In this session we first give advice on how, in general, one approaches the solving of “physics problems.” We then consider three very different oscillating systems, show how in each the equation of motion can be derived and then solve these equations to obtain the motion of the oscillator.
A torsional oscillator comprises a cylinder with moment of inertia, \(I\), hanging from a light rod with torsional spring constant, \(\kappa\). The cylinder also experiences a drag torque equal to \(-\mu \dot \theta\), when moving with angular velocity \(\dot \theta\). The top of the rod is driven with angular displacement \(\phi(t) = \phi_0 \cos{\omega t}\).
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\begin{align*} \theta(t) &= \frac{\omega_0^2\,\phi_0}{\sqrt{\left(\omega_0^2 - \omega^2\right)^2 + \gamma^2\,\omega^2}}\,\cos\left({\omega\,t - \arctan{\left(\frac{\gamma\,\omega}{\left(\omega_0^2 -\omega^2\right)}\right)}}\right) \end{align*} \begin{align*} \text{where} \hspace{5mm} \gamma =\mu/I \hspace{5mm} \text{and} \hspace{5mm} \omega_0^2 = \kappa/I \end{align*}
\begin{align*} & A(\omega) = \frac{\omega_0^2\,\phi_0}{\sqrt{\left(\omega_0^2 - \omega^2\right)^2 + \gamma^2\,\omega^2}}\hspace{1mm}, \hspace{4mm} & \tan{\delta(\omega)} = \frac{\gamma\,\omega}{\left(\omega_0^2 -\omega^2\right)} \end{align*}
A capacitor (of capacitance C), a resistor (of resistance R) and an inductor (of inductance L) are connected to an AC voltage source \(V = V_0 \sin(\omega t)\) starting at \(t=0\) as shown in the diagram below.
Assuming that both the current and the charge of the capacitor are initially zero, determine the expression for \(V_C(t\ge0)\) with \(\omega=\omega_0=\dfrac{1}{\sqrt{LC}}\) and \(L < 4R^2C\).
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\begin{align*} V_C(t\ge0) = V_{0}[-\dfrac{\omega_{0}}{\omega’}e^{-\frac{\gamma t}{2}} \sin(\omega’t) + \sin(\omega_{0}t)] \end{align*} \begin{align*} \text{where} \hspace{5mm} \omega’ = \sqrt{\frac{1}{LC} - \frac{1}{4R^2C^2}} \hspace{5mm} \text{and} \hspace{5mm} \gamma = \frac{1}{RC} \end{align*}
