Syllabus Edition

First teaching 2014

Last exams 2024

|

The Defining Equation of SHM (DP IB Physics: HL)

Revision Note

Lindsay Gilmour

Last updated

The Defining Equation of SHM

  • Simple harmonic motion (SHM) is a fundamental oscillation which appears naturally in various phenomena, including:
    • The swing of a pendulum
    • A mass on a spring
    • Guitar strings
    • Vibrations of molecules

  • SHM is defined as:

A type of oscillation in which the acceleration on a body is proportional to its displacement, but acts in the opposite direction 

  • The conditions for an object to oscillate in SHM are that it shows:
    • Periodic oscillations
    • Acceleration proportional to its displacement
    • Acceleration in the opposite direction to its displacement

  • SHM is isochronous, meaning that the time period stays constant
    • Therefore frequency must also be constant
  • The period of simple harmonic oscillations can be calculated using the equation: 
T space equals space fraction numerator 2 pi over denominator omega end fraction
  • Where:
    • T = the time period of the oscillator (s)
    • ω = angular frequency (rad s−1)
  • The acceleration of an oscillator can be calculated using the defining equation of SHM:
a space equals space minus omega squared x
  • Where:
    • a = acceleration of the oscillator (m s−2)
    • x = displacement of the oscillator from its equilibrium position (m)

  • The equation shows that:
    • The acceleration reaches its maximum value when the displacement is at a maximum i.e. x = x0
    • The minus sign shows that when the object is displacement to the right, the direction of the acceleration is to the left
  • An object in SHM will have a restoring force acting on it to return it to its equilibrium position
  • This restoring force will be directly proportional, but in the opposite direction, to the displacement of the object from the equilibrium position
    • Note: the restoring force and acceleration act in the same direction

Graph of acceleration and displacement, downloadable AS & A Level Physics revision notes

The acceleration of an object in SHM is directly proportional to the negative displacement

Calculating Displacement of an Oscillator

  • The graph of acceleration against displacement is a straight line through the origin sloping downwards (similar to y = −x)
  • Key features of the graph:
    • The gradient is equal to ω2
    • The maximum and minimum displacement x values are the amplitudes −x0 and +x0

  • A solution to the SHM acceleration equation is the displacement equation:

x space equals space x subscript 0 sin space left parenthesis omega t right parenthesis

  • Where:
    • x = displacement of the oscillator (m)
    • x0 = maximum displacement or amplitude (m)
    • t = time (s)
  • This equation can be used to find the position of an object in SHM with a particular angular frequency and amplitude at a moment in time
    • Note: This version of the equation is only relevant when an object begins oscillating from the equilibrium position (x = 0 at t = 0)

  • The displacement will be at its maximum when sin (⍵t) equals 1 or − 1, when x = x0
  • If an object is oscillating from its amplitude position (x = x0 or x = − x0 at t = 0) then the displacement equation will be:

x space equals space x subscript 0 cos space left parenthesis omega t right parenthesis

  • This is because the cosine graph starts at a maximum, whilst the sine graph starts at 0

Displacement SHM graphs, downloadable AS & A Level Physics revision notes

These two graphs represent the same SHM. The difference is the starting position

Equations & Graphs for SHM

  • A summary of the equations and graphs of simple harmonic motion are shown in the table
  • Note that the equations differ depending on the starting point of the oscillator
    • The derivation of these equations are found lower down the page

Summary table of equations and graphs for displacement, velocity and acceleration

9-1-4-calculating-energy-changes-shm-ib-hl

Worked example

A mass is suspended from a fixed point by means of a spring. The stationary mass is pulled vertically downwards through a distance of 4.3 cm and then released at t = 0. The mass is observed to perform simple harmonic motion with a period of 0.8 s.

Calculate the displacement, x, in cm of the mass at time t = 0.3 s.

Step 1: List the known quantities

    • Maximum displacement, x0 = 4.3 cm
    • Period of oscillation, T = 0.8 s
    • Time interval, t = 0.3 s

Step 2: Write down the SHM displacement equation

    • Since the mass is released at t = 0 at its maximum displacement, the displacement equation will involve the cosine function:

x space equals space x subscript 0 cos space left parenthesis omega t right parenthesis

Step 3: Write the equation linking the angular frequency, ω

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

Step 4: Combine the two equations and substitute in the values:

    • Note: the calculator must be in radians mode

x equals x subscript 0 cos space open parentheses fraction numerator 2 πt over denominator T end fraction close parentheses equals 4.3 cross times cos open parentheses fraction numerator 2 straight pi cross times 0.3 over denominator 0.8 end fraction close parentheses

  

x = −3.041 = −3.0 cm (2 s.f.)

    • The negative value means the mass is 3.0 cm on the opposite side of the equilibrium position to where it started i.e. 3.0 cm above it

Calculating Velocity of an Oscillator

  • The velocity, v, of an oscillation can be found by differentiating the displacement with respect to time::

v equals fraction numerator d x over denominator d t end fraction

  • To differentiate the sine and cosine functions:
    • The derivative of y equals sin space x is fraction numerator d y over denominator d x end fraction equals cos space x
    • The derivative of y equals sin space open parentheses a x close parentheses is fraction numerator d y over denominator d x end fraction equals a cos space open parentheses a x close parentheses
    • The derivative of y equals cos space x is fraction numerator d y over denominator d x end fraction equals negative sin space x
    • The derivative of y equals cos space open parentheses a x close parentheses is fraction numerator d y over denominator d x end fraction equals negative a sin space open parentheses a x close parentheses

  • When the oscillation begins from the equilibrium position:

x equals x subscript 0 sin space open parentheses omega t close parentheses

  • Since the derivative of sin space open parentheses omega t close parentheses is fraction numerator d over denominator d t end fraction open parentheses sin space open parentheses omega t close parentheses close parentheses equals omega cos space open parentheses omega t close parentheses:

v equals x subscript 0 fraction numerator d over denominator d t end fraction open parentheses sin space open parentheses omega t close parentheses close parentheses equals x subscript 0 open parentheses omega cos space open parentheses omega t close parentheses close parentheses

  • Therefore, the velocity when the oscillation begins from the equilibrium position is given by:

v equals omega x subscript 0 cos space open parentheses omega t close parentheses


  • When the oscillation begins from the maximum displacement:

x equals x subscript 0 cos space open parentheses omega t close parentheses

  • Since the derivative of cos space open parentheses omega t close parentheses is fraction numerator d over denominator d t end fraction open parentheses cos space open parentheses omega t close parentheses close parentheses equals negative omega sin space open parentheses omega t close parentheses:

v equals x subscript 0 fraction numerator d over denominator d t end fraction open parentheses cos space open parentheses omega t close parentheses close parentheses equals x subscript 0 open parentheses negative omega sin space open parentheses omega t close parentheses close parentheses

  • Therefore, the velocity when the oscillation begins from the maximum displacement is given by:

v equals negative omega x subscript 0 sin space open parentheses omega t close parentheses


  • Since the maximum value of sin space open parentheses omega t close parentheses or cos space open parentheses omega t close parentheses is 1, maximum speed is given by:

v subscript m a x end subscript equals omega x subscript 0

  • This is the maximum speed of the oscillation regardless of its starting point

Calculating Acceleration of an Oscillator

  • The acceleration, a, of an oscillator can be found by differentiating the displacement with respect to time twice:

a equals fraction numerator d squared x over denominator d t squared end fraction equals fraction numerator d v over denominator d t end fraction

  • When the oscillation begins from the equilibrium position:

x equals x subscript 0 sin space open parentheses omega t close parentheses

v equals omega x subscript 0 cos space open parentheses omega t close parentheses

  • Since the derivative of cos space open parentheses omega t close parentheses is fraction numerator d over denominator d t end fraction open parentheses cos space open parentheses omega t close parentheses close parentheses equals negative omega sin space open parentheses omega t close parentheses:

a equals omega x subscript 0 fraction numerator d over denominator d t end fraction open parentheses cos space open parentheses omega t close parentheses close parentheses equals omega x subscript 0 open parentheses negative omega sin space open parentheses omega t close parentheses close parentheses

  • Therefore, the velocity when the oscillation begins from the equilibrium position is given by:

a equals negative omega squared x subscript 0 sin space open parentheses omega t close parentheses


  • When the oscillation begins from the maximum displacement:

x equals x subscript 0 cos space open parentheses omega t close parentheses

v equals negative omega x subscript 0 sin space open parentheses omega t close parentheses

  • Since the derivative of sin space open parentheses omega t close parentheses is fraction numerator d over denominator d t end fraction open parentheses sin space open parentheses omega t close parentheses close parentheses equals omega cos space open parentheses omega t close parentheses:

a equals negative omega x subscript 0 fraction numerator d over denominator d t end fraction open parentheses sin space open parentheses omega t close parentheses close parentheses equals negative omega x subscript 0 open parentheses omega cos space open parentheses omega t close parentheses close parentheses

  • Therefore, the velocity when the oscillation begins from the maximum displacement is given by:

a equals negative omega squared x subscript 0 cos space open parentheses omega t close parentheses


  • Since the maximum value of sin space open parentheses omega t close parentheses or cos space open parentheses omega t close parentheses is 1, maximum acceleration is given by:

a subscript m a x end subscript equals omega squared x subscript 0

  • This is the maximum acceleration of the oscillator regardless of its starting point

Displacement–Velocity Relation for SHM

  • A useful relation between the displacement and velocity of a simple harmonic oscillator can be derived using the expressions from the equilibrium position:

x equals x subscript 0 sin space open parentheses omega t close parentheses

v equals omega x subscript 0 cos space open parentheses omega t close parentheses

  • The key to deriving this expression is to use the trigonometric identity:

sin squared space open parentheses omega t close parentheses space plus space cos squared space open parentheses omega t close parentheses space equals space 1

  • To get started, square both expressions:

x squared equals x subscript 0 squared space sin squared space open parentheses omega t close parentheses

v squared equals omega squared x subscript 0 squared space cos squared space open parentheses omega t close parentheses

  • According to the trig identity, cos squared space open parentheses omega t close parentheses equals 1 minus sin squared space open parentheses omega t close parentheses so:

v squared equals omega squared x subscript 0 squared space left parenthesis 1 minus sin squared space open parentheses omega t close parentheses right parenthesis

v squared equals omega squared x subscript 0 squared minus omega squared x subscript 0 squared space sin squared space open parentheses omega t close parentheses

v squared equals omega squared x subscript 0 squared minus omega squared open square brackets x subscript 0 squared space sin squared space open parentheses omega t close parentheses close square brackets

v squared equals omega squared x subscript 0 squared minus omega squared x squared

v squared equals omega squared space open parentheses x subscript 0 squared minus x squared close parentheses

  • Squaring both sides of the equation leads to the displacement–velocity relation for SHM:

v equals plus-or-minus omega square root of x subscript 0 squared minus x squared end root

  • Note: The same relation will be achieved if the expressions for the maximum displacement are used:

x equals x subscript 0 cos space open parentheses omega t close parentheses

v equals negative omega x subscript 0 sin space open parentheses omega t close parentheses

Examiner Tip

The defining equation of SHM shows acceleration, as a positive value, and displacement, −x as a negative one. This reminds us that acceleration and displacement are vector quantities and are always in the opposite direction to each other in SHM.

Since displacement is a vector quantity, remember to keep the minus sign in your solutions if they are negative. Getting the marks will depend on keeping your positive and negative numbers distinct from each other! Also remember that your calculator must be in radians mode when using the cosine and sine functions. This is because the angular frequency ⍵ is calculated in rad s-1, not degrees.

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Lindsay Gilmour

Author: Lindsay Gilmour

Expertise: Physics

Lindsay graduated with First Class Honours from the University of Greenwich and earned her Science Communication MSc at Imperial College London. Now with many years’ experience as a Head of Physics and Examiner for A Level and IGCSE Physics (and Biology!), her love of communicating, educating and Physics has brought her to Save My Exams where she hopes to help as many students as possible on their next steps.