Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / Section C

Measure $x$ from the static equilibrium position, so the constant gravitational force has already cancelled the equilibrium spring force. With stationary oil, Newton's second law gives $m\ddot{x}=-b\dot{x}-kx$. The characteristic exponents are $-\gamma \pm i{\omega }_1$, whereFor the stipulated phase of the damped cosine,At $t_n=2\pi n/{\omega }_1$, the velocity is not zero: $\dot{x}=-\gamma A{e}^{-\gamma t_n}$. ThusExpanding $e^{-z}$ at small $z=4\pi \gamma /{\omega }_1$ gives the requested light-damping loss $z$. The exact instantaneous energy also satisfies $dE/dt=-b{\dot{x}}^2\le 0$.

From the static extension, $k=mg/(0.20\mathrm{m})=44.145\mathrm{N}{\mathrm{m}}^{-1}$; the instantaneous drag measurement gives $b=0.10\mathrm{N}\mathrm{s}{\mathrm{m}}^{-1}$. Consequently $\gamma =0.05556{\mathrm{s}}^{-1}$ and ${\omega }_1=7.003{\mathrm{s}}^{-1}$. Two successive downward equilibrium crossings are one damped period apart, so the same energy ratio applies. At the first crossing $E=m{v}^2/2=1.125\times 10^{-3}\mathrm{J}$. ThereforeThe speed has decreased by $2.43\times 10^{-3}\mathrm{m}{\mathrm{s}}^{-1}=0.243\mathrm{cm}{\mathrm{s}}^{-1}$. These values use the exact decay factor, not the first-order approximation.

In series the same current passes through both resistors, while their voltage drops add: $V=I(R_1+R_2)$. In parallel the same voltage drives currents $V/R_1$ and $V/R_2$, which add. Thus $R_s=R_1+R_2$ and $R_p^{-1}=R_1^{-1}+R_2^{-1}$.

Use phasors proportional to $e^{i\omega t}$. Then $Z_L=i\omega L$ and $Z_C=1/(i\omega C)$. The printed diagram places C in parallel with the series RL branch, so the input admittance isIts nonzero zero-phase frequency exists for $R^2<L/C$:At zero frequency, $|Y|=1/R$ and $I_0=V_0/R$. At ${\omega }_1$, $R^2+{\omega }_1^2{L}^2=L/C$, givingFor $0<\omega <{\omega }_1$, $\mathrm{Im}Y<0$, so current lags voltage. At very high frequency, $Y\sim i\omega C$, so current leads by $\pi /2$. Zero phase need not coincide exactly with the minimum current magnitude in this lossy network.

The three measured limits determine the components in order:For the changed resistance, the quoted $0.6\mathrm{kHz}$ is an ordinary frequency, so ${\omega }^{\prime }=2\pi (600{\mathrm{s}}^{-1})$. Keeping L and C fixed gives