Question

Problems 13 through 16 deal with the effect of a sequence of impulses on an undamped oscillator. Suppose that y" ty-f(t), y(0) 0, y (0)0 For each of the following choices for f(t): a. Try to predict the nature of the solution without solving the problem G b. Test your prediction by finding the solution and plotting its graph. C. Determine what happens after the sequence of impulses ends. 20

Answer 1

$y''+y=\sum_{k=1}^{20}\delta(t-k\pi)$

And so $L(y'')+L(y)=\sum_{k=1}^{20}L(\delta(t-k\pi))$

We note that $L(y'')=s^2L(y)-sy(0)-y'(0)=s^2L(y)$

Therefore, $s^2L(y)+L(y)=\sum_{k=1}^{20}e^{-k\pi s}$

Which means $(s^2+1)L(y)=\sum_{k=1}^{20}e^{-k\pi s}$

So $L(y)=\frac{\sum_{k=1}^{20}e^{-k\pi s}}{s^2+1}$

Taking inverse Laplace transform, we have $y(t)=\sum_{k=1}^{20} u(t-k\pi)\sin(t-k\pi)$ which is plotted below:

a K H(x) = {x 20:1.0} 20 20 30 40 50 60 70

$\blacksquare$

Please do rate this answer positively if you found it helpful. Thanks and have a good day!

Answer 2

$y''+y=\sum_{k=1}^{20}\delta(t-k\pi)$

And so $L(y'')+L(y)=\sum_{k=1}^{20}L(\delta(t-k\pi))$

We note that $L(y'')=s^2L(y)-sy(0)-y'(0)=s^2L(y)$

Therefore, $s^2L(y)+L(y)=\sum_{k=1}^{20}e^{-k\pi s}$

Which means $(s^2+1)L(y)=\sum_{k=1}^{20}e^{-k\pi s}$

So $L(y)=\frac{\sum_{k=1}^{20}e^{-k\pi s}}{s^2+1}$

Taking inverse Laplace transform, we have $y(t)=\sum_{k=1}^{20} u(t-k\pi)\sin(t-k\pi)$ which is plotted below:

a K H(x) = {x 20:1.0} 20 20 30 40 50 60 70

$\blacksquare$

Please do rate this answer positively if you found it helpful. Thanks and have a good day!