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SHM Practical Investigation Revision Notes | A Level AQA Physics

Plot T² against m or L to get a straight line, then find k = 4π²/gradient or g = 4π²/gradient. Free AQA notes with 2 worked examples and 12 questions answered.

Updated 12 Sept 2026 · Revise Right Team

SHM Practical Investigation is part of Periodic motion (A-level only) in A Level AQA Physics.

7 min

Reading time

16

Lessons in this topic

A Level AQA

Pathway

Reviewed by Revise Right

Content review

What this lesson covers

Plot T² against m or L to get a straight line, then find k = 4π²/gradient or g = 4π²/gradient. Free AQA notes with 2 worked examples and 12 questions answered.

  • Describe the experimental setup for investigating simple harmonic motion using a mass-spring system and a simple pendulum
  • Identify and justify the independent, dependent, and control variables in each SHM investigation
  • Analyse experimental data to determine the relationship between period, mass, spring constant, and pendulum length
  • Evaluate sources of uncertainty and suggest improvements to the experimental method

Key ideas to keep in view

Use this lesson to stay inside Periodic motion (A-level only) while connecting the detail back to the wider Further mechanics and thermal physics (A-level only) topic.

  • Simple harmonic motion (SHM): Oscillatory motion in which the acceleration is directly proportional to the displacement from the equilibrium position and always directed towards it. Occurs in mass-spring systems and simple pendulums under standard conditions.
  • Period (T): The time taken for one complete oscillation, measured in seconds (s). Determined experimentally by timing multiple oscillations and dividing by the number of oscillations to reduce timing uncertainty.
  • Simple pendulum: A point mass suspended on a light, inextensible string that oscillates with SHM for small angles (less than approximately 10 degrees). Period depends only on string length and gravitational field strength.
  • Mass-spring system: A mass attached to a spring that oscillates vertically (or horizontally) with SHM. Period depends on the mass and the spring constant of the spring.

Revision notes

Read the core explanations from SHM Practical Investigation before testing yourself from memory.

1. Why This Practical Matters

Picture a pendulum clock ticking steadily on a wall, or a car suspension absorbing a bump. Both are governed by the same physics: simple harmonic motion. Required Practical 7 asks you to measure that motion directly — and the data you collect can verify some of the most elegant equations in A-level physics.

The practical has two distinct systems: a mass on a spring and a simple pendulum. Each one tests a different relationship, and examiners expect you to know both setups inside out — variables, method, analysis, and evaluation. This lesson walks through every layer of that.

2. The Physics Behind the Practical

Before touching any apparatus, it helps to know what you are actually measuring and why the equations take the form they do. For a mass-spring system, the restoring force obeys Hooke's law: $F = -kx$, where $k$ is the spring constant and $x$ is the displacement. Combining this with Newton's second law gives acceleration proportional to displacement — the defining condition for SHM.

For a simple pendulum, the restoring force is the component of gravity acting along the arc. For small angles (below about 10°), $\sin\theta \approx \theta$ in radians, which again gives acceleration proportional to displacement. This small-angle approximation is a critical assumption — violate it and the motion is no longer truly SHM.

Notice that the period of a mass-spring system depends on mass and spring constant only — not on the amplitude of oscillation. This is a key feature of SHM and something examiners test directly.

For the pendulum, the period depends on length and gravitational field strength only — not on the mass of the bob and not on the amplitude (provided the angle stays small). This independence from mass surprises many students and is a favourite examiner target.

3. Required Practical 7 — Mass-Spring System

The aim of the mass-spring investigation is to find how the period of oscillation varies with mass, and to use that relationship to determine the spring constant $k$. Setting up the apparatus carefully is the foundation of reliable data.

The choice to plot $T^2$ against $m$ rather than $T$ against $m$ is deliberate. Plotting $T$ against $m$ would give a curve, which is much harder to analyse. Linearising the relationship by squaring both sides of the period equation turns a curve into a straight line — and a straight line gives you a gradient you can use to calculate $k$ directly.

Worked example: Determining Spring Constant from a T² vs m Graph

A student plots $T^2$ against $m$ for a mass-spring system and obtains a straight line with gradient $0.40$ s² kg$^{-1}$. Calculate the spring constant $k$.

  1. $T^2 = \frac{4\pi^2}{k} \cdot m$ — This is obtained by squaring both sides of $T = 2\pi\sqrt{m/k}$. The gradient of $T^2$ vs $m$ equals $\frac{4\pi^2}{k}$. [M1]
  2. $\text{gradient} = \frac{4\pi^2}{k} = 0.40 \text{ s}^2 \text{ kg}^{-1}$ — Read the gradient directly from the graph. [M1]
  3. $k = \frac{4\pi^2}{0.40}$ — Divide $4\pi^2$ by the gradient value.
  4. $4\pi^2 = 4 \times 9.8696... \approx 39.48$ — $\pi^2 \approx 9.870$, so $4 \times 9.870 = 39.48$.
  5. $k = \frac{39.48}{0.40} = 98.7 \text{ N m}^{-1}$ — $39.48 \div 0.40 = 98.7$. Units: $\frac{\text{(dimensionless)}}{\text{s}^2 \text{ kg}^{-1}} = \text{kg s}^{-2} = \text{N m}^{-1}$. [A1]

Answer: $k \approx 98.7 \text{ N m}^{-1}$ (to 3 significant figures)

4. Required Practical 7 — Simple Pendulum

The pendulum investigation follows the same logic as the mass-spring experiment: vary one quantity, measure the period, linearise the relationship, and extract a physical constant from the gradient. Here, the variable is the length of the pendulum, and the constant you extract is $g$, the gravitational field strength.

Measuring the length $L$ accurately is the single biggest source of systematic error in this experiment. The length must be measured from the pivot point to the centre of mass of the bob — not to the top or bottom of the bob. Using a vernier calliper to find the bob's diameter and adding the radius to the string length is the correct technique.

Worked example: Determining g from a T² vs L Graph

A student's $T^2$ vs $L$ graph has a gradient of $4.02$ s² m$^{-1}$. Calculate the value of $g$ obtained from this experiment.

  1. $T^2 = \frac{4\pi^2}{g} \cdot L$ — Squaring $T = 2\pi\sqrt{L/g}$ gives this linear form. Gradient $= \frac{4\pi^2}{g}$. [M1]
  2. $\frac{4\pi^2}{g} = 4.02 \text{ s}^2 \text{ m}^{-1}$ — The gradient is read from the best-fit straight line on the graph. [M1]
  3. $g = \frac{4\pi^2}{4.02}$ — Divide $4\pi^2$ by the gradient.
  4. $4\pi^2 = 4 \times 9.8696 \approx 39.48$ — $\pi^2 \approx 9.870$; $4 \times 9.870 = 39.48$.
  5. $g = \frac{39.48}{4.02} \approx 9.82 \text{ m s}^{-2}$ — $39.48 \div 4.02 = 9.82$. This is close to the accepted value of $9.81$ m s$^{-2}$, suggesting the experiment was well-controlled. [A1]

Answer: $g \approx 9.82 \text{ m s}^{-2}$ (to 3 significant figures)

5. Graphical Analysis and Linearisation

Both investigations rely on the same analytical strategy: linearise the period equation so that a graph of $T^2$ against the variable quantity produces a straight line through the origin. The gradient of that line contains the physical constant you want to determine.

When drawing the best-fit line, use a ruler and aim for the line that minimises the total distance of all points from the line — not a line that passes through every point. Anomalous points should be identified, circled, and excluded from the best-fit line, but they must still be shown on the graph.

To find the gradient, draw a large triangle on the best-fit line — as large as possible to reduce the percentage uncertainty in the gradient calculation. Read off the coordinates of the two triangle vertices from the axes, and calculate $\text{gradient} = \frac{\Delta(T^2)}{\Delta L}$ or $\frac{\Delta(T^2)}{\Delta m}$ as appropriate. Always include units in the gradient.

6. Uncertainty, Reliability, and Evaluation

Evaluating your method is worth significant marks in AQA questions. You need to identify specific sources of uncertainty, explain how they affect the result, and suggest concrete improvements — not vague ones like 'be more careful'.

Source of uncertaintySystem affectedEffect on resultImprovement
Reaction time when starting/stopping stopwatch (~0.2 s)BothIncreases absolute uncertainty in total time; reduced by timing more oscillationsUse a light gate and data logger to remove human reaction time
Measuring L to the centre of the bobPendulumSystematic error if radius of bob is not added; underestimates LMeasure bob diameter with vernier calliper; add radius to string length
Amplitude exceeding 10°PendulumSmall-angle approximation breaks down; period increases slightly above predicted valueUse a protractor or measure horizontal displacement to keep angle below 10°
Spring oscillating non-vertically (swinging as well as bouncing)Mass-springPeriod measured is not purely for vertical SHM; introduces scatterDisplace mass carefully along the vertical axis only; use a guide rod if available
Mass of the springMass-springEffective oscillating mass is greater than the attached mass; $T^2$ vs $m$ line does not pass through originAccount for spring mass by extrapolating the line to find the effective additional mass

Reliability is improved by repeating each timing at least three times and calculating a mean period. Validity is ensured by keeping all control variables constant — particularly the amplitude, which must remain small throughout both investigations.

7. Comparing the Two Systems

Both systems demonstrate SHM, but they test different physical relationships. Understanding the comparison between them is useful for evaluation questions and for checking your conceptual understanding of the period equations.

FeatureMass-spring systemSimple pendulum
Period equation$T = 2\pi\sqrt{m/k}$$T = 2\pi\sqrt{L/g}$
Variable investigatedMass $m$ (spring constant $k$ fixed)Length $L$ (g fixed)
Physical constant extractedSpring constant $k$Gravitational field strength $g$
Graph plotted$T^2$ vs $m$$T^2$ vs $L$
Gradient equals$4\pi^2/k$$4\pi^2/g$
Key assumptionSpring obeys Hooke's law; spring mass negligibleSmall-angle approximation ($\theta < 10°$); string is inextensible and massless
Effect of increasing the variablePeriod increases (heavier mass oscillates more slowly)Period increases (longer pendulum oscillates more slowly)
Effect of changing amplitudeNo effect on period (SHM property)No effect on period provided angle stays below ~10°

8. Exam Technique and Mark-Winning Guidance

AQA questions on Required Practical 7 follow predictable patterns. They ask you to describe the method, identify variables, sketch or interpret a graph, calculate a physical constant from a gradient, or evaluate the experiment. Each of these has a specific structure that earns marks.

When describing the method, be specific about numbers: state that you time 20 oscillations (not 'several'), that the amplitude is kept below 10° (not 'small'), and that you repeat each measurement at least three times to find a mean. Vague method descriptions lose marks.

Frequently asked questions

Which factors affect the period of simple harmonic motion?

The period of a mass-spring system depends on mass and spring constant only, and the period of a simple pendulum on length and gravitational field strength only. Neither depends on amplitude, provided a pendulum's angle stays small, and a pendulum's period does not depend on the mass of the bob.

How do you find the spring constant from a T² against m graph?

Squaring T = 2π√(m/k) gives T² = (4π²/k) × m, so a graph of T² against m is a straight line with gradient 4π²/k, and k = 4π²/gradient. For a gradient of 0.40 s² kg⁻¹, k = 39.48 ÷ 0.40 = 98.7 N m⁻¹.

How do you find g from a T² against L graph?

Squaring T = 2π√(L/g) gives T² = (4π²/g) × L, so a graph of T² against L is a straight line through the origin with gradient 4π²/g, and g = 4π²/gradient. For a gradient of 4.02 s² m⁻¹, g = 39.48 ÷ 4.02 ≈ 9.82 m s⁻², close to the accepted 9.81 m s⁻².

Why do you plot T² instead of T in the SHM practical?

T² is plotted because the period equations include a square root, so plotting T against m or L gives a curve that is much harder to analyse. Squaring both sides linearises the relationship, giving a straight line whose gradient, 4π²/k or 4π²/g, lets the physical constant be calculated directly.

What are the sources of uncertainty in the SHM required practical?

Reaction time when starting and stopping a stopwatch, about 0.2 s, is reduced by timing many oscillations or using a light gate. For a pendulum, not adding the bob's radius to the string length underestimates L, and amplitudes over 10° break the small-angle approximation. For a spring, swinging motion and the spring's own mass introduce errors.

Where do you measure the length of a pendulum from?

The length of a pendulum is measured from the pivot point to the centre of mass of the bob, not to its top or bottom. Measure the bob's diameter with a vernier calliper and add the radius to the string length. Not adding the radius is the single biggest source of systematic error in the experiment.

What is the SHM required practical?

The SHM required practical investigates two systems: a mass on a spring, varying the mass to find how the period changes and determine k, and a simple pendulum, varying the length to determine g. In each case T² is plotted against the variable, giving a straight line whose gradient is 4π²/k or 4π²/g.

Why time 20 oscillations in the SHM practical?

Timing 20 oscillations rather than one reduces the percentage uncertainty in the period, because the reaction-time error is a fixed amount spread over a much longer total time. The total time is then shared between the 20 oscillations to give T.

How does a light gate help in the SHM practical?

A light gate and data logger can time oscillations in the SHM practical, removing the human reaction time of about 0.2 s when starting and stopping a stopwatch. Without one, timing many oscillations and repeating each timing at least three times to find a mean reduces the uncertainty.

Why might a T² against m graph not pass through the origin?

A graph of T² against m for a spring may not pass through the origin because the mass of the spring itself adds to the oscillating mass. The effective oscillating mass is greater than the attached mass, and extrapolating the line finds the effective additional mass.

What are the variables in the SHM practical?

In the mass-spring investigation the independent variable is mass, the dependent variable is period, and the spring constant and amplitude are controlled. In the pendulum investigation the independent variable is length, the dependent variable is period, and the amplitude is kept small throughout.

What should you do with anomalous points on a T² graph?

Anomalous points on a T² graph should be identified, circled and excluded from the best-fit line, but still shown on the graph. The best-fit line is drawn with a ruler to minimise the total distance of all points from it, not forced through every point.

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