Question 5.5.4: Consider a mass of 1 kg attached to a spring with spring con...

From our work with spring-mass systems, we know that the displacement y(t ) of the mass from equilibrium must satisfy the initial-value problem

y″+4y′+13y = 2u(t −π) sin3t ,          y(0) = 1,     y′(0) = 0

Taking Laplace transforms, it follows that

s^{2}Y (s)−sy(0)−y′(0)+4(sY (s)−y(0))+13Y (s) = 2L[u(t −π) sin3t ]                (5.5.14)

We know that L[sin3t] = 3/(s2 +9), and by the second shifting property

L[u(t −π) sin3t] = e^{−πs} L[sin 3(t +π)] (5.5.15)

At this point, we observe by basic trigonometry that sin(3t + 3π) =sin3t cos3π +cos3t sin3π =−sin3t . Hence, from (5.5.15) we have

L[u(t −π) sin3t] = e^{−πs} L[−sin3t]=−e^{−πs} \frac {3}{s^{2} +9}

Returning to (5.5.14) and using the given initial conditions, it follow  that

s^{2}Y (s)−s +4sY (s)−4+13Y (s)=−2e^{−πs} \frac {3}{s^{2} +9}

Factoring,

Y (s)(s^{2} +4s +13) = s +4−2e^{−πs} \frac {3}{s^{2} +9}

Solving for Y (s),

Y (s) = Y_{1}(s)+Y_{2}(s)

= \frac {s +4}{s^{2} +4s +13} −2e^{−πs} \frac {3}{(s^{2} +9)(s^{2} +4s +13)}                (5.5.16)

It remains to find the inverse transform of Y (s); we do so one piece at a time using the linearity of the inverse transform. In both Y_{1}(s) and Y_{2}(s), we will algebraically rearrange the expression in order to help us more easily determine the inverse Laplace transform, using an approach similar to our work in example 5.5.3.

Taking the first term in (5.5.16), we observe that since the denominator does not factor, we need to write it in a more familiar form. Completing the square and separating the numerator enables us to write

Y_{1}(s) = \frac {s +4}{(s +2)^{2} +9}= \frac {s +2}{(s +2)^{2} +9}+ \frac {2}{(s +2)^{2} +9}

and see the structure of Laplace transforms of basic functions. In particular, from the first shifting property and the known Laplace transforms of cos3t and sin3t , it follows that

L^{−1}[Y_{1}(s)] = L^{−1} [\frac {s +2}{(s +2)^{2} +9}+ \frac {2}{(s +2)^{2} +9}]= e^{−2t} cos3t + \frac {2}{3}e^{−2t} sin3t                (5.5.17)

Next we find the inverse transform of the term Y2(s) in (5.5.16). That is, we must determine

L^{−1}[Y_{2}(s)]=L^{−1} [−6e^{−πs} \frac {1}{(s^{2} +9)(s^{2} +4s +13)}]                       (5.5.18)

From the presence of e^{−πs} , we know the second shifting property will be used; in addition, we must algebraically rearrange the remaining part of the expression in order to find the inverse transform. Computing the partial fraction decomposition of the rational function in (5.5.18), we equivalently seek

L^{−1}[Y_{2}(s)] = L^{−1} [\frac {6}{40} e^{−πs} (\frac {s −1}{s^{2} +9} − \frac {s +3}{s^{2} +4s +13})]                  (5.5.19)

One additional rearrangement will enable us to find the desired inverse transform. Completing the square in the second fraction and separating the numerator in each enables us to rewrite (5.5.19) as

L^{−1}[Y_{2}(s)] = \frac {6}{40} L^{−1} [e^{−πs}(\frac {s}{s^{2} +9} − \frac {1}{s^{2} +9} − \frac {s +2}{(s +2)^{2} +9}− \frac {1}{(s +2)^{2} +9}

Applying the inverse of the second shifting property to each of the terms in L^{−1}[Y_{2}(s)], it follows that

L^{−1}[Y_{2}(s)] = \frac {6}{40} u(t −π) [cos 3(t −π)− \frac {1}{3}sin 3(t −π) − e^{−2(t−π)} cos 3(t −π)− \frac {1}{3}e^{−2(t−π)} sin 3(t −π)]                     (5.5.20)

Noting that sin(3t − 3π) = −sin3t and cos(3t − 3π) = −cos3t , we can simplify (5.5.20) to

L^{−1}[Y_{2}(s)] = \frac {3}{20} u(t −π) [−cos3t + \frac{1}{3}sin3t −e^{−2(t−π)}(−cos3t + \frac {1}{3}sin3t )]

Combining our work with L^{−1}[Y_{1}(s)] and L^{−1}[Y_{2}(s)] we have therefore shown that y(t ) = L^{−1}[Y (s)] is the function

y(t ) = e^{−2t} (cos3t + \frac {2}{3}sin3t )+ \frac {3}{20}u(t −π)[−cos3t+ \frac {1}{3}sin3t −e^{−2(t−π)}(−cos3t + \frac {1}{3}sin3t )]

A plot of the function y(t ) is given in figure 5.10, where we see that until the forcing function activates at t = π, we see the standard damped oscillations decaying to zero. When the periodic forcing function turns on, the system demonstrates the repeating oscillations generated by this function.