Push an undamped mass with a force F₀·cos(ωt) and it answers at the frequency of the push, with an amplitude that is an exact fraction, A = F₀/(m·(ω₀² − ω²)). The sign of A says whether the mass moves with the push or against it, and the size of A grows as the push frequency approaches the natural one.
Core ideas
Three ideas: why a cosine is the right guess, what the amplitude is made of, and what its sign records.
The second derivative of cos(ωt) is −ω²·cos(ωt), a multiple of itself, so a cosine of the same frequency as the push can balance it. This is undetermined coefficients, as in “Exponential and Sinusoidal Forcing”. With no damper only even derivatives appear, so no sine term is needed.
Substituting A·cos(ωt) gives A·(k − m·ω²) = F₀. Dividing by m, A = F₀/(m·(ω₀² − ω²)). Every symbol in it is a fraction in the lab, so A is exact. Note the squares: it is ω₀² − ω² and not ω₀ − ω.
If ω < ω₀ the denominator is positive and A has the sign of F₀: the mass peaks when the push peaks. If ω > ω₀ it is negative, and the mass is at its most negative when the push is at its most positive. A fast push on a mass that cannot keep up is answered the wrong way round.
The material
The forced equation, the substitution that fixes the amplitude, the theorem with its check, and pushes that show the three behaviours.
Until now the right-hand side of every oscillator equation was zero and the motion was whatever the start left it. Add a force from outside that alternates with frequency ω and size F₀, and the equation becomes m·x″ + c·x′ + k·x = F₀·cos(ωt). This lesson and the next keep the damper off, c = 0, so that the mass answers the push and nothing carries the answer away. The last lesson of the course puts the damper back.
The second line is the structure from “Homogeneous Plus Particular”: any one solution x_p of the forced equation, plus the free motion of the unforced one. The free part carries the constants C₁ and C₂ and is fixed by the start. The part fixed by the push alone is x_p, and it is the subject of this lesson.
If ω² ≠ ω₀² then x_p = A·cos(ωt) with A = F₀/(m·(ω₀² − ω²)) is a solution of m·x″ + k·x = F₀·cos(ωt), and it is the only solution of the form A·cos(ωt) + B·sin(ωt).
Substitute: m·x_p″ + k·x_p = (k − m·ω²)·A·cos(ωt), and (k − m·ω²)·A = m·(ω₀² − ω²)·A = F₀, so the left side is F₀·cos(ωt).
For a sine term B·sin(ωt) the same substitution gives (k − m·ω²)·B·sin(ωt), which must be zero, and k − m·ω² ≠ 0, so B = 0. Dividing by k − m·ω² is allowed exactly when ω² ≠ ω₀², which is why the next lesson exists.
□
Take m = 1, k = 4 and F₀ = 3, so ω₀² = 4 and ω₀ = 2. The push 3·cos(ωt) is tried at four frequencies, the lab starts the mass at rest at the rest position each time, and the tile reads the amplitude.
For ω = 1, x_p = cos(t), and A = 1. Started at rest at the rest position, the whole motion is x = cos(t) − cos(2t): it has x(0) = 0 and x′(0) = 0, and x″ + 4x = −cos(t) + 4cos(2t) + 4cos(t) − 4cos(2t) = 3cos(t).
The curve the lab draws is that sum of two cosines. The tile reports the first one, the part fixed by the push, and it moves with the push.
For ω = 3, A = −3/5. The push is 3·cos(3t), and the response is −(3/5)·cos(3t): when the force is pulling the mass to the right, the response is to the left. The whole motion from rest is −(3/5)·cos(3t) + (3/5)·cos(2t).
The mass is too sluggish to follow a push this fast. It is still going the old way when the force turns, and the spring reverses it before the next push.
The two extremes make sense of the formula. As ω falls toward zero the push becomes a slow steady pull, and A tends to F₀/k, the stretch Hooke's law gives for that force. As ω rises the amplitude shrinks toward zero from below: inertia wins. Between them, near ω₀, the denominator ω₀² − ω² is small and the amplitude is large, as the pushes at 3/2 and 7/4 show: 12/7 and then 16/5.
One tile will look odd. While the push is on, the lab's period tile reads the beat period of the two cosines when the natural frequency is rational. It means something specific, and the next lesson explains it.
Interactive
Free, damped and forced motion, with every exact quantity
Method
Five moves. The second is the one that decides whether the formula can be used at all.
The mass m, the spring constant k, the size of the push F₀ and its frequency ω, with the damper absent. Compute ω₀² = k/m as a fraction.
If they are equal, stop: the formula divides by zero and the next lesson takes over. Otherwise note which is larger, because that will be the sign.
A = F₀/(k − m·ω²). Do the subtraction in the denominator as a fraction first, then divide. For ω = 3/2: 4 − 9/4 = 7/4, and 3/(7/4) = 12/7.
Positive: the mass moves with the push. Negative: against it. The size is |A|, and it is the larger the closer ω² is to ω₀².
m·(−ω²·A) + k·A must equal F₀. For ω = 3, A = −3/5: −9·(−3/5) + 4·(−3/5) = 27/5 − 12/5 = 3.
The lab presets for ω = 1 and ω = 3: m = 1, k = 4, F₀ = 3, started at rest at the rest position.
Both amplitudes are exact fractions, and the lab prints them as A = 1 and A = −3/5. The first is positive because 1 < 2; the second is negative because 3 > 2. Nothing in either figure is rounded.
You should be able to write x_p = A·cos(ωt) for m·x″ + k·x = F₀·cos(ωt), compute A = F₀/(m·(ω₀² − ω²)) exactly as a fraction, say from its sign whether the mass moves with the push or against it, and check the answer by substituting it.
Something to change about this lesson? Write it down, tick it off when it is handled, and export the list when you want to act on it. Kept in this browser only: nothing is sent anywhere, and nobody is notified.
No recommendations yet for this lesson.