91Ó°ÊÓ

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 > 0then it is bounded and ifk < 0 then it is unbounded.

Root finding formula:

If ${\mathrm{b}}^2<4\mathrm{ac}$. Then $\mathrm{Î\pm }=-\frac{\mathrm{b}}{2\mathrm{a}}$, and $\mathrm{β}=\frac{1}{2\mathrm{a}}\sqrt{4\mathrm{ac}-{\mathrm{b}}^2}$.

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

Since the mass is just released without velocity, one has taken $\mathrm{y}\text{'}\left(0\right)=0$.

Now, form the initial value problem using the above information.

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

Then, find the value of roots.

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

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

$$\begin{gathered} \mathrm{Î\pm }=-\frac{\mathrm{b}}{2\mathrm{m}} \\ =-\frac{\frac{1}{4}}{2×\frac{1}{4}} \\ =-\frac{1}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{β}=\frac{1}{2\mathrm{m}}\sqrt{4\mathrm{mk}-{\mathrm{b}}^2} \\ =2\sqrt{8-\frac{1}{16}} \\ =\frac{2\sqrt{127}}{4} \\ =\frac{\sqrt{127}}{2} \end{gathered}$$

Since the auxiliary equation is $\frac{1}{4}{\mathrm{r}}^2+\frac{1}{4}\mathrm{r}+8=0$. And roots are $\mathrm{r}=-\frac{1}{2}$and$\mathrm{r}=\frac{\sqrt{127}}{2}$

.

By the above information, find the general solution.

The general solution is$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{Î\pm ³Ù}}\left({\mathrm{c}}_1\mathrm{³¦´Ç²õβ³Ù}+{\mathrm{c}}_2\mathrm{²õ¾\pm ²Ôβ³Ù}\right)$

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-\frac{1}{2}\mathrm{t}}\left({\mathrm{c}}_1\cos \frac{\sqrt{127}}{2}\mathrm{t}+{\mathrm{c}}_2\sin \frac{\sqrt{127}}{2}\mathrm{t}\right)$…… (3)

Given the initial conditions are.

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

$$\begin{gathered} \left(\mathrm{t}\right)={\mathrm{e}}^{-\frac{1}{2}\mathrm{t}}\left({\mathrm{c}}_1\cos \frac{\sqrt{127}}{2}\mathrm{t}+{\mathrm{c}}_2\sin \frac{\sqrt{127}}{2}\mathrm{t}\right) \\ \mathrm{y}\left(0\right)={\mathrm{e}}^{-\frac{1}{2}\left(0\right)}\left({\mathrm{c}}_1\cos \left(0\right)+{\mathrm{c}}_2\sin \left(0\right)\right) \\ -1={\mathrm{c}}_1\left(1\right)+0 \\ {\mathrm{c}}_1=-1 \end{gathered}$$

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

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=-\frac{1}{2}{\mathrm{c}}_1{\mathrm{e}}^{-\frac{1}{2}\mathrm{t}}\cos \frac{\sqrt{127}}{2}\mathrm{t}-\frac{\sqrt{127}}{2}{\mathrm{c}}_1{\mathrm{e}}^{-\frac{1}{2}\mathrm{t}}\sin \frac{\sqrt{127}}{2}\mathrm{t}-\frac{1}{2}{\mathrm{c}}_2{\mathrm{e}}^{-\frac{1}{2}\mathrm{t}}\sin \frac{\sqrt{127}}{2}\mathrm{t}+\frac{\sqrt{127}}{2}{\mathrm{c}}_2{\mathrm{e}}^{-\frac{1}{2}\mathrm{t}}\cos \frac{\sqrt{127}}{2}\mathrm{t}$

Then,

$$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{t}\right)=-\frac{1}{2}{\mathrm{c}}_1{\mathrm{e}}^{-\frac{1}{2}\mathrm{t}}\cos \frac{\sqrt{127}}{2}\mathrm{t}-\frac{\sqrt{127}}{2}{\mathrm{c}}_1{\mathrm{e}}^{-\frac{1}{2}\mathrm{t}}\sin \frac{\sqrt{127}}{2}\mathrm{t}-\frac{1}{2}{\mathrm{c}}_2{\mathrm{e}}^{-\frac{1}{2}\mathrm{t}}\sin \frac{\sqrt{127}}{2}\mathrm{t}+\frac{\sqrt{127}}{2}{\mathrm{c}}_2{\mathrm{e}}^{-\frac{1}{2}\mathrm{t}}\cos \frac{\sqrt{127}}{2}\mathrm{t} \\ \mathrm{y}\text{'}\left(0\right)=-\frac{1}{2}{\mathrm{c}}_1{\mathrm{e}}^{-\frac{1}{2}\left(0\right)}\cos \left(0\right)-\frac{\sqrt{127}}{2}{\mathrm{c}}_1{\mathrm{e}}^{-\frac{1}{2}\left(0\right)}\sin \left(0\right)-\frac{1}{2}{\mathrm{c}}_2{\mathrm{e}}^{-\frac{1}{2}\left(0\right)}\sin \left(0\right)+\frac{\sqrt{127}}{2}{\mathrm{c}}_2{\mathrm{e}}^{-\frac{1}{2}\left(0\right)}\cos \left(0\right) \\ 0=-\frac{1}{2}\left(-1\right)+\frac{\sqrt{127}}{2}{\mathrm{c}}_2 \\ {\mathrm{c}}_2=-\frac{1}{2}\left(\frac{2}{\sqrt{127}}\right) \\ =-\frac{1}{\sqrt{127}} \end{gathered}$$

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

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-\frac{1}{2}\mathrm{t}}\left(-\cos \frac{\sqrt{127}}{2}\mathrm{t}-\frac{1}{\sqrt{127}}\sin \frac{\sqrt{127}}{2}\mathrm{t}\right)$…… (4)

Rewrite the equation (4) in the form of $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{Ae}}^{\mathrm{Î\pm ³Ù}}\sin \left(\mathrm{β³Ù}+\mathrm{Ï•}\right)$.

Then, find the amplitude.

$$\begin{gathered} \mathrm{A}=\sqrt{{\mathrm{c}}_1^2{+\mathrm{c}}_2^2} \\ =\sqrt{1+\frac{1}{127}} \\ =\sqrt{\frac{128}{127}} \end{gathered}$$

So, the amplitude is$\sqrt{\frac{128}{127}}$

Then, $$\begin{gathered} \tan \mathrm{Ï•}=\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}=\frac{-1}{-\frac{1}{\sqrt{127}}} \\ =\sqrt{127} \\ \mathrm{Ï•}={\tan }^{-1}\left(\sqrt{127}\right) \end{gathered}$$

To find the equilibrium position. Put $\mathrm{y}\text{'}\left(\mathrm{t}\right)=0$

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{Ae}}^{\mathrm{Î\pm ³Ù}}\sin \left(\mathrm{β³Ù}+\mathrm{Ï•}\right) \\ \mathrm{y}\text{'}\left(\mathrm{t}\right)={\mathrm{´¡Î\pm \pm ð}}^{\mathrm{Î\pm ³Ù}}\sin \left(\mathrm{β³Ù}+\mathrm{Ï•}\right)+{\mathrm{´¡Î²\pm ð}}^{\mathrm{Î\pm ³Ù}}\cos \left(\mathrm{β³Ù}+\mathrm{Ï•}\right) \end{gathered}$$

Then, $0={\mathrm{´¡Î\pm \pm ð}}^{\mathrm{Î\pm ³Ù}}\sin \left(\mathrm{β³Ù}+\mathrm{Ï•}\right)+{\mathrm{´¡Î²\pm ð}}^{\mathrm{Î\pm ³Ù}}\cos \left(\mathrm{β³Ù}+\mathrm{Ï•}\right)$

Now solve the above equation to find the value of t.

$$\begin{gathered} A\mathrm{Î\pm }{e}^{\mathrm{Î\pm ³Ù}}\sin \left(\mathrm{β³Ù}+\mathrm{Ï•}\right)+{\mathrm{´¡Î²\pm ð}}^{\mathrm{Î\pm ³Ù}}\cos \left(\mathrm{β³Ù}+\mathrm{Ï•}\right)=0 \\ {\mathrm{Ae}}^{\mathrm{Î\pm ³Ù}}\left(\mathrm{Î\pm ²õ¾\pm ²Ô}\left(\mathrm{β³Ù}+\mathrm{Ï•}\right)+\mathrm{⳦´Ç²õ}\left(\mathrm{β³Ù}+\mathrm{Ï•}\right)\right)=0 \end{gathered}$$

Since${\mathrm{Ae}}^{\mathrm{Î\pm ³Ù}}>0$. Then,

$\mathrm{Î\pm ²õ¾\pm ²Ô}\left(\mathrm{β³Ù}+\mathrm{Ï•}\right)+\mathrm{⳦´Ç²õ}\left(\mathrm{β³Ù}+\mathrm{Ï•}\right)=0$

Divide $\cos \left(\mathrm{β³Ù}+\mathrm{Ï•}\right)$into both sides.

$$\begin{gathered} \frac{\mathrm{Î\pm ²õ¾\pm ²Ô}\left(\mathrm{β³Ù}+\mathrm{Ï•}\right)}{\cos \left(\mathrm{β³Ù}+\mathrm{Ï•}\right)}+\frac{\mathrm{⳦´Ç²õ}\left(\mathrm{β³Ù}+\mathrm{Ï•}\right)}{\cos \left(\mathrm{β³Ù}+\mathrm{Ï•}\right)}=0 \\ \frac{\mathrm{Î\pm ²õ¾\pm ²Ô}\left(\mathrm{β³Ù}+\mathrm{Ï•}\right)}{\cos \left(\mathrm{β³Ù}+\mathrm{Ï•}\right)}+\mathrm{β}=0 \\ \mathrm{Î\pm ³Ùan}\left(\mathrm{β³Ù}+\mathrm{Ï•}\right)=-\mathrm{β} \\ \tan \left(\mathrm{β³Ù}+\mathrm{Ï•}\right)=-\frac{\mathrm{β}}{\mathrm{Î\pm }} \\ \mathrm{β³Ù}+\mathrm{Ï•}={\tan }^{-1}\left(-\frac{\mathrm{β}}{\mathrm{Î\pm }}\right)+\mathrm{°ìÏ€} \end{gathered}$$

Let us take k = 1. Then,

$$\begin{gathered} \mathrm{β³Ù}+\mathrm{Ï•}={\tan }^{-1}\left(-\frac{\mathrm{β}}{\mathrm{Î\pm }}\right)+\mathrm{Ï€} \\ \mathrm{t}=\frac{{\tan }^{-1}\left(-\frac{\mathrm{β}}{\mathrm{Î\pm }}\right)+\mathrm{Ï€}-\mathrm{Ï•}}{\mathrm{β}} \\ =\frac{{\tan }^{-1}\left(-\frac{\frac{\sqrt{127}}{2}}{-\frac{1}{2}}\right)+\mathrm{Ï€}-\mathrm{Ï•}}{\frac{\sqrt{127}}{2}} \\ =\frac{{\tan }^{-1}\left(\sqrt{127}\right)+3.142-{\tan }^{-1}\left(\sqrt{127}\right)}{\frac{\sqrt{127}}{2}} \\ =\frac{2\left(3.142\right)}{\sqrt{127}} \\ ≈0.558 \end{gathered}$$

So, $\mathrm{t}≈0.558$.

Substitute the value of t in equation (4).

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-\frac{1}{2}\mathrm{t}}\left(-\cos \frac{\sqrt{127}}{2}\mathrm{t}-\frac{1}{\sqrt{127}}\sin \frac{\sqrt{127}}{2}\mathrm{t}\right) \\ \mathrm{y}\left(0.558\right)={\mathrm{e}}^{-\frac{1}{2}\left(0.558\right)}\left(-\cos \frac{\sqrt{127}}{2}\left(0.558\right)-\frac{1}{\sqrt{127}}\sin \frac{\sqrt{127}}{2}\left(0.558\right)\right) \\ =0.7567\left(-1-0\right)=-0.7567 \\ ≈0.7567 \end{gathered}$$

So, the maximum displacement from the equilibrium is $0.7567 \mathrm{m}$.