Consider the equation for free mechanical vibration,$\mathrm{my}+\mathrm{by}\text{'}+\mathrm{ky}=0$ , and assume the motion is critically damped. Let $\mathrm{y}\left(0\right)={\mathrm{y}}_0,\mathrm{y}\text{'}\left(0\right)={\mathrm{v}}_0$and assume ${\mathrm{y}}_0\ne 0$.

1. Prove that the mass will pass through its equilibrium at exactly one positive time if and only if $\frac{-2{\mathrm{my}}_0}{2{\mathrm{mv}}_0+{\mathrm{by}}_0}>0$.
2. Use computer software to illustrate part (a) for a specific choice of m, b, k,${\mathrm{y}}_0$ , and ${\mathrm{v}}_0$. Be sure to include an appropriate graph in your illustration.

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 ifk < 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 $\mathrm{\beta }=\frac{1}{2\mathrm{a}}\sqrt{4\mathrm{ac}-{\mathrm{b}}^2}$

Given that, $\mathrm{my}+\mathrm{by}\text{'}+\mathrm{ky}=0$ and assuming that the motion is critically damped.

Initial conditions are$\mathrm{y}\left(0\right)={\mathrm{y}}_0,\mathrm{y}\text{'}\left(0\right)={\mathrm{v}}_0$ and ${\mathrm{y}}_0\ne 0$.

Then one knows that must be. So, solve the given equation to find its roots.

And the auxiliary equation is${\mathrm{mr}}^2+\mathrm{br}+\mathrm{k}=0$ and its roots are $\mathrm{r}=-\frac{\mathrm{b}}{2\mathrm{m}}$and $\mathrm{r}=-\frac{\mathrm{b}}{2\mathrm{m}}$.

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

Use the initial conditions to find the value of c’s.

$$\begin{gathered} \left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\mathrm{t}}+{\mathrm{c}}_2{\mathrm{te}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\mathrm{t}} \\ \mathrm{y}\left(0\right)={\mathrm{c}}_1{\mathrm{e}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\left(0\right)}+{\mathrm{c}}_2\left(0\right){\mathrm{e}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\left(0\right)} \\ {\mathrm{y}}_0={\mathrm{c}}_1+0 \\ {\mathrm{c}}_1={\mathrm{y}}_0 \end{gathered}$$

Differentiate the equation (2) with respect to t.

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

.

Then,

$$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{t}\right)=-\frac{\mathrm{b}}{2\mathrm{m}}{\mathrm{c}}_1{\mathrm{e}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\mathrm{t}}-\frac{\mathrm{b}}{2\mathrm{m}}{\mathrm{c}}_2{\mathrm{te}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\mathrm{t}} \\ \mathrm{y}\text{'}\left(0\right)=-\frac{\mathrm{b}}{2\mathrm{m}}{\mathrm{c}}_1{\mathrm{e}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\left(0\right)}+{\mathrm{c}}_2{\mathrm{e}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\left(0\right)}-\frac{\mathrm{b}}{2\mathrm{m}}{\mathrm{c}}_2\left(0\right){\mathrm{e}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\left(0\right)} \\ {\mathrm{v}}_0=-\frac{\mathrm{b}}{2\mathrm{m}}{\mathrm{y}}_0+{\mathrm{c}}_2 \\ {\mathrm{c}}_2={\mathrm{v}}_0+\frac{\mathrm{b}}{2\mathrm{m}}{\mathrm{y}}_0 \end{gathered}$$

Substitute the values in equation (2).

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{y}}_0{\mathrm{e}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\mathrm{t}}+\left({\mathrm{v}}_0+\frac{\mathrm{b}}{2\mathrm{m}}{\mathrm{y}}_0\right){\mathrm{te}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\mathrm{t}}$…… (3)

Given that, $\frac{-2{\mathrm{my}}_0}{2{\mathrm{mv}}_0+{\mathrm{by}}_0}>0$.

To prove: the mass will pass through its equilibrium at exactly one positive time if and only if$\frac{-2{\mathrm{my}}_0}{2{\mathrm{mv}}_0+{\mathrm{by}}_0}>0$

Since the motion is critically damped, it doesn’t oscillate, so the mass will pass through its equilibrium only once at most.

And the mass passes through its equilibrium if and only if y(t) = 0 has a unique solution for some t > 0.

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

$$\begin{gathered} 0={\mathrm{y}}_0{\mathrm{e}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\mathrm{t}}+\left({\mathrm{v}}_0+\frac{\mathrm{b}}{2\mathrm{m}}{\mathrm{y}}_0\right){\mathrm{te}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\mathrm{t}} \\ -{\mathrm{y}}_0=\frac{\left({\mathrm{v}}_0+\frac{\mathrm{b}}{2\mathrm{m}}{\mathrm{y}}_0\right){\mathrm{te}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\mathrm{t}}}{{\mathrm{e}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\mathrm{t}}} \\ =\left({\mathrm{v}}_0+\frac{\mathrm{b}}{2\mathrm{m}}{\mathrm{y}}_0\right)\mathrm{t} \end{gathered}$$

Now solve for t.

$$\begin{gathered} \left({\mathrm{v}}_0+\frac{\mathrm{b}}{2\mathrm{m}}{\mathrm{y}}_0\right)\mathrm{t}=-{\mathrm{y}}_0 \\ \mathrm{t}=\frac{-{\mathrm{y}}_0}{\left({\mathrm{v}}_0+\frac{\mathrm{b}}{2\mathrm{m}}{\mathrm{y}}_0\right)} \\ =\frac{-2{\mathrm{my}}_0}{2{\mathrm{mv}}_0+{\mathrm{by}}_0} \end{gathered}$$

So,$\mathrm{t}=\frac{-2{\mathrm{my}}_0}{2{\mathrm{mv}}_0+{\mathrm{by}}_0}$ is the only solution. Then we have proven that y(t) = 0 has a unique solution for t > 0 if and only is $\frac{-2{\mathrm{my}}_0}{2{\mathrm{mv}}_0+{\mathrm{by}}_0}>0$.

Hence, the mass passes through its equilibrium only once if and only if$\frac{-2{\mathrm{my}}_0}{2{\mathrm{mv}}_0+{\mathrm{by}}_0}>0$

Referring to part (a): the mass passes through its equilibrium only once if and only if $\frac{-2{\mathrm{my}}_0}{2{\mathrm{mv}}_0+{\mathrm{by}}_0}>0$.

To draw a graph let us assume the values of m, b, k, ${\mathrm{y}}_0$and ${\mathrm{v}}_0$.

Let $\mathrm{m}=1,\mathrm{b}=2,\mathrm{k}=1,{\mathrm{y}}_0=1,{\mathrm{v}}_0=-5$(red) and $\mathrm{m}=1,\mathrm{b}=2,\mathrm{k}=1,{\mathrm{y}}_0=0.3,{\mathrm{v}}_0=2$(blue).

Using the above information, draw the graph.

The red curve indicates that the required inequality is satisfied and we can see that the mass returns to its equilibrium position only once for t > 0.

The blue curve indicates that the required inequality is not satisfied, and we can see that the mass never returns to its equilibrium position again.