Book edition 9th

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

Pages 616 pages

ISBN 9780321977069

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Short Answer

Vibrating Spring without Damping. A vibrating spring without damping can be modeled by the initial value problem $(11)$ in Example $3$ by taking $b = 0$ .

a) If $m = 10 kg, k = 250 kg / \sec^{2}, y (0) = 0 . 3 m$ , and $y' (0) = - 0 . 1 m / \sec$ , find the equation of motion for this undamped vibrating spring.
b) After how many seconds will the mass in part $(a)$ first cross the equilibrium point?
c) When the equation of motion is of the form displayed in $(9)$ , the motion is said to be oscillatory with frequency $β / 2 π$ . Find the frequency of oscillation for the spring system of part $(a)$ .

The equation of motion for vibrating spring is $y (t) = 0 . 3 \cos (5 t) - 0 . 02 \sin (5 t)$
The mass crosses the equilibrium at $t = 0.3$ seconds
The frequency of the spring is $f = \frac{5}{2 π}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Differentiating the equation.

The differential equation without damping is $my'' + ky = 0$

Given $m = 10 kg$ and $k = 250 kg / \sec^{2}$

Let $y = e^{rt}$

Then $y' (t) = r e^{rt}$

$y'' (t) = r^{2} e^{rt}$

Then the auxiliary equation is $10 r^{2} + 250 = 0$

$\begin{matrix} r^{2} + 25 = 0 \\ r = \pm 5 i \end{matrix}$

Therefore, the general solution is $y (t) = c_{1} \cos (5 t) + c_{2} \sin (5 t)$ .

Step 2: Finding c1 and c2

Given initial conditions are $y (0) = 0.3$ and $y' (0) = - 0 . 1$

Then,

$\begin{array}{l} y (0) = c_{1} \cos (0) + c_{2} \sin (0) \\ c_{1} = 0.3 \end{array}$

And $y' (t) = - 5 c_{1} \sin (5 t) + 5 c_{2} \cos (5 t)$

$\begin{array}{l} y' (0) = - 5 c_{1} \sin (0) + 5 c_{2} \cos (0) \\ 5 c_{2} = - 0.1 \\ c_{2} = - 0.02 \end{array}$

Therefore, the solution is $y (t) = 0 . 3 \cos (5 t) - 0 . 02 \sin (5 t)$ .

Step 3: Finding the time.

When the spring crosses the equilibrium $y (t) = 0$ , so we have to find the $t$

$\begin{array}{r} 0 . 3 \cos (5 t) - 0 . 02 \sin (5 t) = 0 \\ 0 . 3 \cos (5 t) = 0 . 02 \sin (5 t) \\ \tan (5 t) = \frac{0.3}{0.02} \\ 5 t = \arctan (15) \\ 5 t \approx 1.504 \\ t \approx 0.3 \end{array}$

So, at $t = 0.3$ seconds the mass crosses the equilibrium

Step 4: Finding the frequency

Here $β = 5$ , the Frequency of the spring is $f = \frac{5}{2 π}$ .

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Step by Step Solution

Step 1: Differentiating the equation.

Step 2: Finding c1 and c2

Step 3: Finding the time.

Step 4: Finding the frequency

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

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Step by Step Solution

Step 1: Differentiating the equation.

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Step 3: Finding the time.

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