Q4.9-6E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$\begin{array}{l}\mathbf{y}\mathbf{"}\left(t\right)\mathbf{+}4\mathbf{y}\mathbf{\text{'}}\left(t\right)\mathbf{+}\mathbf{ky}\left(t\right)\mathbf{=}\mathbf{0}\mathbf{;}\\ \mathbf{y}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{.}\end{array}$

Find the equation of motion and sketch its graph for k = 2, 4, and 6.

Step-by-Step Solution

Answer

Therefore, the equations of motions are the solution for k = 2 is, $\mathbf{y}\left(t\right)\mathbf{=}\left(\frac{1}{2}\mathbf{+}\frac{1}{\sqrt{2}}\right){\mathbf{e}}^{\left(\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\right)\mathbf{t}}\mathbf{+}\left(\frac{1}{2}\mathbf{-}\frac{1}{\sqrt{2}}\right){\mathbf{e}}^{\left(\mathbf{-}\mathbf{2}\mathbf{-}\sqrt{2}\right)\mathbf{t}}$solution for k = 4 is $\mathbf{y}\mathbf{=}{\mathbf{e}}^{-2t}\mathbf{+}\mathbf{2}{\mathbf{te}}^{-2t}$ and solution for k = 6 is  $\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{-}2\mathbf{t}}\mathbf{cos}\sqrt{2}\mathbf{t}\mathbf{+}\sqrt{2}{\mathbf{e}}^{\mathbf{-}2\mathbf{t}}\mathbf{sin}\sqrt{2}\mathbf{t}$

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:

 

$\begin{array}{c}{\mathbf{F}}_{\mathrm{ext}}\mathbf{=}\left[\mathrm{inertia}\right]\mathbf{y}\mathbf{"}\mathbf{+}\left[\mathrm{damping}\right]\mathbf{y}\mathbf{\text{'}}\mathbf{+}\left[\mathrm{stiffness}\right]\mathbf{y}\\ =\mathrm{my}"+\mathrm{by}\text{'}+\mathrm{ky}\end{array}$   …… (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.

Step 2: Evaluate the equation

Given that,  

 $\begin{array}{c}\mathbf{y}\mathbf{"}\left(t\right)\mathbf{+}4\mathbf{y}\mathbf{\text{'}}\left(t\right)\mathbf{+}\mathbf{ky}\left(t\right)\mathbf{=}\mathbf{0}\mathbf{;}\\ \mathbf{y}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{.}\end{array}$

 

And the value of. $\mathbf{k}\mathbf{=}2\mathbf{,}4\mathbf{,}6$

 

Let us take k = 2. Then,

 

 $\mathbf{y}\mathbf{"}\mathbf{+}4\mathbf{y}\mathbf{\text{'}}\mathbf{+}2\mathbf{y}\mathbf{=}\mathbf{0}$  …… (2)

 

Since the auxiliary equation is${\mathbf{r}}^2\mathbf{+}4\mathbf{r}\mathbf{+}2\mathbf{=}\mathbf{0}$. And roots are $\mathbf{r}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}$and.$\mathbf{r}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{-}\sqrt{2}$

 

Step3: Find the initial conditions

By the above information find the general solution.

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

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

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

$\begin{array}{c}\mathbf{y}\left(t\right)\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{\left(\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\right)\mathbf{t}}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{\left(\mathbf{-}\mathbf{2}\mathbf{-}\sqrt{2}\right)\mathbf{t}}\\ \mathbf{y}\left(0\right)\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{\left(0\right)}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{\left(0\right)}\\ \mathbf{1}\mathbf{=}{\mathbf{c}}_1\left(1\right)\mathbf{+}{\mathbf{c}}_2\left(1\right)\\ {\mathbf{c}}_1\mathbf{+}{\mathbf{c}}_2\mathbf{=}\mathbf{1}\end{array}$

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

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

Then,

$\begin{array}{c}\mathbf{y}\mathbf{\text{'}}\left(t\right)\mathbf{=}\left(\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\right){\mathbf{c}}_1{\mathbf{e}}^{\left(\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\right)\mathbf{t}}\mathbf{+}\left(\mathbf{-}\mathbf{2}\mathbf{-}\sqrt{2}\right){\mathbf{c}}_2{\mathbf{e}}^{\left(\mathbf{-}\mathbf{2}\mathbf{-}\sqrt{2}\right)\mathbf{t}}\\ \mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\left(\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\right){\mathbf{c}}_1{\mathbf{e}}^{\left(\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\right)\times 0}\mathbf{+}\left(\mathbf{-}\mathbf{2}\mathbf{-}\sqrt{2}\right){\mathbf{c}}_2{\mathbf{e}}^{\left(\mathbf{-}\mathbf{2}\mathbf{-}\sqrt{2}\right)\times 0}\\ \mathbf{0}\mathbf{=}\left(\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\right){\mathbf{c}}_1\mathbf{+}\left(\mathbf{-}\mathbf{2}\mathbf{-}\sqrt{2}\right){\mathbf{c}}_2\\ \left(\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\right){\mathbf{c}}_1\mathbf{+}\left(\mathbf{-}\mathbf{2}\mathbf{-}\sqrt{2}\right){\mathbf{c}}_2\mathbf{=}0\end{array}$

Solve the equations,

$\begin{array}{c}\left(\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\right){\mathbf{c}}_1\mathbf{+}\left(\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\right){\mathbf{c}}_2\mathbf{-}\left(\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\right){\mathbf{c}}_1\mathbf{-}\left(\mathbf{-}\mathbf{2}\mathbf{-}\sqrt{2}\right){\mathbf{c}}_2\mathbf{=}\left(\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\right)\\ \left(\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\right){\mathbf{c}}_2\mathbf{-}\left(\mathbf{-}\mathbf{2}\mathbf{-}\sqrt{2}\right){\mathbf{c}}_2\mathbf{=}\left(\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\right)\\ \left(\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\mathbf{+}\mathbf{2}\mathbf{+}\sqrt{2}\right){\mathbf{c}}_2\mathbf{=}\left(\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\right)\\ \mathbf{2}\sqrt{2}{\mathbf{c}}_2\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\\ {\mathbf{c}}_2\mathbf{=}\frac{1}{2}\mathbf{-}\frac{1}{\sqrt{2}}\end{array}$

Then, 

$\begin{array}{c}{\mathbf{c}}_1\mathbf{+}{\mathbf{c}}_2\mathbf{=}\mathbf{1}\\ {\mathbf{c}}_1\mathbf{+}\frac{1}{2}\mathbf{-}\frac{1}{\sqrt{2}}\mathbf{=}\mathbf{1}\\ {\mathbf{c}}_1\mathbf{=}\frac{1}{2}\mathbf{+}\frac{1}{\sqrt{2}}\end{array}$

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

$\mathbf{y}\left(t\right)\mathbf{=}\left(\frac{1}{2}\mathbf{+}\frac{1}{\sqrt{2}}\right){\mathbf{e}}^{\left(\mathbf{-}\mathbf{2}\mathbf{+}\sqrt{2}\right)\mathbf{t}}\mathbf{+}\left(\frac{1}{2}\mathbf{-}\frac{1}{\sqrt{2}}\right){\mathbf{e}}^{\left(\mathbf{-}\mathbf{2}\mathbf{-}\sqrt{2}\right)\mathbf{t}}$

Similarly, solution for k = 4 is, $\mathbf{y}\mathbf{=}{\mathbf{e}}^{-2t}\mathbf{+}\mathbf{2}{\mathbf{te}}^{-2t}$ and solution for k = 6 is

  $\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{-}2\mathbf{t}}\mathbf{cos}\sqrt{2}\mathbf{t}\mathbf{+}\sqrt{2}{\mathbf{e}}^{\mathbf{-}2\mathbf{t}}\mathbf{sin}\sqrt{2}\mathbf{t}$

Step4: Graph of the motion equation

The graph of the equation is.$\mathbf{y}\mathbf{"}\left(t\right)\mathbf{+}4\mathbf{y}\mathbf{\text{'}}\left(t\right)\mathbf{+}\mathbf{ky}\left(t\right)\mathbf{=}\mathbf{0}$$\mathbf{y}\mathbf{"}\left(t\right)\mathbf{+}4\mathbf{y}\mathbf{\text{'}}\left(t\right)\mathbf{+}\mathbf{ky}\left(t\right)\mathbf{=}\mathbf{0}$

 

When k = 2, 4, and 6 means draw the graph.