Solve Harmonic Oscillator w/ Impulsive Force: M & Time

[kosig]

So here is the problem. A mass hanging from a spring is modeled by the operator L(y)=2y"+y'/10+2y (y=0 corresponds to hanging equilibrium). Assume mass starts with y(0)=1 and y'(0)=1. Assume an upward impulsive force of mag M is applied at the first possible time which results in complete end in motion. Determine M and time.

My professor worked it by taking laplace transform of linear operator set equal to the impulsive force which I think he made M*e^(-s*t). But I have been reading about Green's Function which applies to this as well. Could someone help me solve it. Not too picky on method. Also beware, if you try to laplace transform it gets pretty messy, hence my confusion.

 

[Valerie Park]

Solution:The equation of motion is given by:L(y) = 2y'' + y'/10 + 2yWe can use Green's function to solve this problem. The Green's function G(t,t') satisfies the equation:L(G) = δ(t-t')where δ(t-t') is the Dirac delta function.The general solution of the homogeneous equation is given by:yh(t) = A*cos(t) + B*sin(t)Now, we need to find the particular solution of the inhomogeneous equation. This can be done by using the method of undetermined coefficients. We assume a particular solution of the form:yp(t) = C*e^(-t/10)Substituting these solutions into the equation of motion, we obtain:2A*cos(t) + 2B*sin(t) + C*e^(-t/10) + (A*sin(t) - B*cos(t))/10 + 2C*e^(-t/10) = M*δ(t-t')By solving for A, B and C, we obtain the following solution for y:y(t) = M*e^(-t/10)*θ(t-t') + A*cos(t) + B*sin(t)where θ(t-t') is the Heaviside step function.Now, we can use the initial conditions to determine the constants A, B and M.At t = 0, we have:y(0) = 1 = M + A ⇒ A = 1 - MAt t = 0, we have:y'(0) = 1 = M/10 + B ⇒ B = 10 - 10MSubstituting these values into the equation for y, we obtain:y(t) = M*e^(-t/10)*θ(t-t') + (1-M)*cos(t) + (10-10M)*sin(t)For the mass to come to rest, we must have y'(t