Energy within simple harmonic motion

In a nutshell

When an object undergoes simple harmonic motion, the energy within the system is interchanged between kinetic energy and another potential energy store. If there is friction or air resistance in the system, the kinetic energy will also be transferred to the thermal stores of the object and given to the surroundings.

Equations

Description	Equation
kinetic energy	$E=\frac 12 mv^2$
maximum velocity of shm	$v_{max}=\omega A$

Variable definitions

Quantity Name	Symbol	Derived Unit	SI BASE Units
$energy$	$E$	$J$	$kg\,m^{2}\,s^{-2}$
$mass$	$m$	$kg$	$kg$
$velocity$	$v$	$m\,s^{-1}$	$m\,s^{-1}$
$angular \ frequency$	$\omega$	$rad \, s^{-1}$	$s^{-1}$
$amplitude$	$A$	$m$	$m$

Energy within simple harmonic motion

When an object is undergoing simple harmonic motion, there will be a restoring force pushing or pulling the object back towards the midpoint.

In the case of a pendulum or mass on a spring, the restoring force would be the weight of the pendulum bob or masses. However gravity isn't always involved in simple harmonic systems.

Consider an object between two springs, when the object is displaced and released it will oscillate between the two springs and the restoring force will be the elastic tension of the two springs.

Physics; Oscillations; KS5 Year 12; Energy within simple harmonic motion

The energy would be transferred from the elastic potential energy of each spring into the kinetic energy of the object before being transferred to the elastic potential of the other spring.

Energy transfer for vertical simple harmonic motion

When an object is in vertical simple harmonic motion, at least one of the restoring forces will be the weight. Lets consider a simple pendulum.

When the pendulum is displaced from the equilibrium, it gains gravitational potential energy. When the pendulum is released the gravitational potential energy is transferred into kinetic energy as the ball accelerates towards the midpoint.

After passing through the midpoint, the ball starts to decelerate, reducing its velocity. This means that the kinetic energy decreases as it transfers back into gravitational potential energy.

Note: Sometimes an exam question might ask you to consider other energy transfers. This would be transfer from kinetic energy into the thermal store of the pendulum as the pendulum moves through the air.

The graph for energy against displacement for a pendulum, assuming no air resistance would look something like this:

Physics; Oscillations; KS5 Year 12; Energy within simple harmonic motion

1	Energy
2	Displacement
3	Total energy
4	Kinetic energy
5	Gravitational potential energy

Note: The total energy at any point of the displacement will always be constant. As shown by the horizontal green line.

Example

A $150 \, g$ pendulum is displaced horizontally by $1.64 \, m$ and released. The pendulum is measured to have a time period of $4.66 \, s$ . Calculate the total energy of the pendulum assuming no resistive forces acting on the system.

Note: In order to answer this question, the kinetic energy at maximum velocity can be used as all of the energy will be in the kinetic energy store.

Firstly, write down the known values:

$m=150 \, g = 0.150 \, kg \newline \\[0.1in]A=1.64 \, m \newline \\[0.1in]T=4.66 \, s$

Next, write down the equations needed and rearrange if necessary:

$v_{max}=\omega A \rightarrow v_{max}=(\frac {2\pi}{T})A\newline \\[0.1in]E=\frac 12 mv_{max}^2$

Then, substitute the values into the equation for $v_{max}$ :

$v_{max}=(\frac {2\pi}{4.66})\times 1.64 \newline \\[0.1in]v_{max}=2.21125...$

Then, substitute the value for $v_{max}$ into the equation for $E$ :

$E= \frac 12 \times 0.15\times 2.21125...^2 \newline \\[0.1in]E=0.36672...$ 

Make sure to include units and round to the lowest number of significant figures of the values given by the question:

$Total \ energy, E=0.367 \, J$

The total energy of the system is $\underline{0.367 \, J}.$ 