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

Calculating Energy Changes in Simple Harmonic Motion

Equations for Energy in SHM

• Potential energy:

$E_P=\frac{1}{2}m{\omega }^2{x}^2$

• Total energy: 

$E_T=\frac{1}{2}m{\omega }^2{x_0}^2$

• The kinetic energy–displacement relation for SHM is:

$E_K=\frac{1}{2}m{\omega }^2\left({x_0}^2-x^2\right)$

• Where:

  • m = mass (kg)

  • ω = angular frequency (rad s⁻¹)

  • x₀ = amplitude (m)

Calculating Total Energy in SHM

• Using the expression for the velocity v of a simple harmonic oscillator that begins oscillating from its equilibrium position:

$v=\omega x_0\cos \left(\omega t+\Phi \right)$

• Where:

  • phase difference, Φ = 0 

  • v = velocity of oscillator (m s⁻¹)

  • ⍵ = angular frequency (rad s⁻¹)

  • x₀ = amplitude (m)

  • t = time (s)

• The kinetic energy E_K of an oscillator can be written as:

$E_K=\frac{1}{2}m{v}^2$

$E_K=\frac{1}{2}m{\left(\omega x_0\cos \left(\omega t\right)\right)}^2$

$E_K=\frac{1}{2}m{\omega }^2{x_0}^2{\cos }^2\left(\omega t\right)$

• Since the maximum value of $\sin \left(\omega t\right)$ or $\cos \left(\omega t\right)$ is 1, maximum kinetic energy is given by:

$E_{K\left(max\right)}=\frac{1}{2}m{\omega }^2{x_0}^2$

• When the kinetic energy of the system is at a maximum, the potential energy is zero

  • Hence this represents the total energy of the system

• The total energy E_T of a system undergoing simple harmonic motion is, therefore, defined by:

$E_T=\frac{1}{2}m{\omega }^2{x_0}^2$

• Where:

  • E_T = total energy of a simple harmonic system (J)

  • m = mass of the oscillator (kg)

  • ⍵ = angular frequency (rad s⁻¹)

  • x₀ = amplitude (m)

• Note: The same expression for total energy will be achieved if the other expression for velocity is used, for an object that begins oscillation at t = 0 from the amplitude position:

$v=-\omega x_0\sin \left(\omega t\right)$

Calculating Potential Energy in SHM

• An expression for the potential energy of a simple harmonic oscillator can be derived using the expressions for velocity and displacement for an object starting its oscillations when t = 0 in the equilibrium position, so x = 0:

$x=x_0\sin \left(\omega t\right)$

$v=\omega x_0\cos \left(\omega t\right)$

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

${\sin }^2\left(\omega t\right)+{\cos }^2\left(\omega t\right)=1$

• In a simple harmonic oscillation, the total energy of the system is equal to:

Total energy = Kinetic energy + Potential energy

$E_T=E_K+E_P$

• The potential energy of an oscillator can be written as:

$E_P=E_T-E_K$

$E_P=\frac{1}{2}m{\omega }^2{x_0}^2-\frac{1}{2}m{v}^2$

• Substitute in for v:

$E_P=\frac{1}{2}m{\omega }^2{x_0}^2-\frac{1}{2}m{\left(\omega x_0\cos \left(\omega t\right)\right)}^2$

$E_P=\frac{1}{2}m{\omega }^2{x_0}^2-\frac{1}{2}m{\omega }^2{x_0}^2{\cos }^2\left(\omega t\right)$

• Taking out a factor of $\frac{1}{2}m{\omega }^2{x_0}^2$gives:

$E_P=\frac{1}{2}m{\omega }^2{x_0}^2\left(1-{\cos }^2\left(\omega t\right)\right)$

$E_P=\frac{1}{2}m{\omega }^2{x_0}^2{\sin }^2\left(\omega t\right)$

$E_P=\frac{1}{2}m{\omega }^2{\left[x_0\sin \left(\omega t\right)\right]}^2$

• Since $x=x_0\sin \left(\omega t\right)$, the potential energy of the system can be written as:

$E_P=\frac{1}{2}m{\omega }^2{x}^2$

• Since the maximum potential energy occurs at the maximum displacement $\left(x=x_0\right)$ of the oscillation,

$E_{P\left(max\right)}=\frac{1}{2}m{\omega }^2{x_0}^2$

• Therefore, it can be seen that:

$E_T=E_{K\left(max\right)}=E_{P\left(max\right)}$

Kinetic Energy–Displacement Relation for SHM

• Using the displacement–velocity relation for SHM:

$v=\pm \omega \sqrt{{x_0}^2-x^2}$

• Substituting into the equation for kinetic energy:

$E_K=\frac{1}{2}m{v}^2$

$E_K=\frac{1}{2}m{\left(\omega \sqrt{{x_0}^2-x^2}\right)}^2$

• This leads to the kinetic energy–displacement relation for SHM:

$E_K=\frac{1}{2}m{\omega }^2\left({x_0}^2-x^2\right)$