A$\frac{1}{8}$ – kg mass is attached to a spring with stiffness 16 N/m. The damping constant for the system is 2 N-sec/m. If the mass is moved $\frac{3}{4}\mathrm{m}$to the left of the equilibrium and given an initial leftward velocity of 2 m/sec, determine the equation of motion of the mass and give its damping factor, quasiperiodic, and quasi frequency.

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

Given that, A $\frac{1}{8}$– kg mass is attached to a spring with stiffness 16 N/m. The mass is moved $\frac{3}{4}\mathrm{m}$to the left of the equilibrium and given an initial leftward velocity of 2 m/sec.

Use the given information to find the equation.

Then, ${\mathrm{F}}_{\mathrm{ext}}=0,\mathrm{b}=2,\mathrm{m}=\frac{1}{8}\mathrm{kg},\mathrm{k}=16\mathrm{N}/\mathrm{m},\mathrm{y}\left(0\right)=-\frac{3}{4}$and $\mathrm{y}\text{'}\left(0\right)=-2$.

Now form the initial value problem using the above information.

$\frac{1}{8}\mathrm{y}+2\mathrm{y}\text{'}+16\mathrm{y}=0;\mathrm{y}\left(0\right)=-\frac{3}{4},\mathrm{y}\text{'}\left(0\right)=-2$…… (2)

Then, find the value of roots.

$$\begin{gathered} {\mathrm{b}}^2=4 \\ 4\mathrm{mk}=4\times \frac{1}{8}\times 16 \\ =8 \end{gathered}$$

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

$$\begin{gathered} \mathrm{\alpha }=-\frac{\mathrm{b}}{2\mathrm{m}} \\ =-\frac{2}{\frac{1}{4}} \\ =-8 \\ \mathrm{\beta }=\frac{1}{2\mathrm{m}}\sqrt{4\mathrm{mk}-{\mathrm{b}}^2} \\ =4\sqrt{8-4} \\ =4\sqrt{4} \\ =8 \end{gathered}$$

Since, the auxiliary equation is $\frac{1}{8}{\mathrm{r}}^2+2\mathrm{r}+16=0$. And roots are $\mathrm{r}=-8$and$\mathrm{r}=8$.

From the above information, find the general solution.

The general solution is$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{\alpha t}}\left({\mathrm{c}}_1\mathrm{cos\beta t}+{\mathrm{c}}_2\mathrm{sin\beta t}\right)$

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-8\mathrm{t}}\left({\mathrm{c}}_1\cos 8\mathrm{t}+{\mathrm{c}}_2\sin 8\mathrm{t}\right)$…… (3)

Given initial conditions are $\mathrm{y}\left(0\right)=\frac{3}{4},\mathrm{y}\text{'}\left(0\right)=2$.

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

$$\begin{gathered} \left(\mathrm{t}\right)={\mathrm{e}}^{-8\mathrm{t}}\left({\mathrm{c}}_1\cos 8\mathrm{t}+{\mathrm{c}}_2\sin 8\mathrm{t}\right) \\ \mathrm{y}\left(0\right)={\mathrm{e}}^{-8\left(0\right)}\left({\mathrm{c}}_1\cos \left(0\right)+{\mathrm{c}}_2\sin \left(0\right)\right) \\ -\frac{3}{4}={\mathrm{c}}_1\left(1\right)+0 \\ {\mathrm{c}}_1=-\frac{3}{4} \end{gathered}$$

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

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=-8{\mathrm{e}}^{-8\mathrm{t}}{\mathrm{c}}_1\cos 8\mathrm{t}-8{\mathrm{e}}^{-8\mathrm{t}}{\mathrm{c}}_1\sin 8\mathrm{t}-8{\mathrm{e}}^{-8\mathrm{t}}{\mathrm{c}}_2\sin 8\mathrm{t}+8{\mathrm{e}}^{-8\mathrm{t}}{\mathrm{c}}_2\cos 8\mathrm{t}$

Then,

$$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{t}\right)=-8{\mathrm{e}}^{-8\mathrm{t}}{\mathrm{c}}_1\cos 8\mathrm{t}-8{\mathrm{e}}^{-8\mathrm{t}}{\mathrm{c}}_1\sin 8\mathrm{t}-8{\mathrm{e}}^{-8\mathrm{t}}{\mathrm{c}}_2\sin 8\mathrm{t}+8{\mathrm{e}}^{-8\mathrm{t}}{\mathrm{c}}_2\cos 8\mathrm{t} \\ \mathrm{y}\text{'}\left(0\right)=-8{\mathrm{e}}^{-8\left(0\right)}{\mathrm{c}}_1\cos \left(0\right)-8{\mathrm{e}}^{-8\left(0\right)}{\mathrm{c}}_1\sin \left(0\right)-8{\mathrm{e}}^{-8\left(0\right)}{\mathrm{c}}_2\sin \left(0\right)+8{\mathrm{e}}^{-8\left(0\right)}{\mathrm{c}}_2\cos \left(0\right) \\ 2=-8{\mathrm{c}}_1+8{\mathrm{c}}_2\left(1\right) \\ -2=6+8{\mathrm{c}}_2 \\ -2-6=8{\mathrm{c}}_2 \\ {\mathrm{c}}_2=-1 \end{gathered}$$

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

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-8\mathrm{t}}\left(-\frac{3}{4}\cos 8\mathrm{t}-\sin 8\mathrm{t}\right)$…… (4)

Rewrite the equation (4) in the form of $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{Ae}}^{\mathrm{\alpha t}}\sin \left(\mathrm{\beta t}+\varphi \right)$.

Then, find the amplitude.

$$\begin{gathered} \mathrm{A}=\sqrt{{\mathrm{c}}_1^2{+\mathrm{c}}_2^2} \\ =\sqrt{\frac{9}{16}+1} \\ =\sqrt{\frac{25}{16}} \\ =\frac{5}{4} \end{gathered}$$

Hence, the amplitude is $\frac{5}{4}$

Then, damping factor is ${\mathrm{Ae}}^{\mathrm{\alpha t}}=\frac{5}{4}{\mathrm{e}}^{-8\mathrm{t}}$

And, one knows that the period of cos t and sin t is $2\mathrm{\pi }$. Then find the quasiperiod of y.

$$\begin{gathered} 8\mathrm{t}=2\mathrm{\pi } \\ \mathrm{t}=\frac{2\mathrm{\pi }}{8} \\ =\frac{\mathrm{\pi }}{4} \end{gathered}$$

And find the equifrequency,

$$\begin{gathered} \mathrm{f}=\frac{1}{\mathrm{t}} \\ =\frac{4}{\mathrm{\pi }} \end{gathered}$$

Then, the equation of motion is $\mathrm{y}\left(\mathrm{t}\right)=\frac{5}{4}{\mathrm{e}}^{-8\mathrm{t}}\sin \left(8\mathrm{t}+\varphi \right)$.