42.2: Energy
- Page ID
- 92293
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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).