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In this session, we extend the solution of the motion of oscillators with one degree of freedom without damping to the case where damping can no longer be ignored.
In this session, we extend the solution of the motion of oscillators with one degree of freedom without damping to the case where damping can no longer be ignored.
Using a force of 4 newtons, a damped harmonic oscillator is displaced from equilibrium by 0.2 meters. At t = 0 it is released from rest. The resultant displacement of the oscillator, from the equilibrium position, as a function of time, is shown in the figure below. Estimate, as well as you can using the given information:
The simplest way to do this problem is to obtain the equation of motion from conservation of energy.
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Mass and Q of oscillator are approximately 5 kg and 30, respectively.
The circuit shown above consists of a capacitor (\(C\)), an inductor (\(L\)) and a resistor (\(R\)). Initially the switch is open and the charge (\(Q\)) on the capacitor is \(Q(t \leq 0) =Q_0\). At \(t = 0\), the switch is closed.
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1. Initial sign will be reversed only if the circuit is underdamped:
i.e., \(R < 2 \sqrt{\frac{L}{C}}\)
\[ Q(t) = Q_{0} e^{-\frac{\gamma t}{2}} \left(\cos(\omega’t) + \frac{\gamma}{2 \omega’} \sin(\omega’t)\right) \] and
\begin{align*}
I(t) &= \omega’ Q_{0} e^{-\frac{\gamma t}{2}} \sin(\omega’t) + \frac{\gamma}{2} \frac{Q_{0}\gamma}{2 \omega’} e^{-\frac{\gamma t}{2}} \sin(\omega’t)\\
&= Q_{0}\frac{\omega_{0}^{2}}{\omega’}e^{-\frac{\gamma t}{2}} \sin(\omega’t).
\end{align*}
\begin{equation}
\nonumber{\text{where} \hspace{4mm} \gamma = {\frac{R}{L}} \hspace{1mm} \text{,} \hspace{4mm} \omega_{0}^{2} = {\frac{1}{LC}} \hspace{4mm} \text{and} \hspace{4mm}
\omega’=\sqrt{\omega_{0}^{2}-\frac{\gamma^2}{4}}}
\end{equation}
Below are the sketches of \(Q(t)\) and \(I(t)\). Note that \(I(0)=0\) and both \(Q(t)\) and \(I(t)\) oscillate at the same frequency.
\[R = \dfrac{1}{20\pi}\sqrt{\dfrac{L}{C}}\]