Question

A $\frac{1}{4}$– kg mass is attached to a spring with stiffness 8 N/m. The damping constant for the system is $\frac{1}{4}$N-sec/m. If the mass is moved 1 m to the left of equilibrium and released,what is the maximum displacement to the right that it will attain?

Short answer

Therefore, the maximum displacement from equilibrium to the right will attain at 0.7567 m.

Step-by-step solution

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 > 0then 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 $\mathrm{\beta }=\frac{1}{2\mathrm{a}}\sqrt{4\mathrm{ac}-{\mathrm{b}}^2}$.

Evaluate the equation

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.

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

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

$$\begin{gathered} \mathrm{\alpha }=-\frac{\mathrm{b}}{2\mathrm{m}} \\ =-\frac{\frac{1}{4}}{2\times \frac{1}{4}} \\ =-\frac{1}{2} \end{gathered}$$

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

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$\mathrm{r}=\frac{\sqrt{127}}{2}$

.

Find the initial conditions

By the above information, find the general solution.

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

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

Find when the maximum displacement happens in the equilibrium.

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

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

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{Ae}}^{\mathrm{\alpha t}}\sin \left(\mathrm{\beta t}+\varphi \right) \\ \mathrm{y}\text{'}\left(\mathrm{t}\right)={\mathrm{A\alpha e}}^{\mathrm{\alpha t}}\sin \left(\mathrm{\beta t}+\varphi \right)+{\mathrm{A\beta e}}^{\mathrm{\alpha t}}\cos \left(\mathrm{\beta t}+\varphi \right) \end{gathered}$$

Then, $0={\mathrm{A\alpha e}}^{\mathrm{\alpha t}}\sin \left(\mathrm{\beta t}+\varphi \right)+{\mathrm{A\beta e}}^{\mathrm{\alpha t}}\cos \left(\mathrm{\beta t}+\varphi \right)$

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

$$\begin{gathered} A\alpha e^{\mathrm{\alpha t}}\sin \left(\mathrm{\beta t}+\varphi \right)+{\mathrm{A\beta e}}^{\mathrm{\alpha t}}\cos \left(\mathrm{\beta t}+\varphi \right)=0 \\ {\mathrm{Ae}}^{\mathrm{\alpha t}}\left(\mathrm{\alpha sin}\left(\mathrm{\beta t}+\varphi \right)+\mathrm{\beta cos}\left(\mathrm{\beta t}+\varphi \right)\right)=0 \end{gathered}$$

Since${\mathrm{Ae}}^{\mathrm{\alpha t}}>0$. Then,

$\mathrm{\alpha sin}\left(\mathrm{\beta t}+\varphi \right)+\mathrm{\beta cos}\left(\mathrm{\beta t}+\varphi \right)=0$

Divide $\cos \left(\mathrm{\beta t}+\varphi \right)$into both sides.

$$\begin{gathered} \frac{\mathrm{\alpha sin}\left(\mathrm{\beta t}+\varphi \right)}{\cos \left(\mathrm{\beta t}+\varphi \right)}+\frac{\mathrm{\beta cos}\left(\mathrm{\beta t}+\varphi \right)}{\cos \left(\mathrm{\beta t}+\varphi \right)}=0 \\ \frac{\mathrm{\alpha sin}\left(\mathrm{\beta t}+\varphi \right)}{\cos \left(\mathrm{\beta t}+\varphi \right)}+\mathrm{\beta }=0 \\ \mathrm{\alpha tan}\left(\mathrm{\beta t}+\varphi \right)=-\mathrm{\beta } \\ \tan \left(\mathrm{\beta t}+\varphi \right)=-\frac{\mathrm{\beta }}{\mathrm{\alpha }} \\ \mathrm{\beta t}+\varphi ={\tan }^{-1}\left(-\frac{\mathrm{\beta }}{\mathrm{\alpha }}\right)+\mathrm{k\pi } \end{gathered}$$

Let us take k = 1. Then,

$$\begin{gathered} \mathrm{\beta t}+\varphi ={\tan }^{-1}\left(-\frac{\mathrm{\beta }}{\mathrm{\alpha }}\right)+\mathrm{\pi } \\ \mathrm{t}=\frac{{\tan }^{-1}\left(-\frac{\mathrm{\beta }}{\mathrm{\alpha }}\right)+\mathrm{\pi }-\varphi }{\mathrm{\beta }} \\ =\frac{{\tan }^{-1}\left(-\frac{\frac{\sqrt{127}}{2}}{-\frac{1}{2}}\right)+\mathrm{\pi }-\varphi }{\frac{\sqrt{127}}{2}} \\ =\frac{{\tan }^{-1}\left(\sqrt{127}\right)+3.142-{\tan }^{-1}\left(\sqrt{127}\right)}{\frac{\sqrt{127}}{2}} \\ =\frac{2\left(3.142\right)}{\sqrt{127}} \\ \approx 0.558 \end{gathered}$$

So, $\mathrm{t}\approx 0.558$.

Substitute the value of t in equation (4).

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

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