Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 70 / 2

Use the prescribed harmonic convention $e^{i\omega t}$. For a propagating acoustic plane wave, let $k=(\omega /c_0)\cos \theta$ and $\ell =(\omega /c_0)\sin \theta >0$. The incident and reflected pressure amplitudes in the upper half-space have vertical factors $e^{i\ell y}$ and $e^{-i\ell y}$ respectively:The linear homentropic acoustic equations imply $i\omega {\rho }_0{v}_y=-{\partial }_y{p}^{\prime }$. At the surface, the normal velocity is therefore $\ell (R-A)/({\rho }_0\omega )=-V$, while the pressure amplitude is $P=A+R$. The surface acoustic impedance condition $P=ZV$ givesHere $Z_f$ is the normal acoustic impedance; the angle in this question is measured from the horizontal, not the normal. For a passive acoustic impedance, the mean power absorbed per unit area is $\frac{1}{2}\mathrm{Re}(P{V}^{\ast })=\frac{1}{2}\mathrm{Re}Z|V{|}^2\ge 0$. The four limiting cases have distinct meanings:

Define $k_0=\omega /c_0$ and use the outgoing acoustic square-root branchFor positive real frequency reached from below, $\gamma =i\sqrt{k_0^2-k^2}$ on the propagating interval and is positive real for $|k|>k_0$. The outgoing field in the lower fluid has pressure amplitude $B{e}^{\gamma y}$, since $y<0$. If $\eta =\hat{\eta }{e}^{i\omega t-ikx}$, the shared normal velocity is $i\omega \hat{\eta }$. The linear homentropic acoustic equations therefore giveThis lower-fluid wave is outgoing; no additional incoming sound is included in defining the impedance seen by the upper fluid.

Put $K_s=T{k}^2-m{\omega }^2$. The sheet's force balance gives $K_s\hat{\eta }=-P+B$, hence $P=({\rho }_0{\omega }^2/\gamma -K_s)\hat{\eta }$. Its prescribed downward velocity amplitude is $V=-i\omega \hat{\eta }$. Thus the tensioned-sheet acoustic impedance isThe first term is the lower fluid's normal acoustic impedance, and the second is the sheet's inertial and elastic-sheet tension response. For a real propagating angle, $Z_f={\rho }_0{c}_0/\sin \theta$.

At fixed nonzero frequency and fixed wavenumber, $m\to \infty$ gives $|Z|\to \infty$, a zero-velocity, in-phase reflecting boundary. At fixed $k\ne 0$, $T\to \infty$ gives the same reflection limit, but there is an important exception: elastic-sheet tension does not resist the spatially uniform mode $k=0$. At normal incidence, the impedance remains ${\rho }_0{c}_0+im\omega$ however large the elastic-sheet tension is. With $m=0$ this mode is transparent. By contrast, arbitrarily large mass resists even a spatially uniform oscillation. These fixed-frequency limits exclude a simultaneously tuned structural resonance.

If $m=T=0$, $Z=Z_f$ and $R=0$. The identical fluids are effectively joined across a massless, untensioned interface: pressure and normal velocity continue without reflection. This is the matched case, rather than the pressure-release case.

With no incoming wave, both fluids obey the outgoing acoustic square-root branch. Their surface pressure amplitudes areSubstitution in the sheet's force balance yields the dispersion relation for an acoustic membrane wave:The equivalent denominator $\Delta =\gamma (T{k}^2-m{\omega }^2)-2{\rho }_0{\omega }^2$ is useful away from the branch points $\gamma =0$. For real guided modes with $|k|>k_0$, this is the familiar added mass of an evanescent fluid layer formThe normal fields decay away from the sheet; continued by analytic continuation roots may describe radiating or leaky modes.

On $D=0$, $T{k}^2-m{\omega }^2=2{\rho }_0{\omega }^2/\gamma$, so the impedance from the preceding solution isThus the divergence of the reflection coefficient corresponds to a reflection pole of a fluid-loaded membrane. The prescribed incoming amplitude is zero, but a homogeneous fluid-sheet mode can have nonzero amplitude. This is a resonance or guided-mode pole of the analytic scattering problem, not arbitrarily large passive reflection at a real propagating angle. In particular, an undamped guided mode has an evanescent normal field and therefore a complex incidence angle in the plane-wave continuation. For real propagating incidence, $Z_f>0$ and the finite-mass, finite-tension sheet has a purely imaginary structural impedance, so its passive reflection remains bounded.

Use the Fourier transform pair $\hat{q}(k)=\int q(x)e^{ikx}dx$, $q(x)=(2\pi {)}^{-1}\int \hat{q}(k)e^{-ikx}dk$. The point force transforms to $F$. With the dispersion relation $D$ defined above, the sheet equation becomesThe point-force radiation from a fluid-loaded sheet is therefore represented exactly byThe causal contour and outgoing acoustic square-root branch are fixed first with $\mathrm{Im}\omega <0$, then continued to the desired real frequency. This prescription fixes how poles and the branch points are passed.

For the acoustic far field $x=r\cos \theta$, $y=r\sin \theta$, take $0<\theta <\pi$ bounded away from grazing and $k_{0}r\gg 1$. The method of steepest descent saddle point is $k_s=k_0\cos \theta$, with ${\gamma }_s=i{k}_0\sin \theta$. The supplied saddle point rule, including its $\sin \theta$ factor, gives, provided the contour deformation crosses no poles,A convenient simplification, free of division by $T$, isEquivalently, when $T\ne 0$,The expression printed in the PDF is missing sound-speed factors for general dimensional $c_0$. It agrees with this result if $c_0=1$ in fully normalized units; when $c_0$ is retained as an arbitrary sound speed, the numerator needs $c_0^{-3/2}$ and the structural term needs $T/c_0^2$ in the last form. These factors arise respectively from cylindrical spreading, the pressure-density relation, and $k_s=\omega \cos \theta /c_0$.

A direct countercheck is the transparent-sheet limit $m=T=0$. The sheet jump condition then gives ${\hat{p}}_{+}(k,0)=F/2$, so the saddle point rule requiresThe printed expression, interpreted continuously after multiplying out its structural factor, instead gives $F(2c_0{)}^{-1}\sqrt{\omega /(2\pi r)}\sin \theta$ times the same phase. It differs by a factor $c_0^{3/2}$; for example it is eight times too large when $c_0=4$. This limit also verifies the normalization of the corrected density field independently of the sheet's elastic-sheet tension.

To decide about poles, track the roots of $\Delta (k,\omega )$ on the chosen square-root sheet and deform the original causal contour to the steepest descent contour. A root contributes a residue exactly when it lies in the region swept out by that deformation; its sign is fixed by the contour orientation. Branch cuts must be retained throughout this comparison. Which roots are crossed can depend on observation angle, producing a change of the modal contribution when a pole meets the deformation boundary. A saddle point approaching a pole or a grazing endpoint requires an approximation uniform in that limit, rather than the isolated saddle point formula above.

The crossed poles are the free fluid-sheet modes of the preceding solution. Real subsonic roots represent evanescent acoustic surface waves carrying energy along the sheet, with normal decay; complex continuations represent leaky or radiating modes. Their residues must be added to the saddle point sound when the causal contour selects them. The specification “no poles contribute” is therefore a substantive condition on the contour, not permission to ignore zeros of the dispersion relation.