Module 3 · Lesson 3.1
The equation of motion
Four forces, one equation, and three routes to it — Newton, D'Alembert and energy — which must agree because they describe the same thing.
Why this matters
Everything in the rest of this course is a way of solving one equation. Free vibration is it with F = 0. Resonance is it with F harmonic. Newmark's method is it discretised. Modal analysis is many copies of it. The earthquake problem is it with the force replaced by −m üg.
So it is worth deriving carefully, once, and understanding what every term means physically — rather than meeting it as a formula and hoping.
By the end of this lesson you should be able to
- Derive the equation of motion by two independent routes
- Say what each of the four forces opposes
- Check the equation dimensionally
- Set up the equation for a physical structure, including the units conversion
What you should already know
- Newton's second law and free-body diagrams
- D'Alembert's principle (Module 1)
- Stiffness as force per unit displacement
The four forces
Take a mass m on a spring of stiffness k and a dashpot of coefficient c, displaced x from equilibrium and moving. Four forces act on it.
The applied force, F(t). Whatever is being done to the structure from outside. It may be zero.
The stiffness force, kx. The spring resists displacement and pushes back towards equilibrium. It depends on WHERE the mass is and not at all on how fast it is going.
The damping force, c ẋ. Resists velocity, and depends on how fast the mass is moving and not at all on where it is. At the extremes of the swing it is zero; passing through equilibrium it is largest.
The inertia force, m ẍ. Resists acceleration. By D'Alembert's principle we treat −m ẍ as a force acting on the mass, which lets us write equilibrium.
The crucial point is that these three internal forces peak at three different instants. In free vibration the spring force is largest at the extremes, the damping force at the centre, and the inertia force at the extremes too — but with the opposite sign to the spring force, which is exactly why they can balance.
Why it is written as equilibrium
Students often meet the equation as a rearrangement of F = ma and find the sign conventions confusing. It reads better the other way round.
A structure in motion is in equilibrium at every instant, provided the inertia force is included among the forces. Written that way:
inertia force + damping force + stiffness force = applied force
Every term is a force in newtons. Every free-body diagram you have ever drawn still works. Nothing about statics has been abandoned; one more force has been added to the list.
From first principles
The equation of motion for a single-degree-of-freedom system
We want to show: Derive m ẍ + c ẋ + kx = F(t) from Newton's second law, then obtain the same result from D'Alembert's principle, and confirm the two agree.
A mass is being pushed by an external force and pulled back by a spring and a dashpot. Newton says the net force equals mass times acceleration. Rearranged, that becomes a statement that four forces sum to zero — and it is that second reading which lets every technique from statics carry over unchanged.
Try it
The mass, the spring, the damper and the four forces
Set the system moving and watch which force is doing the work at each part of the cycle.
Loading
Above 1 the response no longer oscillates.
- Displacement x (mm)
- Velocity x′ (mm/s ÷ 10)
- Acceleration x″ (mm/s² ÷ 100)
Show the numbers behind this plot
- ωn
- 6.28 rad/s
- fn
- 1 Hz
- Critical damping ccr
- 62.8 kN·s/m
- Regime
- underdamped
- Inertia force m x″
- -7.9 kN
- Damping force c x′
- -5.4e-18 kN
- Stiffness force k x
- 7.9 kN
- Applied force F(t)
- 0 kN
m x″ + c x′ + k x − F(t) = 0 kN — zero to within round-off. The four forces balance at this instant, as they do at every instant.
Things worth doing here
- Set ζ = 0 and release from rest. At the extremes the velocity is zero, so the damping force vanishes and the spring alone is balancing inertia. As it passes through zero the spring force vanishes instead.
- Raise ζ to 1 exactly. The mass returns to equilibrium in the shortest time possible without overshooting — that is what critical damping means, and it is not the same as 'heavily damped'.
- Raise ζ above 1. It returns more slowly, not faster. Extra damping past critical is a hindrance.
- Switch to harmonic forcing at β = 1 and watch the applied force stay in step with the DAMPING force, not the spring force. At resonance the load is doing nothing but feeding the damper.
What this shows: Dynamic equilibrium is not equilibrium of applied loads. Inertia, damping and stiffness forces sum to the applied force at every single instant — including instants when the applied force is zero.
Worked example
Setting up the equation of motion for a water tower
Given
- An elevated tank: 90 tonnes when full, carried on four steel columns
- Columns 12 m tall, fixed at the base, effectively fixed at the tank
- Each column has I = 4.16 × 10⁻⁴ m⁴, E = 210 GPa
- Damping estimated at 2% of critical
Find
The equation of motion for lateral movement, with every coefficient evaluated.
Assumptions
- The tank is rigid compared with the columns, so this is one degree of freedom
- The columns' own mass is neglected — it is a few per cent of 90 tonnes
- The water moves with the tank; sloshing is a separate problem and is ignored here
Practice
A mass of 4 000 kg is on a spring of stiffness 1.6 MN/m with a damping coefficient of 8 000 N·s/m. What is the damping ratio?
Practice
For the same system (m = 4 000 kg, k = 1.6 MN/m, c = 8 000 N·s/m), the mass is at x = 20 mm moving at 0.4 m/s with an acceleration of −7.6 m/s². What applied force F(t) is acting, in kN?
Check yourself
In free vibration, at the instant the mass passes through its equilibrium position, which force is zero?
Practice
A water tank of 45 tonnes sits on a frame whose measured period is 0.62 s. What lateral stiffness does that imply, in MN/m?
Practice
A single-storey frame has four columns, each contributing 12EI/h³. Take E = 205 GPa, I = 1.2 × 10⁻⁴ m⁴ and h = 3.6 m. What is the total lateral stiffness, in MN/m?
Check yourself
In the equation mü + cu̇ + ku = p(t), which term carries the energy that leaves the system permanently?
Worked example
Idealising a water tower as one degree of freedom
Given
- A tank of 260 tonnes carried on four steel legs
- Legs 14 m tall, each behaving as 12EI/h³ with E = 200 GPa and I = 2.4 × 10⁻⁴ m⁴
- A nearby rail line produces a periodic excitation near 0.9 Hz
Find
The natural frequency, and whether the rail excitation is a concern.
Assumptions
- Tank rigid relative to the legs; leg mass negligible against the tank
Worked example
Frequency and period from stiffness and mass
Given
- A water tank on a single braced tower
- Lateral stiffness 3.2 × 10⁶ N/m
- Mass of tank and contents 52 000 kg
- Damping 2 % of critical
Find
The natural circular frequency, frequency and period
Worked example
Which is cheaper — adding stiffness or removing mass?
Given
- The same tower: 3.2 × 10⁶ N/m, 52 000 kg, fn = 1.249 Hz
- A proposal to raise the frequency by 15 % of either quantity
Find
Which change moves the frequency more, and by how much
Summary
- m ẍ + c ẋ + kx = F(t), and every term is a force in newtons
- Inertia opposes acceleration, damping opposes velocity, stiffness opposes displacement
- The three peak at different instants, which is why they can balance
- Newton's second law and D'Alembert's principle give the same equation
- Gravity cancels when x is measured from static equilibrium
- The viscous damping term is calibrated to energy loss, not derived from a mechanism
- Check dimensions every time — it catches the mass-versus-weight error at once
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint