Q12 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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.

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 if k < 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 $\mathrm{\beta }=\frac{1}{2\mathrm{a}}\sqrt{4\mathrm{ac}-{\mathrm{b}}^2}$

Step 2: Evaluate the equation

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.

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

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

 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

.

Step 3: Find the initial conditions

By the above information find the general solution.

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

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-4\mathrm{t}}\left({\mathrm{c}}_1\cos 4\mathrm{t}+{\mathrm{c}}_2\sin 4\mathrm{t}\right)$…… (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 $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{Ae}}^{\mathrm{\alpha t}}\sin \left(\mathrm{\beta t}+\mathrm{\varphi }\right)$.

Step 4: Find, when the maximum displacement happens in the equilibrium.

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

 

Then, 

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

 

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.