PX285 - G4 - attached particles on springs

consider two masses $(m)$ connected to a fixed wall is connected with a spring such that all springs are at rest in equilibrium conjugation, and are at equilibrium extension

first mode

suppose the springs are moved in the same direction

if $x_{1} = x_{2}$ , the central spring is never stressed and plays no role
hence, it can be removed

frequency of either mass connected to spring: $ω = \sqrt{\frac{k}{m}}$
there exists a mode with frequency, $ω$ , and $x_{1} = x_{2}$
this is the symmetric mode with:

\vec{x} = (\begin{matrix} x_{1} \\ x_{2} \end{matrix})

the solution to this will be:

\vec{x} (t) = A (\begin{matrix} 1 \\ 1 \end{matrix}) \exp (i ω t)

this is a mode

second mode

suppose the masses are moving in opposite directions

this is equivalent to cutting the system in half, and fixing it to a wall as the centre of the middle spring is stationary

if the mass is moved a distance, $x_{1}$ , the central (cut) spring is compressed by $2 x_{1}$
the energies:

\begin{aligned} E_{c e n t r e} & = \frac{1}{2} k (2 x_{1})^{2} \\ ⟹ E_{h a l f} & = \frac{1}{2} E_{c e n t r e} = k x_{1}^{2} \end{aligned}