Four springs with the same spring constant and three equal masses are attached in a straight line on a horizontal frictionless surface as illustrated in Figure\(5.32\). Determine the normal frequencies for the system and describe the three normal modes of vibration.

Assume that \(x > 0,\,\,y > 0,\,\,z > 0\) and \(z > y > x\) to derive the differential equations of motions, and they must be true for any \(x,\,y\) and\(z\). The Forces are proportional to the change of the length of the string, so, one has

\(\begin{aligned}{c}{F_1} &= - k \times x,\;\;\,\,\,\;{F_2} &= k\left( {y - x} \right),\;\;{F_3} &= - k\left( {y - x} \right)\\{F_4} &= k\left( {z - y} \right),\;\,{F_5} &= - k\left( {z - y} \right),\;{F_6} &= - k \times z\end{aligned}\)

From the second Newton's law one now has:

\(\begin{aligned}{c}mx'' &= {F_1} + {F_2} &= - kx + k\left( {y - x} \right)\\my'' &= {F_3} + {F_4} &= - k\left( {y - x} \right) + k\left( {z - y} \right)\\mz'' &= {F_5} + {F_6} &= - k\left( {z - y} \right) - kz\end{aligned}\)

One needs to rewrite this system in operator form

\(\begin{aligned}{c}mx'' + 2kx - ky &= 0 \Leftrightarrow \left( {m{D^2} + 2k} \right)(x) - k(y) &= 0\\ - kx + my'' + 2ky - kz &= 0 \Leftrightarrow - k(x) + \left( {m{D^2} + 2k} \right)(y) - k(z) &= 0\\ - ky + mz'' + 2kz &= 0 \Leftrightarrow - k(y) + \left( {m{D^2} + 2k} \right)(z) &= 0\end{aligned}\)

One will solve this system using the method of elimination of variables. The first one will eliminate x and z to find \(y\left( t \right)\) and to do. So, one will multiply the first and the third equation by \({\bf{k}}\) and multiply the second by \(\left( {m{D^2} + 2k} \right)\) and then add all those equations together.

\(\begin{aligned}{c}k\left( {m{D^2} + 2k} \right){\rm{(}}x{\rm{)}} - {k^2}{\rm{(}}y{\rm{)}} &= 0\\ - k\left( {m{D^2} + 2k} \right){\rm{(}}x{\rm{)}} + {\left( {m{D^2} + 2k} \right)^2}{\rm{(}}y{\rm{)}} - k\left( {m{D^2} + 2k} \right){\rm{(}}z{\rm{)}} &= 0\\ - {k^2}{\rm{(}}y{\rm{)}} + k\left( {m{D^2} + 2k} \right){\rm{(}}z{\rm{)}} &= 0\\\left( {{{\left( {m{D^2} + 2k} \right)}^2} - 2{k^2}} \right){\rm{(}}y{\rm{)}} &= 0\\\left( {{m^2}{D^4} + 4mk{D^2} + 4{k^2} - 2{k^2}} \right){\rm{(}}y{\rm{)}} &= 0\\\left( {{m^2}{D^4} + 4mk{D^2} + 2{k^2}} \right){\rm{(}}y{\rm{)}} &= 0\end{aligned}\)

The auxiliary equation to the previous equation is \({m^2}{r^4} + 4mk{r^2} + 2{k^2} = 0\) and its solutions are:

\(\begin{aligned}{c}{r^2} &= \frac{{ - 4km \pm \sqrt {16{m^2}{k^2} - 8{m^2}{k^2}} }}{{2{m^2}}} \Leftrightarrow {r^2} &= ( - 2 \pm \sqrt 2 )\frac{k}{m}\\ \Rightarrow {r_{1,2}} &= \pm \sqrt {( - 2 + \sqrt 2 )\frac{k}{m}} ,\;\;\;{r_{3,4}} &= \pm \sqrt {( - 2 - \sqrt 2 )\frac{k}{m}} \\ \Rightarrow {r_{1,2}} &= \pm i\sqrt {(2 - \sqrt 2 )\frac{k}{m}} ,\;\;\;{r_{3,4}} &= \pm i\sqrt {(2 + \sqrt 2 )\frac{k}{m}} \end{aligned}\)

Therefore, the general solution for \(y\left( t \right)\) is

\(y\left( t \right) = {c_1}\cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) + {c_2}\sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right){\bf{ }} + {c_3}\cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) + {c_4}\sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right)\)

The first equation of the system gives us that\(ky = \left( {m{D^2} + 2k} \right){\rm{(}}x{\rm{)}}\), so one has to solve the following equation:

\(\left( {m{D^2} + 2k} \right){\rm{(}}x{\rm{)}} = {c_1}k\cos \left( {\sqrt {\left( {{\rm{2 - }}\sqrt {\rm{2}} } \right)\frac{k}{m}} t} \right) + {c_2}k\sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) + {c_3}k\cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) + {c_4}k\sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right)\)

First, one will find a homogeneous solution for \(x\left( t \right)\) and then find a particular solution. The auxiliary equation is \(m{r^2} + 2k = 0\) and its solutions are\({r_{1,2}} = \pm \sqrt { - 2\frac{k}{m}} \Leftrightarrow {r_{1,2}} = \pm i\sqrt {2\frac{k}{m}} \)

The homogeneous solution for \(x\left( t \right)\) is\({x_h}\left( t \right) = {d_1}\cos \left( {\sqrt {2\frac{k}{m}} t} \right) + {d_2}\sin \left( {\sqrt {2\frac{k}{m}} t} \right)\)

Assume that a particular solution for x has a form of

\({x_p}\left( t \right) = A\cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) + B\sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right){\bf{ }} + C\cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) + D\sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right)\)

First, one needs to find\({D^2}\left( {{x_p}\left( t \right)} \right)\):

\(\begin{aligned}{c}D\left( {{x_p}\left( t \right)} \right) &= - A\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} \sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) + B\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} \cos \left( {\sqrt {\left. {\left( {2 - \sqrt 2 } \right)\frac{k}{m}t} \right)} } \right. - C\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} \sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right)\\ + D\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} \cos \left( {\sqrt {\left. {\left( {2 + \sqrt 2 } \right)\frac{k}{m}t} \right)} } \right.{D^2}\left( {{x_p}\left( t \right)} \right)\\ &= - A\left( {2 - \sqrt 2 } \right)\frac{k}{m}\cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) - B\left( {2 - \sqrt 2 } \right)\frac{k}{m}\sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) - C\left( {2 + \sqrt 2 } \right)\frac{k}{m}\cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) - D\left( {2 + \sqrt 2 } \right)\frac{k}{m}\sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right)\end{aligned}\)

Now, one has that

\(\begin{aligned}{c}\left( {m{D^2} + 2k} \right)\left( {{x_p}\left( t \right)} \right)\\ &= m\left( { - A\left( {2 - \sqrt 2 } \right)\frac{k}{m}\cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) - B\left( {2 - \sqrt 2 } \right)\frac{k}{m}\sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}t} } \right)} \right. - C\left( {2 + \sqrt 2 } \right)\frac{k}{m}\cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}t} } \right) - D\left( {2 + \sqrt 2 } \right)\frac{k}{m}\sin \left( {\sqrt {\left. {\left. {\left( {2 + \sqrt 2 } \right)\frac{k}{m}t} \right)} \right)} } \right.\\ + 2k\left( {A\cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) + B\sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right)} \right.\left. { + C\cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) + D\sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right)} \right)\\ &= Ak\sqrt 2 \cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) + Bk\sqrt 2 \sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right)\left. {\left. { - Ck\sqrt 2 \cos \sqrt {\left( {2 + \sqrt 2 } \right)} \frac{k}{m}t} \right) - Dk\sqrt 2 \sin \sqrt {\left( {2 + \sqrt 2 } \right)} \frac{k}{m}t} \right)\\ &= {c_1}k\cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) + {c_2}k\sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) + {c_3}k\cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) + {c_4}k\sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right)\\ &= k \times y\left( t \right)\end{aligned}\)

Dividing this equation by k and equation the coefficients one will get:

\(\begin{aligned}{l}A\sqrt 2 &= {c_1},\;\;\;B\sqrt 2 &= {c_2},\;\;\; - C\sqrt 2 &= {c_3},\;\;\;D\sqrt 2 &= {c_4}\\ \Rightarrow A &= \frac{{{c_1}}}{{\sqrt 2 }},\;\;\;B &= \frac{{{c_2}}}{{\sqrt 2 }},\;\;\;C &= - \frac{{{c_3}}}{{\sqrt 2 }},\;\;\;D &= - \frac{{{c_4}}}{{\sqrt 2 }}\end{aligned}\)

Therefore, the particular solution for\(x\left( t \right)\) is

\({x_p}\left( t \right) = \frac{{{c_1}}}{{\sqrt 2 }}\cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) + \frac{{{c_2}}}{{\sqrt 2 }}\sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) - \frac{{{c_3}}}{{\sqrt 2 }}\cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) - \frac{{{c_4}}}{{\sqrt 2 }}\sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right)\)And the general solution is

\(x\left( t \right) = {x_p}\left( t \right) + {x_h}\left( t \right)\).

So, one gets,

\( = \frac{{{c_1}}}{{\sqrt 2 }}\cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) + \frac{{{c_2}}}{{\sqrt 2 }}\sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) - \frac{{{c_3}}}{{\sqrt 2 }}\cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) - \frac{{{c_4}}}{{\sqrt 2 }}\sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) + {d_1}\cos \left( {\sqrt {2\frac{k}{m}} t} \right) + {d_2}\sin \left( {\sqrt {2\frac{k}{m}} t} \right)\)One will find \(z\left( t \right)\) from the second equation of the system which gives us that\(kz = - kx + \left( {m{D^2} + 2k} \right){\rm{(}}y{\rm{)}}\)and similarly as before we will first find the second derivative of\(y\left( t \right)\):

\(D{\rm{(}}y\left( t \right){\rm{)}} = - {c_1}\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} \sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) + {c_2}\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} \cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) - {c_3}\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} \sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) + {c_4}\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} \cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right)\)

\({D^2}{\rm{(}}y\left( t \right){\rm{)}} = - {c_1}\left( {2 - \sqrt 2 } \right)\frac{k}{m}\cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) - {c_2}\left( {2 - \sqrt 2 } \right)\frac{k}{m}\sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) - {c_3}\left( {2 + \sqrt 2 } \right)\frac{k}{m}\cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) - {c_4}\left( {2 + \sqrt 2 } \right)\frac{k}{m}\sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right)\)

So, one has\(kz = - kx + m{D^2}{\rm{(}}y{\rm{)}} + 2ky\)

\(\begin{aligned}{l} &= - k\left( {\frac{{{c_1}}}{{\sqrt 2 }}\cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) + \frac{{{c_2}}}{{\sqrt 2 }}\sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right)} \right. - \frac{{{c_3}}}{{\sqrt 2 }}\cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) - \frac{{{c_4}}}{{\sqrt 2 }}\sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right)\left. { + {d_1}\cos \left( {\sqrt {2\frac{k}{m}} t} \right) + {d_2}\sin \left( {\sqrt {2\frac{k}{m}} t} \right)} \right)\\ + m\left( { - {c_1}\left( {2 - \sqrt 2 } \right)\frac{k}{m}\cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) - {c_2}\left( {2 - \sqrt 2 } \right)\frac{k}{m}\sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right)} \right.\left. { - {c_3}\left( {2 + \sqrt 2 } \right)\frac{k}{m}\cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) - {c_4}\left( {2 + \sqrt 2 } \right)\frac{k}{m}\sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right)} \right)\\ + 2k\left( {{c_1}\cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) + {c_2}\sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right)} \right.\left. { + {c_3}\cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) + {c_4}\sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right)} \right)\end{aligned}\)

Finally, by dividing the previous equation by k we have that a general solution for \(z\left( t \right)\) is

\(\begin{aligned}{l}z\left( t \right) &= \frac{{{c_1}}}{{\sqrt 2 }}\cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) + \frac{{{c_2}}}{{\sqrt 2 }}\sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) - \frac{{{c_3}}}{{\sqrt 2 }}\cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) - \frac{{{c_4}}}{{\sqrt 2 }}\sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right)\\ - {d_1}\cos \left( {\sqrt {2\frac{k}{m}t} } \right) - {d_2}\sin \left( {\sqrt {2\frac{k}{m}} t} \right)\end{aligned}\)

From the equations of motion for \(x,{\bf{ }}y\) and\(z\), one has that the normal angular frequencies are:

\(\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} ,\;\;\;\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} \)and\(\sqrt {2\frac{k}{m}} \).

The normal frequencies are\(\frac{1}{{2\pi }}\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} ,\;\;\;\frac{1}{{2\pi }}\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} ,{\bf{ }}\frac{1}{{2\pi }}\sqrt {2\frac{k}{m}} \)

To describe the normal mode for the first normal frequency one will take \({c_3} = {c_4} = {d_1} = {d_2} = 0\) and then

\(\begin{aligned}{l}x\left( t \right) &= \frac{{{c_1}}}{{\sqrt 2 }}\cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) + \frac{{{c_2}}}{{\sqrt 2 }}\sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right)\\y\left( t \right) = {c_1}\cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) + {c_2}\sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right)\\z\left( t \right) = \frac{{{c_1}}}{{\sqrt 2 }}\cos \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right) + \frac{{{c_2}}}{{\sqrt 2 }}\sin \left( {\sqrt {\left( {2 - \sqrt 2 } \right)\frac{k}{m}} t} \right)\end{aligned}\)

The normal mode for the normal frequency is\(x\left( t \right) = z\left( t \right) = \frac{1}{{\sqrt 2 }}y\left( t \right)\)

Setting that\({c_1} = {c_2} = {d_1} = {d_2} = 0\), one will get that,

\(\begin{aligned}{l}x\left( t \right) &= - \frac{{{c_3}}}{{\sqrt 2 }}\cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) - \frac{{{c_4}}}{{\sqrt 2 }}\sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right)\\y\left( t \right) &= {c_3}\cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) + {c_4}\sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right)\\z\left( t \right) &= - \frac{{{c_3}}}{{\sqrt 2 }}\cos \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right) - \frac{{{c_4}}}{{\sqrt 2 }}\sin \left( {\sqrt {\left( {2 + \sqrt 2 } \right)\frac{k}{m}} t} \right)\end{aligned}\)

Therefore, the normal mode for the normal frequency \(\frac{1}{2}\pi \sqrt {\left( {2 + \sqrt 2 } \right)} k/m\) is

\(x\left( t \right) = z\left( t \right) = - \frac{1}{{\sqrt 2 }}y\left( t \right)\)

For\({c_1} = {c_2} = {c_3} = {c_4} = 0\), one will have that,

\(\begin{aligned}{l}x\left( t \right) &= {d_1}\cos \left( {\sqrt {2\frac{k}{m}} t} \right) + {d_2}\sin \left( {\sqrt {2\frac{k}{m}} t} \right)\\y\left( t \right) \equiv 0\\z\left( t \right) &= - {d_1}\cos \left( {\sqrt {2\frac{k}{m}} t} \right) - {d_2}\sin \left( {\sqrt {2\frac{k}{m}} t} \right)\end{aligned}\)

Therefore, the normal mode for the normal frequency\(\sqrt {\bf{2}} {\bf{k/m}}\)is:

\({\bf{x(t) = - z(t),}}\;\;\;{\bf{y(t)}} \equiv 0\).