Question 5.4.7: Consider a spring mass system wherem =1, k =13, and c =4. As...

Using the delta function, the given problem is a standard damped harmonic oscillator equation with an impulse forcing function. In particular, the displacement y of the mass satisfies the initial-value problem

y″+4y′+13y = δ(t −3),    y(0) = 1,       y′(0) = 0                 (5.4.14)

Before we solve the IVP, we can use our intuition as a guide: we expect the size of the oscillations of the mass to decrease in magnitude until t = 3, at which time we expect the problem to restart as the blow from the hammer will increase the displacement of the mass, from which oscillations should eventually decrease to zero. We begin to solve (5.4.14) by using the Laplace transform in order to see how far our method enables us to progress.

Taking the Laplace transform of both sides of (5.4.14),

L[y″]+4L[y′]+13L[y] = L[δ(t −3)]

From corollary 5.3.5, it follows that

s^{2}L[y]−sy(0)−y′(0)+4sL[y]−4y(0)+13L[y] = L[δ(t −3)]

Using the conditions y(0)=1 and y′(0)=0, as well as the fact that L[δ(t −3)]=e^{−3s} , we now have

s^{2}L[y]−s +4sL[y]−4+13L[y] = e^{−3s}

Solving for L[y] = Y (s), we see that

Y (s)(s^{2} +4s +13) = s +4+e^{−3s}

or

Y (s) = \frac {s +4}{s^{2} +4s +13}+ \frac {e^{−3s}}{s^{2} +4s +13}                  (5.4.15)

It remains for us to learn how to compute the inverse Laplace transform of (5.4.15) in order to find the solution y to the IVP. The following sections are devoted to these ideas. Upon further study, we will be able to show that the function y(t ) that satisfies (5.4.15) is

y = \frac {1}{3}e^{−2t} (3 cos3t +2sin3t )+ \frac {1}{3}u(t −3)e^{−2(t−3)} sin 3(t −3)

A plot of this solution is shown in figure 5.6, where y(t ) demonstrates precisely the type of behavior we expect.