Question

A 1 – kg mass is attached to a spring with a stiffness of 100 N/m. The damping constant for the system is 0.2 N-sec/m. If the mass is pushed rightward from the equilibrium position with the velocity of 1 m/sec,when will it attain its maximum displacement to the right?

Short answer

Therefore, the maximum displacement from equilibrium to the right will happen at 0.157sec.

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 > 0 then it is bounded and ifk < 0then 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 ${\mathrm{F}}_{\mathrm{ext}}=0,\mathrm{b}=0.2,\mathrm{m}=1\mathrm{kg},\mathrm{k}=100\mathrm{N}/\mathrm{m},\mathrm{y}\left(0\right)=0$, and $\mathrm{y}\text{'}\left(0\right)=1$.

Now form the initial value problem using the above information.

$\mathrm{y}+0.2\mathrm{y}\text{'}+100\mathrm{y}=0;\mathrm{y}\left(0\right)=0,\mathrm{y}\text{'}\left(0\right)=1$…… (2)

Then, find the value of roots.

$$\begin{gathered} {\mathrm{b}}^2=0.4 \\ 4\mathrm{mk}=4\times 1\times 100 \\ =400 \end{gathered}$$

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

$$\begin{gathered} \mathrm{\alpha }=-\frac{\mathrm{b}}{2\mathrm{m}} \\ =-\frac{0.2}{2\times 1} \\ =-0.1 \end{gathered}$$

$$\begin{gathered} \mathrm{\beta }=\frac{1}{2\mathrm{m}}\sqrt{4\mathrm{mk}-{\mathrm{b}}^2} \\ =\frac{1}{2}\sqrt{400-0.4} \\ =\frac{\sqrt{399.6}}{2} \\ =\sqrt{99.99} \end{gathered}$$

Since the auxiliary equation is ${\mathrm{r}}^2+0.2\mathrm{r}+100=0$. And roots are$\mathrm{r}=-0.1$ and$\mathrm{r}=\sqrt{99.99}$

.

Find the initial conditions

By the above information, find the general solution.

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

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-0.1\mathrm{t}}\left({\mathrm{c}}_1\cos \sqrt{99.99}\mathrm{t}+{\mathrm{c}}_2\sin \sqrt{99.99}\mathrm{t}\right)$…… (3)

Given initial conditions are $\mathrm{y}\left(0\right)=0,\mathrm{y}\text{'}\left(0\right)=1$.

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

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-0.1\mathrm{t}}\left({\mathrm{c}}_1\cos \sqrt{99.99}\mathrm{t}+{\mathrm{c}}_2\sin \sqrt{99.99}\mathrm{t}\right) \\ \mathrm{y}\left(0\right)={\mathrm{e}}^{-0.1\left(0\right)}\left({\mathrm{c}}_1\cos \left(0\right)+{\mathrm{c}}_2\sin \left(0\right)\right) \\ 0={\mathrm{c}}_1\left(1\right)+0 \\ {\mathrm{c}}_1=0 \end{gathered}$$

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

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=-0.1{\mathrm{c}}_1{\mathrm{e}}^{-0.1\mathrm{t}}\cos \sqrt{99.99}\mathrm{t}-\sqrt{99.99}{\mathrm{c}}_1{\mathrm{e}}^{-0.1\mathrm{t}}\sin \sqrt{99.99}\mathrm{t}-0.1{\mathrm{c}}_2{\mathrm{e}}^{-0.1\mathrm{t}}\sin \sqrt{99.99}\mathrm{t}+\sqrt{99.99}{\mathrm{c}}_2{\mathrm{e}}^{-0.1\mathrm{t}}\cos \sqrt{99.99}\mathrm{t}$

Then,

$$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{t}\right)=-0.1{\mathrm{c}}_1{\mathrm{e}}^{-0.1\mathrm{t}}\cos \sqrt{99.99}\mathrm{t}-\sqrt{99.99}{\mathrm{c}}_1{\mathrm{e}}^{-0.1\mathrm{t}}\sin \sqrt{99.99}\mathrm{t}-0.1{\mathrm{c}}_2{\mathrm{e}}^{-0.1\mathrm{t}}\sin \sqrt{99.99}\mathrm{t}+\sqrt{99.99}{\mathrm{c}}_2{\mathrm{e}}^{-0.1\mathrm{t}}\cos \sqrt{99.99}\mathrm{t} \\ \mathrm{y}\text{'}\left(\mathrm{t}\right)=-0.1{\mathrm{c}}_1{\mathrm{e}}^{-0.1\left(0\right)}\cos \left(0\right)-\sqrt{99.99}{\mathrm{c}}_1{\mathrm{e}}^{-0.1\left(0\right)}\sin \left(0\right)-0.1{\mathrm{c}}_2{\mathrm{e}}^{-0.1\left(0\right)}\sin \left(0\right)+\sqrt{99.99}{\mathrm{c}}_2{\mathrm{e}}^{-0.1\mathrm{t}}\cos \left(0\right) \\ 1=-0.1\left(0\right)+\sqrt{99.99}{\mathrm{c}}_2 \\ {\mathrm{c}}_2=\frac{1}{\sqrt{99.99}} \end{gathered}$$

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

$\mathrm{y}\left(\mathrm{t}\right)=\frac{1}{\sqrt{99.99}}{\mathrm{e}}^{-0.1\mathrm{t}}\sin \sqrt{99.99}\mathrm{t}$…… (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.

Since the mass is moving to the right at the beginning, and the maximum displacement to the right will happen at the first moment when the sin is equal to 1.

After that, the motion is damped, and the mass can never return or go farther than this point.

So, the maximum displacement will occur for $\sin \sqrt{99.99}\mathrm{t}=1$.

Then, solve it for t.

$$\begin{gathered} \sin \sqrt{99.99}\mathrm{t}=1 \\ \sqrt{99.99}\mathrm{t}={\sin }^{-1}\left(1\right) \\ =\frac{\mathrm{\pi }}{2} \\ \mathrm{t}=\frac{\mathrm{\pi }}{2\sqrt{99.99}} \\ \approx 0.157 \end{gathered}$$

So, the maximum displacement to the right from the equilibrium will happen at 0.517 sec.