Specification: AQA A-level Physics (7408) 3.6.1.3 | Paper: 1 | Data and formulae booklet: given in the exam
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Simple pendulum: a small, dense bob on a light, inextensible string. For small angles of displacement (below about 10°) the restoring force is approximately proportional to the displacement and directed towards the equilibrium position, so the bob performs simple harmonic motion.
Time period: the time for one complete oscillation (there and back). For a simple pendulum, T = 2π√(l/g) is independent of the mass of the bob and, for small angles, independent of the amplitude.
Length l: measured from the point of suspension to the centre of mass of the bob. A seconds pendulum has a period of 2.00 s (each swing takes 1 s).
The equation
time period = 2π × √(length ÷ gravitational field strength) T = 2π√(l/g) (small-angle approximation, θ below about 10°)
| Quantity | Symbol | Unit |
|---|---|---|
| time period (time for one complete oscillation) | T | seconds (s) |
| length (suspension point to centre of mass of bob) | l | metres (m) |
| gravitational field strength (acceleration of free fall) | g | metres per second squared (m s⁻²), equivalent to N kg⁻¹ |
Rearranged: T² = 4π²l ÷ g | l = gT² ÷ 4π² | g = 4π²l ÷ T² | A graph of T² against l is a straight line through the origin with gradient 4π² ÷ g, so g = 4π² ÷ gradient. Convert cm to m first.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A simple pendulum has a length of 0.80 m. Calculate its time period for small oscillations, to 2 significant figures. (g = 9.81 m s⁻²)
Show solution
F: T = 2π√(l/g)
I: T = 2π√(0.80 ÷ 9.81)
F: T is already the subject
A: 1.794… s = 1.8 s (2 s.f.)
Q2 (Core) A playground swing behaves as a simple pendulum of length 1.50 m. Calculate (a) its time period and (b) its frequency, both to 3 significant figures. (g = 9.81 m s⁻²)
Show solution
F: T = 2π√(l/g)
I: T = 2π√(1.50 ÷ 9.81)
F: T is already the subject
A: (a) 2.456… s = 2.46 s (3 s.f.)
(b) f = 1 ÷ T = 1 ÷ 2.456… = 0.407 Hz (3 s.f.)
Q3 (Core) A simple pendulum has a time period of 1.20 s. Calculate its length in cm, to 2 significant figures. (g = 9.81 m s⁻²)
Show solution
F: T = 2π√(l/g), so T² = 4π²l ÷ g
I: 1.20² = 4π² × l ÷ 9.81
F: l = gT² ÷ 4π² = 9.81 × 1.44 ÷ 4π²
A: 0.3578… m = 36 cm (2 s.f.)
Watch out: forgetting to square T when rearranging. Square both sides before you make l the subject.
Q4 (Core) A student measures the length of a pendulum as 45.0 cm. She times 20 complete oscillations and records a total time of 27.0 s. Calculate a value for g, to 3 significant figures.
Show solution
Convert first: 45.0 cm = 0.450 m; T = 27.0 ÷ 20 = 1.35 s
F: T = 2π√(l/g), so T² = 4π²l ÷ g
I: 1.35² = 4π² × 0.450 ÷ g
F: g = 4π²l ÷ T² = 4π² × 0.450 ÷ 1.8225
A: 9.747… m s⁻² = 9.75 m s⁻² (3 s.f.)
Watch out: l is measured to the centre of mass of the bob, not to the top of it.
Q5 (Exam) A seconds pendulum takes exactly 1.00 s to swing from one side to the other. Calculate its length, to 3 significant figures. (g = 9.81 m s⁻²)
Show solution
Step 1 – time period
one swing is half an oscillation, so T = 2 × 1.00 = 2.00 s
Step 2 – length
F: T² = 4π²l ÷ g
I: 2.00² = 4π² × l ÷ 9.81
F: l = gT² ÷ 4π² = 9.81 × 4.00 ÷ 4π²
A: 0.9939… m = 0.994 m (3 s.f.)
Watch out: using T = 1.00 s. A complete oscillation is there and back, so the period of a seconds pendulum is 2.00 s.
Q6 (Exam) The seconds pendulum from Q5 (length 0.994 m) is taken to the Moon, where g = 1.62 m s⁻². (a) Calculate its time period on the Moon, to 3 significant figures. (b) By what factor has the period increased, to 3 significant figures? (g on Earth = 9.81 m s⁻²)
Show solution
F: T = 2π√(l/g)
I: (a) T = 2π√(0.994 ÷ 1.62)
F: T is already the subject
A: 4.921… s = 4.92 s (3 s.f.)
(b) T ∝ 1 ÷ √g, so factor = √(9.81 ÷ 1.62) = 2.460… = 2.46 (3 s.f.)
Watch out: the period does not increase by 9.81 ÷ 1.62 ≈ 6. It depends on the square root of g.
Q7 (Exam) A student measures T for a pendulum at several lengths and plots a graph of T² (y-axis) against l (x-axis). The best-fit straight line passes through the origin and the points (0.200 m, 0.804 s²) and (1.000 m, 4.020 s²). Calculate the gradient of the line and hence a value for g, to 3 significant figures.
Show solution
Step 1 – gradient
gradient = Δ(T²) ÷ Δl = (4.020 − 0.804) ÷ (1.000 − 0.200) = 3.216 ÷ 0.800 = 4.02 s² m⁻¹
Step 2 – g from the gradient
F: T² = (4π² ÷ g) × l, which has the form y = mx, so gradient = 4π² ÷ g
I: 4.02 = 4π² ÷ g
F: g = 4π² ÷ 4.02 = 39.47… ÷ 4.02
A: 9.820… m s⁻² = 9.82 m s⁻² (3 s.f.)
Watch out: the gradient is not g. It is 4π² ÷ g, so g = 4π² ÷ gradient. Use a large triangle on the line, not data points that sit off it.
Q8 (Exam) A pendulum of length 40.0 cm is lengthened to 1.60 m. Calculate the time period before and after, to 3 significant figures, and state the effect on the period of changing the mass of the bob. (g = 9.81 m s⁻²)
Show solution
Convert first: 40.0 cm = 0.400 m
F: T = 2π√(l/g)
I: before: T = 2π√(0.400 ÷ 9.81); after: T = 2π√(1.60 ÷ 9.81)
F: T is already the subject
A: before: 1.268… s = 1.27 s; after: 2.537… s = 2.54 s (3 s.f.)
the length is 4 times bigger, so T is √4 = 2 times bigger. Mass does not appear in T = 2π√(l/g), so changing the mass of the bob has no effect on the period
Q9 (Stretch) A pendulum of length 1.20 m is pulled aside to an angle of 8.0° and released. (a) Calculate its time period, to 3 significant figures. (b) Use conservation of energy to calculate the speed of the bob at the lowest point, to 2 significant figures. (c) Check your answer using vmax = ωA, where A is the arc length (calculator in radians). (g = 9.81 m s⁻²; ignore air resistance)
Show solution
Step 1 – (a) time period
F: T = 2π√(l/g)
I: T = 2π√(1.20 ÷ 9.81)
A: 2.197… s = 2.20 s (3 s.f.)
Step 2 – (b) height fallen
Δh = l(1 − cos θ) = 1.20 × (1 − cos 8.0°) = 0.01167… m
Step 3 – speed at the lowest point
½mv² = mgΔh, so v = √(2gΔh) = √(2 × 9.81 × 0.01167…)
A: 0.4786… m s⁻¹ = 0.48 m s⁻¹ (2 s.f.)
Step 4 – (c) SHM check
8.0° = 0.1396… rad, so A = lθ = 1.20 × 0.1396… = 0.1675… m; ω = 2π ÷ T = 2.859… rad s⁻¹
vmax = ωA = 2.859… × 0.1675… = 0.4790… m s⁻¹ = 0.48 m s⁻¹ (2 s.f.)
the answers agree because 8.0° is below about 10°, so the small-angle approximation holds and the motion is very close to SHM
Watch out: using degrees in A = lθ. The arc length formula only works with θ in radians.
Q10 (Stretch) To find g, a student measures a pendulum length of 95.0 ± 0.2 cm and times 20 oscillations as 39.1 ± 0.2 s. Calculate (a) g, to 3 significant figures, (b) the percentage uncertainty in g, to 2 significant figures, and (c) the absolute uncertainty in g.
Show solution
Convert first: 95.0 cm = 0.950 m; T = 39.1 ÷ 20 = 1.955 s
Step 1 – (a) g
F: T² = 4π²l ÷ g, so g = 4π²l ÷ T²
I: g = 4π² × 0.950 ÷ 1.955²
A: 9.812… m s⁻² = 9.81 m s⁻² (3 s.f.)
Step 2 – (b) percentage uncertainties
%l = 0.2 ÷ 95.0 × 100 = 0.210…%; %T = 0.2 ÷ 39.1 × 100 = 0.511…% (dividing by 20 does not change the percentage)
g = 4π²l ÷ T², so %g = %l + 2 × %T = 0.210… + 2 × 0.511… = 1.23…%
A: 1.2% (2 s.f.)
Step 3 – (c) absolute uncertainty
Δg = 9.812… × 1.23… ÷ 100 = 0.121… m s⁻²
A: g = 9.81 ± 0.12 m s⁻²
Watch out: forgetting to double the percentage uncertainty in T. T is squared, so its percentage uncertainty counts twice.
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