5.5: Exercises
- Page ID
- 34546
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Exercise \(\PageIndex{1}\)
In Section 5.2, we encountered the complex frequencies \[\omega_\pm = -i\gamma \pm \sqrt{\omega_0^2 - \gamma^2}.\] For fixed \(\omega_0\) and \(\omega_0 > \gamma\) (under-damping), prove that \(\omega_\pm\) lie along a circular arc in the complex plane.
Exercise \(\PageIndex{2}\)
Derive the general solution for the critically damped harmonic oscillator, Eq. (5.3.16), by following these steps:
- Consider the complex ODE, in the under-damped regime \(\omega_0 > \gamma\). We saw in Section 5.3 that the general solution has the form \[z(t) = \psi_+ \, \exp\left[\left(-\gamma - i \sqrt{\omega_0^2 - \gamma^2}\right)t\right] \; +\; \psi_- \, \exp\left[\left(-\gamma +i\sqrt{\omega_0^2 - \gamma^2}\right)t\right]\] for some complex parameters \(\psi_+\) and \(\psi_-\). Define the positive parameter \(\varepsilon = \sqrt{\omega_0^2 - \gamma^2}\). Re-write \(z(t)\) in terms of \(\gamma\) and \(\varepsilon\) (i.e., eliminating \(\omega_0\)).
- The expression for \(z(t)\) is presently parameterized by the independent parameters \(\psi_+\), \(\psi_-\), \(\varepsilon\), and \(\gamma\). We are free to re-define the parameters, by taking \[\begin{align} \alpha &= \psi_+ + \psi_- \\ \beta &= -i\varepsilon(\psi_+ - \psi_-). \end{align}\] Using these equations, express \(z(t)\) using a new set of independent complex parameters, one of which is \(\varepsilon\). Explicitly identify the other independent parameters, and state whether they are real or complex.
- Expand the exponentials in \(z(t)\) in terms of the parameter \(\varepsilon\). Then show that in the limit \(\varepsilon \rightarrow 0\), \(z(t)\) reduces to the critically-damped general solution (5.3.16).
Exercise \(\PageIndex{3}\)
Repeat the above derivation for the critically-damped solution, but starting from the over-damped regime \(\gamma > \omega_0\).
Exercise \(\PageIndex{4}\)
Let \(z(t)\) be a complex function of a real input \(t\), which obeys the differential equation \[\frac{dz}{dt} = -i\,(\omega_1 - i \gamma)\; z(t),\] where \(\omega_1\) and \(\gamma\) are real. Find the general solution for \(z(t)\), and hence show that \(z(t)\) satisfies the damped oscillator equation \[\left[\frac{d^2}{dt^2} + 2\gamma \frac{d}{dt} + \omega_0^2 \right] z(t) = 0\] for some \(\omega_0^2\). Finally, show that this harmonic oscillator is always under-damped.
- Answer
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The general solution is \[z(t) = A \exp\left[-i(\omega_1 - i \gamma) t\right].\] It can be verified by direct substitution that this is a solution to the differential equation. It contains one free parameter, and the differential equation is first-order, so it must be a general solution. Next, \[\begin{align} \frac{d^2z}{dt^2} + 2 \gamma \frac{dz}{dt} &= (-i)^2(\omega_1 - i\gamma)^2 z(t) - 2i \gamma (\omega_1 - i \gamma) z(t) \\ &= \left[- \omega_1^2 + \gamma^2 + 2i\gamma\omega_1 - 2i \gamma \omega_1 - 2\gamma^2)\right] z(t) \\ &= -\left(\omega_1^2 + \gamma^2\right)z(t).\end{align}\] Hence, \(z(t)\) obeys a damped harmonic oscillator equation with \(\omega_0^2 = \omega_1^2 + \gamma^2.\) This expression for the natural frequency ensures that \(\omega_0^2 > \gamma^2\) (assuming the parameters \(\gamma\) and \(\omega_1\) are both real); hence, the harmonic oscillator is always under-damped.