Forcing and resonance

Chapter 8 - Separate natural motion from response

A nonhomogeneous equation adds an input:

$ay''+by'+cy=g(t).$

Every solution splits into

$\boxed{y=y_h+y_p},$

where $y_h$ is homogeneous natural motion and $y_p$ is one particular response to the forcing. Natural motion is a decaying transient only when all homogeneous modes decay; undamped oscillations persist, and unstable modes can grow.

Match the input's shape

For constant coefficients, undetermined coefficients proposes a particular solution with the same family of shapes as $g(t)$ .

Polynomialg(t)=t²+1

Try a general polynomial of the same degree.

Exponentialg(t)=e^{at}

Try $Ae^{at}$ .

Sinusoidg(t)=cos(ωt)

Try $A\cos(ωt)+B\sin(ωt)$ .

Productsg(t)=te^{at}

Multiply the corresponding trial families.

If the trial already belongs to $y_h$ , multiply it by enough powers of $t$ to make it independent. This is the algebraic signature of resonance.

Worked example: ordinary forcing

Solve $y''+y=3\cos2t$ shape matching

Natural motion. $r^2+1=0$ gives $y_h=C_1\cos t+C_2\sin t$ .
Choose a trial. Since frequency 2 does not appear in $y_h$ , try $y_p=A\cos2t+B\sin2t$ .
Substitute. $y_p''+y_p=(-4A+A)\cos2t+(-4B+B)\sin2t$ .
Match coefficients. $-3A=3$ and $-3B=0$ , so $A=-1$ , $B=0$ .
Combine. $y=C_1\cos t+C_2\sin t-\cos2t$ .
Interpret. The output contains natural frequency 1 and forced frequency 2. Initial conditions would set the natural-motion amplitudes.

Lab: sweep through resonance

Forced oscillator

x''+cx'+x=\cos(\omega t)

Forcing frequency ω 0.70 Damping c 0.18

time responsesteady-state amplitude

Reduce damping, then move $\omega$ near 1. The response curve rises sharply because the input frequency approaches the natural frequency. Damping keeps the steady-state amplitude finite and shifts the exact peak slightly below 1.

Resonant forcing

For

$y''+y=\cos t,$

the ordinary trial $A\cos t+B\sin t$ duplicates $y_h$ . Multiply by $t$ :

$y_p=t(A\cos t+B\sin t).$

Substitution gives one valid choice $y_p=\tfrac12 t\sin t$ . Its envelope grows linearly because an undamped input adds energy in phase with the natural motion.

Trial collision. For $y''-2y'+y=e^t$ , which particular trial is appropriate?

The root $r=1$ repeats twice in the homogeneous equation, so multiply the exponential trial by $t^2$ .

Fade the response

Find a particular solution of $y''+3y'+2y=4e^{-t}$ .

Characteristic roots: $\underline{\hspace{5em}}$ .
Collision? The input $e^{-t}$ [is / is not] a natural mode.
Correct trial: $y_p=\underline{\hspace{6em}}$ .
Substitute and find the coefficient.

Check the response

The roots are $-1,-2$ , so $e^{-t}$ collides once. Try $y_p=Ate^{-t}$ . Substitution yields $A e^{-t}=4e^{-t}$ , so $A=4$ . One particular solution is $4te^{-t}$ .

Transient versus steady state

In a damped stable system, $y_h$ decays. What remains is the steady-state response $y_p$ . This split explains why two experiments with different initial conditions can eventually show the same driven behavior.

Error clinic: applying initial conditions to $y_p$ alone

The particular solution only supplies one response to the input. Initial conditions select constants after $y_h+y_p$ has been assembled.

Practice

Practice - polynomial input
8.1 Find the general solution of $y''+y=t$ .
Detailed solution
Practice - sinusoidal input
8.2 Solve $y''+4y=\cos t$ , $y(0)=0$ , $y'(0)=0$ .
Detailed solution
Integrated - collision
8.3 For each input to $y''+4y= g(t)$ , state the trial without solving: (a) $e^{-t}$ , (b) $\cos2t$ , (c) $t^2$ .
Detailed solution
Challenge - beating
8.4 Use $\cos at-\cos bt=-2\sin((a+b)t/2)\sin((a-b)t/2)$ to explain the slow amplitude modulation when an undamped oscillator is forced near, but not at, its natural frequency.
Detailed solution
Practice - extension
8.5 Solve $y''-2y'+y=e^t$ , $y(0)=0$ , $y'(0)=0$ . Explain the required power of $t$ in the particular trial, and verify the complete solution using the substitution $y=e^t u$ .
Detailed solution
Challenge - transfer
8.6 Solve $y''+2y'+2y=\cos t$ , $y(0)=0$ , $y'(0)=0$ . Find the steady-state amplitude, identify the transient, and explain why two solutions with different initial conditions approach the same periodic response.
Detailed solution

Retrieve the design logic

Explain without formulas:

why a particular solution should resemble the input;
why a collision with natural motion requires a factor of $t$ ; and
why damping changes a resonance peak.

I swept the resonance lab, predicted each trial shape, and explained every collision.

Next: Read coupled states as vector fields, eigen-directions, and phase portraits.

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Differential Equations as a Second Language