Two springs and two masses are attached in a straight line on a horizontal frictionless surface as illustrated in Figure\(5.31\). The system is set in motion by holding the mass \({{\bf{m}}_{\bf{2}}}\) at its equilibrium position and pulling the mass \({{\bf{m}}_{\bf{1}}}\) to the left of its equilibrium position a distance \({\bf{1}}{\rm{ }}{\bf{m}}\)and then releasing both masses. Express Newton's law for the system and determine the equations of motion for the two masses if \({{\bf{m}}_{\bf{1}}}{\bf{ = 1 kg, }}{{\bf{m}}_{\bf{2}}}{\bf{ = 2 kg, }}{{\bf{k}}_{\bf{1}}}{\bf{ = 4 N / m,}}\)and\({{\bf{k}}_{\bf{2}}}{\bf{ = 10 / 3 N/ m}}\).

Assume that \(x > 0,{\bf{ }}y > 0\) and \(y > x\) to derive the differential equations of motions, and they must be true for any x and y. The Forces are proportional to the change of the length of the string, so one has\({F_1} = {k_1}\left( {y - x} \right),\;{F_2} = - {k_1}\left( {y - x} \right),\;{F_3} = - {k_2}y\)

From the second Newton's law one now has:

\(\begin{aligned}{c}{m_1}x'' &= {k_1}\left( {y - x} \right)\\{m_2}y'' &= - {k_1}\left( {y - x} \right) - {k_2}y\end{aligned}\)

In our case\({m_1} = 1\;kg,{m_2} = 2\;kg,{k_1} = 4\;N/m,{k_2} = \frac{{10}}{3}\;N/m\), so the system is:

\(\begin{aligned}{c}x'' &= 4(y - x)\\2y'' &= - 4(y - x) - \frac{{10}}{3}y\\ \Leftrightarrow x'' + 4x - 4y &= 0\\ - 6x + 3y'' + 11y &= 0\end{aligned}\)

One will use the method of elimination of variables to solve this system. Let’s transform this system into operator form:

\(\begin{aligned}{c}\left( {{D^2} + 4} \right){\rm{(}}x{\rm{)}} - 4{\rm{(}}y{\rm{)}} &= 0\\ - 6{\rm{(}}x{\rm{)}} + \left( {3{D^2} + 11} \right){\rm{(}}y{\rm{)}} &= 0\end{aligned}\)

Multiplying the first equation by \(\left( {3{D^2} + 11} \right)\) and the second by 4 and then adding them together one gets:

\(\begin{aligned}{c}\left( {3{D^2} + 11} \right)\left( {{D^2} + 4} \right){\rm{(}}x{\rm{)}} - 4\left( {3{D^2} + 11} \right){\rm{(}}y{\rm{)}} &= 0\\ - 24{\rm{(}}x{\rm{)}} + 4\left( {3{D^2} + 11} \right){\rm{(}}y{\rm{)}} &= 0\\\left( {\left( {3{D^2} + 11} \right)\left( {{D^2} + 4} \right) - 24} \right){\rm{(}}x{\rm{)}} &= 0 \Leftrightarrow \left( {3{D^4} + 23{D^2} + 20} \right){\rm{(}}x{\rm{)}} &= 0\end{aligned}\)

The auxiliary equation for the previous equation is\(3{r^4} + 23{r^2} + 20 = \left( {{r^2} + 1} \right)\left( {{r^2} + \frac{{20}}{3}} \right) = 0\)and its solutions are\({r_{1,2}} = \pm i\)and\({r_{3,4}} = \pm \sqrt {\frac{{20}}{3}} i = \pm \frac{{2\sqrt {15} }}{3}\), therefore, the general solution for x is\(x\left( t \right) = {c_1}\cos t + {c_2}\sin t + {c_3}\cos \left( {\frac{{2\sqrt {15} }}{3}t} \right) + {c_4}\sin \left( {\frac{{2\sqrt {15} }}{3}t} \right)\)

From the first equation of the system, one has that\(4y = \left( {{D^2} + 4} \right)x\), so one will first find \({D^2}{\rm{(}}x{\rm{)}}\) and then use it to obtain \(y\left( t \right)\)

\(\begin{aligned}{c}D{\rm{(}}x{\rm{)}} &= - {c_1}\sin t + {c_2}\cos t - \frac{{2\sqrt {15} }}{3}{c_3}\sin \left( {\frac{{2\sqrt {15} }}{3}t} \right) + \frac{{2\sqrt {15} }}{3}{c_4}\cos \left( {\frac{{2\sqrt {15} }}{3}t} \right)\\{D^2}{\rm{(}}x{\rm{)}} &= - {c_1}\cos t - {c_2}\sin t - \frac{{20}}{3}{c_3}\cos \left( {\frac{{2\sqrt {15} }}{3}t} \right) - \frac{{20}}{3}{c_4}\sin \left( {\frac{{2\sqrt {15} }}{3}t} \right)\\y\left( t \right) &= \left( {\frac{1}{4}{D^2} + 1} \right){\rm{(}}x{\rm{)}}\\y\left( t \right) &= - \frac{{{c_1}}}{4}\cos t - \frac{{{c_2}}}{4}\sin t - \frac{5}{3}{c_3}\cos \left( {\frac{{2\sqrt {15} }}{3}t} \right) - \frac{5}{3}{c_4}\sin \left( {\frac{{2\sqrt {15} }}{3}t} \right) + {c_1}\cos t + {c_2}\sin t + {c_3}\cos \left( {\frac{{2\sqrt {15} }}{3}t} \right) + {c_4}\sin \left( {\frac{{2\sqrt {15} }}{3}t} \right)\\ &= \frac{3}{4}{c_1}\cos t + \frac{3}{4}{c_2}\sin t - \frac{2}{3}{c_3}\cos \left( {\frac{{2\sqrt {15} }}{3}t} \right) - \frac{2}{3}{c_4}\sin \left( {\frac{{2\sqrt {15} }}{3}t} \right)\end{aligned}\)

Let’s now find the constants\({c_i},i = \overline {1,4} \). The initial conditions are

\(x\left( 0 \right) = - 1,\;x'\left( 0 \right) = 0,\;y\left( 0 \right) = 0,\;y'\left( 0 \right) = 0\).

First, one needs to find\(D\left( y \right)\):

\(D{\rm{(}}y{\rm{)}} = - \frac{3}{4}{c_1}\sin t + \frac{3}{4}{c_2}\cos t + \frac{{4\sqrt {15} }}{9}\sin \left( {\frac{{2\sqrt {15} }}{3}t} \right) - \frac{{4\sqrt {15} }}{9}\cos \left( {\frac{{2\sqrt {15} }}{3}t} \right)\)

So, one has:

\(\begin{aligned}{l}x\left( 0 \right) &= {c_1} + {c_3} &= - 1,\;\;\;x'\left( 0 \right) &= {c_2} + \frac{{2\sqrt {15} }}{3}{c_4} &= 0\\y\left( 0 \right) &= \frac{3}{4}{c_1} - \frac{2}{3}{c_3} &= 0,\;\;\;y'\left( 0 \right) &= \frac{3}{4}{c_2} - \frac{{4\sqrt {15} }}{9}{c_4} &= 0\end{aligned}\)

Expressing \({c_1}\) in terms of \({c_3}\) from the first equation and substituting it into the third equation we get that\({c_3} = - \frac{9}{{17}}\), and returning this into the first equation gives us that\({c_1} = - \frac{8}{{17}}\). Similarly, from the second and the fourth equations one gets\({c_2} = {c_4} = 0\).

Finally, the equations of motion for the given two masses are

\(\begin{aligned}{l}{\bf{x(t) = - }}\frac{{\bf{8}}}{{{\bf{17}}}}{\bf{cost - }}\frac{{\bf{9}}}{{{\bf{17}}}}{\bf{cos}}\left( {\frac{{{\bf{2}}\sqrt {{\bf{15}}} }}{{\bf{3}}}{\bf{t}}} \right)\\{\bf{y(t) = - }}\frac{{\bf{6}}}{{{\bf{17}}}}{\bf{cost + }}\frac{{\bf{6}}}{{{\bf{17}}}}{\bf{cos}}\left( {\frac{{{\bf{2}}\sqrt {{\bf{15}}} }}{{\bf{3}}}{\bf{t}}} \right)\end{aligned}\)