Q33E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

On a smooth horizontal surface, a mass of m₁ kg isattached to a fixed wall by a spring with spring constantk₁ N/m. Another mass of m₂ kg is attached to thefirst object by a spring with spring constant k₂ N/m. Theobjects are aligned horizontally so that the springs aretheir natural lengths. As we showed in Section 5.6, thiscoupled mass–spring system is governed by the systemof differential equations

 $m_1\frac{d^{2}x}{d{t}^2}+(k_1+k_2)x-k_{2}y=0$

 $m_2\frac{d^{2}y}{d{t}^2}-k_{2}x+k_{2}y=0$

Let’s assume that m₁ = m₂ = 1, k₁ = 3, and k₂ = 2.If both objects are displaced 1 m to the right of theirequilibrium positions (compare Figure 5.26, page 283)and then released, determine the equations of motion forthe objects as follows:

(a)Show that x(2) satisfies the equation $x^{\left(4\right)}\left(t\right)+7x\text{'}\text{'}\left(t\right)+6x\left(t\right)=0$

(b) Find a general solution x(2) to (36).

(c) Substitute x(2) back into (34) to obtain a generalsolution for y(2)

(d) Use the initial conditions to determine the solutions,x(2) and y(2), which are the equations of motion.

Step-by-Step Solution

Answer

Hence, the final answer is 

$x\left(t\right)=\frac{-1}{5}\cos x+\frac{6}{5}\cos \left(\sqrt{6}x\right)$ and

 $y\left(t\right)=\frac{2}{5}\cos x+\frac{3}{5}\cos \left(\sqrt{6}x\right)$

are the equation of motion for the system

Step 1: Motion with spring

They oscillate back and forth about a fixed position. A simple pendulum and a mass on a spring are classic examples of such vibrating motion.

Step 2: Concept of motion of spring will be applied

Let S₁ be a smooth horizontal surface, let m₁ be a mass (kg) attached to S₁ with a spring having spring constant k₁, let m₂ be a mass (kg) attached to S₁ with a spring constant k₂.

We know mass-spring system is governed by the following system of differential equations

    $m_1\frac{d^{2}x}{d{t}^2}+(k_1+k_2)x-k_{2}y=0$                (1)

         $m_2\frac{d^{2}y}{d{t}^2}-k_{2}x+k_{2}y=0$                        (2)

 

(a) To Show: x(t) satisfies

 $x^{\left(4\right)}\left(t\right)+7x\text{'}\text{'}\left(t\right)+6x\left(t\right)=0$

 

We will use elimination method to find x(t).

Let D=d/dt, then system is given by

     $m_1{D}^{2}x+(k_1+k_2)x-k_{2}y=0$              (3)

      $m_2{D}^{2}y-k_{2}x+k_{2}y=0$                           (4)

 

Let m₁=m₂=1 and k₁=3,k₂=2, then the system becomes,

 $\begin{array}{l}{D}^{2}x+5x-2y=0\\ D^{2}y-2x+2y=0\end{array}$

i.e.

                          $(D^2+5)x-2y=0$            (5)

               $(D^2+2)y-2x=0$                        (6)

 

Applying $D^2+2$  onto equation (5) from the left and adding it to the equation got by multiplying equation (6) by 2, we get 

 $\begin{array}{l}(D^2+2)(D^2+5)x-(D^2+2)2y-4x+(D^2+2)2y=0\\ \Rightarrow (D^2+2)(D^2+5)x-4x=0\\ (D^4+2D^2+5D^2+10)x-4x=0\\ (D^4+7D^2+6)x=0\end{array}$

        $x^{\left(4\right)}+7x\text{'}\text{'}+6x=0$                                  (7)

 

 Therefore    x(t) satisfies  $x^{\left(4\right)}+7x\text{'}\text{'}+6x=0$

(b)  To Find: A general solution of equation (7).

The corresponding auxilllary equation of equation (7) is 

 $m^4+7m^2+6=0$

Let m²=k:

 $k^2+7k+6=0$

 

having roots

 $\begin{array}{l}k=\frac{-7\pm \sqrt{49-24}}{2}\\ k=-1,-6\\ m=\sqrt{-1},\sqrt{-6}\\ =\pm i,\pm \sqrt{6}i\end{array}$

Therefore,

 $x\left(t\right)=A_1\cos \left(t\right)+A_2\sin \left(t\right)+A_3\cos \left(\sqrt{6}t\right)+A_4\sin \left(\sqrt{6}t\right)$

 

(c) To Find: general solution of y(t)

 $\begin{array}{l}x\left(t\right)=A_1\cos \left(t\right)+A_2\sin \left(t\right)+A_3\cos \left(\sqrt{6}t\right)+A_4\sin \left(\sqrt{6}t\right)\\ x\text{'}\left(t\right)=A_1-\sin \left(t\right)+A_2\cos \left(t\right)-\sqrt{6}{A}_3\sin \left(\sqrt{6}t\right)+\sqrt{6}{A}_4\cos \left(\sqrt{6}t\right)\\ x\text{'}\text{'}\left(t\right)=-\left[A_1\cos \left(t\right)+A_2\sin \left(t\right)+6A_3\cos \left(\sqrt{6}t\right)+6A_4\sin \left(\sqrt{6}t\right)\right]\end{array}$

 

 

Substituting these values in equation (5) we get,

 $y\left(t\right)=2A_1\cos \left(t\right)+2A_2\sin \left(t\right)-\frac{1}{2}{A}_3\cos \left(\sqrt{6}t\right)-\frac{1}{2}{A}_4\sin \left(\sqrt{6}t\right)$

 

Therefore:  $y\left(t\right)=2A_1\cos \left(t\right)+2A_2\sin \left(t\right)-\frac{1}{2}{A}_3\cos \left(\sqrt{6}t\right)-\frac{1}{2}{A}_4\sin \left(\sqrt{6}t\right)$

 

( d ) The objects are displaced 1m to the right, 

Therefore

 

 $\begin{array}{l}x\left(0\right)=1,\frac{dx\left(t\right)}{dt}=0\\ y\left(0\right)=1,\frac{dy\left(t\right)}{dt}=0\\ \Rightarrow A_1\cos \left(0\right)+A_2\sin \left(0\right)+A_3\cos \left(0\right)+A_4\sin \left(0\right)=1\\ -A_1\sin \left(0\right)+A_2\cos \left(0\right)+-A_3\sqrt{6}\sin \left(0\right)+A_4\sqrt{6}\cos \left(0\right)=0\\ \Rightarrow A_1+A_2=1\\ -2A_1+\frac{1}{2}{A}_3=1\\ -2A_2+\frac{1}{2}\sqrt{6}{A}_4=0\end{array}$

Multiplying equation 8 by 2 and adding it to equation 9:

 $\begin{array}{l}2A_3+\frac{1}{2}{A}_3=2+1\\ \frac{5}{2}{A}_3=3\\ A_3=\frac{6}{5}\end{array}$

Therefore,  $A_3=\frac{6}{5}$

Substituting this value in equation (1):

 $\begin{array}{l}{A}_1+\frac{6}{5}=1\\ \Rightarrow A_1=1-\frac{6}{5}\\ A_1=\frac{-1}{5}\end{array}$

Therefore,  $A_1=\frac{-1}{5}$

 

Equation (10) gives 

$\begin{array}{l}{A}_2=-\sqrt{6}{A}_4\\ \Rightarrow -2\left[\sqrt{6}{A}_4\right]+\frac{1}{2}\sqrt{6}{A}_4=0\\ A_4[-2\sqrt{6}+\frac{\sqrt{6}}{2}]=0\\ \Rightarrow A_4=0\end{array}$

Therefore,  $A_4=0$

 

Hence,  $A_2=0$

 

Therefore,

$x\left(t\right)=\frac{-1}{5}\cos x+\frac{6}{5}\cos \left(\sqrt{6}x\right)$ and $y\left(t\right)=\frac{2}{5}\cos x+\frac{3}{5}\cos \left(\sqrt{6}x\right)$

 

 

Hence, the final answer is 

$x\left(t\right)=\frac{-1}{5}\cos x+\frac{6}{5}\cos \left(\sqrt{6}x\right)$ and $y\left(t\right)=\frac{2}{5}\cos x+\frac{3}{5}\cos \left(\sqrt{6}x\right)$

 

are the equation of motion for the system .