OCR · A-Level · Physics
Course version: H556 · Version from Sept 2015
(b) energy-displacement graphs for a simple harmonic oscillator
Full specification wording
Energy of a simple harmonic oscillator: (b) energy-displacement graphs for a simple harmonic oscillator
Energy–displacement graphs show how kinetic and potential energy vary with displacement x for a simple harmonic oscillator.
Checked against the OCR H556 specification (opens in a new tab) on .
- Point 2 of 2
- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
Energy–displacement graphs show how kinetic and potential energy vary with displacement x for a simple harmonic oscillator. The potential energy curve is a parabola, Ep = ½mω²x², with a minimum of zero at x = 0 and maximum ½mω²A² at x = ±A. The kinetic energy curve is an inverted parabola, Ek = ½mω²(A² − x²), maximum at x = 0 and zero at x = ±A. Adding the two curves at every displacement gives a horizontal straight line at the constant total energy ½mω²A². The graphs are symmetric about x = 0 because displacement can be positive or negative. Reading the graphs shows that energy interchange is continuous and that the total energy depends on amplitude squared.
What earns the marks
- Potential energy varies as Ep = ½mω²x², giving a parabola with minimum zero at x = 0.
- Kinetic energy varies as Ek = ½mω²(A² − x²), giving an inverted parabola with maximum at x = 0.
- At x = ±A, kinetic energy is zero and potential energy equals the total energy ½mω²A².
- The sum of kinetic and potential energy at every displacement is the constant total energy ½mω²A².
- The graphs are symmetric about x = 0 and show that total energy is proportional to A².
- Graphical features such as intercepts, maxima and the constant-sum line can be used to find ω, A or energy values.
Common mistakes and how to put them right
Drawing the potential energy graph as a straight line.
Put it right: It is a parabola because Ep depends on x².
Showing the total energy curve varying with displacement.
Put it right: The total is constant, so it is a horizontal line.
Placing the kinetic energy maximum at x = ±A: kinetic energy is maximum at x = 0.
Forgetting that the graphs are symmetric for negative displacement as well as positive.
Exam tips
- Label the axes with energy and displacement, and mark the key values 0 and ±A clearly.
- Draw kinetic and potential energy on the same axes, then add them to show the constant total.
- Use the intercepts and maxima to read off A and the total energy rather than recalculating from scratch.
Try the lesson
The start of the Energy of a simple harmonic oscillator: (b) energy-displacement graphs for a simple harmonic oscillator lesson, made from checked teaching notes for this specification point.
Listen
Energy–displacement graphs show how kinetic and potential energy vary with displacement x for a simple harmonic oscillator. The potential energy curve is a parabola, Ep = ½mω²x², with a minimum of zero at x = 0 and maximum ½mω²A² at x = ±A. The kinetic energy curve is an inverted parabola, Ek = ½mω²(A² − x²), maximum at x = 0 and zero at x = ±A. Adding the two curves at every displacement gives a horizontal straight line at the constant total energy ½mω²A². The graphs are symmetric about x = 0 because displacement can be positive or negative. Reading the graphs shows that energy interchange is continuous and that the total energy depends on amplitude squared. A strong answer earns marks for these points. Potential energy varies as Ep = ½mω²x², giving a parabola with minimum zero at x = 0. Kinetic energy varies as Ek = ½mω²(A² − x²), giving an inverted parabola with maximum at x = 0. At x = ±A, kinetic energy is zero and potential energy equals the total energy ½mω²A². The sum of kinetic and potential energy at every displacement is the constant total energy ½mω²A². Exam tip: Label the axes with energy and displacement, and mark the key values 0 and ±A clearly.
See it
Put it in order
Put these steps in the order they happen.
- 1. x = ±A: Ep = ½mω²A², Ek = 0
- 2. Displacement increases: Ep rises as a parabola, Ek falls
- 3. x = 0: Ep = 0, Ek maximum
- 4. Total energy proportional to A²
- 5. Graphs symmetric about x = 0
- 6. Sum at every displacement: Ep + Ek = ½mω²A², a horizontal straight line
Key words
Say what each word means before you turn the card.
Energy–displacement relationships
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Parabolic energy curves
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Conservation of total energy
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Amplitude dependence of energy
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See every definition
- Energy–displacement relationships
- Graphs showing how kinetic and potential energy vary with displacement x for a simple harmonic oscillator.
- Parabolic energy curves
- The potential energy curve is a parabola, Ep = ½mω²x², and the kinetic energy curve is an inverted parabola, Ek = ½mω²(A² − x²).
- Conservation of total energy
- Adding the two curves at every displacement gives a horizontal straight line at the constant total energy ½mω²A².
- Amplitude dependence of energy
- The graphs show that total energy is proportional to A².
Label the diagram
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1Box 1
2Box 2
3Box 3
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