With a damper, the response to a steady push settles to a fixed amplitude whose square is an exact fraction, so the divergence of the previous lesson is gone and the peak is finite. The peak is not at the natural frequency: it sits at ω_r² = k/m − c²/(2m²), below it, and for a heavy enough damper it does not exist.
Core ideas
Three ideas: what the response needs once there is a damper, why its amplitude is finite, and where the peak is.
The damper's force −c·x′ turns a cosine into a sine, so the steady response is G·cos(ωt) + H·sin(ωt). The two coefficients solve two linear equations, one from the cosines and one from the sines, and the amplitude of the pair is √(G² + H²).
The result is A² = F₀²/D with D = (k − m·ω²)² + (c·ω)². The first square can be zero, at ω = ω₀, but the second is positive for every ω > 0 once c > 0. So D > 0, A² is finite, and the divergence of the undamped case is gone.
The peak is where D is least. Written in terms of u = ω², D is a parabola, least at u = k/m − c²/(2m²). The damper subtracts from ω₀² and moves the peak down. If that number is zero or negative, there is no peak at all.
The material
The steady response and its amplitude with a derivation, the least value of the denominator, presets on three kinds of curve, and why the peak is not at the natural frequency.
Put the damper back: m·x″ + c·x′ + k·x = F₀·cos(ωt) with c > 0. Every free solution of this equation dies away, as the damping lessons showed, so after a while only the part fixed by the push remains. That is the steady state, and its size is the steady amplitude. The way the motion settles to it is the damped free solution added on, which this lesson does not compute; the lab reports the steady amplitude only.
Substitute into the equation and collect the cosines and the sines separately. Write P = k − m·ω² and Q = c·ω.
The steady response of m·x″ + c·x′ + k·x = F₀·cos(ωt) has amplitude A with A² = F₀²/(P² + Q²), where P = k − m·ω² and Q = c·ω.
Multiply the first equation by P and the second by Q, and subtract: (P² + Q²)·G = F₀·P. Multiply the first by Q and the second by P, and add: (P² + Q²)·H = F₀·Q.
So G = F₀·P/(P² + Q²) and H = F₀·Q/(P² + Q²), and G² + H² = F₀²·(P² + Q²)/(P² + Q²)² = F₀²/(P² + Q²). The amplitude of G·cos(ωt) + H·sin(ωt) is √(G² + H²), as in “Amplitude, Phase and Period”. With c = 0 it reduces to the square of the previous lessons' A.
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With m = 1 and F₀ = 3: P = 5 − 1 = 4, Q = 2, and P² + Q² = 20. The coefficients are G = 3·4/20 = 3/5 and H = 3·2/20 = 3/10, and A² = 9/25 + 9/100 = 9/20.
The lab prints A² = 9/20, an exact fraction. It prints the square and not A, because A is a square root and is not needed for what comes next.
The amplitude is largest where D = P² + Q² is smallest. Expand it in u = ω²: D = m²·u² + (c² − 2mk)·u + k², an upward parabola. It is least at u = (2mk − c²)/(2m²), and there its value is c²·ω_d² with the ω_d² of “Underdamped Motion and the Envelope”.
The last line is the point. The frequency of the peak, the frequency of the free ringing and the natural frequency are three different numbers, in that order. The peak needs u > 0, which is c² < 2mk. When c² ≥ 2mk the parabola is increasing for all u > 0, the amplitude falls steadily from F₀/k at ω = 0, and the lab prints none.
For m = 1, c = 2, k = 5: D = u² − 6u + 25 = (u − 3)² + 16. It is least at u = 3, so ω_r² = 3, with D = 16 and A_max² = 9/16, A_max = 3/4. The natural frequency squared is 5, so the peak is lower than ω₀.
At ω₀² = 5 itself, D = 0 + 4·5 = 20 and A² = 9/20, which is below 9/16. The natural frequency does not give the largest response.
ω_r = √3 is irrational, and the lab takes a rational ω, so a convenient one to try is ω = 7/4, a little above √3. Then P = 5 − 49/16 = 31/16 and Q = 7/2, and A² = 2304/4097. That is just under 9/16: 2304·16 = 36864 while 9·4097 = 36873.
A rational push can come as near to the peak as you like and never reach it. The exact tile tells you how near you are.
For m = 1, c = 4, k = 4, ω_r² = 4 − 8 = −4, and the lab prints none. Here D = (ω² + 4)², so A = F₀/(ω² + 4) falls steadily. The lab preset with c = 3 lies between: it is still underdamped, since 9 < 16, but c² = 9 is above 2mk = 8, and it has no peak either.
Free ringing and a resonant peak are different properties. A damper can be light enough to ring and heavy enough to remove the peak.
Interactive
Free, damped and forced motion, with every exact quantity
Method
Five moves. The third decides whether there is a peak to find.
P = k − m·ω² and Q = c·ω, as fractions. For m = 1, c = 2, k = 5 at ω = 1: P = 4 and Q = 2.
A² = F₀²/(P² + Q²). Square first, then add, then divide: 9/(16 + 4) = 9/20. Adding before squaring is the usual slip.
ω_r² = k/m − c²/(2m²). If it is zero or negative the curve has no peak, and the lab prints none. Here 5 − 2 = 3.
A_max² = F₀²/(c²·ω_d²) with ω_d² = k/m − c²/(4m²). Here ω_d² = 4, so A_max² = 9/16 and A_max = 3/4.
At ω₀ the amplitude squared is F₀²/(c²·ω₀²), here 9/20. It must be below the peak, and ω_r² must be below ω₀².
The lab preset with c = 2, k = 5 and ω = 1, read as a function of the frequency squared.
The peak is lower in frequency than the natural frequency and higher in amplitude than the response at it. The lab prints ω_r² = 3; A_max = 3/4 in the resonant tile. Because ω_r = √3 is not rational, the lab preset uses ω = 7/4 and the tile shows an amplitude squared a hair below 9/16.
You should be able to compute P = k − m·ω² and Q = c·ω, the steady amplitude squared F₀²/(P² + Q²) as an exact fraction, the resonant frequency squared ω_r² = k/m − c²/(2m²), the peak A_max² = F₀²/(c²·ω_d²), and say why there is no peak when c² ≥ 2mk and why the peak is not at ω₀.
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