Mass-Spring Systems & Simple Pendulums Revision Notes | A Level AQA Physics

1. Why Period Formulae Matter

A grandfather clock keeps perfect time not because of its pendulum's weight, but because of its length. A car suspension bounces at a frequency set by the spring stiffness and the car's mass. Both of these are governed by the same two period formulae you will meet in this lesson — and both are examinable at A Level.

The AQA specification requires you to use $T = 2\pi\sqrt{\frac{m}{k}}$ for a mass-spring system and $T = 2\pi\sqrt{\frac{l}{g}}$ for a simple pendulum. You also need to be ready to apply the same reasoning to unfamiliar oscillators when the question supplies the relevant formula. This lesson builds each formula from physical reasoning before moving to calculation and exam technique.

2. The Mass-Spring System

Picture a mass hanging from a vertical spring in equilibrium. Pull the mass down slightly and release it — it oscillates up and down in SHM. The restoring force is provided by the spring: $F = -kx$, where $x$ is displacement from equilibrium and the negative sign confirms the force always acts back towards the centre.

Combining $F = ma$ with $F = -kx$ gives $a = -\frac{k}{m}x$. Comparing this with the SHM condition $a = -\omega^2 x$ reveals that $\omega^2 = \frac{k}{m}$, so $\omega = \sqrt{\frac{k}{m}}$. Since $T = \frac{2\pi}{\omega}$, substituting gives the period formula directly.

Notice what is absent from the formula: amplitude. Doubling the amplitude does not change the period. This is a defining feature of SHM and a favourite examiner target. The period depends only on the inertia of the mass and the stiffness of the spring.

To increase the period, you can increase the mass or decrease the spring constant. To decrease the period, use a stiffer spring or a lighter mass. The square-root relationship means that quadrupling the mass only doubles the period — a subtlety worth noting for analysis questions.

3. Calculating with the Mass-Spring Formula

The formula can be rearranged to find any one of the three quantities — $T$, $m$, or $k$ — given the other two. Squaring both sides gives $T^2 = 4\pi^2\frac{m}{k}$, which is the linearised form used in required practicals. Let's work through a full exam-style calculation.

Worked example: Calculating the period of a mass-spring system

A spring of spring constant 25 N m$^{-1}$ has a mass of 0.100 kg attached to it. Calculate the period of oscillation. [3 marks]

1. $T = 2\pi\sqrt{\dfrac{m}{k}}$ — State the formula first — this earns the method mark even if arithmetic goes wrong later.
2. $T = 2\pi\sqrt{\dfrac{0.100}{25}}$ — Both values are already in SI units (kg and N m$^{-1}$), so no conversion is needed.
3. $\dfrac{0.100}{25} = 0.004$ — 0.100 divided by 25: $0.100 \div 25 = 0.004$. Keep this exact before taking the root.
4. $\sqrt{0.004} = 0.06325...$ — $\sqrt{0.004} = \sqrt{4 \times 10^{-3}} = 2 \times 10^{-1.5} \approx 0.06325$. Use your calculator here.
5. $T = 2\pi \times 0.06325 = 0.3974... \approx 0.40 \text{ s}$ — $2\pi \approx 6.2832$; $6.2832 \times 0.06325 \approx 0.397$ s. Round to 2 significant figures to match the data given.

Answer: $T \approx 0.40$ s

Rearranging for $k$ is equally common in exam questions. Starting from $T^2 = 4\pi^2\frac{m}{k}$, multiply both sides by $k$ and divide by $T^2$ to get $k = \frac{4\pi^2 m}{T^2}$. This rearrangement is also the basis of the required practical for determining $k$ from a graph of $T^2$ against $m$.

4. The Simple Pendulum

Now consider a bob of mass $m$ suspended on a string of length $l$, displaced through a small angle and released. The restoring force is the component of gravity acting along the arc: $F = -mg\sin\theta$. For small angles (below about 10°), $\sin\theta \approx \theta$ in radians, so $F \approx -mg\theta$.

Since arc length $s = l\theta$, the displacement is $x = l\theta$, giving $F \approx -\frac{mg}{l}x$. This has the same form as Hooke's Law with an effective spring constant of $\frac{mg}{l}$. Substituting into the mass-spring formula: $T = 2\pi\sqrt{\frac{m}{mg/l}} = 2\pi\sqrt{\frac{l}{g}}$. The mass cancels entirely.

Two things stand out. First, the mass of the bob has vanished — a heavier bob and a lighter bob on the same length string have identical periods. Second, the formula depends on $g$, which means the same pendulum ticks at a different rate on the Moon. This is why pendulum clocks run slow at high altitude, where $g$ is slightly smaller.

5. Calculating with the Pendulum Formula

The pendulum formula is frequently used in reverse to determine $g$ experimentally — a classic A Level required practical. Squaring both sides gives $T^2 = \frac{4\pi^2}{g}l$, which is linear in $l$. A graph of $T^2$ against $l$ has gradient $\frac{4\pi^2}{g}$, allowing $g$ to be calculated precisely.

Worked example: Finding $g$ from pendulum data

A student measures the period of a pendulum of length 0.640 m and records $T = 1.60$ s. Calculate the value of $g$ from this data. [3 marks]

1. $T = 2\pi\sqrt{\dfrac{l}{g}} \Rightarrow T^2 = \dfrac{4\pi^2 l}{g} \Rightarrow g = \dfrac{4\pi^2 l}{T^2}$ — Squaring both sides removes the square root, making rearrangement straightforward.
2. $T^2 = 1.60 \times 1.60 = 2.56 \text{ s}^2$ — $1.60 \times 1.60 = 2.56$. Always expand the square explicitly to avoid errors.
3. $4\pi^2 = 4 \times 9.8696... = 39.478...$ — $\pi^2 \approx 9.8696$; multiplying by 4 gives approximately 39.48. Keep this unrounded for now.
4. $g = \dfrac{39.478 \times 0.640}{2.56} = \dfrac{25.266}{2.56} = 9.87 \text{ m s}^{-2}$ — Numerator: $39.478 \times 0.640 = 25.266$. Dividing: $25.266 \div 2.56 = 9.87$ m s$^{-2}$. This is close to the accepted value of 9.81 m s$^{-2}$, confirming the result is physically reasonable.

Answer: $g \approx 9.87$ m s$^{-2}$

The small discrepancy from 9.81 m s$^{-2}$ would be attributed to experimental uncertainty — measurement of $l$ to the centre of the bob and timing errors are the main sources. In an exam, if your calculated $g$ is between about 9.5 and 10.0 m s$^{-2}$, the result is physically plausible.

6. Required Practical: Determining $g$ Using a Simple Pendulum

Required practical summary: measure the pendulum length from the pivot to the centre of the bob, keep the amplitude small, and time 10 oscillations for each length to reduce percentage timing uncertainty. Repeat for several lengths, calculate T for each run, then plot T² against l. The gradient equals 4π²/g, so g is found from g = 4π²/gradient.

7. Comparing the Two Systems

Both formulae share the same $2\pi\sqrt{\cdots}$ structure, but the quantities inside the root are different. For the mass-spring system, the ratio $m/k$ captures the balance between inertia and restoring force stiffness. For the pendulum, the ratio $l/g$ captures the balance between the pendulum's length and the strength of gravity providing the restoring force.

The contrast in how mass behaves is the most commonly tested comparison point. In the mass-spring system, mass appears explicitly — more mass means more inertia and a longer period. In the pendulum, mass cancels because the same mass that increases inertia also increases the gravitational restoring force by exactly the same factor.

8. Other Harmonic Oscillators

AQA states that questions may involve other harmonic oscillators — for example, liquid oscillating in a U-tube — but the full formula will always be provided in the question. Your job is to apply the same reasoning you have used for the mass-spring and pendulum systems: identify the variables, substitute correctly, and rearrange if needed.

For instance, a U-tube of liquid with total liquid column length $L$ oscillates with period $T = 2\pi\sqrt{\frac{L}{2g}}$. The structure is identical to the pendulum formula — a length divided by $g$ under the root. If the question gives you this formula and asks you to find $L$ for a given period, you rearrange exactly as you would for the pendulum: square both sides, isolate $L$.

9. Exam Technique and Common Mistakes

AQA calculation questions on these formulae follow a predictable mark structure: one mark for the correct formula (or correct rearrangement), one mark for correct substitution with units, and one accuracy mark for the final answer with appropriate significant figures and units. Writing the formula first — even before substituting — secures the method mark if arithmetic goes wrong.

For 'explain' or 'describe' questions about what affects period, the examiner wants explicit reference to the formula. Saying 'increasing $l$ increases $T$' is stronger than 'a longer pendulum is slower' — the former references the physics directly and is more likely to earn the mark.

For extended-response questions asking you to evaluate an experimental method, structure your answer around: what was measured and how, sources of random and systematic uncertainty, and how reliability was improved (e.g., repeating timings, using many oscillations). Linking each improvement explicitly to the formula — for example, 'timing 10 oscillations reduces the percentage uncertainty in $T$, and since $T$ is squared in $T^2 = 4\pi^2 l/g$, this has a doubled effect on the uncertainty in $g$' — demonstrates the higher-order reasoning that earns top marks.

The chart above shows the expected linear relationship between $T^2$ and $l$. The gradient of the best-fit line through these points is approximately $\frac{4.03 - 0.81}{1.00 - 0.20} = \frac{3.22}{0.80} = 4.025$ s$^2$ m$^{-1}$, giving $g = \frac{4\pi^2}{4.025} \approx 9.80$ m s$^{-2}$ — consistent with the accepted value. In your practical, draw a large triangle on the graph to calculate the gradient and show your working clearly.