OxfordAQA International A Level Physics
OxfordAQA International A Level Physics 3.6 Circular and Periodic Motion Exam Questions
Free OxfordAQA International A Level Physics revision for 3.6 Circular and Periodic Motion: what you need to know, common mistakes and exam-style questions with answers. Covers angular speed, centripetal acceleration and force, simple harmonic motion, displacement, velocity and acceleration graphs, energy in shm and periods.
Practise 3.6 Circular and Periodic Motion with past papers
Once you have learnt this topic, find the 3.6 Circular and Periodic Motion questions in real OxfordAQA International A Level Physics past papers and mark them with the mark scheme. Doing lots of real questions is the quickest way to learn how examiners ask about each idea.
StemUp mark scheme flashcards
A-Level Physics Flashcards for 2027 Onwards
Every answer is written in mark scheme wording, so you learn exactly what examiners give marks for. Then test yourself with the questions on this page.
From £10.99View flashcards
About OxfordAQA International A Level Physics 3.6 Circular and Periodic Motion
Circular and periodic motion opens Unit 3 (PH03), Fields and their consequences, in the OxfordAQA International A2 year. It extends AS mechanics and oscillations into motion in a circle and simple harmonic motion (SHM). You need to describe circular motion with angular speed and centripetal force, then define SHM with the condition a = -omega^2 x and use its equations and graphs. These ideas are used again in satellite orbits, charged particles moving in magnetic fields and rotating coils. Examiners regularly ask for the centripetal force in a named situation, sketches of displacement, velocity and acceleration against time, and calculations of maximum speed or acceleration of an oscillator.
What you need to know
- Angular speed omega = v / r = 2 pi f, measured in rad s-1. One complete revolution is 2 pi radians.
- Centripetal acceleration and force a = v^2 / r = omega^2 r, directed towards the centre. The resultant force needed is F = mv^2 / r = m omega^2 r. It is provided by another force, such as tension, friction or gravity, and does no work because it is perpendicular to the velocity.
- Condition for SHM Acceleration is proportional to displacement and directed towards the equilibrium position: a = -omega^2 x, with omega = 2 pi f.
- SHM equations x = A cos omega t (starting from maximum displacement), v = +/- omega square root of (A^2 - x^2), maximum speed = omega A, maximum acceleration = omega^2 A. Velocity leads displacement by a quarter cycle; acceleration is in antiphase with displacement.
- Energy in SHM Energy changes continuously between kinetic and potential, with the total constant if undamped. Ek is greatest at equilibrium and Ep is greatest at maximum displacement.
- Periods of simple oscillators Mass-spring: T = 2 pi square root of (m / k). Simple pendulum: T = 2 pi square root of (L / g), valid only for small angles. Period is independent of amplitude for SHM.
Common mistakes in 3.6 Circular and Periodic Motion
- Drawing centripetal force as an extra force on a free-body diagram. It is the resultant of the real forces acting, not a separate force.
- Saying an object moving at constant speed in a circle is not accelerating. Its direction changes, so its velocity changes and it accelerates towards the centre.
- Using degrees instead of radians in x = A cos omega t. Calculators must be in radian mode.
- Stating that acceleration is zero at maximum displacement. In SHM acceleration is maximum at maximum displacement and zero at equilibrium.
- Applying the pendulum equation to large amplitude swings, where the small angle approximation fails.
Try three 3.6 Circular and Periodic Motion exam-style questions
1. A car of mass 900 kg travels at 15 m s-1 around a bend of radius 50 m. Calculate the centripetal force and state what provides it. (3 marks)
Show the mark scheme answer
F = mv^2 / r = 900 x 15^2 / 50; F = 4.1 x 10^3 N (4050 N); provided by friction between the tyres and the road.
2. A 0.40 kg mass oscillates on a spring of stiffness 25 N m-1 with amplitude 0.050 m. Calculate its period and maximum speed. (3 marks)
Show the mark scheme answer
T = 2 pi square root of (0.40 / 25) = 0.79 s; omega = square root of (k / m) = 7.9 rad s-1; v max = omega A = 7.9 x 0.050 = 0.40 m s-1.
3. State the two conditions for an object to move with simple harmonic motion. (2 marks)
Show the mark scheme answer
Acceleration is directly proportional to displacement from the equilibrium position; acceleration is always directed towards the equilibrium position (opposite to displacement).
3.6 Circular and Periodic Motion FAQs
What does OxfordAQA International A Level Physics 3.6 Circular and Periodic Motion cover?
It covers angular speed, centripetal acceleration and force, the defining condition for simple harmonic motion, SHM equations and graphs, energy in SHM and the periods of mass-spring systems and pendulums.
Which unit is 3.6 Circular and Periodic Motion examined in?
It is examined in Unit 3 (PH03), Fields and their consequences, which is an A2 unit.
How do I remember the phase relationships in SHM?
Velocity is a quarter of a cycle ahead of displacement, and acceleration is always opposite to displacement. A quick check: at maximum displacement the object is momentarily stationary and its acceleration is largest.
