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Experiment: Simple pendulum, relating length to the period of oscillation

To investigate how the length of a simple pendulum affects its period of oscillation and to use the result to find g.

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Aim

To investigate how the length of a simple pendulum affects its period of oscillation and to use the result to find g.

Variables

  • Manipulated: Length of the pendulum, l
  • Responding: Period of one complete oscillation, T
  • Constant: Mass of the pendulum bob and the angle (amplitude) of swing

Apparatus & materials

  • Pendulum bob
  • Cotton thread
  • Retort stand with clamp
  • Two small pieces of split cork
  • Stopwatch
  • Metre rule
  • Protractor

Procedure

  1. Clamp the thread between two pieces of cork so the pendulum length l, measured from the support to the centre of the bob, is 20.0 cm.
  2. Displace the bob to one side by a small angle (less than 10°) and release it.
  3. Use the stopwatch to time 20 complete oscillations and record the time t; repeat once and average.
  4. Calculate the period from T = t / 20.
  5. Repeat the steps for l = 40.0, 60.0, 80.0 and 100.0 cm.
  6. Record l, t and T, and calculate T² for each length.

Tabulating results

Record length l in cm, the timed 20 oscillations t₁ and t₂ in s, the average t, the period T = t / 20 in s, and the derived value T² in s². Give every reading a column heading with its unit and keep the decimal places consistent.

The graph

Plot T² (y-axis) against l (x-axis). The points lie on a straight line passing through the origin, showing T² is directly proportional to l.

Analysis

Since T = 2π√(l/g), squaring gives T² = (4π²/g) l. The gradient of the T²–l graph equals 4π²/g, so g = 4π² ÷ gradient.

Precautions

  • Keep the swing angle small (below 10°) so the motion stays simple harmonic.
  • Count the oscillations as the bob passes the lowest (equilibrium) point.
  • Make sure the bob swings in one plane, not in a circle.

Measurement · Graph skills

Sample results and what they show

These are example readings, not a mark scheme. For lengths l = 20.0, 40.0, 60.0, 80.0, 100.0 cm the period T rose from about 0.90 s to 2.01 s, giving T² of about 0.81, 1.61, 2.40, 3.24 and 4.04 s².

Two things stand out. First, doubling l from 40.0 cm to 80.0 cm does not double T: T only rises from about 1.27 s to 1.80 s, because T depends on the square root of l.

Second, T² does roughly double from 1.61 s² to 3.24 s², and that is the real pattern, T² is directly proportional to l. Timing 20 oscillations and dividing by 20 keeps the error in each T small, because your reaction-time error is shared across 20 swings.

If one T² value sits well off the trend of the others, re-time that length rather than forcing it onto the line.

Reading the graph and finding the answer

Plot T² / s² on the y-axis against l / m on the x-axis (convert cm to m so g comes out in SI units). The points should fall on a straight line through the origin, confirming T² is proportional to l.

Take the gradient from a large triangle on the best-fit line, using two points that lie on the line rather than raw data points. Reading (0.20 m, 0.81 s²) and (1.00 m, 4.04 s²): gradient = (4.04 − 0.81) s² ÷ (1.00 − 0.20) m = 3.23 s² ÷ 0.80 m = 4.04 s² m⁻¹.

The gradient equals 4π²/g, so g = 4π² ÷ gradient = 39.48 ÷ 4.04 s² m⁻¹ = 9.77 m s⁻², close to the accepted 9.81 m s⁻². A line that misses the origin usually means a systematic length error, such as forgetting the radius of the bob.

Marks examiners look for

The science-process marks here are among the easiest to secure. State the variables cleanly: manipulated l / cm, responding T / s, and a named constant such as the bob mass or the swing angle below 10°.

Keeping the angle small matters because the T² ∝ l relationship only holds for near-simple-harmonic motion.

In the table, head each column as a quantity over its unit (l / cm, t / s, T / s, T² / s²) and keep the decimal places consistent down each column, do not write 0.9 in one row and 1.270 in the next. Count oscillations as the bob passes the lowest point, where it moves fastest and timing is sharpest, and repeat each timing to average out reaction time.

On the graph, draw a single best-fit straight line balanced through the points, then quote g with a unit and a sensible number of significant figures. A conclusion that restates T² ∝ l with the value of g earns the analysis mark.

Source: DSKP KSSM Physics Form 4 and 5 (Versi English) (Bahagian Pembangunan Kurikulum (BPK), KPM)

Written by the spmphysics.com.my editorial team.· Updated 5 Sept 2026

Frequently asked questions

Do I need a lab to practise?
No, the Paper 3 graph and analysis skills can be practised from home with example data.
Why time 20 oscillations instead of just one?
Your reaction time (about 0.2 s) is the same whether you time one swing or twenty. Spread across 20 oscillations it becomes about 0.01 s per period, so the percentage error in T is roughly twenty times smaller. Timing a single swing of about 1 s would leave a 20% error.
Why plot T² against l and not T against l?
Because T = 2π√(l/g), a graph of T against l is a curve, which is hard to use. Squaring gives T² = (4π²/g) l, a straight line through the origin whose gradient is 4π²/g. A straight line is far easier to draw a best fit through and to take a gradient from.
Why must the swing angle stay below about 10°?
The simple-pendulum relationship assumes the restoring force is proportional to the displacement, which is only true for small angles. Beyond about 10° the period grows slightly, so T² would no longer be exactly proportional to l and the value of g would come out too large.

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