The motion of a mass-spring system with damping is governed by

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)+10\mathrm{y}\text{'}\left(\mathrm{t}\right)+\mathrm{ky}\left(\mathrm{t}\right)=0; \\ \mathrm{y}\left(0\right)=1, \mathrm{y}\text{'}\left(0\right)=0. \end{gathered}$$

Find the equation of motion and sketch its graph for k = 20, 25, and 30.

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 < 0 then it is unbounded.

Given that,

$$\begin{gathered} \left(\mathrm{t}\right)+10\mathrm{y}\text{'}\left(\mathrm{t}\right)+\mathrm{ky}\left(\mathrm{t}\right)=0; \\ \mathrm{y}\left(0\right)=1, \mathrm{y}\text{'}\left(0\right)=0 \end{gathered}$$

And the value of $\mathrm{k}=20,25,30$.

Let us take k = 25. Then,

$\mathrm{y}+10\mathrm{y}\text{'}+25\mathrm{y}=0$…… (2)

Since the auxiliary equation is ${\mathrm{r}}^2+10\mathrm{r}+25=0$. And roots are $\mathrm{r}=-5$and $\mathrm{r}=-5$.

From the above information, find the general solution.

The general solution is$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{-5\mathrm{t}}+{\mathrm{c}}_2{\mathrm{te}}^{-5\mathrm{t}}$ …… (3)

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

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

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{-5\mathrm{t}}+{\mathrm{c}}_2{\mathrm{te}}^{-5\mathrm{t}} \\ \mathrm{y}\left(0\right)={\mathrm{c}}_1{\mathrm{e}}^{-5\left(0\right)}+{\mathrm{c}}_2\left(0\right){\mathrm{e}}^{-5\left(0\right)} \\ 1={\mathrm{c}}_1\left(1\right)+0 \\ {\mathrm{c}}_1=1 \end{gathered}$$

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

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=-5{\mathrm{c}}_1{\mathrm{e}}^{-5\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-5\mathrm{t}}-5{\mathrm{c}}_2{\mathrm{te}}^{-5\mathrm{t}}$

Then,

$$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{t}\right)=-5{\mathrm{c}}_1{\mathrm{e}}^{-5\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-5\mathrm{t}}-5{\mathrm{c}}_2{\mathrm{te}}^{-5\mathrm{t}} \\ \mathrm{y}\text{'}\left(0\right)=-5{\mathrm{c}}_1{\mathrm{e}}^{-5\left(0\right)}+{\mathrm{c}}_2{\mathrm{e}}^{-5\left(0\right)}-5{\mathrm{c}}_2\left(0\right){\mathrm{e}}^{-5\left(0\right)} \\ 0=-5{\mathrm{c}}_1+{\mathrm{c}}_2\left(1\right) \\ {\mathrm{c}}_2=5 \end{gathered}$$

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

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-5\mathrm{t}}+5{\mathrm{te}}^{-5\mathrm{t}}$

Similarly, the solution for k = 20 is,$\mathrm{y}=\frac{1+\sqrt{5}}{2}{\mathrm{e}}^{\left(-5+\sqrt{5}\right)\mathrm{t}}+\frac{1+\sqrt{5}}{2}{\mathrm{e}}^{\left(-5-\sqrt{5}\right)\mathrm{t}}$ and solution for k = 30 is $\mathrm{y}={\mathrm{e}}^{-5\mathrm{t}}\cos \sqrt{5}\mathrm{t}+\sqrt{5}{\mathrm{e}}^{-5\mathrm{t}}\sin \sqrt{5}\mathrm{t}$.

The graph of the equation is $\mathrm{y}\left(\mathrm{t}\right)+10\mathrm{y}\text{'}\left(\mathrm{t}\right)+\mathrm{ky}\left(\mathrm{t}\right)=0$.

When k = 20, 25, and 30 means, draw the graph.

Case (1):

Case (2):

Case (3):