EMCH 367 · Controls · Fall 2026
Two carts ride one shake table. Both carry the same mass; the first is held by a single spring–damper pair, the second by a pair on each side. They obey the same equation of motion — the second mount changes nothing about its form, only the two numbers that decide the answer: ωn and ζ.
The instrument
Both systems see exactly the same base motion u(t). System 1’s single mount and each of System 2’s two mounts are set independently, so you can make the second system symmetric, lopsided, or single-sided. Two inputs are available: a steady Sine, and a single violent Pulse — one lurch out and straight back. Both inputs are yours to set: the sine by frequency and amplitude, the pulse by length and height. Watch which system is quieter, then switch the input and watch that answer reverse.
The table itself reaches 25 mm and 0.09 g — the input both systems have to beat.
Both sides act across the same gap, so only the sums reach the equation: kL+kR = 160.0 kN/m · bL+bR = 2.6 kN·s/m
| SYSTEM | ωn | Tn | ζ | r | PEAK DRIFT | PEAK ACCEL |
|---|
The model
Base excitation differs from the force-driven cart in one respect: the input enters through the spring and the damper rather than being applied to the mass. That puts a zero in the transfer function — and it is why the second mount helps in one frequency band and hurts in another.
The spring carries k(x − u) and the damper b(ẋ − u̇), both opposing motion of the mass relative to the cart.
Move every term in x to the left and every term in the base motion to the right. The input appears twice, once through each element.
Applying ℒ term by term, ℒ[ẍ] = s²X(s) and ℒ[u̇] = sU(s), so
with the two parameters that decide everything that follows,
What strains the mount is relative travel z = x − u, not absolute displacement. Substituting x = z + u leaves a familiar second order system driven by ground acceleration:
The two transfer functions disagree about what a good mount is, and the comparison further down is where that disagreement gets settled.
Add a second spring and damper on the other side of the mass — not necessarily matched to the first. Every element still spans the same gap between cart and mass, so all four see the same relative motion and their forces simply add:
This is the same equation with k → kL+kR and b → bL+bR. Nothing about the form of the transfer function changes, so the transmissibility curve below still applies — the system has simply moved to a different place on it:
Only the sums appear. Two elements spanning the same pair of bodies are in parallel, so the response cannot tell kL = kR = 80 from kL = 150, kR = 10. Try it on the sliders above: drag stiffness from one side to the other keeping the total fixed and the traces do not move at all. What asymmetry does change is the static equilibrium the mass settles at, which this model measures from and therefore never shows — and it would matter dynamically the moment a spring went slack, or the mass were free to rotate rather than just translate.
With both sides matched to System 1’s mount, the sums are 2k and 2b, and both parameters rise by the same factor √2:
Doubling the mount is therefore not the same as doubling the damping: it buys stiffness and damping together, and stiffness moves r = ω/ωn down — which is an improvement only if it moves the system away from r = 1.
Frequency response
Evaluating the transmissibility at s = jω gives the curves below. Both systems live on this one chart and the two marked points track the sliders above. Because the input is quoted in rad/s alongside each ωn, r is simply the ratio of two numbers already in the table. With both of System 2’s mounts matched to System 1’s, it sits at exactly r₁/√2 on a curve √2 better damped; change any one slider and it moves wherever its own sums put it. In pulse mode the markers are placed at the pulse’s own frequency, 2π/Tp, which is where a single-cycle pulse carries most of its energy — a useful guide, though a transient is never a single frequency.
Rigid, r < 0.5. The mount is far stiffer than the shaking is fast, so the mass simply rides along with the cart. Transmissibility is near 1: little travel across the mount, but the mass feels the full ground acceleration.
Resonant, r ≈ 1. The peak reaches ≈ 1/(2ζ) — a factor of 10 at ζ = 0.05. This is the band a mount must not be tuned into.
Isolated, r > √2. Past the crossover at r = √2 — where every curve passes through exactly 1, whatever the damping — transmissibility falls below one and the mass moves less than the ground. Notice that in this region more damping makes the isolation worse, because a damper is a stiff path at high frequency.
Where the second mount lands you. Adding it multiplies ωn by √2, so it divides r by √2 and slides the system left. Starting from r > √2 that is a move out of the isolated band and towards resonance — the extra mount makes things worse. Starting from r ≈ 1 it is an escape from resonance and helps a great deal. There is one exact coincidence worth remembering: when System 2 sits precisely at resonance, System 1 is at r = √2, its own crossover, passing the ground through untouched.
Three mountings, one shock
The same 2000 kg mass, and the same 140 mm pulse driving all three at once: one spring–damper pair, two pairs, and a single much softer pair. The dashed markers are the ground’s own peaks — a bar past its marker means that mounting made things worse than bolting the mass down solid.
millimetres · the stroke the spring and damper must allow
g · what the payload feels
| MOUNTING | r | PEAK TRAVEL | PEAK ACCEL | VS. GROUND ACCEL |
|---|
A short shock punishes the second mount. The pulse lasts 0.35 s, about a third of the single mount’s own 1.05 s period, so the mass cannot follow it: of the ground’s 2.30 g, only 0.49 g reaches the mass. Adding the second pair lifts ωn to 8.49 rad/s, which walks the system up the rising flank of the response: what gets through goes from ×0.21 of the ground’s peak to ×0.55. Peak acceleration rises by a factor of 2.6, and the travel across the mount reaches 172 mm, more than the ground’s own 140 mm stroke. The mount is now amplifying the motion it was installed to absorb. Note how little of the change came from the extra damper: ζ rose only from 0.020 to 0.028. Almost all of it is the shift in ωn. Keep stiffening and it gets worse still: this pulse crosses ×1 — the mount transmitting more than a rigid bolt would — at about ωn = 10.7 rad/s, and peaks near ×3.6 at ωn ≈ 23.5 rad/s, where the pulse length is about 1.3 natural periods. For a shock, the whole design rule is pulse length against natural period.
Against a shock, the soft mount wins on both counts. At 2.0 rad/s the pulse is long over before the mass has begun to respond, so it feels 0.11 g — a twentieth of the ground. And unlike the harmonic case, it does not pay for that in stroke: because the mass stays nearly still, the travel across the mount is simply the ground’s own displacement, and it cannot exceed the 140 mm the table moved. Switch the table above to Sine at ω = 6 rad/s and the ordering between the first two mountings inverts. Neither result is wrong — they are answers to different questions, and which one you design for is the actual engineering decision.
Exercises
Soften all three stiffness sliders together to 10 kN/m — System 1’s k, and both of System 2’s, so the second system stays exactly twice the first. System 1 lands at r = 2.68, out in the isolated band, while System 2 is dragged back to r = 1.90. The doubled mount is now the worse one: 33 mm of travel against 29 mm, and 0.04 g against 0.02 g. Explain it from ωn₂ = √2 ωn₁ alone.
Starting from Reset, so both of System 2’s mounts match System 1’s, raise ω until System 2 reads r = 1.00. That happens at ω = 8.9 rad/s, the slider’s closest step to that system’s own ωn = 8.94 rad/s, and being able to read that straight off the table is the whole reason for quoting the input in rad/s. Check that System 1 now reads 1.41, its own crossover, where |X/U| = 1 exactly for every ζ. On the strip chart its red trace matches the blue ground trace in height — but not in phase, which is why the travel across its mount still reads about 50 mm rather than zero.
On Sine at ω = 6.0 rad/s and 25 mm, System 2 travels about 20 mm against System 1’s 161 mm. Switch to Pulse and System 2 becomes the worse one, at 166 mm and nearly 1.4 g against 0.55 g. Explain the reversal using the pulse length against each system’s natural period.
Set kL = 150 and kR = 10, then 30 and 130, then 160 and 0. The total never changes and neither does the trace. Now argue from the free body diagram why the split cannot matter, and name one change to the model that would make it matter.
From Reset, note both peak drifts, then drag the ground amplitude from 25 mm to 50 mm. Every drift and every acceleration doubles, and ωn, ζ and r do not move at all. That is linearity, and it is why the whole design question lives in the ratio of response to input rather than in either one alone — including on Pulse, where a taller pulse raises the ground’s peak acceleration and each system’s in exact proportion.