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First teaching 2023

First exams 2025

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Time Period of a Simple Pendulum (SL IB Physics)

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Katie M

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Katie M

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Time Period of a Simple Pendulum

  • A simple pendulum consists of a string and a bob at the end
    • The bob is a weight, generally spherical and considered a point mass
    • The bob moves from side to side
    • The string is light and inextensible remaining in tension throughout the oscillations
    • The string is attached to a fixed point above the equilibrium position
  • The time period of a simple pendulum for small angles of oscillation is given by:

Period of Pendulum Equation _2

  • Where:
    • T = time period (s)
    • L = length of string (from the pivot to the centre of mass of the bob) (m)
    • g = gravitational field strength (N kg-1)

A simple pendulum

  • The time period of a pendulum does depend on the gravitational field strength, meaning its period would be different on the Earth and the Moon

Small Angle Approximation

  • This formula for time period is limited to small angles (θ < 10°) and therefore small amplitudes of oscillation from the equilibrium point
  • The restoring force of the pendulum is the weight component acting along the arc of the circle towards the equilibrium position
  • It is resolved to act at an angle θ to the horizontal
  • When considering SHM because of small angle approximation it is assumed the restoring force acts along the horizontal
  • So sin θθ 

9-1-4-pendulum-resolved-forces-v2

Forces on a pendulum when it is displaced. Assuming θ < 10°, the small angle approximation can be used to describe the time period of a simple pendulum such as this.

Worked example

A swinging pendulum with a length of 80.0 cm has a maximum angle of displacement of 8°.

Determine the angular frequency of the oscillation.

Answer:

Step 1: List the known quantities

  • Length of the pendulum, L = 80 cm = 0.8 m
  • Acceleration due to gravity, g = 9.81 m s−2

Step 2: Write down the relationship between angular frequency, ω, and period, T

T space equals space fraction numerator 2 pi over denominator omega end fraction

Step 3: Write down the equation for the time period of a simple pendulum

T space equals space 2 pi square root of L over g end root

  • This equation is valid for this scenario since the maximum angle of displacement is less than 10°

Step 4: Equate the two equations and rearrange for ω

fraction numerator 2 straight pi over denominator omega end fraction space equals space 2 pi square root of L over g end root space space space space space rightwards double arrow space space space space space omega space equals space square root of g over L end root

Step 5: Substitute the values to calculate ω

omega space equals space square root of fraction numerator 9.81 over denominator 0.8 end fraction end root= 3.50 rad s−1

Angular frequency:  ω = 3.5 rad s−1

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Katie M

Author: Katie M

Expertise: Physics

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.