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First exams 2025

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Equations for Simple Harmonic Motion (SHM) (HL) (HL IB Physics)

Revision Note

Katie M

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Katie M

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Equations for Simple Harmonic Motion

Summary of SHM Equations

  • For a body that begins oscillating from its equilibrium position (i.e. x space equals space 0 when t space equals space 0), its displacement, velocity and acceleration can be described by the equations:

x space equals space x subscript 0 sin space omega t

v space equals space omega x subscript 0 cos space omega t

a space equals space minus omega squared x subscript 0 sin space omega t

  • For a body that begins oscillating from its amplitude position (i.e. x space equals space x subscript 0 when t space equals space 0), its displacement, velocity and acceleration can be described by the equations:

x space equals space x subscript 0 cos space omega t

v space equals space minus omega x subscript 0 sin space omega t

a space equals space minus omega squared x subscript 0 cos space omega t

  • The variation of an oscillator's velocity with its displacement x is defined by:

v equals plus-or-minus omega square root of open parentheses x subscript 0 squared space minus space x squared close parentheses end root

  • This equation shows that the larger the amplitude x subscript 0 of an oscillation, the greater the distance it must travel in a given time period
    • Hence, the faster it travels, the closer it is to the equilibrium position
  • In the above equations, the variables are as follows:
    • x = displacement of the oscillator (m)
    • x subscript 0 = maximum displacement, or amplitude (m)
    • v = velocity of the oscillator (m s−1)
    • a = acceleration of the oscillator (m s−2)
    • omega = angular frequency (rad s−1)
    • t = time (s)

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

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

The Origin of the Displacement Equations

  • The SHM displacement equation for an object oscillating from its equilibrium position (x = 0 at t = 0) is:

x = x0 sin (⍵t + Φ)

  • Where:
    • Φ = phase difference (radians) = 0

  • Because:
    • The graph of = sin (t) starts from amplitude x0 = 0 when the pendulum is in the equilibrium position at = 0
    • The displacement is at its maximum when sin(⍵t) equals 1 or −1, when x = x0
  • Use the IB revision notes on the graphs of trigonometric functions to aid your understanding of trigonometric graphs

sin-and-cos-graphs

The graph of y = cos (x) has maximum displacement when x = 0

  • The SHM displacement equation for an object oscillating from its amplitude position (x = x0 at t = 0) is:

xx0 cos (⍵t + Φ)

  • The displacement will be at its maximum when cos(⍵t) equals 1 or −1, when x = x0
  • This is because the cosine graph starts at a maximum, whereas the sine graph starts at 0

These two graphs represent the same SHM. The difference is the starting position. In these graphs, the phase difference is Φ = 0.

The Origin of the Velocity Equations in Trigonometric Form

  • The trigonometric equation for the velocity of an object starting from its equilibrium position (x = 0 at t = 0) is:

ωx0 cos (⍵t + Φ)

  • Where:
    • Φ = phase difference (radians) = 0
  • This comes from the fact that velocity is the rate of change of displacement
    • It is the differential of the relevant displacement equation from above: x = x0 sin (⍵t + Φ)
  • The trigonometric equation for the velocity of an object starting from its amplitude position (x = x0 at t = 0) is:

−ω x0 sin (⍵t + Φ

  • This is the differential of the relevant displacement equation from above: xx0 cos (⍵t + Φ

The Origin of the Displacement-Velocity Relation

  • The velocity of an object in simple harmonic motion varies as it oscillates back and forth and is given by the equation:

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

  • ± = ‘plus or minus’.
    • The value can be negative or positive
  • This comes from the fact that acceleration is the rate of change of velocity
    • When the defining equation of simple harmonic motion is integrated using a differential equation the above equation for velocity is obtained
  • This equation shows that when an oscillator has a greater amplitude x0, it has to travel a greater distance in the same time and hence has greater speed v

 

Equations for Calculating Energy Changes in SHM

  • The revision note on Calculating Energy Changes in SHM explains the origin of these two equations for calculating energy changes in simple harmonic motion
  • Potential energy:

E subscript P space equals space 1 half m omega squared x squared

  • Total energy at the amplitude of oscillation: 

E subscript T space equals space 1 half m omega squared x subscript 0 squared

  • Where:
    • = mass (kg)
    • ω = angular frequency (rad s−1)
    • x0 = amplitude (m)

Worked example

The graph shows the potential energy, EP, for a particle oscillating with SHM. The particle has mass 45 g.

9-1-3-worked-example-1

(a)
Use the graph to determine the amplitude and the period of the oscillation.
(b)
Calculate the maximum speed which the particle achieves.
 

Answer:

(a)

Step 1: Use the graph to determine the maximum potential energy of the particle

9-1-3-worked-example-1-solution

  • Maximum potential energy, EPmax = 60 mJ = 60 × 10−3 J

Step 2: Determine the amplitude of the oscillation

  • The amplitude of the motion x subscript 0 is the maximum displacement
  • At the maximum displacement, the particle is at its highest point, hence this is the position of maximum potential energy
  • From the graph, when EP = 60 mJ:

x subscript 0 squared = 8.0 cm = 0.08 m

Amplitude: x subscript 0 space equals space square root of 0.08 end root space equals space 0.28 space straight m

Step 3: Write down the equation for the potential energy of an oscillator and rearrange for angular velocity ω

E subscript P equals 1 half m omega squared x subscript 0 squared

omega squared space equals space fraction numerator 2 E subscript P over denominator m x subscript 0 squared end fraction space space space space space rightwards double arrow space space space space space omega space equals space square root of fraction numerator 2 E subscript P over denominator m x subscript 0 squared end fraction end root

Step 4: Substitute the known values and calculate ω

  • Mass of the particle, m = 45 g = 45 × 10−3 kg

omega space equals space square root of fraction numerator 2 cross times left parenthesis 60 cross times 10 to the power of negative 3 end exponent right parenthesis over denominator left parenthesis 45 cross times 10 to the power of negative 3 end exponent right parenthesis cross times 0.08 end fraction end root space equals space 5.77 space rad space straight s to the power of negative 1 end exponent

Step 5: Determine the time period of the oscillation

T space equals space fraction numerator 2 pi over denominator omega end fraction space equals space fraction numerator 2 pi over denominator 5.77 end fraction = 1.1 s

(b)

  • The maximum speed of the oscillator is equal to

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

v subscript m a x end subscript = 5.77 × 0.28 = 1.6 m s−1

Worked example

A student investigated the behaviour of a 200 g mass oscillating on a spring, and produced the graph shown.

9-1-3-worked-example-2

(a)
Determine the values for amplitude and time period
(b)
Hence find the maximum kinetic energy of the oscillating mass.
 

Answer:

(a) 

  • Read the values of amplitude and time period from the graph 

 9-1-3-worked-example-2-solution

  • Amplitude, x subscript 0 = 0.3 m
  • Time period, T = 2.0 s

(b)

Step 1: Write down the equation for the maximum speed of an oscillator

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

Step 2: Write down the equation relating angular speed and time period

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

Step 3: Combine the two equations and calculate the maximum speed

v subscript m a x end subscript space equals space fraction numerator 2 straight pi x subscript 0 over denominator T end fraction space equals space fraction numerator 2 straight pi cross times 0.3 over denominator 2.0 end fraction

v subscript m a x end subscript = 0.942 m s−1

Step 4: Use the maximum speed to calculate the maximum kinetic energy of the oscillating mass

  • Mass of the oscillator, m = 200 g = 0.2 kg
E subscript K m a x end subscript space equals space 1 half m v subscript m a x end subscript squared
 
EKmax = 0.5 × 0.2 × 0.9422 = 0.1 J

Examiner Tip

There are a large number of equations associated with SHM. Most of them are given in the data booklet which you will be given to use in the exam

Make sure you are familiar with the equations, as you will probably need to use several different ones to solve the longer questions.

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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.