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). The sliders set k and b per element, so System 2 always carries twice the stiffness and twice the damping of System 1. Two inputs are available: a steady Sine, and a single violent Pulse — the table lurches 140 mm and comes straight back, peaking at 2.3 g. Watch which system is quieter, then switch the input and watch that answer reverse.
| SYSTEM | ωn | fn | 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 an identical spring and damper on the other side of the mass. Both new elements span the same gap between cart and mass, so they see the same relative motion and simply add:
This is the same equation with k → 2k and b → 2b. 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:
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; the two marked points track the sliders above, and System 2 always sits at r₁/√2 on a curve √2 better damped. In pulse mode the markers are placed at the pulse’s own frequency, 1/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 1.0 s period, so the mass cannot follow it: of the ground’s 2.30 g, only 0.56 g reaches the mass. Adding the second pair lifts fn to 1.41 Hz, which walks the system up the rising flank of the response: what gets through goes from ×0.24 of the ground’s peak to ×0.64. Peak acceleration rises by a factor of 2.6, and the travel across the mount reaches 178 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 fn = 1.7 Hz, and peaks near ×3.4 at fn ≈ 3.4 Hz, where the pulse length is a little over one natural period. For a shock, the whole design rule is pulse length against natural period.
Against a shock, the soft mount wins on both counts. At 0.33 Hz 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 0.95 Hz 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
Drag the stiffness down until System 1 sits at r > 2. Now System 2 is the one closer to resonance and it moves more than System 1. Explain it from ωn₂ = √2 ωn₁ alone.
Tune the ground frequency until System 2 reads r = 1.00. Check that System 1 now reads 1.41 and that its transmissibility is exactly 1 — it is passing the ground straight through while its neighbour resonates.
On Sine at 0.95 Hz, System 2 travels about 22 mm against System 1’s 166 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.
At r ≈ 1, more damping always helps. Set the ground frequency so that r ≈ 3, then raise b and watch both amplitudes go up. Explain that from the bs + k numerator.