Simple harmonic motion (SHM) (A-level only) — AQA A-Level Physics
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Simple harmonic motion (SHM) is a specific type of oscillatory motion where the acceleration of an object is directly proportional to its displacement from a fixed equilibrium point and is always directed towards that point. In AQA A-Level Physics, you must understand the defining equation a = -ω²x, the associated energy transfers, and how to apply these to systems like mass-spring oscillators and simple pendulums.
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Simple harmonic motion (SHM) (A-level only) explained
Simple harmonic motion (SHM) (A-level only) covers 2 specification points, including Analysis of characteristics of simple harmonic… and Appreciation that the v − t graph is derived….
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Analysis of characteristics of simple harmonic motion (SHM).
Simple harmonic motion is oscillatory motion where the acceleration is directly proportional to displacement from equilibrium and always directed towards that equilibrium point, expressed as a = −ω²x. The minus sign shows the restoring acceleration opposes displacement. Characteristics to analyse include amplitude A, period T, frequency f, angular frequency ω = 2πf = 2π/T, and the phase relationships: displacement and acceleration are in antiphase, while velocity leads displacement by a quarter cycle. For example, for x = A cos(ωt), differentiating gives v = −Aω sin(ωt) and a = −Aω² cos(ωt) = −ω²x, confirming SHM. Graphs of x, v and a against time share the same period, with v maximum at equilibrium and zero at maximum displacement, and a maximum at maximum displacement and zero at equilibrium.
Appreciation that the v − t graph is derived from the gradient of the x −t graph and that the a − t graph is derived from the gradient of the v − t graph.
Displacement, velocity and acceleration are linked by successive gradients. Plotting displacement x against time t, the gradient at any instant equals the velocity v, making the v–t graph the gradient function of the x–t graph. Differentiating again, the gradient of the v–t graph equals the acceleration a. For SHM, x = A cos(ωt) gives v = −Aω sin(ωt) and a = −Aω² cos(ωt) = −ω²x. Graphically, a maximum or minimum of x occurs where v is zero; the steepest x–t slope occurs where v is maximum; and a is opposite in sign to x. This explains why acceleration always points towards equilibrium, opposing velocity only when moving away from equilibrium.
Your focus
- State and apply the defining condition for SHM, a = −ω²x.
- Describe how displacement, velocity and acceleration vary and relate in phase over one cycle.
- Interpret graphs and equations of SHM to determine amplitude, period, frequency and angular frequency.
Show all 6 objectives
- Determine instantaneous velocity from the gradient of a displacement–time graph and instantaneous acceleration from the gradient of a velocity–time graph.
- Construct or interpret linked x–t, v–t and a–t graphs for a simple harmonic oscillator.
- Explain how the gradient relationships lead to a = −ω²x and to acceleration always being directed towards the equilibrium position.
Which command word is this?
A concise, factual answer with no explanation required. For example, 'State what is meant by simple harmonic motion' expects: 'Acceleration is proportional to displacement and directed towards equilibrium.'
Simple harmonic motion (SHM) (A-level only) exam tips
Topic Overview
Simple harmonic motion (SHM) is a fundamental topic in A-level Physics that describes oscillations where the restoring force is proportional to displacement. It underpins many physical systems, from pendulums and springs to molecular vibrations and electromagnetic waves. Understanding SHM is crucial for analysing energy transfers in oscillating systems and for later topics such as resonance and damping.
In the AQA A-Level specification, SHM is a key part of the 'Further Mechanics and Thermal Physics' module. You will learn the defining equation a = -ω²x, derive expressions for velocity and displacement, and study the interchange between kinetic and potential energy. This topic also connects to practical work on investigating oscillations and to real-world applications like car suspensions and building stability.
Key Concepts
- →Definition of SHM: acceleration is directly proportional to displacement from equilibrium and always directed towards equilibrium, expressed as a = -ω²x.
- →Key equations: x = A cos(ωt) or x = A sin(ωt), v = ±ω√(A² - x²), v_max = ωA, a_max = ω²A, and T = 2π/ω.
- →Energy in SHM: total energy is constant and equals ½mω²A²; kinetic energy is maximum at equilibrium and zero at maximum displacement, while potential energy is the opposite.
- →Phase relationships: velocity leads displacement by π/2 radians, and acceleration is π radians out of phase with displacement.
- →Examples of SHM: mass-spring system (T = 2π√(m/k)) and simple pendulum (T = 2π√(L/g)) for small angles.
Marking Points
- States the defining condition a = −ω²x, with acceleration proportional to displacement and directed towards equilibrium.
- Identifies amplitude, period, frequency and angular frequency, and links ω = 2πf = 2π/T.
- Describes the phase relationship between displacement, velocity and acceleration over one cycle.
- Explains that velocity is maximum at equilibrium and zero at maximum displacement, while acceleration is maximum at maximum displacement and zero at equilibrium.
- Interprets or sketches graphs of x, v and a against time, recognising the same period and the phase differences.
- Uses x = A cos(ωt) or x = A sin(ωt) and differentiates to obtain velocity and acceleration expressions.
- State that the gradient of a displacement–time graph at a given instant gives the instantaneous velocity.
- State that the gradient of a velocity–time graph at a given instant gives the instantaneous acceleration.
- Apply the chain of gradients to SHM: where x is maximum, gradient is zero so v = 0; where x crosses zero, gradient is steepest so v is maximum in magnitude.
- Relate the sign of the v–t gradient to the direction of acceleration, showing that a is directed towards the equilibrium position and is opposite in sign to x.
- Use the standard SHM expressions x = A cos(ωt), v = −Aω sin(ωt) and a = −Aω² cos(ωt) to confirm the graphical relationships and the result a = −ω²x.
Examiner Tips
- 💡Start by quoting a = −ω²x and then link each characteristic to this defining relationship.
- 💡When describing graphs, state the phase difference in fractions of a cycle or in radians rather than vaguely saying they are shifted.
- 💡Check whether the question gives period or frequency before calculating ω, and show the conversion clearly.
- 💡Sketch x–t, v–t and a–t on the same time axis so that maxima, minima and zero crossings line up; this makes the gradient relationships visible.
- 💡When asked to compare graphs, quote the gradient relationship explicitly, for example state that the value of v at a time equals the gradient of x–t at that time.
- 💡Always define SHM in terms of acceleration, not force, unless the question specifically asks about force. The AQA mark scheme expects 'acceleration is proportional to displacement and directed towards equilibrium'.
- 💡When using the equation v = ±ω√(A² - x²), remember that the ± indicates direction. In calculations, you can ignore the sign if only the magnitude is required, but state the direction if asked.
- 💡For graph questions, label axes with quantities and units, and ensure the curves have the correct phase relationships. Use radians for phase differences if required.
Common Mistakes
- Omitting the minus sign in a = −ω²x; the correction is to include it to show acceleration is directed towards equilibrium.
- Stating that velocity and acceleration are in phase; the correction is to recognise that velocity leads displacement by a quarter cycle and acceleration is in antiphase with displacement.
- Confusing period with frequency when finding ω; the correction is to use ω = 2π/T for period and ω = 2πf for frequency, keeping the two distinct.
- Assuming acceleration always opposes velocity. Correction: in SHM, acceleration points towards equilibrium, so it opposes velocity when moving away from equilibrium, but acts in the same direction when moving towards it.
- Confusing the value of the graph with its gradient: for example, reading the height of the x–t graph as velocity. Correction: velocity is the gradient of the x–t graph, not its displacement value.
- Treating the gradient as constant over a curved section. Correction: for SHM the gradient changes continuously, so the instantaneous gradient must be found at a point, for example by drawing a tangent.
- Students often think that acceleration is maximum at maximum displacement but in the same direction as displacement. In fact, acceleration is maximum at maximum displacement but directed towards equilibrium (opposite to displacement).
- Many believe that the period of a pendulum depends on its mass or amplitude. For small angles, the period is independent of mass and amplitude, depending only on length and g.
- Some confuse the terms 'frequency' and 'angular frequency'. Angular frequency ω is 2π times the frequency f, and it appears in the SHM equations.
Revision Plan
- 1Day 1-2: Learn the definition of SHM and the equation a = -ω²x. Practice identifying SHM from given equations or graphs.
- 2Day 3-4: Derive and memorise the equations for displacement, velocity, and acceleration as functions of time. Solve problems using these equations.
- 3Day 5-6: Study energy in SHM. Learn the expressions for kinetic and potential energy and how they vary with displacement. Practice sketching energy-displacement graphs.
- 4Day 7-8: Investigate the mass-spring and simple pendulum systems. Derive their periods and solve numerical problems.
- 5Day 9-10: Review past exam questions on SHM, focusing on graph interpretation and multi-step calculations. Use mark schemes to refine answers.
Exam Question Types
- 📋Definition and explanation questions: e.g., 'State what is meant by simple harmonic motion.' These require precise wording: acceleration proportional to displacement, directed towards equilibrium.
- 📋Calculation questions: e.g., 'Calculate the maximum speed of an oscillator given its amplitude and period.' Use v_max = ωA or v_max = 2πfA.
- 📋Graph sketching and interpretation: e.g., 'Sketch graphs of displacement, velocity, and acceleration against time for an object in SHM.' Ensure correct phase relationships and labels.
- 📋Practical and data analysis questions: e.g., 'Describe an experiment to determine the spring constant using SHM.' Focus on measurements, variables, and graph plotting (T² against m).
Command Word Expectations (AQA)
A concise, factual answer with no explanation required. For example, 'State what is meant by simple harmonic motion' expects: 'Acceleration is proportional to displacement and directed towards equilibrium.'
Provide a reasoned account, linking cause and effect. For example, 'Explain why the acceleration is zero at the equilibrium position' expects: 'At equilibrium, displacement is zero, so from a = -ω²x, acceleration is zero.'
Show working, substitute values into the correct equation, and give the answer with appropriate units and significant figures. Marks are awarded for correct substitution and final answer.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A mass of 0.50 kg is attached to a spring of spring constant 20 N m⁻¹ and performs simple harmonic motion with amplitude 0.10 m. Calculate the maximum speed and the maximum acceleration of the mass.
- 1.Step 1: Identify given values: m = 0.50 kg, k = 20 N m⁻¹, A = 0.10 m. The angular frequency ω = √(k/m) = √(20/0.50) = √40 = 6.32 rad s⁻¹.
- 2.Step 2: Use the formula for maximum speed: v_max = ωA = 6.32 × 0.10 = 0.632 m s⁻¹.
- 3.Step 3: Use the formula for maximum acceleration: a_max = ω²A = (6.32)² × 0.10 = 40 × 0.10 = 4.0 m s⁻².
Question: A simple pendulum of length 1.2 m is displaced by a small angle and released. Calculate its period of oscillation. (g = 9.81 m s⁻²)
- 1.Step 1: Identify given values: L = 1.2 m, g = 9.81 m s⁻². For a simple pendulum, T = 2π√(L/g).
- 2.Step 2: Substitute values: T = 2π√(1.2/9.81) = 2π√(0.1223) = 2π × 0.3497 = 2.197 s.
- 3.Step 3: Round to an appropriate number of significant figures: T ≈ 2.2 s.