23.10: Solution to the Underdamped Simple Harmonic Oscillator
- Page ID
- 25899
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Consider the underdamped simple harmonic oscillator equation),
\[\frac{d^{2} x}{d t^{2}}+\frac{b}{m} \frac{d x}{d t}+\frac{k}{m} x=0 \nonumber\]
When \((b / m)^{2}<4 k / m\) we show that the equation has a solution of the form
\[x(t)=x_{\mathrm{m}} e^{-\alpha t} \cos (\gamma t+\phi) \nonumber \]
Solution: Let’s suppose the function x(t) has the form
\[x(t)=A \operatorname{Re}\left(e^{z t}\right) \nonumber \]
where z is a number (possibly complex) and A is a real number. Then
\[\frac{d x}{d t}=z A e^{z t} \nonumber \]
\[\frac{d^{2} x}{d t^{2}}=z^{2} A e^{z t} \nonumber \]
We now substitute Equations (23.C.3), (23.C.4), and (23.C.5), into Equation (23.C.1) resulting in
\[z^{2} A e^{z t}+\frac{b}{m} z A e^{z t}+\frac{k}{m} A e^{z t}=0 \nonumber \]
Collecting terms in Equation (23.C.6) yields
\[\left(z^{2}+\frac{b}{m} z+\frac{k}{m}\right) A e^{z t}=0 \nonumber \]
The condition for the solution is that
\[z^{2}+\frac{b}{m} z+\frac{k}{m}=0 \nonumber \]
This quadratic equation has solutions
\[z=\frac{-(b / m) \pm\left((b / m)^{2}-4 k / m\right)^{1 / 2}}{2} \nonumber \]
When \((b / m)^{2}<4 k / m\), the oscillator is called underdamped, and we have two solutions for z , however the solutions are complex numbers. Let
\[\gamma=\left(k / m-(b / 2 m)^{2}\right)^{1 / 2} \nonumber \]
and
\[\alpha=b / 2 m \nonumber \]
Recall that the imaginary number \(i=\sqrt{-1}\). The two solutions are then \(z_{1}=-\alpha+i \gamma t\) and \(z_{2}=-\alpha-i \gamma t\). Because our system is linear, our general solution is a linear combination of these two solutions,
\[x(t)=A_{1} e^{-\alpha+i \gamma t}+A_{2} e^{-\alpha-i \gamma t}=\left(A_{1} e^{i \gamma t}+A_{2} e^{-i \gamma t}\right) e^{-\alpha t} \nonumber \]
where \(A_{1}\) and \(A_{2}\) are constants. We shall transform this expression into a more familiar equation involving sine and cosine functions with help from the Euler formula,
\[e^{\pm i \gamma t}=\cos (\gamma t) \pm i \sin (\gamma t) \nonumber \]
Therefore we can rewrite our solution as
\[x(t)=\left(A_{1}(\cos (\gamma t)+i \sin (\gamma t))+A_{2}(\cos (\gamma t)-i \sin (\gamma t))\right) e^{-\alpha t} \nonumber \]
A little rearrangement yields
\[x(t)=\left(\left(A_{1}+A_{2}\right) \cos (\gamma t)+i\left(A_{1}-A_{2}\right) \sin (\gamma t)\right) e^{-\alpha t} \nonumber \]
Define two new constants \(C=A_{1}+A_{2}\) and \(D=i\left(A_{1}-A_{2}\right)\). Then our solution looks like
\[x(t)=(C \cos (\gamma t)+D \sin (\gamma t)) e^{-\alpha t} \nonumber \]
Recall from Example 23.5 that we can rewrite
\[C \cos (\gamma t)+D \sin (\gamma t)=x_{\mathrm{m}} \cos (\gamma t+\phi) \nonumber \]
where
\[x_{\mathrm{m}}=\left(C^{2}+D^{2}\right)^{1 / 2}, \text { and } \phi=\tan ^{-1}(D / C) \nonumber \]
Then our general solution for the underdamped case (Equation (23.C.16)) can be written as
\[x(t)=x_{\mathrm{m}} e^{-\alpha t} \cos (\gamma t+\phi) \nonumber \]
There are two other possible cases which we shall not analyze: when \((b / m)^{2}>4 k / m\), a case referred to as overdamped, and when \((b / m)^{2}=4 k / m\), a case referred to as critically damped.