Past exam of the physics A course of the University of Cambridge / 2015 / Paper 2 / Section D / D13

For a small longitudinal disturbance, let the equilibrium pressure and mass density be $p_0,{\rho }_0$, their perturbations be $p^{\prime },{\rho }^{\prime }$, and the fluid velocity be $v$. Linearizing conservation of mass and the inviscid momentum equation in one dimension givesWhen compression is adiabatic, the equation of state gives $p^{\prime }=c_s^2{\rho }^{\prime }$ with $c_s^2=(\partial p/\partial \rho {)}_s$. Differentiating and eliminating $v$ yieldsfor an ideal gas with heat-capacity ratio $\gamma =C_p/C_v$ and molar mass $M_{\mathrm{mol}}$. Compression raises temperature as well as density, explaining the factor $\gamma$ compared with the isothermal value $\sqrt{p_0/{\rho }_0}$.

A rightgoing plane sound wave has pressure and particle velocity in phase, $p^{\prime }={\rho }_0{c}_{s}v$, and mean intensity $⟨p^{\prime }v⟩={\hat{p}}^2/(2{\rho }_0{c}_s)={\rho }_0{c}_s{\hat{v}}^2/2$ for peak amplitudes. Particle displacement is a quarter-cycle out of phase with particle velocity: particles oscillate locally while the compression pattern transports energy. This continuum linear description requires wavelength much larger than the mean free path, small pressure perturbations and negligible viscous attenuation. Adiabatic propagation requires heat exchange across the compressed region to be slow on the oscillation time; strong thermal coupling would instead favour an isothermal response.

The Q factor measures stored energy relative to loss: $Q=2\pi E/(\text{energy lost per cycle})$, or $Q={\omega }_{0}E/P_{\mathrm{loss}}$ in the weak-loss limit. For viscous damping $m\ddot{x}+b\dot{x}+kx=0$, write $\gamma =b/(2m)$, ${\omega }_0=\sqrt{k/m}$ and ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}$. The conventional oscillator parameter isIn free ringdown, fit the amplitude envelope $A(t)=A_0{e}^{-\gamma t}$. Its decay time is ${\tau }_A=1/\gamma$, whereas the cycle-averaged energy envelope decays on ${\tau }_E=1/(2\gamma )$; thus $Q={\omega }_0{\tau }_A/2={\omega }_0{\tau }_E$. Equivalently the logarithmic decrement between same-sign peaks is $\delta =\ln (A_n/A_{n+1})=2\pi \gamma /{\omega }_d$. It gives $Q\simeq \pi /\delta$ for weak damping, or the exact parameter relation $Q=\sqrt{1/4+(\pi /\delta {)}^2}$. Fit multiple peaks above the noise floor, measure the damped period, and verify an exponential envelope rather than assuming all amplitude loss is viscous.

For a driven measurement, allow transients to settle and scan frequency at fixed force amplitude. Near resonance and at high $Q$, the response is Lorentzian in displacement power, with half-power points at displacement amplitudes $1/\sqrt{2}$ of the maximum. Their angular-frequency separation gives $Q\simeq {\omega }_0/\Delta \omega$. Halving the amplitude is not a half-power criterion.

The absorbed-power bandwidth of a viscously damped oscillator can be found exactly. Sinceits maximum is at $\Omega ={\omega }_0$, and the half-power frequencies satisfy $|k-m{\Omega }^2|=b\Omega$. HenceFor broad resonances, displacement and absorbed-power peaks differ, so the measured observable must be specified. Frequency-dependent drive, backgrounds or several overlapping modes can otherwise masquerade as a changed Q factor.

For propagation along $z$, a transverse wave has two perpendicular displacement components. With $\varphi =kz-\omega t$, write $u_x=A\cos \varphi$, $u_y=B\cos (\varphi +\delta )$. Their trajectory at a fixed point obeysThe general nondegenerate trajectory is an ellipse: elliptical polarization. For $\delta =0$ or $\pi$, the components remain in phase or antiphase and the trajectory collapses to a line. A single nonzero component is also linearly polarized. Equal amplitudes $A=B$ with $\delta =\pm \pi /2$ give circular polarization. Unequal quadrature amplitudes give an ellipse aligned with the chosen axes, while general phase differences rotate its axes.

For the displayed convention and $\delta =+\pi /2$, at $z=0$ the vector starts at $+x$ and moves towards $+y$ as time increases. Reversing $\delta$ reverses its rotation. Right- and left-handed labels require a stated viewing and phase convention, so rotation is defined explicitly here. A fixed ellipse is a coherent relationship between transverse components; unpolarized light instead has fluctuating orientation and phase relationships on the measurement timescale.

A linear polarizer transmits a linearly polarized beam with intensity proportional to ${\cos }^2\theta$, allowing its direction to be measured by an analyser. circular polarization transmits equal intensity through any linear analyser, but that test alone does not distinguish it from unpolarized light. A quarter-wave plate converts a suitable coherent circular state to linear polarization, allowing the combination to discriminate them. For light the electric field defines the usual polarization; the magnetic field remains perpendicular to it and to propagation. Longitudinal sound in a homogeneous gas lacks the two transverse degrees of freedom needed for these polarization states.