Total Energy of SHM (College Board AP® Physics 1: Algebra-Based)

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Dan Mitchell-Garnett

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Total energy of SHM

Calculating total energy

  • Only two types of energy store are considered in SHM systems:

    • Kinetic energies

    • Potential energies

  • The total energy of the system is the sum of these energies:

E subscript t o t a l end subscript space equals space U space plus space K

  • Where:

    • E subscript t o t a l end subscript is the total energy of the system, measured in straight J

    • U is the total potential energy of the system, measured in straight J

    • K is the total kinetic energy of the system, measured in straight J

  • It is assumed that no work is done by resistive forces such as drag or friction

    • This means no energy enters or leaves the system

  • For a system showing SHM, total energy remains constant

Different systems

  • The potential energy in the system will differ for different systems

Horizontal object-spring system

  • When oscillations are horizontal, the gravitational potential is unchanged

    • The only changing potential energy is the elastic potential energy of the spring

  • This is a spring-object system, as only these two components need to be considered

Vertical object-spring-Earth system

  • If an object-spring system is oscillating vertically, the gravitational potential energy is changing, as well as the elastic potential energy

    • The system must therefore include the Earth too, as it provides the field in which gravitational potential energy varies

  • The potential energy then becomes the sum of elastic and gravitational potential energies

    • Elastic potential must be defined with displacement from the spring's natural length

    • Gravitational potential can be calculated with h space equals space 0 at any point, the total energy will remain constant

Worked Example

At a time of t subscript 1, a system in simple harmonic motion is at maximum displacement and has a potential energy of 350 J.

At a later time of t subscript 2, the same system has a kinetic energy of 230 J.

Determine the system's potential energy at t subscript 2 and justify your answer.

Answer:

Step 1: Analyze the scenario

  • At t subscript 1, the system is at maximum displacement

    • At maximum displacement, the velocity of the system is zero

  • At t subscript 1, the potential energy of the system is 350 J

  • At t subscript 2, the position is unknown but kinetic energy is 230 J

Step 2: Consider the energy at t subscript 1

  • At t subscript 1, the velocity of the system is zero

    • This means that kinetic energy is also zero

  • Recall the equation for the total energy of the system:

E subscript t o t a l end subscript space equals space U space plus space K

E subscript t o t a l end subscript space equals space 350 space plus space 0

  • The total energy of the system is 350 J

Step 3: Consider the energy at t subscript 2

  • Recall that the total energy of a system in SHM is constant

  • At t subscript 2, the kinetic energy of the system is 230 J

  • Substitute the kinetic energy at t subscript 2 and the total energy from t subscript 1 into the total energy equation:

E subscript t o t a l end subscript space equals space U space plus space K

350 space equals space U space plus space 230

  • The potential energy of the system at t subscript 2 is therefore 120 J

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Dan Mitchell-Garnett

Author: Dan Mitchell-Garnett

Expertise: Physics Content Creator

Dan graduated with a First-class Masters degree in Physics at Durham University, specialising in cell membrane biophysics. After being awarded an Institute of Physics Teacher Training Scholarship, Dan taught physics in secondary schools in the North of England before moving to Save My Exams. Here, he carries on his passion for writing challenging physics questions and helping young people learn to love physics.