A $\frac{1}{4}$– kg mass is attached to a spring with stiffness 8 N/m. The damping constant for the system is 2 N-sec/m. If the mass is pushed 50 cm to the left of equilibrium and given a leftward velocity of 2 m/sec,when will the mass attains its maximum displacement to the left?

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.

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.

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, ${\mathrm{F}}_{\mathrm{ext}}=0,\mathrm{b}=2,\mathrm{m}=\frac{1}{4}\mathrm{kg},\mathrm{k}=8\mathrm{N}/\mathrm{m},\mathrm{y}\left(0\right)=-0.5$and $\mathrm{y}\text{'}\left(0\right)=-2$.

Now form the initial value problem using the above information.

$\frac{1}{4}\mathrm{y}+2\mathrm{y}\text{'}+8\mathrm{y}=0;\mathrm{y}\left(0\right)=-0.5,\mathrm{y}\text{'}\left(0\right)=-2$…… (2)

Then, find the value of roots.

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

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

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

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

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

.

By the above information find the general solution.

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

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-4\mathrm{t}}\left({\mathrm{c}}_1\cos 4\mathrm{t}+{\mathrm{c}}_2\sin 4\mathrm{t}\right)$…… (3)

Given initial conditions are $\mathrm{y}\left(0\right)=-0.5,\mathrm{y}\text{'}\left(0\right)=-2$.

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

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-4\mathrm{t}}\left({\mathrm{c}}_1\cos 4\mathrm{t}+{\mathrm{c}}_2\sin 4\mathrm{t}\right) \\ \mathrm{y}\left(0\right)={\mathrm{e}}^{-4\left(0\right)}\left({\mathrm{c}}_1\cos \left(0\right)+{\mathrm{c}}_2\sin \left(0\right)\right) \\ -0.5={\mathrm{c}}_1\left(1\right)+0 \\ {\mathrm{c}}_1=-0.5 \end{gathered}$$

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

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=-4{\mathrm{c}}_1{\mathrm{e}}^{-4\mathrm{t}}\cos 4\mathrm{t}-4{\mathrm{c}}_1{\mathrm{e}}^{-4\mathrm{t}}\sin 4\mathrm{t}-4{\mathrm{c}}_2{\mathrm{e}}^{-4\mathrm{t}}\sin 4\mathrm{t}+4{\mathrm{c}}_2{\mathrm{e}}^{-4\mathrm{t}}\cos 4\mathrm{t}$

Then,

$$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{t}\right)=-4{\mathrm{c}}_1{\mathrm{e}}^{-4\mathrm{t}}\cos 4\mathrm{t}-4{\mathrm{c}}_1{\mathrm{e}}^{-4\mathrm{t}}\sin 4\mathrm{t}-4{\mathrm{c}}_2{\mathrm{e}}^{-4\mathrm{t}}\sin 4\mathrm{t}+4{\mathrm{c}}_2{\mathrm{e}}^{-4\mathrm{t}}\cos 4\mathrm{t} \\ \mathrm{y}\text{'}\left(0\right)=-4{\mathrm{c}}_1{\mathrm{e}}^{-4\mathrm{t}}\cos 4\mathrm{t}-4{\mathrm{c}}_1{\mathrm{e}}^{-4\mathrm{t}}\sin 4\mathrm{t}-4{\mathrm{c}}_2{\mathrm{e}}^{-4\mathrm{t}}\sin 4\mathrm{t}+4{\mathrm{c}}_2{\mathrm{e}}^{-4\mathrm{t}}\cos 4\mathrm{t} \\ -2=-4\left(-0.5\right)+4{\mathrm{c}}_2 \\ {\mathrm{c}}_2=-1 \end{gathered}$$

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

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-4\mathrm{t}}\left(-0.5\cos 4\mathrm{t}-\sin 4\mathrm{t}\right)$…… (4)

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

To find the value of t, put $\mathrm{y}\text{'}\left(\mathrm{t}\right)=0$.

Then,

role="math" localid="1664082682670" $$\begin{gathered} -4\left(-0.5\right){\mathrm{e}}^{-4\mathrm{t}}\cos 4\mathrm{t}-4\left(-0.5\right){\mathrm{e}}^{-4\mathrm{t}}\sin 4\mathrm{t}+4{\mathrm{e}}^{-4\mathrm{t}}\sin 4\mathrm{t}-4{\mathrm{e}}^{-4\mathrm{t}}\cos 4\mathrm{t}=0 \\ {\mathrm{e}}^{-4\mathrm{t}}\left(2\cos 4\mathrm{t}-4\cos 4\mathrm{t}\right)+4{\mathrm{e}}^{-4\mathrm{t}}\left(0.5\sin 4\mathrm{t}+\sin 4\mathrm{t}\right)=0 \\ -2{\mathrm{e}}^{-4\mathrm{t}}\cos 4\mathrm{t}+6{\mathrm{e}}^{-4\mathrm{t}}\sin 4\mathrm{t}=0 \\ 2{\mathrm{e}}^{-4\mathrm{t}}\left(-\cos 4\mathrm{t}+3\sin 4\mathrm{t}\right)=0 \end{gathered}$$

Then, solve it for t.

$$\begin{gathered} -\frac{\cos 4\mathrm{t}}{\cos 4\mathrm{t}}+3\frac{\sin 4\mathrm{t}}{\cos 4\mathrm{t}}=0 \\ 3\tan 4\mathrm{t}=1 \\ \tan 4\mathrm{t}=\frac{1}{3} \\ 4\mathrm{t}={\tan }^{-1}\left(\frac{1}{3}\right) \\ 4\mathrm{t}=0.32 \\ \mathrm{t}=\frac{0.32}{4} \\ \approx 0.08 \end{gathered}$$

So, the maximum displacement to the left from the equilibrium will happen at 0.08 sec.