Period of Simple Pendulum (AQA A Level Physics): Revision Note

Exam code: 7408

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Reviewed by: Tim

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Period of 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:

T = 2πLg

  • 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 pendulum bob on a string is hanging from a fixed point, with the length measured from the pivot down to the centre of the bob. The dashed vertical line shows the equilibrium line.
The length of a simple pendulum is measured from the pivot to the centre of mass of the bob
  • 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 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 x 

  • When considering SHM because of small angle approximation it is assumed the restoring force acts along the horizontal

  • So sinθ ≅ θ

A displaced pendulum bob with its weight resolved into a component along the string and a restoring component along the arc of the swing. The string of length L makes an angle θ with the dashed vertical equilibrium line. The bob’s weight mg acts downwards, where its restoring component, mg sin θ, acts along the arc towards equilibrium, with x horizontal.
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

Calculate the time period of a simple pendulum on the Moon, if on Earth it has a time period of 7 s. g on the moon is 1/6 of that on Earth.

[2]

Answer:

Step 1: List the known quantities

  • Gravitational field stength on Earth, gE = 9.81 N kg−1

  • Gravitational field stength on moon, gm = gE6 = 1.635 N kg−1

  • Time period, TE = 7 s

  • Length of pendulum L will be the same for both

Step 2: Write down the time period equation for Earth and moon

TE = 2πLgE  = 7 s

Tm = 2πLgm  = 2πLgE6 = 2πLgE × 6 [1 mark]

Step 3: Calculate the time period

Tm = TE × 6 = 7 × 6 = 17.1 s [1 mark]

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Ashika

Author: Ashika

Expertise: Physics Content Creator

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.

Tim

Reviewer: Tim

Expertise: Content Creator

Timothy graduated with a first class degree in Mathematics and Physics from the University of Warwick. After working as a postgraduate researcher, Timothy has worked as a content creator for various online revision platforms, creating physics resources for a range of levels and exam boards.