91Ó°ÊÓ

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.

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)

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

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

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

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

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

The corresponding auxilllary equation of equation (7) is

$m^4+7m^2+6=0$

Let m²=k:

$k^2+7k+6=0$

having roots

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

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

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

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):

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

Therefore, $A_1=\frac{−1}{5}$

Equation (10) gives

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

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 .