Q4.9-1E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A 2 – kg mass is attached to a spring with stiffness k = 50 N/m. The mass is displaced $\frac{1}{4}\mathbf{m}$to the left of the equilibrium point and given a velocity of 1 m/sec to the left. Neglecting damping, find the equation of motion of the mass along with the amplitude, period, and frequency. How long after release does the mass pass through the equilibrium position? 

Step-by-Step Solution

Answer

Therefore, the equilibrium position will release after a long time of$\mathbf{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{449}$ . And the equation of motion of the mass along with the amplitude, period, and frequency is$\mathbf{y}\left(t\right)\mathbf{=}\mathbf{-}\frac{1}{4}\mathbf{cos}\mathbf{5}\mathbf{t}\mathbf{-}\frac{1}{5}\mathbf{sin}\mathbf{5}\mathbf{t}$.

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:

$\begin{array}{c}{\mathbf{F}}_{\mathrm{ext}}\mathbf{=}\left[\mathrm{inertia}\right]\mathbf{y}\mathbf{"}\mathbf{+}\left[\mathrm{damping}\right]\mathbf{y}\mathbf{\text{'}}\mathbf{+}\left[\mathrm{stiffness}\right]\mathbf{y}\\ \mathbf{=}\mathbf{my}\mathbf{"}\mathbf{+}\mathbf{by}\mathbf{\text{'}}\mathbf{+}\mathbf{ky}\dots \left(1\right)\end{array}$

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.

Step 2: Evaluate the equation.

Given that, A 2 – kg mass is attached to a spring with stiffness k = 50 N/m. The mass is displaced $\frac{1}{4}\mathbf{m}$ to the left of the equilibrium point and given a velocity of 1 m/sec to the left.

 

Use the given information to find the equation.

 

Then, ${\mathbf{F}}_{\mathrm{ext}}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{b}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{m}\mathbf{=}2\mathbf{kg}\mathbf{,}\mathbf{k}\mathbf{=}\mathbf{50}\mathbf{N}\mathbf{/}\mathbf{m}\mathbf{,}\mathbf{y}\left(0\right)\mathbf{=}\mathbf{-}\frac{1}{4}$and$\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{1}$.

 

Now form the initial value problem using the above information.

 $\mathbf{2}\mathbf{y}\mathbf{"}\mathbf{+}\mathbf{0}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{50}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\left(0\right)\mathbf{=}\mathbf{-}\frac{1}{4}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{1}\dots \left(2\right)$

   

 Then, simplify the equation (2) to find the value of r.

 $\begin{array}{c}2y"+0y\text{'}+50y=0\\ y"+25y=0\end{array}$

 

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

 

Step3: Find the initial conditions

By the above information find the general solution.

 

The general solution is  $\mathbf{y}\left(t\right)\mathbf{=}{\mathbf{c}}_1\mathbf{cos}\mathbf{5}\mathbf{t}\mathbf{+}{\mathbf{c}}_2\mathbf{sin}\mathbf{5}\mathbf{t}$…… (3)

Given initial conditions are$\mathbf{y}\left(0\right)\mathbf{=}\mathbf{-}\frac{1}{4}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{1}$.

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

 $\begin{array}{c}\mathbf{y}\left(t\right)\mathbf{=}{\mathbf{c}}_1\mathbf{cos}\mathbf{5}\mathbf{t}\mathbf{+}{\mathbf{c}}_2\mathbf{sin}\mathbf{5}\mathbf{t}\\ \mathbf{y}\left(0\right)\mathbf{=}{\mathbf{c}}_1\mathbf{cos}\left(5\times 0\right)\mathbf{+}{\mathbf{c}}_2\mathbf{sin}\left(5\times 0\right)\\ \mathbf{-}\frac{1}{4}\mathbf{=}{\mathbf{c}}_1\left(1\right)\mathbf{+}\mathbf{0}\\ {\mathbf{c}}_1\mathbf{=}\mathbf{-}\frac{1}{4}\end{array}$

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

 $\mathbf{y}\mathbf{\text{'}}\left(t\right)\mathbf{=}\mathbf{-}\mathbf{5}{\mathbf{c}}_1\mathbf{sin}\mathbf{5}\mathbf{t}\mathbf{+}\mathbf{5}{\mathbf{c}}_2\mathbf{cos}\mathbf{5}\mathbf{t}$

 

Then,

$\begin{array}{c}\mathbf{y}\mathbf{\text{'}}\left(t\right)\mathbf{=}\mathbf{-}\mathbf{5}{\mathbf{c}}_1\mathbf{sin}\mathbf{5}\mathbf{t}\mathbf{+}\mathbf{5}{\mathbf{c}}_2\mathbf{cos}\mathbf{5}\mathbf{t}\\ \mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{5}{\mathbf{c}}_1\mathbf{sin}\left(5\times 0\right)\mathbf{+}\mathbf{5}{\mathbf{c}}_2\mathbf{cos}\left(5\times 0\right)\\ \mathbf{-}\mathbf{1}\mathbf{=}\mathbf{0}\mathbf{+}\mathbf{5}{\mathbf{c}}_2\left(1\right)\\ {\mathbf{c}}_2\mathbf{=}\mathbf{-}\frac{1}{5}\end{array}$

 

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

 $\mathbf{y}\left(t\right)\mathbf{=}\mathbf{-}\frac{1}{4}\mathbf{cos}\mathbf{5}\mathbf{t}\mathbf{-}\frac{1}{5}\mathbf{sin}\mathbf{5}\mathbf{t}$

.

Step4: Find the value of t.

Then, find the amplitude.

 $\begin{array}{c}\mathbf{A}\mathbf{=}\sqrt{{\mathbf{c}}_1^2\mathbf{+}{\mathbf{c}}_2^2}\\ \mathbf{=}\sqrt{\frac{1}{16}\mathbf{+}\frac{1}{25}}\\ \mathbf{=}\sqrt{\frac{4\mathbf{1}}{400}}\\ \mathbf{=}\frac{\sqrt{41}}{20}\end{array}$

So, the amplitude is.$\frac{\sqrt{41}}{20}$

And, one knows that the period of cos t and sin t is$\mathbf{2}\mathbf{\pi }$. Then find the period of y.

 $\begin{array}{c}5\mathbf{t}\mathbf{=}\mathbf{2}\mathbf{\pi }\\ \mathbf{t}\mathbf{=}\frac{2\pi }{5}\end{array}$

And find the frequency$\mathbf{f}\mathbf{=}\frac{1}{t}\mathbf{=}\frac{5}{2\mathbf{\pi }}$,.

 

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

Then,$0\mathbf{=}\mathbf{-}\frac{1}{4}\mathbf{cos}\mathbf{5}\mathbf{t}\mathbf{-}\frac{1}{5}\mathbf{sin}\mathbf{5}\mathbf{t}$ .

 

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

$\begin{array}{c}\mathbf{-}\frac{1}{4}\mathbf{cos}\mathbf{5}\mathbf{t}\mathbf{-}\frac{1}{5}\mathbf{sin}\mathbf{5}\mathbf{t}\mathbf{=}\mathbf{0}\\ \mathbf{-}\frac{1}{4}\mathbf{cos}\mathbf{5}\mathbf{t}\mathbf{=}\frac{1}{5}\mathbf{sin}\mathbf{5}\mathbf{t}\\ \mathbf{-}\frac{5}{4}\mathbf{=}\frac{\sin 5t}{\cos 5t}\\ \mathbf{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{449}\end{array}$

 

Therefore, t = 0.449