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23.10: Solution to the Underdamped Simple Harmonic Oscillator

  • Page ID
    25899
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    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.


    This page titled 23.10: Solution to the Underdamped Simple Harmonic Oscillator is shared under a not declared license and was authored, remixed, and/or curated by Peter Dourmashkin (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.