Q4.9-8E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A 20 – kg mass is attached to a spring with a stiffness 200 N/m. The damping constant for the system is 140 N-sec/m. If the mass is pulled 25 cm to the right of the equilibrium and given an initial leftward velocity of 1 m/sec, when will it first return to its equilibrium position?

 

Step-by-Step Solution

Answer

Therefore, the mass will never return to its equilibrium position. 

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;

 

 $\begin{array}{c}{\mathbf{F}}_{\mathrm{ext}}\mathbf{=}\left[\mathrm{inertia}\right]\mathbf{y}\mathbf{"}\mathbf{+}\left[\mathrm{damping}\right]\mathbf{y}\mathbf{\text{'}}\mathbf{+}\left[\mathrm{stiffness}\right]\mathbf{y}\\ =\mathrm{my}"+\mathrm{by}\text{'}+\mathrm{ky}\end{array}$  …… (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.${\mathbf{b}}^2\mathbf{<}\mathbf{4}\mathbf{ac}$ Then,$\mathbf{\alpha }\mathbf{=}\mathbf{-}\frac{b}{2a}$ and.$\mathbf{\beta }\mathbf{=}\frac{1}{2a}\sqrt{\mathbf{4}\mathbf{ac}\mathbf{-}{\mathbf{b}}^2}$

Step 2: Evaluate the equation

Given that, ${\mathbf{F}}_{\mathrm{ext}}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{b}\mathbf{=}140\mathbf{,}\mathbf{m}\mathbf{=}20\mathbf{kg}\mathbf{,}\mathbf{k}\mathbf{=}200\mathbf{N}\mathbf{/}\mathbf{m}\mathbf{,}\mathbf{y}\left(0\right)\mathbf{=}0.25$and$\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{1}$.

 

Now form the initial value problem using the above information.

 

$\mathbf{20}\mathbf{y}\mathbf{"}\mathbf{+}\mathbf{140}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{200}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{1}$   …… (2)

 

Then, find the value of roots.

 $\begin{array}{c}{\mathbf{b}}^2\mathbf{=}19600\\ \mathbf{4}\mathbf{mk}\mathbf{=}\mathbf{4}\mathbf{\times }20\mathbf{\times }200\\ \mathbf{=}16000\end{array}$

 

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

 

Since the auxiliary equation is.$20{\mathbf{r}}^2\mathbf{+}140\mathbf{r}\mathbf{+}200\mathbf{=}\mathbf{0}$ And roots are$\mathbf{r}\mathbf{=}\mathbf{-}2$ and$\mathbf{r}\mathbf{=}\mathbf{-}\mathbf{5}.$

 

Step3: find the initial conditions

By the above information find the general solution.

 

The general solution is $\mathbf{y}\left(t\right)\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{-2t}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{-5t}$…… (3)

 

Given initial conditions are$\mathbf{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{1}$

 

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

 $\begin{array}{c}\mathbf{y}\left(t\right)\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{-2t}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{-5t}\\ \mathbf{y}\left(0\right)\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{\mathbf{-}\mathbf{2}\left(0\right)}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{\mathbf{-}\mathbf{5}\left(0\right)}\\ \mathbf{0}\mathbf{.}\mathbf{25}\mathbf{=}{\mathbf{c}}_1\left(1\right)\mathbf{+}{\mathbf{c}}_2\left(1\right)\\ {\mathbf{c}}_1\mathbf{+}{\mathbf{c}}_2\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}\end{array}$

 

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

 

$\mathbf{y}\mathbf{\text{'}}\left(t\right)\mathbf{=}\mathbf{-}2{\mathbf{e}}^{\mathbf{-}2\mathbf{t}}{\mathbf{c}}_1\mathbf{-}5{\mathbf{e}}^{\mathbf{-}5\mathbf{t}}{\mathbf{c}}_2$

 

Then,

 

 $\begin{array}{c}\mathbf{y}\mathbf{\text{'}}\left(t\right)\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{e}}^{-2t}{\mathbf{c}}_1\mathbf{-}\mathbf{5}{\mathbf{e}}^{-5t}{\mathbf{c}}_2\\ \mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\left(0\right)}{\mathbf{c}}_1\mathbf{-}\mathbf{5}{\mathbf{e}}^{\mathbf{-}\mathbf{5}\left(0\right)}{\mathbf{c}}_2\\ \mathbf{-}\mathbf{1}\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{c}}_1\mathbf{-}\mathbf{5}{\mathbf{c}}_2\\ \mathbf{2}{\mathbf{c}}_1\mathbf{+}\mathbf{5}{\mathbf{c}}_2\mathbf{=}\mathbf{1}\end{array}$

 

Solve the equations,

 $\begin{array}{c}\mathbf{5}{\mathbf{c}}_1\mathbf{+}\mathbf{5}{\mathbf{c}}_2\mathbf{-}\mathbf{2}{\mathbf{c}}_1\mathbf{-}\mathbf{5}{\mathbf{c}}_2\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{-}\mathbf{1}\\ \mathbf{3}{\mathbf{c}}_1\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}\\ {\mathbf{c}}_1\mathbf{=}\frac{1}{12}\end{array}$

Then,

$\begin{array}{c}\frac{1}{12}\mathbf{+}{\mathbf{c}}_2\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}\\ {\mathbf{c}}_2\mathbf{=}\frac{1}{6}\end{array}$ 

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

 

$\mathbf{y}\left(t\right)\mathbf{=}\frac{1}{12}{\mathbf{e}}^{-2t}\mathbf{+}\frac{1}{6}{\mathbf{e}}^{-5t}$…… (4)

 

Step4: find, when will mass return to the equilibrium position

To find the equilibrium position, put$\mathbf{y}\left(t\right)\mathbf{=}\mathbf{0}$

Then,$0\mathbf{=}\frac{1}{12}{\mathbf{e}}^{-2t}\mathbf{+}\frac{1}{6}{\mathbf{e}}^{-5t}$

 

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

 $\begin{array}{c}\frac{1}{12}{\mathbf{e}}^{-2t}\mathbf{+}\frac{1}{6}{\mathbf{e}}^{-5t}\mathbf{=}\mathbf{0}\\ \frac{1}{6}{\mathbf{e}}^{-5t}\left(\frac{1}{2}{\mathbf{e}}^{3t}\mathbf{+}\mathbf{1}\right)\mathbf{=}\mathbf{0}\end{array}$

 

Since,$\frac{1}{6}{\mathbf{e}}^{-5t}\ne 0$

 

Then,

 $\begin{array}{c}\frac{1}{2}{\mathbf{e}}^{3t}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0}\\ \frac{1}{2}{\mathbf{e}}^{3t}\mathbf{=}\mathbf{-}\mathbf{1}\\ {\mathbf{e}}^{3t}\mathbf{=}\mathbf{-}\mathbf{2}\\ \mathbf{t}\mathbf{=}\mathbf{-}\frac{1}{3}\mathbf{ln}\left|2\right|\end{array}$

 

Therefore,.$\mathbf{t}\mathbf{=}\mathbf{-}\frac{1}{3}\mathbf{ln}\left|2\right|\Rightarrow t\approx -0.10034$ Since time is negative, therefore the mass will never return to the equilibrium position.