Q9 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A 2 – kg mass is attached to a spring with a stiffness of 40 N/m. The damping constant for the system is $8\sqrt{5}$N-sec/m. If the mass is pulled 10 cm to the right of the equilibrium and given an initial rightward velocity of 2 m/sec, what is the maximum displacement from equilibrium that it will attain?

Step-by-Step Solution

Answer

Therefore, the maximum displacement from equilibrium will attain at 0.242 m.

Step 1: General form

The Mass–Spring Oscillator:

A damped mass-spring oscillator consists of a mass m attached to a spring fixed at one end, as shown in Figure 4.1. Devise a differential equation that governs the motion of this oscillator, taking into account the forces acting on it due to the spring elasticity, damping friction, and possible external influences.

Mass–spring oscillator equation:

 

   $$\begin{gathered} {\mathrm{F}}_{\mathrm{ext}}=\left[\mathrm{inertia}\right]\mathrm{y}+\left[\mathrm{damping}\right]\mathrm{y}\text{'}+\left[\mathrm{stiffness}\right]\mathrm{y} \\ =\mathrm{my}+\mathrm{by}\text{'}+\mathrm{ky} \end{gathered}$$…… (1)

 

The rule for the bounded equation: Just based on stiffness, we can decide whether it is bounded or not if stiffness k > 0 then it is bounded, and if k < 0 then it is unbounded.

 

Root finding formula:

 

If. Then, $\mathrm{\alpha }=-\frac{\mathrm{b}}{2\mathrm{a}}$ and $\mathrm{\beta }=\frac{1}{2\mathrm{a}}\sqrt{4\mathrm{ac}-{\mathrm{b}}^2}$.

Step 2: Evaluate the equation

Given that, ${\mathrm{F}}_{\mathrm{ext}}=0,\mathrm{b}=8\sqrt{5},\mathrm{m}=2\mathrm{kg},\mathrm{k}=40\mathrm{N}/\mathrm{m},\mathrm{y}\left(0\right)=0.10$ and $\mathrm{y}\text{'}\left(0\right)=2$.

 

Now form the initial value problem using the above information.

 

  $2\mathrm{y}+8\sqrt{5}\mathrm{y}\text{'}+40\mathrm{y}=0;\mathrm{y}\left(0\right)=0.10,\mathrm{y}\text{'}\left(0\right)=2$…… (2)

 

Then, find the value of roots.

 

$$\begin{gathered} {\mathrm{b}}^2=320 \\ 4\mathrm{mk}=4\times 2\times 40 \\ =320 \end{gathered}$$

So, ${\mathrm{b}}^2=4\mathrm{mk}$. Then find the roots.

 

Since the auxiliary equation is $2{\mathrm{r}}^2+8\sqrt{5}\mathrm{r}+40=0$. And roots are $\mathrm{r}=-2\sqrt{5}$ and $\mathrm{r}=-2\sqrt{5}$

Step 3: Find the initial conditions

From the above information, find the general solution.

 

The general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{-2\sqrt{5}\mathrm{t}}+{\mathrm{c}}_2{\mathrm{te}}^{-2\sqrt{5}\mathrm{t}}$…… (3)

Given initial conditions are $\mathrm{y}\left(0\right)=0.10,\mathrm{y}\text{'}\left(0\right)=2$.

Now, substitute the initial conditions to find the value of c.

 $$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{-2\sqrt{5}\mathrm{t}}+{\mathrm{c}}_2{\mathrm{te}}^{-2\sqrt{5}\mathrm{t}} \\ \mathrm{y}\left(0\right)={\mathrm{c}}_1{\mathrm{e}}^{-2\sqrt{5}\left(0\right)}+{\mathrm{c}}_2\left(0\right){\mathrm{e}}^{-2\sqrt{5}\left(0\right)} \\ 0.10={\mathrm{c}}_1\left(1\right)+0 \\ {\mathrm{c}}_1=0.10 \end{gathered}$$

Find the derivative of equation (3). And implement the initial condition.

 $\mathrm{y}\text{'}\left(\mathrm{t}\right)=-2\sqrt{5}{\mathrm{c}}_1{\mathrm{e}}^{-2\sqrt{5}\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-2\sqrt{5}\mathrm{t}}-2\sqrt{5}{\mathrm{c}}_2{\mathrm{te}}^{-2\sqrt{5}\mathrm{t}}$

Then,

 $$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{t}\right)=-2\sqrt{5}{\mathrm{c}}_1{\mathrm{e}}^{-2\sqrt{5}\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-2\sqrt{5}\mathrm{t}}-2\sqrt{5}{\mathrm{c}}_2{\mathrm{te}}^{-2\sqrt{5}\mathrm{t}} \\ \mathrm{y}\text{'}\left(0\right)=-2\sqrt{5}{\mathrm{c}}_1{\mathrm{e}}^{-2\sqrt{5}\left(0\right)}+{\mathrm{c}}_2{\mathrm{e}}^{-2\sqrt{5}\left(0\right)}-2\sqrt{5}{\mathrm{c}}_2\left(0\right){\mathrm{e}}^{-2\sqrt{5}\left(0\right)} \\ 2=-0.2\sqrt{5}+{\mathrm{c}}_2 \\ {\mathrm{c}}_2=2+0.2\sqrt{5} \end{gathered}$$

Now substitute the value of c in equation (3).

$\mathrm{y}\left(\mathrm{t}\right)=\left(0.1\right){\mathrm{e}}^{-2\sqrt{5}\mathrm{t}}+\left(2+0.2\sqrt{5}\right){\mathrm{te}}^{-2\sqrt{5}\mathrm{t}}$.........(1)

Step 4: Find, when the maximum displacement happens in the equilibrium.

Since one has repeated roots, the solution does not oscillate. Then one needs to find when the maximum displacement happens from the equilibrium and return the obtained value for t in the solution to get the maximum displacements. 

 

In other words, to find the time when this maximum happens, one must find the first derivative and equalize it with zero.

 

To find the equilibrium position, put $\mathrm{y}\text{'}\left(\mathrm{t}\right)=0$.

Then, $0=-0.2\sqrt{5}{\mathrm{e}}^{-2\sqrt{5}\mathrm{t}}+\left(2+0.2\sqrt{5}\right){\mathrm{e}}^{-2\sqrt{5}\mathrm{t}}-2\sqrt{5}\left(2+0.2\sqrt{5}\right){\mathrm{te}}^{-2\sqrt{5}\mathrm{t}}$.

Now solve the above equation to find the value of t.

 $$\begin{gathered} 0=-0.2\sqrt{5}{\mathrm{e}}^{-2\sqrt{5}\mathrm{t}}+\left(2+0.2\sqrt{5}\right){\mathrm{e}}^{-2\sqrt{5}\mathrm{t}}-2\sqrt{5}\left(2+0.2\sqrt{5}\right){\mathrm{te}}^{-2\sqrt{5}\mathrm{t}} \\ =-0.2\sqrt{5}{\mathrm{e}}^{-2\sqrt{5}\mathrm{t}}+2{\mathrm{e}}^{-2\sqrt{5}\mathrm{t}}+0.2\sqrt{5}{\mathrm{e}}^{-2\sqrt{5}\mathrm{t}}-4\sqrt{5}{\mathrm{te}}^{-2\sqrt{5}\mathrm{t}}+2{\mathrm{te}}^{-2\sqrt{5}\mathrm{t}} \\ =2{\mathrm{e}}^{-2\sqrt{5}\mathrm{t}}\left(1-2\sqrt{5}\mathrm{t}+\mathrm{t}\right) \\ =1-\left(2\sqrt{5}+1\right)\mathrm{t} \\ \left(2\sqrt{5}+1\right)\mathrm{t}=1 \\ \mathrm{t}=\frac{1}{2\sqrt{5}+1}\times \frac{2\sqrt{5}-1}{2\sqrt{5}-1} \\ =\frac{2\sqrt{5}-1}{19} \\ \approx 0.183 \end{gathered}$$

So, $\mathrm{t}\approx 0.183$ .

 

Substitute the value of t in equation (4).

 $$\begin{gathered} \left(\mathrm{t}\right)=\left(0.1\right){\mathrm{e}}^{-2\sqrt{5}\mathrm{t}}+\left(2+0.2\sqrt{5}\right){\mathrm{te}}^{-2\sqrt{5}\mathrm{t}} \\ \mathrm{y}\left(0.183\right)=\left(0.1\right){\mathrm{e}}^{-2\sqrt{5}\left(0.183\right)}+\left(2+0.2\sqrt{5}\right)\left(0.183\right){\mathrm{e}}^{-2\sqrt{5}\left(0.183\right)} \\ =0.441\left(0.1+2\left(0.183\right)+0.2\sqrt{5}\left(0.183\right)\right) \\ \approx 0.242 \end{gathered}$$

 So, the maximum displacement from the equilibrium is 0.242m.