Physics · Simulation and virtual lab Free
Simple pendulum: find g in Kigali
Swing a pendulum, time it with a light gate and use your results to measure the acceleration due to gravity, g.
- You will learn to
- Measure the period of a pendulum by timing 10 complete oscillations.
- Show that T² is proportional to L, and that T does not depend on the mass.
- Calculate g from the gradient of a graph of T² against L.
- You will measure
- Period T (s) for different lengths; then g in m/s².
- Time needed
- About 30–40 min
- Equipment
- Nothing extra: a phone or computer. Real version: string, bob, stand, metre rule, stopwatch.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To find how the period of a simple pendulum depends on its length, angle and mass, and to measure g in Kigali.
What to do
- Keep the angle at 10°, the mass at 100 g and the place on Kigali.
- Keep the timer on Light gate. Set the length to 0.20 m. Tap Release, then Time 10 oscillations. One oscillation is a complete swing: across and back. When the timing ends, T = time ÷ 10 is shown: tap Record.
- Repeat for lengths 0.40, 0.60, 0.80, 1.00, 1.20 and 1.50 m.
- Look at the graph of T² against L: it is a straight line through the origin. The box below the graph gives g from the gradient.
- Now keep L = 1.00 m and change only the mass, then only the angle (10° to 60°). Record each. What changes, and what does not?
- Choose the Moon or Jupiter and repeat three lengths. Compare the gradients.
- Back on Kigali, choose Hand stopwatch and time L = 1.00 m three times. Why are the results slightly different each time?
Results
Questions
Does the period depend on the mass of the bob?
No. With the same length and angle, a 50 g bob and a 500 g bob have the same period, because the period T = 2π√(L/g) does not contain the mass.
Why do we time 10 oscillations instead of one?
With a hand stopwatch, your reaction time adds an error of about 0.1–0.2 s at the start and at the stop. For one oscillation (about 2 s) that is a large error. For 10 oscillations (about 20 s) the same error is shared by 10, so the period is ten times more accurate. A light gate has almost no reaction error.
Use T = 2π√(L/g) with g = 9.78 m/s² to find the length of a pendulum whose period is 2.00 s (it ticks once each second).
L = gT²/(4π²) = 9.78 × 4.00 ÷ 39.5 ≈ 0.99 m.
What happens to the period when the angle is large (for example 60°)?
It becomes a little longer (about 7 % longer at 60°). The formula T = 2π√(L/g) is only exact for small angles, below about 15°.
Why is the period longer on the Moon?
g on the Moon (1.62 m/s²) is smaller, and T is proportional to 1/√g, so the pendulum swings more slowly: T is about 2.5 times longer.
With air resistance switched on, what happens to the amplitude and to the period?
The amplitude decreases little by little (damped oscillation) because energy is lost to the air, but the period stays almost the same.
For teachers: expected results and how the model works
Expected results in Kigali: L = 0.20 m gives T ≈ 0.90 s; L = 1.00 m gives T ≈ 2.01 s; the gradient of T² against L is about 4.04–4.05 s²/m, giving g ≈ 9.74–9.78 m/s² (a 10° swing makes T about 0.2 % longer than the small-angle formula, so g comes out very slightly low). Ask learners to predict before each change. For S1, use it to practise reading time and calculating averages; for S5, use the T² against L graph and the damping option.
Exam practice Exam set
National-exam style questions on this simulation (17 marks). Try each one on paper first, then open the marking scheme.
A simple pendulum of length 0.80 m makes 20 complete oscillations in 36.0 s. (a) Calculate its period. (b) Calculate the acceleration due to gravity. (4 marks)
Marking scheme: (a) T = 36.0 ÷ 20 = 1.80 s (1). (b) g = 4π²L ÷ T² (1) = 4π² × 0.80 ÷ 1.80² (1) = 9.75 m/s² (1).
Explain why it is better to time 10 or 20 oscillations than one oscillation. (2 marks)
Marking scheme: The reaction-time error of starting and stopping the stopwatch (about 0.2 s) is the same whatever is timed (1), so it becomes a much smaller percentage of a long time; dividing by the number of oscillations gives a more accurate period (1).
A learner plotted T² against L and obtained a straight line through the origin with gradient 4.05 s²/m. (a) Why does the line pass through the origin? (b) Calculate g. (4 marks)
Marking scheme: (a) T² = (4π²/g) L, so T² is directly proportional to L (1, 1). (b) g = 4π² ÷ gradient (1) = 39.48 ÷ 4.05 = 9.75 m/s² (1).
State two precautions to take when timing a pendulum, and explain why the angle of swing should be small. (4 marks)
Marking scheme: Start timing at the centre of the swing, where the bob moves fastest (1); use a fixed reference mark / count carefully from zero (1). The formula T = 2π√(L/g) is only true for small angles (below about 10°) (1); for large angles the period is longer (1).
The same pendulum is taken to the Moon, where g = 1.6 m/s². Calculate its new period if L = 0.80 m, and say how it compares with the period on Earth. (3 marks)
Marking scheme: T = 2π√(0.80 ÷ 1.6) (1) = 2π × 0.707 = 4.44 s (1); about 2.5 times longer than on Earth (1).
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Simple pendulum: find g in Kigali
Aim: To find how the period of a simple pendulum depends on its length, angle and mass, and to measure g in Kigali.
Apparatus (real laboratory): string, bob, stand, metre rule, stopwatch.
What to do (with the simulation on www.donatstem.co.rw)
- Keep the angle at 10°, the mass at 100 g and the place on Kigali.
- Keep the timer on Light gate. Set the length to 0.20 m. Tap Release, then Time 10 oscillations. One oscillation is a complete swing: across and back. When the timing ends, T = time ÷ 10 is shown: tap Record.
- Repeat for lengths 0.40, 0.60, 0.80, 1.00, 1.20 and 1.50 m.
- Look at the graph of T² against L: it is a straight line through the origin. The box below the graph gives g from the gradient.
- Now keep L = 1.00 m and change only the mass, then only the angle (10° to 60°). Record each. What changes, and what does not?
- Choose the Moon or Jupiter and repeat three lengths. Compare the gradients.
- Back on Kigali, choose Hand stopwatch and time L = 1.00 m three times. Why are the results slightly different each time?
Results
| # | L (m) | Time for 10 oscillations (s) | T (s) | T² (s²) |
|---|
| 1 | | | | |
| 2 | | | | |
| 3 | | | | |
| 4 | | | | |
| 5 | | | | |
| 6 | | | | |
| 7 | | | | |
| 8 | | | | |
Graph
Plot T² (s²) on the vertical axis against L (m). Draw the best straight line through the origin and find g = 4π² ÷ gradient.
Questions
- Does the period depend on the mass of the bob? [2]
- Why do we time 10 oscillations instead of one? [2]
- Use T = 2π√(L/g) with g = 9.78 m/s² to find the length of a pendulum whose period is 2.00 s (it ticks once each second). [2]
- What happens to the period when the angle is large (for example 60°)? [2]
- Why is the period longer on the Moon? [2]
- With air resistance switched on, what happens to the amplitude and to the period? [2]
- A simple pendulum of length 0.80 m makes 20 complete oscillations in 36.0 s. (a) Calculate its period. (b) Calculate the acceleration due to gravity. [4]
- Explain why it is better to time 10 or 20 oscillations than one oscillation. [2]
- A learner plotted T² against L and obtained a straight line through the origin with gradient 4.05 s²/m. (a) Why does the line pass through the origin? (b) Calculate g. [4]
- State two precautions to take when timing a pendulum, and explain why the angle of swing should be small. [4]
- The same pendulum is taken to the Moon, where g = 1.6 m/s². Calculate its new period if L = 0.80 m, and say how it compares with the period on Earth. [3]
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Simple pendulum: find g in Kigali
Expected results
Expected results in Kigali: L = 0.20 m gives T ≈ 0.90 s; L = 1.00 m gives T ≈ 2.01 s; the gradient of T² against L is about 4.04–4.05 s²/m, giving g ≈ 9.74–9.78 m/s² (a 10° swing makes T about 0.2 % longer than the small-angle formula, so g comes out very slightly low). Ask learners to predict before each change. For S1, use it to practise reading time and calculating averages; for S5, use the T² against L graph and the damping option.
Answers
- Does the period depend on the mass of the bob? [2]
No. With the same length and angle, a 50 g bob and a 500 g bob have the same period, because the period T = 2π√(L/g) does not contain the mass.
- Why do we time 10 oscillations instead of one? [2]
With a hand stopwatch, your reaction time adds an error of about 0.1–0.2 s at the start and at the stop. For one oscillation (about 2 s) that is a large error. For 10 oscillations (about 20 s) the same error is shared by 10, so the period is ten times more accurate. A light gate has almost no reaction error.
- Use T = 2π√(L/g) with g = 9.78 m/s² to find the length of a pendulum whose period is 2.00 s (it ticks once each second). [2]
L = gT²/(4π²) = 9.78 × 4.00 ÷ 39.5 ≈ 0.99 m.
- What happens to the period when the angle is large (for example 60°)? [2]
It becomes a little longer (about 7 % longer at 60°). The formula T = 2π√(L/g) is only exact for small angles, below about 15°.
- Why is the period longer on the Moon? [2]
g on the Moon (1.62 m/s²) is smaller, and T is proportional to 1/√g, so the pendulum swings more slowly: T is about 2.5 times longer.
- With air resistance switched on, what happens to the amplitude and to the period? [2]
The amplitude decreases little by little (damped oscillation) because energy is lost to the air, but the period stays almost the same.
- A simple pendulum of length 0.80 m makes 20 complete oscillations in 36.0 s. (a) Calculate its period. (b) Calculate the acceleration due to gravity. [4]
(a) T = 36.0 ÷ 20 = 1.80 s (1). (b) g = 4π²L ÷ T² (1) = 4π² × 0.80 ÷ 1.80² (1) = 9.75 m/s² (1).
- Explain why it is better to time 10 or 20 oscillations than one oscillation. [2]
The reaction-time error of starting and stopping the stopwatch (about 0.2 s) is the same whatever is timed (1), so it becomes a much smaller percentage of a long time; dividing by the number of oscillations gives a more accurate period (1).
- A learner plotted T² against L and obtained a straight line through the origin with gradient 4.05 s²/m. (a) Why does the line pass through the origin? (b) Calculate g. [4]
(a) T² = (4π²/g) L, so T² is directly proportional to L (1, 1). (b) g = 4π² ÷ gradient (1) = 39.48 ÷ 4.05 = 9.75 m/s² (1).
- State two precautions to take when timing a pendulum, and explain why the angle of swing should be small. [4]
Start timing at the centre of the swing, where the bob moves fastest (1); use a fixed reference mark / count carefully from zero (1). The formula T = 2π√(L/g) is only true for small angles (below about 10°) (1); for large angles the period is longer (1).
- The same pendulum is taken to the Moon, where g = 1.6 m/s². Calculate its new period if L = 0.80 m, and say how it compares with the period on Earth. [3]
T = 2π√(0.80 ÷ 1.6) (1) = 2π × 0.707 = 4.44 s (1); about 2.5 times longer than on Earth (1).