Energy interchange and damping

Key idea: H2 Physics lessons on free oscillations, simple harmonic motion, energy interchange, damping, forced response and resonance.

  • GCE A-Level H2 Physics 2027

Learn the idea

Big question: How does energy move during SHM, and what does damping change?

In ideal SHM, kinetic and potential energy interchange while total energy stays constant; total energy is proportional to A². Resistive forces transfer mechanical energy to internal energy, so amplitude falls. Light damping allows many cycles, critical damping returns fastest without overshoot, and heavy damping returns more slowly without oscillating.

Track kinetic and potential energy by position

For ideal SHM, total mechanical energy is constant: E = ½mω²A². With the equilibrium potential chosen as zero, potential energy is ½mω²x² and kinetic energy is ½mω²(A² − x²).

At x = ±A the oscillator stops momentarily, so all energy is potential. At x = 0 its speed is Aω and all energy is kinetic. Because energy depends on squared sine or cosine, each energy form repeats twice during one displacement cycle.

Check your understanding: At x = A/2, what fractions of total energy are potential and kinetic?

Potential fraction is x²/A² = 1/4; kinetic fraction is 3/4.

Distinguish light, critical and heavy damping

A resistive force transfers mechanical energy to internal energy, so successive amplitudes fall. Light damping allows repeated crossings of equilibrium. Critical damping gives the fastest return without overshoot; heavy damping also avoids overshoot but returns more slowly.

Damping changes the total-energy envelope, not the rule that energy is conserved across all stores. In a graph, do not call every non-oscillatory trace critical: compare its settling time with the fastest possible non-overshooting response.

Check your understanding: Which damping is best for a door closer that must shut quickly without oscillating?

Approximately critical damping, because it returns fastest without overshoot.

Energy interchange in simple harmonic motionA graph against displacement shows upward-curving potential energy, downward-curving kinetic energy and constant total energy. A second graph against time shows kinetic and potential energy alternating twice per oscillation cycle.Energy against displacementtotalpotentialkinetic−A0+AEnergy against timePEKE0T/2TPosition accountx = ±A: v = 0, KE = 0, PE = totalx = 0: speed maximum, KE = total, PE = 0Cycle accountEnergy swaps every quarter-cycleEach energy curve has period T/2
Scroll diagram horizontally to read all labels.
At the extremes all the ideal oscillator's energy is potential; at equilibrium it is kinetic. Both energy curves repeat every half-period because they depend on squared quantities.
Light, critical and heavy damping comparedThree displacement-time sketches. Light damping crosses equilibrium repeatedly with decreasing amplitude. Critical damping returns to equilibrium fastest without crossing it. Heavy damping also does not cross equilibrium but returns more slowly.Light dampingunderdampedrepeated crossingsCritical dampingfastest without overshootquickest settlingHeavy dampingoverdampedslower settlingxtime
Scroll diagram horizontally to read all labels.
Only light damping oscillates. Critical damping is the boundary case that returns to equilibrium fastest without overshoot; heavy damping is slower.

Key ideas to keep

  • At equilibrium kinetic energy is maximum, not total energy.
  • Damping reduces amplitude and total mechanical energy, not necessarily the natural frequency by a large amount.
  • Critical damping is a boundary case, not the strongest possible damping.

Worked example

Calculate the energy account at one displacement

Question: A 0.30 kg oscillator has ω = 5.0 rad s⁻¹ and amplitude 0.080 m. Find total, potential and kinetic energy at x = 0.050 m, and its speed there.

  1. Step 1: Find the fixed total

    Why: Amplitude sets the ideal oscillator's complete mechanical-energy store.

    Working: E = ½(0.30)(5.0²)(0.080²) = 0.0240 J.

  2. Step 2: Find the position-dependent store

    Why: SHM potential energy grows with x².

    Working: U = ½(0.30)(5.0²)(0.050²) = 0.00938 J; K = 0.0240 − 0.00938 = 0.0146 J.

  3. Step 3: Recover speed

    Why: Kinetic energy gives speed magnitude independently of direction.

    Working: v = √(2K/m) = √[2(0.0146)/0.30] = 0.312 m s⁻¹.

Answer: E = 0.0240 J, U = 0.00938 J, K = 0.0146 J and speed = 0.312 m s⁻¹.

Check: U + K equals the total and the speed is below vmax = Aω = 0.400 m s⁻¹.

Question

A 0.40 kg oscillator with ω = 5.0 rad s⁻¹ has amplitude 0.080 m. Find its energy at an extreme and the kinetic-energy fraction at x = 0.040 m. Explain the lightly damped change over many cycles.

Check the worked solution

E = ½mω²A² = 0.0320 J. Since U/E = x²/A² = 0.25, K/E = 0.75 and K = 0.0240 J. With light damping, energy is transferred gradually to the environment, so successive amplitudes and total mechanical energy fall while oscillations continue.

Practise with support

Try this

An ideal oscillator has total energy 0.090 J and amplitude 0.12 m. Find its potential and kinetic energies at x = 0.080 m, then name the fastest non-oscillatory damping regime.

Hint: For SHM, U/E = x²/A².

Check your answer

U = 0.090(0.080/0.12)² = 0.040 J and K = 0.050 J. Critical damping is the fastest return to equilibrium without oscillation or overshoot.

Practise independently

Your turn

Explain the energy sequence during one ideal SHM cycle beginning at positive maximum displacement, then compare underdamped, critically damped and overdamped return.

Check your answer

At the positive extreme, potential energy is maximum and kinetic energy zero. Towards equilibrium, potential converts to kinetic; at equilibrium kinetic is maximum. The exchange reverses towards the negative extreme and repeats. Underdamping oscillates with decaying amplitude, critical damping returns fastest without overshoot, and overdamping returns without oscillation but more slowly.

Common mistakes

Common mistake

Kinetic energy is greatest where the restoring force is greatest.

What is wrong with this reasoning?

Show better thinking

Kinetic energy is greatest at equilibrium. Potential energy and restoring-force magnitude are greatest at the extremes.

Common mistake

Every non-oscillatory return is critically damped.

What is wrong with this reasoning?

Show better thinking

Critical and heavy damping are both non-oscillatory, but critical damping gives the fastest return without overshoot; heavy damping returns more slowly.

Exam guidance

Link each graph feature to an energy statement rather than only naming the damping type.

Exam-style practice [8 marks]

Sketch kinetic, potential and total energy against displacement for ideal SHM. Then compare lightly, critically and heavily damped displacement–time traces and give one suitable application of critical damping.

Plan before you answer

  • Use squared displacement for energy shapes.
  • Keep total energy horizontal in the ideal graph.
  • Classify damping by crossings and settling time.
Mark your answer and compare the model

Marking points

Tick each point only if your answer states it clearly.

Model answer

Potential energy rises as x² from zero at equilibrium to the total at ±A. Kinetic energy is total minus potential, so it is greatest at equilibrium and zero at the extremes; the ideal total is constant. Light damping gives decaying oscillations, critical damping returns fastest without overshoot and heavy damping returns more slowly. Resistance transfers mechanical energy internally. A near-critically damped door closer shuts quickly without repeated swinging.

Check what stayed with you

Recall question 1

How does total SHM energy depend on amplitude?

Check the answer

It is proportional to A².

Recall question 2

Where is kinetic energy greatest?

Check the answer

At equilibrium.

Recall question 3

How do critical and heavy damping differ?

Check the answer

Both avoid oscillation, but critical damping returns fastest while heavy damping is slower.

Try this next

Continue to the next lesson in this topic.

Forced oscillations, resonance and response

Syllabus and review details

This lesson covers the listed H2 Physics 9478 outcomes. The official topic states no explicit exclusions and defines natural frequency as the frequency of a system in free oscillation. Ideal SHM uses a linear restoring relation and no environmental energy exchange; damping and steady forced response are introduced only when stated.

  • GCE A-Level H2 PhysicsTopic 9(h) / Topic 9(i) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027