Dynamic measurements, model identification and response limits

Displacement, velocity and acceleration emphasise different motion

A displacement sensor reports motion relative to its stated reference; an accelerometer reports a directional acceleration response within its calibrated behaviour. These are different quantities. For a sinusoid of displacement amplitude X and angular frequency ω, the acceleration amplitude is ω²X. Equal displacement at higher frequency can therefore mean substantially more acceleration.

Report whether a quoted amplitude is peak, peak-to-peak or root-mean-square. A centred pure sinusoid has peak-to-peak 2X and RMS X/√2, but those conversions do not apply unchanged to arbitrary waveforms. Sensor saturation, insufficient bandwidth or an unreported filter can distort an apparently precise record.

Samples do not uniquely reveal motion between them

Sampling rate fs means samples per second here. Frequencies separated by a sampling-related ambiguity can produce the same sample sequence: aliasing. For an ideally band-limited signal, the sampling rate must exceed twice its highest frequency to allow reconstruction. Real measurement needs appropriate anti-alias filtering before sampling and practical margins; merely drawing more interpolated points afterward adds no information.

The duration of the record also matters. Several cycles can help estimate a period, while a short record may mix transient and steady behaviour. Preserve the time base, filter settings, trigger and synchronisation between force and response. A phase comparison is unreliable if one channel has an unknown delay.

A fitted parameter needs a model and independent challenge

Identifying a natural frequency or effective damping means selecting model parameters from response evidence. A dominant peak can reflect the excitation or sensor as well as a structural mode. Check which point and direction were observed and whether the excitation was known. Additional mass, flexible supports and cross-axis motion can change the interpretation.

Calibration against a known dynamic reference addresses the measurement chain; it does not automatically validate a mechanical model. Fitting a response and testing it against separate loads or frequencies are different activities. Keep response limits tied to the actual requirement, not just a spectral peak or a successful fit at one condition.

Turn a response record into a candidate parameter

Consider a new constructed free-response record, separate from the aliasing case. A displacement sensor observes one direction of a 2.00 kg effective moving mass after release. Successive positive peaks occur at 0.00, 0.50 and 1.00 s. For this teaching exercise, the acquisition chain is assumed to preserve the motion over the relevant band, the motion is dominated by one mode, and damping is small enough that the observed period can approximate the undamped period. Those are conditions to examine, not conclusions proved by three peaks.

The time between successive positive peaks is 0.50 s, giving f≈1/0.50=2.00 Hz and ω≈2πf=4π rad/s. Invert the earlier relation ωn=√(k/m): the candidate stiffness is k≈m(2π/T)²=2.00×(4π)²≈316 N/m. This is parameter identification: response evidence and a selected model have supplied a parameter, rather than merely showing a spectral peak.

Suppose the period estimate has a supplied bound of ±0.01 s, while mass is treated as exact for this exercise. The two endpoint periods are 0.49 and 0.51 s. Because k is proportional to 1/T², the shorter period gives the larger stiffness: about 329 N/m at 0.49 s and 304 N/m at 0.51 s. These are conditional bounds from the stated inputs, not a measured confidence interval or a complete uncertainty budget. Real mass, timing, damping and model uncertainty would also need attention.

Now make a prediction before looking at a separate comparison. If the effective mass becomes 8.00 kg while the same stiffness and one-mode approximation remain valid, the mass is four times larger and the period should double to about 1.00 s. The original period bounds imply 0.98–1.02 s under that changed-mass model. A separate constructed record instead has period 0.80 s with an assumed bound ±0.01 s, or 0.79–0.81 s. Those intervals do not overlap. The original candidate fails this particular comparison under the assumed bounds.

The mismatch does not tell us uniquely what changed. A mounting or stiffness change, an incorrect effective mass, another mode, or a measurement error could matter. Refitting k to the second record would be a new identification step, not independent confirmation of the first fit. Retain the first prediction and disagreement. Even agreement in period would leave amplitude, decay, direction and other operating conditions to check. All records here are invented to teach this reasoning; they do not validate a physical assembly.

Expose a sampling ambiguity

Compare x=cos(2π×6t) mm with x=cos(2π×2t) mm, sampled at fs=8 samples/s. At t=0, 0.125, 0.250, 0.375, 0.500 s, both give 1, 0,−1, 0, 1 mm. The solid and dashed curves differ between the sample circles. An analysis using only those samples cannot establish which continuous signal generated them.

With amplitude 1 mm=0.001 m, the 6 Hz acceleration amplitude is(2π×6)²×0.001≈1.42 m/s². The 2 Hz value is about 0.158 m/s², nine times smaller. Mistaking the aliased frequency therefore changes a derived acceleration substantially. A suitable acquisition design must prevent the ambiguity; post-processing the same sparse record cannot recover the missing evidence.

Try the reasoning

1. Why cannot increasing display smoothness repair aliasing?

   Show the answer and reasoning

   Interpolated display points do not add the information missed during acquisition.

2. Are 1 mm peak and 1 mm RMS the same sinusoidal motion?

   Show the answer and reasoning

   No. For a centred sinusoid, RMS is peak/√2, so the amplitude convention changes the meaning.

3. Does a well-calibrated accelerometer by itself prove a one-mode structural model?

   Show the answer and reasoning

   No. Excitation, location, direction, bandwidth and the model’s omitted modes still need evaluation.

4. In the constructed first record, someone uses 0.25 s between a positive and a negative extreme as the full period. What stiffness error results?

   Show the answer and reasoning

   Opposite extremes are only half a cycle apart. Using 0.25 s would double the inferred frequency and multiply k=m(2π/T)² by four, giving about 1263 N/m instead of 316 N/m. Read corresponding peaks or account for the fraction of a cycle.

5. Keep the first candidate stiffness but use an effective mass of 4.50 kg. What period is predicted, and is fitting that new period again an independent check?

   Show the answer and reasoning

   The mass ratio is 4.50/2.00=2.25, so period is multiplied by √2.25=1.5, giving 0.75 s. Comparing that prior prediction with a suitable separate record can challenge the model. Choosing k again from that same new record is fitting, not independent confirmation.

What to carry forward

A dynamic result needs a defined quantity, calibration, time base, bandwidth and model scope.