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System Dynamics · Module 2 of 10

Mechanical System Modeling

Most machines are masses, springs, and dampers in translation or rotation. Add a damper to the oscillator and the response decays; gear it down and the inertia it presents changes by the square of the ratio.

Readiness check

This module builds full mechanical models. Tick only what you can do closed-notes.

Write the spring-mass natural frequency ω_n = √(k/m).
Recall a damper force F = c v.
Draw a free-body diagram and sum forces.
Recall torque τ = J α for rotation.
Relate gear speeds through a gear ratio.

0 or 1 weak itemsContinue with this module.

2 weak itemsRevisit modeling and ω_n in Module 1.

3 or more weak itemsReview rigid-body kinetics in Dynamics and Vibrations, Module 8.

The core idea

A mechanical system is a free-body diagram turned into a differential equation. The mass-spring-damper has a natural frequency from stiffness and inertia and a damping ratio from the damper, and rotational systems follow the same form with torque, inertia, and angle.

m ẍ + c ẋ + k x = F(t)ω_n = √(k/m), ζ = c/(2√(k m))ω_d = ω_n√(1 − ζ²)

Mechanical modeling applies Newton's law to a free-body diagram: the sum of forces equals mass times acceleration. For the canonical mass-spring-damper this gives m ẍ + c ẋ + k x = F(t), a second-order equation whose two parameters carry all the dynamics. The natural frequency ω_n = √(k/m) is the undamped oscillation rate; the damping ratio ζ = c/(2√(km)) measures how fast the motion decays, and it decides whether the response is underdamped (ζ < 1), critically damped (ζ = 1), or overdamped (ζ > 1). An underdamped system oscillates at the damped frequency ω_d = ω_n√(1 − ζ²). Rotational systems use the identical form with inertia J, torsional stiffness, and rotary damping, and gears transform inertia by the square of their ratio.

The skill works when: you write the free-body equation and read ω_n and ζ straight from its coefficients.

The skill breaks down when: a gear ratio is applied to the first power instead of the square, or ζ is read before normalising the equation.

The concept. The damper adds the c ẋ term to the oscillator. Its size, relative to the mass and stiffness, sets the damping ratio and decides whether the free response rings, returns smoothly, or crawls back.

The skills, taught in order

Five skills build the translational model, its damping, and the rotational systems that gears connect.

2.1 Translational elements

Translational systems use three elements: a mass with F = m a, a spring with F = k x, and a damper with F = c v. Summing forces on each mass through a free-body diagram, with Newton's law, produces the governing differential equations.

2.2 The mass-spring-damper equation

For one mass the result is m ẍ + c ẋ + k x = F(t). Dividing by m puts it in standard form ẍ + 2ζω_nẋ + ω_n²x = F/m, from which the two characteristic parameters are read directly.

2.3 Damping ratio and damped frequency

The damping ratio ζ = c/(2√(km)) classifies the response. Underdamped systems (ζ < 1) oscillate and decay at ω_d = ω_n√(1 − ζ²); critically damped (ζ = 1) return fastest without overshoot; overdamped (ζ > 1) return slowly. Most machines are lightly underdamped.

Damping ratio	Class	Free response
ζ = 0	undamped	oscillates forever
0 < ζ < 1	underdamped	oscillates and decays
ζ = 1	critically damped	fastest return, no overshoot
ζ > 1	overdamped	slow, no oscillation

The four damping regimes. The damping ratio alone, not the absolute damper value, sets the character of the response.

2.4 Rotational systems

Rotation mirrors translation: inertia J replaces mass, torsional stiffness replaces spring stiffness, and rotary damping replaces the damper, with τ = J α. The governing equation J θ̈ + c θ̇ + k θ = τ(t) has the same form and the same ω_n and ζ.

2.5 Gears and reflected inertia

A gear train changes the inertia a motor sees. Reflecting a load inertia J_L through a gear ratio N (load speed reduced by N) divides it by N²: J_reflected = J_L/N². The square comes from energy conservation, and it is why gearing eases a motor's inertial load.

Engineering connection: sizing a motor for a geared load rests entirely on reflected inertia, and the damping ratio decides whether the machine settles cleanly or rings.

Worked example 1: damping ratio of a mass-spring-damper

A system has m = 1 kg, k = 100 N/m, and c = 4 N·s/m. Find the natural frequency, the damping ratio, and the damped frequency.

Figure 1. With ζ = 0.2 the response oscillates and decays under an exponential envelope. It rings at the damped frequency, slightly below the undamped natural frequency.

ProblemFind ω_n, ζ, and ω_d for the system in Figure 1.
Given / findm = 1 kg, k = 100 N/m, c = 4 N·s/m. Find ω_n, ζ, ω_d.
AssumptionsLinear elements; the model is the standard mass-spring-damper.
Modelω_n = √(k/m), ζ = c/(2√(km)), ω_d = ω_n√(1 − ζ²).
Equationsω_n = √(k/m)ζ = c/(2√(k m))ω_d = ω_n√(1 − ζ²)
Solveω_n = √(100/1) = 10 rad/s. ζ = 4/(2√(100 × 1)) = 4/20 = 0.2. ω_d = 10√(1 − 0.04) = 10√0.96 = 9.80 rad/s.
Checkζ = 0.2 is well below 1, so the system is underdamped and must oscillate, consistent with a real ω_d. The damped frequency sits just under ω_n, as light damping requires.
ConclusionThe system rings at 9.80 rad/s while decaying. Its damping ratio of 0.2 is typical of a lightly damped structure.

Result. ω_n = 10 rad/s, ζ = 0.2, ω_d = 9.80 rad/s.

Worked example 2: reflected inertia through a gear

A motor of inertia J_m = 0.01 kg·m² drives a load of inertia J_L = 0.4 kg·m² through a gear ratio N = 5 (the load turns five times slower). Find the total inertia the motor sees.

Figure 2. The gear divides the load inertia by the square of the ratio before adding it to the motor's own. A 25-fold reduction turns a 0.4 load into 0.016 at the motor shaft.

ProblemFind the equivalent inertia at the motor shaft for the drive in Figure 2.
Given / findJ_m = 0.01 kg·m², J_L = 0.4 kg·m², N = 5. Find J_eq.
AssumptionsIdeal, rigid, lossless gears; the load turns N times slower than the motor.
ModelReflect the load inertia by dividing by N², then add the motor inertia.
EquationsJ_reflected = J_L/N²J_eq = J_m + J_L/N²
SolveJ_L/N² = 0.4/25 = 0.016 kg·m². J_eq = 0.01 + 0.016 = 0.026 kg·m².
CheckThe reflected load (0.016) is comparable to the motor's own inertia (0.01), a well-matched drive. Without gearing the motor would face the full 0.4, twenty-five times larger.
ConclusionGearing reduces the inertia a motor must accelerate by the square of the ratio, which is why drives are geared and why the square matters.

Result. J_eq = 0.026 kg·m² at the motor shaft.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Reading ζ before normalising	Damping ratio off by a factor of m	"Did I divide the equation by m?"	Put it in ẍ + 2ζω_nẋ + ω_n²x form first.
Gear ratio to the first power	Reflected inertia too large by N	"Is the inertia divided by N or N²?"	Reflected inertia goes as 1/N².
Damped equals natural frequency	ω_d reported equal to ω_n	"Is there any damping?"	ω_d = ω_n√(1 − ζ²), below ω_n.
Wrong damping class	Oscillation predicted for ζ > 1	"Is ζ above or below 1?"	Only ζ < 1 oscillates.

Practice ladder

Level 1 · Direct skill

A system has m = 2 kg, k = 50 N/m, c = 0. Find ω_n and the damping class.

Show answer

ω_n = √(50/2) = 5 rad/s. With c = 0, ζ = 0: undamped, oscillates forever.

Level 2 · Mixed concept

Now add c = 10 N·s/m to that 2 kg, 50 N/m system. Find ζ and the damping class.

Show answer

ζ = c/(2√(km)) = 10/(2√100) = 10/20 = 0.5. Underdamped: it oscillates and decays.

Level 3 · Independent problem

A motor (J_m = 0.02 kg·m²) drives a load (J_L = 0.9 kg·m²) through N = 3. Find the inertia at the motor and compare it to the ungeared load.

Show answer

J_eq = 0.02 + 0.9/9 = 0.02 + 0.1 = 0.12 kg·m². The ungeared load (0.9) is 7.5 times larger, so gearing greatly eases the motor.

Transfer task | Real engineering

A positioning stage overshoots and rings after each move. Using ζ, describe one change to the mechanism that would reduce the ringing, and its trade-off.

What good work looks like

Raise the damping (larger c) to increase ζ toward 1, reducing overshoot; the trade-off is a slower response. Tying the fix to ζ, and naming the speed cost, is the point.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I normalised the equation before reading ζ."

"Give me three damper values; I will classify the response of each."

"Solve this mechanical system." Writing the free-body equation is the skill.

"What is the reflected inertia?" Applying the 1/N² rule yourself is the point.

Portfolio task

Model a real mechanism as a mass-spring-damper or its rotational form, compute ω_n and ζ, and predict whether it rings, then compare to how it actually behaves.

Must include: the governing equation, ω_n, ζ, a predicted damping class, and an observation.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the mass-spring-damper equation.

m ẍ + c ẋ + k x = F(t).

2. Give ω_n and ζ from its coefficients.

ω_n = √(k/m) and ζ = c/(2√(km)).

3. What is the damped frequency?

ω_d = ω_n√(1 − ζ²), always below ω_n.

4. How does a gear reflect inertia?

Divided by the square of the ratio: J_L/N².

5. Which damping class returns fastest without overshoot?

Critically damped, ζ = 1.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive ω_n and ζ from a free-body diagram.

+3 daysModel three new mechanisms.

+7 daysCarry the same form into electrical systems, Module 3.

+30 daysReuse reflected inertia when sizing a geared motor.

Textbook mapping

This module follows Karnopp, Margolis, and Rosenberg, System Dynamics: Modeling, Simulation, and Control of Mechatronic Systems, 5th edition. Use these references to read further.

Topic in this module	Where to read more
Translational and rotational elements	Karnopp, Margolis & Rosenberg, Chapter 2
The mass-spring-damper and damping ratio	Karnopp, Margolis & Rosenberg, Chapter 3
Gears and reflected inertia	Karnopp, Margolis & Rosenberg, Chapter 4

Chapter numbers refer to the 5th edition. The mechanical relations are standard, so any recent edition will align closely.