speeds.
Common mix-up

A constant ideal period does not mean unchanged speed, force or energy. Always specify what is held fixed.

CHECK THE IDEA

To triple amplitude at fixed k, must energy triple?

Compare with an explanation

No. E∝A², so energy must become nine times as large.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Change amplitude while holding m=1 kg and k=20 N/m fixed. Compare each state with the 0.20 m reference. Predict the factors for maximum speed, maximum acceleration, energy and period before moving the slider.

On narrow screens, swipe or scroll diagrams sideways to read all labels.

Energy changes with amplitude squared0+Reference A=0.2 m0.4Selected trial energy0.4Energy (J) · same scale for every bar · full half-axis 1.6

A=0.2 m (1× reference); E=0.4 J (1×). v_max=0.894 m/s; a_max=4 m/s². T=1.405 s stays fixed. Both maxima scale by 1×.

Same period, different amplitudePosition x (m)Time (s)0-0.40.351-0.20.70201.0540.21.4050.4

Separate ideal horizontal spring trials with m=1 kg and k=20 N/m held fixed. Reference amplitude is 0.2 m. Each trial receives its own starting energy; the slider does not inject work during an ongoing trajectory.

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.

1. At fixed m and k, doubling amplitude makes maximum acceleration…

Show answer and reasoning

Twice as large. a_max=(k/m)A is linear in A.

2. An oscillator loses mechanical energy to resistance. Its constant-energy ideal model…

Show answer and reasoning

Needs additional energy accounting. Mechanical energy becomes other forms; total energy is not destroyed.

Original written challenge

4 points · self-check · not an official AP question

An ideal horizontal oscillator’s amplitude increases from 0.10 m to 0.30 m with k=40 N/m and the same mass. (a) Find initial and final energy. (b) Find the added energy. (c) Give the maximum-speed factor. (d) State the period change and justify it.

This response is not submitted or saved. Copy it before leaving.

Compare with the answer and four-point rubric
  1. 1 point: E_i=0.20 J and E_f=1.80 J.
  2. 1 point: Added energy=1.60 J.
  3. 1 point: Maximum speed triples because it is proportional to A at fixed m,k.
  4. 1 point: Period is unchanged because T=2π√(m/k) and neither m nor k changed.

Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.

Recall the ideas without notes Review →

Retrieve it before you reveal it.

RECALL 1At fixed m,k, which quantities double when A doubles?

Maximum speed and maximum acceleration magnitude.

RECALL 2How does energy change when A doubles?

It quadruples.

RECALL 3How can a real oscillation’s amplitude decay without destroying energy?

Mechanical energy is converted into thermal or other energy forms.

Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.

More amplitude: more energy, same ideal period

  • Fixed m,k: T independent of A; v_max∝A; a_max∝A; E∝A².
  • Ideal added energy=½k(A_2²−A_1²).

Remember: A constant ideal period does not mean unchanged speed, force or energy. Always specify what is held fixed.

Conditions: Separate ideal horizontal spring trials with m=1 kg and k=20 N/m held fixed. Reference amplitude is 0.2 m. Each trial receives its own starting energy; the slider does not inject work during an ongoing trajectory.

Refresh Kid · Unit 7 · Objectives 7.3.A, 7.4.A · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 7.4, objectives 7.3.A, 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.

OPTIONAL LIVE SUPPORT

Want to work through this with a tutor?

Bring your question about More amplitude: more energy, same ideal period. Your explanation and answers remain free to access.

Request a physics tutor →Ask about this lesson on WhatsAppThe team can confirm teacher availability and next steps.