Q5.6-7E

Question

Suppose the displacement functions $x (t)$ and $y (t)$ for a coupled mass-spring system (similar to the one discussed in Problem 6) satisfy the initial value problem

$\begin{array}{rcl} x'' (t) + 5 x (t) - 2 y (t) = 0 \\ y'' (t) + 2 y (t) - 2 x (t) = 3 \sin 2 t \\ x (0) = x' (0) = 0 \\ y (0) = 1, y' (0) = 0 \end{array}$

Solve for $x (t)$ and $y (t)$

Step-by-Step Solution

Verified

Answer

The solution for $x (t) = \frac{2}{5} cost + \frac{4}{5} sint - \frac{2}{5} \cos \sqrt{6} t + \frac{\sqrt{6}}{5} \sin \sqrt{6} t - \sin 2 t$ and

$y (t) = \frac{4}{5} cost + \frac{8}{5} sint + \frac{1}{5} \cos \sqrt{6} t - \frac{\sqrt{6}}{10} \sin \sqrt{6} t - \frac{1}{2} \sin 2 t$ .

1Using the elimination procedure

Given system can be written in the following form:

$\begin{array}{rcl} (D^{2} + 5) [x] - 2 [y] = 0 \\ - 2 [x] + (D^{2} + 2) [y] = 3 \sin 2 t \end{array}$

By $L_{1} = D^{2} + 5, L_{2} = - 2, L_{3} = - 2, L_{4} = D^{2} + 2, f_{1} = 0, f_{2} = \sin 2 t$ , using the elimination procedure,

One obtains $(L_{1} L_{4} - L_{2} L_{3}) [x] = L_{4} [f_{1}] - L_{2} [f_{2}]$

i.e.

$((D^{2} + 5) (D^{2} + 2) - 4) [x] = (D^{2} + 2) [0] - (- 2) [3 \sin 2 t] (D^{4} + 7 D^{2} + 6) [x] = 0 + 6 \sin 2 t ((D^{2} + 1) (D^{2} + 6)) [x] = 6 \sin 2 t \dots (1)$

2Finding derivatives

The auxiliary equation is $(r^{2} + 1) (r^{2} + 6) = 0$ and its roots $r_{1, 2} = \pm i, r_{3, 4} = \pm i \sqrt{6}$ .

Therefore, the solution to the corresponding homogeneous equation is

$x_{h} (t) = c_{1} cost + c_{2} sint + c_{3} \cos \sqrt{6} t + c_{4} \sin \sqrt{6} t$ .

Let's check if $x (t) = - \sin 2 t$ is a particular solution for (1).

First, one will find derivatives

$\begin{array}{rcl} x' (t) = - 2 \cos 2 t \\ x'' (t) = 4 \sin 2 t \\ x''' (t) = 8 \cos 2 t \\ x^{(4)} (t) = - 16 \sin 2 t \end{array}$

3Finding x

Since

$x^{(4)} (t) + 7 x'' (t) + 6 x (t) = - 16 \sin 2 t + 7 \times 4 \sin 2 t + 6 (- \sin 2 t) = 6 \sin 2 t$

$x_{p} (t) = - \sin 2 t$ is a particular solution to 1 and the general solution is

$x (t) = x_{h} (t) + x_{p} (t) = c_{1} cost + c_{2} sint + c_{3} \cos \sqrt{6} t + c_{4} \sin \sqrt{6} t - \sin 2 t$ .

From the first equation of the system, $x'' + 5 x - 2 y = 0$ one can find a general solution for $y (t)$ but the first one needs to find $x''$

$x' (t) = - c_{1} sint + c_{2} cost - \sqrt{6} c_{3} \sin \sqrt{6} t + \sqrt{6} c_{4} \cos \sqrt{6} t - 2 \cos 2 t x'' (t) = - c_{1} cost - c_{2} sint - 6 c_{3} \cos \sqrt{6} t - 6 c_{4} \sin \sqrt{6} t + 4 \sin 2 t$

4Simplification

Then simplify,

$2 y = x'' + 5 x = - c_{1} cost - c_{2} sint - 6 c_{3} \cos \sqrt{6} t - 6 c_{4} \sin \sqrt{6} t + 4 \sin 2 t + 5 c_{1} cost + 5 c_{2} sint + 5 c_{3} \cos \sqrt{6} t + 5 c_{4} \sin \sqrt{6} t - 5 \sin 2 t = 4 c_{1} cost + 4 c_{2} sint - c_{3} \cos \sqrt{6} t - c_{4} \sin \sqrt{6} t - \sin 2 t$

Therefore, $y (t) = 2 c_{1} cost + 2 c_{2} sint - \frac{c_{3}}{2} \cos \sqrt{6} t - \frac{c_{4}}{2} \sin \sqrt{6} t - \frac{1}{2} \sin 2 t$

It remains to find constants from initial conditions.

5Finding x t , y t

Let's compute $y'$

$y (t) = - 2 c_{1} sint + 2 c_{2} cost + \frac{c_{3} \sqrt{6}}{2} \sin \sqrt{6} t - \frac{c_{4} \sqrt{6}}{2} \cos \sqrt{6} t - \cos 2 t$

$0 = x (0) = c_{1} + c_{3} 0 = x' (0) = c_{2} + \sqrt{6} c_{4} - 2 1 = y (0) = 2 c_{1} - \frac{c_{3}}{2} 0 = y' (0) = 2 c_{2} - \frac{\sqrt{6}}{2} c_{4} - 1$

By solving this system, one obtains $c_{1} = \frac{2}{5}, c_{2} = \frac{4}{5}, c_{3} = - \frac{2}{5}, c_{4} = \frac{\sqrt{6}}{5}$

Finally, $x (t) = \frac{2}{5} cost + \frac{4}{5} sint - \frac{2}{5} \cos \sqrt{6} t + \frac{\sqrt{6}}{5} \sin \sqrt{6} t - \sin 2 t$

And $y (t) = \frac{4}{5} cost + \frac{8}{5} sint + \frac{1}{5} \cos \sqrt{6} t - \frac{\sqrt{6}}{10} \sin \sqrt{6} t - \frac{1}{2} \sin 2 t$

Q5.6-6E

Q5.6-8E

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