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    Simple harmonic systems (A-level only) — AQA A-Level Physics

    Teaching note checked against the official specification.

    Simple harmonic motion (SHM) is a special type of periodic motion where the acceleration is directly proportional to the displacement from the equilibrium position and is always directed towards that equilibrium point. In AQA A-Level Physics, you must understand the defining equation a = -omega squared x, the sinusoidal graphs of displacement, velocity and acceleration, and apply energy conservation to oscillating systems such as mass-spring and simple pendulum.

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    Simple harmonic systems (A-level only) explained

    Simple harmonic systems (A-level only) covers 2 specification points, including Effects of damping on oscillations.

    Read the full explanation

    and Investigation into simple harmonic motion…. Open a point below for its explanation and revision guidance.

    Effects of damping on oscillations.

    Damping removes energy from an oscillating system, usually transferring it to the surroundings by doing work against resistive forces such as friction or air resistance. The amplitude therefore decreases over time, while the period is only slightly affected for light damping. In light damping, the amplitude decays gradually and the oscillations remain approximately simple harmonic. In heavy damping, the system returns slowly to equilibrium without oscillating. In critical damping, the system returns to equilibrium in the shortest possible time without overshooting, which is useful in door closers and moving-coil meters. Greater damping reduces the number of visible oscillations and increases the rate of energy loss.

    Required practical 7: Investigation into simple harmonic motion using a mass–spring system and a simple pendulum.

    You investigate two oscillators that approximate simple harmonic motion. For the mass–spring system, time many oscillations for each added mass to find the period T, then plot T² against mass m; the gradient is 4π²/k, giving the spring constant k. For the pendulum, vary the length l, measure T for small angles, and plot T² against l; the gradient is 4π²/g, giving g. In both cases, repeat readings, use a fiducial marker at the equilibrium position, and start timing as the bob or mass passes it. Keep amplitudes small so the motion stays close to SHM, and identify the equilibrium position carefully.

    Your focus

    1. Describe the effects of light, heavy and critical damping on the amplitude and period of an oscillation.
    2. Explain damping in terms of energy transfer by work done against resistive forces.
    3. Apply the concept of critical damping to a practical system where a fast return to equilibrium without overshoot is required.
    Show all 6 objectives
    1. Carry out a mass–spring oscillation experiment and determine the spring constant from a T² against m graph.
    2. Carry out a simple pendulum experiment and determine g from a T² against l graph.
    3. Evaluate sources of uncertainty and explain how repetition and timing many oscillations improve reliability.

    Which command word is this?

    A brief answer with no explanation required. For example, 'State the condition for simple harmonic motion.' You should write: 'Acceleration is proportional to displacement and directed towards equilibrium.' No further detail is needed.

    Simple harmonic systems (A-level only) exam tips

    Topic Overview

    Simple harmonic systems are a fundamental topic in AQA A-Level Physics that describes oscillatory motion where the restoring force is proportional to displacement. You will study the defining equation a = -omega squared x, the relationships between displacement, velocity and acceleration, and how these vary sinusoidally with time. This topic also covers the energy transfers in SHM and applies the theory to important practical systems such as a mass on a spring and a simple pendulum.

    Understanding SHM is crucial because it appears in many areas of physics, from waves and sound to molecular vibrations and electrical circuits. It builds on your knowledge of forces, energy and circular motion, and it develops your ability to use calculus and trigonometric functions in a physical context. Mastery of SHM is essential for tackling more advanced topics like resonance and damping, and it frequently appears in exam questions that test both conceptual understanding and mathematical application.

    Key Concepts
    • →Definition of SHM: acceleration is directly proportional to displacement from equilibrium and is always directed towards that equilibrium point, expressed as a = -omega squared x.
    • →The equations for displacement, velocity and acceleration as functions of time: x = A cos(omega t), v = -A omega sin(omega t), a = -A omega squared cos(omega t).
    • →The relationship between period T, frequency f and angular frequency omega: omega = 2 pi f = 2 pi / T.
    • →Energy in SHM: total energy is constant and equals 1/2 m omega squared A squared; kinetic energy is maximum at equilibrium and zero at maximum displacement, while potential energy is the opposite.
    • →Practical SHM systems: a mass on a spring has T = 2 pi times square root of (m / k), and a simple pendulum has T = 2 pi times square root of (L / g) for small angles.
    Marking Points
    • State that damping is caused by resistive forces doing work on the oscillator, so mechanical energy is dissipated to the surroundings and the amplitude decreases with time.
    • Describe light damping as a gradual exponential-like decrease in amplitude with the oscillator still completing many oscillations and the period remaining approximately constant.
    • Describe heavy damping as a slow return to equilibrium with no oscillation, and critical damping as the fastest return to equilibrium without overshooting.
    • Explain practical consequences, such as critical damping in door closers, car suspension or moving-coil meters, where overshoot or prolonged oscillation is undesirable.
    • Compare energy loss: greater damping removes a larger fraction of energy per cycle, so the amplitude falls more rapidly and fewer oscillations are observed.
    • Times a large number of oscillations and divides by that number to obtain a mean period, reducing the effect of human reaction time.
    • Uses a fiducial marker at the equilibrium position and starts timing as the oscillator passes it, so each timing begins from the same phase.
    • For the mass–spring system, plots T² against m and uses gradient = 4π²/k to determine the spring constant k.
    • For the pendulum, plots T² against l and uses gradient = 4π²/g to determine g.
    • Keeps the amplitude small so the small-angle approximation holds and the motion remains close to simple harmonic.
    • Repeats measurements and calculates a mean, and considers uncertainties in length, time and mass.
    Examiner Tips
    • 💡Sketch displacement–time curves for light, heavy and critical damping on the same axes, labelling the equilibrium line and showing that only the lightly damped curve crosses it repeatedly.
    • 💡Use the phrase 'returns to equilibrium in the shortest time without overshooting' when defining critical damping, and give one named application to show understanding.
    • 💡When explaining energy transfer, refer to work done against resistive forces and the resulting decrease in amplitude rather than saying that energy is simply 'lost'.
    • 💡State clearly how you reduce timing uncertainty, for example by timing 20 oscillations and dividing by 20.
    • 💡When describing a graph, name the axes and explain how the gradient gives the required constant.
    • 💡Mention at least one control variable, such as keeping the amplitude small or using the same spring.
    • 💡Always start your answer by writing the defining equation a = -omega squared x and explicitly state that the minus sign indicates acceleration is directed towards the equilibrium position. This secures the definition marks.
    • 💡When sketching graphs of displacement, velocity and acceleration against time, ensure you label the axes with the correct quantity and unit, and show the phase relationships: velocity leads displacement by pi/2 radians, and acceleration is in antiphase with displacement.
    • 💡In calculations, use standard form and give your answer to an appropriate number of significant figures, usually the same as the least precise value in the question. Always include units in your final answer.
    Common Mistakes
    • Thinking damping always changes the period dramatically. Correction: for light damping the period is almost unchanged; the main observable effect is a decrease in amplitude.
    • Confusing heavy damping with critical damping. Correction: heavy damping returns to equilibrium slowly without oscillating, whereas critical damping returns in the shortest time without overshooting.
    • Believing that damping stops the oscillator instantly. Correction: damping gradually removes energy, so the amplitude decays over many cycles in the lightly damped case.
    • Timing only one oscillation, which makes reaction time a large fraction of the measured period; correct by timing many oscillations and dividing.
    • Measuring the pendulum length to the top of the bob rather than to its centre, which gives a systematic error in l; correct by measuring to the centre of the bob.
    • Using a large amplitude for the pendulum, so the motion is not simple harmonic; correct by keeping the amplitude small.
    • Students often think that acceleration is maximum at equilibrium because velocity is maximum. In fact, acceleration is zero at equilibrium and maximum at maximum displacement, because acceleration is proportional to displacement.
    • Many students believe that the period of a simple pendulum depends on the mass of the bob. The period depends only on the length of the pendulum and the acceleration due to gravity, provided the angle is small.
    • Students sometimes confuse the terms 'frequency' and 'angular frequency'. Frequency f is measured in hertz (cycles per second), while angular frequency omega is measured in radians per second and equals 2 pi f.
    Revision Plan
    1. 1Start by learning the definition of SHM and the equation a = -omega squared x. Practice identifying SHM from given equations or graphs, and ensure you can explain the significance of the negative sign.
    2. 2Derive and memorise the expressions for displacement, velocity and acceleration as functions of time. Sketch these graphs for at least two full cycles, noting the phase relationships and the points where each quantity is zero or maximum.
    3. 3Study the energy changes in SHM. Learn the equations for kinetic energy, potential energy and total energy, and practice sketching energy-displacement graphs. Solve problems involving energy conservation.
    4. 4Apply SHM to the mass-spring and simple pendulum systems. Derive their periods and solve numerical problems. Understand the assumptions and limitations of each model.
    5. 5Attempt past paper questions on SHM, focusing on both definitions and calculations. Mark your answers using the mark scheme and note where you lose marks. Revise any weak areas.
    Exam Question Types
    • 📋Definition and explanation questions: e.g., 'State what is meant by simple harmonic motion.' These require precise wording including the proportionality to displacement and the direction towards equilibrium. Practice writing the definition concisely.
    • 📋Graph sketching and interpretation: e.g., 'Sketch a graph of velocity against displacement for an object in SHM.' You must recall the shapes (ellipse for v-x, sine/cosine for x-t) and label key points such as amplitude and maximum velocity.
    • 📋Calculation questions: e.g., 'Calculate the maximum acceleration of a mass-spring system given mass, spring constant and amplitude.' These involve using omega = square root of (k/m) and a_max = omega squared A. Show all steps clearly.
    • 📋Practical and experimental questions: e.g., 'Describe an experiment to determine the spring constant using SHM.' You need to know how to measure period and use T = 2 pi times square root of (m/k) to find k, including how to reduce uncertainties.
    Command Word Expectations (AQA)
    State

    A brief answer with no explanation required. For example, 'State the condition for simple harmonic motion.' You should write: 'Acceleration is proportional to displacement and directed towards equilibrium.' No further detail is needed.

    Explain

    You must give reasons or mechanisms. For example, 'Explain why the acceleration is zero at the equilibrium position.' You should say: 'At equilibrium, displacement is zero, and since a = -omega squared x, acceleration is zero.' Include the equation and the link.

    Calculate

    You must show your working and give the final answer with units and appropriate significant figures. For example, 'Calculate the period of a pendulum of length 0.80 m.' Use T = 2 pi times square root of (L/g) and show substitution. A bare answer without working may lose method marks.

    How Students Lose Marks (Examiner Pitfalls)
    Pitfall: Students often state that acceleration is 'proportional to displacement' without including the crucial negative sign or the word 'opposite' to indicate direction. This loses the mark for direction.
    ❌ Weak Answer (Loses Marks):Acceleration is proportional to displacement.
    Example improved answer:Acceleration is directly proportional to displacement from the equilibrium position and is always directed towards that equilibrium position, hence a = -omega squared x.
    Examiner Tip: Always write the defining equation a = -omega squared x and explicitly state that the minus sign shows acceleration is in the opposite direction to displacement. Use the phrase 'directed towards the equilibrium position'.
    Pitfall: In energy calculations for SHM, students frequently forget that total energy is constant and that kinetic energy is maximum at equilibrium while potential energy is maximum at maximum displacement. They also mix up the formulas for kinetic and potential energy.
    ❌ Weak Answer (Loses Marks):At maximum displacement, kinetic energy is maximum and potential energy is zero.
    Example improved answer:At maximum displacement, velocity is zero so kinetic energy is zero; the total energy is entirely potential energy. At equilibrium, displacement is zero so potential energy is zero and kinetic energy is maximum, equal to the total energy.
    Examiner Tip: Sketch energy-displacement graphs for both kinetic and potential energy. Remember that total energy E = 1/2 m omega squared A squared, and that kinetic energy = 1/2 m omega squared (A squared - x squared).
    Step-by-Step Worked Solutions

    Question: A mass of 0.50 kg attached to a spring oscillates with simple harmonic motion. The amplitude is 0.12 m and the period is 0.80 s. Calculate the maximum speed and the speed when the displacement is 0.06 m.

    1. 1.Step 1: Identify given values: m = 0.50 kg, A = 0.12 m, T = 0.80 s. Calculate angular frequency omega = 2 pi / T = 2 pi / 0.80 = 7.854 rad per s.
    2. 2.Step 2: Maximum speed occurs at equilibrium (x = 0): v_max = omega A = 7.854 times 0.12 = 0.9425 m per s.
    3. 3.Step 3: Speed at displacement x = 0.06 m: v = omega times square root of (A squared - x squared) = 7.854 times square root of (0.12 squared - 0.06 squared) = 7.854 times square root of (0.0144 - 0.0036) = 7.854 times square root of 0.0108 = 7.854 times 0.10392 = 0.816 m per s.
    Final Answer: Maximum speed = 0.94 m per s (2 s.f.); speed at x = 0.06 m = 0.82 m per s (2 s.f.).

    Question: A simple pendulum of length 1.20 m is displaced by a small angle and released. Calculate its period of oscillation. State one assumption made in this calculation.

    1. 1.Step 1: For a simple pendulum, period T = 2 pi times square root of (L / g), where L = 1.20 m and g = 9.81 m per s squared.
    2. 2.Step 2: Substitute values: T = 2 pi times square root of (1.20 / 9.81) = 2 pi times square root of 0.12232 = 2 pi times 0.34974 = 2.198 s.
    3. 3.Step 3: State assumption: The oscillations are small (less than about 10 degrees) so that the motion approximates simple harmonic motion.
    Final Answer: Period = 2.2 s (2 s.f.). Assumption: small angle of swing so that sin theta is approximately equal to theta in radians.
    Active Recall Memory Test
    What is the defining equation of simple harmonic motion?
    Key Fact: a = -omega squared x, where a is acceleration, omega is angular frequency, and x is displacement from equilibrium.
    At what point in SHM is the speed of the oscillator maximum?
    Key Fact: At the equilibrium position (x = 0), where all energy is kinetic.
    State the formula for the period of a simple pendulum and the condition for it to be valid.
    Key Fact: T = 2 pi times square root of (L / g), valid for small angles (less than about 10 degrees).
    How does the total energy of an SHM system depend on amplitude?
    Key Fact: Total energy is proportional to the square of the amplitude: E = 1/2 m omega squared A squared.
    Frequently Asked Questions
    What is simple harmonic motion in A-level physics?
    Simple harmonic motion (SHM) is a type of oscillatory motion where the acceleration of the object is directly proportional to its displacement from the equilibrium position and is always directed towards that equilibrium point. This is summarised by the equation a = -omega squared x, where omega is the angular frequency. In AQA A-Level Physics, you need to understand the mathematical relationships, graphs, and energy changes involved in SHM, as well as apply them to systems like mass-spring oscillators and pendulums.
    How do you calculate the period of a mass-spring system?
    The period T of a mass-spring system in SHM is given by T = 2 pi times the square root of (m / k), where m is the mass attached to the spring and k is the spring constant. This formula assumes the spring obeys Hooke's law and the mass of the spring is negligible. You can use this to find the period if you know m and k, or rearrange it to find k from a measured period. Remember to convert mass to kilograms and use consistent units.
    What is the difference between displacement and amplitude in SHM?
    Displacement in SHM is the distance of the oscillating object from its equilibrium position at any given moment, and it varies sinusoidally with time. Amplitude, on the other hand, is the maximum displacement from equilibrium, a constant value for a given oscillation. So, displacement can be any value between -A and +A, while amplitude is always positive and represents the peak value. In equations, displacement is often denoted by x, and amplitude by A.
    Why is acceleration negative in simple harmonic motion?
    In SHM, acceleration is always directed towards the equilibrium position, which is opposite to the direction of displacement when the object is displaced from equilibrium. This is why the defining equation includes a negative sign: a = -omega squared x. If displacement is positive (to the right), acceleration is negative (to the left), and vice versa. This restoring acceleration is what causes the object to oscillate back and forth.
    How does energy change during simple harmonic motion?
    During SHM, energy is continuously transferred between kinetic energy and potential energy. At the equilibrium position, kinetic energy is maximum and potential energy is zero. At maximum displacement, kinetic energy is zero and potential energy is maximum. The total energy remains constant (assuming no damping) and is equal to 1/2 m omega squared A squared. This total energy is proportional to the square of the amplitude, so doubling the amplitude quadruples the total energy.
    What assumptions are made for a simple pendulum to undergo SHM?
    For a simple pendulum to approximate simple harmonic motion, the angle of swing must be small (typically less than about 10 degrees). This allows the approximation sin theta is approximately equal to theta (in radians), which makes the restoring force proportional to displacement. Other assumptions include: the string is inextensible and massless, the bob is a point mass, and there is no air resistance or friction. Under these conditions, the period is given by T = 2 pi times the square root of (L / g), independent of the mass of the bob.