describe the interchange between kinetic and potential energy during simple harmonic motion

Published by Patrick Mutisya · 14 days ago

Simple Harmonic Oscillations – Energy Interchange

Simple Harmonic Oscillations

Objective

Describe the interchange between kinetic and potential energy during simple harmonic motion (SHM).

Key Concepts

  • SHM is characterised by a restoring force proportional to displacement: \$F = -kx\$.

  • The motion can be described by the displacement equation

    \$x(t)=A\cos(\omega t+\phi)\$

    where \$A\$ is amplitude, \$\omega=\sqrt{k/m}\$ is angular frequency, and \$\phi\$ is the phase constant.

  • Velocity and acceleration are the time‑derivatives of displacement:

    \$v(t)=\frac{dx}{dt}=-A\omega\sin(\omega t+\phi),\$

    \$a(t)=\frac{dv}{dt}=-\omega^{2}x(t).\$

Energy in SHM

The total mechanical energy \$E\$ of an ideal SHM system is the sum of kinetic energy \$Ek\$ and elastic potential energy \$Ep\$:

\$E = Ek + Ep = \frac12 m v^{2} + \frac12 k x^{2}.\$

Because \$k = m\omega^{2}\$, the expression can be rewritten as

\$E = \frac12 m\omega^{2}\bigl(A^{2}\bigr) = \frac12 k A^{2},\$

which shows that the total energy is constant and depends only on the amplitude \$A\$.

Energy Interchange During One Cycle

Position \$x\$Velocity \$v\$Kinetic Energy \$E_k\$Potential Energy \$E_p\$Energy State
\$x = 0\$ (equilibrium)\$\pm A\omega\$ (maximum)\$\frac12 m (A\omega)^{2} = \frac12 k A^{2}\$ (maximum)\$0\$ (minimum)All energy is kinetic.
\$x = \pm\frac{A}{\sqrt{2}}\$\$\pm\frac{A\omega}{\sqrt{2}}\$\$\frac14 k A^{2}\$\$\frac14 k A^{2}\$Energy equally split between kinetic and potential.
\$x = \pm A\$ (extrema)\$0\$ (minimum)\$0\$ (minimum)\$\frac12 k A^{2}\$ (maximum)All energy is potential.

Qualitative Description

  1. At the extreme positions (\$x = \pm A\$) the mass momentarily stops (\$v=0\$). All the mechanical energy is stored as elastic potential energy in the spring.

  2. As the mass moves toward the equilibrium position, the spring releases its stored energy, converting potential energy into kinetic energy. The speed increases, reaching its maximum at the equilibrium point.

  3. When the mass passes through equilibrium (\$x=0\$) the kinetic energy is at its maximum and the potential energy is zero. The system continues moving due to inertia.

  4. Beyond equilibrium the spring is compressed (or stretched) in the opposite sense, and the kinetic energy begins to be reconverted into potential energy until the next extreme position is reached.

  5. This continual exchange repeats, producing the sinusoidal motion characteristic of SHM.

Mathematical Demonstration of Energy Conservation

Starting from the displacement equation, substitute \$v = \dot{x}\$ into the kinetic energy term:

\$\$E_k = \frac12 m\bigl(\dot{x}\bigr)^{2}

= \frac12 m\bigl(-A\omega\sin(\omega t+\phi)\bigr)^{2}

= \frac12 mA^{2}\omega^{2}\sin^{2}(\omega t+\phi).\$\$

The potential energy is

\$\$E_p = \frac12 k x^{2}

= \frac12 k\bigl(A\cos(\omega t+\phi)\bigr)^{2}

= \frac12 m\omega^{2}A^{2}\cos^{2}(\omega t+\phi).\$\$

Adding the two gives

\$\$E = \frac12 m\omega^{2}A^{2}\bigl(\sin^{2}(\omega t+\phi)+\cos^{2}(\omega t+\phi)\bigr)

= \frac12 m\omega^{2}A^{2}

= \frac12 k A^{2},\$\$

which is independent of time, confirming that energy is conserved and merely interchanges between kinetic and potential forms.

Suggested Diagram

Suggested diagram: A mass‑spring system showing displacement \$x\$, amplitude \$A\$, equilibrium position, and arrows indicating the direction of velocity and restoring force at key points (extrema and equilibrium).