Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 70 1 b Solution Created 2026-10-03 Updated 2026-10-07
In the Ffowcs Williams-Hawkings equation, the other source types are a surface acoustic monopole associated with acoustic thickness noise, and a volume acoustic quadrupole involving the Lighthill stress tensor. On an impermeable material surface, fluid and surface normal velocities agree. There is no through-surface mass-flux source; the remaining thickness source is . For a rigid body, the leading acoustic compact-source approximation to that source hasThus there is no leading net-volume acoustic monopole. To neglect thickness radiation beyond that leading cancellation, assume negligible volume displacement, as for ideal thin blades, or that its higher multipoles are small compared with the retained acoustic loading noise. Rigidity alone does not make a moving finite-volume body's local thickness source identically zero.
The volume acoustic quadrupole may be neglected for low Mach number motion when exterior turbulent or nonlinear stresses do not provide a competing strong source. We also assume small linear acoustics perturbations, a uniform reference sound speed, and negligible relevant viscous and entropy sources. These are source-strength approximations, particularly important if a loading contribution itself cancels by symmetry. Under them, the retained acoustic dipole is the force exerted by the object on the fluid, with the sign used in the previous solution.
Let , , and . In the acoustic far field, is large compared with the object and . For a source of size with and small surface Mach number, source-dependent delays and the Doppler factor can be neglected to leading order. The surface integral then contains just the total force . Differentiating its retarded time, rather than its spreading factor, gives the radiating termThe sign follows from . Differentiating or the direction instead produces the lower-order near field. A constant total force does not radiate at this leading compact order.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 70 1 c Solution Created 2026-10-03 Updated 2026-10-07
The rotation introduces angular frequency and, after combining the two blades, harmonics such as . The acoustic compact-source approximation requires the propagation time to be small compared with . ThusIt also bounds every blade element's Mach number by . Consequently , and its leading value is one. The separate acoustic far field condition is .
Choose the positive rotation sense so that a first blade at phase has radial and tangential unit vectorsIntegrating the given line force from to yields . Its axial component is constant, while . Since , the compact acoustic dipole sound from this blade isThe other blade has phase and contributes the opposite rotating force. At a common compact retarded time, their total force is , soThis cancellation calls for the next source-delay correction; it does not mean that the complete moving-source field vanishes.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 70 2 d Solution Created 2026-10-03 Updated 2026-10-07
Use the Fourier transform pair , . The point force transforms to . With the dispersion relation 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 , then continued to the desired real frequency. This prescription fixes how poles and the branch points are passed.
For the acoustic far field , , take bounded away from grazing and . The method of steepest descent saddle point is , with . The supplied saddle point rule, including its factor, gives, provided the contour deformation crosses no poles,A convenient simplification, free of division by , isEquivalently, when ,The expression printed in the PDF is missing sound-speed factors for general dimensional . It agrees with this result if in fully normalized units; when is retained as an arbitrary sound speed, the numerator needs and the structural term needs in the last form. These factors arise respectively from cylindrical spreading, the pressure-density relation, and .
A direct countercheck is the transparent-sheet limit . The sheet jump condition then gives , so the saddle point rule requiresThe printed expression, interpreted continuously after multiplying out its structural factor, instead gives times the same phase. It differs by a factor ; for example it is eight times too large when . 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 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.