Oscillations: SHM, damping and resonance
Key idea: Connect measured periodic motion to the defining SHM model, its phase and energy relationships, then distinguish damping from driven resonance.
Before you start: Circular Motion objective chainEnergy & Fields objective chain
By the end, you can
- Describe and investigate free oscillations, and use amplitude, period, frequency, angular frequency, phase and phase difference.
- Recognise SHM from a = −ω²x and use its displacement, velocity, acceleration and graphical relationships.
- Describe kinetic–potential energy interchange and compare light, critical and heavy damping in practical systems.
- Interpret forced-response curves, damping effects and useful or harmful resonance.
Starting-point self-check
1. Check your starting point
Attempt all four groups without notes and mark the first definition, phase sign, energy location or response-curve feature you could not justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Energy interchange and damping 9(h)–(i)
Question 1
A 0.50 kg ideal oscillator has ω = 4.0 rad s⁻¹ and amplitude 0.10 m. Find total, potential and kinetic energy at x = 0.060 m, then state what damping changes.
Check the model response
E = ½mω²A² = 0.040 J. U = ½mω²x² = 0.0144 J and K = E − U = 0.0256 J. Damping transfers mechanical energy to the environment, so amplitude decreases; the return pattern depends on the degree of damping.
repair
2. Repair the eight common breaks
Use only the repair matching an error, then redraw the equilibrium line, phase cycle, energy bars or response axes before retrying.
Energy interchange and damping 9(h)–(i)
Check this idea
Misconception: Kinetic energy is greatest where the restoring force is greatest.
Repair: Kinetic energy is greatest at equilibrium. Potential energy and restoring-force magnitude are greatest at the extremes.
Check this idea
Misconception: Every non-oscillatory return is critically damped.
Repair: Critical and heavy damping are both non-oscillatory, but critical damping gives the fastest return without overshoot; heavy damping returns more slowly.
worked example
3. Follow four worked models
Follow how each solution fixes the timing reference, equilibrium direction, energy boundary and driving frequency before calculating.
Energy interchange and damping 9(h)–(i)
Model 1
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 model response
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.
guided practice
4. Guided practice
Use each hint only to select a time-to-phase conversion, SHM equation, energy location or response-curve comparison.
Energy interchange and damping 9(h)–(i)
Question 1
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 the model response
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.
independent practice
5. Independent practice
Solve without repair notes and state the equilibrium reference, phase convention and energy or driving assumptions used.
Energy interchange and damping 9(h)–(i)
Question 1
Explain the energy sequence during one ideal SHM cycle beginning at positive maximum displacement, then compare underdamped, critically damped and overdamped return.
Check the model response
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.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Energy interchange and damping 9(h)–(i)
Question 1
A 0.25 kg oscillator has ω = 12 rad s⁻¹ and amplitude 0.040 m. Find total energy and the energy split at x = 0.020 m. Explain why critical damping suits a car suspension.
Check the model response
E = ½mω²A² = 0.0288 J. U/E = x²/A² = 0.25, so U = 0.0072 J and K = 0.0216 J. Critical damping returns the suspension to equilibrium as quickly as possible without continued bouncing or overshoot.
Re-test practice
7. Delayed re-test practice
Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Energy interchange and damping 9(h)–(i)
Question 1
An oscillator's amplitude falls from 6.0 cm to 3.0 cm while its ω remains approximately constant. State the factor change in mechanical energy and distinguish this underdamped decay from critical damping.
Check the model response
Since E ∝ A², halving amplitude reduces energy to one quarter. An underdamped system continues to cross equilibrium with decaying amplitude; a critically damped system returns fastest without oscillating.