Q4 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

The motion of a mass-spring system with damping is governed by 

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)+\mathrm{by}\text{'}\left(\mathrm{t}\right)+64\mathrm{y}\left(\mathrm{t}\right)=0; \\ \mathrm{y}\left(0\right)=1, \mathrm{y}\text{'}\left(0\right)=0. \end{gathered}$$

Find the equation of motion and sketch its graph for b = 0,10,16, and 20.

Step-by-Step Solution

Answer

Therefore, the equations of motion are the solution for b = 0 is $\mathrm{y}=\cos 8\mathrm{t}$, the solution for $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-5\mathrm{t}}\cos \sqrt{39}\mathrm{t}+\frac{5}{\sqrt{39}}{\mathrm{e}}^{-5\mathrm{t}}\sin \sqrt{39}\mathrm{t}$ b = 10 is, solution for b = 16 is $\mathrm{y}=\left(1+5\mathrm{t}\right){\mathrm{e}}^{-8\mathrm{t}}$and solution for b = 10 is $\mathrm{y}=-\frac{1}{3}{\mathrm{e}}^{-16\mathrm{t}}+\frac{4}{3}{\mathrm{e}}^{-4\mathrm{t}}$.

A sketch of the graph is shown below.

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)

Step 2: Evaluate the equation

Given that, 

 $$\begin{gathered} \left(\mathrm{t}\right)+\mathrm{by}\text{'}\left(\mathrm{t}\right)+64\mathrm{y}\left(\mathrm{t}\right)=0; \\ \mathrm{y}\left(0\right)=1, \mathrm{y}\text{'}\left(0\right)=0 \end{gathered}$$

And the value of $\mathrm{b}=0,10,16,20$.

Let us take b = 10. Then,

 

$\mathrm{y}+10\mathrm{y}\text{'}+64\mathrm{y}=0$…… (2)

 

Since the auxiliary equation is ${\mathrm{r}}^2+10\mathrm{r}+64=0$. And roots are $\mathrm{r}=-5+\sqrt{39}\mathrm{i}$ and $\mathrm{r}=-5-\sqrt{39}\mathrm{i}$$\mathrm{r}=-5-\sqrt{39}\mathrm{i}$ 

Step 3: Find the initial conditions

By the above information, find the general solution.

The general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{-5\mathrm{t}}\cos \sqrt{39}\mathrm{t}+{\mathrm{c}}_2{\mathrm{e}}^{-5\mathrm{t}}\sin \sqrt{39}\mathrm{t}$   …… (3)

Given the initial conditions are $\mathrm{y}\left(0\right)=1,\mathrm{y}\text{'}\left(0\right)=0$.

 

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

 $$\begin{gathered} \left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{-5\mathrm{t}}\cos \sqrt{39}\mathrm{t}+{\mathrm{c}}_2{\mathrm{e}}^{-5\mathrm{t}}\sin \sqrt{39}\mathrm{t} \\ \mathrm{y}\left(0\right)={\mathrm{c}}_1{\mathrm{e}}^{-5\left(0\right)}\cos \left(0\right)+{\mathrm{c}}_2{\mathrm{e}}^{-5\left(0\right)}\sin \left(0\right) \\ 1={\mathrm{c}}_1\left(1\right)+0 \\ {\mathrm{c}}_1=1 \end{gathered}$$

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

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=-5{\mathrm{c}}_1{\mathrm{e}}^{-3\mathrm{t}}\cos \sqrt{39}\mathrm{t}-\sqrt{39}{\mathrm{c}}_1{\mathrm{e}}^{-5\mathrm{t}}\sin \sqrt{39}\mathrm{t}-5{\mathrm{c}}_2{\mathrm{e}}^{-5\mathrm{t}}\sin \sqrt{39}\mathrm{t}+\sqrt{39}{\mathrm{c}}_2{\mathrm{e}}^{-5\mathrm{t}}\cos \sqrt{39}\mathrm{t}$

Then,

$$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{t}\right)=-5{\mathrm{c}}_1{\mathrm{e}}^{-3\mathrm{t}}\cos \sqrt{39}\mathrm{t}-\sqrt{39}{\mathrm{c}}_1{\mathrm{e}}^{-5\mathrm{t}}\sin \sqrt{39}\mathrm{t}-5{\mathrm{c}}_2{\mathrm{e}}^{-5\mathrm{t}}\sin \sqrt{39}\mathrm{t}+\sqrt{39}{\mathrm{c}}_2{\mathrm{e}}^{-5\mathrm{t}}\cos \sqrt{39}\mathrm{t} \\ \mathrm{y}\text{'}\left(0\right)=-5{\mathrm{c}}_1{\mathrm{e}}^{-5\left(0\right)}\cos \left(0\right)-\sqrt{39}{\mathrm{c}}_1{\mathrm{e}}^{-5\left(0\right)}\sin \left(0\right)-5{\mathrm{c}}_2{\mathrm{e}}^{-5\left(0\right)}\sin \left(0\right)+\sqrt{39}{\mathrm{c}}_2{\mathrm{e}}^{-5\left(0\right)}\cos \left(0\right) \\ 0=-5{\mathrm{c}}_1+\sqrt{39}{\mathrm{c}}_2\left(1\right) \\ {\mathrm{c}}_2=\frac{5}{\sqrt{39}} \end{gathered}$$

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

 $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-5\mathrm{t}}\cos \sqrt{39}\mathrm{t}+\frac{5}{\sqrt{39}}{\mathrm{e}}^{-5\mathrm{t}}\sin \sqrt{39}$

Similarly, the solution for b = 0 is $\mathrm{y}=\cos 8\mathrm{t}$, solution for b = 16 is $\mathrm{y}={\mathrm{e}}^{-8\mathrm{t}}+5{\mathrm{te}}^{-8\mathrm{t}}$and solution for b = 20 is $\mathrm{y}=-\frac{1}{3}{\mathrm{e}}^{-16\mathrm{t}}+\frac{4}{3}{\mathrm{e}}^{-4\mathrm{t}}$.

Step 4: Graph of the motion equation

The graph of the equation is $\mathrm{y}\left(\mathrm{t}\right)+\mathrm{by}\text{'}\left(\mathrm{t}\right)+64\mathrm{y}\left(\mathrm{t}\right)=0$.

When b = 0, 10, 16, and 20 means, draw the graph.

 

Case (1):

Case (2):

Case (3):

Case (4):