Laplace Transformation spring question

Question

Here is the question:

https://i.stack.imgur.com/yJyCO.jpg

I can't seem to get the answer. Are those values in the writing like 1N/m even relevant?

Can someone give me some direction? Thanks!

Answer 1

From that image, it just seems to me that they want you to use Laplace Transforms to solve the system

$$\left\{\begin{aligned} y_1^{\prime\prime}(t) &= y_2(t) - 2y_1(t) \\ y_2^{\prime\prime}(t) &= y_1(t) - 2y_2(t)\end{aligned}\right.$$

subject to the initial conditions $y_1(0) = 1$, $y_1^{\prime}(0) = \sqrt{3}$, $y_2(0)= 1$, $y_2^{\prime}(0) = -\sqrt{3}$. All that information at the beginning was more or less giving you the background information that was necessary to construct that system of equations.

Now, let $\mathcal{L}\{y_1(t)\}=X(s)$ and $\mathcal{L}\{y_2(t)\} = Y(s)$. Then, applying Laplace Transforms to the equations in your system leaves us with

$$\left\{\begin{aligned} s^2X(s) - sy_1(0) - y_1^{\prime}(0) &= Y(s) - 2X(s)\\ s^2Y(s) - sy_2(0) -y_2^{\prime}(0) &= X(s) - 2Y(s)\end{aligned}\right. \implies \left\{\begin{aligned} (s^2+2)X(s) -s-\sqrt{3} &= Y(s) \\ (s^2+2)Y(s) -s+\sqrt{3} &= X(s)\end{aligned}\right.$$

Let's substitute the first equation into the second. Then it follows that

$$(s^2+2)\left[(s^2+2)X(s) - s - \sqrt{3}\right] -s +\sqrt{3} = X(s)\\ \implies (s^2+2)^2X(s) -s(s^2+2) -\sqrt{3}(s^2+2) - s +\sqrt{3} = X(s) \\ \implies ((s^2+1)^2-1)X(s) = s(s^2+3)+\sqrt{3}(s^2+1)\\ \implies X(s) = \frac{s(s^2+3)}{(s^2+2)^2-1} + \frac{\sqrt{3}(s^2+1)}{(s^2+2)^2-1}$$

Now noting that $(s^2+2)^2-1 = (s^2+3)(s^2+1)$ by difference of squares, we see that

$$X(s) = \frac{s}{s^2+1} + \frac{\sqrt{3}}{s^2+3}$$

and hence $$y_1(t) = \mathcal{L}^{-1}\{X(s)\} = \cos t + \sin(\sqrt{3}t).$$

Now, to find $y_2(t)$, we substitute our solution into the first equation and solve for $y_2(t)$; this leaves you with

$$-\cos(t) -3\sin(\sqrt{3}t) = y_2(t) - 2(\cos t + \sin(\sqrt{3}t))\implies y_2(t) = \cos(t) -\sin(\sqrt{3}t)$$

(Note: You could have also shown this by using what we computed for $X(s)$ by plugging it into the second equation and solve for $Y(s)$; it's easy to see that

$$\begin{aligned}Y(s) &= \frac{s}{(s^2+2)(s^2+1)}+\frac{\sqrt{3}}{(s^2+2)(s^2+3)} +\frac{s}{s^2+2} - \frac{\sqrt{3}}{s^2+2} \\ &= \frac{s}{s^2+1} -\frac{\sqrt{3}}{s^2+3}\end{aligned}$$

and thus $y_2(t) = \mathcal{L}^{-1}\{Y(s)\} = \cos t - \sin(\sqrt{3}t)$.)