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Theme C — Wave Behaviour

C.1.4 Time Period of a Simple Pendulum — IB Physics HL Revision Notes & Topical Questions

5 min read 11 parts Lesson 6 of 37 1 quiz

📚 IB Physics HL Hub / Topical Revision
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C.1.4 Time Period of a Simple Pendulum — IB PHYSICS SL notes: Use T = 2π√(l/g) for a simple pendulum and explain the small-angle condition.Simple pendulum — a small dense mass (the bob) on a light inextensible string of length l from a fixed pointThe string stays in tension throu…

  • Restoring force
  • Small angle
  • What the symbols mean
  • Seconds pendulum
  • Angular frequency
  • Reducing uncertainty

Section 1 of 11The simple pendulum

Simple pendulum
  • Simple pendulum a small dense mass (the bob) on a light inextensible string of length l from a fixed point
  • The string stays in tension throughout the oscillation.
  • The bob is treated as a point mass and swings to and fro along a circular arc.
Length
  • Length l is measured from the pivot to the centre of mass of the bob
Forces
  • Forces the bob feels its weight mg and the string tension
  • The tension is perpendicular to the motion, so it provides the centripetal force and does no work.
  • The component of weight along the arc controls the motion.

Section 2 of 11Why the motion is SHM

Restoring force
  • Restoring force the component of weight along the arc is mg sin θ, directed back towards the lowest point
Small angle
  • Small angle for θ in radians sin θ ≈ θ, and the arc displacement is x = lθ
  • F = −mgθ = −(mg/l)x.
  • This is proportional to displacement and directed to equilibrium, so ω2 = g/l.
Accuracy
  • Accuracy at 10° the approximation is accurate to about 0.5%, and at 30° the error is about 5%
Condition
  • Condition the formula holds for angles below about 10°
  • Beyond this the period grows with amplitude.

Section 3 of 11The period formula

Equation
  • T = 2π√(l/g)
    What the symbols mean: T = period (s) · l = length of the pendulum (m) · g = 9.81 m s−2
  • l is in m, g in m s−2 and T in s.
Mass
  • T does not depend on the mass of the bob
Amplitude
  • T does not depend on amplitude for small angles
Gravity
  • T depends on the gravitational field strength
  • The period differs between Earth and the Moon.
  • On the Moon, where g is about 1.6 m s−2, the same pendulum has a period about 2.5 times longer.
Seconds pendulum
  • Seconds pendulum a length of about 1 m gives a period of about 2 s
Angular frequency
  • ω = √(g/l)

Section 4 of 11Pendulum image: data and uncertainty questions

Realistic 3D model of a simple pendulum with pivot, string, centre of bob, length L, metre ruler and stopwatch.
Simple pendulum apparatus: measure L from the suspension point to the centre of the bob. Illustration not to scale; the angle is enlarged for clarity. Use an initial angle below about 10 degrees in the experiment.

Use the existing apparatus model above. The following data are hypothetical practice data; do not read lengths or angles from the illustration.

1. The metal bob in the image is replaced by a denser bob of the same radius. The suspension-to-centre length and small release angle are unchanged. What happens to the period? (1 mark)

OptionStatement
AIt increases in proportion to mass
BIt decreases in proportion to mass
CIt is unchanged in the ideal pendulum model
DIt doubles
Show answer and explanation

C. The ideal small-angle formula T = 2π√(L/g) contains no mass. Changing mass alone does not change the period.

2. Why should the bob be released through a small angle in this experiment? (1 mark)

OptionStatement
ASo sin θ is approximately θ in radians, making the restoring acceleration approximately proportional to displacement
BSo the string becomes slack at the bottom
CSo gravity becomes negligible
DSo all timing uncertainty disappears
Show answer and explanation

A. For small angles, sin θ ≈ θ in radians. The tangential acceleration then has the form a = −(g/L)x, giving approximately simple harmonic motion.

3. Determine g from a graph. (3 marks)

A best-fit graph of T² on the vertical axis against L on the horizontal axis has gradient 3.98 s² m−1. Calculate g to 3 significant figures.

Show worked solution and marking points
  1. From T² = (4π²/g)L, gradient = 4π²/g. (1 mark)
  2. Rearrange to g = 4π²/gradient. (1 mark)
  3. g = 4π² / 3.98 = 9.92 m s−2, to 3 significant figures. (1 mark)

4. Convert a total time and its uncertainty. (3 marks)

A student records a time of (40.2 ± 0.2) s for 20 complete oscillations. Treat the count as exact. Find the period, its absolute uncertainty and its percentage uncertainty.

Show worked solution and marking points
  1. T = 40.2 / 20 = 2.01 s. (1 mark)
  2. ΔT = 0.2 / 20 = 0.01 s, so T = (2.01 ± 0.01) s. (1 mark)
  3. Percentage uncertainty = 100 × 0.01 / 2.01 = 0.50%, to 2 significant figures. (1 mark)

5. Propagate measurement uncertainty. (3 marks)

For g = 4π²L/T², measurements are L = (0.750 ± 0.005) m and T = (1.74 ± 0.02) s. Use the maximum percentage uncertainty rule to estimate the percentage uncertainty in g.

Show worked solution and marking points
  1. Fractional uncertainty in g ≈ ΔL/L + 2ΔT/T. (1 mark)
  2. Percentage uncertainty ≈ 100(0.005/0.750 + 2 × 0.02/1.74). (1 mark)
  3. The result is about 3.0%. This is a maximum uncertainty estimate using the stated rule. (1 mark)

6. Diagnose a length offset. (3 marks)

The student measures from the suspension point to the top of the spherical bob, recording Lm. A graph of T² against Lm has gradient 4.00 s² m−1 and vertical intercept 0.0200 s². Assume the only systematic error is the missing bob radius r. Find r.

Show worked solution and marking points
  1. The true length is L = Lm + r, so T² = (4π²/g)Lm + (4π²/g)r. (1 mark)
  2. Intercept = gradient × r, so r = intercept / gradient. (1 mark)
  3. r = 0.0200 / 4.00 = 0.00500 m = 5.00 mm.
    • The positive intercept is consistent with measuring a length that is too short.
    • (1 mark)

Section 5 of 11Experiment to find g

1
Clamp a thread firmly at a fixed point with a small dense bob on the end.
2
Measure l from the pivot to the centre of the bob with a ruler.
3
Displace the bob by a small angle and release it.
4
Time 20 oscillations, starting the count as the bob passes the centre (fiducial mark).
5
Repeat for five or more lengths.
6
Plot T2 against l and find the gradient.
Relation
  • T2 = (4π2/g)l, so the graph is a straight line through the origin
Result
  • Result g = 4π2/gradient
Reducing uncertainty
  • Reducing uncertainty use a long thread, a small dense bob and many oscillations

Section 6 of 11Sources of error

Sources of error
Systematic errors
  • Systematic errors measuring l to the top of the bob instead of the centre, too large an amplitude, or an elliptical swing instead of a plane swing
Random error
  • Random error reaction time, reduced by timing many oscillations
Non-zero intercept
  • Non-zero intercept a T2–l line with a non-zero intercept usually means a systematic error in l, such as missing the bob radius
  • The line is shifted horizontally.
Propagation
  • Propagation g ∝ 1/T2, so a 1% uncertainty in T gives a 2% uncertainty in g

Section 7 of 11Related effects

Mapping gravity
  • Mapping gravity pendulums were once used to map variations in g across the Earth
Accelerating lift
  • Accelerating lift a pendulum in a lift accelerating upward feels a larger effective g, so its period is shorter
Pendulum clock
  • Pendulum clock a rise in temperature expands the rod, so l increases, T grows and the clock runs slow
  • On a mountain top, where g is slightly smaller, the period also lengthens.
Energy
  • Energy the swing exchanges gravitational potential and kinetic energy (C.1.5)

Section 8 of 11More from the specification

1Simple Harmonic Motion Dynamics
  • In simple harmonic motion (SHM), acceleration is directly proportional to displacement from the equilibrium position and directed in the opposite, negative direction (a = -ω²x).
2Pendulum vs Mass-Spring Oscillators
  • The time period of a simple pendulum depends on length and gravity (T = 2π√(L/g)),
  • whereas a mass-spring system depends on mass and spring constant (T = 2π√(m/k)).
    What the symbols mean: T = period (s) · m = mass on the spring (kg) · k = spring constant (N m−1)
3Energy Transformation and Maximum Velocity
  • Velocity reaches its maximum magnitude as an oscillating object passes through the central equilibrium position (x = 0),
  • where potential energy is minimum.
4Kinematic Graphs of SHM
  • Graphical analysis shows a π/2 phase difference between displacement and velocity graphs,
  • and a π phase difference between displacement and acceleration.
5Position
  • Location relative to a reference point.
6Harmonic
  • Resonant mode of a stationary wave, with frequency a whole-number multiple of the fundamental where allowed.
7Constant
  • Not changing.
8Velocity
  • Rate of change of displacement.
9Equilibrium position
  • Position where the net force is zero.
10Displacement-time graph
  • A graph of displacement against time whose gradient is velocity.
11Velocity-time graph
  • A graph of velocity against time whose gradient is acceleration and area is displacement.
12Spring constant
  • Stiffness k in N/m.

Section 9 of 11More from the specification

  • Elastic potential
    • It is stretched, squashed or bent.

Section 10 of 11Key terms

TermMeaning
Simple pendulumA small mass on a light inextensible string oscillating through small angles.
BobThe mass at the end of the pendulum.
Small-angle approximationsin θ ≈ θ (radians) for small θ.
Seconds pendulumA pendulum with a period of 2 s, about 1 m long.
Fiducial markA fixed reference point used to time the swings.

Section 11 of 11Exam technique

1
Thinking a heavier bob swings with a longer period — T does not depend on mass.
2
Using a large starting angle — the SHM formula is valid only for small angles.
3
Measuring l to the top of the bob rather than its centre.
4
Timing a single oscillation instead of many.
5
Using a pendulum length in centimetres without converting to metres.

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