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First teaching 2014

Last exams 2024

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Calculating Energy Changes in SHM (DP IB Physics: HL)

Revision Note

Lindsay Gilmour

Last updated

Calculating Energy Changes in SHM

  • There are many equations to learn in the topic of simple harmonic motion
  • The key equations are summarised below:

9-1-3-equations-for-shm-1-ib-hl

Pd0TczGl_9-1-3-equations-for-shm-2

9-1-3-equations-for-shm-3

A summary of the equations related to simple harmonic motion. The green stars indicate equations which are not included in the IB data booklet

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 hence, the period of oscillation.
(b)
Calculate the maximum speed which the particle achieves.

Part (a)

Step 1: List the known quantities

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

Step 2: 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 3: Determine the amplitude of the oscillation

      • The amplitude of the motion, x0, 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 = EPmax:

x02 = 8.0 cm = 0.08 m

      • Therefore, the amplitude is:

x0 = √0.08 = 0.28 m

Step 4: Write down the equation for the potential energy of an oscillator

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

Step 5: Rearrange the equation for angular velocity, ω:

omega squared equals fraction numerator 2 E subscript P over denominator m x subscript 0 squared end fraction

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

Step 6: Substitute the known values and calculate ω

omega equals 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

ω = 5.77 rad s−1

Step 7: Write down the relation between period, T, and angular velocity, ω

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

Step 8: Determine the time period of the oscillation

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

Part (b)

Step 1: List the known quantities

      • Angular velocity, ω = 5.77 rad s−1
      • Amplitude, x0 = 0.28 m

Step 2: Write down the equation for the maximum speed of an oscillator

vmax = ωx0

Step 3: Substitute in the known quantities 

vmax = 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.

Part (a) 

Step 1: Read the values of amplitude and time period from the graph 

 9-1-3-worked-example-2-solution

Step 2: State the values of amplitude and time period

      • Amplitude, x0 = 0.3 m
      • Time period, T = 2.0 s

Part (b)

Step 1: List the known quantities

      • Mass of the oscillator, m = 200 g = 0.2 kg

Step 2: Write down the equation for the maximum speed of an oscillator

vmax = ωx0

Step 3: Write down the equation relating angular speed and time period

omega equals fraction numerator 2 pi over denominator T end fraction

Step 4: Combine the two equations and calculate the maximum speed

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

vmax = 0.942 m s−1

Step 5: Use the maximum speed to calculate the maximum kinetic energy of the oscillating mass

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

Worked example

A ball of mass 23 g is held between two fixed points A and B by two stretch helical springs, as shown in the diagram below.

Worked example horizontal mass on spring, downloadable AS & A Level Physics revision notesThe ball oscillates along the line AB with simple harmonic motion of frequency 4.8 Hz and amplitude 1.5 cm.

Calculate the total energy of the oscillations.

Step 1: Write down all known quantities

    • Mass, m = 23 g = 23 × 10–3 kg
    • Amplitude, x0 = 1.5 cm = 0.015 m
    • Frequency, f = 4.8 Hz

Step 2: Write down the equation for the total energy of SHM oscillations:

E equals 1 half m omega squared x subscript 0 squared


Step 3: Write an expression for the angular frequency

omega space equals space 2 pi f

Step 4: Substitute values into the SHM energy equation

E equals 1 half m left parenthesis 2 straight pi f right parenthesis squared x subscript 0 squared

E = 0.5 × (23 × 10−3) × (2π × 4.8)2 × (0.015)2

E = 2.354 × 10−3 = 2.4 mJ (2 s.f.)

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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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.