Q5.6-6E

Question

Referring to the coupled mass-spring system discussed in Example , suppose an external force $E (t) = 37 \cos 3 t$ is applied to the second object of mass $1 kg$ . The displacement functions $x (t)$ and $y (t)$ now satisfy the system

$(16) (2 x'' (t) + 6 x (t) - 2 y (t) = 0, (17) (y'' (t) + 2 y (t) - 2 x (t) = 37 \cos 3 t$

(a) Show that $x (t)$ satisfies the equation $(18) x^{(4)} (t) + 5 x'' (t) + 4 x (t) = 37 \cos 3 t$

(b) Find a general solution $x (t)$ to the equation (18). [Hint: Use undetermined coefficients with $x_{p} = Acos 3 t + Bsin 3 t$ .]

(c) Substitute $x (t)$ back into (16) to obtain a formula for $y (t)$ .

(d) If both masses are displaced 2m to the right of their equilibrium positions and then released, find the displacement functions $x (t)$ and $y (t)$ .

Step-by-Step Solution

Verified

Answer

(a) It has been shown that $x (t)$ satisfies the equation $(18) x^{(4)} (t) + 5 x'' (t) + 4 x (t) = 37 \cos 3 t$ .

(b) $x (t) = c_{1} cost + c_{2} sint + c_{3} \cos 2 t + c_{4} \sin 2 t + \frac{37}{40} \cos 3 t$

(d) The displacement functions x and y are

$x (t) = \frac{23}{8} cost - \frac{9}{5} \cos 2 t + \frac{37}{40} \cos 3 t y (t) = \frac{23}{4} cost + \frac{9}{5} \cos 2 t - 3 \times \frac{37}{40} \cos 3 t$

1Solving the given equation

Let's rewrite the given system in operator form:

$(2 D^{2} + 6) [x] - 2 [y] = 0 - 2 [x] + (D^{2} + 2) [y] = 37 \cos 3 t$

One wants to obtain the equation for x so we will eliminate y from the system. To do so one will multiply the first equation by $(D^{2} + 2)$ and the second by 2 and then add them together:

$\begin{array}{rcl} (D^{2} + 6) (D^{2} + 2) [x] - 2 (D^{2} + 2) [y] = 0 \\ - 4 [x] + 2 (D^{2} + 2) [y] = 2 \times 37 \cos 3 t \\ ((2 D^{2} + 6) (D^{2} + 2) - 4) [x] = 2 \times 37 \cos 3 t \end{array}$

$(2 D^{4} + 4 D^{2} + 6 D^{2} + 12 - 4) [x] = 2 \times 37 \cos 3 t 2 (D^{4} + 5 D^{2} + 4) [x] = 2 \times 37 \cos 3 t (D^{4} + 5 D^{2} + 4) [x] = 37 \cos 3 t$

Rewriting this back in standard form we have that $x (t)$ satisfies

$x^{(4)} (t) + 5 x'' (t) + 4 x (t) = 37 \cos 3 t$ .

2Finding roots

To find a general solution, one will first find a homogeneous solution for $x (t)$ .

The auxiliary equation is:

$r^{4} + 5 r^{2} + 4 = 0$ , and its roots are:

$r^{4} + 5 r^{2} + 4 = (r^{2} + 1) (r^{2} + 4) = 0 r^{2} = - 1, r^{2} = - 4 r_{1, 2} = \pm i, r_{3, 4} = \pm 2 i$

Therefore, the homogeneous solution for x is:

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

3Finding derivatives

One will find a particular solution by using the method of undetermined coefficients and assume that a particular solution has a form of $x_{p} (t) = Acos 3 t + Bsin 3 t$ .

One needs the second and the fourth derivative of $x_{p} (t)$ :

$x' = - 3 Asin 3 t + 3 Bcos 3 t, x'' = - 9 Acos 3 t - 9 Bsin 3 t x''' = 27 Asin 3 t - 27 Bcos 3 t, x^{(4)} = 81 Acos 3 t + 81 Bsin 3 t$

4Substituting The values

Now, one has that,

$x_{p}^{(4)} {(t) + 5 x}_{p}^{}'' (t) + 4 x_{p} (t) = 81 Acos 3 t + 81 Bsin 3 t + 5 (- 9 Acos 3 t - 9 Bsin 3 t) + 4 (Acos 3 t + Bsin 3 t) = 40 Acos 3 t + 40 Bsin 3 t = 37 \cos 3 t \Rightarrow 40 A = 37, 40 B = 0 \Leftrightarrow A = \frac{37}{40}, B = 0$

So, the particular solution for x is $x_{p} (t) = 37 / 40 \cos 3 t$ and the general solution for x is $x (t) = x_{h} + x_{p} = c_{1} cost + c_{2} sint + c_{3} \cos 2 t + c_{4} \sin 2 t + \frac{37}{40} \cos 3 t$ .

5Finding derivatives

The first equation of the given system gives us that $y (t) = x'' (t) + 3 x (t)$

So, one need to find the second derivative of x and substitute it into the previous equation.

$\begin{array}{rcl} x' (t) = - c_{1} sint + c_{2} cost - 2 c_{3} \sin 2 t + 2 c_{4} \cos 2 t - 3 \times \frac{37}{40} \sin 3 t \ h f i l l \\ x'' (t) = - c_{1} cost - c_{2} sint - 4 c_{3} \cos 2 t - 4 c_{4} \sin 2 t - 9 \times \frac{37}{40} \cos 3 t \end{array}$

Substituting this into $y (t) = x'' (t) + 3 x (t)$

One has,

$y (t) = - c_{1} cost - c_{2} sint - 4 c_{3} \cos 2 t - 4 c_{4} \sin 2 t - 9 \times \frac{37}{40} \cos 3 t + 3 (c_{1} cost + c_{2} sint + c_{3} \cos 2 t + c_{4} \sin 2 t + \frac{37}{40} \cos 3 t) y (t) = 2 c_{1} cost + 2 c_{2} sint - c_{3} \cos 2 t - c_{4} \sin 2 t - 3 \times \frac{37}{20} \cos 3 t$

6Finding derivatives

Both masses are displaced 2m to the right of their equilibrium position, so $x (0) = y (0) = 2$ , and since they are released, one has that $x' (0) = y' (0) = 0$ .

First, one will find the first derivative of y:

$y' (t) = - 2 c_{2} sint + 2 c_{2} cost + 2 c_{3} \sin 2 t - 2 c_{4} \cos 2 t - 9 \times \frac{37}{40} \sin 3 t$

So, from this initial condition,one has that

$\begin{array}{rcl} x (0) = c_{1} \times 1 + c_{2} \times 0 + c_{3} \times 1 + c_{4} \times 0 + \frac{37}{40} \times 1 \\ = & c_{1} + c_{3} + \frac{37}{40} \\ = & 2 \\ y (0) = 2 c_{1} \times 1 + 2 c_{2} \times 0 - c_{3} \times 1 - c_{4} \times 0 - 3 \times \frac{37}{20} \times 1 \\ = & 2 c_{1} - c_{3} - 6 \times \frac{37}{40} \\ = & 2 \end{array}$

7Substituting the values

Adding the previous two equations together one will get

$\begin{array}{rcl} 3 c_{1} - 5 \times \frac{37}{40} & = & 4 \\ c_{1} & = & \frac{23}{8} \end{array}$

Substituting this into the second equation one will have $c_{2} = 2 c_{1} - 6 \times \frac{37}{40} - 2 \Leftrightarrow c_{3} = - \frac{9}{5}$

The third and the fourth initial condition give us

$x' (0) = - c_{1} \times 0 + c_{2} \times 1 - 2 c_{3} \times 0 + 2 c_{4} \times 1 - 3 \times \frac{37}{40} \times 0 = c_{2} + 2 c_{4} = 0 y' (0) = - 2 c_{1} \times 0 + 2 c_{2} \times 1 + 2 c_{3} \times 0 - 2 c_{4} \times 1 - 9 \times \frac{37}{40} \times 0 = 2 c_{2} - 2 c_{4} = 0$

8Finding c 2 , c 4

Adding those equations together one gets that $3 c_{2} = 0 \Rightarrow c_{2} = 0$

Substituting this into the first equation one has that $2 c_{4} = 0 \Rightarrow c_{4} = 0$

Finally, one has that the displacement functions x and y is:

$x (t) = \frac{23}{8} cost - \frac{9}{5} \cos 2 t + \frac{37}{40} \cos 3 t y (t) = \frac{23}{4} cost + \frac{9}{5} \cos 2 t - 3 \times \frac{37}{40} \cos 3 t$

Q4E

Q5.6-7E

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