Q32E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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 $\left(a\right)$.

Step-by-Step Solution

Answer

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 $\mathbf{f}\mathbf{=}\frac{5}{2\pi }$ 

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   $\mathrm{k}=250 \mathrm{kg}/{\sec }^2$

 

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  $10{\mathrm{r}}^2+250=0$

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

 

Therefore, the general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\cos \left(5\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(5\mathrm{t}\right)$.

Step 2: Finding c 1 and c 2

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

 

Then,

  $\begin{array}{l}\mathrm{y}\left(0\right)={\mathrm{c}}_1\cos \left(0\right)+{\mathrm{c}}_2\sin \left(0\right)\\ {\mathrm{c}}_1=0.3\end{array}$

 

And $\mathrm{y}\text{'}\left(\mathrm{t}\right)=-5{\mathrm{c}}_1\sin \left(5\mathrm{t}\right)+5{\mathrm{c}}_2\cos \left(5\mathrm{t}\right)$

 $\begin{array}{l}\mathrm{y}\text{'}\left(0\right)=-5{\mathrm{c}}_1\sin \left(0\right)+5{\mathrm{c}}_2\cos \left(0\right)\\ 5{\mathrm{c}}_2=-0.1\\ {\mathrm{c}}_2=-0.02\end{array}$

 

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

Step 3: Finding the time.

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

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

 
 

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  $\mathbf{f}\mathbf{=}\frac{5}{2\pi }$.