Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

Answers without the blur.

Just sign up for free and you're in.

Short Answer

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 $y (t)$ .

(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

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Solving 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

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

Finding 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:

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

Finding 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 $x_{p} (t)$ :

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

Substituting 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 $x (t) = x_{h} + x_{p} = c_{1} cost + c_{2} sint + c_{3} \cos 2 t + c_{4} \sin 2 t + \frac{37}{40} \cos 3 t$ .

Finding 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,

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

Finding 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

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

Substituting 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

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

Finding c2,c4

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:

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

Find Study Materials for

Create Study Materials

Select your language

Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Step by Step Solution

Solving the given equation

Finding roots

Finding derivatives

Substituting The values

Finding derivatives

Finding derivatives

Substituting the values

Finding c2,c4

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Select your language

Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Step by Step Solution

Solving the given equation

Finding roots

Finding derivatives

Substituting The values

Finding derivatives

Finding derivatives

Substituting the values

Finding c2,c4

Most popular questions for Math Textbooks

Want to see more solutions like these?

Recommended explanations on Math Textbooks

Decision Maths

Calculus

Probability and Statistics

Statistics

Mechanics Maths

Geometry

Pure Maths