Materials Science · Architected materials and metamaterials

Acoustic and electromagnetic wave responses

Derive how layer impedance and phase create a directional wave gap, compare an infinite periodic prediction with finite transmission, and distinguish Bragg interference from a local resonator’s dynamic response.

Useful background: Composites, fibres and interfaces; Elasticity, plasticity and tensile strength. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Explain engineered wave response with frequency and geometry limits.
Construct a stated acoustic or normal-incidence electromagnetic layer model and obtain its periodic-mode condition.
Read a dispersion diagram and finite power spectrum without equating evanescence, reflection and absorption.
Derive a frequency-dependent apparent inertia from positive masses and an internal spring, keeping its local-resonance scope explicit.
Test wavelength, boundary and averaging assumptions before transferring a wave response to another frequency, direction or device.

An array can redirect or inhibit waves

Imagine a repeated stack of two layers. Each interface returns part of an incident wave, and each layer delays its phase. At some frequencies the returned contributions reinforce; at others they partly cancel. Geometry, architecture and effective response introduced why arrangement can change a response. Here the question is how an arrangement selects wave modes, rather than its static stiffness.

Periodic scattering and local resonance can modify acoustic or electromagnetic propagation. A band gap is a frequency range lacking certain propagating modes under the structure’s conditions. It is not proof that every wave and direction is blocked.

A harmonic wave repeats f times per second, with frequency f in hertz (Hz). Its angular frequency is ω=2πf in radians per second. In a uniform lossless constituent with phase speed c_j, wavelength is λ_j=c_j/f and phase wavenumber k_j=2π/λ_j=ω/c_j. A layer of thickness d_j accumulates phase δ_j=k_jd_j, in radians. A quarter-wavelength thickness therefore gives δ_j=π/2. Phase is the stage of an oscillation, not an energy fraction.

Use a deliberately restricted one-dimensional model. Interfaces are planar and laterally unlimited; incidence is normal, along +x. Constituents are homogeneous, linear, reciprocal and lossless, with positive real properties taken constant over the illustrated frequency range. The acoustic version is a scalar fluid-like pressure wave, with no shear mode. The electromagnetic version uses transverse E_y and H_z in isotropic nonmagnetic dielectrics. These are two applications of the same layer algebra, not a claim that pressure is an electric field.

For a forward wave, acoustic pressure p and x-directed particle velocity u obey p=Z_j u, with Z_j=ρ_jc_j and mass density ρ_j. Z has units Pa·s/m. For the electromagnetic wave, E_y=Z_jH_z, with wave impedance Z_j=Z_vac/n_j and c_j=c₀/n_j for the stipulated nonmagnetic dielectric index n_j. Here c₀ is the vacuum speed of light and Z_vac is the vacuum wave impedance. Its Z is in ohms. A backward wave reverses the sign of the velocity or magnetic-field contribution. Those signs are what allow an interface to reflect.

Pressure and normal particle velocity are continuous across the ideal fluid interface. Tangential electric and magnetic fields are continuous across the stated dielectric interface without an added surface source. At a single interface, matching forward-plus-reflected fields gives an amplitude reflection coefficient r=(Z₂−Z₁)/(Z₂+Z₁). A stack needs the intervening phases as well as these impedance contrasts.

Carry the fields through one repeat

Choose a positive reference impedance Z₀, used for the identical media outside the eventual finite stack, and write z_j=Z_j/Z₀. The state pair is [p,Z₀u] for acoustics or [E_y,Z₀H_z] for electromagnetism. Both components of a pair have matching units. Use time dependence e^(−iωt), where i²=−1; taking the real part gives the physical oscillation. Forward spatial dependence is e^(+ik_jx).

A forward and a backward wave can be added to express either field at a layer’s entrance and exit. Eliminating their two amplitudes gives the propagation relation ψ_right=P_j ψ_left, with P_j=[[cosδ_j, i z_j sinδ_j],[i sinδ_j/z_j, cosδ_j]]. A matrix here is compact bookkeeping for two coupled equations: the first exit component is cosδ_j times the first entrance component plus i z_j sinδ_j times the second; the other row gives the second exit component.

For one repeat, first cross layer 1 and then layer 2. The period is ℓ=d₁+d₂ and its matrix is P=P₂P₁; matrix order follows the physical sequence. Each layer has determinant cos²δ_j+sin²δ_j=1, so detP=1. The continuous physical fields already account for the interface conditions: no extra interface matrix is inserted into this particular state convention.

A Bloch mode repeats its internal pattern from one period to the next, multiplied by e^(iKℓ). K is the Bloch wavenumber of the periodic arrangement, distinct from each constituent k_j. Thus that multiplier is an eigenvalue of P. Write Λ for that dimensionless multiplier, to distinguish it from wavelength. Since detP=1, its equation is Λ²−2FΛ+1=0, where F=tr(P)/2. Setting Λ=e^(iKℓ) gives cos(Kℓ)=F.

Multiplying the two layer matrices gives F=cosδ₁cosδ₂−½(Z₂/Z₁+Z₁/Z₂)sinδ₁sinδ₂. In this lossless scalar problem F is real. For |F|<1 there is a real Bloch wavenumber and an allowed propagation band. Values |F|=1 mark band edges. If |F|>1, K is complex and no propagating Bloch mode of this specified direction and type exists. A growing and a decaying spatial solution are still available; this is not temporal creation of energy.

A layer phase can select a forbidden interval

Take an invented quarter-wave repeat with impedance ratio Z₂/Z₁=3 and d₁/c₁=d₂/c₂=1/(4f₀). The design frequency f₀ makes δ₁=δ₂=π/2. At frequency f, both phases are δ=(π/2)(f/f₀). These stipulated constant-property inputs make F=cos²δ−(5/3)sin²δ=1−(8/3)sin²δ.

The first forbidden interval has F<−1. Solving 1−(8/3)sin²δ<−1 gives sin²δ>3/4, hence 2/3<f/f₀<4/3 around the first design frequency. This is a directional gap of the infinite periodic model. The edges are separate limiting states, not a claim that every finite specimen has zero transmission there. At f₀, F=−5/3 and the two period multipliers are −3 and −1/3.

Choose the mode decaying toward +x. Then Kℓ=π+i ln3, since cos(π+iα)=−coshα. Its amplitude decreases by 1/3 per period. The real π gives a sign reversal from one repeat to the next. Spatial decay in this lossless gap is evanescence: it need not be absorption. A finite barrier generally combines both spatial solutions to meet both external boundaries.

The first figure’s band diagram plots allowed frequencies against real Kℓ/π from 0 to 1, one half of the one-dimensional reciprocal period. There is no real-K branch in the shaded interval. It is a dispersion diagram, not a drawing of the layers or a measured transmission spectrum. The two physical examples in the table share this normalized result because their phase and impedance ratios match; their dimensional frequencies and layer thicknesses differ.

Invented model equivalents, not specifications of named materials. Normal incidence only; constant positive lossless properties over the plotted range. The finite-spectrum calculation uses identical external media with Z₀=Z₁.
Quantity	Scalar acoustic layers	Nonmagnetic dielectric layers
Constituent inputs	c₁=c₂=2000 m/s; ρ₁=1000, ρ₂=3000 kg/m³	n₁=3, n₂=1; Z_j=Z_vac/n_j
Layer thicknesses	d₁=d₂=5.00 mm	d₁=100 nm; d₂=300 nm
Period ℓ	10.0 mm	400 nm
Design frequency/wavelength	f₀=100 kHz	Vacuum λ₀=1.20 μm; f₀≈249.827 THz
Phase at f₀	π/2 in each layer	π/2 in each layer
External media for finite spectrum	Same Z₁ on both sides	Same n=3 medium on both sides

A finite spectrum is a boundary problem

For N repeats, use P_N=P^N=[[A,B],[C,D]] in the same state convention. Normalize the incident primary-field amplitude to one. With identical real external impedance Z₀, the entrance state is [1+r,1−r] and the exit state is [t,t], where r and t are the reflected and transmitted amplitudes. Thus t=A(1+r)+B(1−r) and t=C(1+r)+D(1−r). Solving these two equations, using detP_N=1, gives t=2/(A+D−B−C).

Power transmission is T=|t|² for these identical external media; reflection is R=|r|². Here the lossless balance is R+T=1 and absorbed power is zero. For different input/output impedances, a field-amplitude square alone is not the transmitted power fraction. Other directions, shear waves, surface modes, scattering or uncollected light would also change an actual measurement account. Reflection, absorption and transmission develops that distinction.

At the design frequency the one-period matrix is diagonal: P=[[-3,0],[0,−1/3]]. Raising it to N gives t=2/[(-3)^N+(-1/3)^N], hence T_N=4/(3^N+3^(−N))². One period transmits 0.3600, whereas three transmit about 0.00547195, or 0.547195%. The remaining power is reflected in this model. Neither number is zero, despite f₀ being inside the infinite-model gap.

Do not replace the finite result by the decaying Bloch amplitude squared. For three periods that shortcut gives (1/3)^6=1/729≈0.00137174, which omits the boundary coupling and the second spatial solution. Outside the gap, finite-stack interference can also create transmission peaks. The lower panel therefore compares complete boundary-matched spectra for one and three periods, rather than labelling every dip an absorbing resonance.

Original lossless normal-incidence bilayer calculation. One repeat has two layers with normalized impedances z1=1 and z2=3 and equal phases delta=(pi/2)(f/f0). The layer sketch is schematic, not a dimensional scale. In the upper chart the horizontal axis is real Bloch phase K ell divided by pi, from zero to one; the vertical axis is frequency f/f0, from zero to two. Allowed branches end at frequencies 2/3 and 4/3 at the zone edge, leaving that open interval without a real-K branch of the stipulated scalar mode. The lower chart uses frequency f/f0 horizontally and finite power transmission T vertically. Its blue solid curve is one period and violet dashed curve is three periods, both with identical Z0=Z1 external media. At f=f0, their power transmissions are 0.36 and 0.005471945956, respectively. The shaded interval is the infinite-model gap, not zero finite transmission. Reflection is one minus transmission and absorption is zero in this model. The frequency-zero endpoints are low-frequency limits, not optical DC measurements. No oblique, shear, lossy or all-angle result is shown.

A local oscillator selects frequency by another mechanism

Bragg interference uses phase accumulated over the repeat. A local resonance instead involves an internal motion with its own frequency. Consider a separate one-dimensional mechanical cell: a positive outer mass M has displacement U, and a positive inner mass m has displacement V, joined by a positive linear spring k_r. It is not the fluid or dielectric layer stack. Motion is small, harmonic and undamped for the derivation; amplitudes U,V use the same e^(−iωt) convention.

For the inner mass, −mω²V=k_r(U−V). Define ω_r²=k_r/m, the inner resonance with the outer displacement prescribed. Solving gives V/U=ω_r²/(ω_r²−ω²). At low frequency the inner mass follows the outer one. Just above ω_r it moves in opposition, with a large amplitude ratio in the ideal undamped model. The formula is singular at ω=ω_r; it does not predict a finite steady amplitude at that exact resonance.

The external driving force F_d on the cell satisfies F_d=−ω²(MU+mV). After eliminating V, write F_d=−ω²M_eff(ω)U, where M_eff=M+mω_r²/(ω_r²−ω²). This is an apparent dynamic inertia referred to the outer motion. The low-frequency limit is M+m, but for ω_r<ω<ω_r√(1+m/M) the apparent coefficient is negative. A larger number in a static weighing experiment is not being changed into negative physical mass.

Every actual mass and spring remains positive. The kinetic energy is ½M U̇²+½m V̇² and the internal spring energy is ½k_r(V−U)² for real time-dependent displacements. These are nonnegative. Eliminating an internal coordinate in a harmonic force relation can produce a signed, frequency-dependent coefficient without making the complete energy negative or defeating conservation. Damping would change phase and make the reduced response complex; the pole and drawn real-valued curve are ideal limits.

A local resonator alone does not establish a propagating-medium gap. To see how coupling matters, place such outer masses on a uniform line, connected to neighbours by positive springs K_h at spacing a. For an outer Bloch motion U_j=Ue^(iqja), the host-spring force is −4K_h sin²(qa/2)U_j. Combining it with the same internal reduction gives 4K_h sin²(qa/2)=ω²M_eff(ω).

For real q the left side is nonnegative. Negative M_eff therefore excludes a real propagating q in that interval for this chain. Positive M_eff alone is not enough: the right side must also lie between 0 and 4K_h. Thus the negative-inertia interval is a sufficient forbidden interval here, not a derivation of the entire band gap, a finite transmission curve or a complete three-dimensional acoustic response. Local resonance can be placed below the frequency scale of host-cell Bragg interference, but its coupling and spatial-averaging assumptions still need checking.

Separate original one-dimensional positive-mass internal-resonator model. A positive external force acts on an outer mass M=0.0100 kilogram with coordinate U, connected by a spring kr=986.9604401 newtons per metre to inner mass m=0.00250 kilogram with coordinate V. All arrows define positive displacement along the same axis; the schematic does not depict the real harmonic motion at one time or the bilayer geometry. The inner resonance with the outer motion prescribed is fr=100 hertz. The chart plots M_eff/M=1+0.25/[1−(f/fr)^2] against f/fr, from zero to 1.5, with the displayed vertical range clipped at minus four and plus four. Separate branches omit the undamped pole at f/fr=1. The negative-coefficient interval is 1<f/fr<sqrt(1.25), about 1.118034, shaded orange with excluded endpoints. At f/fr=1.05 the marked coefficient is minus 1.43902439. The low-frequency limit is 1.25; at the upper interval boundary the coefficient is zero. The actual physical mass remains positive 0.01250 kilogram, and the complete kinetic and spring energies are nonnegative. The signed coefficient is referred to the outer displacement after eliminating the inner coordinate. It is not a complete finite transmission spectrum or the entire chain gap.

Different physical descriptions can be useful

Effective parameters describe an averaged response where scale separation and the model permit it. Near resonances, dispersion and losses can be strong. A negative effective parameter is a property of a defined dynamic model, not a claim that ordinary conservation laws have failed. Negative refraction in a periodic crystal does not, by itself, establish a negative scalar effective refractive index; the propagation and averaging model must justify that inference.

In the equal-phase layered example, a long-wavelength expansion makes this distinction quantitative. For small δ, F≈1−(8/3)δ². With cos(Kℓ)≈1−(Kℓ)²/2, the low-frequency branch has Kℓ≈4δ/√3. For the acoustic layers with ℓ=2d and common speed c, this gives the phase-speed approximation c_eff≈√3c/2≈1732.05 m/s. At 1.00 kHz its approximate wavelength is 1.73205 m, much greater than the 0.0100 m period.

That phase-speed limit is an approximation to the stated branch, not a complete measured set of effective density and modulus. At 100 kHz, each layer has δ=π/2 and the exact trace is −5/3: no real propagating K exists. Extending the low-frequency c_eff unchanged into that gap would erase the mechanism being studied. A small geometric period by itself is also not sufficient near a strong internal resonance; the retained degrees of freedom, frequency dependence and possible spatial dispersion matter.

An observed bending of energy toward an unusual side of an interface requires a direction- and mode-resolved propagation model. The one-dimensional normal-incidence calculations here do not demonstrate negative refraction, a negative scalar electromagnetic index or an omnidirectional gap. They show why the original qualification above matters: a dynamic or averaged quantity must name the field, direction, branch, frequency and boundary conditions that define it.

Test the mode and account for the outgoing power

A reported gap can come from an infinite periodic calculation, a finite numerical specimen or an actual transmission measurement. They provide different kinds of evidence. A cantilever-in-mass study derived an internal-resonator model and compared it with two- and three-dimensional finite-element calculations; fabrication showed a printable geometry. Agreement between those calculations is not an independent experimental transmission spectrum. Here every numerical layer, mass, spring and spectrum is an original teaching model, not a value taken from that design.

A useful comparison states the specimen size and repeat count, incident frequency range, direction and polarisation or acoustic mode, surrounding media, constraints and measured outgoing channels. Sweep beyond a single low-transmission point and compare the result with the proposed mode or interference prediction. In an actual finite structure, side leakage, defects, contacts, support motion and conversion into other modes can remain. A narrow detector or one displacement sensor does not automatically measure all outgoing power.

Loss and dispersion are properties of real constituents and resonators as well as of the architecture. Changing dimensions can shift an ideal spectrum while also changing dissipation, fabrication error or the material state. The present derivations qualify neither a broadband soundproof barrier nor an optical device. Their value is a transferable prediction: specify a model, compute the permitted modes and boundary response, and identify observations that could show that model is incomplete.

Worked examples

Work through the case

An acoustic lattice blocks a narrow frequency band. Does it establish complete broadband soundproofing?

Identify the blocked modes and frequency interval.
Check incidence, finite size and leakage paths.
Assess the remaining transmission spectrum.

No. The demonstrated effect has a bounded frequency and geometry scope.

The same phase design can use very different lengths

Use the two invented layer systems. Verify their quarter-wave phase and find the first infinite-model forbidden interval in dimensional units. Does equal phase require equal physical thickness?

For sound, λ_j=c_j/f₀=2000/100000=0.0200 m=20.0 mm. Each 5.00 mm layer is λ_j/4. Their impedances are Z₁=1000×2000=2.00×10⁶ Pa·s/m and Z₂=3000×2000=6.00×10⁶ Pa·s/m, so the ratio is 3.
The normalized gap gives (2/3)×100=66.6667 kHz to (4/3)×100=133.333 kHz, excluding the two band-edge values from the open gap. This is for the stated normal pressure mode of the infinite periodic model.
For light, δ_j=2πn_jd_j/λ₀. Both optical thicknesses n_jd_j are 300 nm=λ₀/4, so each δ_j=π/2. The physical layers are 100 and 300 nm, not equal. With c₀=299792458 m/s, f₀=c₀/(1.20×10⁻⁶)≈2.49827×10¹⁴ Hz=249.827 THz.
The same normalized band edges correspond to about 166.551 and 333.103 THz under the stipulated constant indices. This is not a measured broadband mirror specification; real dispersion/loss, angle and surrounding media need their own evidence.

The acoustic layers are each 5.00 mm, while the electromagnetic layers are 100 and 300 nm. Matching phase and impedance ratios gives the same normalized mode diagram, not the same wavelength, dimensions or device.

Read the mode at the centre of the gap

At f=f₀ in the ratio-3 quarter-wave model, calculate F, the two period multipliers and the decay of the +x-decaying Bloch amplitude after three periods. Is its squared value a finite-stack transmission?

At δ=π/2, F=1−8/3=−5/3. Solve Λ²−2FΛ+1=0: Λ=F±√(F²−1)=−5/3±4/3, giving −3 and −1/3.
The decaying multiplier is −1/3. Three translations give (−1/3)³=−1/27 in amplitude, relative to the same mode one started with. The negative sign carries the phase reversal.
Its amplitude-square ratio is 1/729≈0.00137174. Kℓ=π+i ln3 gives the same decay. This calculation concerns one Bloch spatial solution, not a field matched to an incident wave and an outgoing medium.
A finite stack combines spatial solutions to satisfy both faces. Its incident-to-transmitted power must be calculated with those boundary conditions; the next case does so.

F=−5/3; multipliers −3 and −1/3. The decaying mode’s amplitude becomes −1/27 after three periods, but its 1/729 square is not automatically the finite transmission.

A lossless gap mainly returns power

For the same centre frequency, compare one and three periods with identical external Z₀=Z₁ media. Find power T and R and decide whether the missing transmitted power was absorbed.

At the centre, P_N is diagonal with entries (−3)^N and (−1/3)^N. The boundary equations give T_N=4/(3^N+3^(−N))². This uses total forward/backward field matching, not just one decaying Bloch factor.
For N=1, T=4/(3+1/3)²=0.3600 and R=1−T=0.6400. For N=3, T=4/(27+1/27)²=0.00547195 and R=0.994528.
As percentages, the three-period model transmits about 0.547195% and reflects 99.4528%. Absorption is zero because the stipulated model is lossless and these are its complete outgoing channels.
An actual low-transmission measurement alone would not identify reflection, absorption or conversion into unmeasured channels. Repeat count and surrounding impedance also belong with the result.

One period: T=0.36, R=0.64. Three periods: T≈0.00547195, R≈0.994528. This model reflects rather than absorbs the untransmitted power.

Positive masses can give a negative reduced coefficient

For the separate internal-resonator cell use M=0.0100 kg, m=0.00250 kg and f_r=100 Hz. Find k_r, the inner/outer displacement ratio and M_eff at 105 Hz. Find the negative-inertia interval. Has the static total mass become negative?

ω_r=2π×100 rad/s and k_r=mω_r²≈986.960 N/m. The mass ratio m/M=0.25. These are invented linear-model inputs, not a fabricated resonator specification.
At 105 Hz, (f/f_r)²=1.1025. Therefore V/U=1/(1−1.1025)=−9.75610. The inner displacement is opposed to the outer one in this ideal harmonic response. Choose an amplitude within whatever small-motion regime justifies the model; the ratio does not qualify an arbitrary excursion.
M_eff/M=1+0.25/(1−1.1025)=−1.43902, giving M_eff≈−0.0143902 kg when referred to U. The actual total mass is still M+m=0.01250 kg. Its kinetic and spring energy contributions remain nonnegative.
The negative interval is 100<f<100√1.25≈111.803 Hz. The pole at 100 Hz and the zero at the upper bound are not points inside this strict interval. In the stated positive-spring chain this interval cannot have a real propagating q; it is not the entire gap or a calculated finite attenuation.

k_r≈986.960 N/m, V/U≈−9.75610 and M_eff≈−0.0143902 kg at 105 Hz. The static mass stays +0.01250 kg; the negative dynamic interval is approximately 100–111.803 Hz, with both endpoints excluded.

An approximation that works at low frequency can fail at the gap

For the acoustic bilayer, use the low-frequency c_eff≈1732.05 m/s. Compare its wavelength at 1.00 kHz with the 0.0100 m period. Can that same constant real speed describe the 100 kHz gap centre?

At 1.00 kHz, λ_eff≈1732.05/1000=1.73205 m. The period/wavelength ratio is 0.0100/1.73205≈0.00577350, much smaller than one in this particular example.
The exact trace at that frequency is about 0.99934208, giving real Kℓ≈0.0362765. This is consistent with the small-phase branch used for the approximation; it is not a proof of every possible homogeneous constitutive parameter.
At 100 kHz the exact F=−5/3 lies outside [−1,1]. A constant positive real c_eff would instead force real K=ω/c_eff and miss the forbidden interval.
A phase-speed estimate also does not separately determine effective density and compressibility. Near internal resonances, the retained internal coordinate and frequency dependence require additional care even if the physical cells are small.

At 1.00 kHz the approximate wavelength spans about 173 periods. At the 100 kHz centre the exact periodic model has complex K, so the low-frequency constant-speed extrapolation is invalid.

Try the reasoning

Why report wavelength relative to unit-cell size?
Show the answer and reasoning
It affects propagation and effective-medium assumptions.
Averages may fail when the scales are comparable.
Can resonance improve one response while adding loss?
Show the answer and reasoning
Yes.
Strong dynamic response often involves dissipation and dispersion.
Keep a periodic two-layer layout, but set Z₂=Z₁ in the same lossless normal-incidence model. The phases can still differ. Does periodicity alone produce the Bragg gap derived above?
Show the answer and reasoning
No. The trace becomes cosδ₁cosδ₂−sinδ₁sinδ₂=cos(δ₁+δ₂), whose magnitude never exceeds one.
The interfaces have no impedance reflection in this scalar model. A periodic label alone does not create a forbidden interval. This is not a statement about arbitrary elastic modes or oblique electromagnetic incidence.
Halve both layer thicknesses without changing the stipulated wave speeds or impedances. What happens to f₀ and the acoustic first-gap edges? Has this made the gap deeply subwavelength relative to its own layer wavelengths?
Show the answer and reasoning
f₀ doubles to 200 kHz; the first-gap edges double to about 133.333 and 266.667 kHz. The thickness/wavelength ratios at corresponding frequencies stay the same.
Keeping δ_j=2πfd_j/c_j fixed requires frequency to double when d_j halves. The quarter-wave Bragg mechanism has been rescaled, not changed into a deeply subwavelength one. Real dispersion, losses or fabrication changes could invalidate the unchanged-property assumption.
At the same f₀ calculate T for N=2 using the supplied matched-boundary formula. What are R and absorbed fraction in this lossless model?
Show the answer and reasoning
T=4/(9+1/9)²≈0.0481856; R≈0.951814; absorbed fraction zero.
About 4.81856% is transmitted and 95.1814% reflected. A gap in the infinite mode model does not require zero finite transmission, and low T does not require dissipated power.
In the separate resonator, reduce k_r to one quarter while holding M and m fixed. Find f_r and the negative-inertia interval. Does this alone predict a complete finite-stack transmission spectrum?
Show the answer and reasoning
f_r halves to 50 Hz. The negative interval becomes 50<f<55.9017 Hz; no complete transmission spectrum follows.
f_r=(1/2π)√(k_r/m), so the frequency scale halves. The mass ratio and fractional negative interval stay the same. Host coupling, external boundaries and damping remain separate inputs; the bilayer spectrum is another model.
For M=0.0100 kg, m=0.00250 kg at f=125 Hz, find M_eff/M. Can its sign alone prove propagation in the stated chain? Compare host springs K_h=500 and 1000 N/m.
Show the answer and reasoning
M_eff/M=0.555556>0, but ω²M_eff≈3426.95 N/m. No real q exists for K_h=500 N/m; for K_h=1000 N/m, qa≈2.36524 radians is a real solution.
The chain requires 0≤ω²M_eff≤4K_h. At 125 Hz, M_eff≈0.00555556 kg and ω=2π×125 rad/s. The first host has maximum left side 2000 N/m, below 3426.95 N/m. The second has maximum 4000 N/m, so sin²(qa/2)=3426.95/4000 and qa=2arcsin√(3426.95/4000)≈2.36524. Thus host coupling can change propagation while the internal coefficient is the same. This still supplies no finite transmission or loss.
At one frequency an actual passive, nonconverting sample has complete collected power fractions R=0.990 and T=0.005, with uncertainties already small enough for this illustrative calculation. What absorbed fraction follows? Does that one point establish the same mechanism as the lossless three-period model?
Show the answer and reasoning
The supplied complete balance gives absorbed fraction 1−0.990−0.005=0.005. It does not establish a unique gap mechanism or a complete spectrum.
Most power is reflected, with 0.5% absorbed in this stipulated account. The ideal layer calculation has zero absorption, so matching a small T alone would not validate every model assumption. If collection were incomplete, the residual would also include uncollected channels.
A finite layered specimen has low transmission at normal incidence for one electromagnetic polarisation. Can that result establish an all-angle gap or a negative scalar refractive index?
Show the answer and reasoning
No. It tests the stated incidence, polarisation, specimen and frequency range only.
Oblique incidence changes phases, impedances and possible modes. Negative refraction needs direction-resolved phase and energy evidence plus a justified propagation model; a transmission dip is not that evidence.

What to carry forward

State modes, frequency and boundaries.
Use effective parameters within their model range.
Interface contrast and accumulated phase must be combined; the unit-cell mode equation and a finite spectrum answer different questions.
In a lossless finite model a stop band can return energy rather than absorb it. Count complete outgoing channels before assigning loss.
A negative apparent dynamic inertia can follow from positive constituents after eliminating internal motion; frequency, coupling and averaging assumptions define its meaning.

References

Teaching resources and research supporting this lesson’s explanations and examples.