Second Order Harmonic Motion Help

Question

A 1kg mass is attached to a spring (k=4kg.s^2) and the system is allowed to come to rest. The spring mass system is attached to come to res.t The spring mass system is attached to a machine that supplies an external driving force f(t)=4coswt Newtons. the system is started from equilibrium, the mass having no initial displacement nor velocity. Ignore any dampening forces.

(a) Find the position of the mass as a function of time.

Answer 1

A mass on a spring does not come to rest without damping so the problem is inconsistent. Let's assume that this is just an ideal spring with a mass hanging at equilibrium so that we can neglect the effect of gravity that then experiences a forcing function. The equation for the y position would be:

ma = -k(y-y_e) + 4cos(ωt)

my_s'' + k(y_s) = 4cos(ωt) where y_s is deviation from equilibrium in y

rewrite as y" + (k/m)y = (4/m)cos(ωt)

This is an inhomogeneous 2nd order differential equation with constant coefficients.

The homogeneous solution is Asin(sqrt(k)t) + Bcos(sqrt(k)t)

The particular solution for sqrt(k/m) ≠ ω can be obtained by the method of undetermined coefficients:

y_p is of the form Ccos(ωt)

Solving for particular solution by subbing into the equation:t

-Cω²cos(ωt) + C(k/m)cos(ωt) = 4cos(ωt)/m C = (4/m)/(k/m-ω²)

y = Acos(sqrt(k/m)t)+Bsin(sqrt(k)t) + ((4/m)/(k-ω²))cos(ωt) y and y' = 0 at t=0

y'=0 implies B = 0

y=0 : Acos(sqrt(k/m)t) + ((4/m)/(k/m-ω²))cos(ωt) = so A = -((4/m)/(k-ω²))cos(ωt)) / cos(sqrt(k)t))

As ω approaches sqrt(k/m), the system amplitude of oscillation blows up because the forcing frequency matches the resonance frequency of the system. Strictly speaking you need to solve for that condition separately using undetermined coefficient for tcos(sqrt(k/m)t) which will grow with time.

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