Question

Suppose the two springs in the coupled mass–spring system discussed in Problem 33 are switched, giving the new data \({\bf{m1}}{\rm{ }} = {\rm{ }}{\bf{m2}}{\rm{ }} = {\rm{ }}{\bf{1}},{\rm{ }}{\bf{k1}}{\rm{ }} = {\rm{ }}{\bf{2}},{\rm{ }}{\bf{and}}{\rm{ }}{\bf{k2}}{\rm{ }} = {\rm{ }}{\bf{3}}\) If both objects are now displaced 1 m to the right of their equilibrium positions and then released, determine the equations of motion of the two objects.

Short answer

\(\begin{array}{*{20}{l}}{x(t) \approx {A_1}\cos (0.915x) + {A_3}\cos (2.676x)}\\{y(t) \approx (1.387){A_1}\cos (0.915x) - (0.72){A_3}\cos (2.676x)}\end{array}\)

Step-by-step solution

Take the coupled mass-spring system of differential equation

We know, coupled mass-spring system of differential equation is given by,

\(\begin{array}{l}{m_1}\frac{{{d^2}x}}{{d{t^2}}} + \left( {{k_1} + {k_2}} \right)x - {k_2}y = 0\\{m_2}\frac{{{d^2}y}}{{d{t^2}}} - {k_2}x + {k_2}y = 0\end{array}\)

Let \(D = \frac{d}{{dx}}\) then the given system of differential equation become

\(\begin{array}{l}{m_1}{D^2}x + \left( {{k_1} + {k_2}} \right)x - {k_2}y = 0\\{m_2}{D^2}y - {k_2}x + {k_2}y = 0\end{array}\)

Now putting the values \({m_1} = {m_2} = 1,{k_1} = 2,{k_2} = 3\) and we have

\(\begin{array}{*{20}{r}}{{D^2}x + (2 + 3)x - 3y = 0}\\{{D^2}y - 3x + 3y = 0}\end{array}\)

\( \Rightarrow {D^2}x + 5x - 3y = 0\) \((1)\)

\({D^2}y - 3x + 3y = 0\) \((2)\)

\( \Rightarrow \left( {{D^2} + 5} \right)x - 3y = 0\) \((3)\)

\(\left( {{D^2} + 3} \right)y - 3x = 0\) \((4)\)

From the left applying (D+3) in equation(3) and multiplying equation(4) by 3, we get

\(\left( {{D^2} + 3} \right)\left( {{D^2} + 5} \right)x - 3\left( {{D^2} + 3} \right)y = 0\) \((5)\)

\(3\left( {{D^2} + 3} \right)y - 9x = 0\) \((6)\)

Now adding equation(5) and equation(6) we get

\(\begin{array}{*{20}{r}}{\left( {{D^2} + 3} \right)\left( {{D^2} + 5} \right)x - 3\left( {{D^2} + 3} \right) + 3\left( {{D^2} + 3} \right) - 9x = 0}\\{\left( {{D^2} + 3} \right)\left( {{D^2} + 5} \right)x - 9x = 0}\\{\left( {\left( {{D^2} + 3} \right)\left( {{D^2} + 5} \right) - 9} \right)x = 0}\\{\left( {{D^4} + 5{D^2} + 3{D^2} + 15 - 9} \right)x = 0}\\{\left( {{D^4} + 8{D^2} + 6} \right)x = 0}\end{array}\)

i.e \({x^4}(t) + 8{x^2}(t) + 6x(t) = 0\)

Find and solve the auxiliary equation

the auxiliary equation of the above differential equation is

\({m^4} + 8{m^2} + 6 = 0\)

Putting \({m^2} = k\)

\(\begin{array}{l} \Rightarrow {k^2} + 8k + 6 = 0\\ \Rightarrow k = \frac{{ - 8 \pm 2\sqrt {10} }}{2}\\ \Rightarrow k \approx - 0.837,k \approx - 7.162\end{array}\)

Since

\(\begin{array}{*{20}{c}}k&{ = {m^2}}\\{ \Rightarrow {{(k)}^{\frac{1}{2}}}}&{ = m}\\{ \Rightarrow {m_1}}&{ = 0.9152i}\\{{m_2}}&{ = - 0.915i}\\{{m_3}}&{ = 2.676i}\\{{m_4}}&{ = - 2.676i}\end{array}\)

Find general solution \(x(t)\) is

Therefore general solution \(x(t)\)is

\(\)\(\begin{array}{*{20}{c}}{x(t) \approx }&{{C_1}{e^{{a_1}x}}\left( {\cos \left( {{b_1}x} \right) + i\sin \left( {{b_1}x} \right)} \right) + {C_2}{e^{{a_2}x}}\left( {\cos \left( {{b_2}x} \right) + i\sin \left( {{b_2}x} \right)} \right)}\\{}&{ + {C_3}{e^{{a_3}x}}\left( {\cos \left( {{b_3}x} \right) + i\sin \left( {{b_3}x} \right)} \right) + {C_4}{e^{{a_4}x}}\left( {\cos \left( {{b_4}x} \right) + i\sin \left( {{b_4}x} \right)} \right)}\end{array}\)

\(\begin{array}{*{20}{c}} \approx &{{C_1}{e^{0.x}}(\cos (0.915x) + i\sin (0.915x)) + {C_2}{e^{0.x}}(\cos ( - 0.915x) + i\sin ( - 0.915x))}\\{}&{ + {C_3}{e^{0.x}}(\cos (2.676x) + i\sin (2.676x)) + {C_4}{e^{0.x}}(\cos ( - 2.676x) + i\sin ( - 2.676x))}\end{array}\)

\(\begin{array}{*{20}{c}} \approx &{{C_1}(\cos (0.915x) + i\sin (0.915x)) + {C_2}(\cos (0.915x) - i\sin (0.915x)) + {C_3}(\cos (2.676x)}\\{}&{ + i\sin (2.676x)) + {C_4}(\cos (2.676x) - i\sin (2.676x))}\end{array}\)

\(\begin{array}{*{20}{c}} \approx &{\left( {{C_1} + {C_2}} \right)\cos (0.915x) + i\left( {{C_1} - {C_2}} \right)\sin (0.915x) + \left( {{C_3} + {C_4}} \right)\cos (2.676x)}\\{}&{ + i\left( {{C_3} - {C_4}} \right)\sin (2.676x)}\end{array}\) \((7)\)

Let

\(\begin{array}{*{20}{l}}{{A_1} = {C_1} + {C_2}}\\{{A_2} = i\left( {{C_1} - {C_2}} \right)}\\{{A_3} = {C_3} + {C_4}}\\{{A_4} = i\left( {{C_3} - {C_4}} \right)}\end{array}\)

Then by putting the values A1,A2,A3, and A4 in (7), we get

\(x(t) \approx {A_1}\cos (0.915x) + {A_2}\sin (0.915x) + {A_3}\cos (2.676x) + {A_4}\sin (2.676x)\)

This is the general solution for \(x(t)\)

Find general solution for \(y(t)\)

By substituting \(x(t)\)in (3) we get,

\(0 \approx \left( {{D^2} + 5} \right)\left( {{A_1}\cos (0.915x) + {A_2}\sin (0.915x) + {A_3}\cos (2.676x) + {A_4}\sin (2.676x)} \right) - 3y(t)\)

\(\begin{array}{*{20}{c}}{0 \approx }&{{D^2}{A_1}\cos (0.915x) + {D^2}{A_2}\sin (0.915x) + {D^2}{A_3}\cos (2.676x) + {D^2}{A_4}\sin (2.676x)}\\{}&{ + 5{A_1}\cos (0.915x) + 5{A_2}\sin (0.915x) + 5{A_3}\cos (2.676x) + 5{A_4}\sin (2.676x) - 3y(t)}\end{array}\)

\(\begin{array}{*{20}{c}}0&{ \approx {A_1}{D^2}(\cos (0.915x)) + {A_2}{D^2}(\sin (0.915x)) + {A_3}{D^2}(\cos (2.676x)) + {A_4}{D^2}(\sin (2.676x)) + 5{A_1}\cos (0.915x)}\\{}&{ + 5{A_2}\sin (0.915x) + 5{A_3}\cos (2.676x) + 5{A_4}\sin (2.676x) - 3y(t)}\end{array}\)\(\begin{array}{l}0 \approx - {A_1}(0.915)D(\sin (0.915x)) + {A_2}(0.915)D(\cos (0.915x)) - {A_3}(2.676)D(\sin (2.676x))\\ + {A_4}(2.676)D(\cos (2.676x)) + 5{A_1}\cos (0.915x) + 5{A_2}\sin (0.915x) + 5{A_3}\cos (2.676x) + 5{A_4}\sin (2.676x) - 3y(t)\end{array}\)\(\begin{array}{*{20}{c}}0&{ \approx - {A_1}{{(0.915)}^2}\cos (0.915x) - {A_2}{{(0.915)}^2}\sin (0.915x) - {A_3}{{(2.676)}^2}\cos (2.676x) - {A_4}{{(2.676)}^2}\sin (2.676x)}\\{}&{ + 5{A_1}\cos (0.915x) + 5{A_2}\sin (0.915x) + 5{A_3}\cos (2.676x) + 5{A_4}\sin (2.676x) - 3y(t)}\end{array}\) \(\begin{array}{*{20}{c}}0&{ \approx \left( { - {A_1}(0.837) + 5{A_1}} \right)\cos (0.915x) + \left( { - {A_2}(0.837) + 5{A_2}} \right)\sin (0.915x) + \left( { - {A_3}(7.160) + 5{A_3}} \right)\cos (2.676x)}\\{}&{ + \left( { - {A_4}(7.160) + 5{A_4}} \right)\sin (2.676x) - 3y(t)}\end{array}\)\(\begin{array}{*{20}{c}}0&{ \approx ( - (0.837) + 5){A_1}\cos (0.915x) + ( - (0.837) + 5){A_2}\sin (0.915x) + ( - (7.160) + 5){A_3}\cos (2.676x)}\\{}&{ + ((7.160) + 5){A_4}\sin (2.676x) - 3y(t)}\end{array}\) \(\begin{array}{*{20}{l}}{0 \approx (4.163){A_1}\cos (0.915x) + (4.163){A_2}\sin (0.915x) - (2.16){A_3}\cos (2.676x) - (2.16){A_4}\sin (2.676x) - 3y(y)}\\{0 \approx 4.163{A_1}\cos (0.915x) + 4.163{A_2}\sin (0.915x) - 2.16{A_3}\cos (2.676x) - 2.16{A_4}\sin (2.676x) - 3y(t)}\end{array}\) \( \Rightarrow 4.163{A_1}\cos (0.915x) + 4.163{A_2}\sin (0.915x) - 2.16{A_3}\cos (2.676x) - 2.16{A_4}\sin (2.676x) \approx 3y(t)\)Then

\(y(t) \approx 1.387{A_1}\cos (0.915x) + 1.387{A_2}\sin (0.915x) - 0.72{A_3}\cos (2.676x) - 0.72{A_4}\sin (2.676x)\)

This is the general solution for \(y(t)\)

Step 4:Find equation of motion for the system

Here given that, the object were displaced 1m to the right. Therefore

\(\begin{array}{*{20}{l}}{x(0) = 1,\frac{{dx(0)}}{{dt}} = 0}\\{y(0) = 1,\frac{{dy(0)}}{{dt}} = 0}\end{array}\)

\( \Rightarrow {A_1}\cos (0) + {A_2}\sin (0) + {A_3}\cos (0) + {A_4}\sin (0) \approx 1\) \((8)\)

\({A_1}(0.915)\sin (0) + {A_2}(0.915)\cos (0) - {A_3}(2.676)\sin (0) + {A_4}(2.676)\cos (0) \approx 0\) \((9)\)

\((1.387){A_1}\cos (0) + (1.387){A_2}\sin (0) - (0.72){A_3}\cos (0) - (0.72){A_4}\sin (0) \approx 1\) \((10)\)

\( - (1.387)(0.915){A_1}\sin (0) + (1.387)(0.915){A_2}\cos (0) + (0.72)(2.676){A_3}\sin (0) - (0.72)(2.676){A_4}\cos (0) \approx 0\) \((11)\)

From (8) \({A_1} + {A_3} = 1\) \((12)\)

From(9) \({A_2}(0.915) + {A_4}(2.676) \approx 0\) \((13)\)

From(10) \((1.387){A_1} - (0.72){A_3} \approx 1\) \((14)\)

Form(11) \((1.269){A_2} - (1.269){A_4} \approx 0\) \((15)\)

Form(12) \({A_1} = 1 - {A_3}\)

Putting the value of \({A_1}\) in (3)

\(\begin{array}{*{20}{c}}{\left( {1 - {A_3}} \right)(1.387) - {A_3}(0.72)}&{ \approx 1}\\{1.387 - {A_3}(1.387) - {A_3}(0.72)}&{ \approx 1}\\{1.387 - {A_3}(2.107)}&{ \approx 1}\\{{A_3} \approx 0.183}&{}\end{array}\)

Putting the value of \({A_3}\) in (12)

\(\begin{array}{l}{A_1} + 0.183 \approx 1\\{A_1} \approx 1 - 0.183 \approx 0.817\end{array}\)

From(15)

\(\begin{array}{l}(1.269)\left( {{A_2} - {A_4}} \right) \approx 0\\{A_2} - {A_4} \approx 0\\{A_2} \approx {A_4}\end{array}\)

Putting \({A_2} \approx {A_4}\) in (13)

\(\begin{array}{*{20}{r}}\begin{array}{l}{A_4}(0.915) + {A_4}(2.676) \approx 0\\{A_4}(3.591) \approx 0\end{array}\\{}\end{array}\)

Then

\(\begin{array}{*{20}{l}}{{A_4} \approx 0}\\{{A_2} \approx 0}\end{array}\)

Hence

\(\begin{array}{*{20}{l}}{x(t) \approx {A_1}\cos (0.915x) + {A_3}\cos (2.676x)}\\{y(t) \approx (1.387){A_1}\cos (0.915x) - (0.72){A_3}\cos (2.676x)}\end{array}\)

are the equation of motion for the system.