Modeling

A driven damped oscillator, from physical setup to interpretation

Four stages: state the system, reduce it to something dimensionless, solve, then check the solution against what the physics says should happen.

1. The system

A mass m on a spring of stiffness k, damped with coefficient c, driven by a periodic force:

mx″+cx′+k*x=(F_0)*cos(ω*t)

Identical in form to an RLC circuit with L*q+R*q+q/C=(V_0)*cos(ω*t), which is worth noticing — one solution covers both.

2. Reduction

Divide by m and name the groupings that carry meaning:

x″+2*γx′+(ω_0)^2*x=(ƒ_0)*cos(ω*t),γ=c/(2*m),(ω_0)=√(,k/m),(ƒ_0)=(F_0)/m

Three parameters became two that matter: the ratio γ/(ω_0) decides the character of the response, and ω/(ω_0) decides where on the response curve you sit. Everything below is a statement about those two ratios.

3. Transient

r^2+2*γ*r+(ω_0)^2=0⇒r=−γ±√(,γ^2−(ω_0)^2)

Damping doesn't only shrink the oscillation, it slows it: (ω_d)<(ω_0). The t in the critical case isn't bookkeeping — a repeated root gives one solution, and one solution cannot meet two initial conditions.

Every regime carries e^(−γ*t), so with any damping at all this decays. Hence transient.

4. Steady state

Write (x_p)=Re*(Z*e^(i*ω*t)) and substitute:

(−ω^2+2*i*γ*ω+(ω_0)^2)*Z=(ƒ_0)⟹Z=(ƒ_0)/((ω_0)^2−ω^2+2*i*γ*ω)

(x_p)=R*cos(ω*t−δ),R=(ƒ_0)/√(,((ω_0)^2−ω^2)^2+4*γ^2*ω^2),tan(δ)=(2*γ*ω)/((ω_0)^2−ω^2)

The complex route is worth the setup: it turns a trigonometric coefficient-matching problem into one division, and the amplitude and phase drop out as modulus and argument.

5. Resonance

Maximise R by minimising the denominator. With u=ω^2,

d()/d(u)*[((ω_0)^2−u)^2+4*γ^2*u]=−2*((ω_0)^2−u)+4*γ^2=0⟹u=(ω_0)^2−2*γ^2

(ω_res)=√(,(ω_0)^2−2*γ^2)

The peak sits slightly below the natural frequency, and disappears entirely once γ≥(ω_0)/√(,2) — past that, damping has flattened the curve and the largest response is at ω=0.

Undamped limit. Set γ=0 and drive at exactly (ω_0). Now cos((ω_0)*t) solves the homogeneous equation, so the usual ansatz returns 0=(ƒ_0)*cos((ω_0)*t). Multiply by t:

(x_p)=A*t*sin((ω_0)*t)⟹(x_p)″+(ω_0)^2*(x_p)=2*A*(ω_0)*cos((ω_0)*t)⟹A=(ƒ_0)/(2*(ω_0))

(x_p)=(ƒ_0)/(2*(ω_0))*t*sin((ω_0)*t)

Amplitude grows without bound. The general rule behind the trick: if the ansatz already solves the homogeneous equation, multiply by t, and again for a repeated root. The system cannot respond in a mode it already sustains for free.

6. Does the model behave

Limits. (ƒ_0)→0 gives R→0. γ→∞ gives R→0. ω→0 gives R→(ƒ_0)/(ω_0)^2=(F_0)/k, the static deflection of the spring under a constant force — which is the check worth doing, because it ties the dynamic answer back to Hooke's law.

Dimensions. [γ]=[(ω_0)]=s^(−1), [(ƒ_0)]=m*s^(−2), so R has units m*s^(−2)/s^(−2)=m.

Where the model stops being true. Linear damping is an approximation — real drag goes as v^2 at speed. Hooke's law fails at large extension. And the unbounded growth at γ=0 is a comment about the model, not about any real spring, which either yields or breaks first.