Q18 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Consider the equation for free mechanical vibration, $\mathrm{my}+\mathrm{by}\text{'}+\mathrm{ky}=0$ , and assume the motion is over-damped. Suppose $\mathrm{y}\left(0\right)>0$and $\mathrm{y}\text{'}\left(0\right)>0$. Prove that the mass will never pass through its equilibrium at any positive time.

Step-by-Step Solution

Answer

Therefore, the mass will never pass through its equilibrium at any positive time is true.

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.

The 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 ${\mathrm{b}}^2<4\mathrm{ac}$. 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 $\mathrm{my}+\mathrm{by}\text{'}+\mathrm{ky}=0.$

 

Assuming that the motion is over-damped

 

Initial conditions are $\mathrm{y}\left(0\right)>0$  and $\mathrm{y}\text{'}\left(0\right)>0$.

 

Then one knows that must be ${\mathrm{b}}^2>4\mathrm{mk}$. So, the roots to the auxiliary equation are two different negative reals ${\mathrm{r}}_1$ and ${\mathrm{r}}_2$.

 

Since the general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{{\mathrm{r}}_1\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{{\mathrm{r}}_2\mathrm{t}}$    …… (2)

 

Let us assume that ${\mathrm{r}}_1<{\mathrm{r}}_2$. Then, denote $\mathrm{y}\left(0\right)={\mathrm{y}}_0,\mathrm{y}\text{'}\left(0\right)={\mathrm{v}}_0$ and ${\mathrm{y}}_0>0,{\mathrm{v}}_0>0$

 

Use the initial conditions to find the value of constant.

 

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{{\mathrm{r}}_1\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{{\mathrm{r}}_2\mathrm{t}} \\ \mathrm{y}\left(0\right)={\mathrm{c}}_1{\mathrm{e}}^{{\mathrm{r}}_1\left(0\right)}+{\mathrm{c}}_2{\mathrm{e}}^{{\mathrm{r}}_2\left(0\right)} \\ {\mathrm{y}}_0={\mathrm{c}}_1+{\mathrm{c}}_2 \\ {\mathrm{c}}_1+{\mathrm{c}}_2={\mathrm{y}}_0 \end{gathered}$$

Differentiate the equation (2) with respect to t.

 $\mathrm{y}\text{'}\left(\mathrm{t}\right)={\mathrm{r}}_1{\mathrm{c}}_1{\mathrm{e}}^{{\mathrm{r}}_1\mathrm{t}}+{\mathrm{r}}_2{\mathrm{c}}_2{\mathrm{e}}^{{\mathrm{r}}_2\mathrm{t}}$

Then,

 $$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{t}\right)={\mathrm{r}}_1{\mathrm{c}}_1{\mathrm{e}}^{{\mathrm{r}}_1\mathrm{t}}+{\mathrm{r}}_2{\mathrm{c}}_2{\mathrm{e}}^{{\mathrm{r}}_2\mathrm{t}} \\ \mathrm{y}\text{'}\left(0\right)={\mathrm{r}}_1{\mathrm{c}}_1{\mathrm{e}}^{{\mathrm{r}}_1\left(0\right)}+{\mathrm{r}}_2{\mathrm{c}}_2{\mathrm{e}}^{{\mathrm{r}}_2\left(0\right)} \\ {\mathrm{v}}_0={\mathrm{r}}_1{\mathrm{c}}_1+{\mathrm{r}}_2{\mathrm{c}}_2 \\ {\mathrm{r}}_1{\mathrm{c}}_1+{\mathrm{r}}_2{\mathrm{c}}_2={\mathrm{v}}_0 \end{gathered}$$

Now solve the equations.

 $$\begin{gathered} {\mathrm{c}}_1=\frac{{\mathrm{v}}_0-{\mathrm{r}}_2{\mathrm{y}}_0}{{\mathrm{r}}_1-{\mathrm{r}}_2} \\ {\mathrm{c}}_2=\frac{{\mathrm{r}}_1{\mathrm{y}}_0-{\mathrm{v}}_0}{{\mathrm{r}}_1-{\mathrm{r}}_2} \end{gathered}$$

 Substitute the values in equation (2).

 

 $\mathrm{y}\left(\mathrm{t}\right)=\frac{{\mathrm{v}}_0-{\mathrm{r}}_2{\mathrm{y}}_0}{{\mathrm{r}}_1-{\mathrm{r}}_2}{\mathrm{e}}^{{\mathrm{r}}_1\mathrm{t}}+\frac{{\mathrm{r}}_1{\mathrm{y}}_0-{\mathrm{v}}_0}{{\mathrm{r}}_1-{\mathrm{r}}_2}{\mathrm{e}}^{{\mathrm{r}}_2\mathrm{t}}$ …… (3)

 

Step 3: Evaluate the general solution

 Let us assume that the mass passes through its equilibrium position for some positive time. That is, y(t) = 0 has a unique solution for some t > 0.

 

Let us find it. Put y(t) = 0 in equation (3).

 

$$\begin{gathered} 0=\frac{{\mathrm{v}}_0-{\mathrm{r}}_2{\mathrm{y}}_0}{{\mathrm{r}}_1-{\mathrm{r}}_2}{\mathrm{e}}^{{\mathrm{r}}_1\mathrm{t}}+\frac{{\mathrm{r}}_1{\mathrm{y}}_0-{\mathrm{v}}_0}{{\mathrm{r}}_1-{\mathrm{r}}_2}{\mathrm{e}}^{{\mathrm{r}}_2\mathrm{t}} \\ -{\mathrm{e}}^{\left({\mathrm{r}}_2-{\mathrm{r}}_1\right)\mathrm{t}}=\frac{\frac{{\mathrm{v}}_0-{\mathrm{r}}_2{\mathrm{y}}_0}{{\mathrm{r}}_1-{\mathrm{r}}_2}}{\frac{{\mathrm{r}}_1{\mathrm{y}}_0-{\mathrm{v}}_0}{{\mathrm{r}}_1-{\mathrm{r}}_2}}=\frac{{\mathrm{v}}_0-{\mathrm{r}}_2{\mathrm{y}}_0}{{\mathrm{r}}_1{\mathrm{y}}_0-{\mathrm{v}}_0} \\ {\mathrm{e}}^{\left({\mathrm{r}}_2-{\mathrm{r}}_1\right)\mathrm{t}}=\frac{{\mathrm{v}}_0-{\mathrm{r}}_2{\mathrm{y}}_0}{{\mathrm{v}}_0-{\mathrm{r}}_1{\mathrm{y}}_0} \\ \left({\mathrm{r}}_2-{\mathrm{r}}_1\right)\mathrm{t}=\ln \left(\frac{{\mathrm{v}}_0-{\mathrm{r}}_2{\mathrm{y}}_0}{{\mathrm{v}}_0-{\mathrm{r}}_1{\mathrm{y}}_0}\right) \\ \mathrm{t}=\frac{1}{{\mathrm{r}}_2-{\mathrm{r}}_1}\ln \left(\frac{{\mathrm{v}}_0-{\mathrm{r}}_2{\mathrm{y}}_0}{{\mathrm{v}}_0-{\mathrm{r}}_1{\mathrm{y}}_0}\right) \end{gathered}$$

So,  $\mathrm{t}=\frac{1}{{\mathrm{r}}_2-{\mathrm{r}}_1}\ln \left(\frac{{\mathrm{v}}_0-{\mathrm{r}}_2{\mathrm{y}}_0}{{\mathrm{v}}_0-{\mathrm{r}}_1{\mathrm{y}}_0}\right)$is the only solution. 

Step 4: Evaluate the value

Let one assume that t > 0 and ${\mathrm{r}}_1<{\mathrm{r}}_2$, so we can take it as $\ln \left(\frac{{\mathrm{v}}_0-{\mathrm{r}}_2{\mathrm{y}}_0}{{\mathrm{v}}_0-{\mathrm{r}}_1{\mathrm{y}}_0}\right)>0$. one gets,

$$\begin{gathered} \ln \left(\frac{{\mathrm{v}}_0-{\mathrm{r}}_2{\mathrm{y}}_0}{{\mathrm{v}}_0-{\mathrm{r}}_1{\mathrm{y}}_0}\right)>0 \\ \frac{{\mathrm{v}}_0-{\mathrm{r}}_2{\mathrm{y}}_0}{{\mathrm{v}}_0-{\mathrm{r}}_1{\mathrm{y}}_0}>1 \\ \frac{{\mathrm{v}}_0-{\mathrm{r}}_2{\mathrm{y}}_0}{{\mathrm{v}}_0-{\mathrm{r}}_1{\mathrm{y}}_0}-1>0 \\ \frac{{\mathrm{v}}_0-{\mathrm{r}}_2{\mathrm{y}}_0-{\mathrm{v}}_0+{\mathrm{r}}_1{\mathrm{y}}_0}{{\mathrm{v}}_0-{\mathrm{r}}_1{\mathrm{y}}_0}>0 \\ \frac{{\mathrm{y}}_0\left({\mathrm{r}}_1-{\mathrm{r}}_2\right)}{{\mathrm{v}}_0-{\mathrm{r}}_1{\mathrm{y}}_0}>0 \end{gathered}$$

Since  ${\mathrm{y}}_0>0$and ${\mathrm{r}}_1<{\mathrm{r}}_2$by the assumption, that the previous inequality is true, must be

 $$\begin{gathered} {\mathrm{v}}_0-{\mathrm{r}}_1{\mathrm{y}}_0<0 \\ {\mathrm{v}}_0<{\mathrm{r}}_1{\mathrm{y}}_0 \end{gathered}$$

 But the above result is not true because ${\mathrm{v}}_0>0$. Therefore, our assumption that 

y(t) = 0 has a solution for t > 0 was wrong.