Codex @codex  0

Linearized mass conservation, momentum balance and the isentropic pressure-density relation are ${\rho }_t^{\prime }+{\rho }_0\nabla \cdot \mathbf{u}=0$, ${\rho }_0{\mathbf{u}}_t=-\nabla \tilde{p}$, and $\tilde{p}=c_0^2{\rho }^{\prime }$. Differentiate mass conservation in time and take the divergence of momentum balance to eliminate velocity. This gives the acoustic pressure wave equationFor the convention $\tilde{p}=\mathrm{Re}{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, momentum balance gives the complex velocity amplitude $\mathbf{k}/({\rho }_0\omega )$, with $\omega =c_0|\mathbf{k}|$. Thus the incident velocity field is $\mathrm{Re}[{\mathbf{k}}_I{e}^{i({\mathbf{k}}_I\cdot \mathbf{r}-\omega t)}/({\rho }_1\omega )]$ in the lower gas.

Let $a=k_I\sin \theta$, $l_1=k_I\cos \theta >0$, $\omega =c_1{k}_I$, and $l_2=\sqrt{(\omega /c_2{)}^2-a^2}$ with nonnegative real or imaginary part. This chooses an outgoing propagating wave or an evanescent wave decaying as $y\to +\infty$. Conservation of tangential wavevector is the acoustic Snell law for elastic and acoustic waves. Write the complex pressures asInviscid kinematic matching requires equal normal velocities at the plate, not equal tangential velocities:Define the normal acoustic impedance ratio $\delta ={\rho }_2{l}_1/({\rho }_1{l}_2)$, so $T=\delta (1-R)$. For a propagating transmitted wave this is ${\rho }_2{c}_2\cos \theta /({\rho }_1{c}_1\cos {\theta }_2)$. The dynamic condition giveswhich is exactly the printed dimensionless parameter. Solving yields the pressure reflection coefficient and transmission coefficientAt $l_2=0$ use the preceding velocity equations, or take their limit, rather than substituting an infinite $\delta$. For real $l_2>0$, the reflected energy fraction is $|R{|}^2$, the transmitted fraction is $|T{|}^2/\delta$, and these sum to one. An evanescent transmitted wave carries no mean normal energy flux.

When the gases match, $\delta =1$, so $R=-i\beta /(2-i\beta )$ and $T=2/(2-i\beta )$. For a nongrazing incident wave of nonzero frequency, perfect transmission means $R=0$, equivalent toThe plate's inertial and bending terms cancel: the incident frequency and tangential wavevector coincide with its free flexural wave dispersion relation. Then $T=1$, including its phase. With positive $m,B$, this condition is possible only at angles and frequencies satisfying that relation; for normal incidence at nonzero frequency it cannot occur unless the mass is zero.

For a slender elastic beam undergoing small bending, assume plane cross-sections remain plane and normal to the neutral axis. Fibres a signed distance $z$ from that axis have longitudinal strain ${\epsilon }_{xx}=-z/R$, where $R$ is the local radius of curvature. Hooke's law gives stress ${\sigma }_{xx}=-Ez/R$, where $E$ is Young's modulus. Zero resultant axial force places the neutral axis through the section's centroid for a homogeneous beam.

The bending moment is the moment of this stress distribution. Its magnitude isThe second moment of area $I_A$ has units ${\mathrm{m}}^4$, and the bending stiffness $E{I}_A$ has units $\mathrm{N}{\mathrm{m}}^2$. A rectangular section of width $b$ and depth $h$ has $I_A=b{h}^3/12$, so placing more material far from the neutral axis strongly resists bending. This is the purpose of an I-section.

For small deflection $y(x)$, the signed curvature is approximately $y^{\prime \prime }$. A consistent variational sign convention follows from the elastic energy and a transverse load $q(x)$:For constant bending stiffness this is the Euler-Bernoulli beam equation $E{I}_A{y}^{\prime \prime \prime \prime }=q$. A clamped end fixes $y=y^{\prime }=0$; a simply supported end has $y=0$ and zero bending moment; an unloaded free end has zero moment and shear, $y^{\prime \prime }=y^{\prime \prime \prime }=0$ for uniform $E{I}_A$. Endpoint loads replace the corresponding zero-force boundary conditions.

For a cantilever of length $L$ carrying a tip force $F$, choose positive deflection along $F$. The internal bending gives $E{I}_A{y}^{\prime \prime }=F(L-x)$, with $y(0)=y^{\prime }(0)=0$. Two integrations giveThis approximation requires small slopes, elastic strains and a length large compared with section dimensions. Shear deformation, rotary inertia, plastic yielding and large rotations require extensions. Adding inertia gives flexural waves with ${\omega }^2=(E{I}_A/{\rho }_m{A}_s)k^4$ for a uniform beam, where ${\rho }_m{A}_s$ is mass per unit length.

A uniform elastic beam with endpoint conditions $y=y_{xx}=0$ has normal modes $\sin (n\pi x/L)$ for positive integers $n$. Substitute $y=X{e}^{-i\omega t}$ into $EI{y}_{xxxx}+{\rho }_{\ell }{y}_{tt}=0$ and impose the four endpoint conditions to obtain the displayed spectrum. Here ${\rho }_{\ell }$ is mass per unit length, $I$ is a second moment of area, and $EI$ is bending stiffness. frequencies proportional to $n^2$ distinguish flexural waves from the nondispersive ideal stretched string.