When the push has exactly the natural frequency, A·cos(ωt) cannot be a solution, and the answer that works is a sine multiplied by t: the amplitude grows in a straight line and never settles. A push just beside the natural frequency produces beats, a fast oscillation inside a slow swelling and shrinking envelope whose period is 2π/|ω₀ − ω|.
Core ideas
Three ideas: why the guess fails, what replaces it, and what the motion looks like when the push is near the natural frequency and not on it.
When ω² = ω₀², the substitution gives m·(−ω₀²·A) + k·A = 0 for every A, so A·cos(ω₀t) and B·sin(ω₀t) both make the left side zero and neither can equal F₀·cos(ω₀t). It is the situation of a repeated root in “Repeated Roots”: the guess is already a free solution, so it is multiplied by t.
Try B·t·sin(ω₀t). Its second derivative contains 2B·ω₀·cos(ω₀t), the very cosine the push needs, and the other terms cancel against the spring. The result is B = F₀/(2m·ω₀), and the amplitude B·t grows in proportion to t. It is finite at every moment and has no bound.
From rest, the response to a push at ω is A·(cos(ωt) − cos(ω₀t)), two cosines of nearby frequency. The identity for the difference of two cosines turns it into a fast sine inside a slow one. The slow one is an envelope that swells and shrinks, and the swelling repeats every 2π/|ω₀ − ω|.
The material
The failure of the amplitude formula, the resonant solution with its check, the beats just beside it, and how the two descriptions join.
The amplitude A = F₀/(m·(ω₀² − ω²)) of the previous lesson grows as ω approaches ω₀, and at ω = ω₀ it has a zero in the denominator. That is not a large answer. It is no answer: the form A·cos(ωt) does not solve the equation at all when the push has the natural frequency.
The cure is the one used for a repeated root: multiply by t. Try x_p = B·t·sin(ω₀t) and differentiate with the product rule.
For ω = ω₀, the function x_p = (F₀/(2m·ω₀))·t·sin(ω₀t) solves m·x″ + k·x = F₀·cos(ω₀t).
Since k = m·ω₀², the last term of the display above is B·(k − m·ω₀²)·t·sin(ω₀t) = 0, so m·x_p″ + k·x_p = 2m·B·ω₀·cos(ω₀t).
Setting 2m·B·ω₀ = F₀ gives B = F₀/(2m·ω₀), and then the left side is F₀·cos(ω₀t).
□
Here ω₀ = 2, and the push is 3·cos(2t). Then B = 3/(2·1·2) = 3/4 and x_p = (3/4)·t·sin(2t). The check: x_p″ = 3·cos(2t) − 3t·sin(2t), so x_p″ + 4x_p = 3·cos(2t), the push exactly.
Started at rest at the rest position, the whole motion is x_p, because x_p and its rate are both zero at t = 0. The lab prints it in the forced-response tile.
The amplitude of this motion is 3t/4: it is 3 at t = 4 and 30 at t = 40. It grows without bound, and it is finite at every moment. It does not leap to infinity when the push is switched on; it climbs by the same amount in every unit of time. The model has no damper to stop the climb, and a real structure has one, or breaks.
Move the push to ω = 3/2 and keep m = 1, k = 4, F₀ = 3. The previous lesson gives A = 12/7, and from rest the motion is x = A·(cos(ωt) − cos(ω₀t)). The identity cos(θ) − cos(ψ) = 2·sin((θ + ψ)/2)·sin((ψ − θ)/2), which follows from the addition and subtraction formulas for the cosine that “Amplitude, Phase and Period” took as given, rewrites it as a product.
In a motion 2A·sin(p·t)·sin(q·t) with q much smaller than p, the factor 2A·sin(q·t) is a slowly varying envelope for the fast factor. The size of the oscillation swells and shrinks, and the interval between one swell and the next is the beat period, π/q.
Here q = (ω₀ − ω)/2, so the beat period is 2π/|ω₀ − ω|.
For ω = 3/2 that is 2π/(1/2) = 4π, about 12.5664 and rounded, which is what the lab prints in the period tile when the push is on. The envelope (24/7)·sin(t/4) itself repeats every 8π, but the size of the swing, which is its absolute value, repeats every 4π. The largest the swing reaches is 2A = 24/7, twice the steady amplitude.
The two descriptions join. Move ω toward ω₀: the swell height 2A grows without bound and the beat period 2π/|ω₀ − ω| lengthens without bound, so the first swell stretches until it never ends. While t is small, sin((ω₀ − ω)t/2) is close to (ω₀ − ω)t/2, and the envelope is close to F₀·t/(m·(ω₀ + ω)), which tends to the resonant ramp F₀·t/(2m·ω₀) as ω → ω₀. The lab preset for ω = 7/4 shows the swell taller and slower than at 3/2; the preset at ω = 2 is the limit. That the ramp is the limit is what the lab demonstrates, and it is not proved here.
Interactive
Free, damped and forced motion, with every exact quantity
Method
Five moves. The first decides which of the two descriptions applies.
If they are equal the motion is resonant. If they are close but unequal it beats, and the amplitude of the previous lesson applies.
B = F₀/(2m·ω₀), and the solution is B·t·sin(ω₀t). For m = 1, k = 4, F₀ = 3: ω₀ = 2 and B = 3/4. When ω₀ is a surd, keep it as a surd in B.
Find x_p″ by the product rule and confirm that m·x_p″ + k·x_p is exactly the push. The t·sin(ω₀t) terms must cancel and the cosine must remain.
2π/|ω₀ − ω|, using the difference of frequencies, not of squares. For ω₀ = 2 and ω = 3/2 it is 4π, rounded to 12.5664 with ≈.
The swing reaches 2·|A| at the top of each beat, with A from the previous lesson. For ω = 3/2 that is 24/7.
The lab presets for ω = 2 and ω = 3/2: m = 1, k = 4, F₀ = 3, started at rest at the rest position.
At resonance the amplitude 3t/4 has no bound and no jump: it is 3 at t = 4. Beside it, with ω = 3/2, the swing is bounded by 24/7 and repeats every 4π, about 12.5664 and rounded. The first is the limit of the second as the push is tuned onto the spring.
You should be able to show that A·cos(ωt) fails when ω = ω₀, write x_p = (F₀/(2m·ω₀))·t·sin(ω₀t) and check it by substitution, say that its amplitude grows linearly and is finite at every moment, and compute the beat period 2π/|ω₀ − ω| as a symbol and as a rounded number marked ≈.
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