Scale, bandwidth, loss and constituent limits

An unusual response has a working window

Two resonant units can have the same natural frequency but different useful ranges. One moves strongly at its peak yet exceeds a power budget near that peak; another moves less but meets a required frequency interval. To choose between them, first define the response, drive and requirement. A picture of a large peak is not enough.

Cell geometry, constituent response, dissipation and manufacturing variation set the useful range. A dynamic effect can shift when scaled because wave lengths and resonances change. A statically similar shape need not have the same frequency response.

Acoustic and electromagnetic wave responses derives a periodic-mode and finite-transmission example; Reconfigurable structures and instabilities follows a changing equilibrium. Here the focus is a different question: how damping, size and variation limit a measured dynamic response. The following one-mode calculation is not a complete band-gap or assembly model.

Specify a driven mode before comparing peaks

Represent one small-motion structural mode by a positive effective mass m, positive stiffness k and positive viscous damping coefficient b. The displacement x is measured relative to a fixed support, and a force F(t)=F₀ cosωt acts in that coordinate. Use m ẍ+b ẋ+kx=F₀ cosωt, where m is in kg, k in N/m and b in N·s/m. The force amplitude F₀ is held constant as frequency is changed.

This is a linear, time-invariant approximation over a stated frequency and amplitude range. Transients have decayed; there is one relevant mode, the support is fixed, and no contact, changing stiffness, instability or other external output-work channel is included. The effective modal quantities are not a claim that every member or constituent moves uniformly.

Write ω₀=√(k/m), f₀=ω₀/(2π), r=ω/ω₀=f/f₀ and ζ=b/(2√(km)). Here ω is angular frequency in rad/s and f is frequency in Hz. ζ is a dimensionless damping ratio. Using complex amplitude x̂ with x(t)=Re[x̂e^(iωt)], substitution gives x̂=F₀/(k−mω²+i bω), where i²=−1. Its phase matters, as well as its magnitude.

The displacement amplitude is X=|x̂|=(F₀/k)/√[(1−r²)²+(2ζr)²]. F₀/k is the static displacement scale. At r=1, X=(F₀/k)/(2ζ). For ζ<1/√2 the maximum displacement is actually at r_X=√(1−2ζ²), slightly below 1 for weak damping. A peak identified from displacement is therefore not automatically at the same frequency as every other response peak.

Use original hypothetical inputs m=0.0100 kg, k=1000 N/m and F₀=0.0100 N. Then f₀≈50.3292 Hz and F₀/k=10.0 μm. Compare ζ=0.025 and ζ=0.10, obtained with b≈0.158114 and 0.632456 N·s/m respectively. These are exact teaching-model choices, not fitted properties of named materials.

Define which curve has a half-power width

Multiply the equation of motion by ẋ. The mass and spring terms are the time derivatives of ½m ẋ² and ½kx². Over a complete steady cycle their average changes are zero, so the mean power supplied by the force equals the dashpot loss: P̄=b⟨ẋ²⟩=½bω²X²≥0. This is a power in watts, not an efficiency or a wave transmission fraction.

Substituting X gives P̄=F₀² bω²/[2((k−mω²)²+(bω)²)]. For this constant-force, constant-b model the maximum occurs exactly at ω=ω₀, with P_max=F₀²/(2b). Increasing b lowers that peak because the motion driven by the same force becomes smaller. If X were prescribed instead, ½bω²X² would increase with b at fixed ω and X. The drive condition belongs with any damping comparison.

For actual mean dissipated power, P̄/P_max=4ζ²r²/[(1−r²)²+4ζ²r²]. Half maximum requires (1−r²)²=4ζ²r². The two positive frequency ratios are r_−=√(1+ζ²)−ζ and r_+=√(1+ζ²)+ζ. Thus Δf=(r_+−r_−)f₀=2ζf₀ exactly for this model. A broader power curve accompanies a lower peak under the fixed-force comparison.

This definition uses half of the maximum P̄, not half the displacement, half the displacement squared or half an unspecified measured signal. Because P̄ includes ω²X², its half-power points generally differ from half-maximum points of X². If another quantity is called “bandwidth”, state its threshold and normalisation.

The ζ=0.025 power peak is about 316.228 μW, with half-power frequencies 49.0867 and 51.6032 Hz, a width 2.51646 Hz. The ζ=0.10 peak is about 79.0569 μW, with half-power frequencies 45.5473 and 55.6132 Hz, a width 10.0658 Hz. None of those widths alone tells whether a specified displacement and power budget are met.

Screen the full frequency interval against the function

For an invented motion-readout requirement, demand X≥40.0 μm and P̄≤100 μW at every frequency in 0.95≤r≤1.05. Equality is permitted, and the arithmetic treats the model inputs as exact. The power limit is an illustrative budget, not a measured temperature or a material allowable. The operating window is the intersection of the frequencies passing both requirements.

At r=1, the lightly damped unit moves 200 μm but dissipates 316.228 μW, so it fails the power limit despite its large displacement. The moderately damped unit moves 50.0 μm and dissipates 79.0569 μW. At the required endpoints it moves about 46.8261 and 42.7936 μm; its largest displacement inside the interval is about 50.2519 μm. All exceed the minimum, and its power nowhere exceeds its 79.0569 μW maximum. It passes the supplied nominal interval.

The first figure shows both complete responses and the resulting accepted frequency sets. Solving X≥40 μm for ζ=0.025 gives 0.868215≤r≤1.115214. Its power budget excludes the open interval about 0.963914<r<1.037437, leaving two separated useful flanks. A large response at each end of the required band does not repair the failed centre.

For ζ=0.10 the power peak is below the budget at every frequency. Its displacement condition gives 0.910315≤r≤1.063639, one useful interval containing the nominal requirement. That useful interval and the half-power interval answer different questions; their boundaries need not agree. Changing the minimum response, the budget or the drive can change the selection; no universal optimum damping has been established.

Geometric similarity must include the damping and drive

Let s be a uniform multiplier on every length of a geometrically similar elastic cell. Assume unchanged density ρ and elastic modulus E over the new frequency range, the same dimensionless mode shape and the same support type. Then modal mass scales as m_s=s³m. A geometrically similar elastic stiffness scales as k_s=s k: for an axial member EA/L, area scales as s² and length as s; a bending stiffness EI/L³ has the same s factor because I scales as s⁴. This is not the result of changing only one thickness.

It follows that ω₀,s=ω₀/s. Uniform enlargement lowers the natural frequency; shrinking raises it. To preserve the damping ratio, however, b_s must scale as s²b because √(k_s m_s)=s²√(km). Unchanged constituent identity alone does not ensure that damping law. Compare response at the corresponding scaled frequency, not at an unchanged dimensional frequency.

To preserve X divided by cell size at the corresponding r and ζ, also use F₀,s=s²F₀. Then the static scale F₀,s/k_s is s times the original, so X_s=sX. Under those particular similarity conditions P̄_s=s²P̄. If the relevant material volume scales as s³, mean power per that volume scales as 1/s. This is a conditional power-density comparison; surface heat transfer and thermal conduction also change, so no temperature rise follows from it alone.

Consider a different damping mechanism. An isolated sphere moving slowly in a large Newtonian fluid has quasisteady Stokes drag b=6πηR, where η is dynamic viscosity and R is sphere radius. If its drag dominates and R scales as s in an unchanged fluid, b_s=s b instead of s²b. With the same structural m_s and k_s, ζ_s=ζ/s. This case additionally assumes negligible convective and oscillatory fluid inertia, negligible added-fluid-mass correction and no important nearby-wall effect. A real oscillating structure must satisfy those assumptions before using that drag law.

Start from the ζ=0.10 unit. At s=0.10 the ideal similarity choice has ζ=0.10, but the stipulated Stokes-dominated choice has ζ=1.00. Both have m_s=10⁻⁵ kg, k_s=100 N/m, f₀,s≈503.292 Hz and the scaled force amplitude 0.000100 N. At their natural frequency the similarity case moves 5.00 μm; the drag case moves only 0.500 μm. The latter also has no displacement resonance peak above its low-frequency response. A broadened dissipated-power curve has not preserved the original dimensionless function.

A nominal band can fail a bounded population

Manufacturing and material variations can shift a resonance as well as change its loss. From f₀∝√(k/m), a first-order small-change estimate is Δf₀/f₀≈½(Δk/k−Δm/m). A shared geometric change can correlate k and m, so independently adding their percentage effects or root-summing them without a statistical model can be wrong. This linear sensitivity is not an exact rule for large changes or altered mode shapes.

Use a separate supplied population to test a margin. All units have k=1000 N/m, the same force amplitude and ζ=0.10, but their natural frequencies f_i lie anywhere from 0.98f₀ to 1.02f₀, where f₀ is the original nominal value. Their parameters are consistent: m_i=k/(2πf_i)² and b_i=2ζk/(2πf_i). This is a stipulated correlated family with the same static compliance and damping ratio, not a distribution inferred from unspecified printing tolerances.

For each unit the displacement passes when 0.910315≤f/f_i≤1.063639. To guarantee that all supplied units pass, intersect those dimensional intervals. The largest lower edge is 0.910315×1.02≈0.928521 times nominal f₀; the smallest upper edge is 1.063639×0.98≈1.042366 times nominal f₀. The second figure shows the nominal curves and this reduced common window.

The family’s largest possible power peak is only 1.02×79.0569≈80.6381 μW, still below the budget. Yet the full required interval 0.95–1.05 of nominal f₀ is not guaranteed: at its upper end, the f_i=0.98f₀ unit has r_i=1.05/0.98 and X≈38.4019 μm. The nominal unit passed there, but this permitted unit fails the minimum. Shortening the requirement to 0.95–1.04 would pass this supplied family under the model.

A population spectrum can also mix different resonances and phases. An averaged broad envelope is not automatically a single unit’s damping linewidth, nor does it prove that every unit works over that envelope. The observed coordinate, averaging rule, manufacturing record and parameter variation belong with a bandwidth claim.

The scaled mode still depends on its constituents

The scaling above holds E, density and the chosen damping description fixed in their specified roles. Real material response can depend on frequency, temperature, strain amplitude and history. For linear uniaxial harmonic response, a complex modulus E*(ω)=E′(ω)+iE″(ω) separates storage and loss under the e^(iωt) convention. With ε=ε_a cosωt, stress is E′ε_a cosωt−E″ε_a sinωt; a passive loss component E″≥0 gives positive net work per cycle.

For a separate uniform-strain material element, dissipated work per volume is W_d/V=πE″ε_a² and mean power per volume is P̄/V=(ω/2)E″ε_a². These are not extra losses to add blindly to the dashpot model: constituent loss may already contribute to its effective b. A real architecture also has a nonuniform strain field, so the local strain can differ greatly from displacement divided by an overall cell size.

As an original illustrative material-point calculation, take E″=0.20 GPa, ε_a=0.001 and f=50 Hz. It gives 628.319 J/m³ per cycle and 31.4159 kW/m³ mean dissipated power. At 500 Hz with the same E″ and strain, the power density would be ten times larger. If separately supplied measurements at that frequency instead gave twice E″, it would be twenty times larger. None of this supplies a temperature without a heat-flow account.

A shifted resonance can sample a different E′ and E″, so the original k and b may no longer be valid constants. A single-mode fixed-b half-power formula then need not describe the full curve. Strong local motion can also activate nonlinear stiffness, contact, yield, fatigue or heating before a linear model’s large resonance is reached. An ideal divergence as loss tends to zero is a warning that omitted effects matter, not a material performance prediction.

Compare complete performance

Report bandwidth, peak response, losses, mass and boundary conditions. Improving a narrow peak can worsen useful integrated performance. Include fabrication defects and finite sample size when transferring an ideal periodic calculation to a real component.

Measure or model the intended function with the same drive amplitude, coordinate, supports, temperature and specimen size. A finite array can have boundary resonances, leakage, coupling and localised modes absent from a single-cell calculation. Increasing repeat count is not automatically equivalent to averaging independent copies of this oscillator.

Agreement between an analytical model and a finite-element calculation checks consistency under their assumptions; it does not by itself measure dissipation or validate every material parameter. A fabricated geometry also needs an actual response record. A small transmitted amplitude or a large internal displacement alone does not establish absorbed power. The complete power balance and outgoing channels matter.

An electromagnetic design may likewise have material dispersion and loss, but the mechanical oscillator’s damping law cannot simply be assigned to it without a justified field model. Use the previous wave-response lesson for its specified mode and boundary calculation. The present examples establish conditional comparisons and a way to test margins; they do not establish a universal best bandwidth, a material bound for every architecture or a qualified device.

Worked examples

Try the reasoning

1. Does a high peak response establish broad bandwidth?

   Show the answer and reasoning

   No.

   Peak magnitude and frequency range are distinct.

2. Why include losses?

   Show the answer and reasoning

   They affect attenuation, efficiency and heating.

   An ideal lossless model may misstate usable performance.

3. Enlarge all lengths of the base ζ=0.10 unit by s=2 under the same-ζ similarity assumptions. Scale force to preserve displacement divided by cell size. Find m_s, k_s, f₀,s, b_s, F₀,s and the natural-frequency displacement. Does keeping the old 50.3292 Hz drive preserve r?

   Show the answer and reasoning

   m_s=0.0800 kg; k_s=2000 N/m; f₀,s≈25.1646 Hz; b_s≈2.52982 N·s/m; F₀,s=0.0400 N; X at the new f₀ is 100 μm. The old drive gives r=2, not 1.

   Use m_s=s³m, k_s=s k, f₀,s=f₀/s, b_s=s²b and F₀,s=s²F₀. The static displacement becomes 20 μm and X=20/(2×0.10)=100 μm. A frequency comparison must scale the drive as well as the geometry.

4. A ζ=0.10 unit is driven at its natural frequency. Its force amplitude doubles from 0.0100 to 0.0200 N while the linear model parameters stay fixed. Does it still pass the 40 μm minimum and 100 μW maximum there?

   Show the answer and reasoning

   It moves 100 μm but dissipates about 316.228 μW, so it fails the power budget.

   Displacement is linear in F₀ and mean loss is quadratic in F₀. Doubling force doubles 50 μm and multiplies 79.0569 μW by four. More response does not necessarily make the unit acceptable; the larger amplitude must also remain within the assumed linear regime.

5. At a fixed ω and prescribed displacement amplitude X, a controller doubles b while holding X fixed. What happens to mean dissipated power? Why does that differ from the falling fixed-force power peak?

   Show the answer and reasoning

   The mean power doubles. With prescribed X, the controller must supply the larger damping loss; under fixed force the motion would instead fall.

   At prescribed X use P̄=½bω²X². At the natural-frequency peak under prescribed F₀ use P_max=F₀²/(2b). These comparisons impose different input conditions and therefore do not contradict each other.

6. A small stiffness change is +2% and a mass change is −2%. Compare the first-order natural-frequency change with the exact change in √(k/m). Then consider a different correlated change: both k and m increase by exactly 2%.

   Show the answer and reasoning

   The first case gives about +2% to first order and +2.0204% exactly. Equal +2% changes in k and m leave f₀ unchanged exactly.

   The sensitivity gives ½(0.02−(−0.02))=0.02. The exact ratio is √(1.02/0.98)≈1.02020406. If both scale by 1.02, their ratio cancels. A list of separate tolerances is insufficient to specify their correlations or a population spectrum.

7. Two independently measured complex displacement amplitudes at the same frequency are 100 μm at phase −30° and 100 μm at phase −150°. A reported coordinate is defined as their arithmetic complex average. What is its amplitude, and is it the average of the two amplitudes?

   Show the answer and reasoning

   The complex average is −i50 μm, with amplitude 50 μm. The average of the two magnitudes is 100 μm, a different quantity.

   The real parts ±100 cos30° cancel; both imaginary parts are −50 μm, so their average remains −50 μm. The measured or averaged coordinate must be defined before interpreting its envelope. This arithmetic does not model mechanical coupling or determine energy from the average coordinate alone.

8. For the separate uniform material point, E″=0.20 GPa and ε_a=0.001 at 50 Hz. First double ε_a without changing E″ or f. In another comparison keep the original strain but use 500 Hz with a separately supplied E″=0.40 GPa. Give the loss-power changes and explain whether they can simply be added to an already fitted modal b.

   Show the answer and reasoning

   Doubling strain gives four times the loss, about 125.664 kW/m³. The higher-frequency, doubled-E″ case gives twenty times the original loss, about 628.319 kW/m³. Adding either to a fitted modal loss could double-count the same dissipation.

   Use P̄/V=πfE″ε_a². The first multiplier is 2²=4; the second is 10×2=20. Local strain and the frequency-dependent loss modulus need their own evidence, and the modal damping may already include that constituent mechanism.

9. An instrument reports a width at half the maximum of X². May that width automatically be labelled the half-power width of this dashpot?

   Show the answer and reasoning

   No. The actual dashpot power is proportional to ω²X², not just X².

   Its peak and half-maximum frequencies must be determined for that actual measured quantity. Near a narrow resonance, replacing ω² by a constant can be an approximation, but it must be justified and labelled rather than used as an exact identity.

What to carry forward

• Report the entire operating window.
• Check scale transfer and real losses.
• State the drive, observed coordinate and threshold before comparing peaks or bandwidths.
• A usable frequency range is the intersection of all supplied requirements; separated passing flanks do not cover a failed centre.
• Geometric frequency scaling does not ensure similar damping, forcing, constituent response or temperature.
• A robust common band is an intersection over the stated variation, not an averaged broad envelope.