AQA A-Level Physics: Further mechanics and thermal physics (7408) – Revision Notes

These notes condense section 3.6 Further mechanics and thermal physics of the AQA AS and A-level Physics specification (7407/7408, version 1.4, AS and A-level exams June 2016 onwards): 3.6.1 Periodic motion and 3.6.2 Thermal physics. The whole section is A-level only; Paper 1 assesses 6.1 and Paper 2 assesses 6.2. It is core content, not one of the options (sections 3.9 to 3.13) studied for Paper 3 Section B. For full explanations and worked examples, use the study guide.

Test yourself with the practice questions, track outcomes on the printable checklist, and find weak spots with the free diagnostics. All AQA A-level Physics topics are on the course hub.

Key equations

Equation	Symbols	Use
ω = v/r = 2πf	ω angular speed (rad s⁻¹)	Circular motion, SHM
a = v²/r = ω²r	a centripetal acceleration	Circular motion
F = mv²/r = mω²r	F resultant force to centre	Circular motion
a = −ω²x	x displacement from equilibrium	Defines SHM
x = A cos ωt	A amplitude	SHM, timed from x = A
v = ±ω√(A² − x²)	v velocity at displacement x	SHM
v_max = ωA, a_max = ω²A		SHM
T = 2π√(m/k)	k spring constant	Mass-spring
T = 2π√(l/g)	l length to centre of bob	Pendulum (small angles)
Q = mcΔθ	c specific heat capacity	Temperature change
Q = ml	l specific latent heat	Change of state
pV = nRT = NkT	n moles, N molecules	Ideal gas, T in kelvin
W = pΔV	ΔV change in volume	Work at constant pressure
pV = ⅓Nm(c_rms)²	c_rms root mean square speed	Kinetic theory
½m(c_rms)² = (3/2)kT = (3/2)RT/N_A		Mean molecular kinetic energy

3.6.1.1 Circular motion

• 2π rad = 360°. θ (rad) = arc length/radius.
• Constant speed in a circle means changing velocity, so there is an acceleration towards the centre and a resultant (centripetal) force towards the centre.
• The centripetal force is supplied by real forces. Never add it as an extra arrow.
• Estimates: pick sensible r and f (or v), then a = ω²r and F = mω²r.

3.6.1.2 Simple harmonic motion

Condition: acceleration proportional to displacement and directed towards the equilibrium position.

Method: SHM calculation

1. Find ω = 2πf = 2π/T.
2. Maximum speed ωA at equilibrium; maximum acceleration ω²A at the extremes.
3. Speed at x: ω√(A² − x²). Acceleration at x: −ω²x.
4. Position at time t: x = A cos ωt (radians mode), only if timing starts at x = +A.

Graph links

From	To	How
x–t	v–t	Gradient
v–t	a–t	Gradient
x–t (cosine)	a–t	Same shape, inverted, scaled by ω²

3.6.1.3 Simple harmonic systems

• Periods do not depend on amplitude. Mass-spring T ∝ √m; pendulum T ∝ √l, and independent of the mass of the bob.
• Pendulum SHM relies on the small-angle approximation sin θ ≈ θ.
• Energy: E_k + E_p = constant (undamped). E_k maximum at x = 0; E_p maximum at x = ±A. Against time, each energy has half the period of the oscillation.
• Damping: light (amplitude decays, period about the same); critical (back to equilibrium fastest, no oscillation); heavy (slow return, no oscillation).
• Required practical 7: time many oscillations, repeat, time from the equilibrium position; plot T² against m or l and use the gradient (4π²/k or 4π²/g).

3.6.1.4 Forced vibrations and resonance

Must-know distinctions: free vs forced

Free vibration	Forced vibration
No driving force after release	Periodic driving force
Oscillates at the natural frequency	Oscillates at the driving frequency

• Resonance: driving frequency = natural frequency, giving maximum amplitude.
• Increasing damping: peak lower, broader, slightly lower in frequency.
• Examples: swings, car suspension, strings and air columns resonating at their stationary-wave frequencies.

3.6.2.1 Thermal energy transfer

• Internal energy = sum of randomly distributed kinetic and potential energies of the particles.
• First law (qualitative): internal energy rises when the body is heated or work is done on it.
• During a change of state: potential energies change, kinetic energies don’t, so temperature is constant.
• Continuous flow: power = (mass per second) × c × Δθ.

Method: mixtures

1. Write energy lost by the hot part = energy gained by the cold part(s).
2. Include latent heat for any melting or boiling.
3. Use the same final temperature T in every term and solve.

Must-know distinction: temperature measures mean kinetic energy per particle; internal energy also includes potential energy and depends on how many particles there are.

3.6.2.2 Ideal gases

Law (fixed mass)	Constant	Straight-line graph
Boyle	T	p against 1/V
Charles	p	V against T (through 0 K)
Pressure law	V	p against T (through 0 K)

• Absolute zero: extrapolated intercept at about −273 °C. Add 273 to convert °C to K.
• k = R/N_A; N = nN_A; n = mass/molar mass; molecular mass = molar mass/N_A.
• Work done by a gas expanding at constant pressure = pΔV (area under a p–V graph).
• Required practical 8: Boyle (constant temperature, change pressure slowly) and Charles (constant pressure, water bath).

3.6.2.3 Kinetic theory

• Brownian motion: random motion of visible particles, evidence for molecules.
• Gas laws are empirical; kinetic theory is derived from assumptions.
• Assumptions: many identical molecules, random motion, elastic collisions, negligible molecular volume, no intermolecular forces except in collisions, negligible collision time.
• Derivation chain: Δp = 2mu → time between hits 2L/u → force mu²/L → sum over N → mean u² = ⅓(c_rms)² → p = F/L².
• c_rms = √(mean of c²), which is not the same as mean speed.
• Ideal gas internal energy = total kinetic energy = (3/2)NkT.

Quick self-test

1. A turntable rotates at 45 revolutions per minute. Find its angular speed.
2. A cyclist rides round a circular track of radius 20 m at 8.0 m s⁻¹. Find the centripetal acceleration.
3. A 0.20 kg ball on a string moves in a horizontal circle of radius 0.50 m at ω = 6.0 rad s⁻¹. Find the centripetal force.
4. An oscillator has A = 0.10 m and T = 2.0 s. Find v_max and a_max.
5. Find the period of a simple pendulum of length 1.00 m (g = 9.81 m s⁻²).
6. The mass on a spring is made four times larger. What happens to the period?
7. How much energy heats 0.50 kg of water from 20 °C to 100 °C? (c = 4200 J kg⁻¹ K⁻¹)
8. How much energy boils away 0.20 kg of water at 100 °C? (l = 2.26 × 10⁶ J kg⁻¹)
9. A gas occupies 0.0249 m³ at 1.0 × 10⁵ Pa and 300 K. How many moles are there? (R = 8.31 J mol⁻¹ K⁻¹)
10. Three molecules have speeds 200, 400 and 600 m s⁻¹. Find their mean speed and c_rms.
11. Find the mean kinetic energy of a gas molecule at 400 K. (k = 1.38 × 10⁻²³ J K⁻¹)
12. A gas expands from 2.0 × 10⁻³ m³ to 3.0 × 10⁻³ m³ at a constant 1.2 × 10⁵ Pa. Find the work done.

Answers

1. ω = 2π × 45/60 = 4.7 rad s⁻¹
2. a = 8.0²/20 = 3.2 m s⁻²
3. F = mω²r = 0.20 × 36 × 0.50 = 3.6 N
4. ω = π rad s⁻¹; v_max = 0.31 m s⁻¹; a_max = 0.99 m s⁻²
5. T = 2π√(1.00/9.81) = 2.0 s
6. T ∝ √m, so the period doubles.
7. Q = 0.50 × 4200 × 80 = 1.7 × 10⁵ J
8. Q = 0.20 × 2.26 × 10⁶ = 4.5 × 10⁵ J
9. n = pV/RT = 2490/2493 = 1.0 mol
10. Mean = 400 m s⁻¹; c_rms = √[(200² + 400² + 600²)/3] = 432 m s⁻¹
11. (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 400 = 8.3 × 10⁻²¹ J
12. W = pΔV = 1.2 × 10⁵ × 1.0 × 10⁻³ = 120 J

Where marks are usually lost

• Not converting revolutions per minute to hertz before using ω = 2πf.
• Using x = A cos ωt with the calculator in degrees.
• Defining SHM without both parts: proportional to displacement and directed towards equilibrium.
• Drawing energy-time graphs with the same period as the motion instead of half.
• Describing resonance as “the system vibrates a lot” without saying driving frequency equals natural frequency.
• Forgetting the meltwater warming term in an ice-mixture calculation.
• Using °C in pV = nRT, or grams where kilograms are needed.
• Treating mean speed and c_rms as the same quantity.
• Skipping the “mean u² = ⅓(c_rms)²” step, or not explaining it, in the kinetic theory derivation.

Official syllabus

AQA AS and A-level Physics (7407/7408) specification, version 1.4, July 2026, AS and A-level exams June 2016 onwards, published by AQA. Section 3.6 Further mechanics and thermal physics (A-level only).