Mathematics 11–12 · Year 11
Pendulum: period against length and rearranging T = 2 pi sqrt(L/g)
Formulas and equations (Mathematics Standard, Year 11: substituting into the formula, and changing the subject only in the form L = g T^2 / (4 pi^2), because the Standard syllabus limits changing the subject to y = ax^2 + c); Working with functions: Direct and inverse variation (Mathematics Advanced, Year 11)
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The idea
A formula becomes a relationship the learner tests by measurement: doubling the length does not double the period, and plotting L against T^2 gives a straight line whose gradient contains g.
What you need
- 1.5 m of string, a 50 g hooked mass, a retort stand clamped to the bench with the bob hanging past the bench edge, a metre rule
- A stopwatch to time 10 swings (do not tape a phone to the bob: a phone outweighs the 50 g mass, so it becomes the bob, moves the centre of mass away from the measured L and, for a 15 cm phone at L = 0.25 m, lengthens the period by about 1.5 per cent)
How to do it
- Set the length L (pivot to the centre of the bob) to 0.25 m, pull the bob less than 10 degrees aside, time 10 full swings and divide by 10; three trials, keep the mean.
- Repeat for L = 0.50, 0.75 and 1.00 m.
- Substitute each L into T = 2 pi sqrt(L/g) with g = 9.80 m/s^2 and compare with the measured T.
- Plot T against L (a curve) and L against T^2 (a line through the origin); find the gradient and compute g = 4 pi^2 x gradient.
- Mathematics Advanced: rearrange T = 2 pi sqrt(L/g) to make L the subject, L = g T^2 / (4 pi^2). Standard: start from L = g T^2 / (4 pi^2), which has the form y = a x^2 + c with c = 0, and make T the subject to recover the formula. Then find the length that gives a period of exactly 2.00 s.
What you should see
With g = 9.80 m/s^2 the periods are 1.004, 1.419, 1.738 and 2.007 s for L = 0.25, 0.50, 0.75 and 1.00 m; doubling L (0.25 to 0.50 m, or 0.50 to 1.00 m) multiplies T by sqrt 2 = 1.414. Timing 10 swings with a hand stopwatch fixes T to about 0.02 s, 2.0 per cent of 1.004 s and 1.0 per cent of 2.007 s. A 2.00 s period needs L = 0.993 m, and a measured T = 2.00 s at L = 1.000 m gives g = 9.87 m/s^2. The learner knows it worked when the L against T^2 plot is straight through the origin and the recovered g is within 3 per cent of 9.80 m/s^2.
What changes
- What you change
- string length L
- What you measure
- period T
- What you keep the same
- swing below 10 degrees
- same bob
- swings counted from the same extreme
Common misconceptions
Each of these ideas is wrong, and the activity is a chance to test it.
- A heavier bob swings slower; the mass is not in the formula and the measured period does not change.
- Doubling the length doubles the period; it multiplies the period by sqrt 2.
Safety card
Hazards
- a swinging mass striking a face
- the retort stand toppling
Controls
- clamp the stand to the bench
- stand clear of the swing plane
Note
No chemicals and no heat are used, so the NSW Department of Education Chemical Safety in Schools package does not apply; ordinary classroom supervision.
Curriculum references
The NSW syllabus outcomes and Australian Curriculum v9 codes this activity supports. They are references, not a verified or complete curriculum alignment.
- Mathematics Standard 11–12 Syllabus (2024), Year 11 focus area Formulas and equations; Year 11 taught from Term 1 2026, Year 12 from Term 4 2026, first HSC examination 2027 (the 2017 syllabus is still taught to Year 12 until then); page read 2026-09-22MST-11-01
- Mathematics Advanced 11–12 Syllabus (2024), Year 11 focus area Working with functions; Year 11 taught from Term 1 2026, Year 12 from Term 4 2026, first HSC examination 2027 (the 2017 syllabus is still taught to Year 12 until then); page read 2026-09-22MAV-11-02
- Australian Curriculum v9No Australian Curriculum v9 code is listed.
Sources
The pages the author read to write this activity.