42.2: Energy
- Page ID
- 92293
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The kinetic energy \(K\) of a particle of mass \(m\) moving with speed \(v\) is defined to be the work required to accelerate the particle from rest to speed \(v\); this is found to be
\[K=\frac{1}{2} m v^{2} .\]
From Hooke's law, the potential energy \(U\) of a simple harmonic oscillator particle at position \(x\) can be shown to be
\[U=\frac{1}{2} k x^{2}\]
The total mechanical energy \(E=K+U\) of a simple harmonic oscillator can be found by observing that when \(x= \pm A\), we have \(v=0\), and therefore the kinetic energy \(K=0\) and the total energy is all potential. Since the potential energy at \(x= \pm A\) is \(U=k A^{2} / 2\) (by Eq.\(\PageIndex{12}\), the total energy must be
\[E=\frac{1}{2} k A^{2}\]
Since total energy is conserved, the energy \(E\) is constant and does not change throughout the motion, although the kinetic energy \(K\) and potential energy \(U\) do change.
In a simple harmonic oscillator, the energy sloshes back and forth between kinetic and potential energy, as shown in Fig. \(\PageIndex{1}\). At the endpoints of its motion \((x= \pm A)\), the oscillator is momentarily at rest, and the energy is entirely potential; when passing through the equilibrium position \((x=0)\), the energy is entirely kinetic. In between, kinetic energy is being converted to potential energy or vice versa.
We can find the velocity \(v\) of a simple harmonic oscillator as a function of position \(x\) (rather than time \(t\) ) by writing an expression for the conservation of energy:
\[E =K+U \]
\[\frac{1}{2} k A^{2} =\frac{1}{2} m v^{2}+\frac{1}{2} k x^{2}\]
Solving for \(v\), we find
\[v(x)= \pm A \sqrt{\frac{k}{m}} \sqrt{1-\frac{x^{2}}{A^{2}}}\]
This can be simplified somewhat by using Eq. (39.10) to give
\[v(x)= \pm A \omega \sqrt{1-\frac{x^{2}}{A^{2}}}\]
where \(A \omega\) is, by inspection of Eq. (39.1.10), the maximum speed of the oscillator (the speed it has while passing through the equilibrium position).