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Found in: Page 173

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Vibrating Spring without Damping. A vibrating spring without damping can be modeled by the initial value problem $\left(11\right)$ in Example ${\mathbf{3}}$ by taking ${\mathbf{b}}{\mathbf{=}}{\mathbf{0}}$. a) If ${\mathbf{m}}{\mathbf{=}}{\mathbf{10}}{\mathbf{}}{\mathbf{kg}}{\mathbf{,}}{\mathbf{k}}{\mathbf{=}}{\mathbf{250}}{\mathbf{}}{\mathbf{kg}}{\mathbf{/}}{{\mathbf{sec}}}^{2}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}{\mathbf{.}}{\mathbf{3}}{\mathbf{}}{\mathbf{m}}$, and ${\mathbf{y}}{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{0}{\mathbf{.}}{\mathbf{1}}{\mathbf{}}{\mathbf{m}}{\mathbf{/}}{\mathbf{sec}}$, find the equation of motion for this undamped vibrating spring.b) After how many seconds will the mass in part $\left(a\right)$ first cross the equilibrium point?c) When the equation of motion is of the form displayed in $\left(9\right)$, the motion is said to be oscillatory with frequency ${\mathbf{\beta }}{\mathbf{/}}{\mathbf{2}}{\mathbf{\pi }}$. Find the frequency of oscillation for the spring system of part $\left(a\right)$.

1. The equation of motion for vibrating spring is $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{cos}\mathbf{\left(}\mathbf{5}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{sin}\mathbf{\left(}\mathbf{5}\mathbf{t}\mathbf{\right)}$
2. The mass crosses the equilibrium at $\mathbf{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3}$ seconds
3. The frequency of the spring is $\mathbf{f}\mathbf{=}\frac{5}{2\pi }$
See the step by step solution

## Step 1: Differentiating the equation.

The differential equation without damping is $\mathrm{my}\text{'}\text{'}+\mathrm{ky}=0$

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

Let $\mathrm{y}={\mathrm{e}}^{\mathrm{rt}}$

Then $\mathrm{y}\text{'}\left(\mathrm{t}\right)=\mathrm{r}{\mathrm{e}}^{\mathrm{rt}}$

$\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)={\mathrm{r}}^{2}{\mathrm{e}}^{\mathrm{rt}}$

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

$\begin{array}{c}{\mathrm{r}}^{2}+25=0\\ \mathrm{r}=±5\mathrm{i}\end{array}$

Therefore, the general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_{1}\mathrm{cos}\left(5\mathrm{t}\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(5\mathrm{t}\right)$.

## Step 2: Finding c1 and  c2

Given initial conditions are $\mathrm{y}\left(0\right)=0.3$ and $\mathrm{y}\text{'}\left(0\right)=-0.1$

Then,

$\begin{array}{l}\mathrm{y}\left(0\right)={\mathrm{c}}_{1}\mathrm{cos}\left(0\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(0\right)\\ {\mathrm{c}}_{1}=0.3\end{array}$

And $\mathrm{y}\text{'}\left(\mathrm{t}\right)=-5{\mathrm{c}}_{1}\mathrm{sin}\left(5\mathrm{t}\right)+5{\mathrm{c}}_{2}\mathrm{cos}\left(5\mathrm{t}\right)$

$\begin{array}{l}\mathrm{y}\text{'}\left(0\right)=-5{\mathrm{c}}_{1}\mathrm{sin}\left(0\right)+5{\mathrm{c}}_{2}\mathrm{cos}\left(0\right)\\ 5{\mathrm{c}}_{2}=-0.1\\ {\mathrm{c}}_{2}=-0.02\end{array}$

Therefore, the solution is $\mathrm{y}\left(\mathrm{t}\right)=0.3\mathrm{cos}\left(5\mathrm{t}\right)-0.02\mathrm{sin}\left(5\mathrm{t}\right)$ .

## Step 3: Finding the time.

When the spring crosses the equilibrium $\mathrm{y}\left(\mathrm{t}\right)=0$ , so we have to find the $\mathbf{t}$

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

So, at $\mathbf{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3}$ seconds the mass crosses the equilibrium

## Step 4: Finding the frequency

Here $\mathbf{\beta }\mathbf{=}\mathbf{5}$, the Frequency of the spring is $\mathbf{f}\mathbf{=}\frac{5}{2\pi }$.