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Forcing and resonance

Chapter 8 - Separate natural motion from response

A nonhomogeneous equation adds an input:

Every solution splits into

where (y_h) is transient natural motion and (y_p) is one particular response to the forcing.

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
  1. Natural motion. \(r^2+1=0\) gives \(y_h=C_1\cos t+C_2\sin t\).
  2. Choose a trial. Since frequency 2 does not appear in \(y_h\), try \(y_p=A\cos2t+B\sin2t\).
  3. Substitute. \(y_p''+y_p=(-4A+A)\cos2t+(-4B+B)\sin2t\).
  4. Match coefficients. \(-3A=3\) and \(-3B=0\), so \(A=-1\), \(B=0\).
  5. Combine. \(y=C_1\cos t+C_2\sin t-\cos2t\).
  6. 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\(x''+cx'+x=\cos(\omega t)\)

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

the ordinary trial (A\cos t+B\sin t) duplicates (y_h). Multiply by (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}).

  1. Characteristic roots: (\underline{\hspace{5em}}).
  2. Collision? The input (e^{-t}) [is / is not] a natural mode.
  3. Correct trial: (y_p=\underline{\hspace{6em}}).
  4. 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

  1. Practice - polynomial input

    8.1 Find the general solution of \(y''+y=t\).

    Detailed solution

  2. Practice - sinusoidal input

    8.2 Solve \(y''+4y=\cos t\), \(y(0)=0\), \(y'(0)=0\).

    Detailed solution

  3. 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

  4. 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

Retrieve the design logic

Explain without formulas:

  1. why a particular solution should resemble the input;
  2. why a collision with natural motion requires a factor of (t); and
  3. why damping changes a resonance peak.

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