5.3.2 - Energy in simple harmonic motion

Energy Transfer During SHM

Think of a mass on a spring sliding through its equilibrium position. As long as the oscillation is ideal and there is no significant damping, the total mechanical energy stays constant. What changes is the form of the energy.

At the equilibrium position, $x=0$, the oscillator has its greatest speed. If potential energy is measured from the equilibrium position, the oscillator's mechanical energy is kinetic energy at this instant. At a turning point, $x=+A$ or $x=-A$, the oscillator is instantaneously at rest. Its kinetic energy is zero, and the energy is stored as potential energy.

For a horizontal spring system this potential energy is elastic potential energy in the spring. For a vertical spring oscillator, the restoring potential energy about equilibrium is the sum of elastic and gravitational potential energies; these individual stores are not separately minimum at the equilibrium point. For a small-angle pendulum it is gravitational potential energy. The storage mechanism is different, but the SHM energy idea is the same: kinetic energy and potential energy continually interchange while the total stays constant.

The word "ideal" matters. If resistive forces remove energy from the system, the amplitude changes; that belongs to damping and is handled in the next lesson. Here, the energy-displacement graphs are for an undamped oscillator.

Energy Equations From SHM

The previous SHM lesson gave the velocity-displacement relationship

Here $v$ is velocity in $\text{m s}^{-1}$, $\omega$ is angular frequency in $\text{rad s}^{-1}$, $A$ is amplitude in m, and $x$ is displacement from equilibrium in m. The sign tells you direction of travel; the energy depends on $v^2$, so the sign does not affect the kinetic energy.

Kinetic energy is

Substitute the SHM expression for $v^2$:

so

The total energy can be found at equilibrium, where $x=0$ and the speed is maximum. Then $v_\text{max}=\omega A$, so

The potential energy is the rest of the total energy:

For a mass-spring oscillator, $m\omega^2$ is equal to the spring constant $k$, so the same potential-energy result can be written as $E_p=\frac12kx^2$. Here $x$ is displacement from equilibrium. For a horizontal spring it is also extension from the relaxed spring; for a vertical spring the equilibrium extension has already been subtracted and $E_p$ is the combined restoring potential energy. The useful habit is to connect the energy equations to $E_k=\frac12mv^2$, $v=\pm\omega\sqrt{A^2-x^2}$, and conservation of energy.

Worked example:

A $0.080\ \text{kg}$ mass oscillates with amplitude $0.045\ \text{m}$ and frequency $1.6\ \text{Hz}$. Find the total energy and the kinetic energy when $x=0.030\ \text{m}$.

First find angular frequency:

Total energy:

Potential energy at $x=0.030\ \text{m}$:

The total and potential energies round to $8.2\times10^{-3}\,\mathrm J$ and $3.6\times10^{-3}\,\mathrm J$, but retain unrounded values in the subtraction. So the kinetic energy is

The kinetic energy is $4.5\times10^{-3}\,\mathrm J$ to 2 significant figures.