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    <name>
      <text>20.4.1.5 Damped Spring Example</text>
    </name>
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      <text><![CDATA[<h4 class="HELM_example">Example</h4>
<p>A damped spring, constrained to move in one direction, such as might be found in a railway buffer, is subjected to an impulse of duration 5 seconds. The spring constant divided by the mass causing the impulse is 10 m\(^{-2}\) s\(^{-2}\) and the frictional
    force divided by this mass is 2 m\(^{-2}\)s\(^{-2}\).</p>
<ol class="HELM_parts">
    <li>
        <p>Write down the equation governing the motion in terms of the displacement \(x\) m and time \(t\) seconds including the impulse \(u(t)\).</p>
    </li>
    <li>
        <p>Write down the initial conditions on the displacement \((x)\) and velocity.</p>
    </li>
    <li>
        <p>Solve the equation for displacement as a function of time.</p>
    </li>
    <li>
        <p>Draw a graph of the oscillations for \(t=0\) to \(10\)s.</p>
    </li>
</ol>

<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
    <li>
        <p>Since the system involves a restoring force and friction, after dividing through by the mass, the equation of motion may be written: \[\displaystyle \dfrac{d^2 x}{dt^2} + 2 \dfrac{dx}{dt} + 10 x = u(t) - u(t-5)\] where the right-hand side represents
            the impulse being switched on at \(t=0\) s and switched off at \(t=5\) s.</p>
    </li>
    <li>
        <p>Since the system starts from rest \(x(0)=x'(0)=0\).</p>
    </li>
    <li>
        <p>Taking the Laplace Transform of each term of the differential equation gives \[\displaystyle {\cal L}\left[\dfrac{d^2 x}{dt^2}\right] + 2{\cal L}\left[ \dfrac{dx}{dt} \right] + 10 {\cal L}\left[x\right] = {\cal L}\left[u(t)\right] - {\cal L}\left[u(t-5)\right]\]
            i.e. \(s^2 X(s) - s\,x(0) - x'(0) + 2 (s \, X(s) - x(0) ) + 10 X(s) = \frac{1}{s} - \frac{1}{s} \text{e}^{-5s}\)</p>
        <p>but as \(x(0) = x'(0) = 0\), this simplifies to \(\ \displaystyle s^2 X(s) + 2 \; s \; X(s) + 10 X(s) = \frac{1}{s}\left[1-e^{-5s}\right]\) \[\begin{eqnarray*}
\text{i.e.}  \qquad X(s)  &=& \frac{1}{s(s^2 + 2s + 10)}\left[ 1  - \text{e}^{-5s}\right]\\  &=&
\left[\frac{1}{10}\cdot\frac{1}{s} - \frac{1}{10}\cdot
\frac{s+2}{s^2+2s+10}\right]\left[ 1 - \text{e}^{-5s}\right]\  \text{(using partial fractions)}
%\\ &=& \left[\frac{1}{10}\frac{1}{s} - \frac{1}{10}
%\frac{s+1+1}{(s+1)^2 + 3^2 }\right]\left[ 1 - \text{e}^{-5s}\right] 
\\ &=& \left[\frac{1}{10}\cdot\frac{1}{s} - \frac{1}{10}\cdot
\frac{s+1}{(s+1)^2 + 3^2} - \frac{1}{30}\cdot \frac{3}{(s+1)^2 + 3^2}
\right]\left[ 1 - \text{e}^{-5s}\right] \\ &=& \frac{1}{10}\cdot\frac{1}{s} - \frac{1}{10}\cdot \frac{s+1}{(s+1)^2 + 3^2} - \frac{1}{30}\cdot
\frac{3}{(s+1)^2 + 3^2} \\ &&-
\frac{1}{10}\cdot\frac{1}{s} \text{e}^{-5s} + \frac{1}{10}\cdot \frac{s+1}{(s+1)^2 + 3^2}\text{e}^{-5s} + \frac{1}{30}\cdot \frac{3}{(s+1)^2 +
3^2}\text{e}^{-5s}
\end{eqnarray*}\]
        </p>
    </li>
    <li> <center><img src="@@PLUGINFILE@@/20-4-fig-16-damped_osc.svg" alt="Figure shows a graph with a damped oscillation at 0.1m followed by another damped oscillation at 0m after the impulse. " width="333" height="180" class="img-responsive atto_image_button_text-bottom"></center>
        <p>According to the graph the damped spring has a damped oscillation about a displacement of 0.1 m after the start of the impulse and a damped oscillation about a displacement of zero after the impulse has finished.</p>
    </li>
</ol>]]></text>
<file name="20-4-fig-16-damped_osc.svg" path="/" encoding="base64">[encoded data, 13672 characters]</file>
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