Explain how the kinetic energy and potential energy of an object undergoing simple harmonic motion vary with time over one complete oscillation, and identify the time within one cycle at which the rate of energy transfer between kinetic and potential energy is greatest.
Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).
An object oscillates with simple harmonic motion about its equilibrium position with period T.
Model answer (4 marks)
Kinetic energy is maximum when the displacement is zero (at the equilibrium position) and is zero when the displacement is at its maximum (the amplitude). Potential energy is maximum at the maximum displacement and is zero at the equilibrium position, so the two energies vary in opposite phase. Both KE and PE vary sinusoidally with time and each completes two full cycles during one oscillation, i.e. they have a period of T/2. The rate of energy transfer between KE and PE is greatest when the object passes through the equilibrium position (t = T/4, 3T/4, …), where the slope of the energy–time curves is steepest.
Examiner tips
- Use the correct terms – kinetic energy, potential energy, equilibrium position, amplitude, period T/2; show the opposite phase relationship; identify the steepest slope at t = T/4, 3T/4; keep answer concise to fit 4 marks.
Common mistakes
- Confusing the time of maximum energy transfer with the time of maximum displacement; not recognising that KE and PE each have a period of T/2; using non‑UK spelling such as "potential energy" instead of "potential energy" (UK spelling is the same but avoid American spellings).
Mark scheme (4 marks)
- Kinetic energy is maximum at the equilibrium position (zero displacement) and minimum (zero) at maximum displacement (amplitude).
- Potential energy is maximum at maximum displacement (amplitude) and minimum (zero) at the equilibrium position, so it varies in the opposite sense to kinetic energy.
- Both kinetic and potential energy vary periodically with a period of T/2 (i.e. they each complete two full cycles per oscillation), since both are proportional to the square of a sinusoidal function.
- The rate of energy transfer between KE and PE is greatest when the object is at the equilibrium position (or equivalently at T/4, 3T/4, etc. after release from rest at amplitude), because this is where the gradient of each energy–time graph is steepest / the speed is changing most rapidly.
Key terms in this question
simple harmonic motion · kinetic energy · potential energy · rate of energy transfer
Related
- All IB DP Physics Higher Level (2023 syllabus) revision notes →
- How to answer a "Explain" question →
- Decode the mark scheme abbreviations →
More Simple harmonic motion questions
- Explain why the total mechanical energy of an object undergoing simple harmonic …
- Explain why the motion of a simple pendulum released from a small angle approxim…
- Explain how the velocity and acceleration of an object undergoing simple harmoni…
- Explain how the restoring force acting on a mass–spring system gives rise to sim…