Past exam of the physics A course of the University of Cambridge / 2019 / Paper 2 / Section B / B9

A waveguide confines the transverse structure of a wave while permitting propagation along a chosen route. It can deliver energy to a distant receiver without the same geometric spreading as an unguided beam, protect a signal from unwanted coupling, and select transverse normal modes. It also makes propagation paths and coupling to devices more controllable. Confinement does not eliminate wave absorption or scattering loss, and it can introduce wave dispersion and cutoff frequencies of a membrane waveguide or corresponding cutoffs for other wave types.

Two examples are an optical fiber, which guides light by its core-and-cladding refractive index structure, and a hollow electromagnetic waveguide used to carry microwaves between a source and an antenna. A duct for acoustic waves is a further example. Each uses different material and boundary conditions, but the common idea is controlled transverse confinement.

For the trial displacement $z=\sin (qy)e^{i(kx-\omega t)}$, direct differentiation givesThe two-dimensional wave equation is therefore satisfied ifHere $k$ is the longitudinal wavenumber, $q$ the transverse wavenumber, and $c_m$ the speed of an unconstrained membrane wave. The physical displacement is the real part, multiplied by an arbitrary displacement amplitude. $T$ has units $\mathrm{N}{\mathrm{m}}^{-1}$ and $\sigma$ is surface mass density in $\mathrm{kg}{\mathrm{m}}^{-2}$, so $T/\sigma$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$ as required.

The fixed edge at $y=0$ is automatically satisfied by $\sin (qy)$. At $y=a$, the displacement must vanish at every $x,t$, so $\sin (qa)=0$. Thus the allowed transverse normal modes haveThe case $n=0$ gives zero displacement for this sine solution and is not a physical propagating mode. Negative $n$ only reverses the amplitude sign and does not create a distinct transverse mode. This discrete transverse spectrum is a consequence of the Dirichlet boundary conditions, rather than quantization of the longitudinal wavenumber in an indefinitely long guide.

For the $n$th transverse mode, substitute $q_n=n\pi /a$ into the membrane waveguide dispersion relation:Longitudinal propagation requires real nonzero $k$, henceBelow the cutoff frequency of a membrane waveguide, $k=i\kappa$, with $\kappa =\sqrt{(n\pi /a{)}^2-\sigma {\omega }^2/T}$, and a bounded excitation decays as $e^{-\kappa x}$ rather than carrying a travelling wave along an infinite guide. At equality $k=0$, the displacement has no longitudinal phase advance and the group velocity is zero; it is a cutoff oscillation, not a propagating mode carrying finite longitudinal power.

For positive angular frequency, the first three branches areEach branch is even in $k$, has minimum ${\omega }_n(0)=n\pi c_m/a$, and starts with zero slope. At large $|k|$, all approach the straight asymptote $\omega =c_m|k|$. Their group velocities are $v_{g,n}=c_m^{2}k/{\omega }_n$, whereas their longitudinal phase velocities are $v_{p,n}={\omega }_n/k$ for $k\ne 0$, giving $v_{p,n}{v}_{g,n}=c_m^2$.

Resistance to bending adds a restoring contribution that grows more rapidly with spatial curvature than the tension contribution. Short-wavelength modes are therefore stiffened especially strongly, raising their frequencies and changing the large-wavenumber behaviour.

For example, a thin plate of flexural rigidity $D_b$ has $\sigma z_{tt}=T{\nabla }^{2}z-D_b{\nabla }^{4}z$. When its edge conditions admit the same sine transverse mode, bending stiffness in a membrane waveguide givesThus the cutoffs increase, and bending-dominated propagation has $\omega \propto k^2$ at large $|k|$, rather than approaching $c_m|k|$. Its phase velocity and group velocity then grow with $|k|$. If “rigidly fixed” means a genuinely clamped plate edge, the slope must also vanish and the mode shapes change; the qualitative stiffening and stronger wave dispersion remain, but simple $q_n=n\pi /a$ need not remain exact.