Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / Section C

In the uniform-Earth chord oscillator, neglect the mass removed by the narrow tunnel and any rotation of the Earth. Inside a uniform sphere, the shell theorem gives $\mathbf{F}=-(4\pi G\rho m/3)\mathbf{r}$. On the constrained chord $\mathbf{r}=(d,y)$, its along-tunnel component isThe transverse component is balanced by the tunnel constraint. With release from $y_s=\sqrt{R^2-d^2}$ at rest,The far end is $-y_s$, so $t_{\mathrm{transit}}=\pi /{\omega }_0\approx 2530\mathrm{s}=42.2\min$, independent of chord offset in this ideal model. Here $\rho =5515\mathrm{kg}{\mathrm{m}}^{-3}$ and $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ give ${\omega }_0=1.24171\times 10^{-3}{\mathrm{s}}^{-1}$.

With linear drag, $\ddot{y}+(b/m)\dot{y}+{\omega }_0^{2}y=0$. Inserting a complex exponential $e^{-(\gamma +i\omega )t}$ givesso in the underdamped regimeReal physical solutions combine conjugate modes. Release from rest specifically gives $y=y_s{e}^{-\gamma t}[\cos \omega t+(\gamma /\omega )\sin \omega t]$; a lone exponential with prefactor $y_s$ does not impose that initial velocity.

The critical gravity-tunnel settling time follows from critical damping, the fastest non-oscillatory release:Interpret the 1% tolerance relative to the initial displacement from equilibrium. With $s={\omega }_{0}t$, $(1+s)e^{-s}=0.01$ has $s=6.63835$. HenceThe equilibrium is reached only asymptotically; using a pure $e^{-{\omega }_{0}t}$ would miss the critical-damping polynomial factor and underestimate this time.

For the series-fed parallel LC voltage response, use phasors with time dependence $e^{i\omega t}$. The parallel branch has impedance $iX$, where $X=\omega L/(1-{\omega }^{2}LC)$. ThereforeAt the ideal parallel resonance ${\omega }_0=1/\sqrt{LC}$ its input impedance diverges. The input-current phase changes sign on crossing this pole; expressions there are interpreted by limits.

The capacitor voltage equals the parallel-branch voltage. Define $q=R(1-{\omega }^{2}LC)/(\omega L)$. ThenAt low frequency $H\sim i\omega L/R$; at high frequency $H\sim -i/(\omega RC)$; at resonance $H=1$. Thus it is a band-pass voltage filter or resonant frequency selector. Its phase decreases continuously from $+\pi /2$ through zero to $-\pi /2$. Half-power points satisfy $|q|=1$ and

For the supplied component values and $\omega =2\pi (50){\mathrm{s}}^{-1}$, $X=421.745\Omega$ and $V_{C,\mathrm{pk}}=99.3046\mathrm{V}$. The element peak currents areInductor and capacitor currents are opposite phasors; their magnitudes subtract here rather than add. The resistor carries the input current, withThese results assume ideal components and sinusoidal steady state.