OxfordAQA International AS and A-level Physics: Oscillating systems, damping and resonance (9630) – Revision Notes

These notes condense sections 3.5.1 (Oscillating systems) and 3.5.2 (Forced vibrations and resonance) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, exams from May/June 2020 onwards. Both sections are shared AS and A-level content. The derivation of the period formulae is in 3.6.2, which is International A-level only. For full explanations use the study guide, and test yourself with the practice questions. The course hub and printable checklist show the rest of the course. Not sure where to start? Try the free diagnostics. Related: measurements and their errors.

Key definitions

• Period T: the time taken by one full oscillation. Frequency f = 1/T.
• Amplitude A: maximum displacement from equilibrium.
• Free vibration: displaced once, then left to oscillate at its natural frequency.
• Forced vibration: driven by a periodic force; it vibrates at the driver’s frequency.
• Damping: energy transferred from the oscillator to the surroundings by resistive forces, so the amplitude falls.
• Resonance: the driver’s frequency matches the natural frequency, so the amplitude peaks and the driver transfers energy at its greatest rate.

Formulae

Result	Equation	Notes
Mass-spring period	T = 2π√(m/k)	k in N m⁻¹; independent of amplitude
Simple pendulum period	T = 2π√(l/g)	Small angles; l to centre of bob; independent of mass
Rearranged	k = 4π²m / T², g = 4π²l / T²	Gradient of T² against m is 4π²/k; against l is 4π²/g
Frequency	f = 1/T	Hz
International A-level only	a = −ω²x, T = 2π/ω	Condition for SHM
International A-level only	x = A cos ωt, v = ±ω√(A² − x²)	Maximum speed ωA, maximum acceleration ω²A
International A-level only	Total energy = ½mω²A²	For a spring, ½kA²

Method in steps: using the period formulae

1. Write down which system it is and which quantity is fixed.
2. Check units: kg, m, N m⁻¹, s.
3. Substitute into T = 2π√(m/k) or T = 2π√(l/g), or rearrange first.
4. For “what happens when…” questions, use proportionality: T² ∝ m (spring) and T² ∝ l (pendulum).
5. Give an answer with a sensible number of significant figures and the unit.

Worked reminder. A 0.60 kg mass on a spring with k = 24 N m⁻¹: T = 2π√(0.60/24) = 2π × 0.158 = 0.99 s. A pendulum needs T = 1.8 s: l = gT²/4π² = 9.81 × 3.24 / 39.48 = 0.81 m.

Energy in an oscillation

Position	Ek	Ep	Total
Maximum displacement	zero	maximum	constant
Equilibrium	maximum	minimum	constant

• Against displacement: Ep is a U-shaped parabola, Ek an inverted parabola, total a horizontal line meeting both tops.
• Against time: total is horizontal. Ek and Ep each go through two cycles for every cycle of the motion, are never negative, and are a quarter of a period out of step.
• With damping, the total falls with time and the curves shrink.

Worked reminder. A 0.40 kg trolley on a horizontal spring (k = 200 N m⁻¹) is pulled 5.0 cm and released. Ep at the start = ½kA² = ½ × 200 × 0.050² = 0.25 J, which equals the maximum Ek. Maximum speed = √(2 × 0.25 / 0.40) = 1.1 m s⁻¹.

Damping

Type	Behaviour	Example
Light	Oscillates; amplitude falls each cycle; period almost unchanged	Playground swing left alone
Critical	Returns to equilibrium fastest with no oscillation	Car suspension, door closer
Heavy	No oscillation; slow return	Door closer set too stiff

If the resistive force grows with speed, the amplitude falls by the same fraction each cycle. Mechanical energy becomes thermal energy.

Required practical 4

Pendulum: vary l, time 10 oscillations from a fiducial marker at equilibrium, find T, plot T² against l. Gradient 4π²/g.

Mass-spring: vary m, time 10 oscillations, plot T² against m. Gradient 4π²/k.

Good practice:

• Time multiple oscillations and repeat; take the mean.
• Start and stop at equilibrium with a fiducial marker, where the speed is highest and the timing error smallest.
• Use a set square and plumb line to check the string is vertical and to set a small angle.
• Keep the pendulum angle small.
• Use at least six values and a line of best fit.
• Percentage uncertainty in T² is double that in T.
• For the spring, mass of the spring itself means the line does not pass through the origin.

Worked reminder (damping). A lightly damped oscillator has amplitude 6.0 cm. After 10 oscillations it is 3.0 cm. If the energy is proportional to amplitude squared, the fraction of energy left is (3.0/6.0)² = 0.25, so 75% has been transferred to thermal energy of the surroundings. A second 10 oscillations would leave the amplitude at 1.5 cm if the same fraction is lost each time.

What to write in a “describe” answer on damping. Name the resistive force. State that energy is transferred out of the oscillator. State that the amplitude decreases with time. State what happens to the period (almost no change for light damping). Name the type if asked: light, critical or heavy.

Forced vibrations and resonance

1. A driver applies a periodic force at frequency f.
2. After the start-up, the system vibrates at f.
3. Amplitude rises as f approaches the natural frequency f₀ and peaks at resonance.
4. More damping gives a lower, broader peak, slightly below f₀. Light damping gives a tall, sharp peak.

Examples: a child on a swing, a loose car panel rattling at one engine speed, footbridges, vehicle suspension, musical instruments, and stationary waves on strings and in air columns.

Stationary-wave resonance. A string fixed at both ends resonates at f = v/2L, then 2f, 3f and so on. An air column closed at one end resonates when the length is λ/4, then 3λ/4, 5λ/4.

Worked reminder. A string of length 1.0 m with wave speed 50 m s⁻¹: lowest resonance f = 50 / 2.0 = 25 Hz; next 50 Hz. A closed tube resonates at its shortest length of 0.15 m: λ = 0.60 m, so f = 340 / 0.60 = 567 Hz.

Must-know distinctions

Compare	Point
Natural vs driving frequency	Natural depends on the system. Driving comes from the driver. Resonance needs them equal.
Free vs forced	Free: natural frequency. Forced: driver’s frequency.
Light vs critical damping	Light: oscillates. Critical: shortest time, no oscillation.
Energy vs motion frequency	Ek and Ep cycle at twice the frequency of the motion.
Amplitude vs period	Amplitude does not change T (small angles for a pendulum).

Quick self-test

1. Calculate T for a 0.25 kg mass on a spring of stiffness 36 N m⁻¹.
2. Calculate T for a pendulum of length 0.25 m.
3. The mass on a spring is multiplied by 9. By what factor does T change?
4. A pendulum is lengthened to four times its length. By what factor does T change?
5. State how light damping affects the period and the amplitude.
6. A pendulum has a period of 0.80 s. State the frequency of the motion and of its kinetic energy graph.
7. A pendulum’s T² against l graph is a line with gradient 4.05 s² m⁻¹. Find g.
8. A string of length 1.2 m has wave speed 90 m s⁻¹. Find its lowest resonant frequency.
9. State what happens to the height and width of a resonance peak when damping is increased.
10. A tube closed at one end is 0.20 m long and resonates at its lowest frequency. The speed of sound is 340 m s⁻¹. Find the frequency.
11. International A-level only: a 0.50 kg mass on a spring with k = 18 N m⁻¹ has amplitude 0.050 m. Find ω, the maximum speed and the total energy.

Answers

1. T = 2π√(0.25/36) = 0.52 s
2. T = 2π√(0.25/9.81) = 1.0 s
3. T ∝ √m, so T is multiplied by 3.
4. T ∝ √l, so T is multiplied by 2.
5. The period is almost unchanged. The amplitude falls with time.
6. Motion: 1/0.80 = 1.25 Hz. Kinetic energy: twice that, 2.5 Hz.
7. g = 4π² / 4.05 = 9.7 m s⁻² (9.75).
8. λ = 2.4 m, so f = 90 / 2.4 = 37.5 Hz.
9. The peak is lower and broader.
10. λ = 4 × 0.20 = 0.80 m, so f = 340 / 0.80 = 425 Hz.
11. ω = √(18/0.50) = 6.0 rad s⁻¹. Maximum speed = ωA = 6.0 × 0.050 = 0.30 m s⁻¹. Total energy = ½kA² = ½ × 18 × 0.050² = 0.0225 J.

Where marks are usually lost

• Taking the string length alone instead of the distance to the centre of the bob.
• Stating that T depends on amplitude or on the mass of a pendulum bob.
• Not squaring or square-rooting correctly when rearranging T = 2π√(m/k).
• Describing Ek and Ep graphs against time without saying they are at twice the motion’s frequency.
• Saying that resonance occurs at “high” driving frequencies instead of equal frequencies.
• Describing damping as “slowing the oscillation” without stating that the amplitude falls and why (energy loss).
• Mixing up the effect of more damping on peak height and on sharpness.
• Quoting a graph gradient without the 4π² step, or without units.
• Giving an answer for the closed tube using λ/2 rather than λ/4.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Sections 3.5.1 and 3.5.2, with the derivation in 3.6.2.