A machine applies a force cos(2t) to the normalized oscillator y″ + y′ + 4y = cos(2t). Here y is displacement and t is time. Try the displacement y = sin(2t)/2: its second derivative cancels 4y, while its first derivative supplies cos(2t). The displacement reaches a peak one quarter-cycle after the force. But its initial velocity is 1, not zero. A periodically driven motion and a motion released from rest are different questions.
Predict
The undamped natural angular frequency is 2. Does positive damping make the largest displacement response occur exactly at 2? Does every underdamped oscillator have a positive-frequency displacement peak? Predict the comparison between damping 1 and damping 3 before using the controls.
Experiment
Change the driving frequency
Normalized teaching model y″ + c y′ + 4y = f(t). Time and displacement use fixed teaching units. Analytic values, temporary controls; no numerical time integration or saved evidence.
Gain = 0.50000 displacement/force; lag = 1.57080 radians. Sinusoidal particular: 0.00000 cos(ωt) + 0.50000 sin(ωt).
Displacement-gain maximum over ω ≥ 0: frequency 1.87083, gain 0.51640. This frequency differs from the natural frequency 2.
At t = 0: zero-IC displacement 0.00000, velocity 0.00000. Sinusoidal particular 0.00000; free correction 0.00000. Positive damping makes the free correction decay.
Fixed plot window: time 0–16, displacement −5 to 5. Solid navy: zero-IC total. Dashed red: sinusoidal particular when defined. The periodic plot has no startup correction.
Predict before changing a control, then record an equation-based explanation in the lesson practice. The plot samples 321 times; smooth-looking samples do not prove an equation or a uniform bound.
First inspect the default at time zero: the sinusoidal particular needs a free correction to match rest. Compare frequencies 1.5, 2 and 2.5 with damping 1. Then choose damping 3 and compare its static response with the oscillatory cases. The continuous maximizing frequency is reported separately; a sparse menu of frequencies need not contain the exact maximum. Finally set damping 0 and frequency 2. Explain why the dashed sinusoid disappears instead of being drawn with a very large finite amplitude.
Understand
Two coefficients encode amplitude and phase
For a mass m > 0, damping c > 0 and spring stiffness k > 0 driven by force amplitude F > 0, the linear model is m y″ + c y′ + k y = F cos(ωt), with angular frequency ω ≥ 0. The units of m, c and k make every term a force. For ω > 0, substitute a cos(ωt) + b sin(ωt) and equate the two independent coefficients:
(k−mω2)a+cωb=F,(k−mω2)b−cωa=0.
Let D denote the positive squared denominator. Solving gives:
D=(k−mω2)2+c2ω2,a=DF(k−mω2),b=DFcω.
Writing the same real motion as A cos(ωt − δ) gives amplitude A = F/√D and lag δ = atan2(cω, k − mω²). The two-argument atan2 chooses the quadrant from both signs. Below the natural frequency the lag lies between zero and π/2; above it the lag lies between π/2 and π. At ω = 0 the input is constant, the particular is F/k, and phase is only a convenient zero convention, not a measurable cycle lag.
Complex notation packages the algebra: multiplying e to the iωt by H(iω) = 1/(k − mω² + icω), then taking the real part, gives the same real coefficients. The imaginary unit i satisfies i² = −1. The gain |H(iω)| is displacement amplitude per unit force. It is not an energy gain.
Worked phase check above the natural frequency
For m = 1, c = 1, k = 4, ω = 3, the denominator is −5 + 3i. Its reciprocal is (−5 − 3i)/34. Taking the real part after multiplication by cos(3t) + i sin(3t) gives −5cos(3t)/34 + 3sin(3t)/34. The sine coefficient is positive even though the reciprocal’s imaginary part is negative. Amplitude is 1/√34 and lag is in quadrant II. A principal arctan of −3/5 alone gives the wrong quadrant.
Find a maximum, do not guess it from the natural frequency
Maximizing gain is minimizing D. Set x = ω² ≥ 0. Then D = m²x² + (c² − 2mk)x + k² is a strictly convex quadratic. Its vertex lies in the allowed positive region exactly when c² is less than 2mk:
ωpeak={k/m−c2/(2m2),0,0<c2<2mk,c2≥2mk.
This is the displacement maximum for fixed force amplitude. The natural frequency √(k/m), the free damped frequency √(k/m − c²/(4m²)) when underdamped, and this maximizing driving frequency answer different questions. At m = 1, k = 4, c = 1 they are 2, √15/2 and √(7/2). The maximum gain is 2/√15. At c = 3 the free motion still oscillates, but the largest displacement gain is the static value 1/4. The stricter peak threshold explains this counterexample.
Velocity amplitude is ω times displacement amplitude. Its squared gain has numerator x, so differentiating gives a sign controlled by k² − m²x². Thus velocity amplitude, and average dissipated power c times its squared amplitude divided by 2, peak at √(k/m) for c > 0. A claim about “the resonance” must specify which output is being maximized and which input amplitude is fixed.
Restore initial conditions and handle zero damping
For the default example, the zero-initial-state solution is:
y(t)=21sin(2t)−152e−t/2sin(215t).
The second term is a homogeneous solution with initial displacement zero and initial velocity −1. Both roots have negative real part, so it decays. At c = 0 that conclusion fails: nonresonant free oscillations persist. At c = 0, ω = 2 the denominator vanishes and no sinusoidal particular exists. The zero-IC solution is t sin(2t)/4, a secular response with growing envelope. This ideal linear model omits nonlinear limits, actuator limits and damage; the calculation is not a physical prediction at arbitrarily large amplitudes.
Apply
Explain why changing damping from 1 to 3 removes the positive displacement peak without removing oscillation from the free response. Then repair the statement “a sinusoidal particular automatically satisfies zero initial conditions.” Use both an initial-value check and a homogeneous correction. The six practice tasks also change the measured output from displacement to velocity and power.
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For m y″+c y′+k y=F cos(ωt), m,k,F>0,c>0, derive the cosine and sine coefficients of the sinusoidal particular. Explain their units.
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For y″+y′+4y=cos3t, find the particular, amplitude and lag. Why does arctan(3/−5) alone give a misleading phase?
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For m,k>0,c>0 prove the displacement-gain maximum occurs at ω=sqrt(k/m−c²/(2m²)) when c²<2mk, and at ω=0 otherwise.
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For m=1,k=4 compare c=1 and c=3. Give the natural frequency, damped free frequency and maximizing driving frequency. Explain why underdamped motion does not ensure a positive displacement-response peak.
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Find the zero-initial-state solution of y″+y′+4y=cos2t and explain why (1/2)sin2t alone is insufficient.
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For c>0, a unit sinusoidal force drives m y″+c y′+k y=f. Show velocity amplitude peaks at ω=√(k/m), unlike displacement in general. Give average power and its maximizing frequency.
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Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
